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{"id":0,"problem":"Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$. Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet?","answer":27.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_1","question":"Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$. Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet?"}
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{"id":1,"problem":"Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$. What is $x+y$?","answer":36.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_10","question":"Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$. What is $x+y$?"}
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{"id":2,"problem":"What is the degree measure of the acute angle formed by lines with slopes $2$ and $\\frac{1}{3}$?","answer":45.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_11","question":"What is the degree measure of the acute angle formed by lines with slopes $2$ and $\\frac{1}{3}$?"}
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{"id":3,"problem":"What is the value of\n\\[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \\dots + 18^3 - 17^3?\\]","answer":3159.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_12","question":"What is the value of\n\\[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \\dots + 18^3 - 17^3?\\]"}
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{"id":4,"problem":"In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?","answer":36.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_13","question":"In a table tennis tournament every participant played every other participant exactly once. Although there were twice as many right-handed players as left-handed players, the number of games won by left-handed players was $40\\%$ more than the number of games won by right-handed players. (There were no ties and no ambidextrous players.) What is the total number of games played?"}
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{"id":5,"problem":"How many complex numbers satisfy the equation $z^5=\\overline{z}$, where $\\overline{z}$ is the conjugate of the complex number $z$?","answer":7.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_14","question":"How many complex numbers satisfy the equation $z^5=\\overline{z}$, where $\\overline{z}$ is the conjugate of the complex number $z$?"}
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{"id":7,"problem":"Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\\tfrac{\\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":21.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_16","question":"Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\\tfrac{\\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":8,"problem":"Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\\frac{1}{2^m}$.\nWhat is the probability that Flora will eventually land at 10? Write the answer as a simplified fraction $\\frac{m}{n}$, find $m+n$","answer":3.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_17","question":"Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance $m$ with probability $\\frac{1}{2^m}$.\nWhat is the probability that Flora will eventually land at 10? Write the answer as a simplified fraction $\\frac{m}{n}$, find $m+n$"}
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{"id":10,"problem":"What is the product of all solutions to the equation\n\\[\\log_{7x}2023\\cdot \\log_{289x}2023=\\log_{2023x}2023\\]","answer":1.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_19","question":"What is the product of all solutions to the equation\n\\[\\log_{7x}2023\\cdot \\log_{289x}2023=\\log_{2023x}2023\\]"}
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{"id":11,"problem":"The weight of $\\frac{1}{3}$ of a large pizza together with $3 \\frac{1}{2}$ cups of orange slices is the same as the weight of $\\frac{3}{4}$ of a large pizza together with $\\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza? The answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m-n$?","answer":4.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_2","question":"The weight of $\\frac{1}{3}$ of a large pizza together with $3 \\frac{1}{2}$ cups of orange slices is the same as the weight of $\\frac{3}{4}$ of a large pizza together with $\\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza? The answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m-n$?"}
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{"id":12,"problem":"Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.\n1\n1 1\n1 3 1\n1 5 5 1\n1 7 11 7 1\nEach row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?","answer":5.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_20","question":"Rows 1, 2, 3, 4, and 5 of a triangular array of integers are shown below.\n1\n1 1\n1 3 1\n1 5 5 1\n1 7 11 7 1\nEach row after the first row is formed by placing a 1 at each end of the row, and each interior entry is 1 greater than the sum of the two numbers diagonally above it in the previous row. What is the units digits of the sum of the 2023 numbers in the 2023rd row?"}
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{"id":13,"problem":"If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$. For example, if $\\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$, but if $\\overline{AC}$ and $\\overline{CB}$ are edges and $\\overline{AB}$ is not an edge, then $d(A, B) = 2$. Let $Q$, $R$, and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). Find the probability that $d(Q, R) > d(R, S)$. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":29.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_21","question":"If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$. For example, if $\\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$, but if $\\overline{AC}$ and $\\overline{CB}$ are edges and $\\overline{AB}$ is not an edge, then $d(A, B) = 2$. Let $Q$, $R$, and $S$ be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). Find the probability that $d(Q, R) > d(R, S)$. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":14,"problem":"Let $f$ be the unique function defined on the positive integers such that \\[\\sum_{d\\mid n}d\\cdot f\\left(\\frac{n}{d}\\right)=1\\] for all positive integers $n$. What is $f(2023)$?","answer":96.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_22","question":"Let $f$ be the unique function defined on the positive integers such that \\[\\sum_{d\\mid n}d\\cdot f\\left(\\frac{n}{d}\\right)=1\\] for all positive integers $n$. What is $f(2023)$?"}
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{"id":15,"problem":"How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation\n\\[(1+2a)(2+2b)(2a+b) = 32ab?\\]","answer":1.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_23","question":"How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation\n\\[(1+2a)(2+2b)(2a+b) = 32ab?\\]"}
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{"id":16,"problem":"Let $K$ be the number of sequences $A_1$, $A_2$, $\\dots$, $A_n$ such that $n$ is a positive integer less than or equal to $10$, each $A_i$ is a subset of $\\{1, 2, 3, \\dots, 10\\}$, and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$, inclusive. For example, $\\{\\}$, $\\{5, 7\\}$, $\\{2, 5, 7\\}$, $\\{2, 5, 7\\}$, $\\{2, 5, 6, 7, 9\\}$ is one such sequence, with $n = 5$.What is the remainder when $K$ is divided by $10$?","answer":5.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_24","question":"Let $K$ be the number of sequences $A_1$, $A_2$, $\\dots$, $A_n$ such that $n$ is a positive integer less than or equal to $10$, each $A_i$ is a subset of $\\{1, 2, 3, \\dots, 10\\}$, and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$, inclusive. For example, $\\{\\}$, $\\{5, 7\\}$, $\\{2, 5, 7\\}$, $\\{2, 5, 7\\}$, $\\{2, 5, 6, 7, 9\\}$ is one such sequence, with $n = 5$.What is the remainder when $K$ is divided by $10$?"}
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{"id":17,"problem":"There is a unique sequence of integers $a_1, a_2, \\cdots a_{2023}$ such that\n\\[\\tan2023x = \\frac{a_1 \\tan x + a_3 \\tan^3 x + a_5 \\tan^5 x + \\cdots + a_{2023} \\tan^{2023} x}{1 + a_2 \\tan^2 x + a_4 \\tan^4 x \\cdots + a_{2022} \\tan^{2022} x}\\]whenever $\\tan 2023x$ is defined. What is $a_{2023}?$","answer":-1.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_25","question":"There is a unique sequence of integers $a_1, a_2, \\cdots a_{2023}$ such that\n\\[\\tan2023x = \\frac{a_1 \\tan x + a_3 \\tan^3 x + a_5 \\tan^5 x + \\cdots + a_{2023} \\tan^{2023} x}{1 + a_2 \\tan^2 x + a_4 \\tan^4 x \\cdots + a_{2022} \\tan^{2022} x}\\]whenever $\\tan 2023x$ is defined. What is $a_{2023}?$"}
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{"id":18,"problem":"How many positive perfect squares less than $2023$ are divisible by $5$?","answer":8.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_3","question":"How many positive perfect squares less than $2023$ are divisible by $5$?"}
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{"id":19,"problem":"How many digits are in the base-ten representation of $8^5 \\cdot 5^{10} \\cdot 15^5$?","answer":18.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_4","question":"How many digits are in the base-ten representation of $8^5 \\cdot 5^{10} \\cdot 15^5$?"}
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{"id":20,"problem":"Janet rolls a standard $6$-sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3$? The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":265.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_5","question":"Janet rolls a standard $6$-sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3$? The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":21,"problem":"Points $A$ and $B$ lie on the graph of $y=\\log_{2}x$. The midpoint of $\\overline{AB}$ is $(6, 2)$. What is the positive difference between the $x$-coordinates of $A$ and $B$? The final answer can be written in the form $m \\sqrt{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":9.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_6","question":"Points $A$ and $B$ lie on the graph of $y=\\log_{2}x$. The midpoint of $\\overline{AB}$ is $(6, 2)$. What is the positive difference between the $x$-coordinates of $A$ and $B$? The final answer can be written in the form $m \\sqrt{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":22,"problem":"A digital display shows the current date as an $8$-digit integer consisting of a $4$-digit year, followed by a $2$-digit month, followed by a $2$-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?","answer":9.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_7","question":"A digital display shows the current date as an $8$-digit integer consisting of a $4$-digit year, followed by a $2$-digit month, followed by a $2$-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?"}
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{"id":23,"problem":"Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently?","answer":7.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12A_Problems\/Problem_8","question":"Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently?"}
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{"id":25,"problem":"Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice? The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":7.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_1","question":"Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice? The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":26,"problem":"In the $xy$-plane, a circle of radius $4$ with center on the positive $x$-axis is tangent to the $y$-axis at the origin, and a circle with radius $10$ with center on the positive $y$-axis is tangent to the $x$-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect? The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":7.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_10","question":"In the $xy$-plane, a circle of radius $4$ with center on the positive $x$-axis is tangent to the $y$-axis at the origin, and a circle with radius $10$ with center on the positive $y$-axis is tangent to the $x$-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect? The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":27,"problem":"Calculate the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m^2+n^2$?","answer":13.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_11","question":"Calculate the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m^2+n^2$?"}
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{"id":28,"problem":"For complex number $u = a+bi$ and $v = c+di$ (where $i=\\sqrt{-1}$), define the binary operation\n$u \\otimes v = ac + bdi$\nSuppose $z$ is a complex number such that $z\\otimes z = z^{2}+40$. What is $|z|^2$?","answer":50.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_12","question":"For complex number $u = a+bi$ and $v = c+di$ (where $i=\\sqrt{-1}$), define the binary operation\n$u \\otimes v = ac + bdi$\nSuppose $z$ is a complex number such that $z\\otimes z = z^{2}+40$. What is $|z|^2$?"}
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{"id":29,"problem":"A rectangular box $P$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $P$ is $13$, the areas of all $6$ faces of $P$ is $\\frac{11}{2}$, and the volume of $P$ is $\\frac{1}{2}$. Find the length of the longest interior diagonal connecting two vertices of $P$. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":13.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_13","question":"A rectangular box $P$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $P$ is $13$, the areas of all $6$ faces of $P$ is $\\frac{11}{2}$, and the volume of $P$ is $\\frac{1}{2}$. Find the length of the longest interior diagonal connecting two vertices of $P$. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":30,"problem":"For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?","answer":5.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_14","question":"For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots?"}
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{"id":32,"problem":"In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$","answer":11.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_16","question":"In the state of Coinland, coins have values $6,10,$ and $15$ cents. Suppose $x$ is the value in cents of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of $x?$"}
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{"id":33,"problem":"Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\\circ,$ Find the area of $ABC$. The final answer can be simplified in the form $m \\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ without square factore. What is $m+n$?","answer":18.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_17","question":"Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\\circ,$ Find the area of $ABC$. The final answer can be simplified in the form $m \\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ without square factore. What is $m+n$?"}
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{"id":36,"problem":"Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy? ","answer":50.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_2","question":"Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy? "}
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{"id":40,"problem":"When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$?","answer":11.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_23","question":"When $n$ standard six-sided dice are rolled, the product of the numbers rolled can be any of $936$ possible values. What is $n$?"}
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{"id":41,"problem":"Suppose that $a$, $b$, $c$ and $d$ are positive integers satisfying all of the following relations.\n\\[abcd=2^6\\cdot 3^9\\cdot 5^7\\]\n\\[\\text{lcm}(a,b)=2^3\\cdot 3^2\\cdot 5^3\\]\n\\[\\text{lcm}(a,c)=2^3\\cdot 3^3\\cdot 5^3\\]\n\\[\\text{lcm}(a,d)=2^3\\cdot 3^3\\cdot 5^3\\]\n\\[\\text{lcm}(b,c)=2^1\\cdot 3^3\\cdot 5^2\\]\n\\[\\text{lcm}(b,d)=2^2\\cdot 3^3\\cdot 5^2\\]\n\\[\\text{lcm}(c,d)=2^2\\cdot 3^3\\cdot 5^2\\]\nWhat is $\\text{gcd}(a,b,c,d)$?","answer":3.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_24","question":"Suppose that $a$, $b$, $c$ and $d$ are positive integers satisfying all of the following relations.\n\\[abcd=2^6\\cdot 3^9\\cdot 5^7\\]\n\\[\\text{lcm}(a,b)=2^3\\cdot 3^2\\cdot 5^3\\]\n\\[\\text{lcm}(a,c)=2^3\\cdot 3^3\\cdot 5^3\\]\n\\[\\text{lcm}(a,d)=2^3\\cdot 3^3\\cdot 5^3\\]\n\\[\\text{lcm}(b,c)=2^1\\cdot 3^3\\cdot 5^2\\]\n\\[\\text{lcm}(b,d)=2^2\\cdot 3^3\\cdot 5^2\\]\n\\[\\text{lcm}(c,d)=2^2\\cdot 3^3\\cdot 5^2\\]\nWhat is $\\text{gcd}(a,b,c,d)$?"}
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{"id":43,"problem":"A $3-4-5$ right triangle is inscribed in circle $A$, and a $5-12-13$ right triangle is inscribed in circle $B$. Find the ratio of the area of circle $A$ to the area of circle $B$. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?","answer":194.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_3","question":"A $3-4-5$ right triangle is inscribed in circle $A$, and a $5-12-13$ right triangle is inscribed in circle $B$. Find the ratio of the area of circle $A$ to the area of circle $B$. The final answer can be written in the form $\\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?"}
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{"id":44,"problem":"Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?","answer":1625.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_4","question":"Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint?"}
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{"id":45,"problem":"You are playing a game. A $2 \\times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \\times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A \"turn\" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?","answer":4.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_5","question":"You are playing a game. A $2 \\times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \\times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A \"turn\" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensure that at least one of your guessed squares is covered by the rectangle?"}
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{"id":46,"problem":"When the roots of the polynomial \n\\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \\cdot \\cdot \\cdot (x-10)^{10}\\]\nare removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?","answer":6.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_6","question":"When the roots of the polynomial \n\\[P(x) = (x-1)^1 (x-2)^2 (x-3)^3 \\cdot \\cdot \\cdot (x-10)^{10}\\]\nare removed from the number line, what remains is the union of $11$ disjoint open intervals. On how many of these intervals is $P(x)$ positive?"}
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{"id":47,"problem":"For how many integers $n$ does the expression\\[\\sqrt{\\frac{\\log (n^2) - (\\log n)^2}{\\log n - 3}}\\]represent a real number, where log denotes the base $10$ logarithm?","answer":901.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_7","question":"For how many integers $n$ does the expression\\[\\sqrt{\\frac{\\log (n^2) - (\\log n)^2}{\\log n - 3}}\\]represent a real number, where log denotes the base $10$ logarithm?"}
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{"id":48,"problem":"How many nonempty subsets $B$ of ${0, 1, 2, 3, \\cdots, 12}$ have the property that the number of elements in $B$ is equal to the least element of $B$? For example, $B = {4, 6, 8, 11}$ satisfies the condition.","answer":144.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_8","question":"How many nonempty subsets $B$ of ${0, 1, 2, 3, \\cdots, 12}$ have the property that the number of elements in $B$ is equal to the least element of $B$? For example, $B = {4, 6, 8, 11}$ satisfies the condition."}
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{"id":49,"problem":"What is the area of the region in the coordinate plane defined by\n$| | x | - 1 | + | | y | - 1 | \\le 1$?","answer":8.0,"url":"https:\/\/artofproblemsolving.com\/wiki\/index.php\/2023_AMC_12B_Problems\/Problem_9","question":"What is the area of the region in the coordinate plane defined by\n$| | x | - 1 | + | | y | - 1 | \\le 1$?"}
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{"question": "A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45\u00b0 to 60\u00b0. After how much more time will this car reach the base of the tower?", "options": ["A)5(\u221a3 + 1)", "B)6(\u221a3 + \u221a2)", "C)7(\u221a3 \u2013 1)", "D)8(\u221a3 \u2013 2)", "E)None of these"], "rationale": "Explanation :\nLet the height of the building be h. Initially, he was at an angle of 450. tan 45 = h/distance between car and tower. h = distance between car and tower (since tan 45 = 1).\nNow, after 10 minutes, it travelled a certain distance, and angle changed to 600.\ntan 60 = h/x x = h/\u221a3\nSo, in 10 minutes, it has travelled a distance of h \u2013 x = h - h/\u221a3.\n10 minutes = h *( 1 \u2013 1\u221a3)\nh can be travelled in 10 / (1 \u2013 1\u221a3).\nTo travel a distance of x, which is h/\u221a3, it takes :\nh = 10 / (1 \u2013 1/\u221a3)\nh / \u221a3 = 10/ \u221a3 * (1 \u2013 1/\u221a3). Multiply numerator and denominator by 1 + \u221a3 ( conjugate of 1 - \u221a3). We get, x = h/\u221a3 = 10 (1 + \u221a3) / 2 = 5* (1 + \u221a3)\nSo, it takes 5(1 + \u221a3) minutes to reach the base of the tower.\nAnswer : A", "correct": "A"}
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{"question": "The original price of an item is discounted 22%. A customer buys the item at this discounted price using a $20-off coupon. There is no tax on the item, and this was the only item the customer bought. If the customer paid $1.90 more than half the original price of the item, what was the original price of the item?", "options": ["A)$61", "B)$65", "C)$67.40", "D)$70", "E)$78.20"], "rationale": "Let x be the original price of the item\nDiscounted price = 0.78x\nPayment made by the customer after using the $20 coupon = 0.78x - 20\n0.78x - 20 = x/2 + 1.9\nx = 78.20\nAnswer: E", "correct": "E"}
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{"question": "Find out which of the following values is the multiple of X, if it is divisible by 9 and 12?", "options": ["A)36", "B)15", "C)17", "D)5", "E)7"], "rationale": "9=3*3\n12=3*4\nThe number should definitely have these factors 3*3*4\n36 is the number that has these factors\nSo, 36 is the multiple of X\nAnswer is A", "correct": "A"}
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{"question": "If the probability that Stock A will increase in value during the next month is 0.56, and the probability that Stock B will increase in value during the next month is 0.74. What is the greatest value for the probability that neither of these two events will occur?", "options": ["A)0.22", "B)0.26", "C)0.37", "D)0.46", "E)0.63"], "rationale": "The probability that stock A does not increase is 0.44, and the probability that stock B does not increase is 0.26. Now, how can the probability that both do not increase be more than individual probability of not increasing for each? So the probability that both do not increase can not be more than 0.26. Basically the probability that both do not increase is between 0 and 0.26.", "correct": "B"}
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{"question": "A trader sold an article at a profit of 20% for Rs.360. What is the cost price of the article?", "options": ["A)270", "B)300", "C)280", "D)320", "E)315"], "rationale": "Cost Price = Selling Price / (100+Profit%) \u00d7 100 => 360 / (100+20) \u00d7 100 => 360 / 120 \u00d7 100 = Rs.300\nOption B", "correct": "B"}
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{"question": "20 marbles were pulled out of a bag of only white marbles, painted black, and then put back in. Then, another 20 marbles were pulled out, of which 1 was black, after which they were all returned to the bag. If the percentage of black marbles pulled out the second time represents their percentage in the bag, how many marbles in total Q does the bag currently hold?", "options": ["A)40", "B)200", "C)380", "D)400", "E)3200"], "rationale": "We know that there are 20 black marbles in the bag and this number represent 1/20 th of the number of all marbles in the bag, thus there are total Q of 20*20=400 marbles.\nAnswer: D.", "correct": "D"}
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{"question": "Find the total no. of distinct bike no.'s that can beformed using 2 letters followed by 2 no.'s. How many letters need to be distinct?", "options": ["A)74453", "B)64543", "C)74325", "D)65000", "E)97656"], "rationale": "Out of 26 alphabets two distinct letters can be chosen in 26P2 ways. Coming to the numbers part, there are 10 ways to choose the first digit and similarly, there are another 10 ways to choose the second digit. Hence, there are in total 10X10 = 100 ways.\nCombined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.\nD", "correct": "D"}
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{"question": "A train running at a speed of 100 miles/hour, takes 10 hours to reach its destination. After covering quarter of the distance, it starts raining and the train has to be slowed to speed of 75 miles/hour. What is the total journey duration?", "options": ["A)10", "B)11.5", "C)12.5", "D)13.5", "E)15"], "rationale": "Distance to destination = 100 X 10 = 1000 miles.\nDistance remaining when it starts to rain = 1000 - 250 = 750 miles.\nSpeed for remaining distance = 75 miles / hour.\nTime taken to cover remaining distance = 750 / 75 = 10 hours.\nTotal duration of the journey = 2.5 + 10 = 12.5 hours.\nThe correct option is C.", "correct": "C"}
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{"question": "Of the 200 students in a school, at least 45% attended the prom night and at least 35% took part in the debating session. What is the maximum number of students who could have neither attended the prom night nor the debating session?", "options": ["A)27", "B)81", "C)90", "D)99", "E)110"], "rationale": "To maximize the number of students who did neither, we should minimize the number of students who debated or attended the prom.\nLet's assume that all 35% of students who debated also attended the prom.\nThen 35% did both, 10% only attended prom, and 55% did neither.\n0.55*200 = 110\nThe answer is E.", "correct": "E"}
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{"question": "A sales person gets a 10% commission on each sale he makes. How many sales of $250 each must he make in order to reach a salary of at least $1000?", "options": ["A)15", "B)24", "C)25", "D)40", "E)52"], "rationale": "10% of 250 = 25.\nTotal salary required = 1000\nEarning from single sale = 25\n# of sales = 1000/25 =40\nSo 40 sales\nD is the correct choice", "correct": "D"}
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{"question": "A company produces 420 units of a particular computer component every month, at a production cost to the company of $110 per component, and sells all of the components by the end of each month. What is the minimum selling price per component that will guarantee that the yearly profit (revenue from sales minus production costs) will be at least $626,400 ?", "options": ["A)226", "B)230", "C)240", "D)260", "E)280"], "rationale": "450*12(x-110)=626400\nwhere x is a selling cost of one item\nx-110, is a profit from one item\n450 - number of items produced and sold per month\n12 - is a number of month in a year\nSimplifying the equation will lead to x-110=116, then x = 230\nB", "correct": "B"}
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{"question": "At a certain factory, 10 percent of the staplers produced on Monday were defective and 2 percent of the non-defective staplers were rejected by mistake. If 72 of the non-defective staplers were rejected, what was the number of staplers produced that day?", "options": ["A)4,000", "B)4,200", "C)4,500", "D)4,800", "E)5,000"], "rationale": "We're told that 10% of staplers in a factory are defective.\nX = Total staplers\n0.1X = defective staplers\n0.9X = normal staplers\nNext, we're told that 2% of the normal staplers were rejected by mistake and that this = 72 staplers.\n0.9X(0.02) = 72\n0.018X = 72\n18X = 72,000\nX = 4,000\nFinal Answer:\nA", "correct": "A"}
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{"question": "Machine A puts out a yo-yo every 6 minutes. Machine B puts out a yo-yo every 9 minutes. After how many minutes will they have produced 10 yo-yos?", "options": ["A)24 minutes", "B)32 minutes", "C)36 minutes", "D)64 minutes", "E)72 minutes"], "rationale": "A's speed = 3 yo-yos every 18 minutes\nB's speed = 2 yo-yos every 18 minutes\nA + B's speed = 3 + 2 = 5 yo-yos every 18 minutes\nboth together will finish 10 yo-yos in 36 minutes\ncorrect option is C", "correct": "C"}
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{"question": "Add: +45 and -30", "options": ["A)-30", "B)+30", "C)0", "D)15", "E)-15"], "rationale": "45 - 30 = 15\nANSWER : D", "correct": "D"}
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{"question": "In how many ways can the letters of the word \"PROBLEC\" be rearranged to make 7 letter words such that none of the letters repeat?", "options": ["A)2!", "B)3!", "C)7!", "D)8!", "E)9!"], "rationale": "There are seven positions to be filled.\nThe first position can be filled using any of the 7 letters contained in PROBLEM.\nThe second position can be filled by the remaining 6 letters as the letters should not repeat.\nThe third position can be filled by the remaining 5 letters only and so on.\n758\nTherefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! Ways.\nC", "correct": "C"}
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{"question": "Let A and B be independent events with P (A) = 0.2 and P(B) = 0.8. Find P(A/B)?", "options": ["A)0.2", "B)0.4", "C)0.6", "D)1.2", "E)1.5"], "rationale": "P(A/B) = P (A n B)/P(B)\nHere, P (A n B) = 0.16\nP(A/B) = 0.16/0.8 = 0.2\nA", "correct": "A"}
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{"question": "Consider there is an staircase elevator and you are coming down. If you walk 20 steps and stop, then you reach bottom in 10 minutes. If you walk 10 steps and stop, you reach to the ground in 20 minutes. What is the speed of the elevator?", "options": ["A)1 step/minute", "B)2 step/minute", "C)3 step/minute", "D)4 step/minute", "E)None of the above"], "rationale": "Let total number of steps in the elevator be n and let the speed be e\nElevator covered n-20 steps in 10 mins\n(n-20)/e=10.................1\nElevator covered n-10 steps in 20 mins\n(n-10)/e=20......................2\nFrom (1) and (2)\nn=30\ne=1 step/min\nHence (A) is correct answer.", "correct": "A"}
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{"question": "Last year, a Home Appliance Store sold an average(arithmetic mean) of 42 microwave ovens per month. In the first 10 months of this year,the store has sold an average(arithmetic mean) of only 20 microwave ovens per month. What was the average number of microwave ovens sold per month during the entire 22 months period ?", "options": ["A)21", "B)30", "C)31", "D)32", "E)None of the above"], "rationale": "42\u00d712+20\u00d710 /12+10=504+200/22=704/22=32\nAnswer D", "correct": "D"}
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{"question": "An exam is given in a certain class. The average (arithmetic mean) of the highest score and the lowest score is equal to x. If the average score for the entire class is equal to y and there are z students in the class, where z > 5, then in terms of x, y, and z, what is the average score for the class excluding the highest and lowest scorers?", "options": ["A)(zy \u2013 2x)/z", "B)(zy \u2013 2)/z", "C)(zx \u2013 y)/(z \u2013 2)", "D)(zy \u2013 2x)/(z -2)", "E)(zy \u2013 x)/(z + 2)"], "rationale": "Highest: H\nLowest: L\nNumber of students in the class: Z\nNumber of students in the class excluding the highest and lowest : Z-2\nAverage of Highest and Lowest: (H + L)/2=X => H+L=2X\nAverage of Entire Class : (H+L+Others)/Z=Y => Others= ZY-2X\nAverage of the others in the class: (ZY-2X)/(Z-2)\nAnswer: D", "correct": "D"}
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{"question": "[5 + ? \u00d7 19 - 15 - 7]/[13 \u00d7 13 - 156] = 6", "options": ["A)4", "B)4.5", "C)5", "D)5.5", "E)6.5"], "rationale": "(? \u00d7 19 - 17)/(169 - 156) = 6\n=> ? \u00d7 19 - 17 = 13 \u00d7 6 = 76\n=> ? \u00d7 19 = 78 + 17 = 95\n? = 95/19 = 5\nAnswer: Option C", "correct": "C"}
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{"question": "A grocer makes a 25% profit on the selling price for each bag of flour it sells. If he sells each bag for $100 and makes $3,000 in profit, how many bags did he sell?", "options": ["A)12", "B)16", "C)24", "D)30", "E)40"], "rationale": "Profit on one bag: 100*1.25= 125\nNumber of bags sold = 3000/125 = 24\nAnswer is C.", "correct": "C"}
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{"question": "Alex and Jacob works at a toy shop that make toys. Alex takes 7 hours to make a toy, and Jacob takes 9 hours to make a toy. During a month, both of them makes 35 toys in total. If both of them have worked for almost similar number of hours how many toys have been prepared by Jacob?", "options": ["A)15", "B)16", "C)17", "D)18", "E)19"], "rationale": "Lets say Alex has worked for x hrs., and Jacob has worked for y hrs. So, number of toys prepared by Alex is x/7, and Jacob is y/9. Since total number of toys prepared by both of them is 35.\n=> x/7 + y/9 = 35.\n=> 9x + 7y = (35)(63)\n=> 7y = (35)(63) - 9x\n=> y = (5)(63) - (9/7)x\n=> y = 315 - (9/7)x\n=> x is to be a multiple of 7. Also, we need to minimize the difference between x & y. Here are some possible values,\nx = 126, y = 315 - (9/7)126 = 153\nx = 133, y = 315 - (9/7)133 = 144\nx = 140, y = 315 - (9/7)140 = 135\nx = 147, y = 315 - (9/7)147 = 126\nAs we can see minimum difference between x and y is when x is 140 hrs. and y is 135 hrs. Thus total toys created by Jacob = y/9 = 135/9 = 15.\nAnswer: A", "correct": "A"}
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{"question": "John likes to have lightly flavored tea every evening. In a 50% strong milk tea, he replaces 15% of it with milk twice. Then, he replaces 10 percent of the resultant solution with more milk.\nWhat is the final concentration of tea John drinks?", "options": ["A)15.38%", "B)42%", "C)39.86%", "D)22.35%", "E)32.51%"], "rationale": "Imagine starting out with 100 ml of 50% milk tea.\nIn step 1, 15% of the tea is replaced with milk. Thus, 85% of the original tea remains. Since this is done twice, we have a concentration of 50x0.85x0.85% (=36.125%) of tea solution.\nFinally, 10% of this solution is replaced with milk again. So, the final concentration of tea is 36.125*0.9%\nThis equals 32.51% of tea solution.\nAnswer: E", "correct": "E"}
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{"question": "In a class 1/16 of the students study math, 1/10 of the students study bio, 1/8 of the students study english. The total number of students is a 4 digit number. Find the diffrence between maximum number of students and minimum number of students.", "options": ["A)8880", "B)8870", "C)8890", "D)7890", "E)6780"], "rationale": "LCM of 16,10,8 = 80\nthe largest 4 digit number divisible by 80 = 9920\nThe smallest 4 digit number divisible by 80 = 1040\nSo, required difference = 9920-1040= 8880\nANSWER:A", "correct": "A"}
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{"question": "On a normal day Bill usually averages about 15 mph when riding his bicycle. On a windy day, his speed is reduced by 4 mph. How far can Bill travel on a windy day in 21 minutes? Round to the nearest hundredth.", "options": ["A)2 miles", "B)2.25 miles", "C)3.25 miles", "D)3.85 miles", "E)2.85 miles"], "rationale": "15 mph - 4 mph= 11 mph\n11 mph x (21/60)= 3.85 miles\nAnswer D", "correct": "D"}
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{"question": "A retailer sold an appliance for 40 percent above cost, which represented a gross profit of $20.00. For what price did the retailer sell the appliance?", "options": ["A)$27.30", "B)$51.00", "C)$63.00", "D)$70.00", "E)$91.00"], "rationale": "Let the cost be A. Then the selling price is A+0.4*A.\nSo the profit is 0.4 * A.\n0.4*A=20 ---> A=50.\nSo the selling price is 50+20=70.\nThe answer is (D).", "correct": "D"}
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{"question": "At 6% per annum simple interest, Rahul borrowed Rs. 500. What amount will he pay to clear the debt after 4 years", "options": ["A)750", "B)700", "C)620", "D)600", "E)None of these"], "rationale": "We need to calculate the total amount to be paid by him after 4 years, so it will be Principal + simple interest.\nSo,\n=>500+500\u22176\u22174 /100=>Rs.620\nOption C", "correct": "C"}
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{"question": "A computer routine was developed to generate two numbers (x,y) the first being a random number between 0 and 100 inclusive, and the second being less than or equal to the square root of the first. Each of the following pair satisfies the routine except", "options": ["A)(99,10)", "B)(85,9)", "C)(50,7)", "D)(1,1)", "E)(1,0)"], "rationale": "99 is generated\nWe don't know what the square root of 99 is because we would need a calculator, but we know the square root of 100 is 10, so the square root of 99 has to be less than 10.\nANSWER:A", "correct": "A"}
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{"question": "A jeep travels a certain distance taking 6 hours in the forward journey. During the return journey, it increased its speed by 12km/hr and took 4 hours. What is the distance travelled by the jeep?", "options": ["A)126km", "B)144km", "C)127km", "D)228km", "E)128km"], "rationale": "Let 'x' be the distance and 'y' be the speed of the forward journey. Then, we have 6v=d and 4(v+12)=d\n=> v=d/6 and v=d/4 - 12\n=> d/6 = d/4 - 12\n=> d/12 = 12\n=> d=144\nAnswer: B", "correct": "B"}
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{"question": "When I was 2 years old, my brother was half my age. Now I am 60 years old, how old is my brother?", "options": ["A)A)59", "B)B)69", "C)C)79", "D)D)89", "E)E)99"], "rationale": "Half of 2 is 1. =>2+58=60-> 1+58=59\nAnswer A", "correct": "A"}
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{"question": "The original retail price of an appliance was 60 percent more than its wholesale cost. If the appliance was actually sold for 20 percent less than the original retail price, then it was sold for what percent more than its wholesale cost?", "options": ["A)20%", "B)28%", "C)36%", "D)40%", "E)42%"], "rationale": "wholesale cost = 100;\noriginal price = 100*1.6 = 160;\nactual price = 160*0.8 = 128.\nAnswer: B.", "correct": "B"}
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||||
{"question": "On a map, the length of the road from Town F to Town G is measured to be 20 inches. On this map, 1/4 inch represents an actual distance of 10 miles. What is the actual distance, in miles, from Town F to Town G along this road?", "options": ["A)800", "B)720", "C)960", "D)1140", "E)1160"], "rationale": "Here we are given a ratio: 1/4 inch on the map = 10 miles, so 1 inch on the map = 40 miles. If the map-distance between the towns is 20 inches, then the actual distance must be 20 x 40 = 800\nAnswer: A.", "correct": "A"}
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{"question": "When folded into two equal halves a rectangular sheet had a perimeter of 48cm for each part folded along one set of sides and the same is 66cm when folded along the other set of sides. Find the area of the sheet.", "options": ["A)1584", "B)1120", "C)792", "D)1320", "E)1200"], "rationale": "Let the sheet be folded along its breadth and its perimeter = 48cm\nTherefore, (l/2 + b) = 48 ... (i)\nNow, let the sheet be folded along its length, and the perimeter = 66cm\n(l + b/2)= 66 \u2026... (ii)\nSolving (i) and (ii), we get,\nl = 56cm, b = 20cm\nArea = l*b\nArea = 1120 cm2\nANSWER IS B", "correct": "B"}
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{"question": "Suppose you can travel from a place M to a place N by 3 buses, from place N to place O by 4 buses, from place O to place P by 1 buses and from place P to place Q by 3 buses. In how many ways can you travel from M to Q ?", "options": ["A)24", "B)36", "C)72", "D)84", "E)None"], "rationale": "The bus from M to N can be selected in 3 ways. The bus from N to O can be selected in 4 ways. The bus from O to P can be selected in 1 way. The bus from P to Q can be selected in 3 ways. So, by the General Counting Principle, one can travel from M to Q in 3*4*1*3= 36 ways\nAnswer B", "correct": "B"}
|
||||
{"question": "A rectangular solid, 3 x 4 x 15, is inscribed in a sphere, so that all eight of its vertices are on the sphere. What is the diameter of the sphere?", "options": ["A) 13.3542", "B) 15.8113", "C) 18.3451", "D) 19.5667", "E) 20.8888"], "rationale": "In an inscribed rectangle in a sphere, we will have a line joining the opposite vertices as the diameter.\nAccording to the Pythagoras theorem, sides 3, 4 give diagonal as 5 ==> with 5 and 15, we get 5sqrt(10).\n5sqrt(10) or 15.8113 is the diameter of the sphere.\nanswer = B", "correct": "B"}
|
||||
{"question": "A starts travel towards south 3km, then travel 5 km towards east, and again travels 3 km to north, and finally travels 2km towards west. In the end how far from is A from home?", "options": ["A)3km", "B)2km", "C)4km", "D)5km", "E)6km"], "rationale": "3s,5e,3n,2w\n5-2=3e\n3-3=0\n3km\nANSWER:A", "correct": "A"}
|
||||
{"question": "While selling a watch, a shopkeeper gives a discount of 5%. If he gives a discount of 7%, he earns Rs. 15 less as profit. The marked price of the watch is:", "options": ["A)Rs. 697.50", "B)Rs. 712.50", "C)Rs. 787.50", "D)Rs. 750", "E)Rs. 780"], "rationale": "If he increases the discount by 2%, then his profit is 15 less. Let the marked price be X.\n.02x = 15\nx = 750 marked price\nANSWER:D", "correct": "D"}
|
||||
{"question": "A student instead of finding the value of 7/8 of a number, found the value of 7/18 of the number. If his answer differed from the actual one by 770, find the that number.", "options": ["A)1584", "B)2520", "C)1728", "D)1656", "E)None"], "rationale": "According to the question,\n=> [7/8 - 7/18 ]x = 770\n=> 7*10*x /18*8 = 770\n=> x = 11*18*8\n=> 1584.\nAnswer : A", "correct": "A"}
|
||||
{"question": "The monthly salary S of a shop assistant is the sum of a fixed salary of $500 plus 5% of all monthly sales. What should the monthly sales be so that her monthly salary reaches $1500?", "options": ["A)$50000", "B)$40000", "C)$30000", "D)$20000", "E)None of these"], "rationale": "Let S be the total monthly salary and x be the monthly sales, hence\nS = 500 + 5% * x\nFind sales x so that S = 1500, hence\n1500 = 500 + 5% * x = 500 + 0.05 x\nSolve for x\nx = (1500 - 500) / 0.05 = $20000\nAnswer D", "correct": "D"}
|
||||
{"question": "An aeroplane flies along the four sides of a square at the speeds of 200, 400, 600 and 800km/hr. Find the average speed of the plane around the field?", "options": ["A)384", "B)562", "C)458", "D)156", "E)452"], "rationale": "Let the each side of the square is x km\naverage speed of plane is y km/hr\n(x/200)+(x/400)+(x/600)+(x/800) = 4x/y\n25x/2400 = 4x/y\ny= 384 km/hr\nAnswer is A", "correct": "A"}
|
||||
{"question": "Jack buys 18 sharpeners (white and brown) for rs. 100. If he pays 1 rupee more for each white than brown sharpeners. How many of white and how many brown sharpeners did he buy?", "options": ["A)10,8", "B)9,8", "C)7,8", "D)5,6", "E)11,12"], "rationale": "Total cost=100\nnumber of sharp=18\ncost of white=cost of brown+1\n100/18=5.5...-(1)\nalso 100%18=10...-(2)\nas cost of white is 1 more than that of brown\nfrom 1 int. value will be 5\nnow remainder is 10 so 10 sharp. will be of cost (5+1)\n=> 10*(5+1)+8*5\n=>10*6+8*5\n=60+40\n100\nwhite=10\nbrown=8\nANSWER:A", "correct": "A"}
|
||||
{"question": "Hoses A and B spout water at different constant rates, and hose A can fill a certain pool in 8 hours. Hose A filled the pool alone for the first 2 hours and the two hoses, working together, then finished filling the pool in another 3 hours. How many hours would it have taken hose B, working alone, to fill the entire pool?", "options": ["A)8", "B)15", "C)12", "D)6", "E)3"], "rationale": "Since hose A can fill the pool in 8 hours, then in 2 + 3 = 5 hours it will fill 5/8th of the pool. Thus the remaining 3/8th is filled by hose B in 3 hours. This means that hose B,working alone, to fill the entire pool will need 3*8/3 = 8 hours.\nAnswer: A.", "correct": "A"}
|
||||
{"question": "If 120 is reduced to 96, what is the reduction percent?", "options": ["A)30%", "B)40%", "C)20%", "D)10%", "E)5%"], "rationale": "reduction = 120 \u2013 96 = 24\n\u2234 Reduction percent = (24/120)\u00d7100% =20%\nAnswer:C", "correct": "C"}
|
||||
{"question": "I know a 5 digit number having a property that with a 1 after it, it is three times as large as it would be with a 1 before it.\nWhat is that number?", "options": ["A)42857", "B)32456", "C)76523", "D)24567", "E)43566"], "rationale": "Let the number be x\n10x +1 = 3(100,000 + x)\n=> x = 42857.", "correct": "A"}
|
||||
{"question": "At Daifu university, 24% of all students are members of both a chess club and a swim team. If 20% of members of the swim team are not members of the chess club, what percentage of all Daifu students are members of the swim team?", "options": ["A)20%", "B)30%", "C)40%", "D)50%", "E)60%"], "rationale": "Assume there are total of 100 students. 24 students are members of both clubs. We are told that:20% of members of the swim team are not members of the chess club, thus if S is a # of members of the swim team then 0.2S is # of members of only the swim teem:\n24+0.2S=S --> S=30.\nAnswer: B.", "correct": "B"}
|
||||
{"question": "If the population of a city increases by 5 % annually, what will be the population of the city in 2 years time if its current population is 78000?", "options": ["A)81900", "B)85995", "C)85800", "D)90000", "E)None of these"], "rationale": "The % change in population of city in two years time is 1.05*1.05 = 1.1025 = 10.25%\nTherefore, after 2 years the population of the city will be 1.1025 * 78000 = 85995\nANSWER B", "correct": "B"}
|
||||
{"question": "Two cars start at the same time from opposite ends of a highway that is 50 miles long. One car is riding at 12 mph and the second car is riding at 13 mph. How long after they begin will they meet?", "options": ["A) 1", "B) 1.25", "C) 1.50", "D) 1.75", "E) 2"], "rationale": "Time they will meet = total distance/ relative speed= 50/12+13 = 50/25 = 2\nAnswer is E", "correct": "E"}
|
||||
{"question": "A shopkeeper employed a servant at a monthly salary of 1500. In addition to it, he agreed to pay him a commission of 15% on the monthly sale. How much sale in Rupees should the servant do if he wants his monthly income as 6000?", "options": ["A)30000", "B)415000", "C)31500", "D)50000", "E)None of these"], "rationale": "Servant\u2019s commission amount\n= 6000 \u2013 1500 = 4500\ni.e.,15% = 4500\nor,100% = 4500\u204415 \u00d7 100 = 30000\nAnswer A", "correct": "A"}
|
||||
{"question": "A man borrows Rs.360 If he pays it back in 12 monthly installments of Rs.31.50, what is his interest rate?", "options": ["A)1.5%", "B)4.5%", "C)10%", "D)5%", "E)12%"], "rationale": "Instead of paying monthly 360/12 = 30Rs, the man pays 31.50Rs. Therefore, the interest rate is 1.5/30 = 0.5/10 = 5/100 = 5%.\nAnswer D", "correct": "D"}
|
||||
{"question": "The price of a product is reduced by 30% . By what percentage should it be increased to make it 100%", "options": ["A)41.86%", "B)42.86%", "C)43.86%", "D)44.86%", "E)45.86%"], "rationale": "If initial price is Rs 100 and reduced price is Rs 70.\nThen, to make it 100 again, price should increase by 100*30/70= 300/7 % or 42.86% approx\nANSWER:B", "correct": "B"}
|
||||
{"question": "I have a money pouch containing Rs. 700. There are equal number of 25 paise coins, 50 paise coins and one rupee coins.\nHow many of each are there?", "options": ["A)453", "B)651", "C)400", "D)487", "E)286"], "rationale": "25 paise + 50 paise + 100 paise = 175 paise and Rs. 700 = 70,000 paise\n70,000/175 = 400", "correct": "C"}
|
||||
{"question": "A man spends Rs. 3500 per month and saves 12 1/2% of his income. His monthly income is ?", "options": ["A)Rs. 4400", "B)Rs. 4270", "C)Rs. 4000", "D)Rs. 3937.50", "E)None of these"], "rationale": "87 1/2% of P = 3500\n\u21d2 {(175/2) x P} / 100 = 3500\n\u2235 P = (3500 x 2 x 100) / 175 = 4000\nCorrect Option: C", "correct": "C"}
|
||||
{"question": "Five dozen toys are packed in a box and 98 boxes are kept in a tempo. How many tempos can lift 29400 toys in one round ?", "options": ["A)4", "B)5", "C)7", "D)6", "E)8"], "rationale": "Five dozen = 5 x 12 = 60\n\u21d2 No of toys can be kept in 1 box = 60\n\u2234 No of toys can be kept in 98 boxes = 60 x 98 = 5880\n\u2234 29400 toys can be lifted by = 29400 / 5880 = 5 tempos\nOption: B", "correct": "B"}
|
||||
{"question": "There are 10 oranges in a basket. Find the no. of ways in which 2 oranges are chosen from the basket?", "options": ["A)45", "B)90", "C)120", "D)150", "E)180"], "rationale": "Required number of ways = 10C2 = 10*9/2 = 45\nAnswer is A", "correct": "A"}
|
||||
{"question": "A company contracts to paint 3 houses. Mr.Brown can paint a house in 6 days while Mr.Black would take 8 days and Mr.Blue 12 days. After 8 days Mr.Brown goes on vacation and Mr. Black begins to work for a period of 6 days. How many days will it take Mr.Blue to complete the contract?", "options": ["A)7", "B)8", "C)10", "D)11", "E)12"], "rationale": "let x is amount of work to be done to paint one house.\nSo Brown's one day work is x/6, black's can do x/8 work in\none day and blue is x/12.\nTotal houses is 3, so tatal work to be done is 3x.\n3x= 8*(x/6) + 6*(x/8) + y*(x/12)\nfinally y = 11.\nblue will complete the remaining work in 11 days.\nANSWER:D", "correct": "D"}
|
||||
{"question": "Train A leaves a station every 16 minutes and Train B leaves every 17 minutes. If both trains just left the station simultaneously, how long until they do so again?", "options": ["A)272 minutes", "B)304 minutes", "C)190 minutes", "D)70 minutes", "E)35 minutes"], "rationale": "We have to find the LCM:\n17 is a prime number which means the LCM of 16 and 17 has to be 16*17=272\nCorrect answer is A.", "correct": "A"}
|
||||
{"question": "A hollow cube of size 5cm is taken, with the thickness of 1cm. It is made of smaller cubes of size 1cm .If the outer surface of the cube is painted how many faces of the smaller cubes remain unpainted?", "options": ["A)438", "B)550", "C)500", "D)450", "E)498"], "rationale": "Volume of Big Cube considering it is not hollow = L3 = 5*5*5 = 125 cm3\nSize of hollow cube (considering 1 cm thickness on two faces of large cube = 5 - 2 = 3cm\nVolume of hollow cube = 3*3*3 = 27 cm3\nSo Total Volume filled up by smaller cubes = Volume of Larger Cube - Volume of hollow cube\n= 125 - 27\n= 98 cm3\nVolume of 1 small cube = 1*1*1 = 1 cm3\nTotal number of small cubes in the larger cube = 98 / 1 = 98\nand Number of faces of 98 small cubes (6 faces each cube has) = 98*6 = 588 faces\nTotal Surface area of 6 faces of larger cube painted = 6*L2 = 6*5*5 = 150cm2\nSurface area of one face of small cube = 1*1 = 1cm2\nNumber of faces of small cube painted = 150/1 = 150 faces\nHence number of faces of the smaller cubes remain unpainted= 588-150\n= 438\nanswer.A", "correct": "A"}
|
||||
{"question": "In a chocolate store, all chocolates are either vanilla or cocoa flavored only. 10% of the chocolates are cocoa flavored, 90% of the rest are squashed. What percentage of the chocolates are both vanilla flavored and not squashed?", "options": ["A)1%", "B)2%", "C)5%", "D)9%", "E)10%"], "rationale": "If 10% of chocolates are cocoa flavored, then 90% are vanilla flavored.\n90% of 90% are squashed, i.e. 81% are squashed.\nVanilla flavored and non squashed= 90-81= 9%\nD is the answer", "correct": "D"}
|
||||
{"question": "There is well of depth 30m and frog is at bottom of the well. He jumps 3m up one day and falls back 2m down the same day. How many days will it take for the frog to come out of the well?", "options": ["A)25 days", "B)26 days", "C)27 days", "D)28 days", "E)29 days"], "rationale": "frog jumps 3 m up day & falls back 2 m down at night\nso,frog will be 3-2=1 m up in a day.\nThus, in 27 days it will be 27 m up\non 28 th day it will be at top i.e 27+3 = 30 m & will not fall down.\nANSWER:D", "correct": "D"}
|
||||
{"question": "The sum of the 5 consecutive two digit odd numbers when divided by 10 becomes a perfect square, which of the following can be one of these 5 numbers?", "options": ["A)47", "B)91", "C)41", "D)67", "E)44"], "rationale": "perfect square:- 1,4,9,16,25,36\nsum=square*10=10,40,90,160,250,360\nsum of 4 odd consecutive numbers is multiple of 4\nso the only number left are 40,160,360\nsum/4=40/4=10 is not possible\nsum/4=360/4=90 is not possible\nsum/4=160/4=40 is the only option available i.e 41\nANSWER:C", "correct": "C"}
|
||||
{"question": "In a class, 8% of total students are interested in Football. 4/5 of total students are interested in Cricket. 10% of total students are interested in Basketball and remaining 20 students are not interested in any games. How many students are there in the class?", "options": ["A)850", "B)800", "C)900", "D)950", "E)1000"], "rationale": "Let x is total no. of students\n8x/100+4x/5+10x/100+20=x\nBy solving this\nx=1000\nANSWER:E", "correct": "E"}
|
||||
{"question": "Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years. What is definitely the difference between R and Q's age?", "options": ["A)22", "B)27", "C)29", "D)Cannot be determined", "E)None of the above"], "rationale": "R - Q = R - T\nQ = T.\nAlso R + T = 50; R + Q = 50\nSo, (R - Q) cannot be determined.\nAnswer:D", "correct": "D"}
|
||||
{"question": "Calculate the maximum distance you can travel with $8.50 on a cab which charges $3.50 for the first quarter-mile and 10 cents for each additional quarter mile.", "options": ["A)11.75 miles", "B)12.75 miles", "C)17.75 miles", "D)14.75 miles", "E)10.75 miles"], "rationale": "Subtract the fee for te first quarter mile $8.50 - $3.50 = $5\nCalculate how many extra additional quarter miles---$5/10 cents => 50 quarter miles => 12.5 miles\nTotal distance is 12.5 miles + 1/4 (first quarter mile)\n12.75 miles\nAnswer: B", "correct": "B"}
|
||||
{"question": "In IPL season, Sachin current batting average is 51. In the finals, he scores 78 runs, then is batting average will be 54. Find out the total number of matches played by Sachin in this season.", "options": ["A)6", "B)8", "C)9", "D)10", "E)11"], "rationale": "Let total number of matches = x\nthen, total runs 54*x\ntotal runs before final = 51*(x-1)\nruns in the final match\n54*x - 51*(x-1) = 78\nx= 9\nANSWER:C", "correct": "C"}
|
||||
{"question": "Amy is organizing her bookshelves and finds that she has 10 different types of books. She then codes each book with either a single letter or a pair of two different letters. If each type of book is uniquely represented by either a single letter or pair of letters, what is the smallest number of letters Amy will need to create the codes for all 10 types of books? (Assume the order of letters in a pair does not matter.)", "options": ["A)3", "B)4", "C)5", "D)10", "E)20"], "rationale": "The question asks for the smallest value of n, such that (n + nC2) = 10 (n represents the number of letters. In this equation, n by itself is for single-letter codes and nC2 is for two-letter codes).\nAt this point, you'd need to pick numbers, since there's really no easy way to solve nC2 = (10 \u2013 n) without a calculator.\nLooking at the answer choices, you can eliminate 10 and 20, so you can quickly narrow down the values you need to test. (i.e. (10 \u2013 n) suggests n can not be less than 10.)\nAs a general rule, whenever you're asked for the smallest value that satisfies a condition, start by testing the smallest number in the answers. Conversely, if you're asked for the largest value, start with the greatest answer.\nPlug-in n=4 to (n + nC2) = (4 + 4C2) = 4 + (4x3 /2) = (4 + 6) = 10 ANS:D", "correct": "D"}
|
||||
{"question": "A rectangular piece of 150 sq m has a length which is 1m more than the 4 times the breadth. What is the perimeter of the piece?", "options": ["A)60 m", "B)61 m", "C)62 m", "D)63 m", "E)64 m"], "rationale": "Let its breadth be = x m.\nSo length will be = (4x+1) m.\nNow,\nx * (4x+1) = 150\nor, 4x^2+x-150 = 0\nor, (4x+25)(x-6) = 0\nEither 4x = -25 or x = 6\nAs breadth can not take negetive value so x = 6\nSo its length is 4*6+1 = 25\nSo perimeter will be 2*(25+6)=62 mLet its breadth be = x m.\nSo length will be = (4x+1) m.\nNow,\nx * (4x+1) = 150\nor, 4x^2+x-150 = 0\nor, (4x+25)(x-6) = 0\nEither 4x = -25 or x = 6\nAs breadth can not take negetive value so x = 6\nSo its length is 4*6+1 = 25\nSo perimeter will be 2*(25+6)=62 m\nANSWER:C", "correct": "C"}
|
||||
{"question": "One gram of a certain health food contains 9 percent of the minimum daily requirement of vitamin E and 8 percent of the minimum daily requirement of vitamin A. If vitamins E and A are to be obtained from no other source, how many grams of the health food must be eaten daily to provide at least the minimum daily requirement of both vitamins?", "options": ["A)8.5", "B)10.5", "C)12.5", "D)14.5", "E)16.5"], "rationale": "100% / 8% = 12.5\n12.5 grams of the health food provides 12.5(8%) = 100% of the vitamin A requirement and more than 100% of the vitamin E requirement.\nThe answer is C.", "correct": "C"}
|
||||
{"question": "Assistants are needed to prepare for preparation. Each helper can make either 2 large cakes or 35 small cakes/hr. The kitchen is available for 3 hours and 20 large cakes & 700 small cakes are needed. How many helpers are required?", "options": ["A)8", "B)10", "C)12", "D)15", "E)19"], "rationale": "20 large cakes will require the equivalent of 10 helpers working for one hour. 700 small cakes will require the equivalent of 20 helpers working for one hour. This means if only one hour were available we would need 30 helpers. But since three hours are available we can use 10 helpers.\nB", "correct": "B"}
|
||||
{"question": "R, S, T, and U are points on a line, and U is the midpoint of line segment ST. If the lengths of line segments RS, RT, and ST are 5, 17, and 22, respectively. What is the length of line segment RU?", "options": ["A)6", "B)7", "C)8", "D)9", "E)10"], "rationale": "Since SR + RT = 22 = ST, then R is somewhere between S and T.\nSince ST is 22, then SU is 11 because U is the midpoint of ST.\nSince SR < SU, then R is somewhere between S and U.\nThen SR + RU = SU.\n5 + RU = 11\nRU = 6\nThe answer is A.", "correct": "A"}
|
||||
{"question": "Six pita breads contain the same amount of falafel as do two rolls. Three rolls contain the same amount of falafel as five baguettes do. Two baguettes contain the same amount of falafel as how many pita breads?", "options": ["A)12/25", "B)3/2", "C)3", "D)2", "E)25/3"], "rationale": "6P = 2R\n3R = 5B\n2B = ?P\nThus, P : R : B = 18 : 6 : 12\nP : B = 18 : 12\n= 3 : 2\nThus P = 3\nAnswer : C", "correct": "C"}
|
||||
{"question": "A shopkeeper in order to promote his new shop put a discount of 20% on all the items for one day. Now he must sell the items at original price the other day. By what percentage must he increase the price to original?", "options": ["A)21%", "B)20%", "C)25%", "D)33%", "E)18%"], "rationale": "Suppose every item is priced at $100. On 20% discount, the price will become $80. Now he must add $20 to each item for original price which is 25% of $80.", "correct": "C"}
|
||||
{"question": "The bus fare for two persons for travelling between Agra and Aligarh id four-thirds the train fare between the same places for one person. The total fare paid by 6 persons travelling by bus and 8 persons travelling by train between the two places is Rs.1512. Find the train fare between the two places for one person?", "options": ["A)126", "B)77", "C)88", "D)66", "E)54"], "rationale": "Let the train fare between the two places for one person be Rs.t\nBus fare between the two places for two persons Rs.4/3 t\n=> 6/2 (4/3 t) + 8(t) = 1512\n=> 12t = 1512 => t = 126.\nAnswer:A", "correct": "A"}
|
||||
{"question": "A rectangle has a length of 8 centimeters and a width of 3 centimeters. Find the perimeter.", "options": ["A)18cm", "B)22cm", "C)20cm", "D)30cm", "E)28cm"], "rationale": "Perimeter = 2(8 cm) + 2(3 cm) = 16 cm + 6 cm = 22 cm\nanswer:B.", "correct": "B"}
|
||||
{"question": "Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it ?", "options": ["A)520", "B)720", "C)920", "D)None", "E)Cannot be determined"], "rationale": "We have to arrange 6 books. The number of permutations is 6*5*4*3*2*1= 720\nAnswer : B", "correct": "B"}
|
||||
{"question": "A straight picket fence is composed of x pickets each of which is 1/2 inch wide. If there are 6 inches of space between each pair of pickets, which of the following represents the length of fence in feet?", "options": ["A)13x/2", "B)13x/2 - 6", "C)13x/24", "D)(13x+1)/24", "E)(13x-12)/24"], "rationale": "Number of pickets = x\nSize of pickets = 1/2\nlength of pickets = 1/2x\nIf there are x pickets, it implies that there are x -1 spaces between the picket\nLength of space = 6\ntotal number of length = 1/2 x + 6(x-1) in inches\ntotal length in feet =( 1/2 x + 6(x-1))/12\nSimplify to get (13X-12)/24\nANSWER:E", "correct": "E"}
|
||||
{"question": "A ship went on a voyage. After it had traveled 180 miles a plane started with 10 times the speed of the ship. Find the distance when they meet from starting point.", "options": ["A)238", "B)289", "C)200", "D)287", "E)187"], "rationale": "Let the speed of the ship = m miles/hr. and plane took 't' hours to meet the ship\nThen, m\u00d7t is the distance ship traveled after plane started\nSo we have, mt + 180 = 10mt\n\u21d2 9mt = 180\n\u21d2 mt = 20\nHence distance = 180 + 20 = 200 miles\nAnswer:C", "correct": "C"}
|
||||
{"question": "In a large forest, 300 deer were caught, tagged, and returned during 2001. During 2002, 500 deer were caught at random, of which only 20 had tags from the previous year. If the percent of deer in the forest that had tags during the second year and were caught in the 500 deer sample is representative of the percent of the total deer population in the forest with tags, what is the total deer population in the forest (assuming no change in population between 2001 and 2002)?", "options": ["A)300", "B)500", "C)5000", "D)6000", "E)7500"], "rationale": "Let N = the total number of deer in the forest.\nDuring the first year, the percent of deer in the entire population with tags was: 300/N\n20/500 is the percent of deer caught during the second year that had tags. Since this sample percent matches the percent for the entire population (i.e., the total number of tagged deer divided by the total number of deer), the two ratios are equal.\nEquating these two percents:\nSample = Population\n(20/500)=(300/N)\nN = (300/1)*(500/20)\nN=7500\nAnswer E", "correct": "E"}
|
||||
{"question": "In a railway station, there are two trains going. One in the harbor line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbor line starts at 5.02 A.M. A man goes to the station every day to catch the first train that comes. What is the probability of the man catching the first train?", "options": ["A)0.9", "B)0.8", "C)0.6", "D)0.65", "E)1.5"], "rationale": "For each 10 min interval, if man comes in first 2 min, he'll catch the 1st train, if he comes in next 8 min, he'll catch the 2nd train.\nHence, for harbor line = (2/10) = 0.2 and for main line 0.8.\nAnswer:B", "correct": "B"}
|
||||
{"question": "The average (arithmetic mean) of the weight of 10 vehicles is 12.2 tons. The average weight of the group of vehicles increased by 2.6 tons after a new heavy duty truck was added to the group? What is the weight in tons of the heavy duty truck?", "options": ["A)40.8", "B)41.6", "C)42.2", "D)43.5", "E)44.8"], "rationale": "The new average is 14.8 tons.\nOn average, the ten trucks are 2.6 tons below the average for a total weighting of 26 tons.\nTherefore, the added truck must be 14.8 + 26 = 40.8 tons\nThe answer is A.", "correct": "A"}
|
||||
{"question": "Boomtown urban planners expect the city\u2019s population to increase by 10% per year over the next two years. If that projection were to come true, the population two years from now would be exactly double the population of one year ago. Which of the following is closest to the percent population increase in Boomtown over the last year?", "options": ["A)20%", "B)40%", "C)50%", "D)65%", "E)75%"], "rationale": "Population now - 100;\nPopulation one year from now - 110;\nPopulation two years from now - 121;\nSince the population two years from now (121) is exactly double the population one year ago then the population one year ago was 121/2=60.5.\nNow, the question asks about the population increase over the last year, so from 60.5 (last year) to 100 (now): percent increase=difference/original*100=(100-60.5)/60.5*100=39.5/60.5*100=~2/3*100=~65%.\nAnswer: D.", "correct": "D"}
|
||||
{"question": "Arjun and Sajal are friends, each has some money. If Arun gives $30 to Sajal, the Sajal will have twice the money left with Arjun. But, if Sajal gives $10 to Arjun, Arjun will have thrice as much as is left with Sajal. How much money does each have?", "options": ["A)62, 35", "B)62, 34", "C)34, 62", "D)42, 62", "E)62, 42"], "rationale": "Suppose Arun has $X and Sajal has $Y. then,\n2(x-30)= y+30 => 2x-y =90 \u2026(i)\nand x +10 =3(y-10) => x-3y = - 40 \u2026(ii)\nSolving (i) and (ii), we get x =62 and y =34.\nArun has $62 and Sajal has $34.\nAnswer B.", "correct": "B"}
|
||||
{"question": "Julie\u2019s yard is rectangular. One side of the yard is 100 feet wide. The total area of the yard is 3,000 square feet. What is the length of the other side of the yard?", "options": ["A)30 feet", "B)20 feet", "C)10 feet", "D)50 feet", "E)60 feet"], "rationale": "Area = length x width. Divide area by width to find the missing side.\n3000 \u00f7100 = 30\nThe other side is 30 feet.\nCorrect answer A", "correct": "A"}
|
||||
{"question": "The greatest common factor of two positive integers is 11. The least common multiple of these two integers is 7700. If one of the integers is 350, what is the other?", "options": ["A)242", "B)308", "C)352", "D)412", "E)456"], "rationale": "GCF*LCM = product of 2 numbers\n11*7700 = product of 2 numbers\nother number = 11*7700/350 = 242\nAnswer is A", "correct": "A"}
|
||||
{"question": "A square piece of cloth is trimmed by 4 feet on one edge to form a rectangular piece, which is then cut diagonally in half to create two triangles. If the area of each of triangle is 70 square feet, what was the perimeter (in feet) of the original piece of square cloth?", "options": ["A)56", "B)58", "C)60", "D)62", "E)64"], "rationale": "Let x be the length of one side of the original square.\nThe area of the rectangle is x(x-4)=140.\nx=14.\nThe perimeter of the square was 4*14=56 feet.\nThe answer is A.", "correct": "A"}
|
||||
{"question": "The length of the ribbon was originally 30 cm. It was reduced in the ratio 5 : 3. What is its length now?", "options": ["A)18", "B)30", "C)6", "D)15", "E)12"], "rationale": "Length of ribbon originally = 30 cm\nLet the original length be 5x and reduced length be 3x.\nBut 5x = 30 cm\nx = 30/5 cm = 6 cm\nTherefore, reduced length = 3 cm\n= 3 \u00d7 6 cm = 18 cm\nAnswer:A", "correct": "A"}
|
||||
{"question": "M = abc is a three digit number and N = cba, if M > N and M - N + 396c = 990. Then how many values of M are more than 300.", "options": ["A)20", "B)30", "C)40", "D)200", "E)None"], "rationale": "From the given data,\nabc \u2013 cba + 396c = 990\n100a + 10b + c \u2013 (100c + 10b + a) + 396c = 990\n99a \u2013 99c + 396c = 990\nObserve that each term is divisible by 99. So on dividing the above expression by 99, we get\na \u2013 c + 4c = 10\na + 3c = 10\nFor c = 1, a = 7\nc = 2, a = 4\nc = 3, a = 1\n'b' can take any value from 0 to 9\nWe have to find the value of M more than 300. So minimum value of 'a' should be 4.\nSo total possibilities are 402, 412, ...., 492 = 10 values\n701, 711, ....., 791 = 10 values\nSo total values = 20.\nCorrect option: A", "correct": "A"}
|
||||
{"question": "there are more than 501 students in a school such that 20% of them exactly took physics and 28% of them exactly took math. What could be the least possible no of students in the school?", "options": ["A)550", "B)570", "C)600", "D)700", "E)none of these"], "rationale": "20% means 1/5 and 28% means 7/25,taking the lcm of the denominators 5 and 25 we get 25,the least multiple of 25 which is greater than 501 is 525. So, answer is none\nANSWER:E", "correct": "E"}
|
||||
{"question": "If Raj was one-third as old as Rahim 5 years back and Raj is 17 years old now, How old is Rahim now?", "options": ["A)37", "B)41", "C)40", "D)42", "E)43"], "rationale": "Raj\u2019s age today = 17 decades,\nHence, 5 decades back, he must be 12 years old.\nRahim must be 36 years old, Because (3\u00d712).\n5 years back Rahim must be 41 years old today. Because (36+5).", "correct": "B"}
|
||||
{"question": "A cow is tethered in the middle of a field with a 14 feet long rope. If the cow grazes 10 sq.ft. per day, then approximately what time will be taken by the cow to graze the whole field?", "options": ["A)51 days", "B)61 days", "C)71 days", "D)81 days", "E)91 days"], "rationale": "Area of the field grazed = [22/7*14*14]sq.ft. = 616 sq.ft.\nNumber of days taken to graze the field = 616/10 days\n=> 61 days\nANSWER:B", "correct": "B"}
|
||||
{"question": "A book was sold for Rs 27.50 with a profit of 10%. If it were sold for Rs. 25.75, then would have been percentage of profit and loss ?", "options": ["A)2% Profit", "B)3% Profit", "C)2% Loss", "D)3% Loss", "E)4% Loss"], "rationale": "S.P.=(100+gain%100\u2217C.P)\nSo, C.P. = (100/110\u221725.75)\nWhen S.P. = 25.75 then\nProfit=25.75\u221225=Re.0.75\nProfit%=0.75/25\u2217100=3%\nAnswer is B", "correct": "B"}
|
||||
{"question": "In how many ways can a teacher in a kindergarten school arrange a group of 3 children (Susan, Tim and Zen) on 3 identical chairs in a straight line so that Susan is on the left of Tim?", "options": ["A)7", "B)3", "C)2", "D)1", "E)6"], "rationale": "Total ways in which 3 children can be arranged on 3 chairs = 3*2*1 = 6\nBut in half cases Susan will be left of Tim and in other half of cases Tim will be on left of Susan\ni.e. Desired cases in which Susan is on the left of Tim = (1/2)*6 = 3\nB", "correct": "B"}
|
||||
{"question": "The telephone bill of a certain establishment is party fixed and partly varies as the number of calls consumed. When in a certain month 540 calls made the bill is Rs.1800. In another month 620 calls are consumed then the bill becomes Rs.2040. In another month 500 units are consumed due to more\nholidays. The bill for that month would be :", "options": ["A)Rs.1560", "B)Rs.1680", "C)Rs.1840", "D)Rs.1950", "E)Rs.1690"], "rationale": "Let the fixed amount be Rs. X and the cost of each unit be Rs. Y.\nThen, 540y + x = 1800 \u2026. And 620y + x = 2040\nOn subtracting (i) from (ii), we get 80y = 240 -> y = 3\nPutting y = 3 in (i) we get :\n540 * 3 + x = 1800 x = (1800-1620) = 180\n. : Fixed charges = Rs.180, Charge per unit = Rs.3.\nTotal charges for consuming 500 units = 180 +(500*3) = Rs.1680\nAnswer:B", "correct": "B"}
|
||||
{"question": "Two balls A and B rotate along a circular track. Ball A makes 2 full rotations in 26 minutes. Ball B makes 5 full rotation in 35 minutes. If they start rotating now from the same point, when will they be at the same starting point again?", "options": ["A)1 hour and 31 minutes", "B)2 hour and 31 minutes", "C)3 hour and 31 minutes", "D)4 hour and 31 minutes", "E)5 hour and 31 minutes"], "rationale": "If ball A makes 2 rotations in 26 minutes, it makes 1 rotation in 13 minutes. If ball B makes 5 rotations in 35 minutes, it makes 1 rotation in 7 minutes.\nThe two balls start rotating now and makes several rotations before they are at the SAME starting points. Ball A would have done a WHOLE number X of rotations and ball B would have done a WHOLE number Y of rotations. Also they would have rotated during the same period of time T. Hence\nT = 13 X = 7 Y\nHence 13 X = 7 Y\nSolve the above for X\nX = 7 Y / 13\nWe want the time when they are FIRST at the same starting point. Therefore X and Y are the smallest whole numbers of the equation X = 7 Y / 13. The smallest value of Y that gives X as a whole number is 13. Hence\nX = 7 (13) / 13 = 7\nTime T is given by\nT = 13 X = 13 * 7 = 91 minutes = 1 hour and 31 minutes\ncorrect answer A", "correct": "A"}
|
||||
{"question": "A bookshelf contains 45 books, 30 of which are hardcover and 20 of which are fiction. What is the maximum number of books that are both hardcover and fiction?", "options": ["A)10", "B)15", "C)18", "D)20", "E)30"], "rationale": "Total Books = 45\nHard Cover = 30\nNon hardcover = 15\nFiction = 20\nNon-Fiction = 25\nMaximum number of Hardcover fiction will be 20( Assuming All the Fiction Books are Hard Cover )\nHence, the correct answer will be (D)", "correct": "D"}
|
||||
{"question": "A newspaper costs $4 on Sunday and $1 the rest of the days of the week. If a hotel orders twice as many papers on Sunday as it does the rest of the days of the week and pays $210 per week for newspapers, how many newspapers does it buy on Monday?", "options": ["A)15", "B)30", "C)45", "D)60", "E)75"], "rationale": "Number of paper bought on monday = x\n# of paper bought on sunday = 2x\nTotal cost = 210 = 6*x(rest of the day cost)+8*x (sunday cost)\n14x = 210\nx = 15\nAns A", "correct": "A"}
|
||||
{"question": "A number of friends decided to go on a picnic and planned to spend Rs. 96 on eatables. Four of them, however, did not turn up. As a consequence, the remaining ones had to contribute Rs. 4 extra, each. The number of those who attended the picnic was", "options": ["A)8", "B)12", "C)16", "D)24", "E)25"], "rationale": "Let the number of persons be x. Then,\n96/x-4-96/x=4 => x=12\nSo, required number =x-4=8.\nAnswer is A", "correct": "A"}
|
||||
{"question": "A wire in the shape of rectangle of length 27 cm and breadth 17 cm is rebent to form a square. What will be the measure of each side?", "options": ["A)9", "B)11", "C)22", "D)25", "E)31"], "rationale": "Perimeter of rectangle = 2 (27 + 17) cm\n= 88cm\nPerimeter of square of side x cm = 4x\nTherefore, perimeter of rectangle = Perimeter of Square\n88 cm = 4x\nx = 22\nTherefore, each side of square = 22 cm\nANSWER : OPTION C", "correct": "C"}
|
||||
{"question": "A man divides Rs 8600 among 5 sons, 4 daughters and 2 nephews. If each daughter receives four times as much as each nephew, and each son receives five as much as each nephew. How much does each daughter receive ?", "options": ["A)Rs 400", "B)Rs 500", "C)Rs 600", "D)Rs 700", "E)Rs 800"], "rationale": "If each nephew got Rs x, then\n2x+16x+25x = 8600\nx= 200\nEach daughter got 4*200 = Rs 800\nANSWER:E", "correct": "E"}
|
||||
{"question": "Silu and Meenu were walking on the road.\nSilu said, \"I weigh 51 Kgs. How much do you weigh?\"\nMeenu replied that she wouldn't reveal her weight directly as she is overweight.\nBut she said, \"I weigh 29 Kgs plus half of my weight. \"How much does Meenu weigh?", "options": ["A)12", "B)28", "C)27", "D)58", "E)91"], "rationale": "It is given that Meenu weighs 29 Kgs plus half of her own weight.\nIt means that 29 Kgs is the other half. So she weighs 58 Kgs.\nSolving mathematically, let's assume that her weight is A Kgs.\nA = 29 + A/2\n2 \u00d7 A = 58 + A\nA = 58 Kgs.\nAnswer:D", "correct": "D"}
|
||||
{"question": "Roy was suffering from severe headaches. He went to see his doctor and the doctor gave him 5 tablets asking him to take one tablet every 15 minutes.\nHow much time will it take Roy to consume all the 5 tablets?", "options": ["A)45 Min", "B)75 Min", "C)90 Min", "D)120 Min", "E)60 Min"], "rationale": "Tablet 1 will be taken in 0 min.\nTablet 2 will be taken in 15 min.\nTablet 3 will be taken in 30 min.\nTablet 4 will be taken in 45 min.\nTablet 5 will be taken in 60 min.", "correct": "E"}
|
||||
{"question": "In a bag of red and green sweets, the ratio of red sweets to green sweets is 3:4. If the bag contains 120 green sweets, how many red sweets are there?", "options": ["A)90", "B)80", "C)95", "D)100", "E)85"], "rationale": "Let x = red sweets\nWrite the items in the ratio as a fraction.\nred/green=3/4=x/120\n3 \u00d7 120 = 4 \u00d7 x\n360 = 4x\nx=360/4=90\nAnswer:A", "correct": "A"}
|
||||
{"question": "A club consists of members whose ages are in A.P. The common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250, then number of members in the club are :", "options": ["A)18", "B)20", "C)25", "D)26", "E)27"], "rationale": "Let, n be the number of members in the club.Then,\n250 =(n x [2\u00d77+(n+1)\u00d73/12]) / 2.\nn =25.\nHence, the number of members in the club is 25.\nAnswer : C", "correct": "C"}
|
||||
{"question": "M men agree to purchase a gift for Rs. D. If 3 men drop out how much more will each have to contribute towards the purchase of the gift?", "options": ["A)D/(M-3)", "B)MD/3", "C)M/(D-3)", "D)3D/(M2-3M)", "E)None of these"], "rationale": "Initial contribution = D/m\nAfter 3 men drop out, then the contribution = D/M-3\nthe extra amount to pay = (D/m-3)-D/m\n=D(m-m+3)/)(m^2-3*m)\n= 3D/(m^2-3*m)\nANSWER:D", "correct": "D"}
|
||||
{"question": "At what price should the Karan mark a sewing machine that costs him Rs. 1200/- so that even after offering a 20% discount, he makes 20% profit?", "options": ["A)1,879", "B)1,875", "C)1,876", "D)1,872", "E)1,800"], "rationale": "Cost of a sewing machine = Rs. 1200/-\nBy giving 20% discount on the marked price of a sewing machine, the cost price is :\n100/80 * 1200 = Rs. 1500/- By making a profit of 20% on the cost price of a sewing machine, the marked price of the sewing machine is:\n120/100 \u00c3\u2014 1500 = Rs. 1,800/-\nANSWER: 3", "correct": "E"}
|
||||
{"question": "Train \u2018A\u2019 leaves Mumbai Central for Lucknow at 11 am, running at the speed of 40 kmph. Train \u2018B\u2019 leaves Mumbai Central for Lucknow by the same route at 2 pm on the same day, running at the speed of 72 kmph. At what time will the two trains meet each other?", "options": ["A)12 am on the next day", "B)5 am on the next day", "C)5 pm on the next day", "D)2 pm on the next day", "E)None of these"], "rationale": "Distance covered by train A before the train B leaves\nMumbai Central = 40 \u00d7 3 = 120 km\nTime taken to cross each other = 120\u204412 = 10 hours\nRequired time = 2pm + 10 = 12 am on the next day\nAnswer A", "correct": "A"}
|
||||
{"question": "Mark told John \"If you give me half your money I will have Rs.75. John said, \"if you give me one third of your money, I will have Rs.75/-. How much money did John have ?", "options": ["A)22", "B)60", "C)28", "D)26", "E)18"], "rationale": "Let the money with Mark and John are M and J, respectively.\nNow\nM + J/2 = 75\nM/3 + J = 75\nSolving we get M = 45, and J = 60.\nAnswer:B", "correct": "B"}
|
||||
{"question": "The number of water lilies on a certain lake doubles every two days. If there is exactly one water lily on the lake, it takes 60 days for the lake to be fully covered with water lilies. In how many days will the lake be fully covered with lilies, if initially there were 64 water lilies on it?", "options": ["A)15", "B)28", "C)30", "D)53", "E)59"], "rationale": "Starting from 1 Water Lilly it takes 60 days.\nIf there are already two present, it can be taken as the first day is over.\nIt will take 59 more days.\nNotice that we are told thatthe number of water lilies on a certain lake doubles every two days, thus if initially there were 64 water lilies instead of one, we can consider that 7 days are over and therefore only 53 days are left.\nAnswer: D.", "correct": "D"}
|
||||
{"question": "x men working x hours per day can do x units of a work in x days. How much work can be completed by y men working y hours per day in y days?", "options": ["A)x2/y2 units", "B)y3/x2 units", "C)x3/y2 units", "D)y2/x2 units", "E)None of these"], "rationale": "Amount of work completed by 1 man in 1 day, working 1 hours a day = x/x3=1/x2\nAmount of work y men in y days, working y hours a day = y3 \u00d7 (1/x2) = y3/x2 . Answer : Option B", "correct": "B"}
|
||||
{"question": "ABCDE is a regular pentagon with F at its center. How many different quadrilaterals can be formed by joining 4 of the points A,B,C,D,E and F?", "options": ["A)12", "B)10", "C)5", "D)15", "E)20"], "rationale": "The number of polygons with k sides that can be formed by joining them is nCk\nfor quadrilaterals k=4\nit has 6 sides n=6\n6C4=15\nAnswer is D", "correct": "D"}
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||||
{"question": "Points A, B, C, D lie in this order on the circumference of a circle. Minor arc AC is 160\u00b0, and minor arc BD is 150\u00b0. If B bisects minor arc AC, then what is the measure of minor arc AD?", "options": ["A)80\u00b0", "B)130\u00b0", "C)140\u00b0", "D)160\u00b0", "E)220\u00b0"], "rationale": "B bisects minor arc AC means Arc BC is 80 degrees. Now , we have arc BD = 150, therefore CD = 70.\nNow, Arc AC = 160 , CD = 70 => Arc AD = 360-230 = 130\nAnswer B", "correct": "B"}
|
||||
{"question": "If 75 percent of the employees of a certain company take a winter vacation, 40 percent take a winter and a summer vacation, and 20 percent take neither a winter nor a summer vacation, what Q percent of the employees take a summer vacation but not a winter vacation?", "options": ["A)5%", "B)15%", "C)25%", "D)35%", "E)45%"], "rationale": "Winter = 75\nBoth = 40\nNeither = 20\nWinter + Summer - Both + Neither = 100\n75 + Summer - 40 + 20 = 100\nSummer = 45\nSummer but not winter Q= Summer only = Summer - Both(i.e. summer overlap with winter) = 45 - 40 = 5\nAnswer: A", "correct": "A"}
|
||||
{"question": "The cross-section of a canal is shaped like a trapezium. If the canal is 10 m wide at the top and 6 m wide at the bottom and the area of cross-section is 640 square meters, the depth of cannel is?", "options": ["A)26", "B)28", "C)21", "D)80", "E)23"], "rationale": "1/2 * d (10 + 6)\n= 640\nd = 80\nAnswer: D", "correct": "D"}
|
||||
{"question": "During one season, a tennis team won 20 matches and lost 30% of their matches. What was the number of matches that the team lost?", "options": ["A)70", "B)30", "C)3", "D)7", "E)5"], "rationale": "Knowing that the team lost 30 % of their matches, it has won 70 % of their matches\nTotal matches = 20 / (70/ 100) = 14\nHence number of matches that the team lost = 20 x 14/100 = 3=C", "correct": "C"}
|
||||
{"question": "A point on the edge of a fan blade that is rotating in a plane 10 centimeters from the center of the fan. What is the distance traveled, in centimeters, by this point after 30 seconds when the fan runs at the rate of 300 revolutions per minutes?", "options": ["A)750pi", "B)1500pi", "C)1875pi", "D)3000pi", "E)7500pi"], "rationale": "60 seconds - 300 revolutions\n30 seconds - 150 revolutions\ndistance travelled in 1 revolution = 2*pi*r\ndistance travelled in 150 revolutions = 300*pi*r\n= 3000pi\nAnswer is D.", "correct": "D"}
|
||||
{"question": "If n is such that 36 \u2264 n \u2264 72, then x = (n2 + 2\u221an(n + 4) + 16) / (n+ 4\u221an+ 4) satisfies", "options": ["A)20 < x < 54", "B)23 < x < 58", "C)25 < x < 64", "D)28 < x < 60", "E)None of these"], "rationale": "36 \u2264 n \u2264 72\nx = (n2 + 2\u221an(n + 4) + 16) / (n+ 4\u221an+ 4)\nPut x = 36,\nx = (362 + 2\u221a36(36 + 4) + 16) / (36+ 4\u221a36+ 4)\ni.e which is least value for n = 28.\nAnswer : D", "correct": "D"}
|
||||
{"question": "At its maximum speed, a space shuttle can travel 700m high in 40 seconds. It will also take 5 seconds to pass a point. What then is the length of the space shuttle?", "options": ["A)50 m", "B)75 m", "C)100 m", "D)125 m", "E)150 m"], "rationale": "Let the length of the space shuttle be x metres and its speed be y m/sec. Then, x / y = 1 \u21d2 y = x / 5\n\u2234 (x + 700) / 40 = x / 5 \u21d4 x = 100 m. Answer C", "correct": "C"}
|
||||
{"question": "A starts a business with Rs.40,000. After 2 months, B joined him with Rs.60,000. C joined them after some more time with Rs.120,000. At the end of the year, out of a total profit of Rs.375,000, C gets Rs.150,000 as his share. How many months after B joined the business, did C join?", "options": ["A)2 months", "B)4 months", "C)23 months", "D)24 months", "E)84 months"], "rationale": "Assume that C was there in the business for x months\nA:B:C = 40000*12 : 60000*10 : 120000*x\n= 40*12 : 60*10 : 120x = 40 : 5*10 : 10x\n=8 : 10 : 2x\n= 4 : 5 : x\nC's share = 375000*x/(9+x) = 150000\n=> 375x/(9+x) = 150\n=> 15x = 6(9+x)\n=> 5x = 18 + 2x\n=> 3x = 18\n=> x = 18/3 = 6\nIt means C was there in the business for 6 months. Given that B joined the business\nafter 2 months. Hence C joined after 4 months after B joined\nAnswer is B", "correct": "B"}
|
||||
{"question": "A paper is in a square form whose one side is 20 cm. Two semi circles are drawn on its opposites as diameters. If these semi circles are cut down what is the area of the remaining paper?", "options": ["A)8.75", "B)8.79", "C)8.75", "D)8.71", "E)8.72"], "rationale": "(5 * 3.5)/2 = 8.75\nAnswer:C", "correct": "C"}
|
||||
{"question": "An athlete runs M miles in 4 hours, then rides a bike N miles in the same number of hours. Which of the following represents the average speed, in miles per hour, for these two activities combined?", "options": ["A)M + N / 8", "B)2M + N / 8", "C)M + N / 4", "D)M + 3N / 8", "E)M + N / 5"], "rationale": "M + N / 8\nformular for avg speed is total distance / total time\nTime spent running = 4 and the time spent biking = 4\ntotal time is 4 + 4 = 8\nTotal distance is M+ N\nThus A", "correct": "A"}
|
||||
{"question": "8 man work for 6 days to complete a work. How many men are required to complete same work in 1/2 day.", "options": ["A)93 men", "B)94 men", "C)95 men", "D)96 men", "E)97 men"], "rationale": "To complete a work for 6 days, 8 men are required.\nFor completing a work in 1 day = 6*8\n= 48 men\nFor completing a work in half a day (1/2) = 48*2\n= 96 men\nANSWER:D", "correct": "D"}
|
||||
{"question": "64 boys and 40 girls form a group for social work. During their membership drive, the same number of boys and girls joined the group. How many members does the group have now, if the ratio of boys to girls is 4:3?", "options": ["A)277", "B)288", "C)200", "D)277", "E)168"], "rationale": "Let us say x boys and x girls joined the group.\n(64 + x)/(40 + x) = 4/3\n192 + 3x = 160 + 4x => x = 32\nNumber of members in the group = 64 + x + 40 + x\n= 104 + 2x = 168.\nAnswer:E", "correct": "E"}
|
||||
{"question": "A cyclist travels at 12 miles per hour. How many minutes will it take to travel 48 miles?", "options": ["A)1", "B)240", "C)30", "D)60", "E)120"], "rationale": "At 12 miles per hour, to cover 48 miles the cyclist will need 4 hours or 240 minutes.\nAnswer: B.", "correct": "B"}
|
||||
{"question": "Kevin drove from A to B at a constant speed of 70 mph. Once he reached B, he turned right around with pause, and returned to A at a constant speed of 90 mph. Exactly 3 hours before the end of his trip, he was still approaching B, only 70 miles away from it. What is the distance between A and B?", "options": ["A)180", "B)90", "C)270", "D)360", "E)None of the above"], "rationale": "In the last 70 miles of his approach to B, Kevin was traveling at 70 mph, so he traveled that distance in 1 hr, or 60 minutes. That means, when he arrived at B, 60 minutes had elapsed, and he took (3 hr) \u2013 (1 hr) = 2 hr to drive the distance D at 90 mph.\nD = RT = (90 mph)[ (2 hr] = 180 mi\nAnswer = (A)", "correct": "A"}
|
||||
{"question": "30 is subtracted from a number, it is reduced to its one third. What is the value of 50% of that number?", "options": ["A)22.5", "B)84", "C)21", "D)24", "E)25"], "rationale": "2/3 x = 30 => x = 45\n45 * 1/2 = 22.5\nANSWER:A", "correct": "A"}
|
||||
{"question": "If a man rows at the rate of 4 kmph in still water and his rate against the current is 2 kmph, then the man's rate along the current is:", "options": ["A)15 kmph", "B)6 kmph", "C)12 kmph", "D)14 kmph", "E)6 kmph"], "rationale": "The speed of the current is 4-2=2 kmph. Thus, if the man navigates along the current his speed is 6kmph. Answer: E", "correct": "E"}
|
||||
{"question": "The sum of the digits of a three digit number is 17, and the sum of the squares of its digits is 109. If we subtract 495 from the number, we shall get a number consisting of the same digits written in the reverse order. Find the number.", "options": ["A)368", "B)377", "C)288", "D)997", "E)112"], "rationale": "Sum of the squares should be equal to 109. Only Options B and D satisfying. When we subtract 495, only 863 becomes 368\nAnswer:A", "correct": "A"}
|
||||
{"question": "X and Y are two alloys which were made by mixing zinc and copper in the ratio 6:9 and 7:11, respectively. If 40 grams of alloy X and 60 grams of alloy Y are melted and mixed to form alloy Z, what is the ratio of zinc and copper in the alloy Z ?", "options": ["A)69:91", "B)59:91", "C)59:90", "D)59:91", "E)69:101"], "rationale": "The ratio of zinc and copper in mixture 1 is 6/9 and in mixture 2 is 7/11.\n40 grams of mixture 1 contains 6*40/15=16 grams of zinc and 24 grams of copper\n60 grams of mixture 2 contains 7*60/18=77/3 grams of zinc and 110/3 grams of copper\nThus, ratio =(16+77/3)/(24+110/3) =59/91\nANSWER:B", "correct": "B"}
|
||||
{"question": "The nefarious bandit Hoopsmot decides to go in with his criminal partner Smolapon to purchase a number of senators. Hoopsmot contributes $16,000 to their bribery pool, and Smolapon contributes just $4,000. Their total allows them to influence 30 senators. How many senators of these can be considered Hoopsmot's?", "options": ["A)18", "B)20", "C)22", "D)24", "E)26"], "rationale": "A = 16000\nB = 4000\nA share 16 parts & B share 4 parts\nTotal 20 parts -----> 30\n----> 1 part -------> 1.5\nA share = 16 parts -----> 24\nD", "correct": "D"}
|
||||
{"question": "The difference between the squares of two numbers is 256000 and the sum of the numbers is 1000. The numbers are", "options": ["A)600, 400", "B)628, 372", "C)640, 360", "D)None of these", "E)Cannot be determined"], "rationale": "Let the numbers be x and y.\nThen, x^2 - y^2 = 256000 and x + y = 1000.\nOn dividing we get : x - y = 256.\n\u2039=\u203aSolving x + y = 1000 and x - y = 256,\n\u2039=\u203awe get : x = 628 and y = 372.\nAnswer B", "correct": "B"}
|
||||
{"question": "An astronaut weighing 211 pounds on Earth would weigh 182 pounds on Venus. The weight of the astronaut on Venus would be approximately what percent of the astronaut\u2019s weight on Earth?", "options": ["A)50%", "B)60%", "C)70%", "D)86%", "E)90%"], "rationale": "Weight of astronaut on Earth = 211 pounds\nWeight of astronaut on Venus = 182 pounds\nWeight of astronaut on Venus as a percentage of Weight of astronaut on Earth = (182/211)*100 = 86%\nAnswer D", "correct": "D"}
|
||||
{"question": "A man walks at 5 kmph for 6 hrs and at 4 kmph for 12 hrs. His average speed is", "options": ["A)4 1/3 km/h", "B)7 2/3 km/h", "C)9 \u00bd km/h", "D)8 km/h", "E)81 km/h"], "rationale": "Avg speed = total distance/total time\n= 5*6 + 4*12 / 18\n=4 1/3 km/h", "correct": "A"}
|
||||
{"question": "[(272 - 32) (124 + 176)] / (17 x 15 - 15) = ?", "options": ["A)0", "B)2.25", "C)300", "D)400", "E)None of these"], "rationale": "Given expression = [(272 - 32) (124 + 176)] / (17 x 15 - 15)\n= (240 x 300 ) / 240\n= 300\nCorrect Option: C", "correct": "C"}
|
||||
{"question": "Everyone in the family earns money each month. If the total income of a family per month is $9000 and the median income is $3000, how many members are there in the family?", "options": ["A)2", "B)3", "C)4", "D)5", "E)6"], "rationale": "There must be more than two members.\nIf there are four members, then the middle two average $3000 for a total of $6000, and the highest earner must earn at least $3000 which puts the total at $9000 minimum. The lowest earner pushes the total past $9000 so there can not be four family members.\nThere must be three family members.\nThe answer is B.", "correct": "B"}
|
||||
{"question": "The bus fare of one adult is Rs. 140 from Ranchi to Patna and bus fare of a child is half the fare of one adult between the same places. What is the total bus fare of 4 adults and 3 children between same places?", "options": ["A)Rs. 666", "B)Rs. 670", "C)Rs. 700", "D)Rs. 570", "E)Rs. 770"], "rationale": "Fare for Adult = Rs. 140.\nFare of Child = Half of the Adult = Rs. 70.\nSo,\nTotal fare = 4 *140 + 3 *70 = 560 +210 = Rs. 770.\nANSWER : E", "correct": "E"}
|
||||
{"question": "An organization decided to raise Rs. 6 lakh by collecting equal contribution from each of its employees. If each of them had contributed Rs. 60 extra, the contribution would have been Rs. 6.24 lakh. How many employees are there in that organization?", "options": ["A)300", "B)200", "C)400", "D)100", "E)500"], "rationale": "Required number of employees = (624000 - 600000)/60=24000/60=400\nAnswer is C.", "correct": "C"}
|
||||
{"question": "If there are 5,000 voters out of which 20% are not eligible to vote and there are two candidates contesting. The winning candidate won by 15% of the votes. What is the total number of votes he got ?", "options": ["A)3267", "B)2678", "C)2797", "D)2300", "E)2781"], "rationale": "Number of voters eligible for voting = 5000 \u00d7 0.8 = 4000\nNumber of extra votes gotten by the winning candidate = 4000 \u00d7 0.15 = 600\nLet the number of votes won by winning candidate = x.\n\u21d2 x \u2013 (4000 \u2013 x) = 600\n\u21d2 x = 2300\nAnswer: D", "correct": "D"}
|
||||
{"question": "For bringing each copper coin from the bottom of a river, a coin-diver gets 20 cents, and for each brass coin she gets 25 cents. If after one dive, she got $3.40. What is the minimum number of copper coins that she brought?", "options": ["A)4", "B)3", "C)2", "D)1", "E)0"], "rationale": "Let's subtract $0.20 until we find a multiple of $0.25.\n$3.40 - $0.20*2 = $3.00, which is a multiple of $0.25.\nThe answer is C.", "correct": "C"}
|
||||
{"question": "Ram and Krishna start from A and B, respectively, at the same time and travel towards each other at constant speeds of 20m/s and 40m/s, respectively, along the same route. Ram meets Krishna at point C on the road after 10 seconds. Find the total distance between A to B.", "options": ["A)700 meters", "B)1000 meters", "C)700 kilometers", "D)555 meters", "E)600 meters"], "rationale": "Vr=20m/s, Vk=40m/s\ndistance A-C = 20*10=200m\ndistance B-C = 40*10=400m\nTherefore, distance A-C = 200+400=600m.\noption E", "correct": "E"}
|
||||
{"question": "Car \u2018X\u2019 covers a distance of 320 kms in 8 hours and car \u2018Y\u2019 covers a distance of 415 kms in 5 hrs. What is the difference in the speed of the two cars?", "options": ["A)42kms/hr", "B)41km/hr", "C)43kms/hr", "D)45kms/hr", "E)None of these"], "rationale": "The speed of Car \u2019X\u2019=320kms/8hr=40kms/hr\nThe speed of car \u2019Y\u2019=415kms/5hr=83kms/hr\nthe difference is 43km/hr\nANSWER:C", "correct": "C"}
|
||||
{"question": "Winson runs from his home to his school at an average speed of 10 miles/hr, and then walks home along the same route at an average speed of 5 miles/hr. If the whole journey took one hour, how many miles is his home from his school?", "options": ["A)9", "B)6", "C)4", "D)3", "E)2"], "rationale": "Suppose x is the distance then\ngoing time + coming time = total time = 1 hour\nx/10 + x/5 = 1\nx = 1.5=2 miles\nAnswer E.", "correct": "E"}
|
||||
{"question": "A sporting goods store carries only yellow and white golf balls. At the beginning of the day it had 600 golf balls in stock, and by the end of the day it had sold 80% of its inventory of golf balls. If the store sold an equal number of yellow and white golf balls, and in doing so sold all of its white golf balls, how many yellow golf balls did the store have to begin the day?", "options": ["A)80", "B)120", "C)240", "D)320", "E)360"], "rationale": "Since the store sold an equal number of white and yellow balls, 80%/2 = 40% of the inventory at the start of the day was white balls. Then 60% of the inventory consisted of yellow balls.\n0.6(600) = 360\nThe answer is E.", "correct": "E"}
|
||||
{"question": "A flagstaff 17.5 metre high casts a shadow of length 40.25 metre. The height of building, which casts a shadow of length 28.75 metre under similar conditions will be :", "options": ["A)12 metre", "B)12.5 metre", "C)13.5 metre", "D)14 metre", "E)15 metre"], "rationale": "Less shadow, Less Height (Direct Proportion)\nSo, let height of building be x metre\nthen,\n40.25:17.5::28.75:x\n=>x=17.5\u221728.75/ 40.25\n=>x=12.5\nOption B", "correct": "B"}
|
||||
{"question": "Two cars are travelling from the same starting point in the same direction, having started their commute at the same time. The first car travels at a steady rate of 55 mph, while the second travels at a steady rate of 52 mph. How much time will pass before the cars are 15 miles away from each other?", "options": ["A)3 hours", "B)5 hours", "C)6 hours", "D)4 hours", "E)7 hours"], "rationale": "Relative Speed: 55-52=3 mph\nDistance:15 miles\nTime: distance/speed=15/3= 5 hours\nCorrect answer is B", "correct": "B"}
|
||||
{"question": "The events A and B are independent. The probability that event A occurs is 0.6, and the probability that at least one of the events A or B occurs is 0.96. What is the probability that event B occurs?", "options": ["A)0.5", "B)0.6", "C)0.7", "D)0.8", "E)0.9"], "rationale": "Let x be the probability that B does not occur.\nP(A and B do not occur) = 1 - 0.96 = 0.04\n0.4x = 0.04\nx=0.1\nP(B occurs) = 1 - x = 0.9\nThe answer is E.", "correct": "E"}
|
||||
{"question": "The ratio of the volumes of a cube to that of the sphere which will fit inside the cube is?", "options": ["A)2: \u03c0", "B)7:2", "C)8:2", "D)6: \u03c0", "E)8:3"], "rationale": "a3 : a3/8 * 4/3 \u03c0 => 6: \u03c0\nAnswer: Option D", "correct": "D"}
|
||||
{"question": "My wall contains 8 red colour ties, 13 violet colour ties,10 blue colour ties, 5 pink colour ties, 4 green colour ties. If electricity is gone and I want at least two ties of same colour then how many ties I should take out from my rack?", "options": ["A)2", "B)3", "C)4", "D)5", "E)6"], "rationale": "5 ties will get you one of different colored ties in the worst case. Thus, one more tie and you will have at least one pair. Thus, 6 is the correct answer.\nANSWER:E", "correct": "E"}
|
||||
{"question": "Find 25/12*5", "options": ["A)2.5498", "B)0.4167", "C)3.3987", "D)8.5497", "E)5.6312"], "rationale": "Answer=25/12*5\n=25/60=0.4167\nOption B is correct", "correct": "B"}
|
||||
{"question": "The value of log2 4 is:", "options": ["A)2", "B)4", "C)6", "D)8", "E)12"], "rationale": "Let log2 4 = n.\nlog2 4 = 2.\nAnswer: Option A", "correct": "A"}
|
||||
{"question": "Calculate the percentage gain of a merchant who purchased 90 kg of oranges for Rs. 450 and sold the whole lot at the rate of Rs. 7.50 per kg.", "options": ["A)50 %", "B)60 %", "C)55 %", "D)70 %", "E)58%"], "rationale": "C.P. of 1 kg = 450/90 = Rs. 5\nS.P. of 1 kg = Rs. 7.50\nGain = 7.50-5 = 2.50\nGain % = 2.50/5 * 100 = 50%. Answer: A", "correct": "A"}
|
||||
{"question": "A train M leaves City A at 5 am and reaches City B at 9am. Another train N leaves City B at 7am and reaches City A at 1030am. At what time do the 2 trains cross one another?", "options": ["A)1 hr 23 min", "B)1 hr 15 min", "C)1 hr 8 min", "D)56 min", "E)55 min"], "rationale": "Let the distance between the cities be x\nThey meet after y hrs after 7am\nM covers x in 4hrs\nN covers x in 3 1/2 i.e 7/2 hrs\nspeed of M =x/4\nspeed of N = 2x/7\nDistance covered by M in y+2 hrs + Distance covered by N in\ny hrs is x\nx/4 (y+2) +2x/7(y)=x\ny=14/15hr or 56 min\nAnswer : D.", "correct": "D"}
|
||||
{"question": "Janice bikes at 10 miles per hour, while Jennie bikes at 20. How long until they have collectively biked 1 mile?", "options": ["A)1 minute", "B)2 minutes", "C)3 minutes", "D)4 minutes", "E)5 minutes"], "rationale": "Janice's speed = 1/6 miles per minute\nJennie's speed = 1/3 miles per minute\nJanice + Jennie's speed= (1/6 + 1/3) = 1/2 miles per minute\nBoth together will finish the mile in 2 minutes\ncorrect option is B", "correct": "B"}
|
||||
{"question": "In an exam, a candidate secured 504 marks of the maximum mark of M. If the maximum mark M is converted into 800 marks, he would have secured 420 marks. What is the value of M?", "options": ["A)278", "B)2890", "C)270", "D)2702", "E)960"], "rationale": "504/M = 420/800\n(504 * 800) / 420 = M\nM = 960\nAnswer:E", "correct": "E"}
|
||||
{"question": "If Jill needed to buy 10 bottles of soda for a party in which 8 people attended, how many bottles of soda will she need to buy for a party in which 12 people are attending?", "options": ["A)6", "B)8", "C)10", "D)12", "E)14"], "rationale": "We can set up a proportion to solve:\n10 bottles / 8 people = x bottles / 12 people.\nCross-multiply to solve a proportion:\n(10)(12) = (8)(x)\n120 = 8x\n10 = x\nAnswer :C.", "correct": "C"}
|
||||
{"question": "Two ants are standing side-by-side. One ant, which is 4 inches tall, casts a shadow that is 10 inches long. The other ant is 6 inches tall. Compute, in inches, the length of the shadow that the taller ant casts.", "options": ["A)36", "B)28", "C)42", "D)15", "E)20"], "rationale": "The ratio of shadow to height is constant, so if x is the length of the shadow, then\n4/10 = 6/x and x = 15 .\ncorrect answer D", "correct": "D"}
|
||||
{"question": "The height of a room to its semi-perimeter is 2:5. It costs Rs.260 to paper the walls of the room with paper 50cm wide at Rs.2 per meter allowing an area of 15 sq.m for doors and windows. The height of the room is:", "options": ["A)2.6m", "B)3.9m", "C)4m", "D)4.2m", "E)4.4m"], "rationale": "Let, height= 2x metres & (length+ breadth)= 5x metres.\nLength of paper= (260/2)m= 130m.\nTherefore, area of paper= (130*50/100)= 65m2\nArea of 4 walls= (65+15)=80m2\n2(length+breadth)*height=80.\nTherefore, 2*5x*2x=80 or x2=4 or x=2\nTherefore, height of the room= 4m\nANSWER:C", "correct": "C"}
|
||||
{"question": "The sum of k consecutive integers is 51. If the least integer is -50, then k =", "options": ["A)40", "B)62", "C)82", "D)92", "E)102"], "rationale": "The difference is consistent with each integers , therefore the series can be A.P.\nSum of A.P. = A + (N-1) D\nA=First term\nD=Difference between each integer\nN=number of terms\nSum = A + (N - 1 ) D\n51= -50 + N - 1\nN = 102\nAnswer = E", "correct": "E"}
|
||||
{"question": "In a survey of students, each student selected from a list of 10 songs the 2 songs that the student liked best. If each song was selected 5 times, how many students were surveyed?", "options": ["A)96", "B)48", "C)32", "D)25", "E)18"], "rationale": "Each out of 10 songs was selected 5 times --> the total number of selections = 10*5 = 50.\nEach student selected 2 songs --> the total number of students = 50/2 = 25.\nAnswer: D.", "correct": "D"}
|
||||
{"question": "If one of the roots of the quadratic equation x^2 + mx + 22 = 0 is 1.5, then what is the value of m?", "options": ["A)-23.5", "B)-17.5", "C)-10.5", "D)-16.2", "E)Cannot be determined"], "rationale": "Here x=1.5 must satisfy the equation\n=> 1.5^2 + 1.5m + 22 = 0\n=> m=-16.2\nANSWER:D", "correct": "D"}
|
||||
{"question": "At an election meeting 10 speakers are to address the meeting. The only protocol to be observed is that whenever they speak the pm should speak before the mp and the mp should speak before the mla. In how many ways can the meeting be held?", "options": ["A)10!/3", "B)10!/6", "C)10!/2", "D)10!/4", "E)10!/5"], "rationale": "10 speakers can be arranged in 10! ways. Protocol to be observed only one possibility from 3! is appropriate. So, total number of ways=10!/3!=10!/6\nANS:B", "correct": "B"}
|
||||
{"question": "Anna is able to buy 5 more articles for $300 after the price of each article decreased by 15%. What is the new selling price of each article?", "options": ["A)$8", "B)$10", "C)$13.6", "D)$22.9", "E)$40"], "rationale": "p = old price.\nn = the number of items for $300 for p.\npn = (0.85p)(n + 5) --> n = 0.85(n + 5) --> n = 17.\nNew price = 300/(n + 5) = 13.6.\nAnswer: C.", "correct": "C"}
|
||||
{"question": "In a row of children Neha is 12th from left end and Radha is 6th from right end. When Radha is shifted to left by 2 places and Neha is shifted to right by 2 places there 6 children between Radha and Neha. How many children are there in the row?", "options": ["A)23", "B)27", "C)26", "D)28", "E)29"], "rationale": "After moving 2 positions to the right Neha is 14 positions from the left, and after moving 2 positions to the left, Radha is on the 8th position from the right. If there are 6 children between them, the total number of children is 14+6+8 = 28\nANSWER:D", "correct": "D"}
|
||||
{"question": "10kg of a mixture contains 30% sand and 70% clay. In order to make the mixture contain equal quantities of clay and sand how much of the mixture is to be removed and replaced with pure sand?", "options": ["A)10/7", "B)20/7", "C)30/7", "D)40/7", "E)50/7"], "rationale": "The mixture contains 3kg sand and 7 kg clay.\nFor the mixture to be in equal quantities, there should be 2 kg of clay removed.\nClay and sand are in the ratio 7:3\nSo part of sand to be removed = 2*3/7 = 6/7\nSo total mixture to be removed = 2 + 6/7 = 20/7\nANSWER:B", "correct": "B"}
|
||||
{"question": "A man spends 70% of his income. If his income increases by 20%, then what will be his new expenditure?", "options": ["A)58.3%", "B)62.5%", "C)63.5%", "D)64.5%", "E)65.5%"], "rationale": "Let Rs 100 be the income\nExpenditure=Rs70\nIncreased income=Rs120\nExpenditure in amount is same.\nSo, expenditure % =70/120 *100=58.3%\nANSWER:A", "correct": "A"}
|
||||
{"question": "What is the greatest number of identical bouquets that can be made out of 28 white and 98 red tulips if no flowers are to be left out? (Two bouquets are identical whenever the number of red tulips in the two bouquets is equal and the number of white tulips in the two bouquets is equal.)", "options": ["A)4", "B)7", "C)10", "D)14", "E)21"], "rationale": "The greatest common divisor of 28 and 98 is 14.\nWe can make 14 identical bouquets with 2 white tulips and 7 red tulips in each bouquet.\nThe answer is D.", "correct": "D"}
|
||||
{"question": "Sharon works for 5 hours to earn enough tips to buy an ice cream cake, while Karen works for 4. After how many hours will they be able to buy the cake together?", "options": ["A)1 hour", "B)2 hours", "C)3 hours", "D)4 hours", "E)5 hours"], "rationale": "Sharon's earnings = 1/5 cake per hour\nKaren's earnings = 1/4 cake per hour\nSharon + Karen's earnings= 9/20\nThey will be able to buy the cake in just over 2 hours\ncorrect option is C", "correct": "C"}
|
||||
{"question": "If x<0, y>0, and |x^3| > |y^2|, which of the following must be true?", "options": ["A)x > y", "B)y^2 > x^2", "C)-x^3 < y^2", "D)\u2013x < y", "E)x < \u2013y"], "rationale": "Let\u2019s go through each answer choice: (A) can never be true, since no negative is greater than a positive. (B) doesn\u2019t have to be true \u2013 consider what would happen if x = -2 and y = 1. (C) can never be true, as x^3 must be negative, and y^2 must be positive. (D) can never be true, since if x < 0, -x is the same thing as |x|, and |x| > y. (E) can be manipulated by multiplying both sides by -1, which gives us \u2013x > y. Remember that x < 0, so \u2013x = |x|, and y is positive, so |y| = y. Thus \u2013x^3 > y^2 is the same statement as |x^3| > |y^2|, and (B) must be true.", "correct": "B"}
|
||||
{"question": "Printer A and Printer B can each print 1\u20442 page per second. How long will it take both printers working together to print 100 pages?", "options": ["A)25 seconds", "B)50 seconds", "C)100 seconds", "D)200 seconds", "E)400 seconds"], "rationale": "Total work = Printer A + Printer B = 2 Printer A\n100= 2 * 1/2 * T => T=100 seconds\nAnswer: C", "correct": "C"}
|
||||
{"question": "Two ants are moving from their farms towards each other. Ant A is moving at a speed of 9 cm per hour and ant B is moving at a speed of 6 cm per hour. If the farms are 75 cm away from each other, what will be the distance (in cm) that ant A travels until meeting ant B?", "options": ["A)45", "B)48", "C)51", "D)54", "E)57"], "rationale": "The two ants move a total of 15 cm per hour.\nThe time it takes until they meet is 75/15=5 hours.\nIn that time, the distance that ant A travels is 5*9=45 cm.\nThe answer is A.", "correct": "A"}
|
||||
{"question": "Roberts has a property worth of $1023.65. But in a record his property worth is written as greatest positive even integer less than or equal to his property worth and it is divisible by 100. Find the difference between actual property and recorded property worth?", "options": ["A)23.65", "B)1000", "C)35.62", "D)2.65", "E)1023.65"], "rationale": "Since Robert property worth is written as greatest positive even integer less than or equal to his property worth and it is divisible by 100 then it is =1000 (greatest positive even integer less than or equal to his property worth and it is divisible by 100 is 1000).\nHence the difference = 1023.65 - 1000 = 23.65\nAnswer: A.", "correct": "A"}
|
||||
{"question": "A man spend 810 in buying trouser at Rs 70 each and shirt at 30 each. What will be the ratio of trouser and shirt when the maximum number of trouser is purchased?", "options": ["A)9 Trousers", "B)8 Trousers", "C)10 Trousers", "D)7 Trousers", "E)11 Trousers"], "rationale": "Lets assume S as price of shirt and T as price of trousers, we have bellow equation:\n70 T + 30 S = 810\nSimplifying we get : 7T + 3S = 81\nT = ( 81 - 3*S )/7\nWe need to find the least value of S which will make (81 - 3*S) divisible by 7\nSimplifying by taking 3 as common factor 3*(27-S) / 7\nLooking at the above equation its not difficult to find out least value of S is 6 so that 27- 3S becomes divisible by S\nHence, T = (81-3*S)/7 = (81-3*6)/7 = 63/7 = 9\nANSWER:A", "correct": "A"}
|
||||
{"question": "If a subscription for 15 issues of a magazine costs $42.00 and represents a saving of 25 percent of the cover prices, what is the cover price per issue?", "options": ["A)$7.73", "B)$6.73", "C)$5.73", "D)$4.73", "E)$3.73"], "rationale": "Let subscription per magazine = x\n15x = 42\n=> x= 2.8\nLet cover price per magazine = c\nSince there is a 25% saving on cover prices\n0.75c=x\n=> 0.75c = 2.8\n=>c= 3.73\nAnswer E", "correct": "E"}
|
||||
{"question": "Christopher and Jonathan were having bets. They decide that a coin will be flipped twenty times and each time it turns heads, Christopher will give $2 to Jonathan and each time it turns out to be Tails, Jonathan will give 3$ to Christopher. After flipping for twenty times none of the both won or lost any amount.\nHow many times did the coin landed on Heads ?", "options": ["A)10", "B)23", "C)16", "D)18", "E)12"], "rationale": "The amount won and lost by both is equal.\nThus 2x = 3(20-x) --- x in the number of times heads came\nX = 12", "correct": "E"}
|
||||
{"question": "Allen starts from X, goes to Y. At the same time Bob starts from Y and goes towards X. Once Allen reaches Y he changes his direction and returns to X. Once Bob reaches X, he changes his direction and returns to Y. Throughout Allen travels at 54 kmph and Bob travels at 78kmph. By the time they meet for the second time, Bob covers 48 km more than Allen. Find the distance between X and Y.", "options": ["A)144km", "B)72 km", "C)126km", "D)84 km", "E)48km"], "rationale": "Total distance 126km\nin an hour both Allen and Bob covered 126km\nthat is 54+78=132\nthey meet for the first time Bob covered more KM than Allen. 78-54=24.\nso when they meet for the second time Bob covered 24*2= 48 more km (ANSWER E)", "correct": "E"}
|
||||
{"question": "Tires of a certain brand, when purchased new, last for four years. A customer can choose to purchase the new tires at a cost of $180 per tire or can have his current tires repaired at a cost of $40 per tire, a repair that will make the current tires last for one additional year. The average cost per year of the new tires is what percent greater than the cost of repairing the current tires?", "options": ["A)8%", "B)10%", "C)12.5%", "D)16.7%", "E)25%"], "rationale": "Average cost of new tire = $45/tire\ncost of repairing the current tire = $40/tire\nnew tire is $5 more per tire.\ni e. 5/40=1/8=12.5%\nANSWER:C", "correct": "C"}
|
||||
{"question": "A rope 20 meters long is cut into two pieces. If the length of one piece of rope is 3 meters shorter than the length of the other, what is the length, in meters, of the longer piece of rope?", "options": ["A)7.5", "B)8.9", "C)9.9", "D)11.5", "E)11.7"], "rationale": "Length of the rope = 20 meters.\nAssume length of longer piece = x meters.\nLength of shorter piece = x - 3\nWe know that x + x - 3 = 20\n2x = 23\nLength of the longer piece = x = 11.5 meters\nCorrect Option: D", "correct": "D"}
|
||||
{"question": "Jerry purchased a 1-year $5,000 bond that paid an annual interest rate of 12% compounded every six months. How much interest had this bond accrued at maturity?", "options": ["A)$5102", "B)$618", "C)$216", "D)$202", "E)$200"], "rationale": "A=P(1+r/q)nq .Here q is no of times interest is compounded in a yr so it is = 2. Amount comes out to be 5618 .Hence interest is 5618-5000=618. >>B", "correct": "B"}
|
||||
{"question": "Decipher the following multiplication table:\nM A D\nB E\n-------------\nM A D\nR A E\n-------------\nA M I D", "options": ["A)9 2 0 0", "B)9 2 0 9", "C)9 2 0 1", "D)9 2 0 7", "E)9 2 2 2"], "rationale": "It is clear that E = 1 as MAD\u00d7E=MAD\nFrom the hundred's line, M + A = 10 + M or 1 + M + A = 10 + M\nAs A = 10 not possible, A = 9\nSo I = 0.\nand From the thousand's line R + 1 = A. So R = 8.\nM 9 D\nB 1\n-------------\nM 9 D\n8 9 1\n-------------\n9 M 0 D\n-------------\nAs B\u00d7D = 1, B and D takes 3, 7 in some order.\nIf B = 7 and D = 3, then M93\u00d77 = _51 is not satisfied. So B = 3 and D = 7.\n2 9 7\n3 1\n-------------\n2 9 7\n8 9 1\n-------------\n9 2 0 7\n-------------\nAnswer:D", "correct": "D"}
|
||||
{"question": "Sachin was twice as old as Ajay 10 years back. How old is Ajay today if Sachin will be 40 years old in 10 years", "options": ["A)18", "B)25", "C)15", "D)20", "E)21"], "rationale": "Explanation:\nSachin's age today = 30 years.\nSachin's age 10 years back = 20 years.\nAjay's age 10 years back = 10 years.\nAjay's age today = 20 years\nAnswer: Option D", "correct": "D"}
|
||||
{"question": "What will be the cost of gardening 1-metre \u2013 broad boundary around a rectangular plot having perimeter of 340 metres at the rate of 10 per square metre?", "options": ["A)3400", "B)1700", "C)3440", "D)Cannot be determined", "E)None of these"], "rationale": "Let l and b be the length and breadth of rectangular plot respectively.\n\u2234 According to the question,we have\n2(l + b) = 340 \u21d2 l + b = 170\nNow, (l + 2) and (b + 2) be the length and breadth of plot with boundary.\n\u2234 Required area = (l + 2) (b + 2) \u2013 lb\n= lb + 2l + 2b + 4 \u2013 lb\n= 2(l + b) + 4 = 344\n\u2234 Required cost = 344 \u00d7 10 = 3440\nAnswer C", "correct": "C"}
|
||||
{"question": "Last year, 34 percent of Ace Book Company's sales revenue came from the sale of novels. Of the remaining revenue, 1/3 was from the sale of biographies. The company's revenue from the sale of novels was approximately, how many times its revenue from the sale of biographies?", "options": ["A)1.3", "B)1.5", "C)2.1", "D)2.5", "E)3.1"], "rationale": "Percentage of revenue from novels = 34%\nRemaining revenue = 66%\nSale of biographies = 1/3 of 66% = 22%\nSale of novels / sale of biographies\n= 34/22\napprox 1.5\nAnswer B", "correct": "B"}
|
||||
{"question": "A bee bypasses 0.05% of flowers it flies by because it can sense they don't have any nectar in them. How many flowers will the bee fly by to bypass 8 flowers?", "options": ["A)2000", "B)4000", "C)8000", "D)16000", "E)32000"], "rationale": "Let the number of flowers to be flown by be x.\nThen, .05% of x=8\n(5/100)*(1/100)*x=8\nx=16000\nAnswer is D", "correct": "D"}
|
||||
{"question": "Fernando purchased a university meal plan that allows him to have a total of 3 lunches and 3 dinners per week. If the cafeteria is closed on weekends and Fernando always goes home for a dinner on Friday nights, how many options does he have to allocate his meals?", "options": ["A)5C3*4C3", "B)5C4*4C2", "C)5C2*4C4", "D)5C6*4C5", "E)4C3*4C3"], "rationale": "He can allocate his 3 free lunches on any 3 days from 5 (excluding weekends), so in 5C3 ways.\nHe can allocate his 3 free dinners on any 3 days from 4 (excluding weekends and Friday), so in 4C3 ways.\nTotal = 5C3*4C3 ways\nANS:A", "correct": "A"}
|
||||
{"question": "What should come in place of the question mark(?) in each of the following questions ?\na2 - b2/(a + b)2 (?)=(a - b)2", "options": ["A)(a + b)(a - b)", "B)(a - b)2", "C)(a + b)2", "D)a3 + b3", "E)None of these"], "rationale": "(a - b)2 x (a + b)2 / a2 - b2 = (a - b)2 x (a + b)2 / (a + b)(a - b) = (a + b) (a - b)\nAnswer : Option A", "correct": "A"}
|
||||
{"question": "A number is as much greater than 36 as is less than 86. Find the Number.", "options": ["A)60", "B)56", "C)51", "D)61", "E)41"], "rationale": "Let the number be x. Then, X-36 = 86-X\n2X = 86+36 = 122, x = 61.\nThe answer is option D) 61.", "correct": "D"}
|
||||
{"question": "A certain phone manufacturer ships its products in crates. A crate consists of p pallets, and each pallet holds 1250 phones. If the manufacturer ships 4 crates, how many phones are shipped?", "options": ["A)1000p", "B)1500p", "C)2000p", "D)2500p", "E)30000"], "rationale": "1 pallet has 1250 phones, so p pallets hold 1250p phones\n1 crate has 1250p phones, so 4 will have 1250p * 3 = 2500p", "correct": "D"}
|
||||
{"question": "A can construct a wall in 40 min and B can construct the wall in 45 min. How many hours is needed to contruct a wall if both the person working together.", "options": ["A)20 min", "B)22 min", "C)23 min", "D)21 min", "E)20 min"], "rationale": "A's one minute work = 1/40\nB's one minute work = 1/45\n(A+B)'s one minute work = 1/40 + 1/45 = 85/40*45 = 17/360\nso, (A+B)will do work together in 360/17 min = 21 3/17 minutes\n21 min approximately\nANSWER:D", "correct": "D"}
|
||||
{"question": "An express electric train takes exact three seconds to enter tunnel which is 1 mile long.\nIf train is traveling at 120 mile an hour, how long will it take to pass completely through the tunnel ?", "options": ["A)43 seconds", "B)39 seconds", "C)20 seconds", "D)33 seconds", "E)55 seconds"], "rationale": "The train takes 30 seconds to travel 1 mile, plus 3 seconds for the complete train to pass any point, making a total of 33 seconds.", "correct": "D"}
|
||||
{"question": "A merchant receives an invoice for a motor boat for $20 000 with terms 4/30, n/100. What is the highest simple interest rate at which he can afford to borrow money in order to take advantage of the discount?", "options": ["A)247.67", "B)237.67", "C)227.67", "D)215.67", "E)None of these"], "rationale": "Explanation:\nSuppose the merchant will take advantage of the cash discount of 4% of $20 000 = $800 by paying the bill within 30 days from the date of invoice. He needs to borrow $20 000 = $800 = $19 200. He would borrow this money on day 30 and repay it on day 100 (the day the original invoice is due) resulting in a 70-day loan. The interest he should be willing to pay on borrowed money should not exceed the cash discount $800.\nr=I/pt=21.73%\nThe highest simple interest rate at which the merchant can afford to borrow money is 21.73%. This is a break-even rate. If he can borrow money, say at a rate of 15%, he should do so. He would borrow $19 200 for 70 days at 15%. Maturity value of the loan is $19 200(1+0.15(70/365))=$19 752.33\nsavings would be $20 000 \u2212 $19 752.33 = $247.67\nAnswer: A", "correct": "A"}
|
||||
{"question": "There are ten lime soda bottles on a table in a restaurant. They are to be served among two different groups of customers consisting of 5 members each. How many ways are there to create these 2 groups?", "options": ["A)90", "B)105", "C)126", "D)252", "E)525"], "rationale": "Number of ways to select 5 bottles out of 10= 10!/5!5!= 252\nD is the answer", "correct": "D"}
|
||||
{"question": "A special cereal mixture contains rice, wheat and corn in the ratio of 2:3:5. If a bag of the mixture contains 3 pounds of rice, how much corn does it contain?", "options": ["A)6.5", "B)7.5", "C)7", "D)6", "E)None of the above"], "rationale": "Let x = amount of corn\nrice/corn=2/5=3/x\n2 \u00d7 x = 3 \u00d7 5\n2x = 15\nx=7.5\nAnswer:B", "correct": "B"}
|
||||
{"question": "You can purchase one soda and two energy bars for 150 cents, or two sodas and three energy bars for 300\ncents. If the costs of the items do not change, compute the cost in cents of six sodas and seven bars.", "options": ["A)500", "B)600", "C)750", "D)800", "E)900"], "rationale": "The cost of adding one soda and one energy bar is 150 cents. We need to purchase six sodas and seven bars. We know two sodas and three bars costs 300 cents, so to that purchase, we'd need to add four sodas and four bars - or four sets of items. So, we take the initial 300 + (4 * 150) = 300+600 = 900.\nAnswer - E.", "correct": "E"}
|
||||
{"question": "A pen company produces very fine quality of writing pens. Company knows that on average 10% of the produced pens are always defective so are rejected before packing. Company promises to deliver 7200 pens to its wholesaler at Rs. 10 each. It estimates the overall profit on all the manufactured pens to be 25%. What is the manufactured cost of each pen?", "options": ["A)Rs. 6", "B)Rs. 7.2", "C)Rs. 5.6", "D)Rs. 8", "E)None of these"], "rationale": "The company is able to deliver 90% of the manufactured pens. Means to produce 7200 pens they must have to produce 8000 pens as 10% are defectives. So, let K be the manufacturing price of each pen.\nTotal income (including 25% profit) = 8000 *K *1.25\nThis same income is obtained by selling 90% manufactured pens at Rs. 10 which is equal to 7200 *10.\nThus,\n8000 *K *1.25 = 7200 *10\nK = Rs. 7.2. [90% of 8000 = 7200]\nAnswer: Option B", "correct": "B"}
|
||||
{"question": "A two digit number exceeds the sum of the digits of that number by 18. If the digit at the unit's place is double the digit in the ten's place, what is the number?", "options": ["A)12", "B)24", "C)42", "D)48", "E)49"], "rationale": "Let the ten's digit be x.\nThen, unit's digit = 2x.\nNumber = 10x + 2x = 12x;\nSum of digits = x + 2x = 3x.\nTherefore, 12x - 3x = 18\n\u2039=\u203a 9x = 18\n\u2039=\u203a x = 2.\nHence, required number = 12x = 24.\nAnswer B", "correct": "B"}
|
||||
{"question": "Suppose for any real number x, [x] denotes the greatest integer less than or equal to x. Let L(x,y) = [x] + [y] + [x + y] and R(x,y) = [2x] + [2y]. Then it is impossible to find any two positive real numbers x and y for which", "options": ["A)L(x,y) = R(x,y)", "B)L(x,y) \u2260 R(x,y)", "C)L(x,y) < R(x,y)", "D)L(x,y) < R(x,y)", "E)None"], "rationale": "x \u2013 1\u2264 [x] \u2264 x\n2x + 2y - 3 \u2264 L(x,y) \u2264 2x + 2y\n=> a \u2013 3 \u2264 L \u2264 a\n2x + 2y -2 \u2264 L(x,y) \u2264 2x + 2y\n=> a \u2013 2 \u2264 R \u2264 a\nTherefore, L \u2264 R.\nAnswer : D", "correct": "D"}
|
||||
{"question": "Three bells ring at intervals of 36 seconds, 40 seconds and 48 seconds, respectively. They start ringing together at a particular time. When they will ring together again?", "options": ["A)After 6 minutes", "B)After 12 minutes", "C)After 18 minutes", "D)After 24 minutes", "E)none"], "rationale": "LCM of 36,40,48 is 720\n720/60=12\nANSWER:B", "correct": "B"}
|
||||
{"question": "An electric pole, 14 metres high, casts a shadow of 10 metres. Find the height of a tree that casts a shadow of 15 metres under similar conditions.", "options": ["A)21", "B)22", "C)20", "D)23", "E)24"], "rationale": "direct proportion x1/y1=x2/y2\n14/10=15/x\n(14/10) 15=x\n21 = x\nAnswer:A", "correct": "A"}
|
||||
{"question": "At my favorite fruit stand, an orange costs 18 dollars, a pineapple costs 27 dollars, and a grape costs 15 dollars. Using the same logic, can you tell how much a mango costs?", "options": ["A)22 dollars", "B)15 dollars", "C)20 dollars", "D)18 dollars", "E)10 dollars"], "rationale": "He cost is equal to 3 dollars for each letter in the fruits name.", "correct": "B"}
|
||||
{"question": "In the coordinate plane, a triangle has vertices at (a,0), (b,0), and (x,y). If a>x>b>0>y, which of the following represents the area of that triangle?", "options": ["A)(ay\u2212by)/2", "B)(ab\u2212ay)/2", "C)(by\u2212ay)/2", "D)(ay+by)/x", "E)(a\u2212b)/2y"], "rationale": "We must plug in the three points that satisfy y<0<b<x<a.\nOnly C satisfies the area of a triangle.\nAnswer:\nC. (by\u2212ay)/2", "correct": "C"}
|
||||
{"question": "A car finishes a journey in 20 hours at the speed of 60 km/hr. If the same distance is to be covered in 10 hours, how much speed does the car gain?", "options": ["A)80 kmph", "B)50 kmph", "C)120 kmph", "D)70 kmph", "E)80 kmph"], "rationale": "20 x 60 = 10 x S2\nS2 = 120 kmph\nAnswer: Option C", "correct": "C"}
|
||||
{"question": "Three friends Alan, Roger and Peter attempt to answer a question on an exam. Alan randomly guesses the answer, giving him a 1/5 probability of guessing correctly. Roger cheats by looking at the paper of the student in front of him, giving him a 2/3 probability of answering correctly. And Peter dutifully performs the calculations, then marks the answer, giving him a 5/6 probability of a correct answer. What is the probability that the question is answered correctly, but not via cheating?", "options": ["A)1/18", "B)1/9", "C)23/90", "D)5/18", "E)13/45"], "rationale": "Prob(Alan) = 1/5\nProb(Roger) without cheating = 2/3-1 = 1/3\nProb(Peter) = 5/6\nTotal Probability = 1/5*1/3*/5/6 = 1/18\nAnswer is A", "correct": "A"}
|
||||
{"question": "The difference between simple interest and C.I. at the same rate for Rs.5000 for 2 years in Rs.72. The rate of interest is?", "options": ["A)10%", "B)12%", "C)6%", "D)8%", "E)4%"], "rationale": "5000 = 72(100/R)2\n5 R2 = 720 => R = 12\nAnswer: Option B", "correct": "B"}
|
||||
{"question": "All 250 files on Sam's hard drive are infected by either a virus or a worm or both. The number of files that are infected only by a worm is 2.5 times the number of files that are infected by both a virus and a worm. If 50% of the files were not infected by a virus, how many of Sam's files were NOT infected by a worm?", "options": ["A)50", "B)70", "C)75", "D)100", "E)125"], "rationale": "n(Total) = 250\nn(only worm) = 125(50% of total)\nn(only worm ) = 2.5 * n(both worm and virus)\nSo,\nn(both worn and virus) = 125/2.5 = 50\nn(Total) = n(only worm) + n(both worm and virus) + n(only virus)\nn(only virus) = 250-125-50 = 75\nHence, the files not infected by worm is n(Only virus) = 75\nANSWER :(Option C)", "correct": "C"}
|
||||
{"question": "A father wants to divide Rs. 5100 between his two sons, Mohan and Sohan who are 23 and 24 at present. He divides the amount in such a way that if their shares are invested at compound interest at 4% p.a. they will receive equal amount on attaining the age of 26 years. Find Mohan's share.", "options": ["A)2400", "B)2500", "C)2600", "D)2700", "E)None of these"], "rationale": "Let, the amount Mohan and Sohan receive be Rs. m and Rs. n, respectively. The amount that they receive 3 years and 2 years after should be equal.\n\u21d2m(1+4/100)3=n(1+4/100)2\n\u21d2m(1+4/100)=n\n\u21d2m(26/25)=n\n\u21d2m/n=25/26\nTherefore, Rs.5100 must be distribued in the ratio 25 : 26\nSo Mohan's share = 5100\u00d725/(25+26)=2500\nAnswer B", "correct": "B"}
|
||||
{"question": "What is 60% of 30% of 1400 grams?", "options": ["A)450 gms", "B)100 gms", "C)252 gms", "D)240 gms", "E)None of these"], "rationale": "60/100 * 30/100 * 1400= 252\nAnswer: C", "correct": "C"}
|
||||
{"question": "A certain liquid passes through a drain at a rate of w/25 gallons every x seconds. At that rate, how many minutes will it take y gallons to pass through the drain?", "options": ["A)y/(1200xy)", "B)20xy/w", "C)5xy/(12w)", "D)w/(3xy)", "E)3y/(wx)"], "rationale": "Time needed for w/25 gallons of liquid to pass through a drain = x seconds\nTime needed for w gallons of liquid to pass through a drain = 25x seconds\nTime needed for y gallons of liquid to pass through a drain = (25x/w)*y = 25xy/w seconds\n= (25xy/w )/60 = 5xy/(12w) mins\nAnswer C", "correct": "C"}
|
||||
{"question": "A coin made of alloy of aluminum and silver measures 2 x 15 mm (it is 2 mm thick and its diameter is 15 mm). If the weight of the coin is 30 grams and the volume of aluminum in the alloy equals that of silver, what will be the weight of a coin measuring 1 x 30 mm made of pure aluminum if silver is twice as heavy as aluminum?", "options": ["A)36 grams", "B)40 grams", "C)42 grams", "D)48 grams", "E)50 grams"], "rationale": "Coin is basically a cylinder.\nSo volume of coin T= pi r^2 h = pi (7.5)^2 * 2\nCoin=Silver+Aluminum\nNow total volume of coin(T) = volume of silver + volume of aluminum\nAlso, volume of silver(Vs)= volume of aluminum(Va)\nT= Va+Vb\nT=2Va\nVa=T/2= pi (7.5)^2 * 2 /2 = pi (7.5)^2\nSilver is twice as heavy as aluminum.\nLet the weight of aluminum in coin be x\nWeight of Silver = 2x\nTotal weight of coin = 30\nx+2x=30\nx=10\nWeight of Aluminum in coin is 10gm\nWright of Silver in coin is 20gm.\nWeight of Aluminum in coin is 10gm and volume is pi (7.5)^2\nNow new Aluminum coin is made with dimension 1x30mm.\nVolume of this new coin = pi (15)^2*1.\nVolume of pi (7.5)^2 contains weight of 10 gm of aluminum\nVolume of pi (15)^2*1 will contain = 10/ pi(7.5)^ * pi (15)^2 * 1= 40gm\nANSWER:B", "correct": "B"}
|
||||
{"question": "If 10 is subtracted from 2/3 of a number the result is equal to sum of 40 and 1/3 of the number. Find the number", "options": ["A)100", "B)160", "C)150", "D)210", "E)220"], "rationale": "Let the number be x. Then,\n2x/3 - 10 = x/3 + 40\n=> x/3 = 50 => x = 150\nAnswer: Option C", "correct": "C"}
|
||||
{"question": "What is the largest integral value of 'k' for which the quadratic equation x2 - 5x + k = 0 will have two real and distinct roots?", "options": ["A)9", "B)7", "C)3", "D)8", "E)12"], "rationale": "Any quadratic equation will have real and distinct roots if the discriminant D > 0\nThe discriminant 'D' of a quadratic equation ax2 + bx + c = 0 is given by b2 - 4ac\nIn this question, the value of D = 52 - 4 * 1 * k\nIf D > 0, then 25 > 4k or k < 6.2.\nTherefore, the highest integral value that k can take is 3.\ncorrect choice is (C)", "correct": "C"}
|
||||
{"question": "900 + 5 \u00d7 12 = ?", "options": ["A)820", "B)202", "C)420", "D)209", "E)960"], "rationale": "900 + 5 \u00d7 12 = ?\nor, ? = 900 + 60 = 960\nAnswer E", "correct": "E"}
|
||||
{"question": "Shweta rides at the rate of 10 km per hour but stops for 10 minutes to take rest at the end of every 15 km. How many hours will she take to cover 100 km", "options": ["A)9 hours.", "B)10 hours.", "C)11 hours.", "D)12 hours.", "E)13 hours."], "rationale": "After every 15 km she will take a rest of 10 minutes so after every 90 minutes she will 10 min break.\nshe will 10 hours to cover 90 km distance and 1 hour to cover remaining 10km.\nSo the answer is 11 hours.\nANSWER:C", "correct": "C"}
|
||||
{"question": "Mr.Sam takes 17 hours to go by train to a certain city and return by car. He loses 4 hours if he goes both ways by train. How long would he have taken if he had traveled by car in both ways?", "options": ["A)22 hrs", "B)18 hrs", "C)16 hrs", "D)20 hrs", "E)13 hrs"], "rationale": "Going one way by train and one way by car, he takes 17 hours.\nGoing both ways by train, he takes 4 hours more => The train takes 4 hours more one way\nTherefore travelling both ways by car, he takes 4 hours less than 17\n=> He takes 17-4 = 13 hours.\nE)", "correct": "E"}
|
||||
{"question": "Jim filled his dog's bowl with dog food. Starting at 8:00 am, Jim's dog ate exactly once an hour, consuming exactly 1/3 of the dog food remaining in the bowl at each feeding session. Approximately, what percent of the original food was in the dog's bowl right before the dog began to eat at noon of the same day?", "options": ["A)20%", "B)25%", "C)30%", "D)35%", "E)40%"], "rationale": "The amount remaining after each feeding session is 2/3 of what was in the bowl.\nThere were four feeding sessions.\nThe amount remaining just before noon was (2/3)^4 = 16/81, which is about 20%.\nThe answer is A.", "correct": "A"}
|
||||
{"question": "John conducted a survey about car color. 60% of the people who took the survey were women. Of the men who were surveyed, 75% preferred red cars over green cars. If 10 men liked green cars more than red, how many people took the survey?", "options": ["A)100", "B)120", "C)50", "D)200", "E)80"], "rationale": "Let N be the number of people who took the survey. The number of men M is given by M = N - 60%N.\nThe number of men G who liked green cars more than red cars is given by\nG = M - 25%M\nGiven that G = 10, solve for N\n40 = (N - 60%N) - 25%(N - 60%N)\nN = 100\nCorrect answer is A.", "correct": "A"}
|
||||
{"question": "How many ways A boy can reach the top of stairs which contain 10 steps, when he can take either one or two steps every time?", "options": ["A)88", "B)89", "C)90", "D)91", "E)92"], "rationale": "case 1:1 1 1 1 1 1 1 1 1 1 > 1!\ncase 2:1 1 1 1 1 1 1 1 2 > 9!/8!\ncase 3:1 1 1 1 1 1 2 2 > 8!/6!*2!\ncase 4:1 1 1 1 2 2 2 > 7!/4!*3!\ncase 5:1 1 2 2 2 2 > 6!/4!*2!\ncase 6:2 2 2 2 2 > 1!\nadd answers of all cases => 1+9+28+35+15+1= 89\nANSWER:B", "correct": "B"}
|
||||
{"question": "IF one gallon of soft drink is made of 40% orange juice and 60% water, how many additional gallons of orange juice must be mixed in order to make the orange juice 60% of the soft drink?", "options": ["A)0.5", "B)1", "C)1.25", "D)1.5", "E)2"], "rationale": "Let x be the quantity to be added\n(0.4+y) / 1+y = 60/100\n=> 4+10y = 6+6y\n=> y = 2/4 = 0.5\nAnswer is A", "correct": "A"}
|
||||
{"question": "What is the units digit of 9^3-7?", "options": ["A)1", "B)3", "C)5", "D)2", "E)4"], "rationale": "The unit's digit of 9^3 = 9\n9-7=2\nAnswer D", "correct": "D"}
|
||||
{"question": "5 horses are in a race. Mr.Jain selects two of horses at random and bets on them. The probability that he selected the winning horse is", "options": ["A)1/5", "B)2/5", "C)3/5", "D)4/5", "E)6/5"], "rationale": "There are 5 horses.\nProbability of winning for each horse = 1/5.\nProbability of winning with 2 selected horses= (1/5)+(1/5)= 2/5.\nAnswer is 2/5.\nANSWER:B", "correct": "B"}
|
||||
{"question": "On dividing 2272 and 875 by a 3-digit number N, we get the same remainder. The sum of the digits of N is:", "options": ["A)10", "B)11", "C)12", "D)13", "E)14"], "rationale": "(2272 - 875) = 1397, is exactly divisible by N.\nNow, 1397 = 11 * 127\nThe required 3-digit number is 127, the sum of whose digit is 10.\nANSWER:A", "correct": "A"}
|
||||
{"question": "On a test the passing students had an average of 83, while the failing students had an average\nof 55. If the overall class average was 76, what percent of the class passed?", "options": ["A)44%", "B)66%", "C)68%", "D)72%", "E)75%"], "rationale": "Let p = proportion that passed. Then 83p + 55(1- p) = 76, so p = 21/28 = 75\ncorrect answer E", "correct": "E"}
|
||||
{"question": "The average wages of a worker during a fortnight comprising of 15 consecutive working days was $90 per day. During the first 7 days, his average wage was $87 per day and the average wage during the last 7 days was $92 per day. What was his wage on the 8th day?", "options": ["A)$83", "B)$92", "C)$90", "D)$97", "E)$104"], "rationale": "The total wage earned during the 15 days that the worker worked = 15 * 90 = $ 1350.\nThe total wage earned during the first 7 days = 7 * 87 = $ 609.\nThe total wage earned during the last 7 days = 7 * 92 = $ 644.\nTotal wage earned during the 15 days = wage during first 7 days + wage on 8th day + wage during the last 7 days.\nOr 1350 = 609 + wage on 8th day + 644\nWage on 8th day = 1350 - 609 - 644 = $ 97.\nAnswer D", "correct": "D"}
|
||||
{"question": "Two numbers are in the ratio 3:5. If 9 is subtracted from each, the new numbers are in the ratio 12:23. The smaller number is?", "options": ["A)21", "B)33", "C)35", "D)42", "E)58"], "rationale": "Let the numbers be 3x and 5x\n3x-9 / 5x-9 = 12/23\n23(3x-9) = 12(5x-9)\n9x = 99\nx = 11\nThe smaller number is = 3*11 = 33\nAnswer is B", "correct": "B"}
|
||||
{"question": "If 6 yrs are subtracted from the present age of Ajay and the remainder is divided by 18, then the present age of Rahul is obtained. If Rahul is 2 yrs younger to Denis whose age is 5 yrs, then what is Ajay's present age?", "options": ["A)40", "B)60", "C)70", "D)80", "E)90"], "rationale": "Present age of Denis =5 years\nPresent age of Rahul =5\u22122=3\nLet present age of Ajay =x\nThen, present age of Rahul =x\u2212618\nx\u2212618=3\u21d2x\u22126=3\u00d718=54\u21d2x=54+6=60\nB", "correct": "B"}
|
||||
{"question": "Anna has 4 books. If she decide to arrange the 4 books in every possible combination and moved just one book every minute, how long would it taken by her ?", "options": ["A)22 minutes", "B)30 minutes", "C)15 minutes", "D)24 minutes", "E)35 minutes"], "rationale": "Number of ways of arranging 4 books = 4 ! = 4 x 3 x 2 x 1 = 24.\nSo, total time taken = 24 minutes\nAnswer: D", "correct": "D"}
|
||||
{"question": "The manufacturer of tyres is offering a 20% discount on the price of its tubeless tyres. Some retailers are offering additional discounts. If a retailer offers an additional 20% discount, then what is the total discount available at that retailer?", "options": ["A)10%", "B)25%", "C)28%", "D)30%", "E)36%"], "rationale": "Discount = 1-0.8*0.8=1-0.64=0.36=36%\nAnswer choice E", "correct": "E"}
|
||||
{"question": "A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, then what is the area of the parking space in square feet?", "options": ["A)46", "B)81", "C)126", "D)252", "E)None"], "rationale": "We have l=9 and l+2b=37\nArea = (l x b)\n=(9 x 14) sq.ft\n= 126 sq.ft.\nAnswer C", "correct": "C"}
|
||||
{"question": "Peter wants to find 10 additional people to form a scavenger hunt team with him. If he has 10 friends who would like to participate, how many choices does he have for forming his team?", "options": ["A)0", "B)1", "C)2", "D)3", "E)4"], "rationale": "Peter has 10 friends out of which he has to select 10 so 10C10= 1 Choice.\nANSWER:B", "correct": "B"}
|
||||
{"question": "The average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weighs 53 kg is replaced by a new man. Find the weight of the new man.", "options": ["A)71", "B)62", "C)43", "D)67", "E)40"], "rationale": "Total weight increased = (1.8 x 10) kg =18 kg.\nWeight of the new man = (53 + 18) kg =71 kg.\nANSWER A", "correct": "A"}
|
||||
{"question": "A box contains a certain number of balls, marked successively from 1 to n. If there are 45 different ways that two balls can be selected from the box such that the ball with number 3 marked on it is not selected, then what is the value of n?", "options": ["A)11", "B)10", "C)9", "D)8", "E)7"], "rationale": "(n-1)C2=45\nn-1=10\nn=11\nThe answer is A.", "correct": "A"}
|
||||
{"question": "If Q, a positive integer, has 5 factors, which of the following must be true about Q?\nI. Q is the square of a prime number.\nII. Q is the fourth power of a prime number.\nIII. Q is the product of two prime numbers.", "options": ["A)I only", "B)III only", "C)II only", "D)I and II only", "E)I and III only"], "rationale": "If Q has 5 factors, we can represent Q = a^4, where a is positive integer more than 1.Let's assume that \"a\" is not a prime number. Let a = kp, where both k and p are positive integers.\nThus, Q = (kp)4=k4\u2217p4(kp)4=k4\u2217p4. Now the number of factors of Q = (4+1)*(4+1) = 25. But as the given condition states that Q has ONLY 5 factors, thus \"a\" can't have any other factor except 1 and itself. Thus, a = prime number.\nStatement I :We can represent Q = (a^2)^2. Thus, we have to prove whether a^2 is a prime number. Take a=2. We can see that it is not a prime number. Thus, this option can't answer a \"MUST be true question\"\nStatement II : Always true as proved above.\nStatement III : Again take a =2. Thus, Q = 64. We don't have this as product of 2 primes.\nThe Answer is, B.", "correct": "B"}
|
||||
{"question": "If 6x - y = 24 and y = 3x, what is the value of x?", "options": ["A)8", "B)9", "C)10", "D)11", "E)12"], "rationale": "6x - 3x = 24\n3x = 24\nx = 8\nThe answer is A.", "correct": "A"}
|
||||
{"question": "You can rent DVDs at a local video store for $4.00 per movie without a membership. However, if you purchase a membership for $7.00 per month, you can rent DVDs for $2.00 each. What is the minimum amount of DVDs you would have to rent to make it worth it to purchase the membership?", "options": ["A)1", "B)2", "C)3", "D)4", "E)5"], "rationale": "Let's compare the cost to rent x CDs.\n4x > 2x+7\n2x > 7\nx > 3.5\nThe minimum number of CDs you would need to rent is 4.\nThe answer is D.", "correct": "D"}
|
||||
{"question": "A bag of cat food weighs 7 pounds and 4 ounces. How much does the bag weigh in ounces?", "options": ["A) 108", "B) 112", "C) 116", "D) 120", "E) 124"], "rationale": "1 pound = 16 ounces.\n7 pounds and 4 ounces = (7 x 16) + 4 = 116 ounces.\nAnswer: C.", "correct": "C"}
|
||||
{"question": "By himself, Jack can clean a yacht in 12 hours. On a particular day, he happens to finish two-thirds of the work. The remaining portion of the work is done by Jill, whose rate of cleaning is just 5% of what Jack can do. How long does it take Jill to finish the remaining work?", "options": ["A)4", "B)8", "C)22", "D)50", "E)20"], "rationale": "Jack did 2/3 of the work, which is 8 hours. So if Jack would finish the job this would take him 4 extra hours. Jill's rate is 5% of what Jack would do in those 4 hours. That means it would take her ten times as much time as Jack put into the job. 5*4 equals 20, answer E.", "correct": "E"}
|
||||
{"question": "Professors borrowed Rs. 5000 from the university at simple interest. After 3 years, the university got Rs. 300 on interest. What was the rate of interest per annum?", "options": ["A)2%", "B)8%", "C)5%", "D)10%", "E)None of these"], "rationale": "(100 * 300 )/(5000*3) = 2%\nAnswer : A", "correct": "A"}
|
||||
{"question": "In a sale, a discount of 20% was available on all the articles. If Vinit purchased an article for Rs.4578 in the sale. What was the actual selling price of the article?", "options": ["A)s.5050", "B)s.5723", "C)s.5040", "D)s.4950", "E)s.4870"], "rationale": "Rs.4578 = 20% of SP\n:. SP = 4578 x 100/80 = Rs.5723\nAnswer: Option B", "correct": "B"}
|
||||
{"question": "If 27 bottles of soda cost A cents, how much will B bottles cost in dollars?", "options": ["A)AB/2700", "B)27/AB", "C)AB/270", "D)2700/AB", "E)100AB/27"], "rationale": "27 bottles cost A cents or A/100 dollars\n1 bottle will cost = A/100/27 = A/2700\nB bottles in dollars = B*A/2700 = AB/2700\nHence, answer will be A.", "correct": "A"}
|
||||
{"question": "A bag contains 11 candy bars: three cost 50 cents each, four cost $1 each and four cost $2\neach. How many ways can 3 candy bars be selected from the 11 candy bars so that the total cost is more than $4?", "options": ["A)8", "B)28", "C)46", "D)66", "E)70"], "rationale": "The ways of choosing 3 candy bars with a total cost over $4 include: choose 3 out of 4 (2 dollars each); choose 2 out of 4 (2 dollars each) and 1 from the other 7. So, the total number of ways is C4\n3 + (7 C4\n2 ) = 46. Incidentally, the total number ways of choosing 3 candy bars out of 11 is C11\n3 = 165. So the probability of them costing more than $4 if they are randomly chosen is\nC4\n3 + (7 C4\n2 )\nC11\n3\n=\n46/165\ncorrect answer C", "correct": "C"}
|
||||
{"question": "At a conference, one team is made up of 4 men and 4 women. Four presenters are chosen to present the team's findings in front of the entire conference. How many different groups of presenters can be chosen from the team if a team cannot be composed of men only or women only? (Two groups of presenters are considered different if at least one presenter is different.)", "options": ["A)120", "B)19", "C)180", "D)420", "E)460"], "rationale": "No of ways = All ways to choose - ways using just men - ways using just women = C(8,4)-C(4,4)-C(4,4) = 21 - 1 - 1 = 19\nAnswer is (B)", "correct": "B"}
|
||||
{"question": "Exactly 2/5th of the children in a certain class are girls. If there are 100 boys in the class, how many girls are in the class?", "options": ["A)50", "B)100", "C)150", "D)200", "E)70"], "rationale": "This means 2/5 of the class are boys\n2x/5=100\nx=250\ntherefore, girls = 150\nAnswer is C", "correct": "C"}
|
||||
{"question": "Two numbers are said to be relatively prime if they share no common positive factors other than 1. Set S contains the set of integers from 1 to 1,000, inclusive. What is the probability that a number chosen from S is relatively prime to 1,000?", "options": ["A)5/7", "B)3/5", "C)4/7", "D)2/5", "E)2/7"], "rationale": "We need all numbers between 1 and 1000 that are co-prime.\nBetween 1 to 10 there are 4 : 1,3,7,9\nTake the period of 10's , we have 100 periods of 10's between 1 to 1000\nSo the total number of co-primes = 400\nNow, the simple part ...\nProbability = 400/1000 (i.e picking a co-prime from the first 1000 numbers )\nAns: 2/5 D", "correct": "D"}
|
||||
{"question": "At the of his one-year investment, Charles received $54,080, including interest and principal from a certain investment. If the investment paid an annual interest of 8 percent compounded semi-annually, which of the following is the amount of money that Charles originally invested?", "options": ["A)$45,000", "B)$50,000", "C)$54,000", "D)$59,000", "E)$62,000"], "rationale": "You are given that a certain investment gave you X $ after 1 year. So the original investment must be <X\nThus you can rule out options D-E as these options will make the original amount > the interest+principle amount\nOption C is very close to the amount after 1st year and 4% will definitely give you >80$ in interests.\nNow you only have 2 options left (A and B)\nPer the question, let x be the original amount invested ---> x(1.04)^2 = 54080 .@NL This clearly shows that answer must be B.\nANSWER:B", "correct": "B"}
|
||||
{"question": "Joe's age, Joe's sister's age and Joe\u2019s fathers age sums up to 100. When Joe is as old as his father, Joe's sister will be twice as old as now. When Joe is as old as his father then his father is twice as old as when his sister was as old as her father. Find the age of Joe's father?", "options": ["A)45", "B)48", "C)50", "D)55", "E)58"], "rationale": "Joe+Sister+Father=100\nAfter x years joe age is equal to his father\nJoe+x = father\nTherefore, Sister+x = 2 * Sister\n=> Sister=x\nJoe+Sister=Father\nTherefore,\n=> 2*Father = 100\n=> Father= 50\nHence (C) is the correct answer", "correct": "C"}
|
||||
{"question": "If one third of 3/4 of a number is 21. Then, find the number?", "options": ["A)84", "B)66", "C)28", "D)19", "E)11"], "rationale": "x * 1/3 * 3/4 =21 => x = 84\nAnswer: A", "correct": "A"}
|
||||
{"question": "If m > 0, y > 0, and x is m percent of 4y, then, in terms of y, m is what percentage of x?", "options": ["A)y/400", "B)4y", "C)50y", "D)2500/y", "E)5000/y"], "rationale": "x = (m/100)*4y\nm = 100x/4y = 25x/y\nm is (2500/y)% of x.\nThe answer is D.", "correct": "D"}
|
||||
{"question": "A man wants to eat fruit for breakfast and vegetable for dinner. He has 6 different types of fruit and 8 different types of vegetables. He can only eat one type at a time. In how many ways can he eat breakfast and dinner.", "options": ["A)54", "B)24", "C)48", "D)20", "E)36"], "rationale": "Number of choices for fruit=6, number of choices for vegetable=8\nThe total number of combinations =8*6=48\nAnswer C", "correct": "C"}
|
||||
{"question": "There is a 90% chance that a registered voter in Burghtown voted in the last election. If five registered voters are chosen at random, what is the approximate likelihood that exactly four of them voted in the last election?", "options": ["A)26.2%", "B)32.8%", "C)43.7%", "D)59.0%", "E)65.6%"], "rationale": "The probability that four of five voted is :\nP(1st one voted) X ... X P(4th one voted) X (5th one NOT voted)\n= 0.9 x 0.9 x 0.9 x 0.9 x 0.1\n= 0.81 x 0.81 x 0.1 = 0.6561\nANSWER: E", "correct": "E"}
|
||||
{"question": "Alice wants to put up fencing around three sides of her rectangular yard and leave one side of 10 meters unfenced. If the yard has an area of 240 square meters, how many meters of fencing does she need?", "options": ["A)58", "B)62", "C)66", "D)70", "E)74"], "rationale": "The sides of the rectangle have a length of 10 and L.\nThe area is 10*L=240 so L=24.\nShe needs fencing with a length of 10+2*24=58 meters.\nThe answer is A.", "correct": "A"}
|
||||
{"question": "John would make the 3-letter codes with diffenrent 5 vowels and 20 consonants with the condition that the middle letter must be vowel and the first letter and the third letter must be different from each other and are both consonant. How many different codes can be made?", "options": ["A)1,980", "B)2,020", "C)2,100", "D)2,200", "E)1,900"], "rationale": "There should be a vowel in the middle of the 3-letter code, which means that 5 letters can be in the middle. Then, 20 letters can be placed in the first letter and 19 letters can be placed in the last letter as they should be different letters. Thus, 20*5*19=1,900.\nTherefore, the answer is E.", "correct": "E"}
|
||||
{"question": "What is the sum of three consecutive integers whose product can be expressed as 727+728+729.", "options": ["A)37", "B)38", "C)36", "D)30", "E)39"], "rationale": "The addition of any 3 consecutive numbers is always divisible by 3\nthat rules out options a,b,c\nNow focus on options 30 and 39\nIf we consider\n12, 13, 14 the sum is 39\nAnswer = E", "correct": "E"}
|
||||
{"question": "The capacity of a tank of dimensions (8 m \u00d7 6 m \u00d7 2.5 m) is", "options": ["A)120 litres", "B)1200 litres", "C)12000 litres", "D)120000 litres", "E)None of these"], "rationale": "Capacity of the bank = Volume of the bank\n= (8x100x6x100x2.5x100/1000) =\nanswer D", "correct": "D"}
|
||||
{"question": "The population of a city is 5265526. If there are 4169516 adults in the city, how many children are there in the city?", "options": ["A)1095961", "B)1065961", "C)1085961", "D)1097961", "E)1096061"], "rationale": "Population of the city = 5265526\nNumber of adults = 4169516\nNumber of children = 5265526-4169516 = 1096061\nAnswer :E", "correct": "E"}
|
||||
{"question": "It takes a worker 9 minutes to drive from home to work at an average rate of 20 kilometers per hour. How many minutes will it take the worker to cycle from home to work along the same route at an average rate of 6 kilometers per hour?", "options": ["A)30", "B)32", "C)35", "D)36", "E)40"], "rationale": "distance = time*speed = (9/60)(20) kilometers\ntime to bike = distance/speed = (9*20)/(60*6) = 30/60 hours = 30 minutes.\nThe answer is A.", "correct": "A"}
|
||||
{"question": "Two friends are eating a jar full of candies. Had P eaten alone, it would have taken him 10 minutes to finish the candies in the jar. Had Q eaten alone, it would have taken her 5 minutes to finish half the jar. Since both of them are eating simultaneously, how many minutes would it take them to empty the jar?", "options": ["A)4", "B)5", "C)6", "D)7", "E)8"], "rationale": "Together they eat 1/10 + 1/10 = 1/5 of the jar per minute.\nThe time to finish the jar is 5 minutes.\nThe answer is B.", "correct": "B"}
|
||||
{"question": "A grocery sells a bag of ice for $1.25, and makes 20% profit. If it sells 500 bags of ice, how much total profit does it make?", "options": ["A)125", "B)150", "C)225", "D)250", "E)275"], "rationale": "Profit per bag = 1.25 * 0.20 = 0.25\nTotal profit = 500 * 0.25 = 125\nAnswer is A.", "correct": "A"}
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{"content": "If $3 a ^ { m + 2 } b$ and $\\frac { 1 } { 2 } ab ^ { n - 1 }$ are similar terms, then $m + n$ is equal to.", "answer": "1", "steps": "$3 a ^ { m + 2 } b$ and $\\frac { 1 } { 2 } ab ^ { n - 1 }$ are like terms. We can obtain $m + 2 = 1$ and $n - 1 = 1$. Solving for $m$ and $n$, we get $m = - 1$ and $n = 2$. Therefore, $m + n = - 1 + 2 = 1$.", "expr_cands": ["3 a ^ { m + 2 } b", "a", "m", "b", "\\frac { 1 } { 2 } ab ^ { n - 1 }", "n", "m + n", "m + 2 = 1", "m = - 1", "n - 1 = 1", "n = 2", "1"], "exprs": ["m + 2 = 1", "n - 1 = 1", "m = - 1", "n = 2", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a ^ { m + 2 } b"}, {"id": "m + 2 = 1"}, {"id": "\\frac { 1 } { 2 } ab ^ { n - 1 }"}, {"id": "$3 a ^ { m + 2 } b$ 与 $\\frac { 1 } { 2 } ab ^ { n - 1 }$ 是同类项"}, {"id": "n - 1 = 1"}, {"id": "m = - 1"}, {"id": "n = 2"}, {"id": "m + n"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "3 a ^ { m + 2 } b", "target": "m + 2 = 1"}, {"rel": "被描述", "source": "3 a ^ { m + 2 } b", "target": "n - 1 = 1"}, {"rel": "等式方程求解", "source": "m + 2 = 1", "target": "m = - 1"}, {"rel": "被描述", "source": "\\frac { 1 } { 2 } ab ^ { n - 1 }", "target": "m + 2 = 1"}, {"rel": "被描述", "source": "\\frac { 1 } { 2 } ab ^ { n - 1 }", "target": "n - 1 = 1"}, {"rel": "限制性描述", "source": "$3 a ^ { m + 2 } b$ 与 $\\frac { 1 } { 2 } ab ^ { n - 1 }$ 是同类项", "target": "m + 2 = 1"}, {"rel": "限制性描述", "source": "$3 a ^ { m + 2 } b$ 与 $\\frac { 1 } { 2 } ab ^ { n - 1 }$ 是同类项", "target": "n - 1 = 1"}, {"rel": "等式方程求解", "source": "n - 1 = 1", "target": "n = 2"}, {"rel": "代入", "source": "m = - 1", "target": "1"}, {"rel": "代入", "source": "n = 2", "target": "1"}, {"rel": "被代入", "source": "m + n", "target": "1"}]}}
|
||||
{"content": "The solution to the equation $y - \\frac { y - 1 } { 2 } = - \\frac { y + 2 } { 5 }$ is ____ ?", "answer": "y = - \\frac { 9 } { 7 }", "steps": "To eliminate the denominator, we have $10 y - 5 ( y - 1 ) = - 2 ( y + 2 )$. Expanding the brackets gives $10 y - 5 y + 5 = - 2 y - 4$. Rearranging terms gives $10 y - 5 y + 2 y = - 4 - 5$, which simplifies to $7 y = - 9$. Dividing both sides by 7 gives $y = - \\frac { 9 } { 7 }$.", "expr_cands": ["y - \\frac { y - 1 } { 2 } = - \\frac { y + 2 } { 5 }", "y", "10 y - 5 ( y - 1 ) = - 2 ( y + 2 )", "y = - \\frac { 9 } { 7 }", "10 y - 5 y + 5 = - 2 y - 4", "10 y - 5 y + 2 y = - 4 - 5", "7 y = - 9", "1"], "exprs": ["y = - \\frac { 9 } { 7 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y - \\frac { y - 1 } { 2 } = - \\frac { y + 2 } { 5 }"}, {"id": "y = - \\frac { 9 } { 7 }"}], "links": [{"rel": "等式方程求解", "source": "y - \\frac { y - 1 } { 2 } = - \\frac { y + 2 } { 5 }", "target": "y = - \\frac { 9 } { 7 }"}]}}
|
||||
{"content": "If $( m + 4 ) ^ 2 + | n - 3 | = 0$, then $\\frac { 1 } { 2 } m - n$ = ____?", "answer": "- 5", "steps": "$\\because ( m + 4 ) ^ 2 + | n - 3 | = 0$, $\\therefore m + 4 = 0$, $n - 3 = 0$, which means $m = - 4$, $n = 3$. Then the original expression equals $- 2 - 3 = - 5$.", "expr_cands": ["( m + 4 ) ^ { 2 } + | n - 3 | = 0", "m", "n", "\\frac { 1 } { 2 } m - n", "m + 4 = 0", "m = - 4", "n - 3 = 0", "n = 3", "- 2 - 3", "- 5"], "content_formula": [[["(", "m", "+", "4", ")", "^", "{", "2", "}", "+", "|", "n", "-", "3", "|", "=", "0"], ["\\frac", "{", "1", "}", "{", "2", "}", "m", "-", "n"]], [[1, 18], [20, 30]]], "steps_formula": [[["(", "m", "+", "4", ")", "^", "{", "2", "}", "+", "|", "n", "-", "3", "|", "=", "0"], ["m", "+", "4", "=", "0"], ["n", "-", "3", "=", "0"], ["m", "=", "-", "4"], ["n", "=", "3"], ["-", "2", "-", "3", "=", "-", "5"]], [[1, 18], [20, 25], [26, 31], [33, 37], [38, 41], [45, 52]]], "exprs": ["m + 4 = 0", "n - 3 = 0", "m = - 4", "n = 3", "- 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m + 4 ) ^ { 2 } + | n - 3 | = 0"}, {"id": "m + 4 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "n - 3 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "m = - 4"}, {"id": "n = 3"}, {"id": "\\frac { 1 } { 2 } m - n"}, {"id": "- 5"}], "links": [{"rel": "被描述", "source": "( m + 4 ) ^ { 2 } + | n - 3 | = 0", "target": "m + 4 = 0"}, {"rel": "被描述", "source": "( m + 4 ) ^ { 2 } + | n - 3 | = 0", "target": "n - 3 = 0"}, {"rel": "等式方程求解", "source": "m + 4 = 0", "target": "m = - 4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m + 4 = 0"}, {"rel": "等式方程求解", "source": "n - 3 = 0", "target": "n = 3"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "n - 3 = 0"}, {"rel": "代入", "source": "m = - 4", "target": "- 5"}, {"rel": "代入", "source": "n = 3", "target": "- 5"}, {"rel": "被代入", "source": "\\frac { 1 } { 2 } m - n", "target": "- 5"}]}}
|
||||
{"content": "Given a quadratic equation in one variable $x$, $x ^ 2 + x + m = 0$, with one root being $x = 1$, what is the other root of this equation?", "answer": "- 2", "steps": "Suppose the quadratic equation in one variable about $x$ is $x ^ 2 + x + m = 0$, and $\\alpha$ is another real root of the equation. Since one real root of the quadratic equation in one variable about $x$ is $1$, we have $\\alpha + 1 = - 1$. Therefore, $\\alpha = - 2$.", "expr_cands": ["x", "x ^ { 2 } + x + m = 0", "m", "x = 1", "1", "\\alpha + 1 = - 1", "alpha = - 2", "\\alpha", "\\alpha = - 2"], "exprs": ["\\alpha + 1 = - 1", "\\alpha = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + x + m = 0"}, {"id": "\\alpha + 1 = - 1"}, {"id": "x = 1"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } + x + m = 0$ 的一个根是 $x = 1$"}, {"id": "设关于 $x$ 的一元二次方程 $x ^ { 2 } + x + m = 0$ 的另一个实数根是 \\alpha"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "\\alpha = - 2"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + x + m = 0", "target": "\\alpha + 1 = - 1"}, {"rel": "等式方程求解", "source": "\\alpha + 1 = - 1", "target": "\\alpha = - 2"}, {"rel": "被描述", "source": "x = 1", "target": "\\alpha + 1 = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } + x + m = 0$ 的一个根是 $x = 1$", "target": "\\alpha + 1 = - 1"}, {"rel": "假设描述", "source": "设关于 $x$ 的一元二次方程 $x ^ { 2 } + x + m = 0$ 的另一个实数根是 \\alpha", "target": "\\alpha + 1 = - 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "\\alpha + 1 = - 1"}]}}
|
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{"content": "The parabola $y = - 5 { x } ^ 2 + 1$ is translated $2$ units upward and $1$ unit to the left, resulting in the parabola _____.", "answer": "y = - 5 ( x + 1 ) ^ { 2 } + 3", "steps": "The parabola $y = - 5 { x } ^ 2 + 1$ is first shifted upward by 2 units, resulting in $y = - 5 { x } ^ 2 + 3$. Then it is shifted left by 1 unit, resulting in $y = - 5 {( x + 1 )} ^ 2 + 3$.", "expr_cands": ["y = - 5 { x } ^ { 2 } + 1", "y", "x", "2", "1", "y = - 5 { x } ^ { 2 } + 3", "1 - 5 x ^ { 2 } = - 5 { x } ^ { 2 } + 3", "y = - 5 { ( x + 1 ) } ^ { 2 } + 3", "3 - 5 x ^ { 2 } = - 5 { ( x + 1 ) } ^ { 2 } + 3", "3 - 5 x ^ { 2 }", "y = - 5 ( x + 1 ) ^ { 2 } + 3"], "exprs": ["y = - 5 { ( x + 1 ) } ^ { 2 } + 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "y = - 5 { ( x + 1 ) } ^ { 2 } + 3"}, {"id": "y = - 5 { x } ^ { 2 } + 1"}, {"id": "1"}, {"id": "将抛物线 $y = - 5 { x } ^ { 2 } + 1$ 向上平移 $2$ 个单位长度"}, {"id": "再向左平移 $1$ 个单位长度"}], "links": [{"rel": "被描述", "source": "2", "target": "y = - 5 { ( x + 1 ) } ^ { 2 } + 3"}, {"rel": "被描述", "source": "y = - 5 { x } ^ { 2 } + 1", "target": "y = - 5 { ( x + 1 ) } ^ { 2 } + 3"}, {"rel": "被描述", "source": "1", "target": "y = - 5 { ( x + 1 ) } ^ { 2 } + 3"}, {"rel": "限制性描述", "source": "将抛物线 $y = - 5 { x } ^ { 2 } + 1$ 向上平移 $2$ 个单位长度", "target": "y = - 5 { ( x + 1 ) } ^ { 2 } + 3"}, {"rel": "限制性描述", "source": "再向左平移 $1$ 个单位长度", "target": "y = - 5 { ( x + 1 ) } ^ { 2 } + 3"}]}}
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{"content": "If the radical $\\sqrt { x - 8 }$ is defined, then the range of real numbers for $x$ is ____ ?", "answer": "x \\ge 8", "steps": "Since the radical $\\sqrt { x - 8 }$ is defined, therefore $x - 8 \\ge 0$, which implies $x \\ge 8$.", "expr_cands": ["\\sqrt { x - 8 }", "x", "x - 8 \\ge 0", "8 \\le x", "x \\ge 8"], "exprs": ["x - 8 \\ge 0", "x \\ge 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x - 8 }"}, {"id": "x - 8 \\ge 0"}, {"id": "根式 $\\sqrt { x - 8 }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 8"}], "links": [{"rel": "被描述", "source": "\\sqrt { x - 8 }", "target": "x - 8 \\ge 0"}, {"rel": "不等式方程求解", "source": "x - 8 \\ge 0", "target": "x \\ge 8"}, {"rel": "限制性描述", "source": "根式 $\\sqrt { x - 8 }$ 有意义", "target": "x - 8 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 8 \\ge 0"}]}}
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{"content": "If $a ^ { m } \\times a ^ { 2 } = a ^ { 7 }$, then the value of $m$ is ____?", "answer": "5", "steps": "According to the multiplication rule of powers with the same base: when multiplying powers with the same base, keep the base the same and add the exponents. We have $m + 2 = 7$, so solving for $m$ gives $m = 5$.", "expr_cands": ["a ^ { m } \\times a ^ { 2 } = a ^ { 7 }", "a", "m", "m + 2 = 7", "m = 5"], "exprs": ["m + 2 = 7", "m = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { m } \\times a ^ { 2 } = a ^ { 7 }"}, {"id": "m + 2 = 7"}, {"id": "m = 5"}], "links": [{"rel": "同取对数", "source": "a ^ { m } \\times a ^ { 2 } = a ^ { 7 }", "target": "m + 2 = 7"}, {"rel": "等式方程求解", "source": "m + 2 = 7", "target": "m = 5"}]}}
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{"content": "If line segment $a$ and $b$ satisfy $\\frac { a } { b } = \\frac { 5 } { 2 }$, then the value of $\\frac { a - b } { b }$ is ____?", "answer": "\\frac { 3 } { 2 }", "steps": "$\\because \\frac { a } { b } = \\frac { 5 } { 2 }$, $\\therefore$ we can assume $a = 5 k$, then $b = 2 k$, $\\therefore \\frac { a - b } { b } = \\frac { 5 k - 2 k } { 2 k } = \\frac { 3 } { 2 }$.", "expr_cands": ["a", "b", "\\frac { a } { b } = \\frac { 5 } { 2 }", "\\frac { a - b } { b }", "a = 5 k", "k", "b = 2 k", "\\frac { 3 } { 2 }"], "exprs": ["a = 5 k", "b = 2 k", "\\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a } { b } = \\frac { 5 } { 2 }"}, {"id": "a = 5 k"}, {"id": "可设 $a = 5 k$"}, {"id": "线段 $a$ , $b$ 满足 $\\frac { a } { b } = \\frac { 5 } { 2 }$"}, {"id": "b = 2 k"}, {"id": "设b=2k"}, {"id": "\\frac { a - b } { b }"}, {"id": "\\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\frac { a } { b } = \\frac { 5 } { 2 }", "target": "a = 5 k"}, {"rel": "被描述", "source": "\\frac { a } { b } = \\frac { 5 } { 2 }", "target": "b = 2 k"}, {"rel": "代入", "source": "a = 5 k", "target": "\\frac { 3 } { 2 }"}, {"rel": "假设描述", "source": "可设 $a = 5 k$", "target": "a = 5 k"}, {"rel": "限制性描述", "source": "线段 $a$ , $b$ 满足 $\\frac { a } { b } = \\frac { 5 } { 2 }$", "target": "a = 5 k"}, {"rel": "限制性描述", "source": "线段 $a$ , $b$ 满足 $\\frac { a } { b } = \\frac { 5 } { 2 }$", "target": "b = 2 k"}, {"rel": "代入", "source": "b = 2 k", "target": "\\frac { 3 } { 2 }"}, {"rel": "假设描述", "source": "设b=2k", "target": "b = 2 k"}, {"rel": "被代入", "source": "\\frac { a - b } { b }", "target": "\\frac { 3 } { 2 }"}]}}
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{"content": "Calculate $( 3 a - b ) ( - 3 a - b )$ equals to ____?", "answer": "b ^ { 2 } - 9 a ^ { 2 }", "steps": "$- b$ is the same term, and its opposite is $3 a$ and $- 3 a$, respectively. Therefore, the result is $( - b ) ^ { 2 } - 9 a ^ { 2 } = b ^ { 2 } - 9 a ^ { 2 }$.", "expr_cands": ["( 3 a - b ) ( - 3 a - b )", "b", "a", "- b", "3 a", "- 3 a", "( - b ) ^ { 2 } - 9 a ^ { 2 }", "b ^ { 2 } - 9 a ^ { 2 }"], "exprs": ["b ^ { 2 } - 9 a ^ { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 3 a - b ) ( - 3 a - b )"}, {"id": "b ^ { 2 } - 9 a ^ { 2 }"}], "links": [{"rel": "展开", "source": "( 3 a - b ) ( - 3 a - b )", "target": "b ^ { 2 } - 9 a ^ { 2 }"}]}}
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{"content": "If $x ^ a = 2$ and $x ^ b = 5$, then the value of $x ^ { 2 a + b }$ is ____?", "answer": "20", "steps": "Since $x ^ a = 2$ and $x ^ b = 5$, it follows that $x ^ { 2 a } = 2 ^ 2 = 4$. Therefore, $x ^ { 2 a + b } = x ^ { 2 a } \\times x ^ b = 4 \\times 5 = 20$.", "expr_cands": ["x ^ { a } = 2", "a", "x", "x ^ { b } = 5", "b", "x ^ { 2 a + b }", "x ^ { 2 a }", "4", "20"], "exprs": ["20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 a + b }"}, {"id": "20"}, {"id": "x ^ { a } = 2"}, {"id": "x ^ { b } = 5"}], "links": [{"rel": "被代入", "source": "x ^ { 2 a + b }", "target": "20"}, {"rel": "代入", "source": "x ^ { a } = 2", "target": "20"}, {"rel": "代入", "source": "x ^ { b } = 5", "target": "20"}]}}
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{"content": "If $\\sqrt { x + { 2 } }$ is undefined, then the range of values for $x$ is ____?", "answer": "x < - 2", "steps": "From the given information, we have $x + 2 < 0$, which implies that $x < - 2$.", "expr_cands": ["\\sqrt { x + { 2 } }", "x", "x + 2 < 0", "x < - 2"], "exprs": ["x + 2 < 0", "x < - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + { 2 } }"}, {"id": "x + 2 < 0"}, {"id": "$\\sqrt { x + { 2 } }$ 无意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x < - 2"}], "links": [{"rel": "被描述", "source": "\\sqrt { x + { 2 } }", "target": "x + 2 < 0"}, {"rel": "不等式方程求解", "source": "x + 2 < 0", "target": "x < - 2"}, {"rel": "限制性描述", "source": "$\\sqrt { x + { 2 } }$ 无意义", "target": "x + 2 < 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x + 2 < 0"}]}}
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{"content": "If $a - b = \\frac { 1 } { 2 }$, then the value of $- 3 ( b - a )$ is ____?", "answer": "\\frac { 3 } { 2 }", "steps": "$\\because a - b = \\frac { 1 } { 2 }$, which means $b - a = - \\frac { 1 } { 2 }$, $\\therefore - 3 ( b - a ) = - 3 * ( - \\frac { 1 } { 2 }) = \\frac { 3 } { 2 }$.", "expr_cands": ["a - b = \\frac { 1 } { 2 }", "a", "b", "- 3 ( b - a )", "b - a", "- \\frac { 1 } { 2 }", "\\frac { 3 } { 2 }"], "exprs": ["\\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 3 ( b - a )"}, {"id": "\\frac { 3 } { 2 }"}, {"id": "a - b = \\frac { 1 } { 2 }"}], "links": [{"rel": "被代入", "source": "- 3 ( b - a )", "target": "\\frac { 3 } { 2 }"}, {"rel": "代入", "source": "a - b = \\frac { 1 } { 2 }", "target": "\\frac { 3 } { 2 }"}]}}
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{"content": "When $x = - 2$, the value of the algebraic expression $x ( 2 - k ) + 4$ is 18. What is the value of the expression when $x = 5$?", "answer": "- 31", "steps": "Substituting $x = - 2$ into $x ( 2 - k ) + 4 = 18$, we get $- 4 + 2 k + 4 = 18$. Solving for $k$, we get $k = 9$. Therefore, the algebraic expression is $- 7 x + 4$. Substituting $x = 5$ into this expression, we get $- 7 \\times 5 + 4 = - 31$.", "expr_cands": ["x = - 2", "x", "x ( 2 - k ) + 4", "k", "18", "x = 5", "x ( 2 - k ) + 4 = 18", "2 k = 18", "- 4 + 2 k + 4 = 18", "k = 9", "- 7 x + 4", "- 7 * 5 + 4", "- 31"], "exprs": ["x ( 2 - k ) + 4 = 18", "- 4 + 2 k + 4 = 18", "k = 9", "- 31"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ( 2 - k ) + 4"}, {"id": "x ( 2 - k ) + 4 = 18"}, {"id": "18"}, {"id": "代数式 $x ( 2 - k ) + 4$ 的值为 $18$"}, {"id": "- 4 + 2 k + 4 = 18"}, {"id": "x = - 2"}, {"id": "k = 9"}, {"id": "- 31"}, {"id": "x = 5"}], "links": [{"rel": "被描述", "source": "x ( 2 - k ) + 4", "target": "x ( 2 - k ) + 4 = 18"}, {"rel": "被代入", "source": "x ( 2 - k ) + 4", "target": "- 31"}, {"rel": "被代入", "source": "x ( 2 - k ) + 4 = 18", "target": "- 4 + 2 k + 4 = 18"}, {"rel": "被描述", "source": "18", "target": "x ( 2 - k ) + 4 = 18"}, {"rel": "限制性描述", "source": "代数式 $x ( 2 - k ) + 4$ 的值为 $18$", "target": "x ( 2 - k ) + 4 = 18"}, {"rel": "等式方程求解", "source": "- 4 + 2 k + 4 = 18", "target": "k = 9"}, {"rel": "代入", "source": "x = - 2", "target": "- 4 + 2 k + 4 = 18"}, {"rel": "代入", "source": "k = 9", "target": "- 31"}, {"rel": "代入", "source": "x = 5", "target": "- 31"}]}}
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{"content": "If the square root of $2 x + 3$ is meaningful in the range of real numbers, then the condition that $x$ satisfies is ____?", "answer": "x \\ge - \\frac { 3 } { 2 }", "steps": "From the given condition, it is known that $2 x + 3 \\geq 0$. Solving for $x$, we get $x \\geq - \\frac { 3 } { 2 }$.", "expr_cands": ["\\sqrt { 2 x + 3 }", "x", "2 x + 3 \\ge 0", "- \\frac { 3 } { 2 } \\le x", "x \\ge - \\frac { 3 } { 2 }"], "exprs": ["2 x + 3 \\ge 0", "x \\ge - \\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 2 x + 3 }"}, {"id": "2 x + 3 \\ge 0"}, {"id": "二次根式 $\\sqrt { 2 x + 3 }$ 在实数范围内有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge - \\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\sqrt { 2 x + 3 }", "target": "2 x + 3 \\ge 0"}, {"rel": "不等式方程求解", "source": "2 x + 3 \\ge 0", "target": "x \\ge - \\frac { 3 } { 2 }"}, {"rel": "限制性描述", "source": "二次根式 $\\sqrt { 2 x + 3 }$ 在实数范围内有意义", "target": "2 x + 3 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 x + 3 \\ge 0"}]}}
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{"content": "The meaningful condition for the fraction $\\frac { x ^ 2 - 16 } { x - 4 }$ is ____?", "answer": "x \\neq 4", "steps": "The fraction $\\frac { x ^ 2 - 16 } { x - 4 }$ is meaningful only if $x - 4 \\neq 0$, which implies $x \\neq 4$.", "expr_cands": ["\\frac { x ^ { 2 } - 16 } { x - 4 }", "x", "x - 4 \\neq 0", "x \\neq 4"], "exprs": ["x - 4 \\neq 0", "x \\neq 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x ^ { 2 } - 16 } { x - 4 }"}, {"id": "x - 4 \\neq 0"}, {"id": "分式 $\\frac { x ^ { 2 } - 16 } { x - 4 }$ 有意义的条件"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 4"}], "links": [{"rel": "被描述", "source": "\\frac { x ^ { 2 } - 16 } { x - 4 }", "target": "x - 4 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 4 \\neq 0", "target": "x \\neq 4"}, {"rel": "限制性描述", "source": "分式 $\\frac { x ^ { 2 } - 16 } { x - 4 }$ 有意义的条件", "target": "x - 4 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 4 \\neq 0"}]}}
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{"content": "If $\\frac { | a | } { a - a ^ 2 } = \\frac { 1 } { a - 1 }$, then the possible values of $a$ are ____?", "answer": "a < 0", "steps": "\\because $\\frac { | a | } { a - { a } ^ { 2 } } = \\frac { | a | } { a ( 1 - a ) }$ , when $a < 0$ , the original expression $= \\frac { 1 } { a - 1 }$.", "expr_cands": ["\\frac { | a | } { a - { a } ^ { 2 } } = \\frac { 1 } { a - 1 }", "a", "\\frac { | a | } { a - { a } ^ { 2 } } = \\frac { | a | } { a ( 1 - a ) }", "a = 2", "a < 0", "\\frac { 1 } { a - 1 }"], "exprs": ["a < 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { | a | } { a - { a } ^ { 2 } } = \\frac { 1 } { a - 1 }"}, {"id": "a < 0"}, {"id": "绝对值恒大于等于0"}], "links": [{"rel": "被描述", "source": "\\frac { | a | } { a - { a } ^ { 2 } } = \\frac { 1 } { a - 1 }", "target": "a < 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a < 0"}]}}
|
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{"content": "Given $x + 3 y - 3 = 0$, what is $3 ^ { x } \\cdot 27 ^ { y }$?", "answer": "27", "steps": "Since $x + 3 y - 3 = 0$, it follows that $x + 3 y = 3$. Therefore, $3 ^ x \\cdot 27 ^ y = 3 ^ { x + 3 y } = 3 ^ 3 = 27$.", "expr_cands": ["x + 3 y - 3 = 0", "y", "x", "3 ^ { x } \\cdot 27 ^ { y }", "x + 3 y = 3", "3 ^ { x + 3 y }", "27"], "exprs": ["x + 3 y = 3", "3 ^ { x + 3 y }", "27"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 3 y - 3 = 0"}, {"id": "x + 3 y = 3"}, {"id": "3 ^ { x } \\cdot 27 ^ { y }"}, {"id": "3 ^ { x + 3 y }"}, {"id": "27"}], "links": [{"rel": "移项", "source": "x + 3 y - 3 = 0", "target": "x + 3 y = 3"}, {"rel": "提取因式参考", "source": "x + 3 y = 3", "target": "3 ^ { x + 3 y }"}, {"rel": "代入", "source": "x + 3 y = 3", "target": "27"}, {"rel": "提取因式", "source": "3 ^ { x } \\cdot 27 ^ { y }", "target": "3 ^ { x + 3 y }"}, {"rel": "被代入", "source": "3 ^ { x + 3 y }", "target": "27"}]}}
|
||||
{"content": "Given a quadratic equation $x ^ 2 - 3 x - 4 = 0$ with two roots $m$ and $n$, what is $m ^ 2 + n ^ 2$?", "answer": "17", "steps": "$\\because$ $m$ and $n$ are the two roots of the quadratic equation $x ^ 2 - 3 x - 4 = 0$, $\\therefore$ $m + n = 3$, $mn = - 4$. Thus, $m ^ 2 + n ^ 2 = ( m + n ) ^ 2 - 2 mn = 9 + 8 = 17$.", "expr_cands": ["x ^ { 2 } - 3 x - 4 = 0", "x", "m", "n", "m ^ { 2 } + n ^ { 2 }", "x = - 1", "x = 4", "m + n = 3", "mn = - 4", "( m + n ) ^ { 2 } - 2 mn", "17"], "exprs": ["m + n = 3", "mn = - 4", "( m + n ) ^ { 2 } - 2 mn", "17"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 x - 4 = 0"}, {"id": "m + n = 3"}, {"id": "一元二次方程 $x ^ { 2 } - 3 x - 4 = 0$ 的两根是 $m$ , $n$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "mn = - 4"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "m ^ { 2 } + n ^ { 2 }"}, {"id": "( m + n ) ^ { 2 } - 2 mn"}, {"id": "17"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 3 x - 4 = 0", "target": "m + n = 3"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 x - 4 = 0", "target": "mn = - 4"}, {"rel": "提取因式参考", "source": "m + n = 3", "target": "( m + n ) ^ { 2 } - 2 mn"}, {"rel": "代入", "source": "m + n = 3", "target": "17"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - 3 x - 4 = 0$ 的两根是 $m$ , $n$", "target": "m + n = 3"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - 3 x - 4 = 0$ 的两根是 $m$ , $n$", "target": "mn = - 4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "m + n = 3"}, {"rel": "提取因式参考", "source": "mn = - 4", "target": "( m + n ) ^ { 2 } - 2 mn"}, {"rel": "代入", "source": "mn = - 4", "target": "17"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "mn = - 4"}, {"rel": "提取因式", "source": "m ^ { 2 } + n ^ { 2 }", "target": "( m + n ) ^ { 2 } - 2 mn"}, {"rel": "被代入", "source": "( m + n ) ^ { 2 } - 2 mn", "target": "17"}]}}
|
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{"content": "Solve the equation: $2 \\sqrt { 3 } x = - \\sqrt { 24 }$, $x$ = ____ ?", "answer": "- \\sqrt { 2 }", "steps": "Since $2 \\sqrt { 3 } x = - \\sqrt { 24 }$, therefore $x = - \\frac { \\sqrt { 24 }} { 2 \\sqrt { 3 }}$, therefore $x = - \\sqrt { 2 }$.", "expr_cands": ["2 \\sqrt { 3 } x = - \\sqrt { 24 }", "x", "x = - \\sqrt { 2 }", "x = - \\frac { \\sqrt { 24 } } { 2 \\sqrt { 3 } }"], "exprs": ["x = - \\sqrt { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 \\sqrt { 3 } x = - \\sqrt { 24 }"}, {"id": "x = - \\sqrt { 2 }"}], "links": [{"rel": "等式方程求解", "source": "2 \\sqrt { 3 } x = - \\sqrt { 24 }", "target": "x = - \\sqrt { 2 }"}]}}
|
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{"content": "If $x = 2$ is a solution of the quadratic equation $x ^ 2 + x - a = 0$, then the value of $a$ is ____?", "answer": "6", "steps": "Substituting $x = 2$ into the equation $x ^ 2 + x - a = 0$ yields $4 + 2 - a = 0$, which gives the solution $a = 6$.", "expr_cands": ["x = 2", "x", "x ^ { 2 } + x - a = 0", "a", "6 - a = 0", "4 + 2 - a = 0", "a = 6"], "exprs": ["4 + 2 - a = 0", "a = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + x - a = 0"}, {"id": "4 + 2 - a = 0"}, {"id": "x = 2"}, {"id": "a = 6"}], "links": [{"rel": "被代入", "source": "x ^ { 2 } + x - a = 0", "target": "4 + 2 - a = 0"}, {"rel": "等式方程求解", "source": "4 + 2 - a = 0", "target": "a = 6"}, {"rel": "代入", "source": "x = 2", "target": "4 + 2 - a = 0"}]}}
|
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{"content": "Given $a - b = 2$ and $a - c = \\frac { 1 } { 2 }$, what is the value of the algebraic expression ${ ( b - c ) } ^ { 2 } + 3 ( b - c ) + \\frac { 9 } { 4 }$?", "answer": "0", "steps": "Since $a - b = 2$, $a - c = \\frac { 1 } { 2 }$, we have $b - c = \\frac { 1 } { 2 } - 2 = - \\frac { 3 } { 2 }$. Therefore, $( b - c ) ^ 2 + 3 ( b - c ) + \\frac { 9 } { 4 } = {( - \\frac { 3 } { 2 })} ^ 2 + 3 * ( - \\frac { 3 } { 2 }) + \\frac { 9 } { 4 } = \\frac { 9 } { 4 } - \\frac { 9 } { 2 } + \\frac { 9 } { 4 } = 0$.", "expr_cands": ["a - b = 2", "b", "a", "a - c = \\frac { 1 } { 2 }", "c", "{ ( b - c ) } ^ { 2 } + 3 ( b - c ) + \\frac { 9 } { 4 }", "b - c = - \\frac { 3 } { 2 }", "( b - c ) ^ { 2 } + 3 ( b - c ) + \\frac { 9 } { 4 }", "0"], "exprs": ["b - c = - \\frac { 3 } { 2 }", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - b = 2"}, {"id": "b - c = - \\frac { 3 } { 2 }"}, {"id": "a - c = \\frac { 1 } { 2 }"}, {"id": "( b - c ) ^ { 2 } + 3 ( b - c ) + \\frac { 9 } { 4 }"}, {"id": "0"}], "links": [{"rel": "联立", "source": "a - b = 2", "target": "b - c = - \\frac { 3 } { 2 }"}, {"rel": "代入", "source": "b - c = - \\frac { 3 } { 2 }", "target": "0"}, {"rel": "联立", "source": "a - c = \\frac { 1 } { 2 }", "target": "b - c = - \\frac { 3 } { 2 }"}, {"rel": "被代入", "source": "( b - c ) ^ { 2 } + 3 ( b - c ) + \\frac { 9 } { 4 }", "target": "0"}]}}
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{"content": "The minimum integer for which $y _ { 1 } < y _ { 2 }$ is ____ , given $y _ { 1 } = - 5 x + \\frac { 1 } { 2 }$ and $y _ { 2 } = \\frac { 1 } { 2 } x + 1$.", "answer": "0", "steps": "$y _ { 1 } = - 5 x + \\frac { 1 } { 2 }$ , $y _ { 2 } = \\frac { 1 } { 2 } x + 1$ , solving the inequality $- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1$ gives $x > - \\frac { 1 } { 11 }$. Therefore, the smallest integer that makes $y _ { 1 } < y _ { 2 }$ is $0$.", "expr_cands": ["y _ { 1 } = - 5 x + \\frac { 1 } { 2 }", "y _ { 1 }", "x", "y _ { 2 } = \\frac { 1 } { 2 } x + 1", "y _ { 2 }", "y _ { 1 } < y _ { 2 }", "- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1", "- \\frac { 1 } { 11 } < x", "x > - \\frac { 1 } { 11 }", "0"], "exprs": ["- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1", "x > - \\frac { 1 } { 11 }", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y _ { 2 } = \\frac { 1 } { 2 } x + 1"}, {"id": "- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1"}, {"id": "y _ { 1 } = - 5 x + \\frac { 1 } { 2 }"}, {"id": "y _ { 1 } < y _ { 2 }"}, {"id": "x > - \\frac { 1 } { 11 }"}, {"id": "0"}, {"id": "函数 $y _ { 1 } = - 5 x + \\frac { 1 } { 2 }$ , $y _ { 2 } = \\frac { 1 } { 2 } x + 1$"}, {"id": "使 $y _ { 1 } < y _ { 2 }$ 的最小整数"}], "links": [{"rel": "代入", "source": "y _ { 2 } = \\frac { 1 } { 2 } x + 1", "target": "- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1"}, {"rel": "不等式方程求解", "source": "- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1", "target": "x > - \\frac { 1 } { 11 }"}, {"rel": "代入", "source": "y _ { 1 } = - 5 x + \\frac { 1 } { 2 }", "target": "- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1"}, {"rel": "被代入", "source": "y _ { 1 } < y _ { 2 }", "target": "- 5 x + \\frac { 1 } { 2 } < \\frac { 1 } { 2 } x + 1"}, {"rel": "被描述", "source": "x > - \\frac { 1 } { 11 }", "target": "0"}, {"rel": "限制性描述", "source": "函数 $y _ { 1 } = - 5 x + \\frac { 1 } { 2 }$ , $y _ { 2 } = \\frac { 1 } { 2 } x + 1$", "target": "0"}, {"rel": "限制性描述", "source": "使 $y _ { 1 } < y _ { 2 }$ 的最小整数", "target": "0"}]}}
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{"content": "The solution of the one-variable linear equation in $x$, $3 xy + \\frac { x } { 2 } = - 4$, is $2$. What is the value of $y$?", "answer": "y = - \\frac { 5 } { 6 }", "steps": "Substituting $x = 2$ into the equation, we get $6 y + 1 = - 4$, so $6 y = - 5$. Therefore, $y = - \\frac { 5 } { 6 }$.", "expr_cands": ["x", "3 xy + \\frac { x } { 2 } = - 4", "y", "2", "x = 2", "6 y + 1 = - 4", "y = - \\frac { 5 } { 6 }", "6 y = - 5"], "exprs": ["x = 2", "6 y + 1 = - 4", "y = - \\frac { 5 } { 6 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "x = 2"}, {"id": "3 xy + \\frac { x } { 2 } = - 4"}, {"id": "x"}, {"id": "关于 $x$ 的一元一次方程 $3 xy + \\frac { x } { 2 } = - 4$ 的解为 $2$"}, {"id": "6 y + 1 = - 4"}, {"id": "y = - \\frac { 5 } { 6 }"}], "links": [{"rel": "被描述", "source": "2", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "6 y + 1 = - 4"}, {"rel": "被描述", "source": "3 xy + \\frac { x } { 2 } = - 4", "target": "x = 2"}, {"rel": "被代入", "source": "3 xy + \\frac { x } { 2 } = - 4", "target": "6 y + 1 = - 4"}, {"rel": "被描述", "source": "x", "target": "x = 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元一次方程 $3 xy + \\frac { x } { 2 } = - 4$ 的解为 $2$", "target": "x = 2"}, {"rel": "等式方程求解", "source": "6 y + 1 = - 4", "target": "y = - \\frac { 5 } { 6 }"}]}}
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{"content": "If real numbers $x$ and $y$ satisfy the condition $2 x - y = 2$, then the value of the algebraic expression ${ x } ^ 2 - xy + \\frac { 1 } { 4 } { y } ^ 2 + 2$ is ____?", "answer": "3", "steps": "Since $2 x - y = 2$, therefore $x - \\frac { 1 } { 2 } y = 1$, thus ${ x } ^ 2 - xy + \\frac { 1 } { 4 } { y } ^ 2 + 2 = {( x - \\frac { 1 } { 2 } y )} ^ 2 + 2 = 1 + 2 = 3$.", "expr_cands": ["x", "y", "2 x - y = 2", "{ x } ^ { 2 } - xy + \\frac { 1 } { 4 } { y } ^ { 2 } + 2", "x - \\frac { 1 } { 2 } y = 1", "{ ( x - \\frac { 1 } { 2 } y ) } ^ { 2 } + 2", "3"], "exprs": ["x - \\frac { 1 } { 2 } y = 1", "{ ( x - \\frac { 1 } { 2 } y ) } ^ { 2 } + 2", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - y = 2"}, {"id": "x - \\frac { 1 } { 2 } y = 1"}, {"id": "{ x } ^ { 2 } - xy + \\frac { 1 } { 4 } { y } ^ { 2 } + 2"}, {"id": "{ ( x - \\frac { 1 } { 2 } y ) } ^ { 2 } + 2"}, {"id": "3"}], "links": [{"rel": "同乘除", "source": "2 x - y = 2", "target": "x - \\frac { 1 } { 2 } y = 1"}, {"rel": "提取因式参考", "source": "x - \\frac { 1 } { 2 } y = 1", "target": "{ ( x - \\frac { 1 } { 2 } y ) } ^ { 2 } + 2"}, {"rel": "代入", "source": "x - \\frac { 1 } { 2 } y = 1", "target": "3"}, {"rel": "提取因式", "source": "{ x } ^ { 2 } - xy + \\frac { 1 } { 4 } { y } ^ { 2 } + 2", "target": "{ ( x - \\frac { 1 } { 2 } y ) } ^ { 2 } + 2"}, {"rel": "被代入", "source": "{ ( x - \\frac { 1 } { 2 } y ) } ^ { 2 } + 2", "target": "3"}]}}
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{"content": "Given $a - b = 2$, what is $2 a - 2 b + 5$ equal to?", "answer": "9", "steps": "\\because $a - b = 2$, \\therefore the original expression = $2 ( a - b ) + 5 = 4 + 5 = 9$.", "expr_cands": ["a - b = 2", "b", "a", "2 a - 2 b + 5", "2 ( a - b ) + 5", "9"], "exprs": ["2 ( a - b ) + 5", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a - 2 b + 5"}, {"id": "2 ( a - b ) + 5"}, {"id": "a - b = 2"}, {"id": "9"}], "links": [{"rel": "提取因式", "source": "2 a - 2 b + 5", "target": "2 ( a - b ) + 5"}, {"rel": "被代入", "source": "2 ( a - b ) + 5", "target": "9"}, {"rel": "提取因式参考", "source": "a - b = 2", "target": "2 ( a - b ) + 5"}, {"rel": "代入", "source": "a - b = 2", "target": "9"}]}}
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{"content": "If $x = 2$ is a solution of the equation $ax = 4$, then the value of $a$ is ____?", "answer": "2", "steps": "Because $x = 2$ is a solution to the equation $ax = 4$, therefore $2 a = 4$, which implies that $a = 2$.", "expr_cands": ["x = 2", "x", "ax = 4", "a", "2 a = 4", "a = 2"], "exprs": ["2 a = 4", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ax = 4"}, {"id": "2 a = 4"}, {"id": "x = 2"}, {"id": "$x = 2$ 是方程 $ax = 4$ 的解"}, {"id": "a = 2"}], "links": [{"rel": "被描述", "source": "ax = 4", "target": "2 a = 4"}, {"rel": "等式方程求解", "source": "2 a = 4", "target": "a = 2"}, {"rel": "被描述", "source": "x = 2", "target": "2 a = 4"}, {"rel": "限制性描述", "source": "$x = 2$ 是方程 $ax = 4$ 的解", "target": "2 a = 4"}]}}
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{"content": "The expression $8 - | 3 m - 9 |$ has a maximum value. What is the maximum value?", "answer": "8", "steps": "When $m = 3$, the expression $8 - | 3 m - 9 |$ has its maximum value of $8$.", "expr_cands": ["8 - | 3 m - 9 |", "m", "3 m - 9 = 0", "m = 3", "8"], "exprs": ["3 m - 9 = 0", "m = 3", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "8 - | 3 m - 9 |"}, {"id": "3 m - 9 = 0"}, {"id": "式子 $8 - | 3 m - 9 |$ 有最大值"}, {"id": "绝对值恒大于等于0"}, {"id": "m = 3"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "8 - | 3 m - 9 |", "target": "3 m - 9 = 0"}, {"rel": "被代入", "source": "8 - | 3 m - 9 |", "target": "8"}, {"rel": "等式方程求解", "source": "3 m - 9 = 0", "target": "m = 3"}, {"rel": "限制性描述", "source": "式子 $8 - | 3 m - 9 |$ 有最大值", "target": "3 m - 9 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "3 m - 9 = 0"}, {"rel": "代入", "source": "m = 3", "target": "8"}]}}
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{"content": "$7$. If the product of $( 3 x - k ) ( 2 - 3 x )$ does not contain a linear term in $x$, find the value of $k$.", "answer": "- 2", "steps": "$( 3 x - k ) ( 2 - 3 x ) = 6 x - 9 x ^ { 2 } - 2 k + 3 kx = - 9 x ^ { 2 } + ( 6 + 3 k ) x - 2 k$, given that there is no constant term in terms of $x$, we have $6 + 3 k = 0$, which gives us $k = - 2$.", "expr_cands": ["7", "( 3 x - k ) ( 2 - 3 x )", "x", "k", "- 9 x ^ { 2 } + ( 6 + 3 k ) x - 2 k", "6 + 3 k = 0", "k = - 2"], "exprs": ["- 9 x ^ { 2 } + ( 6 + 3 k ) x - 2 k", "6 + 3 k = 0", "k = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 3 x - k ) ( 2 - 3 x )"}, {"id": "- 9 x ^ { 2 } + ( 6 + 3 k ) x - 2 k"}, {"id": "x"}, {"id": "6 + 3 k = 0"}, {"id": "$( 3 x - k ) ( 2 - 3 x )$ 乘积中不含 $x$ 的一次项"}, {"id": "k = - 2"}], "links": [{"rel": "提取因式", "source": "( 3 x - k ) ( 2 - 3 x )", "target": "- 9 x ^ { 2 } + ( 6 + 3 k ) x - 2 k"}, {"rel": "被描述", "source": "- 9 x ^ { 2 } + ( 6 + 3 k ) x - 2 k", "target": "6 + 3 k = 0"}, {"rel": "提取因式参考", "source": "x", "target": "- 9 x ^ { 2 } + ( 6 + 3 k ) x - 2 k"}, {"rel": "等式方程求解", "source": "6 + 3 k = 0", "target": "k = - 2"}, {"rel": "限制性描述", "source": "$( 3 x - k ) ( 2 - 3 x )$ 乘积中不含 $x$ 的一次项", "target": "6 + 3 k = 0"}]}}
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{"content": "Given $a = 0.2$ and $b = 0.04$, what is the ratio of $a$ to $b$?", "answer": "5 : 1", "steps": "Since $a = 0.2$ and $b = 0.04$, it follows that $a : b = 0.2 : 0.04 = 5 : 1$.", "expr_cands": ["a = 0.2", "a", "b = 0.04", "b", "a : b", "5.0"], "exprs": ["5.0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a : b"}, {"id": "5.0"}, {"id": "a = 0.2"}, {"id": "b = 0.04"}], "links": [{"rel": "被代入", "source": "a : b", "target": "5.0"}, {"rel": "代入", "source": "a = 0.2", "target": "5.0"}, {"rel": "代入", "source": "b = 0.04", "target": "5.0"}]}}
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{"content": "If $a$ and $b$ are opposite numbers, then $2 a + 2 b - 2016$ = ____?", "answer": "- 2016", "steps": "Since $a$ and $b$ are opposite numbers, we have $a + b = 0$. Therefore, $2 a + 2 b - 2016 = 2 ( a + b ) - 2016 = - 2016$.", "expr_cands": ["a", "b", "2 a + 2 b - 2016", "a + b = 0", "2 ( a + b ) - 2016", "- 2016"], "exprs": ["a + b = 0", "2 ( a + b ) - 2016", "- 2016"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "2 a + 2 b - 2016"}, {"id": "2 ( a + b ) - 2016"}, {"id": "- 2016"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "提取因式参考", "source": "a + b = 0", "target": "2 ( a + b ) - 2016"}, {"rel": "代入", "source": "a + b = 0", "target": "- 2016"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "提取因式", "source": "2 a + 2 b - 2016", "target": "2 ( a + b ) - 2016"}, {"rel": "被代入", "source": "2 ( a + b ) - 2016", "target": "- 2016"}]}}
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{"content": "If the values of algebraic expressions $5 x - 7$ and $4 x + 9$ are opposite, then the value of $x$ is ____?", "answer": "- \\frac { 2 } { 9 }", "steps": "According to the problem, we have $5 x - 7 + 4 x + 9 = 0$, and solving for $x$ gives $x = - \\frac { 2 } { 9 }$.", "expr_cands": ["5 x - 7", "x", "4 x + 9", "5 x - 7 + 4 x + 9 = 0", "x = - \\frac { 2 } { 9 }"], "exprs": ["5 x - 7 + 4 x + 9 = 0", "x = - \\frac { 2 } { 9 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x - 7"}, {"id": "5 x - 7 + 4 x + 9 = 0"}, {"id": "4 x + 9"}, {"id": "代数式 $5 x - 7$ 和 $4 x + 9$ 的值互为相反数"}, {"id": "x = - \\frac { 2 } { 9 }"}], "links": [{"rel": "被描述", "source": "5 x - 7", "target": "5 x - 7 + 4 x + 9 = 0"}, {"rel": "等式方程求解", "source": "5 x - 7 + 4 x + 9 = 0", "target": "x = - \\frac { 2 } { 9 }"}, {"rel": "被描述", "source": "4 x + 9", "target": "5 x - 7 + 4 x + 9 = 0"}, {"rel": "限制性描述", "source": "代数式 $5 x - 7$ 和 $4 x + 9$ 的值互为相反数", "target": "5 x - 7 + 4 x + 9 = 0"}]}}
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{"content": "If $a ^ m = 3$, $a ^ n = \\frac { 1 } { 2 }$, then $a ^ { 2 m - 3 n }$ = ____ ?", "answer": "72", "steps": "Since $a ^ { m } = 3$ and $a ^ { n } = \\frac { 1 } { 2 }$, therefore $a ^ { 2 m - 3 n } = a ^ { 2 m } \\div a ^ { 3 n } = {( a ^ { m })} ^ { 2 } \\div {( a ^ { n })} ^ { 3 } = 3 ^ { 2 } \\div \\frac { 1 } { 8 } = 72$.", "expr_cands": ["a ^ { m } = 3", "a", "m", "{ a } ^ { n } = \\frac { 1 } { 2 }", "n", "a ^ { 2 m - 3 n }", "{ a } ^ { 2 m - 3 n }", "72"], "exprs": ["72"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ a } ^ { 2 m - 3 n }"}, {"id": "72"}, {"id": "a ^ { m } = 3"}, {"id": "{ a } ^ { n } = \\frac { 1 } { 2 }"}], "links": [{"rel": "被代入", "source": "{ a } ^ { 2 m - 3 n }", "target": "72"}, {"rel": "代入", "source": "a ^ { m } = 3", "target": "72"}, {"rel": "代入", "source": "{ a } ^ { n } = \\frac { 1 } { 2 }", "target": "72"}]}}
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{"content": "When $x = 3$, what is the value of the algebraic expression $\\sqrt { 11 - 2 x }$?", "answer": "\\sqrt { 5 }", "steps": "When $x = 3$, $\\sqrt { 11 - 2 x } = \\sqrt { 11 - 2 * 3 } = \\sqrt { 5 }$.", "expr_cands": ["x = 3", "x", "\\sqrt { 11 - 2 x }", "\\sqrt { 5 }"], "exprs": ["\\sqrt { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 11 - 2 x }"}, {"id": "\\sqrt { 5 }"}, {"id": "x = 3"}], "links": [{"rel": "被代入", "source": "\\sqrt { 11 - 2 x }", "target": "\\sqrt { 5 }"}, {"rel": "代入", "source": "x = 3", "target": "\\sqrt { 5 }"}]}}
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{"content": "Given that $x _ 1 = - 1$ is a root of the equation $x ^ 2 + mx - 6 = 0$, the other root of the equation is ____?", "answer": "6", "steps": "Assuming the other root of the equation is $x _ 1$, we can deduce from the given information that $- 1 \\times x _ 1 = - 6$, which means $x _ 1 = 6$.", "expr_cands": ["x _ { 1 } = - 1", "x _ { 1 }", "x ^ { 2 } + mx - 6 = 0", "m", "x", "- 1 \\times x _ { 1 } = - 6", "x_{1} = 6", "x _ { 1 } = 6"], "exprs": ["x _ { 1 }", "- 1 \\times x _ { 1 } = - 6", "x _ { 1 } = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程另一个根为 $x _ { 1 }$"}, {"id": "x _ { 1 }"}, {"id": "x _ { 1 } = - 1"}, {"id": "- 1 \\times x _ { 1 } = - 6"}, {"id": "x ^ { 2 } + mx - 6 = 0"}, {"id": "$x _ { 1 } = - 1$ 是方程 $x ^ { 2 } + mx - 6 = 0$ 的一个根"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "x _ { 1 } = 6"}], "links": [{"rel": "假设描述", "source": "设方程另一个根为 $x _ { 1 }$", "target": "x _ { 1 }"}, {"rel": "假设描述", "source": "设方程另一个根为 $x _ { 1 }$", "target": "- 1 \\times x _ { 1 } = - 6"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "- 1 \\times x _ { 1 } = - 6"}, {"rel": "被描述", "source": "x _ { 1 } = - 1", "target": "- 1 \\times x _ { 1 } = - 6"}, {"rel": "等式方程求解", "source": "- 1 \\times x _ { 1 } = - 6", "target": "x _ { 1 } = 6"}, {"rel": "被描述", "source": "x ^ { 2 } + mx - 6 = 0", "target": "- 1 \\times x _ { 1 } = - 6"}, {"rel": "限制性描述", "source": "$x _ { 1 } = - 1$ 是方程 $x ^ { 2 } + mx - 6 = 0$ 的一个根", "target": "- 1 \\times x _ { 1 } = - 6"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "- 1 \\times x _ { 1 } = - 6"}]}}
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{"content": "If $a$ is a constant, what is the value of $a$ if the product of $( x + a )$ and $( x - 5 )$ does not contain a linear term in $x$?", "answer": "5", "steps": "$\\because$ $( x + a ) ( x - 5 ) = x ^ { 2 } - 5 x + ax - 5 a = x ^ { 2 } + ( a - 5 ) x - 5 a$ , and $\\because$ there is no linear term in the product, $\\therefore$ $a - 5 = 0$, which solves to $a = 5$.", "expr_cands": ["a", "( x + a )", "x", "( x - 5 )", "( x + a ) ( x - 5 )", "x ^ { 2 } + ( a - 5 ) x - 5 a", "a - 5 = 0", "a = 5"], "exprs": ["( x + a ) ( x - 5 )", "x ^ { 2 } + ( a - 5 ) x - 5 a", "a - 5 = 0", "a = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + a )"}, {"id": "( x + a ) ( x - 5 )"}, {"id": "( x - 5 )"}, {"id": "$a$ 为常数"}, {"id": "$( x + a )$ 和 $( x - 5 )$ 的乘积中不含有 $x$ 的一次项"}, {"id": "x ^ { 2 } + ( a - 5 ) x - 5 a"}, {"id": "x"}, {"id": "a - 5 = 0"}, {"id": "a = 5"}], "links": [{"rel": "被描述", "source": "( x + a )", "target": "( x + a ) ( x - 5 )"}, {"rel": "提取因式", "source": "( x + a ) ( x - 5 )", "target": "x ^ { 2 } + ( a - 5 ) x - 5 a"}, {"rel": "被描述", "source": "( x - 5 )", "target": "( x + a ) ( x - 5 )"}, {"rel": "限制性描述", "source": "$a$ 为常数", "target": "( x + a ) ( x - 5 )"}, {"rel": "限制性描述", "source": "$a$ 为常数", "target": "a - 5 = 0"}, {"rel": "限制性描述", "source": "$( x + a )$ 和 $( x - 5 )$ 的乘积中不含有 $x$ 的一次项", "target": "( x + a ) ( x - 5 )"}, {"rel": "限制性描述", "source": "$( x + a )$ 和 $( x - 5 )$ 的乘积中不含有 $x$ 的一次项", "target": "a - 5 = 0"}, {"rel": "被描述", "source": "x ^ { 2 } + ( a - 5 ) x - 5 a", "target": "a - 5 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "x ^ { 2 } + ( a - 5 ) x - 5 a"}, {"rel": "等式方程求解", "source": "a - 5 = 0", "target": "a = 5"}]}}
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{"content": "If $\\sqrt { m - 20 } + \\sqrt { n + 15 } = 0$, then $\\sqrt { m - n + 1 }$ = ____ ?", "answer": "6", "steps": "From the given information, we have $m - 20 = 0$ and $n + 15 = 0$, which implies $m = 20$ and $n = - 15$. Therefore, $\\sqrt { m - n + 1 } = \\sqrt { 20 - ( - 15 ) + 1 } = 6$.", "expr_cands": ["\\sqrt { m - 20 } + \\sqrt { n + 15 } = 0", "m", "n", "\\sqrt { m - n + 1 }", "m - 20 = 0", "m = 20", "n + 15 = 0", "n = - 15", "6"], "exprs": ["m - 20 = 0", "n + 15 = 0", "m = 20", "n = - 15", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { m - 20 } + \\sqrt { n + 15 } = 0"}, {"id": "m - 20 = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "n + 15 = 0"}, {"id": "m = 20"}, {"id": "n = - 15"}, {"id": "\\sqrt { m - n + 1 }"}, {"id": "6"}], "links": [{"rel": "被描述", "source": "\\sqrt { m - 20 } + \\sqrt { n + 15 } = 0", "target": "m - 20 = 0"}, {"rel": "被描述", "source": "\\sqrt { m - 20 } + \\sqrt { n + 15 } = 0", "target": "n + 15 = 0"}, {"rel": "等式方程求解", "source": "m - 20 = 0", "target": "m = 20"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "m - 20 = 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "n + 15 = 0"}, {"rel": "等式方程求解", "source": "n + 15 = 0", "target": "n = - 15"}, {"rel": "代入", "source": "m = 20", "target": "6"}, {"rel": "代入", "source": "n = - 15", "target": "6"}, {"rel": "被代入", "source": "\\sqrt { m - n + 1 }", "target": "6"}]}}
|
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{"content": "If $| m - \\sqrt { 3 } | - \\sqrt { 3 } + m = 0$, then the range of values for $m$ is ____?", "answer": "m \\le \\sqrt { 3 }", "steps": "Since $| m - \\sqrt { 3 } | - \\sqrt { 3 } + m = 0$, it follows that $| m - \\sqrt { 3 } | = - ( m - \\sqrt { 3 } )$. Therefore, $m - \\sqrt { 3 } \\le 0$, which implies that $m \\le \\sqrt { 3 }$.", "expr_cands": ["| m - \\sqrt { 3 } | - \\sqrt { 3 } + m = 0", "m", "| m - \\sqrt { 3 } | = - ( m - \\sqrt { 3 } )", "m - \\sqrt { 3 } \\le 0", "m \\le \\sqrt { 3 }"], "exprs": ["| m - \\sqrt { 3 } | = - ( m - \\sqrt { 3 } )", "m - \\sqrt { 3 } \\le 0", "m \\le \\sqrt { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| m - \\sqrt { 3 } | - \\sqrt { 3 } + m = 0"}, {"id": "| m - \\sqrt { 3 } | = - ( m - \\sqrt { 3 } )"}, {"id": "m - \\sqrt { 3 } \\le 0"}, {"id": "绝对值恒大于等于0"}, {"id": "m \\le \\sqrt { 3 }"}], "links": [{"rel": "移项", "source": "| m - \\sqrt { 3 } | - \\sqrt { 3 } + m = 0", "target": "| m - \\sqrt { 3 } | = - ( m - \\sqrt { 3 } )"}, {"rel": "被描述", "source": "| m - \\sqrt { 3 } | = - ( m - \\sqrt { 3 } )", "target": "m - \\sqrt { 3 } \\le 0"}, {"rel": "不等式方程求解", "source": "m - \\sqrt { 3 } \\le 0", "target": "m \\le \\sqrt { 3 }"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m - \\sqrt { 3 } \\le 0"}]}}
|
||||
{"content": "The minimum integer solution of the inequality $x - 1 \\ge 3$ is ____ ?", "answer": "4", "steps": "$x - 1 \\ge 3$ , $x \\ge 4$ , so the smallest integer solution of the inequality is $4$.", "expr_cands": ["x - 1 \\ge 3", "x", "4 \\le x", "x \\ge 4", "4"], "exprs": ["x \\ge 4", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - 1 \\ge 3"}, {"id": "x \\ge 4"}, {"id": "4"}, {"id": "不等式 $x - 1 \\ge 3$ 的最小整数解"}], "links": [{"rel": "不等式方程求解", "source": "x - 1 \\ge 3", "target": "x \\ge 4"}, {"rel": "被描述", "source": "x \\ge 4", "target": "4"}, {"rel": "限制性描述", "source": "不等式 $x - 1 \\ge 3$ 的最小整数解", "target": "4"}]}}
|
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{"content": "If $\\sqrt { 1 - a }$ is defined, then the range of values for $a$ is ____?", "answer": "a \\le 1", "steps": "From the given condition, we can deduce that $1 - a \\geq 0$. Solving for $a$, we get $a \\leq 1$.", "expr_cands": ["\\sqrt { 1 - a }", "a", "1 - a \\ge 0", "a \\le 1"], "exprs": ["1 - a \\ge 0", "a \\le 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 1 - a }"}, {"id": "1 - a \\ge 0"}, {"id": "$\\sqrt { 1 - a }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a \\le 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { 1 - a }", "target": "1 - a \\ge 0"}, {"rel": "不等式方程求解", "source": "1 - a \\ge 0", "target": "a \\le 1"}, {"rel": "限制性描述", "source": "$\\sqrt { 1 - a }$ 有意义", "target": "1 - a \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "1 - a \\ge 0"}]}}
|
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{"content": "If $- a = 3$, what is the value of $| a | - a$?", "answer": "6", "steps": "Since $- a = 3$, it follows that $a = - 3$. Therefore, $| a | - a = | - 3 | - ( - 3 ) = 3 + 3 = 6$.", "expr_cands": ["- a = 3", "a", "| a | - a", "a = - 3", "6"], "exprs": ["a = - 3", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- a = 3"}, {"id": "a = - 3"}, {"id": "| a | - a"}, {"id": "6"}], "links": [{"rel": "等式方程求解", "source": "- a = 3", "target": "a = - 3"}, {"rel": "代入", "source": "a = - 3", "target": "6"}, {"rel": "被代入", "source": "| a | - a", "target": "6"}]}}
|
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{"content": "$8$. $a$ is the smallest natural number, $b$ is the largest negative integer, $c$ is the rational number with the smallest absolute value. What is the value of $a - b + c$?", "answer": "1", "steps": "According to the given conditions: $a = 0$, $b = - 1$, $c = 0$, therefore $a - b + c = 1$. ", "expr_cands": ["8", "a", "b", "c", "a - b + c", "a = 0", "b = - 1", "c = 0", "1"], "exprs": ["a = 0", "b = - 1", "c = 0", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a = 0"}, {"id": "$8$ , $a$ 是最小的自然数"}, {"id": "b"}, {"id": "b = - 1"}, {"id": "$b$ 是最大的负整数"}, {"id": "c"}, {"id": "c = 0"}, {"id": "$c$ 是绝对值最小的有理数"}, {"id": "绝对值恒大于等于0"}, {"id": "a - b + c"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "a", "target": "a = 0"}, {"rel": "代入", "source": "a = 0", "target": "1"}, {"rel": "限制性描述", "source": "$8$ , $a$ 是最小的自然数", "target": "a = 0"}, {"rel": "被描述", "source": "b", "target": "b = - 1"}, {"rel": "代入", "source": "b = - 1", "target": "1"}, {"rel": "限制性描述", "source": "$b$ 是最大的负整数", "target": "b = - 1"}, {"rel": "被描述", "source": "c", "target": "c = 0"}, {"rel": "代入", "source": "c = 0", "target": "1"}, {"rel": "限制性描述", "source": "$c$ 是绝对值最小的有理数", "target": "c = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "c = 0"}, {"rel": "被代入", "source": "a - b + c", "target": "1"}]}}
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{"content": "Given that $a + 1$ is the arithmetic mean of $a - 1$ and $2 a$, what is the value of $a$?", "answer": "3", "steps": "Because $a + 1$ is the arithmetic mean of $a - 1$ and $2 a$, we have $2 ( a + 1 ) = ( a - 1 ) + 2 a$. Solving for $a$, we get $a = 3$.", "expr_cands": ["a + 1", "a", "a - 1", "2 a", "2 ( a + 1 ) = a - 1 + 2 a", "a = 3"], "exprs": ["2 ( a + 1 ) = a - 1 + 2 a", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a + 1"}, {"id": "2 ( a + 1 ) = a - 1 + 2 a"}, {"id": "a - 1"}, {"id": "2 a"}, {"id": "$a + 1$ 是 $a - 1$ 与 $2 a$ 的等差中项"}, {"id": "a = 3"}], "links": [{"rel": "被描述", "source": "a + 1", "target": "2 ( a + 1 ) = a - 1 + 2 a"}, {"rel": "等式方程求解", "source": "2 ( a + 1 ) = a - 1 + 2 a", "target": "a = 3"}, {"rel": "被描述", "source": "a - 1", "target": "2 ( a + 1 ) = a - 1 + 2 a"}, {"rel": "被描述", "source": "2 a", "target": "2 ( a + 1 ) = a - 1 + 2 a"}, {"rel": "限制性描述", "source": "$a + 1$ 是 $a - 1$ 与 $2 a$ 的等差中项", "target": "2 ( a + 1 ) = a - 1 + 2 a"}]}}
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{"content": "$x + 1$ is the arithmetic square root of 4, what is the value of x?", "answer": "1", "steps": "From the given information, we have $x + 1 = 2$. Solving for $x$, we get $x = 1$.", "expr_cands": ["x + 1", "x", "4", "x + 1 = 2", "x = 1"], "exprs": ["x + 1 = 2", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 1"}, {"id": "x + 1 = 2"}, {"id": "4"}, {"id": "$x + 1$ 是 $4$ 的算术平方根"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "x + 1", "target": "x + 1 = 2"}, {"rel": "等式方程求解", "source": "x + 1 = 2", "target": "x = 1"}, {"rel": "被描述", "source": "4", "target": "x + 1 = 2"}, {"rel": "限制性描述", "source": "$x + 1$ 是 $4$ 的算术平方根", "target": "x + 1 = 2"}]}}
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{"content": "If the value of the algebraic expression $x ^ { 2 } - 2 x - 1$ is $2$ when $x = m$, then the value of the algebraic expression $m ^ { 2 } - 2 m + 2018$ is ____ ?", "answer": "2021", "steps": "Substituting $x = m$ into $x ^ 2 - 2 x - 1 = 2$ yields $m ^ 2 - 2 m - 1 = 2$, which means $m ^ 2 - 2 m = 3$. Therefore, $m ^ 2 - 2 m + 2018 = 3 + 2018 = 2021$.", "expr_cands": ["x = m", "m", "x", "x ^ { 2 } - 2 x - 1", "2", "m ^ { 2 } - 2 m + 2018", "x ^ { 2 } - 2 x - 1 = 2", "x = - 1", "x = 3", "m ^ { 2 } - 2 m - 1 = 2", "m = - 1", "m = 3", "m ^ { 2 } - 2 m = 3", "2021"], "exprs": ["m ^ { 2 } - 2 m - 1 = 2", "m ^ { 2 } - 2 m = 3", "2021"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = m"}, {"id": "m ^ { 2 } - 2 m - 1 = 2"}, {"id": "x ^ { 2 } - 2 x - 1"}, {"id": "2"}, {"id": "$x = m$ 时代数式 $x ^ { 2 } - 2 x - 1$ 的值为 $2$"}, {"id": "m ^ { 2 } - 2 m = 3"}, {"id": "m ^ { 2 } - 2 m + 2018"}, {"id": "2021"}], "links": [{"rel": "被描述", "source": "x = m", "target": "m ^ { 2 } - 2 m - 1 = 2"}, {"rel": "移项", "source": "m ^ { 2 } - 2 m - 1 = 2", "target": "m ^ { 2 } - 2 m = 3"}, {"rel": "被描述", "source": "x ^ { 2 } - 2 x - 1", "target": "m ^ { 2 } - 2 m - 1 = 2"}, {"rel": "被描述", "source": "2", "target": "m ^ { 2 } - 2 m - 1 = 2"}, {"rel": "限制性描述", "source": "$x = m$ 时代数式 $x ^ { 2 } - 2 x - 1$ 的值为 $2$", "target": "m ^ { 2 } - 2 m - 1 = 2"}, {"rel": "代入", "source": "m ^ { 2 } - 2 m = 3", "target": "2021"}, {"rel": "被代入", "source": "m ^ { 2 } - 2 m + 2018", "target": "2021"}]}}
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{"content": "If a linear function is given by $y = kx + 1$, and when $x = 2$, $y = 2$, what is the value of $k$?", "answer": "\\frac { 1 } { 2 }", "steps": "When $x = 2$ and $y = 2$ are substituted into the linear function, we get $2 k + 1 = 2$. Solving for $k$, we get $k = \\frac { 1 } { 2 }$.", "expr_cands": ["y = kx + 1", "k", "y", "x", "x = 2", "y = 2", "2 k + 1 = 2", "k = \\frac { 1 } { 2 }"], "exprs": ["2 k + 1 = 2", "k = \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = kx + 1"}, {"id": "2 k + 1 = 2"}, {"id": "x = 2"}, {"id": "y = 2"}, {"id": "k = \\frac { 1 } { 2 }"}], "links": [{"rel": "被代入", "source": "y = kx + 1", "target": "2 k + 1 = 2"}, {"rel": "等式方程求解", "source": "2 k + 1 = 2", "target": "k = \\frac { 1 } { 2 }"}, {"rel": "代入", "source": "x = 2", "target": "2 k + 1 = 2"}, {"rel": "代入", "source": "y = 2", "target": "2 k + 1 = 2"}]}}
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{"content": "$( 1 )$ Calculate $( x - y ) ^ 3 \\cdot ( y - x ) =$ ____ ?", "answer": "- ( x - y ) ^ { 4 }", "steps": "$( 1 )$ The original expression is equal to ${ ( x - y ) } ^ { 3 } \\cdot [ - ( x - y ) ] = - { ( x - y ) } ^ { 4 }$.", "expr_cands": ["( 1 )", "( x - y ) ^ { 3 } \\cdot ( y - x )", "y", "x", "{ ( x - y ) } ^ { 3 } \\cdot [ - ( x - y ) ]", "- { ( x - y ) } ^ { 4 }"], "exprs": ["{ ( x - y ) } ^ { 3 } \\cdot [ - ( x - y ) ]", "- { ( x - y ) } ^ { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x - y ) ^ { 3 } \\cdot ( y - x )"}, {"id": "{ ( x - y ) } ^ { 3 } \\cdot [ - ( x - y ) ]"}, {"id": "- { ( x - y ) } ^ { 4 }"}], "links": [{"rel": "提取因式", "source": "( x - y ) ^ { 3 } \\cdot ( y - x )", "target": "{ ( x - y ) } ^ { 3 } \\cdot [ - ( x - y ) ]"}, {"rel": "提取因式", "source": "{ ( x - y ) } ^ { 3 } \\cdot [ - ( x - y ) ]", "target": "- { ( x - y ) } ^ { 4 }"}]}}
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{"content": "If the equation $k ( x ^ 2 + 1 ) + x ^ 2 = x ^ { | k | } + 3$ with respect to $x$ is a linear equation, then $k$ = ____?", "answer": "- 1", "steps": "The equation $k ( x ^ 2 + 1 ) + x ^ 2 = x ^ { | k | } + 3$ is a linear equation in one variable. We obtain $| k | = 1$ and $k + 1 = 0$. Solving for $k$, we get $k = - 1$.", "expr_cands": ["x", "k ( x ^ { 2 } + 1 ) + x ^ { 2 } = x ^ { | k | } + 3", "k", "| k | = 1", "k = - 1", "k = 1", "k + 1 = 0"], "exprs": ["| k | = 1", "k + 1 = 0", "k = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "k ( x ^ { 2 } + 1 ) + x ^ { 2 } = x ^ { | k | } + 3"}, {"id": "| k | = 1"}, {"id": "关于 $x$ 的方程 $k ( x ^ { 2 } + 1 ) + x ^ { 2 } = x ^ { | k | } + 3$ 为一元一次方程"}, {"id": "k + 1 = 0"}, {"id": "k = - 1"}], "links": [{"rel": "被描述", "source": "k ( x ^ { 2 } + 1 ) + x ^ { 2 } = x ^ { | k | } + 3", "target": "| k | = 1"}, {"rel": "被描述", "source": "k ( x ^ { 2 } + 1 ) + x ^ { 2 } = x ^ { | k | } + 3", "target": "k + 1 = 0"}, {"rel": "联立", "source": "| k | = 1", "target": "k = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $k ( x ^ { 2 } + 1 ) + x ^ { 2 } = x ^ { | k | } + 3$ 为一元一次方程", "target": "| k | = 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $k ( x ^ { 2 } + 1 ) + x ^ { 2 } = x ^ { | k | } + 3$ 为一元一次方程", "target": "k + 1 = 0"}, {"rel": "联立", "source": "k + 1 = 0", "target": "k = - 1"}]}}
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{"content": "When $a$ = ____ ?, the equation $12 + { x } ^ { a - 1 } = 9$ is a linear equation in one variable $x$.", "answer": "2", "steps": "It can be deduced from the problem that $a - 1 = 1$, so $a = 2$.", "expr_cands": ["a", "x", "12 + { x } ^ { a - 1 } = 9", "a - 1 = 1", "a = 2"], "exprs": ["a - 1 = 1", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "12 + { x } ^ { a - 1 } = 9"}, {"id": "a - 1 = 1"}, {"id": "关于 $x$ 的方程 $12 + { x } ^ { a - 1 } = 9$ 是一元一次方程"}, {"id": "a = 2"}], "links": [{"rel": "被描述", "source": "12 + { x } ^ { a - 1 } = 9", "target": "a - 1 = 1"}, {"rel": "等式方程求解", "source": "a - 1 = 1", "target": "a = 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $12 + { x } ^ { a - 1 } = 9$ 是一元一次方程", "target": "a - 1 = 1"}]}}
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{"content": "Given $3 ^ n = a$, $3 ^ m = b$, what is $3 ^ { m + n - 1 }$?", "answer": "\\frac { ab } { 3 }", "steps": "$\\because$ ${ 3 } ^ { n } = a$, ${ 3 } ^ { m } = b$, $\\therefore$ the original expression $= { 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3 = \\frac { ab } { 3 }$.", "expr_cands": ["3 ^ { n } = a", "a", "n", "3 ^ { m } = b", "b", "m", "3 ^ { m + n - 1 }", "{ 3 } ^ { n } = a", "{ 3 } ^ { m } = b", "{ 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3", "\\frac { ab } { 3 }"], "exprs": ["{ 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3", "\\frac { ab } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 ^ { m + n - 1 }"}, {"id": "{ 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3"}, {"id": "\\frac { ab } { 3 }"}, {"id": "3 ^ { n } = a"}, {"id": "3 ^ { m } = b"}, {"id": "原式 = ${ 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3 = \\frac { ab } { 3 }$"}], "links": [{"rel": "展开", "source": "3 ^ { m + n - 1 }", "target": "{ 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3"}, {"rel": "被描述", "source": "{ 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3", "target": "\\frac { ab } { 3 }"}, {"rel": "被描述", "source": "3 ^ { n } = a", "target": "\\frac { ab } { 3 }"}, {"rel": "被描述", "source": "3 ^ { m } = b", "target": "\\frac { ab } { 3 }"}, {"rel": "限制性描述", "source": "原式 = ${ 3 } ^ { m } \\cdot { 3 } ^ { n } \\div 3 = \\frac { ab } { 3 }$", "target": "\\frac { ab } { 3 }"}]}}
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{"content": "When $m$ = ____ ?, $3 x { y } ^ { 2 m + 5 } { z } ^ { 2 }$ is a sextic monomial.", "answer": "- 1", "steps": "Since $3 x { y } ^ { 2 m + 5 } { z } ^ { 2 }$ is a monomial of degree six, we have $1 + 2 m + 5 + 2 = 6$. Solving for $m$, we get $m = - 1$.", "expr_cands": ["m", "3 x { y } ^ { 2 m + 5 } { z } ^ { 2 }", "z", "y", "x", "1 + 2 m + 5 + 2 = 6", "m = - 1"], "exprs": ["1 + 2 m + 5 + 2 = 6", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x { y } ^ { 2 m + 5 } { z } ^ { 2 }"}, {"id": "1 + 2 m + 5 + 2 = 6"}, {"id": "$3 x { y } ^ { 2 m + 5 } { z } ^ { 2 }$ 是六次单项式"}, {"id": "m = - 1"}], "links": [{"rel": "被描述", "source": "3 x { y } ^ { 2 m + 5 } { z } ^ { 2 }", "target": "1 + 2 m + 5 + 2 = 6"}, {"rel": "等式方程求解", "source": "1 + 2 m + 5 + 2 = 6", "target": "m = - 1"}, {"rel": "限制性描述", "source": "$3 x { y } ^ { 2 m + 5 } { z } ^ { 2 }$ 是六次单项式", "target": "1 + 2 m + 5 + 2 = 6"}]}}
|
||||
{"content": "If $\\frac { 1 } { 2 } y ^ { 2 n - 1 } - 2 = 0$ is a one-variable linear equation in terms of $y$, then $n$ = ____?", "answer": "1", "steps": "From the given information, it is known that $2 n - 1 = 1$. Therefore, $n = 1$.", "expr_cands": ["\\frac { 1 } { 2 } y ^ { 2 n - 1 } - 2 = 0", "y", "n", "2 n - 1 = 1", "n = 1"], "exprs": ["2 n - 1 = 1", "n = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 2 } y ^ { 2 n - 1 } - 2 = 0"}, {"id": "2 n - 1 = 1"}, {"id": "$\\frac { 1 } { 2 } y ^ { 2 n - 1 } - 2 = 0$ 是关于 $y$ 的一元一次方程"}, {"id": "n = 1"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 2 } y ^ { 2 n - 1 } - 2 = 0", "target": "2 n - 1 = 1"}, {"rel": "等式方程求解", "source": "2 n - 1 = 1", "target": "n = 1"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 2 } y ^ { 2 n - 1 } - 2 = 0$ 是关于 $y$ 的一元一次方程", "target": "2 n - 1 = 1"}]}}
|
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{"content": "If $3 x ^ { 3 } - mxy ^ { 2 } = 3 x ( x + 2 y ) ( x - 2 y )$, then the value of $m$ is ____?", "answer": "12", "steps": "$\\because$ $3 x ( x + 2 y ) ( x - 2 y ) = 3 x ( x ^ { 2 } - 4 y ^ { 2 } ) = 3 x ^ { 3 } - 12 xy ^ { 2 } = 3 x ^ { 3 } - mxy ^ { 2 }$ , $\\therefore$ $- 12 = - m$ , i.e. $m = 12$.", "expr_cands": ["3 x ^ { 3 } - mxy ^ { 2 } = 3 x ( x + 2 y ) ( x - 2 y )", "m", "x", "y", "3 x ( x + 2 y ) ( x - 2 y ) = 3 x ^ { 3 } - mxy ^ { 2 }", "- 12 = - m", "m = 12"], "exprs": ["- 12 = - m", "m = 12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { 3 } - mxy ^ { 2 } = 3 x ( x + 2 y ) ( x - 2 y )"}, {"id": "- 12 = - m"}, {"id": "m = 12"}], "links": [{"rel": "移项", "source": "3 x ^ { 3 } - mxy ^ { 2 } = 3 x ( x + 2 y ) ( x - 2 y )", "target": "- 12 = - m"}, {"rel": "等式方程求解", "source": "- 12 = - m", "target": "m = 12"}]}}
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{"content": "Given that $a$ is a real number, what is the value of $a$ if the fractional equation $\\frac { 3 x + a } { x + 2 } = 1$ has no solution?", "answer": "6", "steps": "$\\frac { 3 x + a } { x + 2 } = 1$, multiply both sides of the equation by $x + 2$, we get $3 x + a = x + 2$, rearrange and combine like terms, we get $2 x = 2 - a$. Since the fraction equation $\\frac { 3 x + a } { x + 2 } = 1$ has no solution for $x$, we have $x + 2 = 0$, which gives us $x = - 2$. Therefore, $2 - a = - 4$, which gives us $a = 6$.", "expr_cands": ["a", "\\frac { 3 x + a } { x + 2 } = 1", "x", "x + 2", "3 x + a = x + 2", "2 x = 2 - a", "x + 2 = 0", "x = - 2", "2 - a = - 4", "a = 6"], "exprs": ["3 x + a = x + 2", "x + 2 = 0", "2 x = 2 - a", "x = - 2", "2 - a = - 4", "a = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 x + a } { x + 2 } = 1"}, {"id": "3 x + a = x + 2"}, {"id": "2 x = 2 - a"}, {"id": "x + 2 = 0"}, {"id": "分式方程 $\\frac { 3 x + a } { x + 2 } = 1$ 无解"}, {"id": "关于 $x$ 的分式方程 $\\frac { 3 x + a } { x + 2 } = 1$ 无解"}, {"id": "分式方程无解，则分母为0"}, {"id": "x = - 2"}, {"id": "2 - a = - 4"}, {"id": "a = 6"}], "links": [{"rel": "同乘除", "source": "\\frac { 3 x + a } { x + 2 } = 1", "target": "3 x + a = x + 2"}, {"rel": "被描述", "source": "\\frac { 3 x + a } { x + 2 } = 1", "target": "x + 2 = 0"}, {"rel": "移项", "source": "3 x + a = x + 2", "target": "2 x = 2 - a"}, {"rel": "被代入", "source": "2 x = 2 - a", "target": "2 - a = - 4"}, {"rel": "等式方程求解", "source": "x + 2 = 0", "target": "x = - 2"}, {"rel": "限制性描述", "source": "分式方程 $\\frac { 3 x + a } { x + 2 } = 1$ 无解", "target": "x + 2 = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的分式方程 $\\frac { 3 x + a } { x + 2 } = 1$ 无解", "target": "x + 2 = 0"}, {"rel": "属性描述", "source": "分式方程无解，则分母为0", "target": "x + 2 = 0"}, {"rel": "代入", "source": "x = - 2", "target": "2 - a = - 4"}, {"rel": "等式方程求解", "source": "2 - a = - 4", "target": "a = 6"}]}}
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{"content": "If the monomial $25 x ^ { n } y$ is a fourth degree monomial, then the value of $n$ is ____?", "answer": "3", "steps": "Since the monomial $25 x ^ n y$ is a fourth degree monomial, we have $n + 1 = 4$. Therefore, the value of $n$ is 3.", "expr_cands": ["25 x ^ { n } y", "y", "n", "x", "n + 1 = 4", "n = 3", "3"], "exprs": ["n + 1 = 4", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "25 x ^ { n } y"}, {"id": "n + 1 = 4"}, {"id": "单项式 $25 x ^ { n } y$ 是四次单项式"}, {"id": "3"}, {"id": "$n$ 的值"}], "links": [{"rel": "被描述", "source": "25 x ^ { n } y", "target": "n + 1 = 4"}, {"rel": "被描述", "source": "n + 1 = 4", "target": "3"}, {"rel": "限制性描述", "source": "单项式 $25 x ^ { n } y$ 是四次单项式", "target": "n + 1 = 4"}, {"rel": "限制性描述", "source": "$n$ 的值", "target": "3"}]}}
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{"content": "Given a quadratic equation in one variable $x$, $x ^ 2 - x + m - 1 = 0$, with one root being $0$, what is the value of $m$?", "answer": "1", "steps": "Substituting $x = 0$ into the equation $x ^ 2 - x + m - 1 = 0$, we get $m - 1 = 0$. Solving for $m$, we have $m = 1$.", "expr_cands": ["x", "x ^ { 2 } - x + m - 1 = 0", "m", "0", "x = 0", "m - 1 = 0", "m = 1"], "exprs": ["x = 0", "m - 1 = 0", "m = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "0"}, {"id": "x = 0"}, {"id": "x ^ { 2 } - x + m - 1 = 0"}, {"id": "x"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } - x + m - 1 = 0$ 有一个根是 $0$"}, {"id": "m - 1 = 0"}, {"id": "m = 1"}], "links": [{"rel": "被描述", "source": "0", "target": "x = 0"}, {"rel": "代入", "source": "x = 0", "target": "m - 1 = 0"}, {"rel": "被描述", "source": "x ^ { 2 } - x + m - 1 = 0", "target": "x = 0"}, {"rel": "被代入", "source": "x ^ { 2 } - x + m - 1 = 0", "target": "m - 1 = 0"}, {"rel": "被描述", "source": "x", "target": "x = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - x + m - 1 = 0$ 有一个根是 $0$", "target": "x = 0"}, {"rel": "等式方程求解", "source": "m - 1 = 0", "target": "m = 1"}]}}
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{"content": "If the value of the algebraic expression $1 - 8 x$ is the opposite of $9 x - 3$, then $x$ = ____?", "answer": "2", "steps": "According to the fact that the sum of two numbers that are opposite in sign is zero, we have: $1 - 8 x + 9 x - 3 = 0$. Simplifying and rearranging, we get: $x = 2$.", "expr_cands": ["1 - 8 x", "x", "9 x - 3", "1 - 8 x + 9 x - 3 = 0", "x = 2"], "exprs": ["1 - 8 x + 9 x - 3 = 0", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1 - 8 x"}, {"id": "1 - 8 x + 9 x - 3 = 0"}, {"id": "9 x - 3"}, {"id": "代数式 $1 - 8 x$ 与 $9 x - 3$ 的值互为相反数"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "1 - 8 x", "target": "1 - 8 x + 9 x - 3 = 0"}, {"rel": "等式方程求解", "source": "1 - 8 x + 9 x - 3 = 0", "target": "x = 2"}, {"rel": "被描述", "source": "9 x - 3", "target": "1 - 8 x + 9 x - 3 = 0"}, {"rel": "限制性描述", "source": "代数式 $1 - 8 x$ 与 $9 x - 3$ 的值互为相反数", "target": "1 - 8 x + 9 x - 3 = 0"}]}}
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{"content": "If $m$ and $n$ are reciprocals, then the value of $m ^ 2 n - ( m + 3 )$ is ____?", "answer": "- 3", "steps": "$\\because$ $m$ and $n$ are reciprocals, $\\therefore$ $mn = 1$. $\\because$ $m$ and $n$ are opposite in sign, $\\therefore$ $m ^ 2 n - ( m + 3 ) = m - ( m + 3 ) = m - m - 3 = - 3$.", "expr_cands": ["m", "n", "m ^ { 2 } n - ( m + 3 )", "mn = 1", "- 3"], "exprs": ["mn = 1", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m"}, {"id": "mn = 1"}, {"id": "n"}, {"id": "$m$ 与 $n$ 是互为倒数"}, {"id": "m ^ { 2 } n - ( m + 3 )"}, {"id": "- 3"}], "links": [{"rel": "被描述", "source": "m", "target": "mn = 1"}, {"rel": "代入", "source": "mn = 1", "target": "- 3"}, {"rel": "被描述", "source": "n", "target": "mn = 1"}, {"rel": "限制性描述", "source": "$m$ 与 $n$ 是互为倒数", "target": "mn = 1"}, {"rel": "被代入", "source": "m ^ { 2 } n - ( m + 3 )", "target": "- 3"}]}}
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{"content": "$x = - 1$ is a solution of the equation $\\frac { 1 } { x - 2 } = \\frac { 2 } { x + a }$, the value of $a$ is ____?", "answer": "- 5", "steps": "Substituting $x = - 1$ into $\\frac { 1 } { x - 2 } = \\frac { 2 } { x + a }$ yields $\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }$, solving for $a$ gives $a = - 5$. It can be verified that $a = - 5$ is a solution to the equation $\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }$.", "expr_cands": ["x = - 1", "x", "\\frac { 1 } { x - 2 } = \\frac { 2 } { x + a }", "a", "- \\frac { 1 } { 3 } = \\frac { 2 } { a - 1 }", "\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }", "a = - 5"], "exprs": ["\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }", "a = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x - 2 } = \\frac { 2 } { x + a }"}, {"id": "\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }"}, {"id": "x = - 1"}, {"id": "a = - 5"}], "links": [{"rel": "被代入", "source": "\\frac { 1 } { x - 2 } = \\frac { 2 } { x + a }", "target": "\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }"}, {"rel": "等式方程求解", "source": "\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }", "target": "a = - 5"}, {"rel": "代入", "source": "x = - 1", "target": "\\frac { 1 } { - 1 - 2 } = \\frac { 2 } { - 1 + a }"}]}}
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{"content": "The equation of the line $y = - 3 x + 1$ after translating $2$ units to the right and $1$ unit up is _____.", "answer": "y = - 3 x + 8", "steps": "The line $y = - 3 x + 1$ is translated 2 units to the right, resulting in $y = - 3 ( x - 2 ) + 1$, which simplifies to $y = - 3 x + 7$. Then, it is translated 1 unit upwards, resulting in $y = - 3 x + 7 + 1$, which simplifies to $y = - 3 x + 8$.", "expr_cands": ["y = - 3 x + 1", "x", "y", "2", "1", "y = - 3 ( x - 2 ) + 1", "1 - 3 x = - 3 ( x - 2 ) + 1", "- 3 x + 7", "y = - 3 x + 7 + 1", "7 - 3 x = - 3 x + 7 + 1", "- 3 x + 8"], "exprs": ["y = - 3 ( x - 2 ) + 1", "y = - 3 x + 7 + 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - 3 x + 1"}, {"id": "y = - 3 ( x - 2 ) + 1"}, {"id": "2"}, {"id": "直线 $y = - 3 x + 1$ 向右平移 $2$ 个单位"}, {"id": "y = - 3 x + 7 + 1"}, {"id": "1"}, {"id": "再向上平移 $1$ 个单位"}], "links": [{"rel": "被描述", "source": "y = - 3 x + 1", "target": "y = - 3 ( x - 2 ) + 1"}, {"rel": "被描述", "source": "y = - 3 ( x - 2 ) + 1", "target": "y = - 3 x + 7 + 1"}, {"rel": "被描述", "source": "2", "target": "y = - 3 ( x - 2 ) + 1"}, {"rel": "限制性描述", "source": "直线 $y = - 3 x + 1$ 向右平移 $2$ 个单位", "target": "y = - 3 ( x - 2 ) + 1"}, {"rel": "被描述", "source": "1", "target": "y = - 3 x + 7 + 1"}, {"rel": "限制性描述", "source": "再向上平移 $1$ 个单位", "target": "y = - 3 x + 7 + 1"}]}}
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{"content": "If the square root of $x + 1$ is defined, then the possible values of $x$ are ____?", "answer": "x \\ge - 1", "steps": "According to the problem, we can obtain that $x + 1 \\ge 0$, and solving for $x$ gives $x \\ge - 1$.", "expr_cands": ["\\sqrt { x + 1 }", "x", "x + 1 \\ge 0", "- 1 \\le x", "x \\ge - 1"], "exprs": ["x + 1 \\ge 0", "x \\ge - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + 1 }"}, {"id": "x + 1 \\ge 0"}, {"id": "二次根式 $\\sqrt { x + 1 }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge - 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { x + 1 }", "target": "x + 1 \\ge 0"}, {"rel": "不等式方程求解", "source": "x + 1 \\ge 0", "target": "x \\ge - 1"}, {"rel": "限制性描述", "source": "二次根式 $\\sqrt { x + 1 }$ 有意义", "target": "x + 1 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x + 1 \\ge 0"}]}}
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{"content": "If $x < 0$, $y > 0$, and $x ^ 2 = 9$, $| y | = 2$, then $x + y$ = ____?", "answer": "- 1", "steps": "Since $x < 0$, $y > 0$ and $x ^ 2 = 9$, $| y | = 2$, therefore $x = - 3$, $y = 2$. Thus, $x + y = - 3 + 2 = - 1$.", "expr_cands": ["x < 0", "x", "y > 0", "y", "x ^ { 2 } = 9", "| y | = 2", "x + y", "x = - 3", "x = 3", "y = - 2", "y = 2", "- 1"], "exprs": ["x = - 3", "y = 2", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } = 9"}, {"id": "x = - 3"}, {"id": "x < 0"}, {"id": "| y | = 2"}, {"id": "y = 2"}, {"id": "y > 0"}, {"id": "x + y"}, {"id": "- 1"}], "links": [{"rel": "联立", "source": "x ^ { 2 } = 9", "target": "x = - 3"}, {"rel": "代入", "source": "x = - 3", "target": "- 1"}, {"rel": "联立", "source": "x < 0", "target": "x = - 3"}, {"rel": "联立", "source": "| y | = 2", "target": "y = 2"}, {"rel": "代入", "source": "y = 2", "target": "- 1"}, {"rel": "联立", "source": "y > 0", "target": "y = 2"}, {"rel": "被代入", "source": "x + y", "target": "- 1"}]}}
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{"content": "If the simplest quadratic radical $\\sqrt { 3 a + 4 }$ and $\\sqrt { 25 - 4 a }$ are of the same type, then $a$ = ____?", "answer": "3", "steps": "$\\because$ The simplest quadratic radical $\\sqrt { 3 a + 4 }$ and $\\sqrt { 25 - 4 a }$ are of the same type, $\\therefore$ $3 a + 4 = 25 - 4 a$, solving for $a$, we get $a = 3$.", "expr_cands": ["\\sqrt { 3 a + 4 }", "a", "\\sqrt { 25 - 4 a }", "3 a + 4 = 25 - 4 a", "a = 3"], "exprs": ["3 a + 4 = 25 - 4 a", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 3 a + 4 }"}, {"id": "3 a + 4 = 25 - 4 a"}, {"id": "\\sqrt { 25 - 4 a }"}, {"id": "最简二次根式 $\\sqrt { 3 a + 4 }$ 与 $\\sqrt { 25 - 4 a }$ 是同类二次根式"}, {"id": "a = 3"}], "links": [{"rel": "被描述", "source": "\\sqrt { 3 a + 4 }", "target": "3 a + 4 = 25 - 4 a"}, {"rel": "等式方程求解", "source": "3 a + 4 = 25 - 4 a", "target": "a = 3"}, {"rel": "被描述", "source": "\\sqrt { 25 - 4 a }", "target": "3 a + 4 = 25 - 4 a"}, {"rel": "限制性描述", "source": "最简二次根式 $\\sqrt { 3 a + 4 }$ 与 $\\sqrt { 25 - 4 a }$ 是同类二次根式", "target": "3 a + 4 = 25 - 4 a"}]}}
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{"content": "If the expansion of $( ax + 3 y ) ( x - y )$ does not contain the term $xy$, then the value of $a$ is ____?", "answer": "3", "steps": "$( ax + 3 y ) ( x - y ) = a { x } ^ { 2 } - axy + 3 xy - 3 { y } ^ { 2 } = a { x } ^ { 2 } + ( 3 - a ) xy - 3 { y } ^ { 2 }$, because there is no xy term in the expanded expression, so 3 - a = 0, a = 3.", "expr_cands": ["( ax + 3 y ) ( x - y )", "y", "a", "x", "xy", "a { x } ^ { 2 } + ( 3 - a ) xy - 3 { y } ^ { 2 }", "3 - a = 0", "a = 3"], "exprs": ["a { x } ^ { 2 } + ( 3 - a ) xy - 3 { y } ^ { 2 }", "3 - a = 0", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( ax + 3 y ) ( x - y )"}, {"id": "a { x } ^ { 2 } + ( 3 - a ) xy - 3 { y } ^ { 2 }"}, {"id": "xy"}, {"id": "3 - a = 0"}, {"id": "$( ax + 3 y ) ( x - y )$ 的展开式中不含 $xy$ 项"}, {"id": "a = 3"}], "links": [{"rel": "提取因式", "source": "( ax + 3 y ) ( x - y )", "target": "a { x } ^ { 2 } + ( 3 - a ) xy - 3 { y } ^ { 2 }"}, {"rel": "被描述", "source": "a { x } ^ { 2 } + ( 3 - a ) xy - 3 { y } ^ { 2 }", "target": "3 - a = 0"}, {"rel": "提取因式参考", "source": "xy", "target": "a { x } ^ { 2 } + ( 3 - a ) xy - 3 { y } ^ { 2 }"}, {"rel": "等式方程求解", "source": "3 - a = 0", "target": "a = 3"}, {"rel": "限制性描述", "source": "$( ax + 3 y ) ( x - y )$ 的展开式中不含 $xy$ 项", "target": "3 - a = 0"}]}}
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{"content": "If $2$ is the mean of the ratio of $x$ and $5$, then $x$ = ____ ?", "answer": "\\frac { 4 } { 5 }", "steps": "\\because $2$ is the middle term of the ratio between $x$ and $5$, \\therefore $2 ^ 2 = 5 x$, solving for $x$ gives $x = \\frac { 4 } { 5 }$.", "expr_cands": ["2", "x", "5", "2 ^ { 2 } = 5 x", "x = \\frac { 4 } { 5 }"], "exprs": ["2 ^ { 2 } = 5 x", "x = \\frac { 4 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "2 ^ { 2 } = 5 x"}, {"id": "x"}, {"id": "5"}, {"id": "$2$ 是 $x$ 和 $5$ 的比例中项"}, {"id": "x = \\frac { 4 } { 5 }"}], "links": [{"rel": "被描述", "source": "2", "target": "2 ^ { 2 } = 5 x"}, {"rel": "等式方程求解", "source": "2 ^ { 2 } = 5 x", "target": "x = \\frac { 4 } { 5 }"}, {"rel": "被描述", "source": "x", "target": "2 ^ { 2 } = 5 x"}, {"rel": "被描述", "source": "5", "target": "2 ^ { 2 } = 5 x"}, {"rel": "限制性描述", "source": "$2$ 是 $x$ 和 $5$ 的比例中项", "target": "2 ^ { 2 } = 5 x"}]}}
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{"content": "What is the largest integer solution that makes the inequality $4 x + 3 < x + 6$ true?", "answer": "0", "steps": "Since $4 x - x < 6 - 3$, therefore $3 x < 3$, therefore $x < 1$. Thus, the largest integer solution to the inequality is $0$.", "expr_cands": ["4 x + 3 < x + 6", "x", "4 x - x < 6 - 3", "x < 1", "3 x < 3", "0"], "exprs": ["x < 1", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x + 3 < x + 6"}, {"id": "x < 1"}, {"id": "0"}, {"id": "使不等式 $4 x + 3 < x + 6$ 成立的最大整数解"}], "links": [{"rel": "不等式方程求解", "source": "4 x + 3 < x + 6", "target": "x < 1"}, {"rel": "被描述", "source": "x < 1", "target": "0"}, {"rel": "限制性描述", "source": "使不等式 $4 x + 3 < x + 6$ 成立的最大整数解", "target": "0"}]}}
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{"content": "If the line $y = ( m - 3 ) x$ passes through the second and fourth quadrants, then the range of values for $m$ is ____?", "answer": "m < 3", "steps": "$\\because$ The line $y = ( m - 3 ) x$ passes through the second and fourth quadrants. $\\therefore$ $m - 3 < 0$, $\\therefore$ $m < 3$.", "expr_cands": ["y = ( m - 3 ) x", "y", "x", "m", "m - 3 < 0", "m < 3"], "exprs": ["m - 3 < 0", "m < 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m - 3 ) x"}, {"id": "m - 3 < 0"}, {"id": "直线 $y = ( m - 3 ) x$ 经过第二"}, {"id": "四象限"}, {"id": "m < 3"}], "links": [{"rel": "被描述", "source": "y = ( m - 3 ) x", "target": "m - 3 < 0"}, {"rel": "不等式方程求解", "source": "m - 3 < 0", "target": "m < 3"}, {"rel": "限制性描述", "source": "直线 $y = ( m - 3 ) x$ 经过第二", "target": "m - 3 < 0"}, {"rel": "限制性描述", "source": "四象限", "target": "m - 3 < 0"}]}}
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{"content": "The solution set of the inequality $6 - ( 4 x + 3 ) > 2 x$ is ____?", "answer": "x < \\frac { 1 } { 2 }", "steps": "$6 - ( 4 x + 3 ) > 2 x$, remove parentheses, we get $6 - 4 x - 3 > 2 x$, move terms, we get $- 4 x - 2 x > 3 - 6$, combine like terms, we get $- 6 x > - 3$, divide both sides by $- 6$, we get $x < \\frac { 1 } { 2 }$.", "expr_cands": ["6 - ( 4 x + 3 ) > 2 x", "x", "x < \\frac { 1 } { 2 }", "6 - 4 x - 3 > 2 x", "- 4 x - 2 x > 3 - 6", "- 6 x > - 3", "1"], "exprs": ["x < \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "6 - ( 4 x + 3 ) > 2 x"}, {"id": "x < \\frac { 1 } { 2 }"}], "links": [{"rel": "不等式方程求解", "source": "6 - ( 4 x + 3 ) > 2 x", "target": "x < \\frac { 1 } { 2 }"}]}}
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{"content": "$( x + 5 ) ( x + a )$ has a coefficient of $+ 3$ for $x$. What is the value of $a$?", "answer": "- 2", "steps": "$( x + 5 ) ( x + a ) = { x } ^ { 2 } + ax + 5 x + 5 a = { x } ^ { 2 } + ( a + 5 ) x + 5 a$ . Since the coefficient of $x$ is $+ 3$, we have $a + 5 = 3$. Therefore, $a = - 2$.", "expr_cands": ["( x + 5 ) ( x + a )", "a", "x", "+ 3", "{ x } ^ { 2 } + ( a + 5 ) x + 5 a", "a + 5 = 3", "a = - 2"], "exprs": ["{ x } ^ { 2 } + ( a + 5 ) x + 5 a", "a + 5 = 3", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + 5 ) ( x + a )"}, {"id": "{ x } ^ { 2 } + ( a + 5 ) x + 5 a"}, {"id": "x"}, {"id": "+ 3"}, {"id": "a + 5 = 3"}, {"id": "$( x + 5 ) ( x + a )$ 的结果中"}, {"id": "$x$ 的系数是 $+ 3$"}, {"id": "a = - 2"}], "links": [{"rel": "提取因式", "source": "( x + 5 ) ( x + a )", "target": "{ x } ^ { 2 } + ( a + 5 ) x + 5 a"}, {"rel": "被描述", "source": "{ x } ^ { 2 } + ( a + 5 ) x + 5 a", "target": "a + 5 = 3"}, {"rel": "提取因式参考", "source": "x", "target": "{ x } ^ { 2 } + ( a + 5 ) x + 5 a"}, {"rel": "被描述", "source": "+ 3", "target": "a + 5 = 3"}, {"rel": "等式方程求解", "source": "a + 5 = 3", "target": "a = - 2"}, {"rel": "限制性描述", "source": "$( x + 5 ) ( x + a )$ 的结果中", "target": "a + 5 = 3"}, {"rel": "限制性描述", "source": "$x$ 的系数是 $+ 3$", "target": "a + 5 = 3"}]}}
|
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{"content": "Given that when $x = 1$, the value of the algebraic expression $ax ^ 3 + bx + 5$ is $- 4$, what is the value of the expression when $x = - 1$?", "answer": "14", "steps": "$\\because$ When $x = 1$, $ax ^ 3 + bx + 5 = - 4$, $\\therefore$ $a + b + 5 = - 4$, solving for $a + b = - 9$, $\\therefore$ when $x = - 1$, $ax ^ 3 + bx + 5 = - ( a + b ) + 5 = - ( - 9 ) + 5 = 14$.", "expr_cands": ["x = 1", "x", "ax ^ { 3 } + bx + 5", "a", "b", "- 4", "x = - 1", "ax ^ { 3 } + bx + 5 = - 4", "a + b + 5 = - 4", "a + b = - 9", "- ( a + b ) + 5", "14"], "exprs": ["a + b + 5 = - 4", "- ( a + b ) + 5", "a + b = - 9", "14"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 1"}, {"id": "a + b + 5 = - 4"}, {"id": "ax ^ { 3 } + bx + 5"}, {"id": "- 4"}, {"id": "当 $x = 1$ 时"}, {"id": "代数式 $ax ^ { 3 } + bx + 5$ 的值为 $- 4$"}, {"id": "a + b = - 9"}, {"id": "- ( a + b ) + 5"}, {"id": "x = - 1"}, {"id": "当 $x = - 1$ 时"}, {"id": "代数式 $ax ^ { 3 } + bx + 5$ 的值"}, {"id": "14"}], "links": [{"rel": "被描述", "source": "x = 1", "target": "a + b + 5 = - 4"}, {"rel": "移项", "source": "a + b + 5 = - 4", "target": "a + b = - 9"}, {"rel": "被描述", "source": "ax ^ { 3 } + bx + 5", "target": "a + b + 5 = - 4"}, {"rel": "被描述", "source": "ax ^ { 3 } + bx + 5", "target": "- ( a + b ) + 5"}, {"rel": "被描述", "source": "- 4", "target": "a + b + 5 = - 4"}, {"rel": "限制性描述", "source": "当 $x = 1$ 时", "target": "a + b + 5 = - 4"}, {"rel": "限制性描述", "source": "代数式 $ax ^ { 3 } + bx + 5$ 的值为 $- 4$", "target": "a + b + 5 = - 4"}, {"rel": "代入", "source": "a + b = - 9", "target": "14"}, {"rel": "被代入", "source": "- ( a + b ) + 5", "target": "14"}, {"rel": "被描述", "source": "x = - 1", "target": "- ( a + b ) + 5"}, {"rel": "限制性描述", "source": "当 $x = - 1$ 时", "target": "- ( a + b ) + 5"}, {"rel": "限制性描述", "source": "代数式 $ax ^ { 3 } + bx + 5$ 的值", "target": "- ( a + b ) + 5"}]}}
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{"content": "If $y = \\sqrt { 3 - x }$ is meaningful, then the range of values for $x$ is ____?", "answer": "x \\le 3", "steps": "From the given condition, we have $3 - x \\geq 0$. Solving for $x$, we get $x \\leq 3$.", "expr_cands": ["y = \\sqrt { 3 - x }", "y", "x", "3 - x \\ge 0", "x \\le 3"], "exprs": ["3 - x \\ge 0", "x \\le 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { 3 - x }"}, {"id": "3 - x \\ge 0"}, {"id": "$y = \\sqrt { 3 - x }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\le 3"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { 3 - x }", "target": "3 - x \\ge 0"}, {"rel": "不等式方程求解", "source": "3 - x \\ge 0", "target": "x \\le 3"}, {"rel": "限制性描述", "source": "$y = \\sqrt { 3 - x }$ 有意义", "target": "3 - x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "3 - x \\ge 0"}]}}
|
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{"content": "When $x$ = ____ ?, $2 x + 3$ and $5 + 6 x$ are opposite numbers.", "answer": "- 1", "steps": "According to the problem, we have $2 x + 3 + 5 + 6 x = 0$, and solving for $x$ gives $x = - 1$.", "expr_cands": ["x", "2 x + 3", "5 + 6 x", "2 x + 3 + 5 + 6 x = 0", "x = - 1"], "exprs": ["2 x + 3 + 5 + 6 x = 0", "x = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 3"}, {"id": "2 x + 3 + 5 + 6 x = 0"}, {"id": "5 + 6 x"}, {"id": "$2 x + 3$ 与 $5 + 6 x$ 互为相反数"}, {"id": "x = - 1"}], "links": [{"rel": "被描述", "source": "2 x + 3", "target": "2 x + 3 + 5 + 6 x = 0"}, {"rel": "等式方程求解", "source": "2 x + 3 + 5 + 6 x = 0", "target": "x = - 1"}, {"rel": "被描述", "source": "5 + 6 x", "target": "2 x + 3 + 5 + 6 x = 0"}, {"rel": "限制性描述", "source": "$2 x + 3$ 与 $5 + 6 x$ 互为相反数", "target": "2 x + 3 + 5 + 6 x = 0"}]}}
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{"content": "If $x$ and $y$ are real numbers and satisfy $| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0$, then the value of $x + y$ is ____?", "answer": "- 5", "steps": "Because $| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0$, therefore $2 x + 1 = 0$, $9 + 2 y = 0$, solving gives: $x = - \\frac { 1 } { 2 }$, $y = - \\frac { 9 } { 2 }$, then $x + y = - \\frac { 1 } { 2 } - \\frac { 9 } { 2 } = - 5$.", "expr_cands": ["x", "y", "| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0", "x + y", "2 x + 1 = 0", "x = - \\frac { 1 } { 2 }", "9 + 2 y = 0", "y = - \\frac { 9 } { 2 }", "- 5"], "exprs": ["2 x + 1 = 0", "9 + 2 y = 0", "x = - \\frac { 1 } { 2 }", "y = - \\frac { 9 } { 2 }", "- 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0"}, {"id": "2 x + 1 = 0"}, {"id": "$x$ , $y$ 为实数"}, {"id": "且满足 $| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0$"}, {"id": "绝对值恒大于等于0"}, {"id": "9 + 2 y = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x = - \\frac { 1 } { 2 }"}, {"id": "y = - \\frac { 9 } { 2 }"}, {"id": "x + y"}, {"id": "- 5"}], "links": [{"rel": "被描述", "source": "| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0", "target": "2 x + 1 = 0"}, {"rel": "被描述", "source": "| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0", "target": "9 + 2 y = 0"}, {"rel": "等式方程求解", "source": "2 x + 1 = 0", "target": "x = - \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "$x$ , $y$ 为实数", "target": "2 x + 1 = 0"}, {"rel": "限制性描述", "source": "$x$ , $y$ 为实数", "target": "9 + 2 y = 0"}, {"rel": "限制性描述", "source": "且满足 $| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0$", "target": "2 x + 1 = 0"}, {"rel": "限制性描述", "source": "且满足 $| 2 x + 1 | + \\sqrt { 9 + 2 y } = 0$", "target": "9 + 2 y = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "2 x + 1 = 0"}, {"rel": "等式方程求解", "source": "9 + 2 y = 0", "target": "y = - \\frac { 9 } { 2 }"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "9 + 2 y = 0"}, {"rel": "代入", "source": "x = - \\frac { 1 } { 2 }", "target": "- 5"}, {"rel": "代入", "source": "y = - \\frac { 9 } { 2 }", "target": "- 5"}, {"rel": "被代入", "source": "x + y", "target": "- 5"}]}}
|
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{"content": "The difference between the polynomial $2 x ^ 2 - 4 x + 1$ and $2 - x + x ^ 2$ is _____.", "answer": "x ^ { 2 } - 3 x - 1", "steps": "The difference between $2 x ^ 2 - 4 x + 1$ and $2 - x + x ^ 2$ is: $2 x ^ 2 - 4 x + 1 - ( 2 - x + x ^ 2 ) = 2 x ^ 2 - 4 x + 1 - 2 + x - x ^ 2 = x ^ 2 - 3 x - 1$.", "expr_cands": ["2 x ^ { 2 } - 4 x + 1", "x", "2 - x + x ^ { 2 }", "2 x ^ { 2 } - 4 x + 1 - ( 2 - x + x ^ { 2 } )", "x ^ { 2 } - 3 x - 1"], "exprs": ["x ^ { 2 } - 3 x - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 2 } - 4 x + 1"}, {"id": "x ^ { 2 } - 3 x - 1"}, {"id": "2 - x + x ^ { 2 }"}, {"id": "整式 $2 x ^ { 2 } - 4 x + 1$ 与 $2 - x + x ^ { 2 }$ 的差"}], "links": [{"rel": "被描述", "source": "2 x ^ { 2 } - 4 x + 1", "target": "x ^ { 2 } - 3 x - 1"}, {"rel": "被描述", "source": "2 - x + x ^ { 2 }", "target": "x ^ { 2 } - 3 x - 1"}, {"rel": "限制性描述", "source": "整式 $2 x ^ { 2 } - 4 x + 1$ 与 $2 - x + x ^ { 2 }$ 的差", "target": "x ^ { 2 } - 3 x - 1"}]}}
|
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{"content": "If $\\frac { 2 } { 3 } x ^ { m + 1 } y ^ 4$ and $- \\frac { 2 x ^ 3 y ^ { n + 1 }} { 3 }$ are like terms, then $m ^ n$ = ____?", "answer": "8", "steps": "Because $\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }$ and $- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }$ are like terms, therefore $m + 1 = 3$, $n + 1 = 4$, thus $m = 2$, $n = 3$, and $m ^ { n } = 2 ^ { 3 } = 8$.", "expr_cands": ["\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }", "y", "m", "x", "- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }", "n", "m ^ { n }", "m + 1 = 3", "m = 2", "n + 1 = 4", "n = 3", "8"], "exprs": ["m + 1 = 3", "n + 1 = 4", "m = 2", "n = 3", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }"}, {"id": "m + 1 = 3"}, {"id": "- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }"}, {"id": "$\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }$ 与 $- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }$ 是同类项"}, {"id": "n + 1 = 4"}, {"id": "m = 2"}, {"id": "n = 3"}, {"id": "m ^ { n }"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }", "target": "m + 1 = 3"}, {"rel": "被描述", "source": "\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }", "target": "n + 1 = 4"}, {"rel": "等式方程求解", "source": "m + 1 = 3", "target": "m = 2"}, {"rel": "被描述", "source": "- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }", "target": "m + 1 = 3"}, {"rel": "被描述", "source": "- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }", "target": "n + 1 = 4"}, {"rel": "限制性描述", "source": "$\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }$ 与 $- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }$ 是同类项", "target": "m + 1 = 3"}, {"rel": "限制性描述", "source": "$\\frac { 2 } { 3 } x ^ { m + 1 } y ^ { 4 }$ 与 $- \\frac { 2 x ^ { 3 } y ^ { n + 1 } } { 3 }$ 是同类项", "target": "n + 1 = 4"}, {"rel": "等式方程求解", "source": "n + 1 = 4", "target": "n = 3"}, {"rel": "代入", "source": "m = 2", "target": "8"}, {"rel": "代入", "source": "n = 3", "target": "8"}, {"rel": "被代入", "source": "m ^ { n }", "target": "8"}]}}
|
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{"content": "If $3$ is the middle term of the ratio between $x$ and $6$, then $x$ = ____ ?", "answer": "( 1 * 2 + 1 / 2 )", "steps": "From the given information: $6 x = 3 * 3$, $6 x = 9$, and $x = ( 1 * 2 + 1 / 2 )$.", "expr_cands": ["3", "x", "6", "6 x = 3 * 3", "x = \\frac { 3 } { 2 }", "6 x", "9", "x = ( 1 * 2 + 1 / 2 )"], "exprs": ["6 x = 3 * 3", "x = \\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "6 x = 3 * 3"}, {"id": "x"}, {"id": "6"}, {"id": "$3$ 是数 $x$ 和 $6$ 的比例中项"}, {"id": "x = \\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "3", "target": "6 x = 3 * 3"}, {"rel": "等式方程求解", "source": "6 x = 3 * 3", "target": "x = \\frac { 3 } { 2 }"}, {"rel": "被描述", "source": "x", "target": "6 x = 3 * 3"}, {"rel": "被描述", "source": "6", "target": "6 x = 3 * 3"}, {"rel": "限制性描述", "source": "$3$ 是数 $x$ 和 $6$ 的比例中项", "target": "6 x = 3 * 3"}]}}
|
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{"content": "If $x$ is the arithmetic square root of $64$, then the cube root of $3 x + 3$ is ____?", "answer": "3", "steps": "$\\because x$ is the arithmetic square root of $64$, $\\therefore x = 8$, $\\therefore 3 x + 3 = 24 + 3 = 27$, $\\because 3 ^ 3 = 27$, $\\therefore$ the cube root of $27$ is $3$.", "expr_cands": ["x", "64", "3 x + 3", "x = 8", "27", "3 ^ { 3 }", "3"], "exprs": ["x = 8", "27", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x"}, {"id": "x = 8"}, {"id": "64"}, {"id": "$x$ 是 $64$ 的算术平方根"}, {"id": "3 x + 3"}, {"id": "27"}, {"id": "3"}, {"id": ", $27$ 的立方根是 $3$"}], "links": [{"rel": "被描述", "source": "x", "target": "x = 8"}, {"rel": "代入", "source": "x = 8", "target": "27"}, {"rel": "被描述", "source": "64", "target": "x = 8"}, {"rel": "限制性描述", "source": "$x$ 是 $64$ 的算术平方根", "target": "x = 8"}, {"rel": "被代入", "source": "3 x + 3", "target": "27"}, {"rel": "被描述", "source": "27", "target": "3"}, {"rel": "限制性描述", "source": ", $27$ 的立方根是 $3$", "target": "3"}]}}
|
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{"content": "Given the equation $2 x - 4 = 6 x + a$ has a solution satisfying $| 2 x + 3 | = 0$, find the value of $a$.", "answer": "2", "steps": "Solve $| 2 x + 3 | = 0$ to get $x = - \\frac { 3 } { 2 }$. Since $x = - \\frac { 3 } { 2 }$ is a solution to the equation $2 x - 4 = 6 x + a$, we have $2 * ( - \\frac { 3 } { 2 }) - 4 = 6 * ( - \\frac { 3 } { 2 }) + a$, which gives $a = 2$.", "expr_cands": ["2 x - 4 = 6 x + a", "x", "a", "| 2 x + 3 | = 0", "x = - \\frac { 3 } { 2 }", "- 7 = a - 9", "2 * ( - \\frac { 3 } { 2 } ) - 4 = 6 * ( - \\frac { 3 } { 2 } ) + a", "a = 2"], "exprs": ["x = - \\frac { 3 } { 2 }", "2 * ( - \\frac { 3 } { 2 } ) - 4 = 6 * ( - \\frac { 3 } { 2 } ) + a", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| 2 x + 3 | = 0"}, {"id": "x = - \\frac { 3 } { 2 }"}, {"id": "绝对值恒大于等于0"}, {"id": "2 x - 4 = 6 x + a"}, {"id": "2 * ( - \\frac { 3 } { 2 } ) - 4 = 6 * ( - \\frac { 3 } { 2 } ) + a"}, {"id": "a = 2"}], "links": [{"rel": "被描述", "source": "| 2 x + 3 | = 0", "target": "x = - \\frac { 3 } { 2 }"}, {"rel": "代入", "source": "x = - \\frac { 3 } { 2 }", "target": "2 * ( - \\frac { 3 } { 2 } ) - 4 = 6 * ( - \\frac { 3 } { 2 } ) + a"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x = - \\frac { 3 } { 2 }"}, {"rel": "被代入", "source": "2 x - 4 = 6 x + a", "target": "2 * ( - \\frac { 3 } { 2 } ) - 4 = 6 * ( - \\frac { 3 } { 2 } ) + a"}, {"rel": "等式方程求解", "source": "2 * ( - \\frac { 3 } { 2 } ) - 4 = 6 * ( - \\frac { 3 } { 2 } ) + a", "target": "a = 2"}]}}
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{"content": "If $- \\frac { 3 x ^ { n } y ^ { n } z } { 8 }$ is a monomial of degree nine, then $n$ = ____ ?", "answer": "4", "steps": "Since the monomial $- \\frac { 3 x ^ n y ^ n z } { 8 }$ is a ninth degree monomial, we have $n + n + 1 = 9$. Solving for $n$, we get $n = 4$.", "expr_cands": ["- \\frac { 3 x ^ { n } y ^ { n } z } { 8 }", "y", "x", "n", "z", "n + n + 1 = 9", "n = 4"], "exprs": ["n + n + 1 = 9", "n = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 3 x ^ { n } y ^ { n } z } { 8 }"}, {"id": "n + n + 1 = 9"}, {"id": "单项式 $- \\frac { 3 x ^ { n } y ^ { n } z } { 8 }$ 是一个九次单项式"}, {"id": "n = 4"}], "links": [{"rel": "被描述", "source": "- \\frac { 3 x ^ { n } y ^ { n } z } { 8 }", "target": "n + n + 1 = 9"}, {"rel": "等式方程求解", "source": "n + n + 1 = 9", "target": "n = 4"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { 3 x ^ { n } y ^ { n } z } { 8 }$ 是一个九次单项式", "target": "n + n + 1 = 9"}]}}
|
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{"content": "Given that $2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3$ is a cubic trinomial in terms of $x$ and $y$, the value of $3 n ^ { 2 } - 5 n + 1$ is ____?", "answer": "9", "steps": "Since $2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3$ is a cubic trinomial in terms of $x$ and $y$, we have $| n | = 1$ and $n - 1 \\neq 0$. Solving for $n$, we get $n = - 1$. Therefore, $3 n ^ { 2 } - 5 n + 1 = 9$.", "expr_cands": ["2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3", "x", "y", "n", "3 n ^ { 2 } - 5 n + 1", "| n | = 1", "n = - 1", "n = 1", "n - 1 \\neq 0", "n \\neq 1", "9"], "exprs": ["| n | = 1", "n - 1 \\neq 0", "n = - 1", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3"}, {"id": "| n | = 1"}, {"id": "$2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3$ 是关于 $x$ , $y$ 的三次三项式"}, {"id": "n - 1 \\neq 0"}, {"id": "n = - 1"}, {"id": "3 n ^ { 2 } - 5 n + 1"}, {"id": "9"}], "links": [{"rel": "被描述", "source": "2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3", "target": "| n | = 1"}, {"rel": "被描述", "source": "2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3", "target": "n - 1 \\neq 0"}, {"rel": "联立", "source": "| n | = 1", "target": "n = - 1"}, {"rel": "限制性描述", "source": "$2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3$ 是关于 $x$ , $y$ 的三次三项式", "target": "| n | = 1"}, {"rel": "限制性描述", "source": "$2 x ^ { 2 } y ^ { | n | } - ( n - 1 ) y + 3$ 是关于 $x$ , $y$ 的三次三项式", "target": "n - 1 \\neq 0"}, {"rel": "联立", "source": "n - 1 \\neq 0", "target": "n = - 1"}, {"rel": "代入", "source": "n = - 1", "target": "9"}, {"rel": "被代入", "source": "3 n ^ { 2 } - 5 n + 1", "target": "9"}]}}
|
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{"content": "If the equation $k + \\sqrt { x - 3 } = 5$ has no real roots with respect to $x$, then the range of values for $k$ is ____?", "answer": "k > 5", "steps": "$\\because k + \\sqrt { x - 3 } = 5$ is equivalent to $\\sqrt { x - 3 } = 5 - k$ having no real solutions, $\\therefore 5 - k < 0$, which implies $k > 5$.", "expr_cands": ["x", "k + \\sqrt { x - 3 } = 5", "k", "\\sqrt { x - 3 } = 5 - k", "5 - k < 0", "5 < k", "k > 5"], "exprs": ["\\sqrt { x - 3 } = 5 - k", "5 - k < 0", "k > 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "k + \\sqrt { x - 3 } = 5"}, {"id": "\\sqrt { x - 3 } = 5 - k"}, {"id": "5 - k < 0"}, {"id": "关于 $x$ 的方程 $k + \\sqrt { x - 3 } = 5$ 无实数根"}, {"id": "k > 5"}], "links": [{"rel": "移项", "source": "k + \\sqrt { x - 3 } = 5", "target": "\\sqrt { x - 3 } = 5 - k"}, {"rel": "被描述", "source": "\\sqrt { x - 3 } = 5 - k", "target": "5 - k < 0"}, {"rel": "不等式方程求解", "source": "5 - k < 0", "target": "k > 5"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $k + \\sqrt { x - 3 } = 5$ 无实数根", "target": "5 - k < 0"}]}}
|
||||
{"content": "Given the monomials $3 a ^ { 2 } b ^ { m - 1 }$ and $- 7 a ^ { n } b$ are like terms, what is the value of $m + n$?", "answer": "4", "steps": "$\\because$ The monomial $3 a ^ { 2 } b ^ { m - 1 }$ and $- 7 a ^ { n } b$ are like terms, $\\therefore$ $n = 2$, $m - 1 = 1$, $\\therefore$ $n = 2$, $m = 2$. Hence, $m + n = 4$.", "expr_cands": ["3 a ^ { 2 } b ^ { m - 1 }", "b", "a", "m", "- 7 a ^ { n } b", "n", "m + n", "n = 2", "m - 1 = 1", "m = 2", "4"], "exprs": ["n = 2", "m - 1 = 1", "m = 2", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a ^ { 2 } b ^ { m - 1 }"}, {"id": "n = 2"}, {"id": "- 7 a ^ { n } b"}, {"id": "单项式 $3 a ^ { 2 } b ^ { m - 1 }$ 与 $- 7 a ^ { n } b$ 互为同类项"}, {"id": "m - 1 = 1"}, {"id": "m = 2"}, {"id": "m + n"}, {"id": "4"}], "links": [{"rel": "被描述", "source": "3 a ^ { 2 } b ^ { m - 1 }", "target": "n = 2"}, {"rel": "被描述", "source": "3 a ^ { 2 } b ^ { m - 1 }", "target": "m - 1 = 1"}, {"rel": "代入", "source": "n = 2", "target": "4"}, {"rel": "被描述", "source": "- 7 a ^ { n } b", "target": "n = 2"}, {"rel": "被描述", "source": "- 7 a ^ { n } b", "target": "m - 1 = 1"}, {"rel": "限制性描述", "source": "单项式 $3 a ^ { 2 } b ^ { m - 1 }$ 与 $- 7 a ^ { n } b$ 互为同类项", "target": "n = 2"}, {"rel": "限制性描述", "source": "单项式 $3 a ^ { 2 } b ^ { m - 1 }$ 与 $- 7 a ^ { n } b$ 互为同类项", "target": "m - 1 = 1"}, {"rel": "等式方程求解", "source": "m - 1 = 1", "target": "m = 2"}, {"rel": "代入", "source": "m = 2", "target": "4"}, {"rel": "被代入", "source": "m + n", "target": "4"}]}}
|
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{"content": "The minimum value of the square root of $7 - 2 x$ is ____?", "answer": "0", "steps": "$\\because \\sqrt { 7 - 2 x } \\ge 0$, $\\therefore$ the minimum value of the quadratic radical $\\sqrt { 7 - 2 x }$ is $0$.", "expr_cands": ["\\sqrt { 7 - 2 x }", "x", "\\sqrt { 7 - 2 x } \\ge 0", "x \\le \\frac { 7 } { 2 }", "0"], "exprs": ["\\sqrt { 7 - 2 x } \\ge 0", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 7 - 2 x }"}, {"id": "\\sqrt { 7 - 2 x } \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "0"}, {"id": "二次根式 $\\sqrt { 7 - 2 x }$ 的最小值是 $0$"}, {"id": "二次根式 $\\sqrt { 7 - 2 x }$ 的最小值"}], "links": [{"rel": "被描述", "source": "\\sqrt { 7 - 2 x }", "target": "\\sqrt { 7 - 2 x } \\ge 0"}, {"rel": "被描述", "source": "\\sqrt { 7 - 2 x } \\ge 0", "target": "0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "\\sqrt { 7 - 2 x } \\ge 0"}, {"rel": "限制性描述", "source": "二次根式 $\\sqrt { 7 - 2 x }$ 的最小值是 $0$", "target": "0"}, {"rel": "限制性描述", "source": "二次根式 $\\sqrt { 7 - 2 x }$ 的最小值", "target": "0"}]}}
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{"content": "Given a quadratic equation $x ^ 2 - x + k = 0$ with one root being $1$, the other root is ____?", "answer": "0", "steps": "If the other root of the equation is $x _ 2$, then $1 + x _ 2 = 1$, which gives $x _ 2 = 0$. Therefore, the other root is $0$.", "expr_cands": ["x ^ { 2 } - x + k = 0", "k", "x", "1", "x _ { 2 }", "1 + x _ { 2 } = 1", "x_{2} = 0", "x _ { 2 } = 0", "0"], "exprs": ["x _ { 2 }", "1 + x _ { 2 } = 1", "x _ { 2 } = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程的另一个根是 $x _ { 2 }$"}, {"id": "x _ { 2 }"}, {"id": "x ^ { 2 } - x + k = 0"}, {"id": "1 + x _ { 2 } = 1"}, {"id": "1"}, {"id": "一元二次方程 $x ^ { 2 } - x + k = 0$ 的一根为 $1$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "x _ { 2 } = 0"}], "links": [{"rel": "假设描述", "source": "设方程的另一个根是 $x _ { 2 }$", "target": "x _ { 2 }"}, {"rel": "限制性描述", "source": "设方程的另一个根是 $x _ { 2 }$", "target": "1 + x _ { 2 } = 1"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "1 + x _ { 2 } = 1"}, {"rel": "被描述", "source": "x ^ { 2 } - x + k = 0", "target": "1 + x _ { 2 } = 1"}, {"rel": "等式方程求解", "source": "1 + x _ { 2 } = 1", "target": "x _ { 2 } = 0"}, {"rel": "被描述", "source": "1", "target": "1 + x _ { 2 } = 1"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - x + k = 0$ 的一根为 $1$", "target": "1 + x _ { 2 } = 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "1 + x _ { 2 } = 1"}]}}
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{"content": "If $m$ is a solution of the equation $3 x - 2 = 1$, then the value of the algebraic expression $30 m + 10$ is ____?", "answer": "40", "steps": "Substituting $x = m$, we get $3 m - 2 = 1$. Solving for $m$, we have $3 m = 3$. Therefore, $30 m + 10 = 3 \\times 10 + 10 = 40$.", "expr_cands": ["m", "3 x - 2 = 1", "x", "30 m + 10", "x = m", "3 m - 2 = 1", "m = 1", "3 m = 3", "40"], "exprs": ["x = m", "3 m - 2 = 1", "m = 1", "40"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m"}, {"id": "x = m"}, {"id": "3 x - 2 = 1"}, {"id": "x"}, {"id": "$m$ 是方程 $3 x - 2 = 1$ 的解"}, {"id": "3 m - 2 = 1"}, {"id": "m = 1"}, {"id": "30 m + 10"}, {"id": "40"}], "links": [{"rel": "被描述", "source": "m", "target": "x = m"}, {"rel": "代入", "source": "x = m", "target": "3 m - 2 = 1"}, {"rel": "被描述", "source": "3 x - 2 = 1", "target": "x = m"}, {"rel": "被代入", "source": "3 x - 2 = 1", "target": "3 m - 2 = 1"}, {"rel": "被描述", "source": "x", "target": "x = m"}, {"rel": "限制性描述", "source": "$m$ 是方程 $3 x - 2 = 1$ 的解", "target": "x = m"}, {"rel": "等式方程求解", "source": "3 m - 2 = 1", "target": "m = 1"}, {"rel": "代入", "source": "m = 1", "target": "40"}, {"rel": "被代入", "source": "30 m + 10", "target": "40"}]}}
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{"content": "If $a ^ { x } = 3$, $a ^ { y } = 2$, then the value of $a ^ { 3 x + 2 y }$ is ____?", "answer": "108", "steps": "Since $a ^ { x } = 3$ and $a ^ { y } = 2$, therefore $a ^ { 3 x + 2 y } = ( a ^ { x }) ^ { 3 } * ( a ^ { y }) ^ { 2 } = 3 ^ { 3 } * 2 ^ { 2 } = 108$.", "expr_cands": ["a ^ { x } = 3", "a", "x", "a ^ { y } = 2", "y", "a ^ { 3 x + 2 y }", "108"], "exprs": ["108"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 3 x + 2 y }"}, {"id": "108"}, {"id": "a ^ { x } = 3"}, {"id": "a ^ { y } = 2"}], "links": [{"rel": "被代入", "source": "a ^ { 3 x + 2 y }", "target": "108"}, {"rel": "代入", "source": "a ^ { x } = 3", "target": "108"}, {"rel": "代入", "source": "a ^ { y } = 2", "target": "108"}]}}
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{"content": "Given that $y$ is directly proportional to $x$, when $x = 2$, $y = 8$. What is $x$ when $y = 12$?", "answer": "3", "steps": "Let $y = kx$. Substituting $x = 2$ and $y = 8$ into the equation gives $8 = 2 k$. Solving for $k$, we get $k = 4$. Therefore, the function can be expressed as $y = 4 x$. Substituting $y = 12$ into the equation gives $12 = 4 x$. Solving for $x$, we get $x = 3$.", "expr_cands": ["y", "x", "x = 2", "y = 8", "y = 12", "y = kx", "k", "8 = 2 k", "k = 4", "y = 4 x", "12 = 4 x", "x = 3"], "exprs": ["y = kx", "8 = 2 k", "k = 4", "12 = 4 x", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $y = kx$"}, {"id": "y = kx"}, {"id": "x = 2"}, {"id": "8 = 2 k"}, {"id": "y = 8"}, {"id": "k = 4"}, {"id": "y = 12"}, {"id": "12 = 4 x"}, {"id": "x = 3"}], "links": [{"rel": "假设描述", "source": "设 $y = kx$", "target": "y = kx"}, {"rel": "被代入", "source": "y = kx", "target": "8 = 2 k"}, {"rel": "被代入", "source": "y = kx", "target": "12 = 4 x"}, {"rel": "代入", "source": "x = 2", "target": "8 = 2 k"}, {"rel": "等式方程求解", "source": "8 = 2 k", "target": "k = 4"}, {"rel": "代入", "source": "y = 8", "target": "8 = 2 k"}, {"rel": "代入", "source": "k = 4", "target": "12 = 4 x"}, {"rel": "代入", "source": "y = 12", "target": "12 = 4 x"}, {"rel": "等式方程求解", "source": "12 = 4 x", "target": "x = 3"}]}}
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{"content": "If $\\sqrt { x - 11 } = 5$, then the arithmetic square root of $x$ is ____?", "answer": "6", "steps": "Because $\\sqrt { x - 11 } = 5$, $x$ equals $25 + 11 = 36$, so the arithmetic square root of $x$ is $6$.", "expr_cands": ["\\sqrt { x - 11 } = 5", "x", "x = 36", "6"], "exprs": ["x = 36", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x - 11 } = 5"}, {"id": "x = 36"}, {"id": "6"}, {"id": "$x$ 的算术平方根"}], "links": [{"rel": "等式方程求解", "source": "\\sqrt { x - 11 } = 5", "target": "x = 36"}, {"rel": "被描述", "source": "x = 36", "target": "6"}, {"rel": "限制性描述", "source": "$x$ 的算术平方根", "target": "6"}]}}
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{"content": "Given the fractional equation about $x$, $\\frac { 2 } { x + 4 } = \\frac { m } { x }$, has the same solution as the fractional equation $\\frac { 3 } { 2 x } = \\frac { 1 } { x - 1 }$, then the value of $m ^ 2 - 2 m$ is ____?", "answer": "- \\frac { 48 } { 49 }", "steps": "Solve the fractional equation $\\frac { 3 } { 2 x } = \\frac { 1 } { x - 1 }$ to obtain $x = 3$. After checking that $x = 3$ is indeed a solution, substitute $x = 3$ into $\\frac { 2 } { x + 4 } = \\frac { m } { x }$ to obtain $\\frac { 2 } { 7 } = \\frac { m } { 3 }$. Solving for $m$ yields $m = \\frac { 6 } { 7 }$. Therefore, $m ^ 2 - 2 m = \\left ( \\frac { 6 } { 7 } \\right ) ^ 2 - 2 * \\frac { 6 } { 7 } = - \\frac { 48 } { 49 }$.", "expr_cands": ["x", "\\frac { 2 } { x + 4 } = \\frac { m } { x }", "m", "\\frac { 3 } { 2 x } = \\frac { 1 } { x - 1 }", "m ^ { 2 } - 2 m", "x = 3", "\\frac { 2 } { 7 } = \\frac { m } { 3 }", "m = \\frac { 6 } { 7 }", "- \\frac { 48 } { 49 }"], "exprs": ["x = 3", "\\frac { 2 } { 7 } = \\frac { m } { 3 }", "m = \\frac { 6 } { 7 }", "- \\frac { 48 } { 49 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 } { 2 x } = \\frac { 1 } { x - 1 }"}, {"id": "x = 3"}, {"id": "\\frac { 2 } { x + 4 } = \\frac { m } { x }"}, {"id": "\\frac { 2 } { 7 } = \\frac { m } { 3 }"}, {"id": "关于 $x$ 的分式方程 $\\frac { 2 } { x + 4 } = \\frac { m } { x }$ 与分式方程 $\\frac { 3 } { 2 x } = \\frac { 1 } { x - 1 }$ 的解相同"}, {"id": "m = \\frac { 6 } { 7 }"}, {"id": "m ^ { 2 } - 2 m"}, {"id": "- \\frac { 48 } { 49 }"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 3 } { 2 x } = \\frac { 1 } { x - 1 }", "target": "x = 3"}, {"rel": "被描述", "source": "x = 3", "target": "\\frac { 2 } { 7 } = \\frac { m } { 3 }"}, {"rel": "被描述", "source": "\\frac { 2 } { x + 4 } = \\frac { m } { x }", "target": "\\frac { 2 } { 7 } = \\frac { m } { 3 }"}, {"rel": "等式方程求解", "source": "\\frac { 2 } { 7 } = \\frac { m } { 3 }", "target": "m = \\frac { 6 } { 7 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的分式方程 $\\frac { 2 } { x + 4 } = \\frac { m } { x }$ 与分式方程 $\\frac { 3 } { 2 x } = \\frac { 1 } { x - 1 }$ 的解相同", "target": "\\frac { 2 } { 7 } = \\frac { m } { 3 }"}, {"rel": "代入", "source": "m = \\frac { 6 } { 7 }", "target": "- \\frac { 48 } { 49 }"}, {"rel": "被代入", "source": "m ^ { 2 } - 2 m", "target": "- \\frac { 48 } { 49 }"}]}}
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{"content": "If $( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9$ is a linear equation in one variable $x$, then $m$ = ____?", "answer": "3", "steps": "$\\because$ $( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9$ is a one-variable linear equation in $x$, $\\therefore$ $| 2 m - 5 | = 1$ and $m - 2 \\neq 0$. Solving for $m$, we get $m = 3$.", "expr_cands": ["( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9", "m", "x", "| 2 m - 5 | = 1", "m = 2", "m = 3", "m - 2 \\neq 0", "m \\neq 2"], "exprs": ["| 2 m - 5 | = 1", "m - 2 \\neq 0", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9"}, {"id": "| 2 m - 5 | = 1"}, {"id": "$( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9$ 是关于 $x$ 的一元一次方程"}, {"id": "m - 2 \\neq 0"}, {"id": "m = 3"}], "links": [{"rel": "被描述", "source": "( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9", "target": "| 2 m - 5 | = 1"}, {"rel": "被描述", "source": "( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "| 2 m - 5 | = 1", "target": "m = 3"}, {"rel": "限制性描述", "source": "$( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9$ 是关于 $x$ 的一元一次方程", "target": "| 2 m - 5 | = 1"}, {"rel": "限制性描述", "source": "$( m - 2 ) x ^ { | 2 m - 5 | } - 8 = 9$ 是关于 $x$ 的一元一次方程", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "m - 2 \\neq 0", "target": "m = 3"}]}}
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{"content": "Given that $y$ is directly proportional to $x$, when $x = 2$, $y = 8$. What is $x$ when $y = 16$?", "answer": "4", "steps": "Let $y = kx$. Substituting $x = 2$ and $y = 8$ into the equation gives $8 = 2 k$. Solving for $k$, we get $k = 4$. Therefore, the function can be expressed as $y = 4 x$. Substituting $y = 16$ into the equation gives $16 = 4 x$. Solving for $x$, we get $x = 4$.", "expr_cands": ["y", "x", "x = 2", "y = 8", "y = 16", "y = kx", "k", "8 = 2 k", "k = 4", "y = 4 x", "16 = 4 x", "x = 4"], "exprs": ["y = kx", "8 = 2 k", "k = 4", "y = 4 x", "16 = 4 x", "x = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $y = kx$"}, {"id": "y = kx"}, {"id": "x = 2"}, {"id": "8 = 2 k"}, {"id": "y = 8"}, {"id": "k = 4"}, {"id": "y = 4 x"}, {"id": "y = 16"}, {"id": "16 = 4 x"}, {"id": "x = 4"}], "links": [{"rel": "假设描述", "source": "设 $y = kx$", "target": "y = kx"}, {"rel": "被代入", "source": "y = kx", "target": "8 = 2 k"}, {"rel": "被代入", "source": "y = kx", "target": "y = 4 x"}, {"rel": "代入", "source": "x = 2", "target": "8 = 2 k"}, {"rel": "等式方程求解", "source": "8 = 2 k", "target": "k = 4"}, {"rel": "代入", "source": "y = 8", "target": "8 = 2 k"}, {"rel": "代入", "source": "k = 4", "target": "y = 4 x"}, {"rel": "被代入", "source": "y = 4 x", "target": "16 = 4 x"}, {"rel": "代入", "source": "y = 16", "target": "16 = 4 x"}, {"rel": "等式方程求解", "source": "16 = 4 x", "target": "x = 4"}]}}
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{"content": "The solution to the equation $2 x + 5 a = 3$ is the same as the solution to the equation $2 x + 2 = 0$. What is the value of $a$?", "answer": "1", "steps": "Since $2 x + 2 = 0$, we can solve for $x$ to get $x = - 1$. Also, since the solution to the equation $2 x + 5 a = 3$ is the same as the solution to the equation $2 x + 2 = 0$, which is $x = - 1$, we know that $x = - 1$ is a solution to $2 x + 5 a = 3$. Therefore, we can substitute $x = - 1$ into the equation to get $2 * ( - 1 ) + 5 a = 3$, and solve for $a$ to get $a = 1$.", "expr_cands": ["x", "2 x + 5 a = 3", "a", "2 x + 2 = 0", "x = - 1", "5 a - 2 = 3", "2 * ( - 1 ) + 5 a = 3", "a = 1"], "exprs": ["x = - 1", "2 * ( - 1 ) + 5 a = 3", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 2 = 0"}, {"id": "x = - 1"}, {"id": "2 x + 5 a = 3"}, {"id": "2 * ( - 1 ) + 5 a = 3"}, {"id": "a = 1"}], "links": [{"rel": "等式方程求解", "source": "2 x + 2 = 0", "target": "x = - 1"}, {"rel": "代入", "source": "x = - 1", "target": "2 * ( - 1 ) + 5 a = 3"}, {"rel": "被代入", "source": "2 x + 5 a = 3", "target": "2 * ( - 1 ) + 5 a = 3"}, {"rel": "等式方程求解", "source": "2 * ( - 1 ) + 5 a = 3", "target": "a = 1"}]}}
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{"content": "Given: $x = 5$ is a solution to the equation $3 x - 2 a = 1$ in terms of $x$. Find the value of $a$.", "answer": "7", "steps": "Substituting $x = 5$ into the equation gives $15 - 2 a = 1$, solving for $a$ gives $a = 7$.", "expr_cands": ["x = 5", "x", "3 x - 2 a = 1", "a", "15 - 2 a = 1", "a = 7"], "exprs": ["15 - 2 a = 1", "a = 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 5"}, {"id": "15 - 2 a = 1"}, {"id": "3 x - 2 a = 1"}, {"id": "a = 7"}], "links": [{"rel": "代入", "source": "x = 5", "target": "15 - 2 a = 1"}, {"rel": "等式方程求解", "source": "15 - 2 a = 1", "target": "a = 7"}, {"rel": "被代入", "source": "3 x - 2 a = 1", "target": "15 - 2 a = 1"}]}}
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{"content": "Given $( 2 x ^ 2 - 3 x + a ) ( x + 2 )$, if the result does not contain the term $x$, then $a$ = ____?", "answer": "6", "steps": "$$(2x^2-3x+a)(x+2)=2x^3+x^2+(a-6)x+2a$$Since there is no linear term in the result, we have $a - 6 = 0$, which gives $a = 6$.", "expr_cands": ["( 2 x ^ { 2 } - 3 x + a ) ( x + 2 )", "x", "a", "2 x ^ { 3 } + x ^ { 2 } + ( a - 6 ) x + 2 a", "a - 6 = 0", "a = 6"], "exprs": ["2 x ^ { 3 } + x ^ { 2 } + ( a - 6 ) x + 2 a", "a - 6 = 0", "a = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 2 x ^ { 2 } - 3 x + a ) ( x + 2 )"}, {"id": "2 x ^ { 3 } + x ^ { 2 } + ( a - 6 ) x + 2 a"}, {"id": "x"}, {"id": "a - 6 = 0"}, {"id": "$( 2 x ^ { 2 } - 3 x + a ) ( x + 2 )$ 计算结果中不含 $x$ 项"}, {"id": "a = 6"}], "links": [{"rel": "提取因式", "source": "( 2 x ^ { 2 } - 3 x + a ) ( x + 2 )", "target": "2 x ^ { 3 } + x ^ { 2 } + ( a - 6 ) x + 2 a"}, {"rel": "被描述", "source": "2 x ^ { 3 } + x ^ { 2 } + ( a - 6 ) x + 2 a", "target": "a - 6 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "2 x ^ { 3 } + x ^ { 2 } + ( a - 6 ) x + 2 a"}, {"rel": "等式方程求解", "source": "a - 6 = 0", "target": "a = 6"}, {"rel": "限制性描述", "source": "$( 2 x ^ { 2 } - 3 x + a ) ( x + 2 )$ 计算结果中不含 $x$ 项", "target": "a - 6 = 0"}]}}
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{"content": "In the same Cartesian coordinate system, the equation of the parabola that is symmetric about the $y$-axis with respect to the parabola $y = 3 ( x - 4 ) ^ 2$ is ____?", "answer": "y = 3 ( x + 4 ) ^ { 2 }", "steps": "The equation of the parabola that is symmetric about the $y$-axis with the equation $y = 3 ( x - 4 ) ^ { 2 }$ is $y = 3 ( x + 4 ) ^ { 2 }$.", "expr_cands": ["y = 3 ( x - 4 ) ^ { 2 }", "y", "x", "y = 3 ( x + 4 ) ^ { 2 }", "3 ( x - 4 ) ^ { 2 } = 3 ( x + 4 ) ^ { 2 }"], "exprs": ["y = 3 ( x + 4 ) ^ { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 3 ( x - 4 ) ^ { 2 }"}, {"id": "y = 3 ( x + 4 ) ^ { 2 }"}, {"id": "抛物线 $y = 3 ( x - 4 ) ^ { 2 }$ 关于 $y$ 轴对称的抛物线的解析式为 $y = 3 ( x + 4 ) ^ { 2 }$"}], "links": [{"rel": "被描述", "source": "y = 3 ( x - 4 ) ^ { 2 }", "target": "y = 3 ( x + 4 ) ^ { 2 }"}, {"rel": "限制性描述", "source": "抛物线 $y = 3 ( x - 4 ) ^ { 2 }$ 关于 $y$ 轴对称的抛物线的解析式为 $y = 3 ( x + 4 ) ^ { 2 }$", "target": "y = 3 ( x + 4 ) ^ { 2 }"}]}}
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{"content": "If the solution set of the inequality $2 x - 1 > 0$ is the same as the solution set of the inequality $\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }$ with respect to $x$, then the value of $a$ is ____?", "answer": "3", "steps": "Solve $\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }$ to get $x > \\frac { 1 } { a - 1 }$, solve $2 x - 1 > 0$ to get $x > \\frac { 1 } { 2 }$. Since the solution set of $\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }$ is the same as that of $2 x - 1 > 0$, we have $\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }$. Solving for $a$, we get $a = 3$.", "expr_cands": ["2 x - 1 > 0", "x", "\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }", "a", "x > \\frac { 1 } { a - 1 }", "\\frac { 1 } { 2 } < x", "x > \\frac { 1 } { 2 }", "\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }", "a = 3"], "exprs": ["x > \\frac { 1 } { a - 1 }", "x > \\frac { 1 } { 2 }", "\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }"}, {"id": "x > \\frac { 1 } { a - 1 }"}, {"id": "2 x - 1 > 0"}, {"id": "x > \\frac { 1 } { 2 }"}, {"id": "\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }"}, {"id": "不等式 $2 x - 1 > 0$ 的解集与关于 $x$ 的不等式 $\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }$ 的解集相同"}, {"id": "a = 3"}], "links": [{"rel": "不等式方程部分求解", "source": "\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }", "target": "x > \\frac { 1 } { a - 1 }"}, {"rel": "被描述", "source": "x > \\frac { 1 } { a - 1 }", "target": "\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }"}, {"rel": "不等式方程求解", "source": "2 x - 1 > 0", "target": "x > \\frac { 1 } { 2 }"}, {"rel": "被描述", "source": "x > \\frac { 1 } { 2 }", "target": "\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }"}, {"rel": "等式方程求解", "source": "\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }", "target": "a = 3"}, {"rel": "限制性描述", "source": "不等式 $2 x - 1 > 0$ 的解集与关于 $x$ 的不等式 $\\frac { x + 5 } { 2 } - 1 < \\frac { ax + 2 } { 2 }$ 的解集相同", "target": "\\frac { 1 } { a - 1 } = \\frac { 1 } { 2 }"}]}}
|
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{"content": "Let $x$, $y$, $z$ be real numbers satisfying $x + \\frac { 1 } { y } = 1$ and $y + \\frac { 1 } { z } = 1$. Find the value of $xyz$.", "answer": "- 1", "steps": "Since $x + \\frac { 1 } { y } = 1$ and $y + \\frac { 1 } { z } = 1$, we can solve for $x$ and $z$ in terms of $y$. Specifically, we have $x = 1 - \\frac { 1 } { y } = \\frac { y - 1 } { y }$ and $\\frac { 1 } { z } = 1 - y$, which implies $z = \\frac { 1 } { 1 - y }$. Therefore, we have $xyz = \\frac { y - 1 } { y } \\cdot y \\cdot \\frac { 1 } { 1 - y } = - 1$.", "expr_cands": ["x", "y", "z", "x + \\frac { 1 } { y } = 1", "y + \\frac { 1 } { z } = 1", "xyz", "x = \\frac { y - 1 } { y }", "\\frac { 1 } { z } = 1 - y", "\\frac { 1 } { 1 - y }", "- 1"], "exprs": ["x = \\frac { y - 1 } { y }", "\\frac { 1 } { z } = 1 - y", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + \\frac { 1 } { y } = 1"}, {"id": "x = \\frac { y - 1 } { y }"}, {"id": "y + \\frac { 1 } { z } = 1"}, {"id": "\\frac { 1 } { z } = 1 - y"}, {"id": "xyz"}, {"id": "- 1"}], "links": [{"rel": "等式方程部分求解", "source": "x + \\frac { 1 } { y } = 1", "target": "x = \\frac { y - 1 } { y }"}, {"rel": "代入", "source": "x = \\frac { y - 1 } { y }", "target": "- 1"}, {"rel": "移项", "source": "y + \\frac { 1 } { z } = 1", "target": "\\frac { 1 } { z } = 1 - y"}, {"rel": "代入", "source": "\\frac { 1 } { z } = 1 - y", "target": "- 1"}, {"rel": "被代入", "source": "xyz", "target": "- 1"}]}}
|
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{"content": "If $m = x + | x - 1 |$, then the minimum value of $m$ is ____?", "answer": "1", "steps": "$\\because | x - 1 | \\ge 0$, $\\therefore$ when $| x - 1 | = 0$, $m$ takes the minimum value, which is $x = 1$. $\\therefore$ When $x = 1$, $m = 1$.", "expr_cands": ["m = x + | x - 1 |", "x", "m", "| x - 1 | \\ge 0", "| x - 1 | = 0", "x = 1", "m = 1"], "exprs": ["| x - 1 | \\ge 0", "| x - 1 | = 0", "x = 1", "m = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m = x + | x - 1 |"}, {"id": "| x - 1 | \\ge 0"}, {"id": "绝对值恒大于等于0"}, {"id": "| x - 1 | = 0"}, {"id": "当 $| x - 1 | = 0$ 时 $m$ 取最小值"}, {"id": "x = 1"}, {"id": "m = 1"}], "links": [{"rel": "被描述", "source": "m = x + | x - 1 |", "target": "| x - 1 | \\ge 0"}, {"rel": "被代入", "source": "m = x + | x - 1 |", "target": "m = 1"}, {"rel": "被描述", "source": "| x - 1 | \\ge 0", "target": "| x - 1 | = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "| x - 1 | \\ge 0"}, {"rel": "等式方程求解", "source": "| x - 1 | = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "当 $| x - 1 | = 0$ 时 $m$ 取最小值", "target": "| x - 1 | = 0"}, {"rel": "代入", "source": "x = 1", "target": "m = 1"}]}}
|
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{"content": "The solution to the equation $2 ( m + x ) = 5 x - 6$ is $x = 1$. What is the value of $m$?", "answer": "- \\frac { 3 } { 2 }", "steps": "Substituting $x = 1$ into the equation, we get $2 ( m + 1 ) = 5 - 6$, which gives us $m = - \\frac { 3 } { 2 }$.", "expr_cands": ["2 ( m + x ) = 5 x - 6", "m", "x", "x = 1", "2 ( m + 1 ) = 5 - 6", "m = - \\frac { 3 } { 2 }"], "exprs": ["2 ( m + 1 ) = 5 - 6", "m = - \\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ( m + x ) = 5 x - 6"}, {"id": "2 ( m + 1 ) = 5 - 6"}, {"id": "x = 1"}, {"id": "m = - \\frac { 3 } { 2 }"}], "links": [{"rel": "被代入", "source": "2 ( m + x ) = 5 x - 6", "target": "2 ( m + 1 ) = 5 - 6"}, {"rel": "等式方程求解", "source": "2 ( m + 1 ) = 5 - 6", "target": "m = - \\frac { 3 } { 2 }"}, {"rel": "代入", "source": "x = 1", "target": "2 ( m + 1 ) = 5 - 6"}]}}
|
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{"content": "Given: The two square roots of a positive number are $2 a - 3$ and $a - 2$. Find the value of $a$.", "answer": "\\frac { 5 } { 3 }", "steps": "Because the two square roots of a positive number are $2 a - 3$ and $a - 2$, therefore $2 a - 3 + a - 2 = 0$. Solving for $a$, we get $a = \\frac { 5 } { 3 }$.", "expr_cands": ["2 a - 3", "a", "a - 2", "2 a - 3 + a - 2 = 0", "a = \\frac { 5 } { 3 }"], "exprs": ["2 a - 3 + a - 2 = 0", "a = \\frac { 5 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a - 3"}, {"id": "2 a - 3 + a - 2 = 0"}, {"id": "a - 2"}, {"id": ": 一个正数的两个平方根分别是 $2 a - 3$ 和 $a - 2$"}, {"id": "平方根互为相反数"}, {"id": "a = \\frac { 5 } { 3 }"}], "links": [{"rel": "被描述", "source": "2 a - 3", "target": "2 a - 3 + a - 2 = 0"}, {"rel": "等式方程求解", "source": "2 a - 3 + a - 2 = 0", "target": "a = \\frac { 5 } { 3 }"}, {"rel": "被描述", "source": "a - 2", "target": "2 a - 3 + a - 2 = 0"}, {"rel": "限制性描述", "source": ": 一个正数的两个平方根分别是 $2 a - 3$ 和 $a - 2$", "target": "2 a - 3 + a - 2 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "2 a - 3 + a - 2 = 0"}]}}
|
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{"content": "Given $\\frac { c } { 4 } = \\frac { b } { 5 } = \\frac { a } { 6 } \\neq 0$, what is the value of $\\frac { b + c } { a }$?", "answer": "\\frac { 3 } { 2 }", "steps": "Assuming the original text is in LaTeX format, the translation would be:Let $\\frac { c } { 4 } = \\frac { b } { 5 } = \\frac { a } { 6 } = k$. Then $c = 4 k$, $b = 5 k$, and $a = 6 k$. Therefore, $\\frac { b + c } { a } = \\frac { 5 k + 4 k } { 6 k } = \\frac { 3 } { 2 }$.", "expr_cands": ["\\frac { c } { 4 } = \\frac { b } { 5 } = \\frac { a } { 6 } \\neq 0", "\\frac { b + c } { a }", "c", "b", "a", "\\frac { c } { 4 } = k", "k", "c = 4 k", "b = 5 k", "a = 6 k", "\\frac { 3 } { 2 }"], "exprs": ["c = 4 k", "b = 5 k", "a = 6 k", "\\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $\\frac { c } { 4 } = \\frac { b } { 5 } = \\frac { a } { 6 } = k$ $c = 4 k$ , $b = 5 k$ , $a = 6 k$"}, {"id": "c = 4 k"}, {"id": "b = 5 k"}, {"id": "a = 6 k"}, {"id": "\\frac { b + c } { a }"}, {"id": "\\frac { 3 } { 2 }"}], "links": [{"rel": "假设描述", "source": "设 $\\frac { c } { 4 } = \\frac { b } { 5 } = \\frac { a } { 6 } = k$ $c = 4 k$ , $b = 5 k$ , $a = 6 k$", "target": "c = 4 k"}, {"rel": "假设描述", "source": "设 $\\frac { c } { 4 } = \\frac { b } { 5 } = \\frac { a } { 6 } = k$ $c = 4 k$ , $b = 5 k$ , $a = 6 k$", "target": "b = 5 k"}, {"rel": "假设描述", "source": "设 $\\frac { c } { 4 } = \\frac { b } { 5 } = \\frac { a } { 6 } = k$ $c = 4 k$ , $b = 5 k$ , $a = 6 k$", "target": "a = 6 k"}, {"rel": "代入", "source": "c = 4 k", "target": "\\frac { 3 } { 2 }"}, {"rel": "代入", "source": "b = 5 k", "target": "\\frac { 3 } { 2 }"}, {"rel": "代入", "source": "a = 6 k", "target": "\\frac { 3 } { 2 }"}, {"rel": "被代入", "source": "\\frac { b + c } { a }", "target": "\\frac { 3 } { 2 }"}]}}
|
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{"content": "The solution to the fractional equation $\\frac { 1 } { x - 5 } - \\frac { 10 } { x ^ { 2 } - 10 x + 25 } = 0$ is ____ ?", "answer": "15", "steps": "To eliminate the denominator, we get: $x - 5 - 10 = 0$. Solving for $x$, we get $x = 15$. After checking, we find that $x = 15$ is a solution to the fractional equation.", "expr_cands": ["\\frac { 1 } { x - 5 } - \\frac { 10 } { x ^ { 2 } - 10 x + 25 } = 0", "x", "x - 5 - 10 = 0", "x = 15"], "exprs": ["x - 5 - 10 = 0", "x = 15"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x - 5 } - \\frac { 10 } { x ^ { 2 } - 10 x + 25 } = 0"}, {"id": "x - 5 - 10 = 0"}, {"id": "x = 15"}], "links": [{"rel": "同乘除", "source": "\\frac { 1 } { x - 5 } - \\frac { 10 } { x ^ { 2 } - 10 x + 25 } = 0", "target": "x - 5 - 10 = 0"}, {"rel": "等式方程求解", "source": "x - 5 - 10 = 0", "target": "x = 15"}]}}
|
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{"content": "If $- \\frac { a } { 2 } < - \\frac { a } { 3 }$, then $a$ must satisfy ____?", "answer": "a > 0", "steps": "Multiplying both sides of the original inequality by $- 6$, we get: $3 a > 2 a$. Rearranging and combining like terms, we get: $a > 0$.", "expr_cands": ["- \\frac { a } { 2 } < - \\frac { a } { 3 }", "a", "- 6", "3 a > 2 a", "0 < a", "a > 0"], "exprs": ["a > 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { a } { 2 } < - \\frac { a } { 3 }"}, {"id": "a > 0"}], "links": [{"rel": "不等式方程求解", "source": "- \\frac { a } { 2 } < - \\frac { a } { 3 }", "target": "a > 0"}]}}
|
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{"content": "The polynomial $- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5$ is added to $a { x } ^ { 3 } + 5 x - b$, and the resulting polynomial has no cubic or constant terms. What is the value of $ab$?", "answer": "20", "steps": "From the given problem, we have $( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 + a { x } ^ { 3 } + 5 x - b = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )$. Since the polynomial does not contain a cubic term or a constant term, we have $a - 4 = 0$ and $5 - b = 0$, which implies $a = 4$ and $b = 5$. Therefore, $ab = 4 * 5 = 20$.", "expr_cands": ["- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5", "x", "a { x } ^ { 3 } + 5 x - b", "a", "b", "ab", "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )", "a - 4 = 0", "a = 4", "5 - b = 0", "b = 5", "20"], "exprs": ["( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )", "a - 4 = 0", "5 - b = 0", "a = 4", "b = 5", "20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5"}, {"id": "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )"}, {"id": "a { x } ^ { 3 } + 5 x - b"}, {"id": "多项式 $- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5$ 与 $a { x } ^ { 3 } + 5 x - b$ 相加后"}, {"id": "不含三次项和常数项"}, {"id": "a - 4 = 0"}, {"id": "5 - b = 0"}, {"id": "a = 4"}, {"id": "b = 5"}, {"id": "ab"}, {"id": "20"}], "links": [{"rel": "被描述", "source": "- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5", "target": "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )"}, {"rel": "被描述", "source": "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )", "target": "a - 4 = 0"}, {"rel": "被描述", "source": "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )", "target": "5 - b = 0"}, {"rel": "被描述", "source": "a { x } ^ { 3 } + 5 x - b", "target": "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )"}, {"rel": "限制性描述", "source": "多项式 $- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5$ 与 $a { x } ^ { 3 } + 5 x - b$ 相加后", "target": "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )"}, {"rel": "限制性描述", "source": "多项式 $- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5$ 与 $a { x } ^ { 3 } + 5 x - b$ 相加后", "target": "a - 4 = 0"}, {"rel": "限制性描述", "source": "多项式 $- { x } ^ { 2 } - 4 { x } ^ { 3 } + 5$ 与 $a { x } ^ { 3 } + 5 x - b$ 相加后", "target": "5 - b = 0"}, {"rel": "限制性描述", "source": "不含三次项和常数项", "target": "( - { x } ^ { 2 } - 4 { x } ^ { 3 } + 5 ) + ( a { x } ^ { 3 } + 5 x - b ) = ( a - 4 ) x ^ { 3 } - x ^ { 2 } + ( 5 - b )"}, {"rel": "限制性描述", "source": "不含三次项和常数项", "target": "a - 4 = 0"}, {"rel": "限制性描述", "source": "不含三次项和常数项", "target": "5 - b = 0"}, {"rel": "等式方程求解", "source": "a - 4 = 0", "target": "a = 4"}, {"rel": "等式方程求解", "source": "5 - b = 0", "target": "b = 5"}, {"rel": "代入", "source": "a = 4", "target": "20"}, {"rel": "代入", "source": "b = 5", "target": "20"}, {"rel": "被代入", "source": "ab", "target": "20"}]}}
|
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{"content": "The solution to the fractional equation $\\frac { ax + 2 } { x - 1 } = 1$ for $x$ is $x = 4$. What is the value of $a$?", "answer": "\\frac { 1 } { 4 }", "steps": "To eliminate the denominator, we have $ax + 2 = x - 1$. Substituting $x = 4$, we get $4 a + 2 = 3$, which gives us $a = \\frac { 1 } { 4 }$.", "expr_cands": ["x", "\\frac { ax + 2 } { x - 1 } = 1", "a", "x = 4", "ax + 2 = x - 1", "4 a + 2 = 3", "a = \\frac { 1 } { 4 }"], "exprs": ["ax + 2 = x - 1", "4 a + 2 = 3", "a = \\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { ax + 2 } { x - 1 } = 1"}, {"id": "ax + 2 = x - 1"}, {"id": "x = 4"}, {"id": "4 a + 2 = 3"}, {"id": "a = \\frac { 1 } { 4 }"}], "links": [{"rel": "同乘除", "source": "\\frac { ax + 2 } { x - 1 } = 1", "target": "ax + 2 = x - 1"}, {"rel": "被代入", "source": "ax + 2 = x - 1", "target": "4 a + 2 = 3"}, {"rel": "代入", "source": "x = 4", "target": "4 a + 2 = 3"}, {"rel": "等式方程求解", "source": "4 a + 2 = 3", "target": "a = \\frac { 1 } { 4 }"}]}}
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{"content": "Given that $- 25 a ^ { 2 m } b$ and $2 a ^ 6 b ^ { n + 3 }$ are like terms, what is the value of $m + n$?", "answer": "1", "steps": "According to the problem, we have $2 m = 6$, $n + 3 = 1$. Solving for $m$ and $n$, we get $m = 3$, $n = - 2$. Therefore, $m + n = 1$.", "expr_cands": ["- 25 a ^ { 2 m } b", "b", "a", "m", "2 a ^ { 6 } b ^ { n + 3 }", "n", "m + n", "2 m = 6", "m = 3", "n + 3 = 1", "n = - 2", "1"], "exprs": ["2 m = 6", "n + 3 = 1", "m = 3", "n = - 2", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 25 a ^ { 2 m } b"}, {"id": "2 m = 6"}, {"id": "2 a ^ { 6 } b ^ { n + 3 }"}, {"id": "$- 25 a ^ { 2 m } b$ 和 $2 a ^ { 6 } b ^ { n + 3 }$ 是同类项"}, {"id": "n + 3 = 1"}, {"id": "m = 3"}, {"id": "n = - 2"}, {"id": "m + n"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "- 25 a ^ { 2 m } b", "target": "2 m = 6"}, {"rel": "被描述", "source": "- 25 a ^ { 2 m } b", "target": "n + 3 = 1"}, {"rel": "等式方程求解", "source": "2 m = 6", "target": "m = 3"}, {"rel": "被描述", "source": "2 a ^ { 6 } b ^ { n + 3 }", "target": "2 m = 6"}, {"rel": "被描述", "source": "2 a ^ { 6 } b ^ { n + 3 }", "target": "n + 3 = 1"}, {"rel": "限制性描述", "source": "$- 25 a ^ { 2 m } b$ 和 $2 a ^ { 6 } b ^ { n + 3 }$ 是同类项", "target": "2 m = 6"}, {"rel": "限制性描述", "source": "$- 25 a ^ { 2 m } b$ 和 $2 a ^ { 6 } b ^ { n + 3 }$ 是同类项", "target": "n + 3 = 1"}, {"rel": "等式方程求解", "source": "n + 3 = 1", "target": "n = - 2"}, {"rel": "代入", "source": "m = 3", "target": "1"}, {"rel": "代入", "source": "n = - 2", "target": "1"}, {"rel": "被代入", "source": "m + n", "target": "1"}]}}
|
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{"content": "If rational numbers $m$ and $n$ satisfy $| m - 2 | + ( n + 2019 ) ^ { 2 } = 0$, then $m ^ { - 1 } + n ^ { 0 }$ = ____ ?", "answer": "\\frac { 3 } { 2 }", "steps": "From the given information, we have $m - 2 = 0$ and $n + 2019 = 0$, which implies $m = 2$ and $n = - 2019$. Therefore, $m ^ { - 1 } + n ^ 0 = 1 + \\frac { 1 } { 2 } = \\frac { 3 } { 2 }$.", "expr_cands": ["m", "n", "| m - 2 | + ( n + 2019 ) ^ { 2 } = 0", "m ^ { - 1 } + n ^ { 0 }", "m - 2 = 0", "m = 2", "n + 2019 = 0", "n = - 2019", "\\frac { 3 } { 2 }"], "exprs": ["m - 2 = 0", "n + 2019 = 0", "m = 2", "n = - 2019", "\\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| m - 2 | + ( n + 2019 ) ^ { 2 } = 0"}, {"id": "m - 2 = 0"}, {"id": "有理数 $m$ , $n$ 满足 $| m - 2 | + ( n + 2019 ) ^ { 2 } = 0$"}, {"id": "绝对值恒大于等于0"}, {"id": "n + 2019 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "m = 2"}, {"id": "n = - 2019"}, {"id": "m ^ { - 1 } + n ^ { 0 }"}, {"id": "\\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "| m - 2 | + ( n + 2019 ) ^ { 2 } = 0", "target": "m - 2 = 0"}, {"rel": "被描述", "source": "| m - 2 | + ( n + 2019 ) ^ { 2 } = 0", "target": "n + 2019 = 0"}, {"rel": "等式方程求解", "source": "m - 2 = 0", "target": "m = 2"}, {"rel": "限制性描述", "source": "有理数 $m$ , $n$ 满足 $| m - 2 | + ( n + 2019 ) ^ { 2 } = 0$", "target": "m - 2 = 0"}, {"rel": "限制性描述", "source": "有理数 $m$ , $n$ 满足 $| m - 2 | + ( n + 2019 ) ^ { 2 } = 0$", "target": "n + 2019 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m - 2 = 0"}, {"rel": "等式方程求解", "source": "n + 2019 = 0", "target": "n = - 2019"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "n + 2019 = 0"}, {"rel": "代入", "source": "m = 2", "target": "\\frac { 3 } { 2 }"}, {"rel": "代入", "source": "n = - 2019", "target": "\\frac { 3 } { 2 }"}, {"rel": "被代入", "source": "m ^ { - 1 } + n ^ { 0 }", "target": "\\frac { 3 } { 2 }"}]}}
|
||||
{"content": "Given a quadratic function $y = ax ^ { 2 } + 3 x$, when $x = 2$, $y = 14$, then $a$ = ____?", "answer": "2", "steps": "Substituting $x = 2$ and $y = 14$ into $y = ax ^ 2 + 3 x$, we get $14 = 4 a + 6$. Solving for $a$, we get $a = 2$.", "expr_cands": ["y = ax ^ { 2 } + 3 x", "a", "y", "x", "x = 2", "y = 14", "14 = 4 a + 6", "a = 2"], "exprs": ["14 = 4 a + 6", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ax ^ { 2 } + 3 x"}, {"id": "14 = 4 a + 6"}, {"id": "x = 2"}, {"id": "y = 14"}, {"id": "a = 2"}], "links": [{"rel": "被代入", "source": "y = ax ^ { 2 } + 3 x", "target": "14 = 4 a + 6"}, {"rel": "等式方程求解", "source": "14 = 4 a + 6", "target": "a = 2"}, {"rel": "代入", "source": "x = 2", "target": "14 = 4 a + 6"}, {"rel": "代入", "source": "y = 14", "target": "14 = 4 a + 6"}]}}
|
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{"content": "If the value of the algebraic expression $3 x + 2$ is not less than $0$, then the range of possible values for $x$ is ____?", "answer": "x \\ge - \\frac { 2 } { 3 }", "steps": "It is known from the problem that $3 x + 2 \\geq 0$, which implies that $x \\geq - \\frac { 2 } { 3 }$.", "expr_cands": ["3 x + 2", "x", "0", "3 x + 2 \\ge 0", "- \\frac { 2 } { 3 } \\le x", "x \\ge - \\frac { 2 } { 3 }"], "exprs": ["3 x + 2 \\ge 0", "x \\ge - \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + 2"}, {"id": "3 x + 2 \\ge 0"}, {"id": "0"}, {"id": "代数式 $3 x + 2$ 的值不小于 $0$"}, {"id": "x \\ge - \\frac { 2 } { 3 }"}], "links": [{"rel": "被描述", "source": "3 x + 2", "target": "3 x + 2 \\ge 0"}, {"rel": "不等式方程求解", "source": "3 x + 2 \\ge 0", "target": "x \\ge - \\frac { 2 } { 3 }"}, {"rel": "被描述", "source": "0", "target": "3 x + 2 \\ge 0"}, {"rel": "限制性描述", "source": "代数式 $3 x + 2$ 的值不小于 $0$", "target": "3 x + 2 \\ge 0"}]}}
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{"content": "When $a$ = ____ ?, the equation $\\frac { 3 x } { x - 3 } = 2 - \\frac { a } { 3 - x }$ about $x$ will have an increasing root.", "answer": "9", "steps": "Multiplying both sides of the equation by $x - 3$, we get $3 x = 2 ( x - 3 ) + a$. Since the original equation has a repeated root, the simplest common denominator is $x - 3 = 0$. Solving for $x = 3$, we get $a = 9$.", "expr_cands": ["a", "x", "\\frac { 3 x } { x - 3 } = 2 - \\frac { a } { 3 - x }", "x - 3", "3 x = 2 ( x - 3 ) + a", "x - 3 = 0", "x = 3", "a = 9"], "exprs": ["3 x = 2 ( x - 3 ) + a", "x - 3 = 0", "x = 3", "a = 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 x } { x - 3 } = 2 - \\frac { a } { 3 - x }"}, {"id": "3 x = 2 ( x - 3 ) + a"}, {"id": "x - 3 = 0"}, {"id": "关于 $x$ 的方程 $\\frac { 3 x } { x - 3 } = 2 - \\frac { a } { 3 - x }$ 会产生增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 3"}, {"id": "a = 9"}], "links": [{"rel": "同乘除", "source": "\\frac { 3 x } { x - 3 } = 2 - \\frac { a } { 3 - x }", "target": "3 x = 2 ( x - 3 ) + a"}, {"rel": "被描述", "source": "\\frac { 3 x } { x - 3 } = 2 - \\frac { a } { 3 - x }", "target": "x - 3 = 0"}, {"rel": "被代入", "source": "3 x = 2 ( x - 3 ) + a", "target": "a = 9"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $\\frac { 3 x } { x - 3 } = 2 - \\frac { a } { 3 - x }$ 会产生增根", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 3 = 0"}, {"rel": "代入", "source": "x = 3", "target": "a = 9"}]}}
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{"content": "If $a = - b$, $c = \\frac { 1 } { d }$, $| m | = 2$, then the value of $m ^ 2 - cd + \\frac { a + b } { m }$ is ____?", "answer": "3", "steps": "According to the problem, we have $a + b = 0$, $cd = 1$, and $m = 2$ or $- 2$. Therefore, the original expression is equal to $4 - 1 + 0 = 3$.", "expr_cands": ["a = - b", "b", "a", "c = \\frac { 1 } { d }", "d", "c", "| m | = 2", "m", "m ^ { 2 } - cd + \\frac { a + b } { m }", "a + b = 0", "cd = 1", "m = 2", "- 2", "4 - 1 + 0", "3"], "exprs": ["a + b = 0", "cd = 1", "4 - 1 + 0", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a = - b"}, {"id": "a + b = 0"}, {"id": "c = \\frac { 1 } { d }"}, {"id": "cd = 1"}, {"id": "| m | = 2"}, {"id": "4 - 1 + 0"}, {"id": "m ^ { 2 } - cd + \\frac { a + b } { m }"}, {"id": "3"}], "links": [{"rel": "移项", "source": "a = - b", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "4 - 1 + 0"}, {"rel": "同乘除", "source": "c = \\frac { 1 } { d }", "target": "cd = 1"}, {"rel": "代入", "source": "cd = 1", "target": "4 - 1 + 0"}, {"rel": "代入", "source": "| m | = 2", "target": "4 - 1 + 0"}, {"rel": "计算", "source": "4 - 1 + 0", "target": "3"}, {"rel": "被代入", "source": "m ^ { 2 } - cd + \\frac { a + b } { m }", "target": "4 - 1 + 0"}]}}
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{"content": "If the smallest integer solution of the inequality $5 ( x - 2 ) + 8 < 6 ( x - 1 ) + 7$ is a solution of the equation $2 x - ax = 3$, then the value of $a$ is ____?", "answer": "\\frac { 7 } { 2 }", "steps": "Solve the inequality $5 ( x - 2 ) + 8 < 6 ( x - 1 ) + 7$ to get $x > - 3$. The smallest integer solution is $- 2$. Substituting $x = - 2$ into the equation $- 4 + 2 a = 3$ gives $a = \\frac { 7 } { 2 }$.", "expr_cands": ["5 ( x - 2 ) + 8 < 6 ( x - 1 ) + 7", "x", "2 x - ax = 3", "a", "- 3 < x", "x > - 3", "- 2", "x = - 2", "- 4 + 2 a = 3", "a = \\frac { 7 } { 2 }"], "exprs": ["x > - 3", "x = - 2", "- 4 + 2 a = 3", "a = \\frac { 7 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 ( x - 2 ) + 8 < 6 ( x - 1 ) + 7"}, {"id": "x > - 3"}, {"id": "x = - 2"}, {"id": "最小整数解是 : $- 2$"}, {"id": "2 x - ax = 3"}, {"id": "- 4 + 2 a = 3"}, {"id": "不等式 $5 ( x - 2 ) + 8 < 6 ( x - 1 ) + 7$ 的最小整数解是方程 $2 x - ax = 3$ 的解"}, {"id": "a = \\frac { 7 } { 2 }"}], "links": [{"rel": "不等式方程求解", "source": "5 ( x - 2 ) + 8 < 6 ( x - 1 ) + 7", "target": "x > - 3"}, {"rel": "被描述", "source": "x > - 3", "target": "x = - 2"}, {"rel": "被描述", "source": "x = - 2", "target": "- 4 + 2 a = 3"}, {"rel": "限制性描述", "source": "最小整数解是 : $- 2$", "target": "x = - 2"}, {"rel": "被描述", "source": "2 x - ax = 3", "target": "- 4 + 2 a = 3"}, {"rel": "等式方程求解", "source": "- 4 + 2 a = 3", "target": "a = \\frac { 7 } { 2 }"}, {"rel": "限制性描述", "source": "不等式 $5 ( x - 2 ) + 8 < 6 ( x - 1 ) + 7$ 的最小整数解是方程 $2 x - ax = 3$ 的解", "target": "- 4 + 2 a = 3"}]}}
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{"content": "Given that the square roots of a positive number are $2 a + 1$ and $a - 7$, find the value of ${ a } ^ { 2 } - 2 a + 3$.", "answer": "3", "steps": "Since the two square roots of a positive number are $2 a + 1$ and $a - 7$, we have $2 a + 1 + a - 7 = 0$. Solving for $a$, we get $a = 2$. Therefore, ${ a } ^ 2 - 2 a + 3 = 2 ^ 2 - 2 \\times 2 + 3 = 4 - 4 + 3 = 3$.", "expr_cands": ["2 a + 1", "a", "a - 7", "{ a } ^ { 2 } - 2 a + 3", "2 a + 1 + a - 7 = 0", "a = 2", "3 a = 6", "3"], "exprs": ["2 a + 1 + a - 7 = 0", "a = 2", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a + 1"}, {"id": "2 a + 1 + a - 7 = 0"}, {"id": "a - 7"}, {"id": "一个正数的两个平方根分别为 $2 a + 1$ 和 $a - 7$"}, {"id": "平方根互为相反数"}, {"id": "a = 2"}, {"id": "{ a } ^ { 2 } - 2 a + 3"}, {"id": "3"}], "links": [{"rel": "被描述", "source": "2 a + 1", "target": "2 a + 1 + a - 7 = 0"}, {"rel": "等式方程求解", "source": "2 a + 1 + a - 7 = 0", "target": "a = 2"}, {"rel": "被描述", "source": "a - 7", "target": "2 a + 1 + a - 7 = 0"}, {"rel": "限制性描述", "source": "一个正数的两个平方根分别为 $2 a + 1$ 和 $a - 7$", "target": "2 a + 1 + a - 7 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "2 a + 1 + a - 7 = 0"}, {"rel": "代入", "source": "a = 2", "target": "3"}, {"rel": "被代入", "source": "{ a } ^ { 2 } - 2 a + 3", "target": "3"}]}}
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{"content": "If the two square roots of a positive number are $a + 3$ and $- 2 a$, what is the value of $a$?", "answer": "3", "steps": "Since the two square roots of a positive number are $a + 3$ and $- 2 a$, therefore $a + 3 + ( - 2 a ) = 0$. Solving for $a$, we get $a = 3$.", "expr_cands": ["a + 3", "a", "- 2 a", "a + 3 + ( - 2 a ) = 0", "a = 3"], "exprs": ["a + 3 + ( - 2 a ) = 0", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a + 3"}, {"id": "a + 3 + ( - 2 a ) = 0"}, {"id": "- 2 a"}, {"id": "一个正数的两个平方根是 $a + 3$ 和 $- 2 a$"}, {"id": "平方根互为相反数"}, {"id": "a = 3"}], "links": [{"rel": "被描述", "source": "a + 3", "target": "a + 3 + ( - 2 a ) = 0"}, {"rel": "等式方程求解", "source": "a + 3 + ( - 2 a ) = 0", "target": "a = 3"}, {"rel": "被描述", "source": "- 2 a", "target": "a + 3 + ( - 2 a ) = 0"}, {"rel": "限制性描述", "source": "一个正数的两个平方根是 $a + 3$ 和 $- 2 a$", "target": "a + 3 + ( - 2 a ) = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "a + 3 + ( - 2 a ) = 0"}]}}
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{"content": "If $( m - 2 ) ^ 2 + | n + 3 | = 0$, then the arithmetic square root of $m - n$ is ____?", "answer": "\\sqrt { 5 }", "steps": "From the given information, we have $m - 2 = 0$ and $n + 3 = 0$, which implies $m = 2$ and $n = - 3$. Therefore, $m - n = 2 - ( - 3 ) = 5$, and the arithmetic square root of $5.5$ is $\\sqrt { 5 }$.", "expr_cands": ["( m - 2 ) ^ { 2 } + | n + 3 | = 0", "m", "n", "m - n", "m - 2 = 0", "m = 2", "n + 3 = 0", "n = - 3", "m - n = 5.5", "\\sqrt { 5 }"], "exprs": ["m - 2 = 0", "n + 3 = 0", "m = 2", "n = - 3", "\\sqrt { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 2 ) ^ { 2 } + | n + 3 | = 0"}, {"id": "m - 2 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "n + 3 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "m = 2"}, {"id": "n = - 3"}, {"id": "m - n"}, {"id": "\\sqrt { 5 }"}, {"id": "$m - n$ 的算术平方根"}], "links": [{"rel": "被描述", "source": "( m - 2 ) ^ { 2 } + | n + 3 | = 0", "target": "m - 2 = 0"}, {"rel": "被描述", "source": "( m - 2 ) ^ { 2 } + | n + 3 | = 0", "target": "n + 3 = 0"}, {"rel": "等式方程求解", "source": "m - 2 = 0", "target": "m = 2"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "m - 2 = 0"}, {"rel": "等式方程求解", "source": "n + 3 = 0", "target": "n = - 3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "n + 3 = 0"}, {"rel": "被描述", "source": "m = 2", "target": "\\sqrt { 5 }"}, {"rel": "被描述", "source": "n = - 3", "target": "\\sqrt { 5 }"}, {"rel": "被描述", "source": "m - n", "target": "\\sqrt { 5 }"}, {"rel": "限制性描述", "source": "$m - n$ 的算术平方根", "target": "\\sqrt { 5 }"}]}}
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{"content": "The solution set of the inequality $( m - 4 ) x < 6$ is $x > \\frac { 6 } { m - 4 }$. What is the range of values for $m$?", "answer": "m < 4", "steps": "$\\because$ The solution set of the inequality $( m - 4 ) x < 6$ is $x > \\frac { 6 } { m - 4 }$, $\\therefore$ $m - 4 < 0$, and solving for $m$ gives $m < 4$.", "expr_cands": ["( m - 4 ) x < 6", "m", "x", "x > \\frac { 6 } { m - 4 }", "m - 4 < 0", "m < 4"], "exprs": ["m - 4 < 0", "m < 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 4 ) x < 6"}, {"id": "m - 4 < 0"}, {"id": "x > \\frac { 6 } { m - 4 }"}, {"id": "不等式 $( m - 4 ) x < 6$ 的解集是 $x > \\frac { 6 } { m - 4 }$"}, {"id": "不等式两边都乘或除同一个负数，不等号的方向改变"}, {"id": "m < 4"}], "links": [{"rel": "被描述", "source": "( m - 4 ) x < 6", "target": "m - 4 < 0"}, {"rel": "不等式方程求解", "source": "m - 4 < 0", "target": "m < 4"}, {"rel": "被描述", "source": "x > \\frac { 6 } { m - 4 }", "target": "m - 4 < 0"}, {"rel": "限制性描述", "source": "不等式 $( m - 4 ) x < 6$ 的解集是 $x > \\frac { 6 } { m - 4 }$", "target": "m - 4 < 0"}, {"rel": "属性描述", "source": "不等式两边都乘或除同一个负数，不等号的方向改变", "target": "m - 4 < 0"}]}}
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{"content": "To make the fraction $\\frac { 1 } { x - 5 }$ meaningful, the condition that $x$ must satisfy is ____?", "answer": "x \\neq 5", "steps": "$\\because$ The fraction $\\frac { 1 } { x - 5 }$ is meaningful, $\\therefore$ $x - 5 \\neq 0$, $\\therefore$ $x \\neq 5$.", "expr_cands": ["\\frac { 1 } { x - 5 }", "x", "x - 5 \\neq 0", "x \\neq 5"], "exprs": ["x - 5 \\neq 0", "x \\neq 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x - 5 }"}, {"id": "x - 5 \\neq 0"}, {"id": "要使分式 $\\frac { 1 } { x - 5 }$ 有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 5"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { x - 5 }", "target": "x - 5 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 5 \\neq 0", "target": "x \\neq 5"}, {"rel": "限制性描述", "source": "要使分式 $\\frac { 1 } { x - 5 }$ 有意义", "target": "x - 5 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 5 \\neq 0"}]}}
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{"content": "If $| m | = 9$, $a$ and $b$ are reciprocals, $c$ and $d$ are opposite in sign, then $\\frac { c + d } { m } + ab$ = ____?", "answer": "1", "steps": "According to the problem, we have $m = 3$ or $- 3$, $ab = 1$, and $c + d = 0$. Therefore, the original expression is equal to $0 + 1 = 1$.", "expr_cands": ["| m | = 9", "m", "a", "b", "c", "d", "\\frac { c + d } { m } + ab", "m = 3", "- 3", "ab = 1", "c + d = 0", "0 + 1", "1"], "exprs": ["ab = 1", "c + d = 0", "0 + 1", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| m | = 9"}, {"id": "ab = 1"}, {"id": "a"}, {"id": "b"}, {"id": "$| m | = 9$ , $a$ 与 $b$ 互为倒数"}, {"id": "c"}, {"id": "c + d = 0"}, {"id": "d"}, {"id": "$c$ 与 $d$ 互为相反数"}, {"id": "\\frac { c + d } { m } + ab"}, {"id": "0 + 1"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "| m | = 9", "target": "ab = 1"}, {"rel": "代入", "source": "ab = 1", "target": "0 + 1"}, {"rel": "被描述", "source": "a", "target": "ab = 1"}, {"rel": "被描述", "source": "b", "target": "ab = 1"}, {"rel": "限制性描述", "source": "$| m | = 9$ , $a$ 与 $b$ 互为倒数", "target": "ab = 1"}, {"rel": "被描述", "source": "c", "target": "c + d = 0"}, {"rel": "代入", "source": "c + d = 0", "target": "0 + 1"}, {"rel": "被描述", "source": "d", "target": "c + d = 0"}, {"rel": "限制性描述", "source": "$c$ 与 $d$ 互为相反数", "target": "c + d = 0"}, {"rel": "被代入", "source": "\\frac { c + d } { m } + ab", "target": "0 + 1"}, {"rel": "计算", "source": "0 + 1", "target": "1"}]}}
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{"content": "If the algebraic expression $\\frac { 2 } { a - 2 }$ is meaningful, then the range of real numbers for $a$ is ____?", "answer": "a \\neq 2", "steps": "From the given condition, it is known that $a - 2 \\neq 0$, therefore $a \\neq 2$.", "expr_cands": ["\\frac { 2 } { a - 2 }", "a", "a - 2 \\neq 0", "a \\neq 2"], "exprs": ["a - 2 \\neq 0", "a \\neq 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { a - 2 }"}, {"id": "a - 2 \\neq 0"}, {"id": "代数式 $\\frac { 2 } { a - 2 }$ 有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "a \\neq 2"}], "links": [{"rel": "被描述", "source": "\\frac { 2 } { a - 2 }", "target": "a - 2 \\neq 0"}, {"rel": "不等式方程求解", "source": "a - 2 \\neq 0", "target": "a \\neq 2"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 2 } { a - 2 }$ 有意义", "target": "a - 2 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "a - 2 \\neq 0"}]}}
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{"content": "Given $m ^ { 2 } - 2 m + n ^ { 2 } - \\frac { n } { 3 } + \\frac { 37 } { 36 } = 0$, find the value of $6 n - m ^ { 5 }$.", "answer": "0", "steps": "Since $m ^ { 2 } - 2 m + n ^ { 2 } - \\frac { n } { 3 } + \\frac { 37 } { 36 } = ( m ^ { 2 } - 2 m + 1 ) + ( n ^ { 2 } - \\frac { n } { 3 } + \\frac { 1 } { 36 } ) = ( m - 1 ) ^ { 2 } + ( n - \\frac { 1 } { 6 } ) ^ { 2 } = 0$, therefore $m = 1$, $n = \\frac { 1 } { 6 }$, and $6 n - m ^ { 5 } = 1 - 1 = 0$.", "expr_cands": ["m ^ { 2 } - 2 m + n ^ { 2 } - \\frac { n } { 3 } + \\frac { 37 } { 36 } = 0", "n", "m", "6 n - m ^ { 5 }", "( m - 1 ) ^ { 2 } + ( n - \\frac { 1 } { 6 } ) ^ { 2 } = 0", "m = 1", "n = \\frac { 1 } { 6 }", "0"], "exprs": ["( m - 1 ) ^ { 2 } + ( n - \\frac { 1 } { 6 } ) ^ { 2 } = 0", "m = 1", "n = \\frac { 1 } { 6 }", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m ^ { 2 } - 2 m + n ^ { 2 } - \\frac { n } { 3 } + \\frac { 37 } { 36 } = 0"}, {"id": "( m - 1 ) ^ { 2 } + ( n - \\frac { 1 } { 6 } ) ^ { 2 } = 0"}, {"id": "m = 1"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "n = \\frac { 1 } { 6 }"}, {"id": "6 n - m ^ { 5 }"}, {"id": "0"}], "links": [{"rel": "提取因式", "source": "m ^ { 2 } - 2 m + n ^ { 2 } - \\frac { n } { 3 } + \\frac { 37 } { 36 } = 0", "target": "( m - 1 ) ^ { 2 } + ( n - \\frac { 1 } { 6 } ) ^ { 2 } = 0"}, {"rel": "被描述", "source": "( m - 1 ) ^ { 2 } + ( n - \\frac { 1 } { 6 } ) ^ { 2 } = 0", "target": "m = 1"}, {"rel": "被描述", "source": "( m - 1 ) ^ { 2 } + ( n - \\frac { 1 } { 6 } ) ^ { 2 } = 0", "target": "n = \\frac { 1 } { 6 }"}, {"rel": "代入", "source": "m = 1", "target": "0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "m = 1"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "n = \\frac { 1 } { 6 }"}, {"rel": "代入", "source": "n = \\frac { 1 } { 6 }", "target": "0"}, {"rel": "被代入", "source": "6 n - m ^ { 5 }", "target": "0"}]}}
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{"content": "If $a + b = 2009$, $c + d = - 5$, then the algebraic expression $( a - 2 c ) - ( 2 d - b )$ = ____?", "answer": "2019", "steps": "$\\because a + b = 2009$, $c + d = - 5$, $\\therefore$ the original expression is equal to $a - 2 c - 2 d + b = ( a + b ) - 2 ( c + d ) = 2009 + 10 = 2019$.", "expr_cands": ["a + b = 2009", "b", "a", "c + d = - 5", "c", "d", "( a - 2 c ) - ( 2 d - b )", "( a + b ) - 2 ( c + d )", "2019"], "exprs": ["( a + b ) - 2 ( c + d )", "2019"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 2 c ) - ( 2 d - b )"}, {"id": "( a + b ) - 2 ( c + d )"}, {"id": "a + b = 2009"}, {"id": "c + d = - 5"}, {"id": "2019"}], "links": [{"rel": "提取因式", "source": "( a - 2 c ) - ( 2 d - b )", "target": "( a + b ) - 2 ( c + d )"}, {"rel": "被代入", "source": "( a + b ) - 2 ( c + d )", "target": "2019"}, {"rel": "提取因式参考", "source": "a + b = 2009", "target": "( a + b ) - 2 ( c + d )"}, {"rel": "代入", "source": "a + b = 2009", "target": "2019"}, {"rel": "提取因式参考", "source": "c + d = - 5", "target": "( a + b ) - 2 ( c + d )"}, {"rel": "代入", "source": "c + d = - 5", "target": "2019"}]}}
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{"content": "Given $\\frac { x - 2 } { 3 }$ is equal to $\\frac { 4 - x } { 4 }$, what is the value of $x$?", "answer": "\\frac { 20 } { 7 }", "steps": "According to the problem, we have $\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }$. Multiplying both sides by $12$, we get $4 ( x - 2 ) = 3 ( 4 - x )$. Expanding and simplifying, we get $4 x - 8 = 12 - 3 x$. Rearranging and combining like terms, we get $7 x = 20$. Dividing both sides by $7$, we get $x = \\frac { 20 } { 7 }$.", "expr_cands": ["\\frac { x - 2 } { 3 }", "x", "\\frac { 4 - x } { 4 }", "\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }", "x = \\frac { 20 } { 7 }", "12", "4 ( x - 2 ) = 3 ( 4 - x )", "4 x - 8 = 12 - 3 x", "7 x = 20", "1"], "exprs": ["\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }", "x = \\frac { 20 } { 7 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 2 } { 3 }"}, {"id": "\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }"}, {"id": "\\frac { 4 - x } { 4 }"}, {"id": "$\\frac { x - 2 } { 3 }$ 与 $\\frac { 4 - x } { 4 }$ 的值相等时"}, {"id": "x = \\frac { 20 } { 7 }"}], "links": [{"rel": "被描述", "source": "\\frac { x - 2 } { 3 }", "target": "\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }"}, {"rel": "等式方程求解", "source": "\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }", "target": "x = \\frac { 20 } { 7 }"}, {"rel": "被描述", "source": "\\frac { 4 - x } { 4 }", "target": "\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }"}, {"rel": "限制性描述", "source": "$\\frac { x - 2 } { 3 }$ 与 $\\frac { 4 - x } { 4 }$ 的值相等时", "target": "\\frac { x - 2 } { 3 } = \\frac { 4 - x } { 4 }"}]}}
|
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{"content": "Given that the value of the algebraic expression $a - 2 b$ is $8$, what is the value of the algebraic expression $4 b - 2 a$?", "answer": "- 16", "steps": "$\\because$ The value of the algebraic expression $a - 2 b$ is $8$, $\\therefore$ $a - 2 b = 8$, $\\therefore$ $4 b - 2 a = - 2 ( a - 2 b ) = - 2 * 8 = - 16$.", "expr_cands": ["a - 2 b", "a", "b", "8", "4 b - 2 a", "a - 2 b = 8", "- 2 ( a - 2 b )", "- 16"], "exprs": ["a - 2 b = 8", "- 2 ( a - 2 b )", "- 16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - 2 b"}, {"id": "a - 2 b = 8"}, {"id": "8"}, {"id": "代数式 $a - 2 b$ 的值是 $8$"}, {"id": "4 b - 2 a"}, {"id": "- 2 ( a - 2 b )"}, {"id": "- 16"}], "links": [{"rel": "被描述", "source": "a - 2 b", "target": "a - 2 b = 8"}, {"rel": "提取因式参考", "source": "a - 2 b = 8", "target": "- 2 ( a - 2 b )"}, {"rel": "代入", "source": "a - 2 b = 8", "target": "- 16"}, {"rel": "被描述", "source": "8", "target": "a - 2 b = 8"}, {"rel": "限制性描述", "source": "代数式 $a - 2 b$ 的值是 $8$", "target": "a - 2 b = 8"}, {"rel": "提取因式", "source": "4 b - 2 a", "target": "- 2 ( a - 2 b )"}, {"rel": "被代入", "source": "- 2 ( a - 2 b )", "target": "- 16"}]}}
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{"content": "The equation about $x$, $3 x + 2 ( 3 a + 1 ) = 6 x + a$, has a solution greater than $1$. The range of possible values for $a$ is [].", "answer": "a > \\frac { 1 } { 5 }", "steps": "From $3 x + 2 ( 3 a + 1 ) = 6 x + a$, we get $x = \\frac { 5 a + 2 } { 3 }$. According to the problem, we have $\\frac { 5 a + 2 } { 3 } > 1$, which gives us $a > \\frac { 1 } { 5 }$.", "expr_cands": ["x", "3 x + 2 ( 3 a + 1 ) = 6 x + a", "a", "1", "x = \\frac { 5 a + 2 } { 3 }", "\\frac { 5 a + 2 } { 3 } > 1", "\\frac { 1 } { 5 } < a", "a > \\frac { 1 } { 5 }"], "exprs": ["x = \\frac { 5 a + 2 } { 3 }", "\\frac { 5 a + 2 } { 3 } > 1", "a > \\frac { 1 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + 2 ( 3 a + 1 ) = 6 x + a"}, {"id": "x = \\frac { 5 a + 2 } { 3 }"}, {"id": "1"}, {"id": "\\frac { 5 a + 2 } { 3 } > 1"}, {"id": "关于 $x$ 的方程 $3 x + 2 ( 3 a + 1 ) = 6 x + a$ 的解大于 $1$"}, {"id": "a > \\frac { 1 } { 5 }"}], "links": [{"rel": "等式方程部分求解", "source": "3 x + 2 ( 3 a + 1 ) = 6 x + a", "target": "x = \\frac { 5 a + 2 } { 3 }"}, {"rel": "被描述", "source": "x = \\frac { 5 a + 2 } { 3 }", "target": "\\frac { 5 a + 2 } { 3 } > 1"}, {"rel": "被描述", "source": "1", "target": "\\frac { 5 a + 2 } { 3 } > 1"}, {"rel": "不等式方程求解", "source": "\\frac { 5 a + 2 } { 3 } > 1", "target": "a > \\frac { 1 } { 5 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $3 x + 2 ( 3 a + 1 ) = 6 x + a$ 的解大于 $1$", "target": "\\frac { 5 a + 2 } { 3 } > 1"}]}}
|
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{"content": "If the equation $( a + 1 ) x ^ 2 - 3 ax + 2 a + 17 = 0$ is a linear equation in one variable $x$, then its solution is ____?", "answer": "- 5", "steps": "$\\because$ The equation is a linear equation with one variable, $\\therefore$ we can obtain $a + 1 = 0$, which leads to $a = - 1$. $\\therefore$ The equation becomes $3 x - 2 + 17 = 0$, and the solution is $x = - 5$.", "expr_cands": ["x", "( a + 1 ) x ^ { 2 } - 3 ax + 2 a + 17 = 0", "a", "a + 1 = 0", "a = - 1", "3 x - 2 + 17 = 0", "x = - 5"], "exprs": ["a + 1 = 0", "a = - 1", "3 x - 2 + 17 = 0", "x = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a + 1 ) x ^ { 2 } - 3 ax + 2 a + 17 = 0"}, {"id": "a + 1 = 0"}, {"id": "关于 $x$ 的方程 $( a + 1 ) x ^ { 2 } - 3 ax + 2 a + 17 = 0$ 为一元一次方程"}, {"id": "a = - 1"}, {"id": "3 x - 2 + 17 = 0"}, {"id": "x = - 5"}], "links": [{"rel": "被描述", "source": "( a + 1 ) x ^ { 2 } - 3 ax + 2 a + 17 = 0", "target": "a + 1 = 0"}, {"rel": "被代入", "source": "( a + 1 ) x ^ { 2 } - 3 ax + 2 a + 17 = 0", "target": "3 x - 2 + 17 = 0"}, {"rel": "等式方程求解", "source": "a + 1 = 0", "target": "a = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $( a + 1 ) x ^ { 2 } - 3 ax + 2 a + 17 = 0$ 为一元一次方程", "target": "a + 1 = 0"}, {"rel": "代入", "source": "a = - 1", "target": "3 x - 2 + 17 = 0"}, {"rel": "等式方程求解", "source": "3 x - 2 + 17 = 0", "target": "x = - 5"}]}}
|
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{"content": "If $5 x + 7$ and $11 - 3 x$ are opposite in sign, then $x$ = ____?", "answer": "- 9", "steps": "According to the problem, we have $5 x + 7 + 11 - 3 x = 0$. By rearranging and combining like terms, we get $2 x = - 18$. Solving for $x$, we get $x = - 9$.", "expr_cands": ["5 x + 7", "x", "11 - 3 x", "5 x + 7 + 11 - 3 x = 0", "x = - 9", "2 x = - 18"], "exprs": ["5 x + 7 + 11 - 3 x = 0", "x = - 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x + 7"}, {"id": "5 x + 7 + 11 - 3 x = 0"}, {"id": "11 - 3 x"}, {"id": "$5 x + 7$ 与 $11 - 3 x$ 互为相反数"}, {"id": "x = - 9"}], "links": [{"rel": "被描述", "source": "5 x + 7", "target": "5 x + 7 + 11 - 3 x = 0"}, {"rel": "等式方程求解", "source": "5 x + 7 + 11 - 3 x = 0", "target": "x = - 9"}, {"rel": "被描述", "source": "11 - 3 x", "target": "5 x + 7 + 11 - 3 x = 0"}, {"rel": "限制性描述", "source": "$5 x + 7$ 与 $11 - 3 x$ 互为相反数", "target": "5 x + 7 + 11 - 3 x = 0"}]}}
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{"content": "Given that $m$ is a coefficient, there are two polynomials in $x$ and $y$: $mx ^ 2 - 2 x + y$ and $- 3 x ^ 2 + 2 x + 3 y$. The difference between them does not contain any quadratic terms. Find the value of $m ^ 2 + 3 m - 1$.", "answer": "- 1", "steps": "According to the problem, we have $( mx ^ 2 - 2 x + y ) - ( - 3 x ^ 2 + 2 x + 3 y ) = mx ^ 2 - 2 x + y + 3 x ^ 2 - 2 x - 3 y = ( m + 3 ) x ^ 2 - 4 x - 2 y$. Since the result does not contain a quadratic term, we have $m + 3 = 0$, which implies $m = - 3$. Therefore, $m ^ 2 + 3 m - 1 = ( - 3 ) ^ 2 + 3 ( - 3 ) - 1 = - 1$.", "expr_cands": ["m", "x", "y", "mx ^ { 2 } - 2 x + y", "- 3 x ^ { 2 } + 2 x + 3 y", "m ^ { 2 } + 3 m - 1", "( mx ^ { 2 } - 2 x + y ) - ( - 3 x ^ { 2 } + 2 x + 3 y )", "( m + 3 ) x ^ { 2 } - 4 x - 2 y", "m + 3 = 0", "m = - 3", "- 1"], "exprs": ["( mx ^ { 2 } - 2 x + y ) - ( - 3 x ^ { 2 } + 2 x + 3 y )", "( m + 3 ) x ^ { 2 } - 4 x - 2 y", "m + 3 = 0", "m = - 3", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "mx ^ { 2 } - 2 x + y"}, {"id": "( mx ^ { 2 } - 2 x + y ) - ( - 3 x ^ { 2 } + 2 x + 3 y )"}, {"id": "- 3 x ^ { 2 } + 2 x + 3 y"}, {"id": "关于 $x$ , $y$ 的两个多项式 $mx ^ { 2 } - 2 x + y$ 与 $- 3 x ^ { 2 } + 2 x + 3 y$ 的差中不含二次项"}, {"id": "( m + 3 ) x ^ { 2 } - 4 x - 2 y"}, {"id": "m + 3 = 0"}, {"id": "结果不含二次项"}, {"id": "m = - 3"}, {"id": "m ^ { 2 } + 3 m - 1"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "mx ^ { 2 } - 2 x + y", "target": "( mx ^ { 2 } - 2 x + y ) - ( - 3 x ^ { 2 } + 2 x + 3 y )"}, {"rel": "提取因式", "source": "( mx ^ { 2 } - 2 x + y ) - ( - 3 x ^ { 2 } + 2 x + 3 y )", "target": "( m + 3 ) x ^ { 2 } - 4 x - 2 y"}, {"rel": "被描述", "source": "- 3 x ^ { 2 } + 2 x + 3 y", "target": "( mx ^ { 2 } - 2 x + y ) - ( - 3 x ^ { 2 } + 2 x + 3 y )"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的两个多项式 $mx ^ { 2 } - 2 x + y$ 与 $- 3 x ^ { 2 } + 2 x + 3 y$ 的差中不含二次项", "target": "( mx ^ { 2 } - 2 x + y ) - ( - 3 x ^ { 2 } + 2 x + 3 y )"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的两个多项式 $mx ^ { 2 } - 2 x + y$ 与 $- 3 x ^ { 2 } + 2 x + 3 y$ 的差中不含二次项", "target": "m + 3 = 0"}, {"rel": "被描述", "source": "( m + 3 ) x ^ { 2 } - 4 x - 2 y", "target": "m + 3 = 0"}, {"rel": "等式方程求解", "source": "m + 3 = 0", "target": "m = - 3"}, {"rel": "限制性描述", "source": "结果不含二次项", "target": "m + 3 = 0"}, {"rel": "代入", "source": "m = - 3", "target": "- 1"}, {"rel": "被代入", "source": "m ^ { 2 } + 3 m - 1", "target": "- 1"}]}}
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{"content": "If $y = \\sqrt { x - 3 } + \\sqrt { 3 - x } + 1$, then the value of $2 x + y$ is ____?", "answer": "7", "steps": "Since $y = \\sqrt { x - 3 } + \\sqrt { 3 - x } + 1$, therefore $x - 3 = 0$, solving for $x$ gives $x = 3$, then $y = 1$, so $2 x + y = 7$.", "expr_cands": ["y = \\sqrt { x - 3 } + \\sqrt { 3 - x } + 1", "y", "x", "2 x + y", "x - 3 = 0", "x = 3", "y = 1", "7"], "exprs": ["x - 3 = 0", "x = 3", "y = 1", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 3 } + \\sqrt { 3 - x } + 1"}, {"id": "x - 3 = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x = 3"}, {"id": "y = 1"}, {"id": "2 x + y"}, {"id": "7"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 3 } + \\sqrt { 3 - x } + 1", "target": "x - 3 = 0"}, {"rel": "被代入", "source": "y = \\sqrt { x - 3 } + \\sqrt { 3 - x } + 1", "target": "y = 1"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 3 = 0"}, {"rel": "代入", "source": "x = 3", "target": "y = 1"}, {"rel": "代入", "source": "x = 3", "target": "7"}, {"rel": "代入", "source": "y = 1", "target": "7"}, {"rel": "被代入", "source": "2 x + y", "target": "7"}]}}
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{"content": "Given the equation $2 m + 5$ is the opposite of $2 ( m - \\frac { 1 } { 2 })$, what is the value of $m$?", "answer": "- 1", "steps": "According to the problem, we have $2 m + 5 + 2 ( m - \\frac { 1 } { 2 }) = 0$. Expanding the brackets, we get $2 m + 5 + 2 m - 1 = 0$. Rearranging and combining like terms, we get $4 m = - 4$. Solving for $m$, we get $m = - 1$.", "expr_cands": ["2 m + 5", "m", "2 ( m - \\frac { 1 } { 2 } )", "2 m + 5 + 2 ( m - \\frac { 1 } { 2 } ) = 0", "m = - 1", "2 m + 5 + 2 m - 1 = 0", "4 m = - 4"], "exprs": ["2 m + 5 + 2 ( m - \\frac { 1 } { 2 } ) = 0", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 m + 5"}, {"id": "2 m + 5 + 2 ( m - \\frac { 1 } { 2 } ) = 0"}, {"id": "2 ( m - \\frac { 1 } { 2 } )"}, {"id": "式子 $2 m + 5$ 与 $2 ( m - \\frac { 1 } { 2 } )$ 的值互为相反数"}, {"id": "m = - 1"}], "links": [{"rel": "被描述", "source": "2 m + 5", "target": "2 m + 5 + 2 ( m - \\frac { 1 } { 2 } ) = 0"}, {"rel": "等式方程求解", "source": "2 m + 5 + 2 ( m - \\frac { 1 } { 2 } ) = 0", "target": "m = - 1"}, {"rel": "被描述", "source": "2 ( m - \\frac { 1 } { 2 } )", "target": "2 m + 5 + 2 ( m - \\frac { 1 } { 2 } ) = 0"}, {"rel": "限制性描述", "source": "式子 $2 m + 5$ 与 $2 ( m - \\frac { 1 } { 2 } )$ 的值互为相反数", "target": "2 m + 5 + 2 ( m - \\frac { 1 } { 2 } ) = 0"}]}}
|
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{"content": "If $2 x ^ { 3 - 2 k } + 2 = 4$ is a one-variable linear equation, then $k$ = ____ ?", "answer": "1", "steps": "From the given information, we have $3 - 2 k = 1$. Solving for $k$, we get $k = 1$.", "expr_cands": ["2 x ^ { 3 - 2 k } + 2 = 4", "k", "x", "3 - 2 k = 1", "k = 1"], "exprs": ["3 - 2 k = 1", "k = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 3 - 2 k } + 2 = 4"}, {"id": "3 - 2 k = 1"}, {"id": "$2 x ^ { 3 - 2 k } + 2 = 4$ 是关于 $x$ 的一元一次方程"}, {"id": "k = 1"}], "links": [{"rel": "被描述", "source": "2 x ^ { 3 - 2 k } + 2 = 4", "target": "3 - 2 k = 1"}, {"rel": "等式方程求解", "source": "3 - 2 k = 1", "target": "k = 1"}, {"rel": "限制性描述", "source": "$2 x ^ { 3 - 2 k } + 2 = 4$ 是关于 $x$ 的一元一次方程", "target": "3 - 2 k = 1"}]}}
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{"content": "Given that $- 7$ is a solution to the equation $2 x - 7 = ax$, the value of the algebraic expression $a - \\frac { 3 } { a }$ is ____?", "answer": "2", "steps": "Substituting $x = - 7$ into the equation, we get $- 14 - 7 = - 7 a$. Solving for $a$, we get $a = 3$. Therefore, $a - \\frac { 3 } { a } = 3 - \\frac { 3 } { 3 } = 2$.", "expr_cands": ["- 7", "2 x - 7 = ax", "a", "x", "a - \\frac { 3 } { a }", "x = - 7", "- 14 - 7 = - 7 a", "a = 3", "2"], "exprs": ["x = - 7", "- 14 - 7 = - 7 a", "a = 3", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 7"}, {"id": "x = - 7"}, {"id": "2 x - 7 = ax"}, {"id": "$- 7$ 是方程 $2 x - 7 = ax$ 的解"}, {"id": "- 14 - 7 = - 7 a"}, {"id": "a = 3"}, {"id": "a - \\frac { 3 } { a }"}, {"id": "2"}], "links": [{"rel": "被描述", "source": "- 7", "target": "x = - 7"}, {"rel": "代入", "source": "x = - 7", "target": "- 14 - 7 = - 7 a"}, {"rel": "被描述", "source": "2 x - 7 = ax", "target": "x = - 7"}, {"rel": "被代入", "source": "2 x - 7 = ax", "target": "- 14 - 7 = - 7 a"}, {"rel": "限制性描述", "source": "$- 7$ 是方程 $2 x - 7 = ax$ 的解", "target": "x = - 7"}, {"rel": "等式方程求解", "source": "- 14 - 7 = - 7 a", "target": "a = 3"}, {"rel": "代入", "source": "a = 3", "target": "2"}, {"rel": "被代入", "source": "a - \\frac { 3 } { a }", "target": "2"}]}}
|
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{"content": "The maximum negative integer solution of the inequality $3 x - 2 \\le 5 x + 6$ is ____?", "answer": "x = - 1", "steps": "Since $3 x - 2 \\le 5 x + 6$, we have $3 x - 5 x \\le 6 + 2$, which simplifies to $- 2 x \\le 8$. Therefore, $x \\ge - 4$. Thus, the largest negative integer solution to the inequality is $x = - 1$.", "expr_cands": ["3 x - 2 \\le 5 x + 6", "x", "- 4 \\le x", "3 x - 5 x \\le 6 + 2", "- 2 x \\le 8", "x \\ge - 4", "x = - 1"], "exprs": ["x \\ge - 4", "x = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 2 \\le 5 x + 6"}, {"id": "x \\ge - 4"}, {"id": "x = - 1"}, {"id": "不等式 $3 x - 2 \\le 5 x + 6$ 的最大负整数解"}], "links": [{"rel": "不等式方程求解", "source": "3 x - 2 \\le 5 x + 6", "target": "x \\ge - 4"}, {"rel": "被描述", "source": "x \\ge - 4", "target": "x = - 1"}, {"rel": "限制性描述", "source": "不等式 $3 x - 2 \\le 5 x + 6$ 的最大负整数解", "target": "x = - 1"}]}}
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{"content": "If two simplest quadratic surds $\\sqrt { 3 a - 8 }$ and $\\sqrt { 17 - 2 a }$ can be combined, then the value of $a$ is ____?", "answer": "5", "steps": "From the given information, we have $3 a - 8 = 17 - 2 a$. Solving for $a$, we get $a = 5$.", "expr_cands": ["\\sqrt { 3 a - 8 }", "a", "\\sqrt { 17 - 2 a }", "3 a - 8 = 17 - 2 a", "a = 5"], "exprs": ["3 a - 8 = 17 - 2 a", "a = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 3 a - 8 }"}, {"id": "3 a - 8 = 17 - 2 a"}, {"id": "\\sqrt { 17 - 2 a }"}, {"id": "两个最简二次根式 $\\sqrt { 3 a - 8 }$ 与 $\\sqrt { 17 - 2 a }$ 能够合并"}, {"id": "a = 5"}], "links": [{"rel": "被描述", "source": "\\sqrt { 3 a - 8 }", "target": "3 a - 8 = 17 - 2 a"}, {"rel": "等式方程求解", "source": "3 a - 8 = 17 - 2 a", "target": "a = 5"}, {"rel": "被描述", "source": "\\sqrt { 17 - 2 a }", "target": "3 a - 8 = 17 - 2 a"}, {"rel": "限制性描述", "source": "两个最简二次根式 $\\sqrt { 3 a - 8 }$ 与 $\\sqrt { 17 - 2 a }$ 能够合并", "target": "3 a - 8 = 17 - 2 a"}]}}
|
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{"content": "If $a$ is the smallest positive integer, $b$ is the largest negative integer, and $c$ is the number with the smallest absolute value, then the value of $a ^ { 2015 } - b ^ { 2016 } + c ^ { 2017 }$ is ____?", "answer": "0", "steps": "According to the problem, we have $a = 1$, $b = - 1$, and $c = 0$. Therefore, the original expression is equal to $1 - 1 + 0 = 0$.", "expr_cands": ["a", "b", "c", "a ^ { 2015 } - b ^ { 2016 } + c ^ { 2017 }", "a = 1", "b = - 1", "c = 0", "1 - 1 + 0", "0"], "exprs": ["a = 1", "b = - 1", "c = 0", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a = 1"}, {"id": "$a$ 是最小的正整数"}, {"id": "b"}, {"id": "b = - 1"}, {"id": "$b$ 是最大的负整数"}, {"id": "c"}, {"id": "c = 0"}, {"id": "$c$ 是绝对值最小的数"}, {"id": "a ^ { 2015 } - b ^ { 2016 } + c ^ { 2017 }"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "a", "target": "a = 1"}, {"rel": "代入", "source": "a = 1", "target": "0"}, {"rel": "限制性描述", "source": "$a$ 是最小的正整数", "target": "a = 1"}, {"rel": "被描述", "source": "b", "target": "b = - 1"}, {"rel": "代入", "source": "b = - 1", "target": "0"}, {"rel": "限制性描述", "source": "$b$ 是最大的负整数", "target": "b = - 1"}, {"rel": "被描述", "source": "c", "target": "c = 0"}, {"rel": "代入", "source": "c = 0", "target": "0"}, {"rel": "限制性描述", "source": "$c$ 是绝对值最小的数", "target": "c = 0"}, {"rel": "被代入", "source": "a ^ { 2015 } - b ^ { 2016 } + c ^ { 2017 }", "target": "0"}]}}
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{"content": "Given the function $y = 2 x ^ { 2 a + b } + a + 2 b$ is a direct proportion function, then $a + b$ = ____ ?", "answer": "\\frac { 1 } { 3 }", "steps": "According to the problem, we have $2 a + b = 1$ and $a + 2 b = 0$. Solving for $a$ and $b$, we get $a = \\frac { 2 } { 3 }$ and $b = - \\frac { 1 } { 3 }$. Therefore, $a + b = \\frac { 2 } { 3 } - \\frac { 1 } { 3 } = \\frac { 1 } { 3 }$.", "expr_cands": ["y = 2 x ^ { 2 a + b } + a + 2 b", "a", "y", "x", "b", "a + b", "2 a + b = 1", "a + 2 b = 0", "a = \\frac { 2 } { 3 }", "b = - \\frac { 1 } { 3 }", "\\frac { 1 } { 3 }"], "exprs": ["2 a + b = 1", "a + 2 b = 0", "a = \\frac { 2 } { 3 }", "b = - \\frac { 1 } { 3 }", "\\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 x ^ { 2 a + b } + a + 2 b"}, {"id": "2 a + b = 1"}, {"id": "函数 $y = 2 x ^ { 2 a + b } + a + 2 b$ 是正比例函数"}, {"id": "a + 2 b = 0"}, {"id": "a = \\frac { 2 } { 3 }"}, {"id": "b = - \\frac { 1 } { 3 }"}, {"id": "a + b"}, {"id": "\\frac { 1 } { 3 }"}], "links": [{"rel": "被描述", "source": "y = 2 x ^ { 2 a + b } + a + 2 b", "target": "2 a + b = 1"}, {"rel": "被描述", "source": "y = 2 x ^ { 2 a + b } + a + 2 b", "target": "a + 2 b = 0"}, {"rel": "联立", "source": "2 a + b = 1", "target": "a = \\frac { 2 } { 3 }"}, {"rel": "联立", "source": "2 a + b = 1", "target": "b = - \\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "函数 $y = 2 x ^ { 2 a + b } + a + 2 b$ 是正比例函数", "target": "2 a + b = 1"}, {"rel": "限制性描述", "source": "函数 $y = 2 x ^ { 2 a + b } + a + 2 b$ 是正比例函数", "target": "a + 2 b = 0"}, {"rel": "联立", "source": "a + 2 b = 0", "target": "a = \\frac { 2 } { 3 }"}, {"rel": "联立", "source": "a + 2 b = 0", "target": "b = - \\frac { 1 } { 3 }"}, {"rel": "代入", "source": "a = \\frac { 2 } { 3 }", "target": "\\frac { 1 } { 3 }"}, {"rel": "代入", "source": "b = - \\frac { 1 } { 3 }", "target": "\\frac { 1 } { 3 }"}, {"rel": "被代入", "source": "a + b", "target": "\\frac { 1 } { 3 }"}]}}
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{"content": "If $| x + 1 | + | y - 2 | = 0$, then $x + y$ = ____ ?", "answer": "1", "steps": "From the given information, we have $x + 1 = 0$ and $y - 2 = 0$. Solving for $x$ and $y$, we get $x = - 1$ and $y = 2$. Therefore, $x + y = - 1 + 2 = 1$.", "expr_cands": ["| x + 1 | + | y - 2 | = 0", "y", "x", "x + y", "x + 1 = 0", "x = - 1", "y - 2 = 0", "y = 2", "1"], "exprs": ["x + 1 = 0", "y - 2 = 0", "x = - 1", "y = 2", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x + 1 | + | y - 2 | = 0"}, {"id": "x + 1 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "y - 2 = 0"}, {"id": "x = - 1"}, {"id": "y = 2"}, {"id": "x + y"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "| x + 1 | + | y - 2 | = 0", "target": "x + 1 = 0"}, {"rel": "被描述", "source": "| x + 1 | + | y - 2 | = 0", "target": "y - 2 = 0"}, {"rel": "等式方程求解", "source": "x + 1 = 0", "target": "x = - 1"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x + 1 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "y - 2 = 0"}, {"rel": "等式方程求解", "source": "y - 2 = 0", "target": "y = 2"}, {"rel": "代入", "source": "x = - 1", "target": "1"}, {"rel": "代入", "source": "y = 2", "target": "1"}, {"rel": "被代入", "source": "x + y", "target": "1"}]}}
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{"content": "If $27 ^ { x } = 3 ^ { 12 }$, then $x$ = ____ ?", "answer": "4", "steps": "$27 ^ x$ is equal to $( 3 ^ 3 ) ^ x$ which simplifies to $3 ^ { 3 x }$ and is also equal to $3 ^ { 12 }$. Therefore, $3 x = 12$ and $x = 4$.", "expr_cands": ["27 ^ { x } = 3 ^ { 12 }", "x", "3 ^ { 3 x } = 3 ^ { 12 }", "x = 4", "3 x = 12"], "exprs": ["x = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "27 ^ { x } = 3 ^ { 12 }"}, {"id": "x = 4"}], "links": [{"rel": "等式方程求解", "source": "27 ^ { x } = 3 ^ { 12 }", "target": "x = 4"}]}}
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{"content": "Given: $x ^ 2 - 5 x = 6$, the value of the algebraic expression $10 x - 2 x ^ 2 + 5$ is ____?", "answer": "- 7", "steps": "Because $x ^ 2 - 5 x = 6$, therefore $10 x - 2 x ^ 2 + 5 = - 2 ( x ^ 2 - 5 x ) + 5 = - 2 * 6 + 5 = - 12 + 5 = - 7$.", "expr_cands": ["x ^ { 2 } - 5 x = 6", "x", "10 x - 2 x ^ { 2 } + 5", "x = - 1", "x = 6", "- 2 ( x ^ { 2 } - 5 x ) + 5", "- 7"], "exprs": ["- 2 ( x ^ { 2 } - 5 x ) + 5", "- 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "10 x - 2 x ^ { 2 } + 5"}, {"id": "- 2 ( x ^ { 2 } - 5 x ) + 5"}, {"id": "x ^ { 2 } - 5 x = 6"}, {"id": "- 7"}], "links": [{"rel": "提取因式", "source": "10 x - 2 x ^ { 2 } + 5", "target": "- 2 ( x ^ { 2 } - 5 x ) + 5"}, {"rel": "被代入", "source": "- 2 ( x ^ { 2 } - 5 x ) + 5", "target": "- 7"}, {"rel": "提取因式参考", "source": "x ^ { 2 } - 5 x = 6", "target": "- 2 ( x ^ { 2 } - 5 x ) + 5"}, {"rel": "代入", "source": "x ^ { 2 } - 5 x = 6", "target": "- 7"}]}}
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{"content": "If the equation $x ^ 2 + 6 x - 7 = 0$ is transformed into $( x + m ) ^ 2 = n$, then $n$ = ____?", "answer": "16", "steps": "$x ^ { 2 } + 6 x - 7 = 0$ , $x ^ { 2 } + 6 x = 7$ , $x ^ { 2 } + 6 x + 9 = 7 + 9$ , $( x + 3 ) ^ { 2 } = 16$ , thus $n = 16$.", "expr_cands": ["x ^ { 2 } + 6 x - 7 = 0", "x", "( x + m ) ^ { 2 } = n", "m", "n", "x = - 7", "x = 1", "x ^ { 2 } + 6 x = 7", "x ^ { 2 } + 6 x + 9", "16", "( x + 3 ) ^ { 2 } = 16", "n = 16"], "exprs": ["( x + 3 ) ^ { 2 } = 16", "n = 16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + 6 x - 7 = 0"}, {"id": "( x + 3 ) ^ { 2 } = 16"}, {"id": "将方程 $x ^ { 2 } + 6 x - 7 = 0$ 转化为 $( x + m ) ^ { 2 } = n$"}, {"id": "( x + m ) ^ { 2 } = n"}, {"id": "n = 16"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + 6 x - 7 = 0", "target": "( x + 3 ) ^ { 2 } = 16"}, {"rel": "联立", "source": "( x + 3 ) ^ { 2 } = 16", "target": "n = 16"}, {"rel": "限制性描述", "source": "将方程 $x ^ { 2 } + 6 x - 7 = 0$ 转化为 $( x + m ) ^ { 2 } = n$", "target": "( x + 3 ) ^ { 2 } = 16"}, {"rel": "联立", "source": "( x + m ) ^ { 2 } = n", "target": "n = 16"}]}}
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{"content": "The solution set of the inequality $3 x - 2 a \\leq - 2$ in terms of $x$ is $x \\leq - 1$. What is the value of $a$?", "answer": "- \\frac { 1 } { 2 }", "steps": "$3 x - 2 a \\leq - 2$, moving terms yields $3 x \\leq 2 a - 2$, solving for $x$ gives $x \\leq \\frac { 2 a - 2 } { 3 }$. From the given information, we have $\\frac { 2 a - 2 } { 3 } = - 1$, solving for $a$ gives $a = - \\frac { 1 } { 2 }$.", "expr_cands": ["x", "3 x - 2 a \\le - 2", "a", "x \\le - 1", "3 x \\le 2 a - 2", "x \\le \\frac { 2 a - 2 } { 3 }", "\\frac { 2 a - 2 } { 3 } = - 1", "a = - \\frac { 1 } { 2 }"], "exprs": ["x \\le \\frac { 2 a - 2 } { 3 }", "\\frac { 2 a - 2 } { 3 } = - 1", "a = - \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 2 a \\le - 2"}, {"id": "x \\le \\frac { 2 a - 2 } { 3 }"}, {"id": "x \\le - 1"}, {"id": "\\frac { 2 a - 2 } { 3 } = - 1"}, {"id": "关于 $x$ 的不等式 $3 x - 2 a \\le - 2$ 的解集是 $x \\le - 1$"}, {"id": "a = - \\frac { 1 } { 2 }"}], "links": [{"rel": "不等式方程部分求解", "source": "3 x - 2 a \\le - 2", "target": "x \\le \\frac { 2 a - 2 } { 3 }"}, {"rel": "被描述", "source": "x \\le \\frac { 2 a - 2 } { 3 }", "target": "\\frac { 2 a - 2 } { 3 } = - 1"}, {"rel": "被描述", "source": "x \\le - 1", "target": "\\frac { 2 a - 2 } { 3 } = - 1"}, {"rel": "等式方程求解", "source": "\\frac { 2 a - 2 } { 3 } = - 1", "target": "a = - \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的不等式 $3 x - 2 a \\le - 2$ 的解集是 $x \\le - 1$", "target": "\\frac { 2 a - 2 } { 3 } = - 1"}]}}
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{"content": "If $a - 2 b = 6$, then $4 b - 2 a$ = ____ ?", "answer": "- 12", "steps": "Because $a - 2 b = 6$, therefore $4 b - 2 a = - 2 ( a - 2 b ) = - 2 * 6 = - 12$.", "expr_cands": ["a - 2 b = 6", "a", "b", "4 b - 2 a", "- 2 ( a - 2 b )", "- 12"], "exprs": ["- 2 ( a - 2 b )", "- 12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 b - 2 a"}, {"id": "- 2 ( a - 2 b )"}, {"id": "a - 2 b = 6"}, {"id": "- 12"}], "links": [{"rel": "提取因式", "source": "4 b - 2 a", "target": "- 2 ( a - 2 b )"}, {"rel": "被代入", "source": "- 2 ( a - 2 b )", "target": "- 12"}, {"rel": "提取因式参考", "source": "a - 2 b = 6", "target": "- 2 ( a - 2 b )"}, {"rel": "代入", "source": "a - 2 b = 6", "target": "- 12"}]}}
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{"content": "If $3 + m$ has a square root, then the range of possible values for $m$ is ____?", "answer": "m \\ge - 3", "steps": "Since $3 + m$ has a square root, it follows that $3 + m \\ge 0$, which implies that $m \\ge - 3$.", "expr_cands": ["3 + m", "m", "3 + m \\ge 0", "- 3 \\le m", "m \\ge - 3"], "exprs": ["3 + m \\ge 0", "m \\ge - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 + m"}, {"id": "3 + m \\ge 0"}, {"id": "$3 + m$ 有平方根"}, {"id": "m \\ge - 3"}], "links": [{"rel": "被描述", "source": "3 + m", "target": "3 + m \\ge 0"}, {"rel": "不等式方程求解", "source": "3 + m \\ge 0", "target": "m \\ge - 3"}, {"rel": "限制性描述", "source": "$3 + m$ 有平方根", "target": "3 + m \\ge 0"}]}}
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{"content": "Given that the solution of the one-variable linear equation $4 mx - m = 2$ with respect to $x$ is $x = 1$, then $m$ = ____ ?", "answer": "\\frac { 2 } { 3 }", "steps": "Substituting $x = 1$ into the equation $4 mx - m = 2$ yields: $4 m - m = 2$, which can be solved to obtain $m = \\frac { 2 } { 3 }$.", "expr_cands": ["x", "4 mx - m = 2", "m", "x = 1", "3 m = 2", "4 m - m = 2", "m = \\frac { 2 } { 3 }"], "exprs": ["4 m - m = 2", "m = \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 mx - m = 2"}, {"id": "4 m - m = 2"}, {"id": "x = 1"}, {"id": "m = \\frac { 2 } { 3 }"}], "links": [{"rel": "被代入", "source": "4 mx - m = 2", "target": "4 m - m = 2"}, {"rel": "等式方程求解", "source": "4 m - m = 2", "target": "m = \\frac { 2 } { 3 }"}, {"rel": "代入", "source": "x = 1", "target": "4 m - m = 2"}]}}
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{"content": "If the solution set of the inequality $( a - 2 ) x > a - 2$ is $x > 1$, then the possible values of the variable $a$ are ____?", "answer": "a > 2", "steps": "$\\because$ The solution set of the inequality $( a - 2 ) x > a - 2$ is $x > 1$, $\\therefore$ $a - 2 > 0$, which implies $a > 2$.", "expr_cands": ["( a - 2 ) x > a - 2", "x", "a", "x > 1", "a - 2 > 0", "2 < a", "a > 2"], "exprs": ["a - 2 > 0", "a > 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 2 ) x > a - 2"}, {"id": "a - 2 > 0"}, {"id": "不等式 $( a - 2 ) x > a - 2$ 的解集为 $x > 1$"}, {"id": "a > 2"}], "links": [{"rel": "被描述", "source": "( a - 2 ) x > a - 2", "target": "a - 2 > 0"}, {"rel": "不等式方程求解", "source": "a - 2 > 0", "target": "a > 2"}, {"rel": "限制性描述", "source": "不等式 $( a - 2 ) x > a - 2$ 的解集为 $x > 1$", "target": "a - 2 > 0"}]}}
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{"content": "The result of factoring ${ x } ^ { n } - { y } ^ { n }$ is $( { x } ^ { 2 } + { y } ^ { 2 } ) ( x + y ) ( x - y )$. What is the value of $n$?", "answer": "4", "steps": "$({ x } ^ { 2 } + { y } ^ { 2 }) ( x + y ) ( x - y ) = ({ x } ^ { 2 } + { y } ^ { 2 }) ({ x } ^ { 2 } - { y } ^ { 2 }) = { x } ^ { 4 } - { y } ^ { 4 }$, so the value of $n$ is $4$.", "expr_cands": ["{ x } ^ { n } - { y } ^ { n }", "n", "y", "x", "( { x } ^ { 2 } + { y } ^ { 2 } ) ( x + y ) ( x - y )", "{ x } ^ { 4 } - { y } ^ { 4 }", "4"], "exprs": ["{ x } ^ { 4 } - { y } ^ { 4 }", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( { x } ^ { 2 } + { y } ^ { 2 } ) ( x + y ) ( x - y )"}, {"id": "{ x } ^ { 4 } - { y } ^ { 4 }"}, {"id": "4"}, {"id": "{ x } ^ { n } - { y } ^ { n }"}, {"id": "将 ${ x } ^ { n } - { y } ^ { n }$ 分解因式的结果为 $( { x } ^ { 2 } + { y } ^ { 2 } ) ( x + y ) ( x - y )$"}, {"id": "$n$ 的值"}], "links": [{"rel": "计算", "source": "( { x } ^ { 2 } + { y } ^ { 2 } ) ( x + y ) ( x - y )", "target": "{ x } ^ { 4 } - { y } ^ { 4 }"}, {"rel": "被描述", "source": "{ x } ^ { 4 } - { y } ^ { 4 }", "target": "4"}, {"rel": "被描述", "source": "{ x } ^ { n } - { y } ^ { n }", "target": "4"}, {"rel": "限制性描述", "source": "将 ${ x } ^ { n } - { y } ^ { n }$ 分解因式的结果为 $( { x } ^ { 2 } + { y } ^ { 2 } ) ( x + y ) ( x - y )$", "target": "4"}, {"rel": "限制性描述", "source": "$n$ 的值", "target": "4"}]}}
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{"content": "When $x$ = ____ ?, the value of $5 ( x - 2 ) - 7$ is equal to $8$.", "answer": "5", "steps": "According to the problem, we have $5 ( x - 2 ) - 7 = 8$. Expanding the brackets, we get $5 x - 10 - 7 = 8$. Simplifying and rearranging, we have $5 x = 25$. Solving for $x$, we get $x = 5$.", "expr_cands": ["x", "5 ( x - 2 ) - 7", "8", "5 ( x - 2 ) - 7 = 8", "x = 5", "5 x - 10 - 7 = 8", "5 x = 25"], "exprs": ["5 ( x - 2 ) - 7 = 8", "x = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 ( x - 2 ) - 7"}, {"id": "5 ( x - 2 ) - 7 = 8"}, {"id": "8"}, {"id": "$5 ( x - 2 ) - 7$ 的值等于 $8$"}, {"id": "x = 5"}], "links": [{"rel": "被描述", "source": "5 ( x - 2 ) - 7", "target": "5 ( x - 2 ) - 7 = 8"}, {"rel": "等式方程求解", "source": "5 ( x - 2 ) - 7 = 8", "target": "x = 5"}, {"rel": "被描述", "source": "8", "target": "5 ( x - 2 ) - 7 = 8"}, {"rel": "限制性描述", "source": "$5 ( x - 2 ) - 7$ 的值等于 $8$", "target": "5 ( x - 2 ) - 7 = 8"}]}}
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{"content": "The given fractional equation $\\frac { ax + 6 } { 2 a - x } = 1$ has a solution of $x = 1$. Find the value of $a$.", "answer": "7", "steps": "Substituting $x = 1$ into $\\frac { ax + 6 } { 2 a - x } = 1$, we get $\\frac { a + 6 } { 2 a - 1 } = 1$. Multiplying both sides by $( 2 a - 1 )$, we have $a + 6 = 2 a - 1$. Solving for $a$, we get $a = 7$. Checking, we see that $x = 1$ is indeed a solution to the equation when $a = 7$.", "expr_cands": ["\\frac { ax + 6 } { 2 a - x } = 1", "a", "x", "x = 1", "\\frac { a + 6 } { 2 a - 1 } = 1", "a = 7", "( 2 a - 1 )", "a + 6 = 2 a - 1"], "exprs": ["\\frac { a + 6 } { 2 a - 1 } = 1", "a = 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { ax + 6 } { 2 a - x } = 1"}, {"id": "\\frac { a + 6 } { 2 a - 1 } = 1"}, {"id": "x = 1"}, {"id": "a = 7"}], "links": [{"rel": "被代入", "source": "\\frac { ax + 6 } { 2 a - x } = 1", "target": "\\frac { a + 6 } { 2 a - 1 } = 1"}, {"rel": "等式方程求解", "source": "\\frac { a + 6 } { 2 a - 1 } = 1", "target": "a = 7"}, {"rel": "代入", "source": "x = 1", "target": "\\frac { a + 6 } { 2 a - 1 } = 1"}]}}
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{"content": "If $a ^ { 5 } \\times ( a ^ { n } ) ^ { 3 } = a ^ { 11 }$, then $n$ = ____ ?", "answer": "2", "steps": "$a ^ { 5 } \\times ( a ^ { n } ) ^ { 3 } = a ^ { 5 + 3 n } = a ^ { 11 }$, so we have 5 + 3n = 11, which gives n = 2.", "expr_cands": ["a ^ { 5 } \\times ( a ^ { n } ) ^ { 3 } = a ^ { 11 }", "a", "n", "5 + 3 n = 11", "n = 2"], "exprs": ["5 + 3 n = 11", "n = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 5 } \\times ( a ^ { n } ) ^ { 3 } = a ^ { 11 }"}, {"id": "5 + 3 n = 11"}, {"id": "n = 2"}], "links": [{"rel": "同取对数", "source": "a ^ { 5 } \\times ( a ^ { n } ) ^ { 3 } = a ^ { 11 }", "target": "5 + 3 n = 11"}, {"rel": "等式方程求解", "source": "5 + 3 n = 11", "target": "n = 2"}]}}
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{"content": "If $\\sqrt { 3 m - 6 }$ is defined, then the smallest integer value that $m$ can take is ____?", "answer": "2", "steps": "From the given condition, we have $3 m - 6 \\geq 0$. Solving for $m$, we get $m \\geq 2$. Therefore, the smallest integer value that $m$ can take is $2$.", "expr_cands": ["\\sqrt { 3 m - 6 }", "m", "3 m - 6 \\ge 0", "2 \\le m", "m \\ge 2", "2"], "exprs": ["3 m - 6 \\ge 0", "m \\ge 2", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 3 m - 6 }"}, {"id": "3 m - 6 \\ge 0"}, {"id": "$\\sqrt { 3 m - 6 }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "m \\ge 2"}, {"id": "2"}, {"id": "$m$ 能取的最小整数值"}], "links": [{"rel": "被描述", "source": "\\sqrt { 3 m - 6 }", "target": "3 m - 6 \\ge 0"}, {"rel": "不等式方程求解", "source": "3 m - 6 \\ge 0", "target": "m \\ge 2"}, {"rel": "限制性描述", "source": "$\\sqrt { 3 m - 6 }$ 有意义", "target": "3 m - 6 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "3 m - 6 \\ge 0"}, {"rel": "被描述", "source": "m \\ge 2", "target": "2"}, {"rel": "限制性描述", "source": "$m$ 能取的最小整数值", "target": "2"}]}}
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{"content": "Given that the minimum value of $x$ for $x \\ge 2$ is $a$, and the maximum value of $x$ for $x \\le - 6$ is $b$, what is the value of $a + b$?", "answer": "- 4", "steps": "Because the minimum value of $x$ for $x \\ge 2$ is $a$, $a = 2$; and the maximum value of $x$ for $x \\le - 6$ is $b$, so $b = - 6$. Therefore, $a + b = 2 - 6 = - 4$, and $a + b = - 4$.", "expr_cands": ["x \\ge 2", "x", "a", "x \\le - 6", "b", "a + b", "a = 2", "b = - 6", "- 4"], "exprs": ["a = 2", "b = - 6", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x \\ge 2"}, {"id": "a = 2"}, {"id": "a"}, {"id": "$x \\ge 2$ 的最小值是 $a$ , $x \\le - 6$ 的最大值是 $b$"}, {"id": "x \\le - 6"}, {"id": "b = - 6"}, {"id": "b"}, {"id": "a + b"}, {"id": "- 4"}], "links": [{"rel": "被描述", "source": "x \\ge 2", "target": "a = 2"}, {"rel": "代入", "source": "a = 2", "target": "- 4"}, {"rel": "被描述", "source": "a", "target": "a = 2"}, {"rel": "限制性描述", "source": "$x \\ge 2$ 的最小值是 $a$ , $x \\le - 6$ 的最大值是 $b$", "target": "a = 2"}, {"rel": "限制性描述", "source": "$x \\ge 2$ 的最小值是 $a$ , $x \\le - 6$ 的最大值是 $b$", "target": "b = - 6"}, {"rel": "被描述", "source": "x \\le - 6", "target": "b = - 6"}, {"rel": "代入", "source": "b = - 6", "target": "- 4"}, {"rel": "被描述", "source": "b", "target": "b = - 6"}, {"rel": "被代入", "source": "a + b", "target": "- 4"}]}}
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||||
{"content": "Given $x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 5 = 0$, what is the value of the fraction $\\frac { y } { x } - \\frac { x } { y }$?", "answer": "1.5", "steps": "Since $x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 5 = 0$, we can rewrite it as $( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 0$. Therefore, we have $x = 1$ and $y = 2$. Thus, $\\frac { y } { x } - \\frac { x } { y } = 2 - \\frac { 1 } { 2 } = 1.5$.", "expr_cands": ["x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 5 = 0", "x", "y", "\\frac { y } { x } - \\frac { x } { y }", "x ^ { 2 } - 2 x + 1 + y ^ { 2 } - 4 y + 4 = 0", "( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 0", "x = 1", "y = 2", "\\frac { 3 } { 2 }"], "exprs": ["( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 0", "x = 1", "y = 2", "\\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 5 = 0"}, {"id": "( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 0"}, {"id": "x = 1"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "y = 2"}, {"id": "\\frac { y } { x } - \\frac { x } { y }"}, {"id": "\\frac { 3 } { 2 }"}], "links": [{"rel": "提取因式", "source": "x ^ { 2 } + y ^ { 2 } - 2 x - 4 y + 5 = 0", "target": "( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 0"}, {"rel": "被描述", "source": "( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 0", "target": "x = 1"}, {"rel": "被描述", "source": "( x - 1 ) ^ { 2 } + ( y - 2 ) ^ { 2 } = 0", "target": "y = 2"}, {"rel": "代入", "source": "x = 1", "target": "\\frac { 3 } { 2 }"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "x = 1"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "y = 2"}, {"rel": "代入", "source": "y = 2", "target": "\\frac { 3 } { 2 }"}, {"rel": "被代入", "source": "\\frac { y } { x } - \\frac { x } { y }", "target": "\\frac { 3 } { 2 }"}]}}
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{"content": "If the equation $x + 3 = 2 a$ and $2 x - 6 = 4$ have the same solution for $x$, then $a$ = ____ ?", "answer": "4", "steps": "Solve the equation $2 x - 6 = 4$ to get $x = 5$. Substitute $x = 5$ into $x + 3 = 2 a$ to get $2 a = 8$, which gives $a = 4$ after solving.", "expr_cands": ["x", "x + 3 = 2 a", "a", "2 x - 6 = 4", "x = 5", "8 = 2 a", "2 a = 8", "a = 4"], "exprs": ["x = 5", "2 a = 8", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 6 = 4"}, {"id": "x = 5"}, {"id": "x + 3 = 2 a"}, {"id": "2 a = 8"}, {"id": "a = 4"}], "links": [{"rel": "等式方程求解", "source": "2 x - 6 = 4", "target": "x = 5"}, {"rel": "代入", "source": "x = 5", "target": "2 a = 8"}, {"rel": "被代入", "source": "x + 3 = 2 a", "target": "2 a = 8"}, {"rel": "等式方程求解", "source": "2 a = 8", "target": "a = 4"}]}}
|
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{"content": "The solution set of the inequality $2 x + 4 > 0$ is ____?", "answer": "x > - 2", "steps": "Moving terms around, we get $2 x > - 4$, which can be solved to give $x > - 2$.)", "expr_cands": ["2 x + 4 > 0", "x", "2 x > - 4", "- 2 < x", "x > - 2"], "exprs": ["2 x > - 4", "x > - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 4 > 0"}, {"id": "2 x > - 4"}, {"id": "x > - 2"}], "links": [{"rel": "移项", "source": "2 x + 4 > 0", "target": "2 x > - 4"}, {"rel": "不等式方程求解", "source": "2 x > - 4", "target": "x > - 2"}]}}
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{"content": "The solution to the linear equation $4 x + 1 = 0$ is ____ ?", "answer": "- \\frac { 1 } { 4 }", "steps": "$4 x = - 1$, so $x = - \\frac { 1 } { 4 }$.", "expr_cands": ["4 x + 1 = 0", "x", "4 x = - 1", "x = - \\frac { 1 } { 4 }"], "exprs": ["x = - \\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x + 1 = 0"}, {"id": "x = - \\frac { 1 } { 4 }"}], "links": [{"rel": "等式方程求解", "source": "4 x + 1 = 0", "target": "x = - \\frac { 1 } { 4 }"}]}}
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{"content": "Given $a ^ { 2 } + 2 b = 1$, what is the value of the algebraic expression $2 a ^ { 2 } + 4 b$?", "answer": "2", "steps": "Because $a ^ 2 + 2 b = 1$, therefore $2 a ^ 2 + 4 b = 2 ( a ^ 2 + 2 b ) = 2$.", "expr_cands": ["a ^ { 2 } + 2 b = 1", "b", "a", "2 a ^ { 2 } + 4 b", "2 ( a ^ { 2 } + 2 b )", "2"], "exprs": ["2 ( a ^ { 2 } + 2 b )", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a ^ { 2 } + 4 b"}, {"id": "2 ( a ^ { 2 } + 2 b )"}, {"id": "a ^ { 2 } + 2 b = 1"}, {"id": "2"}], "links": [{"rel": "提取因式", "source": "2 a ^ { 2 } + 4 b", "target": "2 ( a ^ { 2 } + 2 b )"}, {"rel": "被代入", "source": "2 ( a ^ { 2 } + 2 b )", "target": "2"}, {"rel": "提取因式参考", "source": "a ^ { 2 } + 2 b = 1", "target": "2 ( a ^ { 2 } + 2 b )"}, {"rel": "代入", "source": "a ^ { 2 } + 2 b = 1", "target": "2"}]}}
|
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{"content": "If the two distinct real roots of the quadratic equation $x ^ 2 = a$ are $m + 1$ and $2 m - 4$, then $a$ = ____ ?", "answer": "4", "steps": "The original equation can be transformed into $x ^ 2 - a = 0$. Since the two roots of the equation $x ^ 2 - a = 0$ are $m + 1$ and $2 m - 4$, we have $m + 1 + 2 m - 4 = 0$, $( m + 1 ) ( 2 m - 4 ) = - a$. Therefore, $m = 1$ and $a = 4$.", "expr_cands": ["x ^ { 2 } = a", "x", "a", "m + 1", "m", "2 m - 4", "x ^ { 2 } - a = 0", "m + 1 + 2 m - 4 = 0", "m = 1", "( m + 1 ) ( 2 m - 4 ) = - a", "a = 4"], "exprs": ["x ^ { 2 } - a = 0", "m + 1 + 2 m - 4 = 0", "( m + 1 ) ( 2 m - 4 ) = - a", "m = 1", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } = a"}, {"id": "x ^ { 2 } - a = 0"}, {"id": "m + 1 + 2 m - 4 = 0"}, {"id": "m + 1"}, {"id": "2 m - 4"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "一元二次方程 $x ^ { 2 } = a$ 的两个不相等的实数根分别是 $m + 1$ 与 $2 m - 4$"}, {"id": "( m + 1 ) ( 2 m - 4 ) = - a"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "m = 1"}, {"id": "a = 4"}], "links": [{"rel": "移项", "source": "x ^ { 2 } = a", "target": "x ^ { 2 } - a = 0"}, {"rel": "被描述", "source": "x ^ { 2 } - a = 0", "target": "m + 1 + 2 m - 4 = 0"}, {"rel": "被描述", "source": "x ^ { 2 } - a = 0", "target": "( m + 1 ) ( 2 m - 4 ) = - a"}, {"rel": "等式方程求解", "source": "m + 1 + 2 m - 4 = 0", "target": "m = 1"}, {"rel": "被描述", "source": "m + 1", "target": "m + 1 + 2 m - 4 = 0"}, {"rel": "被描述", "source": "m + 1", "target": "( m + 1 ) ( 2 m - 4 ) = - a"}, {"rel": "被描述", "source": "2 m - 4", "target": "m + 1 + 2 m - 4 = 0"}, {"rel": "被描述", "source": "2 m - 4", "target": "( m + 1 ) ( 2 m - 4 ) = - a"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "m + 1 + 2 m - 4 = 0"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } = a$ 的两个不相等的实数根分别是 $m + 1$ 与 $2 m - 4$", "target": "m + 1 + 2 m - 4 = 0"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } = a$ 的两个不相等的实数根分别是 $m + 1$ 与 $2 m - 4$", "target": "( m + 1 ) ( 2 m - 4 ) = - a"}, {"rel": "被代入", "source": "( m + 1 ) ( 2 m - 4 ) = - a", "target": "a = 4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "( m + 1 ) ( 2 m - 4 ) = - a"}, {"rel": "代入", "source": "m = 1", "target": "a = 4"}]}}
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{"content": "If $( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 )$ does not contain terms of ${ x } ^ { 2 }$ and ${ x } ^ { 3 }$ after expansion, then $m + n$ = ____ ?", "answer": "8", "steps": "Because $( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 ) = { x } ^ { 4 } - 3 { x } ^ { 3 } + 4 { x } ^ { 2 } + m { x } ^ { 3 } - 3 m { x } ^ { 2 } + 4 mx + n { x } ^ { 2 } - 3 nx + 4 n = { x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n$. Because there are no $x ^ { 2 }$ and $x ^ { 3 }$ terms in the expansion, we have $4 - 3 m + n = 0$ and $3 - m = 0$, which implies $m = 3$ and $n = 5$. Therefore, $m + n = 3 + 5 = 8$.", "expr_cands": ["( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 )", "m", "n", "x", "{ x } ^ { 2 }", "{ x } ^ { 3 }", "m + n", "{ x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n", "x ^ { 2 }", "x ^ { 3 }", "4 - 3 m + n = 0", "3 - m = 0", "m = 3", "n = 5", "8"], "exprs": ["{ x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n", "4 - 3 m + n = 0", "3 - m = 0", "m = 3", "n = 5", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 )"}, {"id": "{ x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n"}, {"id": "x"}, {"id": "4 - 3 m + n = 0"}, {"id": "$( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 )$ 展开后不含 ${ x } ^ { 2 }$ 和 ${ x } ^ { 3 }$ 项"}, {"id": "3 - m = 0"}, {"id": "m = 3"}, {"id": "n = 5"}, {"id": "m + n"}, {"id": "8"}], "links": [{"rel": "提取因式", "source": "( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 )", "target": "{ x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n"}, {"rel": "被描述", "source": "{ x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n", "target": "4 - 3 m + n = 0"}, {"rel": "被描述", "source": "{ x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n", "target": "3 - m = 0"}, {"rel": "提取因式参考", "source": "x", "target": "{ x } ^ { 4 } - ( 3 - m ) { x } ^ { 3 } + ( 4 - 3 m + n ) { x } ^ { 2 } + ( 4 m - 3 n ) x + 4 n"}, {"rel": "联立", "source": "4 - 3 m + n = 0", "target": "n = 5"}, {"rel": "限制性描述", "source": "$( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 )$ 展开后不含 ${ x } ^ { 2 }$ 和 ${ x } ^ { 3 }$ 项", "target": "4 - 3 m + n = 0"}, {"rel": "限制性描述", "source": "$( { x } ^ { 2 } + mx + n ) ( { x } ^ { 2 } - 3 x + 4 )$ 展开后不含 ${ x } ^ { 2 }$ 和 ${ x } ^ { 3 }$ 项", "target": "3 - m = 0"}, {"rel": "等式方程求解", "source": "3 - m = 0", "target": "m = 3"}, {"rel": "联立", "source": "m = 3", "target": "n = 5"}, {"rel": "代入", "source": "m = 3", "target": "8"}, {"rel": "代入", "source": "n = 5", "target": "8"}, {"rel": "被代入", "source": "m + n", "target": "8"}]}}
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{"content": "The maximum value of the quadratic function $y = - x ^ { 2 } + 1$ is:", "answer": "1", "steps": "When $x = 0$, the maximum value of the quadratic function $y = - x ^ 2 + 1$ is $1$.", "expr_cands": ["y = - x ^ { 2 } + 1", "x", "y", "x = 0", "y = 1", "1"], "exprs": ["x = 0", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - x ^ { 2 } + 1"}, {"id": "x = 0"}, {"id": "二次函数 $y = - x ^ { 2 } + 1$ 的最大值"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "y = - x ^ { 2 } + 1", "target": "x = 0"}, {"rel": "被代入", "source": "y = - x ^ { 2 } + 1", "target": "1"}, {"rel": "代入", "source": "x = 0", "target": "1"}, {"rel": "限制性描述", "source": "二次函数 $y = - x ^ { 2 } + 1$ 的最大值", "target": "x = 0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "x = 0"}]}}
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{"content": "The value of $\\sqrt { ( 1 - a ) ^ { 2 } } ( a \\ge 1 )$ is ____ ?", "answer": "a - 1", "steps": "Since $a \\ge 1$, it follows that $1 - a \\le 0$. Therefore, $\\sqrt {( 1 - a ) ^ 2 } = | 1 - a | = a - 1$.", "expr_cands": ["\\sqrt { ( 1 - a ) ^ { 2 } } ( a \\ge 1 )", "a", "a \\ge 1", "1 - a \\le 0", "1 \\le a", "\\sqrt { ( 1 - a ) ^ { 2 } } = a - 1", "\\sqrt { ( 1 - a ) ^ { 2 } }", "a - 1"], "exprs": ["a - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { ( 1 - a ) ^ { 2 } } ( a \\ge 1 )"}, {"id": "a - 1"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "二次根式有意义，则根式恒大于等于0"}], "links": [{"rel": "被描述", "source": "\\sqrt { ( 1 - a ) ^ { 2 } } ( a \\ge 1 )", "target": "a - 1"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "a - 1"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "a - 1"}]}}
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{"content": "The proposition $( 1 )$ If $\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }$, then $x = 4$ is false. Please provide a counterexample: ____?", "answer": "x = \\frac { 21 } { 5 }", "steps": "Since $\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }$, therefore $3 x - 15 = 6 - 2 x$, therefore $5 x = 21$, solving for $x$ gives $x = \\frac { 21 } { 5 }$. Therefore, the counterexample for the proposition If $\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }$, then $x = 4$ is $x = \\frac { 21 } { 5 }$.", "expr_cands": ["( 1 )", "\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }", "x", "x = 4", "x = \\frac { 21 } { 5 }", "3 x - 15 = 6 - 2 x", "5 x = 21"], "exprs": ["x = \\frac { 21 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }"}, {"id": "x = \\frac { 21 } { 5 }"}, {"id": "命题 \" 如果 $\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }$"}, {"id": "$x = 4$ \" 是假命题的一个反例为 $x = \\frac { 21 } { 5 }$"}], "links": [{"rel": "被描述", "source": "\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }", "target": "x = \\frac { 21 } { 5 }"}, {"rel": "限制性描述", "source": "命题 \" 如果 $\\frac { x - 5 } { 2 } = \\frac { 3 - x } { 3 }$", "target": "x = \\frac { 21 } { 5 }"}, {"rel": "限制性描述", "source": "$x = 4$ \" 是假命题的一个反例为 $x = \\frac { 21 } { 5 }$", "target": "x = \\frac { 21 } { 5 }"}]}}
|
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{"content": "The sum of all non-negative integer solutions of the inequality $11 - 3 x > 1$ is _____.", "answer": "6", "steps": "The solution set of the inequality is $x < ( 3 * 3 + 1 / 3 )$, so the non-negative integer solutions are $0$, $1$, $2$, $3$. Therefore, the sum of all non-negative integer solutions of the inequality $11 - 3 x > 1$ is $0 + 1 + 2 + 3 = 6$.", "expr_cands": ["11 - 3 x > 1", "x", "x < ( 3 * 3 + 1 / 3 )", "0", "1", "2", "3", "x < \\frac { 10 } { 3 }", "0 + 1 + 2 + 3", "6"], "exprs": ["x < \\frac { 10 } { 3 }", "0 + 1 + 2 + 3", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "11 - 3 x > 1"}, {"id": "x < \\frac { 10 } { 3 }"}, {"id": "0 + 1 + 2 + 3"}, {"id": "非负整数解为 $0$ , $1$ , $2$ , $3$"}, {"id": "不等式 $11 - 3 x > 1$ 的所有非负整数解的和"}, {"id": "6"}], "links": [{"rel": "不等式方程求解", "source": "11 - 3 x > 1", "target": "x < \\frac { 10 } { 3 }"}, {"rel": "被描述", "source": "x < \\frac { 10 } { 3 }", "target": "0 + 1 + 2 + 3"}, {"rel": "计算", "source": "0 + 1 + 2 + 3", "target": "6"}, {"rel": "限制性描述", "source": "非负整数解为 $0$ , $1$ , $2$ , $3$", "target": "0 + 1 + 2 + 3"}, {"rel": "限制性描述", "source": "不等式 $11 - 3 x > 1$ 的所有非负整数解的和", "target": "0 + 1 + 2 + 3"}]}}
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||||
{"content": "If $a - b = 2$, $c - a = 3$, then the value of $( b - c ) ^ 2 - 3 ( b - c ) + 4$ is ____?", "answer": "44", "steps": "$\\because a - b = 2$, $c - a = 3$, $\\therefore a - b + c - a = 5$, which means $c - b = 5$. Simplifying, we get $b - c = - 5$. Therefore, the original expression is equal to $25 + 15 + 4 = 44$.", "expr_cands": ["a - b = 2", "b", "a", "c - a = 3", "c", "( b - c ) ^ { 2 } - 3 ( b - c ) + 4", "a - b + c - a = 5", "- a + c + 2 = 5", "c - b = 5", "b - c", "- 5", "25 + 15 + 4", "44"], "exprs": ["a - b + c - a = 5", "c - b = 5", "44"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - b = 2"}, {"id": "a - b + c - a = 5"}, {"id": "c - a = 3"}, {"id": "c - b = 5"}, {"id": "( b - c ) ^ { 2 } - 3 ( b - c ) + 4"}, {"id": "44"}], "links": [{"rel": "联立", "source": "a - b = 2", "target": "a - b + c - a = 5"}, {"rel": "移项", "source": "a - b + c - a = 5", "target": "c - b = 5"}, {"rel": "联立", "source": "c - a = 3", "target": "a - b + c - a = 5"}, {"rel": "代入", "source": "c - b = 5", "target": "44"}, {"rel": "被代入", "source": "( b - c ) ^ { 2 } - 3 ( b - c ) + 4", "target": "44"}]}}
|
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{"content": "If $x ^ 2 + kx + 64 = ( x - 8 ) ^ 2$ holds for all real numbers $x$, then the value of $k$ is ____?", "answer": "- 16", "steps": "Since $x ^ 2 + kx + 64 = ( x - 8 ) ^ 2$, it follows that $x ^ 2 + kx + 64 = x ^ 2 - 16 x + 64$. Therefore, $k = - 16$.", "expr_cands": ["x ^ { 2 } + kx + 64 = ( x - 8 ) ^ { 2 }", "k", "x", "x ^ { 2 } + kx + 64", "x ^ { 2 } - 16 x + 64", "k = - 16"], "exprs": ["k = - 16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + kx + 64 = ( x - 8 ) ^ { 2 }"}, {"id": "k = - 16"}], "links": [{"rel": "移项", "source": "x ^ { 2 } + kx + 64 = ( x - 8 ) ^ { 2 }", "target": "k = - 16"}]}}
|
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{"content": "The maximum integer solution of the inequality $2 x - 6 \\le 8$ is ____?", "answer": "7", "steps": "$\\because$ $2 x - 6 \\le 8$ , $\\therefore$ $x \\le 7$ , so the largest integer solution of the inequality is $7$.", "expr_cands": ["2 x - 6 \\le 8", "x", "x \\le 7", "7"], "exprs": ["x \\le 7", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 6 \\le 8"}, {"id": "x \\le 7"}, {"id": "7"}, {"id": "不等式 $2 x - 6 \\le 8$ 的最大整数解"}, {"id": "该不等式的最大整数解为 $7$"}], "links": [{"rel": "不等式方程求解", "source": "2 x - 6 \\le 8", "target": "x \\le 7"}, {"rel": "被描述", "source": "x \\le 7", "target": "7"}, {"rel": "限制性描述", "source": "不等式 $2 x - 6 \\le 8$ 的最大整数解", "target": "7"}, {"rel": "限制性描述", "source": "该不等式的最大整数解为 $7$", "target": "7"}]}}
|
||||
{"content": "If $2 x + 3 y ^ { 2 } - 2 = 6$, then the value of the algebraic expression $8 x + 12 y ^ { 2 } - 5$ is ____?", "answer": "27", "steps": "Since $2 x + 3 y ^ 2 - 2 = 6$, it follows that $2 x + 3 y ^ 2 = 8$. Therefore, $8 x + 12 y ^ 2 - 5 = 4 ( 2 x + 3 y ^ 2 ) - 5 = 4 * 8 - 5 = 32 - 5 = 27$.", "expr_cands": ["2 x + 3 y ^ { 2 } - 2 = 6", "x", "y", "8 x + 12 y ^ { 2 } - 5", "2 x + 3 y ^ { 2 } = 8", "4 ( 2 x + 3 y ^ { 2 } ) - 5", "27"], "exprs": ["2 x + 3 y ^ { 2 } = 8", "4 ( 2 x + 3 y ^ { 2 } ) - 5", "27"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 3 y ^ { 2 } - 2 = 6"}, {"id": "2 x + 3 y ^ { 2 } = 8"}, {"id": "8 x + 12 y ^ { 2 } - 5"}, {"id": "4 ( 2 x + 3 y ^ { 2 } ) - 5"}, {"id": "27"}], "links": [{"rel": "移项", "source": "2 x + 3 y ^ { 2 } - 2 = 6", "target": "2 x + 3 y ^ { 2 } = 8"}, {"rel": "提取因式参考", "source": "2 x + 3 y ^ { 2 } = 8", "target": "4 ( 2 x + 3 y ^ { 2 } ) - 5"}, {"rel": "代入", "source": "2 x + 3 y ^ { 2 } = 8", "target": "27"}, {"rel": "提取因式", "source": "8 x + 12 y ^ { 2 } - 5", "target": "4 ( 2 x + 3 y ^ { 2 } ) - 5"}, {"rel": "被代入", "source": "4 ( 2 x + 3 y ^ { 2 } ) - 5", "target": "27"}]}}
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{"content": "The product of $( x ^ 2 - mx + 3 ) ( 3 x - 2 )$ does not contain a quadratic term in $x$. What is the value of $m$?", "answer": "- \\frac { 2 } { 3 }", "steps": "Original expression = $3 x ^ { 3 } - 3 mx ^ { 2 } + 9 x - 2 x ^ { 2 } + 2 mx - 6 = 3 x ^ { 3 } - ( 3 m + 2 ) x ^ { 2 } + ( 2 m + 9 ) x - 6$. Since there is no quadratic term in $x$, we have $3 m + 2 = 0$. Therefore, $m = - \\frac { 2 } { 3 }$.", "expr_cands": ["( x ^ { 2 } - mx + 3 ) ( 3 x - 2 )", "x", "m", "3 x ^ { 3 } - 3 mx ^ { 2 } + 9 x - 2 x ^ { 2 } + 2 mx - 6", "3 x ^ { 3 } - ( 3 m + 2 ) x ^ { 2 } + ( 2 m + 9 ) x - 6", "3 m + 2 = 0", "m = - \\frac { 2 } { 3 }"], "exprs": ["3 x ^ { 3 } - 3 mx ^ { 2 } + 9 x - 2 x ^ { 2 } + 2 mx - 6", "3 x ^ { 3 } - ( 3 m + 2 ) x ^ { 2 } + ( 2 m + 9 ) x - 6", "3 m + 2 = 0", "m = - \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x ^ { 2 } - mx + 3 ) ( 3 x - 2 )"}, {"id": "3 x ^ { 3 } - 3 mx ^ { 2 } + 9 x - 2 x ^ { 2 } + 2 mx - 6"}, {"id": "3 x ^ { 3 } - ( 3 m + 2 ) x ^ { 2 } + ( 2 m + 9 ) x - 6"}, {"id": "3 m + 2 = 0"}, {"id": "$( x ^ { 2 } - mx + 3 ) ( 3 x - 2 )$ 的积中不含 $x$ 的二次项"}, {"id": "m = - \\frac { 2 } { 3 }"}], "links": [{"rel": "展开", "source": "( x ^ { 2 } - mx + 3 ) ( 3 x - 2 )", "target": "3 x ^ { 3 } - 3 mx ^ { 2 } + 9 x - 2 x ^ { 2 } + 2 mx - 6"}, {"rel": "提取因式", "source": "3 x ^ { 3 } - 3 mx ^ { 2 } + 9 x - 2 x ^ { 2 } + 2 mx - 6", "target": "3 x ^ { 3 } - ( 3 m + 2 ) x ^ { 2 } + ( 2 m + 9 ) x - 6"}, {"rel": "被描述", "source": "3 x ^ { 3 } - ( 3 m + 2 ) x ^ { 2 } + ( 2 m + 9 ) x - 6", "target": "3 m + 2 = 0"}, {"rel": "等式方程求解", "source": "3 m + 2 = 0", "target": "m = - \\frac { 2 } { 3 }"}, {"rel": "限制性描述", "source": "$( x ^ { 2 } - mx + 3 ) ( 3 x - 2 )$ 的积中不含 $x$ 的二次项", "target": "3 m + 2 = 0"}]}}
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{"content": "If $3 x + ax + y - 6 y$ after combining like terms does not contain the term $x$, then the value of $a$ is ____?", "answer": "- 3", "steps": "$3 x + ax + y - 6 y = ( 3 + a ) x - 5 y$ After combining like terms and eliminating the x term, we get $- 5 y = - 9 y$. Solving for a, we get $a = - 3$.", "expr_cands": ["3 x + ax + y - 6 y", "x", "a", "y", "( 3 + a ) x - 5 y", "3 + a = - 3", "a = - 6"], "exprs": ["( 3 + a ) x - 5 y", "3 + a = - 3", "a = - 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + ax + y - 6 y"}, {"id": "( 3 + a ) x - 5 y"}, {"id": "x"}, {"id": "3 + a = - 3"}, {"id": "$3 x + ax + y - 6 y$ 合并同类项后不含 $x$ 项"}, {"id": "a = - 6"}], "links": [{"rel": "提取因式", "source": "3 x + ax + y - 6 y", "target": "( 3 + a ) x - 5 y"}, {"rel": "被描述", "source": "( 3 + a ) x - 5 y", "target": "3 + a = - 3"}, {"rel": "提取因式参考", "source": "x", "target": "( 3 + a ) x - 5 y"}, {"rel": "等式方程求解", "source": "3 + a = - 3", "target": "a = - 6"}, {"rel": "限制性描述", "source": "$3 x + ax + y - 6 y$ 合并同类项后不含 $x$ 项", "target": "3 + a = - 3"}]}}
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{"content": "If $a$, $b$ are opposite numbers, and $m$, $n$ are reciprocal, what is the value of the algebraic expression $( a + b - 1 ) ^ 3 - \\frac { 1 } { 2 } mn$?", "answer": "- \\frac { 3 } { 2 }", "steps": "Since $a$ and $b$ are opposite numbers, and $m$ and $n$ are reciprocal, therefore $a + b = 0$, $mn = 1$. Thus, $( a + b - 1 ) ^ 3 - \\frac { 1 } { 2 } mn = ( - 1 ) ^ 3 - \\frac { 1 } { 2 } * 1 = - 1 - \\frac { 1 } { 2 } = - \\frac { 3 } { 2 }$.", "expr_cands": ["a", "b", "m", "n", "( a + b - 1 ) ^ { 3 } - \\frac { 1 } { 2 } mn", "a + b = 0", "mn = 1", "- \\frac { 3 } { 2 }"], "exprs": ["a + b = 0", "mn = 1", "- \\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "m"}, {"id": "mn = 1"}, {"id": "n"}, {"id": "$m$ , $n$ 互为倒数"}, {"id": "( a + b - 1 ) ^ { 3 } - \\frac { 1 } { 2 } mn"}, {"id": "- \\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "- \\frac { 3 } { 2 }"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "m", "target": "mn = 1"}, {"rel": "代入", "source": "mn = 1", "target": "- \\frac { 3 } { 2 }"}, {"rel": "被描述", "source": "n", "target": "mn = 1"}, {"rel": "限制性描述", "source": "$m$ , $n$ 互为倒数", "target": "mn = 1"}, {"rel": "被代入", "source": "( a + b - 1 ) ^ { 3 } - \\frac { 1 } { 2 } mn", "target": "- \\frac { 3 } { 2 }"}]}}
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{"content": "When $a = - 2$, what is the value of $a ^ 2 ( a ^ 4 + 4 a ^ 2 + 16 ) - 4 ( a ^ 4 + 4 a ^ 2 + 16 )$?", "answer": "0", "steps": "$a ^ { 2 } ( a ^ { 4 } + 4 a ^ { 2 } + 16 ) - 4 ( a ^ { 4 } + 4 a ^ { 2 } + 16 ) = ( a ^ { 4 } + 4 a ^ { 2 } + 16 ) ( a ^ { 2 } - 4 ) = ( a ^ { 4 } + 4 a ^ { 2 } + 16 ) ( a + 2 ) ( a - 2 )$ because $a = - 2$, therefore $a + 2 = 0$ therefore the original expression equals $0$.", "expr_cands": ["a = - 2", "a", "a ^ { 2 } ( a ^ { 4 } + 4 a ^ { 2 } + 16 ) - 4 ( a ^ { 4 } + 4 a ^ { 2 } + 16 )", "( a ^ { 4 } + 4 a ^ { 2 } + 16 ) ( a + 2 ) ( a - 2 )", "a + 2", "0"], "exprs": ["( a ^ { 4 } + 4 a ^ { 2 } + 16 ) ( a + 2 ) ( a - 2 )", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } ( a ^ { 4 } + 4 a ^ { 2 } + 16 ) - 4 ( a ^ { 4 } + 4 a ^ { 2 } + 16 )"}, {"id": "( a ^ { 4 } + 4 a ^ { 2 } + 16 ) ( a + 2 ) ( a - 2 )"}, {"id": "a = - 2"}, {"id": "0"}], "links": [{"rel": "提取因式", "source": "a ^ { 2 } ( a ^ { 4 } + 4 a ^ { 2 } + 16 ) - 4 ( a ^ { 4 } + 4 a ^ { 2 } + 16 )", "target": "( a ^ { 4 } + 4 a ^ { 2 } + 16 ) ( a + 2 ) ( a - 2 )"}, {"rel": "被代入", "source": "( a ^ { 4 } + 4 a ^ { 2 } + 16 ) ( a + 2 ) ( a - 2 )", "target": "0"}, {"rel": "代入", "source": "a = - 2", "target": "0"}]}}
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{"content": "If the difference between the monomials $- { x } ^ { 3 } { y } ^ { m - 2 }$ and ${ x } ^ { 3 } y$ is still a monomial, then $m$ = ____ ?", "answer": "3", "steps": "The difference between the monomials $- x ^ { 3 } y ^ { m - 2 }$ and $x ^ { 3 } y$ is still a monomial, so we know that they are like terms. Therefore, $- x ^ { 3 } y ^ { m - 2 }$ and $x ^ { 3 } y$ are like terms, and we can conclude that $m - 2 = 1$. Thus, $m = 3$.", "expr_cands": ["- { x } ^ { 3 } { y } ^ { m - 2 }", "y", "m", "x", "{ x } ^ { 3 } y", "- x ^ { 3 } y ^ { m - 2 }", "x ^ { 3 } y", "m - 2 = 1", "m = 3"], "exprs": ["m - 2 = 1", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- { x } ^ { 3 } { y } ^ { m - 2 }"}, {"id": "m - 2 = 1"}, {"id": "{ x } ^ { 3 } y"}, {"id": "单项式 $- { x } ^ { 3 } { y } ^ { m - 2 }$ 与 ${ x } ^ { 3 } y$ 的差仍然是一个单项式"}, {"id": "$- x ^ { 3 } y ^ { m - 2 }$ 与 $x ^ { 3 } y$ 是同类项"}, {"id": "m = 3"}], "links": [{"rel": "被描述", "source": "- { x } ^ { 3 } { y } ^ { m - 2 }", "target": "m - 2 = 1"}, {"rel": "等式方程求解", "source": "m - 2 = 1", "target": "m = 3"}, {"rel": "被描述", "source": "{ x } ^ { 3 } y", "target": "m - 2 = 1"}, {"rel": "限制性描述", "source": "单项式 $- { x } ^ { 3 } { y } ^ { m - 2 }$ 与 ${ x } ^ { 3 } y$ 的差仍然是一个单项式", "target": "m - 2 = 1"}, {"rel": "限制性描述", "source": "$- x ^ { 3 } y ^ { m - 2 }$ 与 $x ^ { 3 } y$ 是同类项", "target": "m - 2 = 1"}]}}
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{"content": "The square root of $x - 5$ is undefined in the real number system, what is the condition that $x$ must satisfy?", "answer": "x < 5", "steps": "$\\because$ The square root of a quadratic expression $\\sqrt { x - 5 }$ is undefined in the real number range. $\\therefore$ $x - 5 < 0$, which means $x < 5$.", "expr_cands": ["\\sqrt { x - 5 }", "x", "x - 5 < 0", "x < 5"], "exprs": ["x - 5 < 0", "x < 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x - 5 }"}, {"id": "x - 5 < 0"}, {"id": "二次根式 $\\sqrt { x - 5 }$ 在实数范围内无意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x < 5"}], "links": [{"rel": "被描述", "source": "\\sqrt { x - 5 }", "target": "x - 5 < 0"}, {"rel": "不等式方程求解", "source": "x - 5 < 0", "target": "x < 5"}, {"rel": "限制性描述", "source": "二次根式 $\\sqrt { x - 5 }$ 在实数范围内无意义", "target": "x - 5 < 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 5 < 0"}]}}
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{"content": "If $( m - 1 ) x ^ { | m | } + 5 = 11$ is a linear equation in one variable $x$, then $m$ = ____ ?", "answer": "- 1", "steps": "From the given condition, we have $| m | = 1$ and $m - 1 \\neq 0$. Solving for $m$, we get $m = - 1$.", "expr_cands": ["( m - 1 ) x ^ { | m | } + 5 = 11", "x", "m", "| m | = 1", "m = - 1", "m = 1", "m - 1 \\neq 0", "m \\neq 1"], "exprs": ["| m | = 1", "m - 1 \\neq 0", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 1 ) x ^ { | m | } + 5 = 11"}, {"id": "| m | = 1"}, {"id": "$( m - 1 ) x ^ { | m | } + 5 = 11$ 是关于 $x$ 的一元一次方程"}, {"id": "m - 1 \\neq 0"}, {"id": "m = - 1"}], "links": [{"rel": "被描述", "source": "( m - 1 ) x ^ { | m | } + 5 = 11", "target": "| m | = 1"}, {"rel": "被描述", "source": "( m - 1 ) x ^ { | m | } + 5 = 11", "target": "m - 1 \\neq 0"}, {"rel": "联立", "source": "| m | = 1", "target": "m = - 1"}, {"rel": "限制性描述", "source": "$( m - 1 ) x ^ { | m | } + 5 = 11$ 是关于 $x$ 的一元一次方程", "target": "| m | = 1"}, {"rel": "限制性描述", "source": "$( m - 1 ) x ^ { | m | } + 5 = 11$ 是关于 $x$ 的一元一次方程", "target": "m - 1 \\neq 0"}, {"rel": "联立", "source": "m - 1 \\neq 0", "target": "m = - 1"}]}}
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{"content": "If $\\frac { x - y } { x + y } = 2$, then $\\frac { x - y } { x + y } - \\frac { 2 ( x + y )} { 3 ( x - y )}$ = ____ ?", "answer": "\\frac { 5 } { 3 }", "steps": "Since $\\frac { x - y } { x + y } = 2$, therefore $\\frac { x + y } { x - y } = \\frac { 1 } { 2 }$, therefore $\\frac { x - y } { x + y } - \\frac { 2 ( x + y )} { 3 ( x - y )} = 2 - \\frac { 2 } { 3 } * \\frac { 1 } { 2 } = \\frac { 5 } { 3 }$.", "expr_cands": ["\\frac { x - y } { x + y } = 2", "x", "y", "\\frac { x - y } { x + y } - \\frac { 2 ( x + y ) } { 3 ( x - y ) }", "\\frac { x + y } { x - y } = \\frac { 1 } { 2 }", "\\frac { 5 } { 3 }"], "exprs": ["\\frac { x + y } { x - y } = \\frac { 1 } { 2 }", "\\frac { 5 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - y } { x + y } = 2"}, {"id": "\\frac { x + y } { x - y } = \\frac { 1 } { 2 }"}, {"id": "\\frac { x - y } { x + y } - \\frac { 2 ( x + y ) } { 3 ( x - y ) }"}, {"id": "\\frac { 5 } { 3 }"}], "links": [{"rel": "同乘除", "source": "\\frac { x - y } { x + y } = 2", "target": "\\frac { x + y } { x - y } = \\frac { 1 } { 2 }"}, {"rel": "代入", "source": "\\frac { x - y } { x + y } = 2", "target": "\\frac { 5 } { 3 }"}, {"rel": "代入", "source": "\\frac { x + y } { x - y } = \\frac { 1 } { 2 }", "target": "\\frac { 5 } { 3 }"}, {"rel": "被代入", "source": "\\frac { x - y } { x + y } - \\frac { 2 ( x + y ) } { 3 ( x - y ) }", "target": "\\frac { 5 } { 3 }"}]}}
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{"content": "Given that $3$ is a root of the quadratic equation $x ^ 2 = p$, what is the other root?", "answer": "x = - 3", "steps": "Substituting $x = 3$ into $x ^ 2 = p$, we get $p = 3 ^ 2 = 9$. Therefore, the original equation is $x ^ 2 = 9$, which can be written as $x ^ 2 - 9 = 0$. Let the other root of the equation be $x$, then $3 x = - 9$. Thus, $x = - 3$.", "expr_cands": ["3", "x ^ { 2 } = p", "p", "x", "x = 3", "9 = p", "p = 9", "x ^ { 2 }", "9", "x ^ { 2 } - 9", "0", "3 x = - 9", "x = - 3"], "exprs": ["x", "p = 9", "3 x = - 9", "x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程的另一根为 $x$"}, {"id": "x"}, {"id": "x ^ { 2 } = p"}, {"id": "p = 9"}, {"id": "3"}, {"id": "$3$ 是一元二次方程 $x ^ { 2 } = p$ 的一个根"}, {"id": "3 x = - 9"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "x = - 3"}], "links": [{"rel": "假设描述", "source": "设方程的另一根为 $x$", "target": "x"}, {"rel": "被描述", "source": "x", "target": "p = 9"}, {"rel": "被描述", "source": "x", "target": "3 x = - 9"}, {"rel": "被描述", "source": "x ^ { 2 } = p", "target": "p = 9"}, {"rel": "被描述", "source": "x ^ { 2 } = p", "target": "3 x = - 9"}, {"rel": "被描述", "source": "p = 9", "target": "3 x = - 9"}, {"rel": "被描述", "source": "3", "target": "p = 9"}, {"rel": "被描述", "source": "3", "target": "3 x = - 9"}, {"rel": "限制性描述", "source": "$3$ 是一元二次方程 $x ^ { 2 } = p$ 的一个根", "target": "p = 9"}, {"rel": "等式方程求解", "source": "3 x = - 9", "target": "x = - 3"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "3 x = - 9"}]}}
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{"content": "Given that the equation $3 x - 2 a = 2 x$ has a solution of $2$ for $x$, the value of the algebraic expression $- a ^ 2 + a - 1$ is ____?", "answer": "- 1", "steps": "Substituting $x = 2$ into the equation, we get $3 \\times 2 - 2 a = 2 \\times 2$, which gives $a = 1$. Substituting $a = 1$ into the algebraic expression $- a ^ 2 + a - 1 = - 1 ^ 2 + 1 - 1 = - 1 + 1 - 1 = - 1$.", "expr_cands": ["x", "3 x - 2 a = 2 x", "a", "2", "- a ^ { 2 } + a - 1", "x = 2", "3 * 2 - 2 a = 2 * 2", "a = 1", "- 1"], "exprs": ["x = 2", "3 * 2 - 2 a = 2 * 2", "a = 1", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "x = 2"}, {"id": "x"}, {"id": "3 x - 2 a = 2 x"}, {"id": "关于 $x$ 的方程 $3 x - 2 a = 2 x$ 的解为 $2$"}, {"id": "3 * 2 - 2 a = 2 * 2"}, {"id": "a = 1"}, {"id": "- a ^ { 2 } + a - 1"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "2", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "3 * 2 - 2 a = 2 * 2"}, {"rel": "被描述", "source": "x", "target": "x = 2"}, {"rel": "被描述", "source": "3 x - 2 a = 2 x", "target": "x = 2"}, {"rel": "被代入", "source": "3 x - 2 a = 2 x", "target": "3 * 2 - 2 a = 2 * 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $3 x - 2 a = 2 x$ 的解为 $2$", "target": "x = 2"}, {"rel": "等式方程求解", "source": "3 * 2 - 2 a = 2 * 2", "target": "a = 1"}, {"rel": "代入", "source": "a = 1", "target": "- 1"}, {"rel": "被代入", "source": "- a ^ { 2 } + a - 1", "target": "- 1"}]}}
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{"content": "Given that $1$ is a solution to the equation $2 x - 7 m = - 12$ in terms of $x$, then $m$ = ____ ?", "answer": "2", "steps": "Substituting $x = 1$ into the equation, we get $2 - 7 m = - 12$, which yields $m = 2$ as the solution.", "expr_cands": ["1", "x", "2 x - 7 m = - 12", "m", "x = 1", "2 - 7 m = - 12", "m = 2"], "exprs": ["x = 1", "2 - 7 m = - 12", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1"}, {"id": "x = 1"}, {"id": "x"}, {"id": "2 x - 7 m = - 12"}, {"id": "$1$ 是关于 $x$ 的方程 $2 x - 7 m = - 12$ 的解"}, {"id": "2 - 7 m = - 12"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "1", "target": "x = 1"}, {"rel": "代入", "source": "x = 1", "target": "2 - 7 m = - 12"}, {"rel": "被描述", "source": "x", "target": "x = 1"}, {"rel": "被描述", "source": "2 x - 7 m = - 12", "target": "x = 1"}, {"rel": "被代入", "source": "2 x - 7 m = - 12", "target": "2 - 7 m = - 12"}, {"rel": "限制性描述", "source": "$1$ 是关于 $x$ 的方程 $2 x - 7 m = - 12$ 的解", "target": "x = 1"}, {"rel": "等式方程求解", "source": "2 - 7 m = - 12", "target": "m = 2"}]}}
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{"content": "If $\\frac { a } { b } = - 3$, find the value of $\\frac { a - b } { 2 a + b }$.", "answer": "\\frac { 4 } { 5 }", "steps": "Because $\\frac { a } { b } = - 3$, therefore the original expression is $\\frac { \\frac { a } { b } - 1 } { \\frac { 2 a } { b } + 1 } = \\frac { - 3 - 1 } { 2 * ( - 3 ) + 1 } = \\frac { 4 } { 5 }$.", "expr_cands": ["\\frac { a } { b } = - 3", "b", "a", "\\frac { a - b } { 2 a + b }", "\\frac { \\frac { a } { b } - 1 } { \\frac { 2 a } { b } + 1 }", "\\frac { 4 } { 5 }"], "exprs": ["\\frac { \\frac { a } { b } - 1 } { \\frac { 2 a } { b } + 1 }", "\\frac { 4 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a - b } { 2 a + b }"}, {"id": "\\frac { \\frac { a } { b } - 1 } { \\frac { 2 a } { b } + 1 }"}, {"id": "\\frac { 4 } { 5 }"}, {"id": "\\frac { a } { b } = - 3"}], "links": [{"rel": "计算", "source": "\\frac { a - b } { 2 a + b }", "target": "\\frac { \\frac { a } { b } - 1 } { \\frac { 2 a } { b } + 1 }"}, {"rel": "被代入", "source": "\\frac { \\frac { a } { b } - 1 } { \\frac { 2 a } { b } + 1 }", "target": "\\frac { 4 } { 5 }"}, {"rel": "代入", "source": "\\frac { a } { b } = - 3", "target": "\\frac { 4 } { 5 }"}]}}
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{"content": "A polynomial subtracted by $- 5 x$ equals $3 x ^ 2 - 5 x + 9$, what is the polynomial?", "answer": "3 x ^ { 2 } - 10 x + 9", "steps": "Because a polynomial subtracted by $- 5 x$ equals $3 x ^ 2 - 5 x + 9$, therefore this polynomial is $3 x ^ 2 - 10 x + 9$.", "expr_cands": ["- 5 x", "x", "3 { x } ^ { 2 } - 5 x + 9", "3 x ^ { 2 } - 5 x + 9", "3 x ^ { 2 } - 5 x + 9 - 5 x", "3 x ^ { 2 } - 10 x + 9"], "exprs": ["3 x ^ { 2 } - 5 x + 9 - 5 x", "3 x ^ { 2 } - 10 x + 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 5 x"}, {"id": "3 x ^ { 2 } - 5 x + 9 - 5 x"}, {"id": "3 { x } ^ { 2 } - 5 x + 9"}, {"id": "一个多项式减去 $- 5 x$ 等于 $3 { x } ^ { 2 } - 5 x + 9$"}, {"id": "这个多项式"}, {"id": "一个多项式减去 $- 5 x$ 等于 $3 x ^ { 2 } - 5 x + 9$"}, {"id": "3 x ^ { 2 } - 10 x + 9"}], "links": [{"rel": "被描述", "source": "- 5 x", "target": "3 x ^ { 2 } - 5 x + 9 - 5 x"}, {"rel": "计算", "source": "3 x ^ { 2 } - 5 x + 9 - 5 x", "target": "3 x ^ { 2 } - 10 x + 9"}, {"rel": "被描述", "source": "3 { x } ^ { 2 } - 5 x + 9", "target": "3 x ^ { 2 } - 5 x + 9 - 5 x"}, {"rel": "限制性描述", "source": "一个多项式减去 $- 5 x$ 等于 $3 { x } ^ { 2 } - 5 x + 9$", "target": "3 x ^ { 2 } - 5 x + 9 - 5 x"}, {"rel": "限制性描述", "source": "这个多项式", "target": "3 x ^ { 2 } - 5 x + 9 - 5 x"}, {"rel": "限制性描述", "source": "一个多项式减去 $- 5 x$ 等于 $3 x ^ { 2 } - 5 x + 9$", "target": "3 x ^ { 2 } - 5 x + 9 - 5 x"}]}}
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{"content": "Given: When $x = 3$, the value of the algebraic expression $ax ^ 5 + bx ^ 3 + cx - 10$ is $7$. What is the value of the polynomial when $x = - 3$?", "answer": "- 27", "steps": "$\\because$ When $x = 3$, the value of the algebraic expression $ax ^ 5 + bx ^ 3 + cx - 10$ is $7$. $\\therefore$ $3 ^ 5 a + 3 ^ 3 b + 3 c - 10 = 7$, which means $3 ^ 5 a + 3 ^ 3 b + 3 c = 17$. Then, when $x = - 3$, the original expression is $- ( 3 ^ 5 a + 3 ^ 3 b + 3 c ) - 10 = - 17 - 10 = - 27$.", "expr_cands": ["x = 3", "x", "ax ^ { 5 } + bx ^ { 3 } + cx - 10", "c", "b", "a", "7", "x = - 3", "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7", "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c = 17", "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10", "- 27"], "exprs": ["3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7", "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10", "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c = 17", "- 27"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 3"}, {"id": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7"}, {"id": "ax ^ { 5 } + bx ^ { 3 } + cx - 10"}, {"id": ": 当 $x = 3$ 时"}, {"id": "代数式 $ax ^ { 5 } + bx ^ { 3 } + cx - 10$ 的值为 $7$"}, {"id": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c = 17"}, {"id": "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10"}, {"id": "x = - 3"}, {"id": "当 $x = - 3$ 时"}, {"id": "该多项式的值"}, {"id": "- 27"}], "links": [{"rel": "被描述", "source": "x = 3", "target": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7"}, {"rel": "移项", "source": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7", "target": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c = 17"}, {"rel": "被描述", "source": "ax ^ { 5 } + bx ^ { 3 } + cx - 10", "target": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7"}, {"rel": "被描述", "source": "ax ^ { 5 } + bx ^ { 3 } + cx - 10", "target": "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10"}, {"rel": "限制性描述", "source": ": 当 $x = 3$ 时", "target": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7"}, {"rel": "限制性描述", "source": "代数式 $ax ^ { 5 } + bx ^ { 3 } + cx - 10$ 的值为 $7$", "target": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c - 10 = 7"}, {"rel": "代入", "source": "3 ^ { 5 } a + 3 ^ { 3 } b + 3 c = 17", "target": "- 27"}, {"rel": "被代入", "source": "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10", "target": "- 27"}, {"rel": "被描述", "source": "x = - 3", "target": "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10"}, {"rel": "限制性描述", "source": "当 $x = - 3$ 时", "target": "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10"}, {"rel": "限制性描述", "source": "该多项式的值", "target": "- ( 3 ^ { 5 } a + 3 ^ { 3 } b + 3 c ) - 10"}]}}
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{"content": "Given a linear equation in one variable $x$, $\\frac { 1 } { 2020 } x + 3 = 10 x + m$, with a solution of $x = - 3$, what is the solution of the linear equation in one variable $y$, $\\frac { 1 } { 2020 } \\times ( 2 y + 1 ) + 3 = 10 ( 2 y + 1 ) + m$?", "answer": "- 2", "steps": "$\\because$ The solution to the one-variable linear equation in $x$, $\\frac { 1 } { 2020 } x + 3 = 10 x + m$, is $x = - 3$. $\\therefore$ The solution to the one-variable linear equation in $y$, $\\frac { 1 } { 2020 } \\times ( 2 y + 1 ) + 3 = 10 ( 2 y + 1 ) + m$, is $2 y + 1 = - 3$, which yields $y = - 2$.", "expr_cands": ["x", "\\frac { 1 } { 2020 } x + 3 = 10 x + m", "m", "x = - 3", "y", "\\frac { 1 } { 2020 } \\times ( 2 y + 1 ) + 3 = 10 ( 2 y + 1 ) + m", "2 y + 1 = - 3", "y = - 2"], "exprs": ["2 y + 1 = - 3", "y = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 2020 } \\times ( 2 y + 1 ) + 3 = 10 ( 2 y + 1 ) + m"}, {"id": "2 y + 1 = - 3"}, {"id": "\\frac { 1 } { 2020 } x + 3 = 10 x + m"}, {"id": "x = - 3"}, {"id": "关于 $x$ 的一元一次方程 $\\frac { 1 } { 2020 } x + 3 = 10 x + m$ 的解为 $x = - 3$"}, {"id": "关于 $y$ 的一元一次方程 $\\frac { 1 } { 2020 } \\times ( 2 y + 1 ) + 3 = 10 ( 2 y + 1 ) + m$ 的解"}, {"id": "y = - 2"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 2020 } \\times ( 2 y + 1 ) + 3 = 10 ( 2 y + 1 ) + m", "target": "2 y + 1 = - 3"}, {"rel": "等式方程求解", "source": "2 y + 1 = - 3", "target": "y = - 2"}, {"rel": "被描述", "source": "\\frac { 1 } { 2020 } x + 3 = 10 x + m", "target": "2 y + 1 = - 3"}, {"rel": "被描述", "source": "x = - 3", "target": "2 y + 1 = - 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元一次方程 $\\frac { 1 } { 2020 } x + 3 = 10 x + m$ 的解为 $x = - 3$", "target": "2 y + 1 = - 3"}, {"rel": "限制性描述", "source": "关于 $y$ 的一元一次方程 $\\frac { 1 } { 2020 } \\times ( 2 y + 1 ) + 3 = 10 ( 2 y + 1 ) + m$ 的解", "target": "2 y + 1 = - 3"}]}}
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{"content": "The solution to the equation $( x + 3 ) ( x + 2 ) - 28 = ( x - 2 ) ( x - 1 )$ is ____?", "answer": "x = 3", "steps": "$( x + 3 ) ( x + 2 ) - 28 = ( x - 2 ) ( x - 1 )$, expanding and simplifying yields $x ^ 2 + 2 x + 3 x + 6 - 28 = x ^ 2 - x - 2 x + 2$, which simplifies to $2 x + 3 x + x + 2 x = 2 - 6 + 28$. Solving for $x$, we get $x = 3$.", "expr_cands": ["( x + 3 ) ( x + 2 ) - 28 = ( x - 2 ) ( x - 1 )", "x", "x = 3", "x ^ { 2 } + 2 x + 3 x + 6 - 28 = x ^ { 2 } - x - 2 x + 2", "2 x + 3 x + x + 2 x = 2 - 6 + 28", "8 x = 24"], "exprs": ["x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + 3 ) ( x + 2 ) - 28 = ( x - 2 ) ( x - 1 )"}, {"id": "x = 3"}], "links": [{"rel": "等式方程求解", "source": "( x + 3 ) ( x + 2 ) - 28 = ( x - 2 ) ( x - 1 )", "target": "x = 3"}]}}
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{"content": "If $xy ^ { 2 n + 3 } - \\frac { 1 } { 5 } y ^ 2 + 6$ is a quintic trinomial, then the value of $n$ is ____?", "answer": "\\frac { 1 } { 2 }", "steps": "Because $xy ^ { 2 n + 3 } - \\frac { 1 } { 5 } y ^ 2 + 6$ is a quintic trinomial, therefore $2 n + 3 = 4$, therefore $n = \\frac { 1 } { 2 }$.", "expr_cands": ["xy ^ { 2 n + 3 } - \\frac { 1 } { 5 } y ^ { 2 } + 6", "y", "n", "x", "2 n + 3 = 4", "n = \\frac { 1 } { 2 }"], "exprs": ["2 n + 3 = 4", "n = \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "xy ^ { 2 n + 3 } - \\frac { 1 } { 5 } y ^ { 2 } + 6"}, {"id": "2 n + 3 = 4"}, {"id": "$xy ^ { 2 n + 3 } - \\frac { 1 } { 5 } y ^ { 2 } + 6$ 是五次三项式"}, {"id": "n = \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "xy ^ { 2 n + 3 } - \\frac { 1 } { 5 } y ^ { 2 } + 6", "target": "2 n + 3 = 4"}, {"rel": "等式方程求解", "source": "2 n + 3 = 4", "target": "n = \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "$xy ^ { 2 n + 3 } - \\frac { 1 } { 5 } y ^ { 2 } + 6$ 是五次三项式", "target": "2 n + 3 = 4"}]}}
|
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{"content": "If the polynomial $3 x ^ m - ( n - 2 ) x + 2$ is a cubic binomial in terms of $x$, then $m + n$ = ____?", "answer": "5", "steps": "Since the polynomial $3 x ^ m - ( n - 2 ) x + 2$ is a cubic binomial, we know that $m = 3$ and $n - 2 = 0$, which means $n = 2$. Therefore, $m + n = 3 + 2 = 5$.", "expr_cands": ["x", "3 x ^ { m } - ( n - 2 ) x + 2", "m", "n", "m + n", "m = 3", "n - 2 = 0", "n = 2", "5"], "exprs": ["m = 3", "n - 2 = 0", "n = 2", "5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { m } - ( n - 2 ) x + 2"}, {"id": "m = 3"}, {"id": "关于 $x$ 的多项式 $3 x ^ { m } - ( n - 2 ) x + 2$ 为三次二项式"}, {"id": "n - 2 = 0"}, {"id": "n = 2"}, {"id": "m + n"}, {"id": "5"}], "links": [{"rel": "被描述", "source": "3 x ^ { m } - ( n - 2 ) x + 2", "target": "m = 3"}, {"rel": "被描述", "source": "3 x ^ { m } - ( n - 2 ) x + 2", "target": "n - 2 = 0"}, {"rel": "代入", "source": "m = 3", "target": "5"}, {"rel": "限制性描述", "source": "关于 $x$ 的多项式 $3 x ^ { m } - ( n - 2 ) x + 2$ 为三次二项式", "target": "m = 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的多项式 $3 x ^ { m } - ( n - 2 ) x + 2$ 为三次二项式", "target": "n - 2 = 0"}, {"rel": "等式方程求解", "source": "n - 2 = 0", "target": "n = 2"}, {"rel": "代入", "source": "n = 2", "target": "5"}, {"rel": "被代入", "source": "m + n", "target": "5"}]}}
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{"content": "If $3 x - 2 y - 7 = 0$, then the value of $4 y - 6 x + 12$ is ____?", "answer": "- 2", "steps": "Since $3 x - 2 y - 7 = 0$, therefore $3 x - 2 y = 7$, therefore $4 y - 6 x + 12 = - 2 ( 3 x - 2 y ) + 12 = - 2 * 7 + 12 = - 14 + 12 = - 2$.", "expr_cands": ["3 x - 2 y - 7 = 0", "y", "x", "4 y - 6 x + 12", "3 x - 2 y = 7", "- 2 ( 3 x - 2 y ) + 12", "- 2"], "exprs": ["3 x - 2 y = 7", "- 2 ( 3 x - 2 y ) + 12", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 2 y - 7 = 0"}, {"id": "3 x - 2 y = 7"}, {"id": "4 y - 6 x + 12"}, {"id": "- 2 ( 3 x - 2 y ) + 12"}, {"id": "- 2"}], "links": [{"rel": "移项", "source": "3 x - 2 y - 7 = 0", "target": "3 x - 2 y = 7"}, {"rel": "提取因式参考", "source": "3 x - 2 y = 7", "target": "- 2 ( 3 x - 2 y ) + 12"}, {"rel": "代入", "source": "3 x - 2 y = 7", "target": "- 2"}, {"rel": "提取因式", "source": "4 y - 6 x + 12", "target": "- 2 ( 3 x - 2 y ) + 12"}, {"rel": "被代入", "source": "- 2 ( 3 x - 2 y ) + 12", "target": "- 2"}]}}
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{"content": "When $x = 2018$, what is the value of the fraction $\\frac { x ^ 2 - 9 } { x + 3 }$?", "answer": "2015", "steps": "When $x = 2018$, $\\frac { x ^ 2 - 9 } { x + 3 } = \\frac {( x + 3 ) ( x - 3 )} { x + 3 } = x - 3 = 2018 - 3 = 2015$.", "expr_cands": ["x = 2018", "x", "\\frac { x ^ { 2 } - 9 } { x + 3 }", "2015"], "exprs": ["2015"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x ^ { 2 } - 9 } { x + 3 }"}, {"id": "2015"}, {"id": "x = 2018"}], "links": [{"rel": "被代入", "source": "\\frac { x ^ { 2 } - 9 } { x + 3 }", "target": "2015"}, {"rel": "代入", "source": "x = 2018", "target": "2015"}]}}
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{"content": "If the expression $3 { x } ^ { 4 } - { x } ^ { 3 } + k { x } ^ { 3 } + { x } ^ { 2 } + 2$ does not contain the term ${ x } ^ { 3 }$, then the value of $k$ is ____?", "answer": "1", "steps": "$3 x ^ { 4 } - x ^ { 3 } + kx ^ { 3 } + x ^ { 2 } + 2 = 3 x ^ { 4 } + ( k - 1 ) x ^ { 3 } + x ^ { 2 } + 2$ , because after combining like terms, there is no cubic term, therefore $k - 1 = 0$ , thus $k = 1$.", "expr_cands": ["3 { x } ^ { 4 } - { x } ^ { 3 } + k { x } ^ { 3 } + { x } ^ { 2 } + 2", "k", "x", "{ x } ^ { 3 }", "3 x ^ { 4 } - x ^ { 3 } + kx ^ { 3 } + x ^ { 2 } + 2", "3 x ^ { 4 } + ( k - 1 ) x ^ { 3 } + x ^ { 2 } + 2", "k - 1 = 0", "k = 1"], "exprs": ["3 x ^ { 4 } + ( k - 1 ) x ^ { 3 } + x ^ { 2 } + 2", "k - 1 = 0", "k = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { 4 } - x ^ { 3 } + kx ^ { 3 } + x ^ { 2 } + 2"}, {"id": "3 x ^ { 4 } + ( k - 1 ) x ^ { 3 } + x ^ { 2 } + 2"}, {"id": "{ x } ^ { 3 }"}, {"id": "k - 1 = 0"}, {"id": "式子 $3 { x } ^ { 4 } - { x } ^ { 3 } + k { x } ^ { 3 } + { x } ^ { 2 } + 2$ 中不含 ${ x } ^ { 3 }$ 项"}, {"id": "k = 1"}], "links": [{"rel": "提取因式", "source": "3 x ^ { 4 } - x ^ { 3 } + kx ^ { 3 } + x ^ { 2 } + 2", "target": "3 x ^ { 4 } + ( k - 1 ) x ^ { 3 } + x ^ { 2 } + 2"}, {"rel": "被描述", "source": "3 x ^ { 4 } + ( k - 1 ) x ^ { 3 } + x ^ { 2 } + 2", "target": "k - 1 = 0"}, {"rel": "提取因式参考", "source": "{ x } ^ { 3 }", "target": "3 x ^ { 4 } + ( k - 1 ) x ^ { 3 } + x ^ { 2 } + 2"}, {"rel": "等式方程求解", "source": "k - 1 = 0", "target": "k = 1"}, {"rel": "限制性描述", "source": "式子 $3 { x } ^ { 4 } - { x } ^ { 3 } + k { x } ^ { 3 } + { x } ^ { 2 } + 2$ 中不含 ${ x } ^ { 3 }$ 项", "target": "k - 1 = 0"}]}}
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{"content": "Given a linear equation in one variable $x$, $mx = 5 x - 2$, with a solution of $x = 2$, what is the value of $m$?", "answer": "4", "steps": "$\\because$ The solution to the one-variable linear equation $mx = 5 x - 2$ with respect to $x$ is $x = 2$, $\\therefore$ $2 m = 10 - 2$, solving for $m$, we get: $m = 4$.", "expr_cands": ["x", "mx = 5 x - 2", "m", "x = 2", "2 m = 10 - 2", "m = 4"], "exprs": ["2 m = 10 - 2", "m = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "mx = 5 x - 2"}, {"id": "2 m = 10 - 2"}, {"id": "x = 2"}, {"id": "m = 4"}], "links": [{"rel": "被代入", "source": "mx = 5 x - 2", "target": "2 m = 10 - 2"}, {"rel": "等式方程求解", "source": "2 m = 10 - 2", "target": "m = 4"}, {"rel": "代入", "source": "x = 2", "target": "2 m = 10 - 2"}]}}
|
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{"content": "Given $5 x ^ { 3 } y ^ { m }$ and $6 x ^ { n } y ^ { 2 }$ can be combined into one term, then the value of $m ^ { n }$ is ____?", "answer": "8", "steps": "$\\because$ $5 x ^ { 3 } y ^ { m }$ and $6 x ^ { n } y ^ { 2 }$ are like terms, $\\therefore$ $n = 3$, $m = 2$, then $m ^ { n } = 8$.", "expr_cands": ["5 x ^ { 3 } y ^ { m }", "y", "m", "x", "6 x ^ { n } y ^ { 2 }", "n", "m ^ { n }", "n = 3", "m = 2", "8"], "exprs": ["n = 3", "m = 2", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x ^ { 3 } y ^ { m }"}, {"id": "n = 3"}, {"id": "6 x ^ { n } y ^ { 2 }"}, {"id": "$5 x ^ { 3 } y ^ { m }$ 与 $6 x ^ { n } y ^ { 2 }$ 可以合并为一项"}, {"id": ", $5 x ^ { 3 } y ^ { m }$ 与 $6 x ^ { n } y ^ { 2 }$ 是同类项"}, {"id": "m = 2"}, {"id": "m ^ { n }"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "5 x ^ { 3 } y ^ { m }", "target": "n = 3"}, {"rel": "被描述", "source": "5 x ^ { 3 } y ^ { m }", "target": "m = 2"}, {"rel": "代入", "source": "n = 3", "target": "8"}, {"rel": "被描述", "source": "6 x ^ { n } y ^ { 2 }", "target": "n = 3"}, {"rel": "被描述", "source": "6 x ^ { n } y ^ { 2 }", "target": "m = 2"}, {"rel": "限制性描述", "source": "$5 x ^ { 3 } y ^ { m }$ 与 $6 x ^ { n } y ^ { 2 }$ 可以合并为一项", "target": "n = 3"}, {"rel": "限制性描述", "source": "$5 x ^ { 3 } y ^ { m }$ 与 $6 x ^ { n } y ^ { 2 }$ 可以合并为一项", "target": "m = 2"}, {"rel": "限制性描述", "source": ", $5 x ^ { 3 } y ^ { m }$ 与 $6 x ^ { n } y ^ { 2 }$ 是同类项", "target": "n = 3"}, {"rel": "限制性描述", "source": ", $5 x ^ { 3 } y ^ { m }$ 与 $6 x ^ { n } y ^ { 2 }$ 是同类项", "target": "m = 2"}, {"rel": "代入", "source": "m = 2", "target": "8"}, {"rel": "被代入", "source": "m ^ { n }", "target": "8"}]}}
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{"content": "The square of the sum of two numbers $a$ and $b$ minus $- 3$ is ____?", "answer": "( a + b ) ^ { 2 } + 3", "steps": "The square of the sum of two numbers $a$ and $b$ minus $- 3$ is $( a + b ) ^ 2 - ( - 3 ) = ( a + b ) ^ 2 + 3$.", "expr_cands": ["a", "b", "- 3", "( a + b ) ^ { 2 } - ( - 3 )", "( a + b ) ^ { 2 } + 3"], "exprs": ["( a + b ) ^ { 2 } - ( - 3 )", "( a + b ) ^ { 2 } + 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "( a + b ) ^ { 2 } - ( - 3 )"}, {"id": "b"}, {"id": "- 3"}, {"id": "$a$ , $b$ 两数和的平方与 $- 3$ 的差"}, {"id": "( a + b ) ^ { 2 } + 3"}], "links": [{"rel": "被描述", "source": "a", "target": "( a + b ) ^ { 2 } - ( - 3 )"}, {"rel": "计算", "source": "( a + b ) ^ { 2 } - ( - 3 )", "target": "( a + b ) ^ { 2 } + 3"}, {"rel": "被描述", "source": "b", "target": "( a + b ) ^ { 2 } - ( - 3 )"}, {"rel": "被描述", "source": "- 3", "target": "( a + b ) ^ { 2 } - ( - 3 )"}, {"rel": "限制性描述", "source": "$a$ , $b$ 两数和的平方与 $- 3$ 的差", "target": "( a + b ) ^ { 2 } - ( - 3 )"}]}}
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{"content": "Given $x$ and $y$ satisfy the linear equation $3 x + y = 6$, if ${ y } < 0$, then the range of possible values for $x$ is ____?", "answer": "x > 2", "steps": "$\\because$ $3 x + y = 6$, $\\therefore$ $y = - 3 x + 6$. Since $y < 0$, we know that $- 3 x + 6 < 0$. Solving for $x$, we get $x > 2$.", "expr_cands": ["x", "y", "3 { x } + { y } = 6", "{ y } < 0", "3 x + y = 6", "y = - 3 x + 6", "y < 0", "- 3 x + 6 < 0", "2 < x", "x > 2"], "exprs": ["y = - 3 x + 6", "- 3 x + 6 < 0", "x > 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 { x } + { y } = 6"}, {"id": "y = - 3 x + 6"}, {"id": "{ y } < 0"}, {"id": "- 3 x + 6 < 0"}, {"id": "x > 2"}], "links": [{"rel": "移项", "source": "3 { x } + { y } = 6", "target": "y = - 3 x + 6"}, {"rel": "代入", "source": "y = - 3 x + 6", "target": "- 3 x + 6 < 0"}, {"rel": "被代入", "source": "{ y } < 0", "target": "- 3 x + 6 < 0"}, {"rel": "不等式方程求解", "source": "- 3 x + 6 < 0", "target": "x > 2"}]}}
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{"content": "Given a linear function $y = ax - 3 ( a \\neq 0 )$, when $x = 1$, $y = 5$. What is the value of $a$?", "answer": "8", "steps": "Substituting $x = 1$ and $y = 5$ into $y = ax - 3 ( a \\neq 0 )$ yields $a - 3 = 5$. Solving for $a$ gives $a = 8$.", "expr_cands": ["y = ax - 3 ( a \\neq 0 )", "a", "y", "x", "x = 1", "y = 5", "5 = a - 3", "a - 3 = 5", "a = 8"], "exprs": ["a - 3 = 5", "a = 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ax - 3 ( a \\neq 0 )"}, {"id": "a - 3 = 5"}, {"id": "x = 1"}, {"id": "y = 5"}, {"id": "a = 8"}], "links": [{"rel": "被代入", "source": "y = ax - 3 ( a \\neq 0 )", "target": "a - 3 = 5"}, {"rel": "等式方程求解", "source": "a - 3 = 5", "target": "a = 8"}, {"rel": "代入", "source": "x = 1", "target": "a - 3 = 5"}, {"rel": "代入", "source": "y = 5", "target": "a - 3 = 5"}]}}
|
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{"content": "Given $a$ and $b$ are opposite numbers, what is the value of $a - ( - b ) + 8$?", "answer": "8", "steps": "$\\because a$ and $b$ are opposite numbers, $\\therefore a + b = 0$, $\\therefore a - ( - b ) + 8 = a + b + 8 = 0 + 8 = 8$.", "expr_cands": ["a", "b", "a - ( - b ) + 8", "a + b = 0", "a + b + 8", "8"], "exprs": ["a + b = 0", "a + b + 8", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "a - ( - b ) + 8"}, {"id": "a + b + 8"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "8"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "计算", "source": "a - ( - b ) + 8", "target": "a + b + 8"}, {"rel": "被代入", "source": "a + b + 8", "target": "8"}]}}
|
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{"content": "When $m$ = ____ ?, the fractional equation $\\frac { x - 2 } { x - 3 } = \\frac { m } { 3 - x }$ will have a repeated root.", "answer": "- 1", "steps": "Eliminating the denominator in the fractional equation, we get $x - 2 = - m$. Since the fractional equation has an extraneous root, we have $x - 3 = 0$, which means $x = 3$. Substituting $x = 3$ into the polynomial equation, we get $m = - 1$.", "expr_cands": ["m", "\\frac { x - 2 } { x - 3 } = \\frac { m } { 3 - x }", "x", "x - 2 = - m", "x - 3 = 0", "x = 3", "m = - 1"], "exprs": ["x - 2 = - m", "x - 3 = 0", "x = 3", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 2 } { x - 3 } = \\frac { m } { 3 - x }"}, {"id": "x - 2 = - m"}, {"id": "x - 3 = 0"}, {"id": "分式方程 $\\frac { x - 2 } { x - 3 } = \\frac { m } { 3 - x }$ 会出现增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 3"}, {"id": "m = - 1"}], "links": [{"rel": "同乘除", "source": "\\frac { x - 2 } { x - 3 } = \\frac { m } { 3 - x }", "target": "x - 2 = - m"}, {"rel": "被描述", "source": "\\frac { x - 2 } { x - 3 } = \\frac { m } { 3 - x }", "target": "x - 3 = 0"}, {"rel": "联立", "source": "x - 2 = - m", "target": "m = - 1"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "限制性描述", "source": "分式方程 $\\frac { x - 2 } { x - 3 } = \\frac { m } { 3 - x }$ 会出现增根", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 3 = 0"}, {"rel": "联立", "source": "x = 3", "target": "m = - 1"}]}}
|
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{"content": "Given that the value of ${ y } ^ { 2 } - 3 y + 5$ is $9$, find the value of $- 3 { y } ^ { 2 } + 9 y + 42$ which is ____ ?", "answer": "30", "steps": "According to the problem, we have ${ y } ^ { 2 } - 3 y + 5 = 9$, so ${ y } ^ { 2 } - 3 y = 4$. Then, $- 3 { y } ^ { 2 } + 9 y + 42 = - 3 ( { y } ^ { 2 } - 3 y ) + 42 = - 3 * 4 + 42 = 30$.", "expr_cands": ["{ y } ^ { 2 } - 3 y + 5", "y", "9", "- 3 { y } ^ { 2 } + 9 y + 42", "{ y } ^ { 2 } - 3 y + 5 = 9", "y = - 1", "y = 4", "{ y } ^ { 2 } - 3 y = 4", "- 3 ( { y } ^ { 2 } - 3 y ) + 42", "30"], "exprs": ["{ y } ^ { 2 } - 3 y + 5 = 9", "{ y } ^ { 2 } - 3 y = 4", "- 3 ( { y } ^ { 2 } - 3 y ) + 42", "30"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ y } ^ { 2 } - 3 y + 5"}, {"id": "{ y } ^ { 2 } - 3 y + 5 = 9"}, {"id": "9"}, {"id": "${ y } ^ { 2 } - 3 y + 5$ 的值为 $9$"}, {"id": "{ y } ^ { 2 } - 3 y = 4"}, {"id": "- 3 { y } ^ { 2 } + 9 y + 42"}, {"id": "- 3 ( { y } ^ { 2 } - 3 y ) + 42"}, {"id": "30"}], "links": [{"rel": "被描述", "source": "{ y } ^ { 2 } - 3 y + 5", "target": "{ y } ^ { 2 } - 3 y + 5 = 9"}, {"rel": "移项", "source": "{ y } ^ { 2 } - 3 y + 5 = 9", "target": "{ y } ^ { 2 } - 3 y = 4"}, {"rel": "被描述", "source": "9", "target": "{ y } ^ { 2 } - 3 y + 5 = 9"}, {"rel": "限制性描述", "source": "${ y } ^ { 2 } - 3 y + 5$ 的值为 $9$", "target": "{ y } ^ { 2 } - 3 y + 5 = 9"}, {"rel": "提取因式参考", "source": "{ y } ^ { 2 } - 3 y = 4", "target": "- 3 ( { y } ^ { 2 } - 3 y ) + 42"}, {"rel": "代入", "source": "{ y } ^ { 2 } - 3 y = 4", "target": "30"}, {"rel": "提取因式", "source": "- 3 { y } ^ { 2 } + 9 y + 42", "target": "- 3 ( { y } ^ { 2 } - 3 y ) + 42"}, {"rel": "被代入", "source": "- 3 ( { y } ^ { 2 } - 3 y ) + 42", "target": "30"}]}}
|
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{"content": "The equation $\\sqrt { x - 2 } = 1$ has a root of ____ ?", "answer": "3", "steps": "Solving the equation: $x ^ 2 - 4 x + 4 = x - 1$. Simplifying, we get: $x ^ 2 - 5 x + 5 = 0$. Using the quadratic formula, we get: $x = \\frac { 5 \\pm \\sqrt { 5 }} { 2 }$. However, since $\\frac { 5 - \\sqrt { 5 }} { 2 }$ is negative, it is extraneous. Therefore, the only solution is $x = \\frac { 5 + \\sqrt { 5 }} { 2 }$. To verify, we substitute $x = \\frac { 5 + \\sqrt { 5 }} { 2 }$ into the original equation and simplify to get both sides equal.", "expr_cands": ["\\sqrt { x - 2 } = 1", "x", "x - 2 = 3", "x = 5", "x = 3"], "exprs": ["x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x - 2 } = 1"}, {"id": "x = 3"}], "links": [{"rel": "等式方程求解", "source": "\\sqrt { x - 2 } = 1", "target": "x = 3"}]}}
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{"content": "The value of the algebraic expression $\\frac { 1 } { x + 2 }$ is $1$ less than the value of $\\frac { 1 + x } { x - 2 }$. Find the value of $x$.", "answer": "- 4", "steps": "According to the problem, we have $\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1$. Simplifying the equation by eliminating the denominators, we get $x - 2 = x ^ 2 + 3 x + 2 - x ^ 2 + 4$. Solving for $x$, we get $x = - 4$. After checking, we find that $x = - 4$ is indeed a solution to the fractional equation.", "expr_cands": ["\\frac { 1 } { x + 2 }", "x", "\\frac { 1 + x } { x - 2 }", "1", "\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1", "x = - 4", "x - 2 = x ^ { 2 } + 3 x + 2 - x ^ { 2 } + 4"], "exprs": ["\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1", "x = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 + x } { x - 2 }"}, {"id": "\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1"}, {"id": "\\frac { 1 } { x + 2 }"}, {"id": "1"}, {"id": "代数式 $\\frac { 1 } { x + 2 }$ 的值比 $\\frac { 1 + x } { x - 2 }$ 的值小 $1$"}, {"id": "x = - 4"}], "links": [{"rel": "被描述", "source": "\\frac { 1 + x } { x - 2 }", "target": "\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1"}, {"rel": "等式方程求解", "source": "\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1", "target": "x = - 4"}, {"rel": "被描述", "source": "\\frac { 1 } { x + 2 }", "target": "\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1"}, {"rel": "被描述", "source": "1", "target": "\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 1 } { x + 2 }$ 的值比 $\\frac { 1 + x } { x - 2 }$ 的值小 $1$", "target": "\\frac { 1 } { x + 2 } = \\frac { 1 + x } { x - 2 } - 1"}]}}
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{"content": "If $( x - 3 ) ^ { 2 } + | x - y + m | = 0$, when $y < 0$, then the range of $m$ is ____?", "answer": "m < - 3", "steps": "From the given information, we have $x - 3 = 0$ and $x - y + m = 0$. Solving for $x$, we get $x = 3$. Substituting this value of $x$ in the second equation, we get $y = m + 3$. Since $y$ is less than zero, we have $m + 3 < 0$, which implies $m < - 3$.", "expr_cands": ["( x - 3 ) ^ { 2 } + | x - y + m | = 0", "y", "x", "m", "y < 0", "x - 3 = 0", "x = 3", "x - y + m = 0", "y = m + 3", "m + 3 < 0", "m < - 3"], "exprs": ["x - 3 = 0", "x - y + m = 0", "x = 3", "y = m + 3", "m + 3 < 0", "m < - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x - 3 ) ^ { 2 } + | x - y + m | = 0"}, {"id": "x - 3 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "x = 3"}, {"id": "x - y + m = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "y = m + 3"}, {"id": "y < 0"}, {"id": "m + 3 < 0"}, {"id": "m < - 3"}], "links": [{"rel": "被描述", "source": "( x - 3 ) ^ { 2 } + | x - y + m | = 0", "target": "x - 3 = 0"}, {"rel": "被描述", "source": "( x - 3 ) ^ { 2 } + | x - y + m | = 0", "target": "x - y + m = 0"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "x - 3 = 0"}, {"rel": "联立", "source": "x = 3", "target": "y = m + 3"}, {"rel": "联立", "source": "x - y + m = 0", "target": "y = m + 3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x - y + m = 0"}, {"rel": "代入", "source": "y = m + 3", "target": "m + 3 < 0"}, {"rel": "被代入", "source": "y < 0", "target": "m + 3 < 0"}, {"rel": "不等式方程求解", "source": "m + 3 < 0", "target": "m < - 3"}]}}
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{"content": "If four distinct positive integers $m$, $n$, $p$, $q$ satisfy $( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4$, then the value of $m + n + p + q$ is ____?", "answer": "20", "steps": "$\\because$ Four distinct positive integers $m$, $n$, $p$, $q$ satisfy $( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4$, $\\therefore$ the possible solutions that satisfy the conditions are: $5 - m = 1$, $5 - n = - 1$, $5 - p = 2$, $5 - q = - 2$. Solving for $m$, $n$, $p$, $q$, we get $m = 4$, $n = 6$, $p = 3$, $q = 7$. Therefore, $m + n + p + q = 20$.", "expr_cands": ["m", "n", "p", "q", "( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4", "m + n + p + q", "5 - m = 1", "m = 4", "5 - n = - 1", "n = 6", "5 - p = 2", "p = 3", "5 - q = - 2", "q = 7", "20"], "exprs": ["5 - m = 1", "5 - n = - 1", "5 - p = 2", "5 - q = - 2", "m = 4", "n = 6", "p = 3", "q = 7", "20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4"}, {"id": "5 - m = 1"}, {"id": "四个互不相同的正整数 $m$ , $n$ , $p$ , $q$ 满足 $( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4$"}, {"id": "5 - n = - 1"}, {"id": "5 - p = 2"}, {"id": "5 - q = - 2"}, {"id": "m = 4"}, {"id": "n = 6"}, {"id": "p = 3"}, {"id": "q = 7"}, {"id": "m + n + p + q"}, {"id": "20"}], "links": [{"rel": "被描述", "source": "( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4", "target": "5 - m = 1"}, {"rel": "被描述", "source": "( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4", "target": "5 - n = - 1"}, {"rel": "被描述", "source": "( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4", "target": "5 - p = 2"}, {"rel": "被描述", "source": "( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4", "target": "5 - q = - 2"}, {"rel": "等式方程求解", "source": "5 - m = 1", "target": "m = 4"}, {"rel": "限制性描述", "source": "四个互不相同的正整数 $m$ , $n$ , $p$ , $q$ 满足 $( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4$", "target": "5 - m = 1"}, {"rel": "限制性描述", "source": "四个互不相同的正整数 $m$ , $n$ , $p$ , $q$ 满足 $( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4$", "target": "5 - n = - 1"}, {"rel": "限制性描述", "source": "四个互不相同的正整数 $m$ , $n$ , $p$ , $q$ 满足 $( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4$", "target": "5 - p = 2"}, {"rel": "限制性描述", "source": "四个互不相同的正整数 $m$ , $n$ , $p$ , $q$ 满足 $( 5 - m ) ( 5 - n ) ( 5 - p ) ( 5 - q ) = 4$", "target": "5 - q = - 2"}, {"rel": "等式方程求解", "source": "5 - n = - 1", "target": "n = 6"}, {"rel": "等式方程求解", "source": "5 - p = 2", "target": "p = 3"}, {"rel": "等式方程求解", "source": "5 - q = - 2", "target": "q = 7"}, {"rel": "代入", "source": "m = 4", "target": "20"}, {"rel": "代入", "source": "n = 6", "target": "20"}, {"rel": "代入", "source": "p = 3", "target": "20"}, {"rel": "代入", "source": "q = 7", "target": "20"}, {"rel": "被代入", "source": "m + n + p + q", "target": "20"}]}}
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{"content": "Given that $a$ is a root of the equation $x ^ 2 - 2 x - 8 = 0$ with respect to $x$, what is the value of $a ^ 2 - 2 a$?", "answer": "8", "steps": "Substituting $x = a$ into the equation $x ^ 2 - 2 x - 8 = 0$ yields $a ^ 2 - 2 a - 8 = 0$, so $a ^ 2 - 2 a = 8$.", "expr_cands": ["a", "x", "x ^ { 2 } - 2 x - 8 = 0", "a ^ { 2 } - 2 a", "x = a", "x = - 2", "x = 4", "a ^ { 2 } - 2 a - 8 = 0", "a = - 2", "a = 4", "a ^ { 2 } - 2 a = 8"], "exprs": ["x = a", "a ^ { 2 } - 2 a - 8 = 0", "a ^ { 2 } - 2 a = 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "x = a"}, {"id": "x"}, {"id": "x ^ { 2 } - 2 x - 8 = 0"}, {"id": "$a$ 是关于 $x$ 方程 $x ^ { 2 } - 2 x - 8 = 0$ 的一个根"}, {"id": "a ^ { 2 } - 2 a - 8 = 0"}, {"id": "a ^ { 2 } - 2 a = 8"}], "links": [{"rel": "被描述", "source": "a", "target": "x = a"}, {"rel": "代入", "source": "x = a", "target": "a ^ { 2 } - 2 a - 8 = 0"}, {"rel": "被描述", "source": "x", "target": "x = a"}, {"rel": "被描述", "source": "x ^ { 2 } - 2 x - 8 = 0", "target": "x = a"}, {"rel": "被代入", "source": "x ^ { 2 } - 2 x - 8 = 0", "target": "a ^ { 2 } - 2 a - 8 = 0"}, {"rel": "限制性描述", "source": "$a$ 是关于 $x$ 方程 $x ^ { 2 } - 2 x - 8 = 0$ 的一个根", "target": "x = a"}, {"rel": "移项", "source": "a ^ { 2 } - 2 a - 8 = 0", "target": "a ^ { 2 } - 2 a = 8"}]}}
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{"content": "The quadratic equation $ax ^ { 2 } + 2 x = 0$ has one root at $1$. What is the value of $a$?", "answer": "- 2", "steps": "$\\because$ One root of the quadratic equation $ax ^ 2 + 2 x = 0$ is $1$, $\\therefore$ $x = 1$ satisfies the quadratic equation $ax ^ 2 + 2 x = 0$ in terms of $x$, $\\therefore$ $a + 2 = 0$, solving for $a$, $a = - 2$.", "expr_cands": ["ax ^ { 2 } + 2 x = 0", "a", "x", "1", "x = 1", "a + 2 = 0", "a = - 2"], "exprs": ["x = 1", "a + 2 = 0", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1"}, {"id": "x = 1"}, {"id": "ax ^ { 2 } + 2 x = 0"}, {"id": "x"}, {"id": "一元二次方程 $ax ^ { 2 } + 2 x = 0$ 的一个根是 $1$"}, {"id": "a + 2 = 0"}, {"id": "a = - 2"}], "links": [{"rel": "被描述", "source": "1", "target": "x = 1"}, {"rel": "代入", "source": "x = 1", "target": "a + 2 = 0"}, {"rel": "被描述", "source": "ax ^ { 2 } + 2 x = 0", "target": "x = 1"}, {"rel": "被代入", "source": "ax ^ { 2 } + 2 x = 0", "target": "a + 2 = 0"}, {"rel": "被描述", "source": "x", "target": "x = 1"}, {"rel": "限制性描述", "source": "一元二次方程 $ax ^ { 2 } + 2 x = 0$ 的一个根是 $1$", "target": "x = 1"}, {"rel": "等式方程求解", "source": "a + 2 = 0", "target": "a = - 2"}]}}
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{"content": "Given that $a$ and $b$ are constants, if the polynomial $( a - 2 ) x ^ 2 + ( 2 b + 1 ) xy - x + y - 7$ in terms of $x$ and $y$ does not contain any quadratic terms, then the algebraic expression $a - 2 b$ is equal to ____?", "answer": "3", "steps": "From the given information, we have $a - 2 = 0$ and $2 b + 1 = 0$. Solving for $a$ and $b$, we get $a = 2$ and $b = - \\frac { 1 } { 2 }$. Therefore, $a - 2 b = 2 - 2 * ( - \\frac { 1 } { 2 }) = 3$.", "expr_cands": ["a", "b", "x", "y", "( a - 2 ) x ^ { 2 } + ( 2 b + 1 ) xy - x + y - 7", "a - 2 b", "a - 2 = 0", "a = 2", "2 b + 1 = 0", "b = - \\frac { 1 } { 2 }", "3"], "exprs": ["a - 2 = 0", "2 b + 1 = 0", "a = 2", "b = - \\frac { 1 } { 2 }", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 2 ) x ^ { 2 } + ( 2 b + 1 ) xy - x + y - 7"}, {"id": "a - 2 = 0"}, {"id": "关于 $x$ , $y$ 的多项式 $( a - 2 ) x ^ { 2 } + ( 2 b + 1 ) xy - x + y - 7$ 不含二次项"}, {"id": "2 b + 1 = 0"}, {"id": "a = 2"}, {"id": "b = - \\frac { 1 } { 2 }"}, {"id": "a - 2 b"}, {"id": "3"}], "links": [{"rel": "被描述", "source": "( a - 2 ) x ^ { 2 } + ( 2 b + 1 ) xy - x + y - 7", "target": "a - 2 = 0"}, {"rel": "被描述", "source": "( a - 2 ) x ^ { 2 } + ( 2 b + 1 ) xy - x + y - 7", "target": "2 b + 1 = 0"}, {"rel": "等式方程求解", "source": "a - 2 = 0", "target": "a = 2"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的多项式 $( a - 2 ) x ^ { 2 } + ( 2 b + 1 ) xy - x + y - 7$ 不含二次项", "target": "a - 2 = 0"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的多项式 $( a - 2 ) x ^ { 2 } + ( 2 b + 1 ) xy - x + y - 7$ 不含二次项", "target": "2 b + 1 = 0"}, {"rel": "等式方程求解", "source": "2 b + 1 = 0", "target": "b = - \\frac { 1 } { 2 }"}, {"rel": "代入", "source": "a = 2", "target": "3"}, {"rel": "代入", "source": "b = - \\frac { 1 } { 2 }", "target": "3"}, {"rel": "被代入", "source": "a - 2 b", "target": "3"}]}}
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{"content": "The equation ${ ( x - 1 ) } ^ { 0 } = 1$ holds true when ____ ?", "answer": "x \\neq 1", "steps": "From the given condition, we have $x - 1 \\neq 0$, which implies that $x \\neq 1$.", "expr_cands": ["{ ( x - 1 ) } ^ { 0 } = 1", "x - 1 \\neq 0", "x \\neq 1", "x"], "exprs": ["x - 1 \\neq 0", "x \\neq 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ ( x - 1 ) } ^ { 0 } = 1"}, {"id": "x - 1 \\neq 0"}, {"id": "式 ${ ( x - 1 ) } ^ { 0 } = 1$ 成立的条件"}, {"id": "多项式零次方项，若底数不为0，则恒等于1"}, {"id": "x \\neq 1"}], "links": [{"rel": "被描述", "source": "{ ( x - 1 ) } ^ { 0 } = 1", "target": "x - 1 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 1 \\neq 0", "target": "x \\neq 1"}, {"rel": "限制性描述", "source": "式 ${ ( x - 1 ) } ^ { 0 } = 1$ 成立的条件", "target": "x - 1 \\neq 0"}, {"rel": "属性描述", "source": "多项式零次方项，若底数不为0，则恒等于1", "target": "x - 1 \\neq 0"}]}}
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{"content": "Given $\\sqrt { a - 1 } + | b - 5 | = 0$, what is the value of $( a - b ) ^ 2$?", "answer": "16", "steps": "Since $\\sqrt { a - 1 } + | b - 5 | = 0$, it follows that $a - 1 = 0$ and $b - 5 = 0$. Solving for $a$ and $b$, we get $a = 1$ and $b = 5$. Therefore, $( a - b ) ^ { 2 } = ( 1 - 5 ) ^ { 2 } = ( - 4 ) ^ { 2 } = 16$.", "expr_cands": ["\\sqrt { a - 1 } + | b - 5 | = 0", "a", "b", "( a - b ) ^ { 2 }", "a - 1 = 0", "a = 1", "b - 5 = 0", "b = 5", "16"], "exprs": ["a - 1 = 0", "b - 5 = 0", "a = 1", "b = 5", "16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { a - 1 } + | b - 5 | = 0"}, {"id": "a - 1 = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "b - 5 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "a = 1"}, {"id": "b = 5"}, {"id": "( a - b ) ^ { 2 }"}, {"id": "16"}], "links": [{"rel": "被描述", "source": "\\sqrt { a - 1 } + | b - 5 | = 0", "target": "a - 1 = 0"}, {"rel": "被描述", "source": "\\sqrt { a - 1 } + | b - 5 | = 0", "target": "b - 5 = 0"}, {"rel": "等式方程求解", "source": "a - 1 = 0", "target": "a = 1"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "a - 1 = 0"}, {"rel": "等式方程求解", "source": "b - 5 = 0", "target": "b = 5"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b - 5 = 0"}, {"rel": "代入", "source": "a = 1", "target": "16"}, {"rel": "代入", "source": "b = 5", "target": "16"}, {"rel": "被代入", "source": "( a - b ) ^ { 2 }", "target": "16"}]}}
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{"content": "The maximum positive integer solution of the inequality $12 - 5 x \\geq 0$ in terms of $x$ is ____?", "answer": "x = 2", "steps": "Moving terms, we get: $- 5 x \\ge - 12$. Dividing both sides by $- 5$ and changing the direction of the inequality, we get: $x \\le \\frac { 12 } { 5 }$. Therefore, the largest positive integer solution to the inequality $12 - 5 x \\ge 0$ is $2$.", "expr_cands": ["x", "12 - 5 x \\ge 0", "- 5 x \\ge - 12", "x \\le \\frac { 12 } { 5 }", "1", "2"], "exprs": ["x \\le \\frac { 12 } { 5 }", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "12 - 5 x \\ge 0"}, {"id": "x \\le \\frac { 12 } { 5 }"}, {"id": "2"}, {"id": "关于 $x$ 的不等式 $12 - 5 x \\ge 0$ 的最大正整数解"}], "links": [{"rel": "不等式方程求解", "source": "12 - 5 x \\ge 0", "target": "x \\le \\frac { 12 } { 5 }"}, {"rel": "被描述", "source": "x \\le \\frac { 12 } { 5 }", "target": "2"}, {"rel": "限制性描述", "source": "关于 $x$ 的不等式 $12 - 5 x \\ge 0$ 的最大正整数解", "target": "2"}]}}
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{"content": "If $y = \\frac { \\sqrt { x - 4 } + \\sqrt { 4 - x } } { 2 } - 2$, then $( x + y ) ^ { y }$ = ____ ?", "answer": "\\frac { 1 } { 4 }", "steps": "Since $y = \\frac { \\sqrt { x - 4 } + \\sqrt { 4 - x }} { 2 } - 2$, it follows that $x - 4 \\geq 0$, $4 - x \\geq 0$, and thus $x = 4$. Therefore, $y = - 2$. Hence, $( x + y ) ^ y = ( 4 - 2 ) ^ { - 2 } = 2 ^ { - 2 } = \\frac { 1 } { 4 }$.", "expr_cands": ["y = \\frac { \\sqrt { x - 4 } + \\sqrt { 4 - x } } { 2 } - 2", "x", "y", "( x + y ) ^ { y }", "x - 4 \\ge 0", "4 \\le x", "4 - x \\ge 0", "x \\le 4", "x = 4", "y = - 2", "\\frac { 1 } { 4 }"], "exprs": ["x - 4 \\ge 0", "4 - x \\ge 0", "x = 4", "y = - 2", "\\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { \\sqrt { x - 4 } + \\sqrt { 4 - x } } { 2 } - 2"}, {"id": "x - 4 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "4 - x \\ge 0"}, {"id": "x = 4"}, {"id": "y = - 2"}, {"id": "( x + y ) ^ { y }"}, {"id": "\\frac { 1 } { 4 }"}], "links": [{"rel": "被描述", "source": "y = \\frac { \\sqrt { x - 4 } + \\sqrt { 4 - x } } { 2 } - 2", "target": "x - 4 \\ge 0"}, {"rel": "被描述", "source": "y = \\frac { \\sqrt { x - 4 } + \\sqrt { 4 - x } } { 2 } - 2", "target": "4 - x \\ge 0"}, {"rel": "被代入", "source": "y = \\frac { \\sqrt { x - 4 } + \\sqrt { 4 - x } } { 2 } - 2", "target": "y = - 2"}, {"rel": "联立", "source": "x - 4 \\ge 0", "target": "x = 4"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 4 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "4 - x \\ge 0"}, {"rel": "联立", "source": "4 - x \\ge 0", "target": "x = 4"}, {"rel": "代入", "source": "x = 4", "target": "y = - 2"}, {"rel": "代入", "source": "x = 4", "target": "\\frac { 1 } { 4 }"}, {"rel": "代入", "source": "y = - 2", "target": "\\frac { 1 } { 4 }"}, {"rel": "被代入", "source": "( x + y ) ^ { y }", "target": "\\frac { 1 } { 4 }"}]}}
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{"content": "Solve the equation $\\frac { x - 6 } { x - 5 } + 1 = \\frac { m } { x - 5 }$ (where $m$ is a constant) for $x$ and find the value of $m$ that produces an increase in the number of roots.", "answer": "- 1", "steps": "To eliminate the denominator, we get: $x - 6 + x - 5 = m$. Since the equation has an extraneous root, we have $x - 5 = 0$, which means $x = 5$. Substituting $x = 5$ into the polynomial equation, we get $m = - 1$.", "expr_cands": ["x", "\\frac { x - 6 } { x - 5 } + 1 = \\frac { m } { x - 5 }", "m", "x - 6 + x - 5 = m", "x - 5 = 0", "x = 5", "m = - 1"], "exprs": ["x - 6 + x - 5 = m", "x - 5 = 0", "x = 5", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 6 } { x - 5 } + 1 = \\frac { m } { x - 5 }"}, {"id": "x - 6 + x - 5 = m"}, {"id": "x - 5 = 0"}, {"id": "解关于 $x$ 的方程 $\\frac { x - 6 } { x - 5 } + 1 = \\frac { m } { x - 5 }$ ( 其中 $m$ 为常数 ) 产生增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 5"}, {"id": "m = - 1"}], "links": [{"rel": "同乘除", "source": "\\frac { x - 6 } { x - 5 } + 1 = \\frac { m } { x - 5 }", "target": "x - 6 + x - 5 = m"}, {"rel": "被描述", "source": "\\frac { x - 6 } { x - 5 } + 1 = \\frac { m } { x - 5 }", "target": "x - 5 = 0"}, {"rel": "被代入", "source": "x - 6 + x - 5 = m", "target": "m = - 1"}, {"rel": "等式方程求解", "source": "x - 5 = 0", "target": "x = 5"}, {"rel": "限制性描述", "source": "解关于 $x$ 的方程 $\\frac { x - 6 } { x - 5 } + 1 = \\frac { m } { x - 5 }$ ( 其中 $m$ 为常数 ) 产生增根", "target": "x - 5 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 5 = 0"}, {"rel": "代入", "source": "x = 5", "target": "m = - 1"}]}}
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{"content": "If $\\sqrt { 2 a - 2 }$ and $| b + 2 |$ are opposite in sign, then $a - b$ = ____ ?", "answer": "3", "steps": "Since $\\sqrt { 2 a - 2 }$ and $| b + 2 |$ are opposite in sign, it follows that $\\sqrt { 2 a - 2 } + | b + 2 | = 0$. Since $\\sqrt { 2 a - 2 } \\ge 0$ and $| b + 2 | \\ge 0$, we have $a = 1$ and $b = - 2$. Therefore, $a - b = 1 + 2 = 3$.", "expr_cands": ["\\sqrt { 2 a - 2 }", "a", "| b + 2 |", "b", "a - b", "\\sqrt { 2 a - 2 } + | b + 2 | = 0", "\\sqrt { 2 a - 2 } \\ge 0", "1 \\le a", "| b + 2 | \\ge 0", "a = 1", "b = - 2", "3"], "exprs": ["\\sqrt { 2 a - 2 } + | b + 2 | = 0", "a = 1", "b = - 2", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 2 a - 2 }"}, {"id": "\\sqrt { 2 a - 2 } + | b + 2 | = 0"}, {"id": "| b + 2 |"}, {"id": "$\\sqrt { 2 a - 2 }$ 与 $| b + 2 |$ 互为相反数"}, {"id": "a = 1"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "b = - 2"}, {"id": "绝对值恒大于等于0"}, {"id": "a - b"}, {"id": "3"}], "links": [{"rel": "被描述", "source": "\\sqrt { 2 a - 2 }", "target": "\\sqrt { 2 a - 2 } + | b + 2 | = 0"}, {"rel": "被描述", "source": "\\sqrt { 2 a - 2 } + | b + 2 | = 0", "target": "a = 1"}, {"rel": "被描述", "source": "\\sqrt { 2 a - 2 } + | b + 2 | = 0", "target": "b = - 2"}, {"rel": "被描述", "source": "| b + 2 |", "target": "\\sqrt { 2 a - 2 } + | b + 2 | = 0"}, {"rel": "限制性描述", "source": "$\\sqrt { 2 a - 2 }$ 与 $| b + 2 |$ 互为相反数", "target": "\\sqrt { 2 a - 2 } + | b + 2 | = 0"}, {"rel": "代入", "source": "a = 1", "target": "3"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "a = 1"}, {"rel": "代入", "source": "b = - 2", "target": "3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b = - 2"}, {"rel": "被代入", "source": "a - b", "target": "3"}]}}
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{"content": "If $\\sqrt { x - 2 }$ is defined, then the smallest integer that $x$ can take is ____ ?", "answer": "2", "steps": "From the given condition, we have $x - 2 \\ge 0$. Solving for $x$, we get $x \\ge 2$. Therefore, the smallest integer that $x$ can take is $2$.", "expr_cands": ["\\sqrt { x - 2 }", "x", "x - 2 \\ge 0", "2 \\le x", "x \\ge 2", "2"], "exprs": ["x - 2 \\ge 0", "x \\ge 2", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x - 2 }"}, {"id": "x - 2 \\ge 0"}, {"id": "$\\sqrt { x - 2 }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 2"}, {"id": "2"}, {"id": "$x$ 可以取的最小整数"}], "links": [{"rel": "被描述", "source": "\\sqrt { x - 2 }", "target": "x - 2 \\ge 0"}, {"rel": "不等式方程求解", "source": "x - 2 \\ge 0", "target": "x \\ge 2"}, {"rel": "限制性描述", "source": "$\\sqrt { x - 2 }$ 有意义", "target": "x - 2 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 2 \\ge 0"}, {"rel": "被描述", "source": "x \\ge 2", "target": "2"}, {"rel": "限制性描述", "source": "$x$ 可以取的最小整数", "target": "2"}]}}
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{"content": "The polynomial that is less than $2 a ^ 2 - 3 a - 7$ by $3 - 2 a ^ 2$ is ____?", "answer": "4 a ^ { 2 } - 3 a - 10", "steps": "The polynomial that is less by $3 - 2 a ^ 2$ than $2 a ^ 2 - 3 a - 7$ is: $2 a ^ 2 - 3 a - 7 - ( 3 - 2 a ^ 2 ) = 4 a ^ 2 - 3 a - 10$.", "expr_cands": ["2 a ^ { 2 } - 3 a - 7", "a", "3 - 2 a ^ { 2 }", "2 a ^ { 2 } - 3 a - 7 - ( 3 - 2 a ^ { 2 } )", "4 a ^ { 2 } - 3 a - 10"], "exprs": ["2 a ^ { 2 } - 3 a - 7 - ( 3 - 2 a ^ { 2 } )", "4 a ^ { 2 } - 3 a - 10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a ^ { 2 } - 3 a - 7"}, {"id": "2 a ^ { 2 } - 3 a - 7 - ( 3 - 2 a ^ { 2 } )"}, {"id": "3 - 2 a ^ { 2 }"}, {"id": "比 $2 a ^ { 2 } - 3 a - 7$ 少 $3 - 2 a ^ { 2 }$ 的多项式"}, {"id": "4 a ^ { 2 } - 3 a - 10"}], "links": [{"rel": "被描述", "source": "2 a ^ { 2 } - 3 a - 7", "target": "2 a ^ { 2 } - 3 a - 7 - ( 3 - 2 a ^ { 2 } )"}, {"rel": "计算", "source": "2 a ^ { 2 } - 3 a - 7 - ( 3 - 2 a ^ { 2 } )", "target": "4 a ^ { 2 } - 3 a - 10"}, {"rel": "被描述", "source": "3 - 2 a ^ { 2 }", "target": "2 a ^ { 2 } - 3 a - 7 - ( 3 - 2 a ^ { 2 } )"}, {"rel": "限制性描述", "source": "比 $2 a ^ { 2 } - 3 a - 7$ 少 $3 - 2 a ^ { 2 }$ 的多项式", "target": "2 a ^ { 2 } - 3 a - 7 - ( 3 - 2 a ^ { 2 } )"}]}}
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{"content": "If the equation $kx ^ { 2 } + 2 x = 2 x ^ { 2 } + 1$ is a quadratic equation in $x$, then the condition that the real number $k$ should satisfy is ____?", "answer": "k \\neq 2", "steps": "From the original equation, we have $( k - 2 ) x ^ 2 + 2 x - 1 = 0$. Since the equation $kx ^ 2 + 2 x = 2 x ^ 2 + 1$ is a quadratic equation in $x$, we have $k - 2 \\neq 0$, which means $k \\neq 2$.", "expr_cands": ["kx ^ { 2 } + 2 x = 2 x ^ { 2 } + 1", "k", "x", "( k - 2 ) x ^ { 2 } + 2 x - 1 = 0", "k - 2 \\neq 0", "k \\neq 2"], "exprs": ["( k - 2 ) x ^ { 2 } + 2 x - 1 = 0", "k - 2 \\neq 0", "k \\neq 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "kx ^ { 2 } + 2 x = 2 x ^ { 2 } + 1"}, {"id": "( k - 2 ) x ^ { 2 } + 2 x - 1 = 0"}, {"id": "k - 2 \\neq 0"}, {"id": "方程 $kx ^ { 2 } + 2 x = 2 x ^ { 2 } + 1$ 是关于 $x$ 的一元二次方程"}, {"id": "k \\neq 2"}], "links": [{"rel": "移项", "source": "kx ^ { 2 } + 2 x = 2 x ^ { 2 } + 1", "target": "( k - 2 ) x ^ { 2 } + 2 x - 1 = 0"}, {"rel": "被描述", "source": "( k - 2 ) x ^ { 2 } + 2 x - 1 = 0", "target": "k - 2 \\neq 0"}, {"rel": "不等式方程求解", "source": "k - 2 \\neq 0", "target": "k \\neq 2"}, {"rel": "限制性描述", "source": "方程 $kx ^ { 2 } + 2 x = 2 x ^ { 2 } + 1$ 是关于 $x$ 的一元二次方程", "target": "k - 2 \\neq 0"}]}}
|
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{"content": "$16$. If $| a - 1 | + | b + 3 | = 0$, then $a + b$ = ____?", "answer": "- 2", "steps": "$\\because$ $| a - 1 | + | b + 3 | = 0$ , $\\therefore$ $a - 1 = 0$ , $b + 3 = 0$ , which yields $a = 1$ , $b = - 3$ , $\\therefore$ $a + b = 1 + ( - 3 ) = - 2$ .", "expr_cands": ["16", "| a - 1 | + | b + 3 | = 0", "b", "a", "a + b", "a - 1 = 0", "a = 1", "b + 3 = 0", "b = - 3", "- 2"], "exprs": ["a - 1 = 0", "b + 3 = 0", "a = 1", "b = - 3", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 1 | + | b + 3 | = 0"}, {"id": "a - 1 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "b + 3 = 0"}, {"id": "a = 1"}, {"id": "b = - 3"}, {"id": "a + b"}, {"id": "- 2"}], "links": [{"rel": "被描述", "source": "| a - 1 | + | b + 3 | = 0", "target": "a - 1 = 0"}, {"rel": "被描述", "source": "| a - 1 | + | b + 3 | = 0", "target": "b + 3 = 0"}, {"rel": "等式方程求解", "source": "a - 1 = 0", "target": "a = 1"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a - 1 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b + 3 = 0"}, {"rel": "等式方程求解", "source": "b + 3 = 0", "target": "b = - 3"}, {"rel": "代入", "source": "a = 1", "target": "- 2"}, {"rel": "代入", "source": "b = - 3", "target": "- 2"}, {"rel": "被代入", "source": "a + b", "target": "- 2"}]}}
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{"content": "If $a + b = 2$, then the algebraic expression $3 - 2 a - 2 b$ = ____ ?", "answer": "- 1", "steps": "Since $a + b = 2$, therefore $3 - 2 a - 2 b = 3 - 2 ( a + b ) = 3 - 2 * 2 = 3 - 4 = - 1$.", "expr_cands": ["a + b = 2", "a", "b", "3 - 2 a - 2 b", "3 - 2 ( a + b )", "- 1"], "exprs": ["3 - 2 ( a + b )", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 - 2 a - 2 b"}, {"id": "3 - 2 ( a + b )"}, {"id": "a + b = 2"}, {"id": "- 1"}], "links": [{"rel": "提取因式", "source": "3 - 2 a - 2 b", "target": "3 - 2 ( a + b )"}, {"rel": "被代入", "source": "3 - 2 ( a + b )", "target": "- 1"}, {"rel": "提取因式参考", "source": "a + b = 2", "target": "3 - 2 ( a + b )"}, {"rel": "代入", "source": "a + b = 2", "target": "- 1"}]}}
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{"content": "Given $y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3$, the arithmetic square root of $xy$ is ____?", "answer": "\\sqrt { 6 }", "steps": "Since $y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3$, therefore $x = 2$, $y = 3$, then $xy = 6$, so the arithmetic square root of $xy$ is $\\sqrt { 6 }$.", "expr_cands": ["y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3", "y", "x", "xy", "x = 2", "y = 3", "6", "\\sqrt { 6 }"], "exprs": ["x = 2", "y = 3", "6", "\\sqrt { 6 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3"}, {"id": "x = 2"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "y = 3"}, {"id": "xy"}, {"id": "6"}, {"id": "\\sqrt { 6 }"}, {"id": "故 $xy$ 的算术平方根是 : $\\sqrt { 6 }$"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3", "target": "x = 2"}, {"rel": "被代入", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3", "target": "y = 3"}, {"rel": "代入", "source": "x = 2", "target": "y = 3"}, {"rel": "代入", "source": "x = 2", "target": "6"}, {"rel": "限制性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x = 2"}, {"rel": "代入", "source": "y = 3", "target": "6"}, {"rel": "被代入", "source": "xy", "target": "6"}, {"rel": "被描述", "source": "6", "target": "\\sqrt { 6 }"}, {"rel": "限制性描述", "source": "故 $xy$ 的算术平方根是 : $\\sqrt { 6 }$", "target": "\\sqrt { 6 }"}]}}
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{"content": "If the equations $x ^ { 2 } + k ^ { 2 } - 16 = 0$ and $x ^ { 2 } - 3 k + 12 = 0$ have the same real root for $x$, then the value of $k$ is ____?", "answer": "4", "steps": "Because the equations $x ^ 2 + k ^ 2 - 16 = 0$ and $x ^ 2 - 3 k + 12 = 0$ have the same real root for $x$, we have $x ^ 2 + k ^ 2 - 16 = x ^ 2 - 3 k + 12$, which gives $k ^ 2 + 3 k - 28 = 0$. Solving this equation, we get $k = 4$ or $- 7$. By substituting $4$ and $- 7$ into the original equations, we find that when $k = - 7$, the equation $x ^ 2 - 3 k + 12 = 0$ has no solution. Therefore, $k = 4$.", "expr_cands": ["x", "x ^ { 2 } + k ^ { 2 } - 16 = 0", "k", "x ^ { 2 } - 3 k + 12 = 0", "x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12", "k ^ { 2 } + 3 k - 28 = 0", "k = - 7", "k = 4", "- 7", "4"], "exprs": ["x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12", "k ^ { 2 } + 3 k - 28 = 0", "k = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + k ^ { 2 } - 16 = 0"}, {"id": "x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12"}, {"id": "x ^ { 2 } - 3 k + 12 = 0"}, {"id": "关于 $x$ 的方程 $x ^ { 2 } + k ^ { 2 } - 16 = 0$ 和 $x ^ { 2 } - 3 k + 12 = 0$ 有相同的实数根"}, {"id": "k ^ { 2 } + 3 k - 28 = 0"}, {"id": "k = 4"}, {"id": "$x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12$ 可得 $k ^ { 2 } + 3 k - 28 = 0$ 解之得 $k = 4$ 或 $- 7$"}, {"id": "当 $k = - 7$ 时"}, {"id": "方程 $x ^ { 2 } - 3 k + 12 = 0$ 无解"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + k ^ { 2 } - 16 = 0", "target": "x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12"}, {"rel": "移项", "source": "x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12", "target": "k ^ { 2 } + 3 k - 28 = 0"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 k + 12 = 0", "target": "x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x ^ { 2 } + k ^ { 2 } - 16 = 0$ 和 $x ^ { 2 } - 3 k + 12 = 0$ 有相同的实数根", "target": "x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12"}, {"rel": "被描述", "source": "k ^ { 2 } + 3 k - 28 = 0", "target": "k = 4"}, {"rel": "限制性描述", "source": "$x ^ { 2 } + k ^ { 2 } - 16 = x ^ { 2 } - 3 k + 12$ 可得 $k ^ { 2 } + 3 k - 28 = 0$ 解之得 $k = 4$ 或 $- 7$", "target": "k = 4"}, {"rel": "限制性描述", "source": "当 $k = - 7$ 时", "target": "k = 4"}, {"rel": "限制性描述", "source": "方程 $x ^ { 2 } - 3 k + 12 = 0$ 无解", "target": "k = 4"}]}}
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{"content": "If $\\frac { 1 } { x } = \\frac { 3 } { y }$, then the value of $( \\frac { y } { x } - \\frac { x } { y }) \\div \\frac { 2 ( x - y ) ^ 2 } { xy - y ^ 2 }$ is ____?", "answer": "- 2", "steps": "$\\because \\frac { 1 } { x } = \\frac { 3 } { y }$, $\\therefore y = 3 x$. $\\because$ the original expression $= \\frac { - ( x - y ) ( x + y )} { xy } \\div \\frac { 2 ( x - y ) ^ 2 } {( x - y ) y } = \\frac { - ( x - y ) ( x + y )} { xy } \\div \\frac { 2 ( x - y )} { y } = - \\frac {( x - y ) ( x + y )} { xy } \\times \\frac { y } { 2 ( x - y )} = - \\frac { x + y } { 2 x }$, when $y = 3 x$, the original expression $= - \\frac { x + 3 x } { 2 x } = - 2$.", "expr_cands": ["\\frac { 1 } { x } = \\frac { 3 } { y }", "x", "y", "( \\frac { y } { x } - \\frac { x } { y } ) \\div \\frac { 2 ( x - y ) ^ { 2 } } { xy - y ^ { 2 } }", "y = 3 x", "y = y", "\\frac { - ( x - y ) ( x + y ) } { xy } \\div \\frac { 2 ( x - y ) ^ { 2 } } { ( x - y ) y }", "- \\frac { x + y } { 2 x }", "3 x", "- \\frac { x + 3 x } { 2 x }", "- 2"], "exprs": ["y = 3 x", "- \\frac { x + y } { 2 x }", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x } = \\frac { 3 } { y }"}, {"id": "y = 3 x"}, {"id": "( \\frac { y } { x } - \\frac { x } { y } ) \\div \\frac { 2 ( x - y ) ^ { 2 } } { xy - y ^ { 2 } }"}, {"id": "- \\frac { x + y } { 2 x }"}, {"id": "- 2"}], "links": [{"rel": "同乘除", "source": "\\frac { 1 } { x } = \\frac { 3 } { y }", "target": "y = 3 x"}, {"rel": "代入", "source": "y = 3 x", "target": "- 2"}, {"rel": "计算", "source": "( \\frac { y } { x } - \\frac { x } { y } ) \\div \\frac { 2 ( x - y ) ^ { 2 } } { xy - y ^ { 2 } }", "target": "- \\frac { x + y } { 2 x }"}, {"rel": "被代入", "source": "- \\frac { x + y } { 2 x }", "target": "- 2"}]}}
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{"content": "If the expression $3 x - 2$ is equal to $\\frac { x + 1 } { 2 }$, then the value of $x$ is ____?", "answer": "1", "steps": "From the given problem, we have $3 x - 2 = \\frac { x + 1 } { 2 }$. Multiplying both sides by $2$ to eliminate the denominator, we get $2 ( 3 x - 2 ) = x + 1$. Expanding the left side, we have $6 x - 4 = x + 1$. Moving all the terms with $x$ to one side, we get $6 x - x = 1 + 4$. Combining like terms, we get $5 x = 5$. Dividing both sides by $5$, we get $x = 1$.", "expr_cands": ["3 x - 2", "x", "\\frac { x + 1 } { 2 }", "3 x - 2 = \\frac { x + 1 } { 2 }", "x = 1", "2 ( 3 x - 2 ) = x + 1", "6 x - 4 = x + 1", "6 x - x = 1 + 4", "5 x = 5", "1"], "exprs": ["3 x - 2 = \\frac { x + 1 } { 2 }", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 2"}, {"id": "3 x - 2 = \\frac { x + 1 } { 2 }"}, {"id": "\\frac { x + 1 } { 2 }"}, {"id": "式子 $3 x - 2$ 与 $\\frac { x + 1 } { 2 }$ 的值相"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "3 x - 2", "target": "3 x - 2 = \\frac { x + 1 } { 2 }"}, {"rel": "等式方程求解", "source": "3 x - 2 = \\frac { x + 1 } { 2 }", "target": "x = 1"}, {"rel": "被描述", "source": "\\frac { x + 1 } { 2 }", "target": "3 x - 2 = \\frac { x + 1 } { 2 }"}, {"rel": "限制性描述", "source": "式子 $3 x - 2$ 与 $\\frac { x + 1 } { 2 }$ 的值相", "target": "3 x - 2 = \\frac { x + 1 } { 2 }"}]}}
|
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{"content": "Given that $8$ is a root of the quadratic equation $x ^ 2 + kx + 16 = 0$ in terms of $x$, the value of $k$ is ____?", "answer": "- 10", "steps": "Substituting $8$ directly into the equation yields ${ 8 } ^ 2 + 8 k + 16 = 0$, which gives the solution $k = - 10$.", "expr_cands": ["8", "x", "{ x } ^ { 2 } + kx + 16 = 0", "k", "{ 8 } ^ { 2 } + 8 k + 16 = 0", "k = - 10"], "exprs": ["{ 8 } ^ { 2 } + 8 k + 16 = 0", "k = - 10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { 2 } + kx + 16 = 0"}, {"id": "{ 8 } ^ { 2 } + 8 k + 16 = 0"}, {"id": "8"}, {"id": "x"}, {"id": "$8$ 是关于 $x$ 的一元二次方程 ${ x } ^ { 2 } + kx + 16 = 0$ 的一个根"}, {"id": "k = - 10"}], "links": [{"rel": "被描述", "source": "{ x } ^ { 2 } + kx + 16 = 0", "target": "{ 8 } ^ { 2 } + 8 k + 16 = 0"}, {"rel": "等式方程求解", "source": "{ 8 } ^ { 2 } + 8 k + 16 = 0", "target": "k = - 10"}, {"rel": "被描述", "source": "8", "target": "{ 8 } ^ { 2 } + 8 k + 16 = 0"}, {"rel": "被描述", "source": "x", "target": "{ 8 } ^ { 2 } + 8 k + 16 = 0"}, {"rel": "限制性描述", "source": "$8$ 是关于 $x$ 的一元二次方程 ${ x } ^ { 2 } + kx + 16 = 0$ 的一个根", "target": "{ 8 } ^ { 2 } + 8 k + 16 = 0"}]}}
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{"content": "If the fraction $\\frac { a + 3 } { a - 2 }$ is meaningful in the real number range, then the range of values for $a$ is ____?", "answer": "a \\neq 2", "steps": "The fraction $\\frac { a + 3 } { a - 2 }$ is meaningful in the real number range, so $a - 2 \\neq 0$. Solving for $a$, we get $a \\neq 2$.", "expr_cands": ["\\frac { a + 3 } { a - 2 }", "a", "a - 2 \\neq 0", "a \\neq 2"], "exprs": ["a - 2 \\neq 0", "a \\neq 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a + 3 } { a - 2 }"}, {"id": "a - 2 \\neq 0"}, {"id": "分式 $\\frac { a + 3 } { a - 2 }$ 在实数范围内有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "a \\neq 2"}], "links": [{"rel": "被描述", "source": "\\frac { a + 3 } { a - 2 }", "target": "a - 2 \\neq 0"}, {"rel": "不等式方程求解", "source": "a - 2 \\neq 0", "target": "a \\neq 2"}, {"rel": "限制性描述", "source": "分式 $\\frac { a + 3 } { a - 2 }$ 在实数范围内有意义", "target": "a - 2 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "a - 2 \\neq 0"}]}}
|
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{"content": "If $x < 0$, then the result of $\\frac { \\sqrt { x ^ 2 }} { x }$ is ____?", "answer": "- 1", "steps": "Because $x < 0$, therefore $\\frac { \\sqrt { x ^ 2 }} { x } = \\frac { - x } { x } = - 1$.", "expr_cands": ["x < 0", "x", "\\frac { \\sqrt { x ^ { 2 } } } { x }", "\\frac { - x } { x }", "- 1"], "exprs": ["- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x < 0"}, {"id": "- 1"}, {"id": "\\frac { \\sqrt { x ^ { 2 } } } { x }"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "分式有意义，则分母不为0"}, {"id": "$\\frac { \\sqrt { x ^ { 2 } } } { x }$ 的结果"}], "links": [{"rel": "被描述", "source": "x < 0", "target": "- 1"}, {"rel": "被描述", "source": "\\frac { \\sqrt { x ^ { 2 } } } { x }", "target": "- 1"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "- 1"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "- 1"}, {"rel": "限制性描述", "source": "$\\frac { \\sqrt { x ^ { 2 } } } { x }$ 的结果", "target": "- 1"}]}}
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{"content": "If $\\sqrt { x ^ 2 } = - x$, what is the condition for this statement to be true?", "answer": "x \\le 0", "steps": "$\\because \\sqrt { x ^ { 2 } } = - x$ , $\\therefore$ $- x \\ge 0$ , which means $x \\le 0$ , $\\therefore$ $x$ must be a non-positive number.", "expr_cands": ["\\sqrt { x ^ { 2 } } = - x", "x", "- x \\ge 0", "x \\le 0"], "exprs": ["- x \\ge 0", "x \\le 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x ^ { 2 } } = - x"}, {"id": "- x \\ge 0"}, {"id": "$\\sqrt { x ^ { 2 } } = - x$ 成立的条件"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\le 0"}], "links": [{"rel": "被描述", "source": "\\sqrt { x ^ { 2 } } = - x", "target": "- x \\ge 0"}, {"rel": "不等式方程求解", "source": "- x \\ge 0", "target": "x \\le 0"}, {"rel": "限制性描述", "source": "$\\sqrt { x ^ { 2 } } = - x$ 成立的条件", "target": "- x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "- x \\ge 0"}]}}
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{"content": "If the degree of the monomial $- 4 xyz ^ 2$ is the same as the degree of the polynomial $2 xy - 3 xy ^ 2 + 4 x ^ my ^ 2$, then $m$ = ____?", "answer": "2", "steps": "$\\because$ The degree of the monomial $- 4 xyz ^ 2$ is the same as the degree of the polynomial $2 xy - 3 xy ^ 2 + 4 x ^ my ^ 2$, $\\therefore$ $2 + 1 + 1 = m + 2$, solving for $m$, we get $m = 2$.", "expr_cands": ["- 4 xyz ^ { 2 }", "y", "z", "x", "2 xy - 3 xy ^ { 2 } + 4 x ^ { m } y ^ { 2 }", "m", "2 + 1 + 1 = m + 2", "m = 2"], "exprs": ["2 + 1 + 1 = m + 2", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 4 xyz ^ { 2 }"}, {"id": "2 + 1 + 1 = m + 2"}, {"id": "2 xy - 3 xy ^ { 2 } + 4 x ^ { m } y ^ { 2 }"}, {"id": "单项式 $- 4 xyz ^ { 2 }$ 的次数与多项式 $2 xy - 3 xy ^ { 2 } + 4 x ^ { m } y ^ { 2 }$ 的次数相同"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "- 4 xyz ^ { 2 }", "target": "2 + 1 + 1 = m + 2"}, {"rel": "等式方程求解", "source": "2 + 1 + 1 = m + 2", "target": "m = 2"}, {"rel": "被描述", "source": "2 xy - 3 xy ^ { 2 } + 4 x ^ { m } y ^ { 2 }", "target": "2 + 1 + 1 = m + 2"}, {"rel": "限制性描述", "source": "单项式 $- 4 xyz ^ { 2 }$ 的次数与多项式 $2 xy - 3 xy ^ { 2 } + 4 x ^ { m } y ^ { 2 }$ 的次数相同", "target": "2 + 1 + 1 = m + 2"}]}}
|
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{"content": "If $a = - 1$, then the value of $- a + 1$ is ____?", "answer": "2", "steps": "When $a = - 1$, the original expression equals $1 + 1 = 2$.", "expr_cands": ["a = - 1", "a", "- a + 1", "1 + 1", "2"], "exprs": ["1 + 1", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a = - 1"}, {"id": "1 + 1"}, {"id": "- a + 1"}, {"id": "2"}], "links": [{"rel": "代入", "source": "a = - 1", "target": "1 + 1"}, {"rel": "计算", "source": "1 + 1", "target": "2"}, {"rel": "被代入", "source": "- a + 1", "target": "1 + 1"}]}}
|
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{"content": "If $a$, $b$ are opposite numbers, and $m$, $n$ are reciprocal, then $a + b + mn ^ { 2 } - ( n + 2 )$ = ____?", "answer": "- 2", "steps": "Since $a$ and $b$ are opposite numbers, and $m$ and $n$ are reciprocal, therefore $a + b = 0$, $mn = 1$. Therefore, $a + b + mn ^ { 2 } - ( n + 2 ) = 0 + mn \\times n - n - 2 = 0 + 1 \\times n - n - 2 = 0 + n - n - 2 = - 2$.", "expr_cands": ["a", "b", "m", "n", "a + b + mn ^ { 2 } - ( n + 2 )", "a + b = 0", "mn = 1", "- 2"], "exprs": ["a + b = 0", "mn = 1", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "m"}, {"id": "mn = 1"}, {"id": "n"}, {"id": "$m$ , $n$ 互为倒数"}, {"id": "a + b + mn ^ { 2 } - ( n + 2 )"}, {"id": "- 2"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "- 2"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "m", "target": "mn = 1"}, {"rel": "代入", "source": "mn = 1", "target": "- 2"}, {"rel": "被描述", "source": "n", "target": "mn = 1"}, {"rel": "限制性描述", "source": "$m$ , $n$ 互为倒数", "target": "mn = 1"}, {"rel": "被代入", "source": "a + b + mn ^ { 2 } - ( n + 2 )", "target": "- 2"}]}}
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{"content": "If $\\frac { y } { x + y } = \\frac { 1 } { 2 }$, then $\\frac { x } { y }$ = ____?", "answer": "1", "steps": "$\\because \\frac { y } { x + y } = \\frac { 1 } { 2 }$, $\\therefore 2 y = x + y$, so $y = x$. Thus, $\\frac { x } { y } = 1$.", "expr_cands": ["\\frac { y } { x + y } = \\frac { 1 } { 2 }", "y", "x", "\\frac { x } { y }", "2 y = x + y", "y = x", "1"], "exprs": ["y = x", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { y } { x + y } = \\frac { 1 } { 2 }"}, {"id": "y = x"}, {"id": "\\frac { x } { y }"}, {"id": "1"}], "links": [{"rel": "等式方程部分求解", "source": "\\frac { y } { x + y } = \\frac { 1 } { 2 }", "target": "y = x"}, {"rel": "代入", "source": "y = x", "target": "1"}, {"rel": "被代入", "source": "\\frac { x } { y }", "target": "1"}]}}
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{"content": "If $x = 2$ is a solution of the equation $x ^ 2 + mx + 2 = 0$ with respect to $x$, then the value of $m$ is ____?", "answer": "- 3", "steps": "Since $x = 2$ is a solution of the quadratic equation $x ^ 2 + mx + 2 = 0$, we have $4 + 2 m + 2 = 0$. Therefore, $m = - 3$.", "expr_cands": ["x = 2", "x", "x ^ { 2 } + mx + 2 = 0", "m", "2 m + 6 = 0", "4 + 2 m + 2 = 0", "m = - 3"], "exprs": ["4 + 2 m + 2 = 0", "m = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + mx + 2 = 0"}, {"id": "4 + 2 m + 2 = 0"}, {"id": "x = 2"}, {"id": "m = - 3"}], "links": [{"rel": "被代入", "source": "x ^ { 2 } + mx + 2 = 0", "target": "4 + 2 m + 2 = 0"}, {"rel": "等式方程求解", "source": "4 + 2 m + 2 = 0", "target": "m = - 3"}, {"rel": "代入", "source": "x = 2", "target": "4 + 2 m + 2 = 0"}]}}
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{"content": "Given: $y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }$ is an inverse proportion function, then $m$ = ____ ?", "answer": "- 2", "steps": "Because $y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }$ is an inverse proportion function, the exponent of $x$ is $m ^ { 2 } - 5 = - 1$, which means $m ^ { 2 } = 4$. Solving for $m$, we get $m = 2$ or $- 2$. Since $m - 2 \\neq 0$, we have $m \\neq 2$, which implies $m = - 2$.", "expr_cands": ["y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }", "m", "y", "x", "m ^ { 2 } - 5 = - 1", "m = - 2", "m = 2", "m ^ { 2 } = 4", "- 2", "m - 2 \\neq 0", "m \\neq 2"], "exprs": ["m ^ { 2 } - 5 = - 1", "m - 2 \\neq 0", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }"}, {"id": "m ^ { 2 } - 5 = - 1"}, {"id": "因为 $y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }$ 是反比例函数"}, {"id": "m - 2 \\neq 0"}, {"id": "m = - 2"}], "links": [{"rel": "被描述", "source": "y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }", "target": "m ^ { 2 } - 5 = - 1"}, {"rel": "被描述", "source": "y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "m ^ { 2 } - 5 = - 1", "target": "m = - 2"}, {"rel": "限制性描述", "source": "因为 $y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }$ 是反比例函数", "target": "m ^ { 2 } - 5 = - 1"}, {"rel": "限制性描述", "source": "因为 $y = ( m - 2 ) x ^ { m ^ { 2 } - 5 }$ 是反比例函数", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "m - 2 \\neq 0", "target": "m = - 2"}]}}
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{"content": "If $| a - 4 | + | b - 5 | = 0$, then $a + b$ = ____?", "answer": "9", "steps": "According to the problem, we have $a - 4 = 0$ and $b - 5 = 0$, so $a = 4$ and $b = 5$. Therefore, $a + b = 4 + 5 = 9$.", "expr_cands": ["| a - 4 | + | b - 5 | = 0", "b", "a", "a + b", "a - 4 = 0", "a = 4", "b - 5 = 0", "b = 5", "9"], "exprs": ["a - 4 = 0", "b - 5 = 0", "a = 4", "b = 5", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 4 | + | b - 5 | = 0"}, {"id": "a - 4 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "b - 5 = 0"}, {"id": "a = 4"}, {"id": "b = 5"}, {"id": "a + b"}, {"id": "9"}], "links": [{"rel": "被描述", "source": "| a - 4 | + | b - 5 | = 0", "target": "a - 4 = 0"}, {"rel": "被描述", "source": "| a - 4 | + | b - 5 | = 0", "target": "b - 5 = 0"}, {"rel": "等式方程求解", "source": "a - 4 = 0", "target": "a = 4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a - 4 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b - 5 = 0"}, {"rel": "等式方程求解", "source": "b - 5 = 0", "target": "b = 5"}, {"rel": "代入", "source": "a = 4", "target": "9"}, {"rel": "代入", "source": "b = 5", "target": "9"}, {"rel": "被代入", "source": "a + b", "target": "9"}]}}
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{"content": "$( 2 m - 4 ) { x } ^ 2 + 3 mx + { m } ^ 2 - 4 = 0$ is a quadratic equation in $x$. If it has a root of $0$, then we can substitute $x = 0$ into the equation and get:$$(2m-4)(0)^2+3m(0)+m^2-4=0$$Simplifying this gives:$$m^2-4=0$$Which can be factored as:$$(m+2)(m-2)=0$$Therefore, $m$ must be either $- 2$ or $2$.", "answer": "- 2", "steps": "According to the problem, we have $m ^ 2 - 4 = 0$ and $2 m - 4 \\neq 0$. Solving for $m$, we get $m = - 2$.", "expr_cands": ["( 2 m - 4 ) { x } ^ { 2 } + 3 mx + { m } ^ { 2 } - 4 = 0", "x", "m", "0", "m ^ { 2 } - 4 = 0", "m = - 2", "m = 2", "2 m - 4 \\neq 0", "m \\neq 2"], "exprs": ["m ^ { 2 } - 4 = 0", "2 m - 4 \\neq 0", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 2 m - 4 ) { x } ^ { 2 } + 3 mx + { m } ^ { 2 } - 4 = 0"}, {"id": "m ^ { 2 } - 4 = 0"}, {"id": "x"}, {"id": "0"}, {"id": "$( 2 m - 4 ) { x } ^ { 2 } + 3 mx + { m } ^ { 2 } - 4 = 0$ 关于 $x$ 的一元二次方程有一根为 $0$"}, {"id": "2 m - 4 \\neq 0"}, {"id": "m = - 2"}], "links": [{"rel": "被描述", "source": "( 2 m - 4 ) { x } ^ { 2 } + 3 mx + { m } ^ { 2 } - 4 = 0", "target": "m ^ { 2 } - 4 = 0"}, {"rel": "被描述", "source": "( 2 m - 4 ) { x } ^ { 2 } + 3 mx + { m } ^ { 2 } - 4 = 0", "target": "2 m - 4 \\neq 0"}, {"rel": "联立", "source": "m ^ { 2 } - 4 = 0", "target": "m = - 2"}, {"rel": "被描述", "source": "x", "target": "m ^ { 2 } - 4 = 0"}, {"rel": "被描述", "source": "0", "target": "m ^ { 2 } - 4 = 0"}, {"rel": "限制性描述", "source": "$( 2 m - 4 ) { x } ^ { 2 } + 3 mx + { m } ^ { 2 } - 4 = 0$ 关于 $x$ 的一元二次方程有一根为 $0$", "target": "m ^ { 2 } - 4 = 0"}, {"rel": "限制性描述", "source": "$( 2 m - 4 ) { x } ^ { 2 } + 3 mx + { m } ^ { 2 } - 4 = 0$ 关于 $x$ 的一元二次方程有一根为 $0$", "target": "2 m - 4 \\neq 0"}, {"rel": "联立", "source": "2 m - 4 \\neq 0", "target": "m = - 2"}]}}
|
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{"content": "Given that $a$ and $b$ are rational numbers and $| a - 1 | + ( b + 2 ) ^ 2 = 0$, what is the value of $a + b$?", "answer": "- 1", "steps": "From the given information, we know that $a = 1$ and $b = - 2$. Therefore, the original expression is equal to $1 - 2 = - 1$.", "expr_cands": ["a", "b", "| a - 1 | + ( b + 2 ) ^ { 2 } = 0", "a + b", "a = 1", "b = - 2", "1 - 2", "- 1"], "exprs": ["a = 1", "b = - 2", "1 - 2", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 1 | + ( b + 2 ) ^ { 2 } = 0"}, {"id": "a = 1"}, {"id": "绝对值恒大于等于0"}, {"id": "b = - 2"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "a + b"}, {"id": "1 - 2"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "| a - 1 | + ( b + 2 ) ^ { 2 } = 0", "target": "a = 1"}, {"rel": "被描述", "source": "| a - 1 | + ( b + 2 ) ^ { 2 } = 0", "target": "b = - 2"}, {"rel": "代入", "source": "a = 1", "target": "1 - 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a = 1"}, {"rel": "代入", "source": "b = - 2", "target": "1 - 2"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "b = - 2"}, {"rel": "被代入", "source": "a + b", "target": "1 - 2"}, {"rel": "计算", "source": "1 - 2", "target": "- 1"}]}}
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{"content": "The value of the fraction $\\frac { x - 1 } { x + 1 }$ is zero. What is the value of $x$?", "answer": "1", "steps": "According to the problem, we have $x - 1 = 0$ and $x + 1 \\neq 0$. Solving for $x$, we get $x = 1$.", "expr_cands": ["\\frac { x - 1 } { x + 1 }", "x", "x - 1 = 0", "x = 1", "x + 1 \\neq 0", "x \\neq - 1"], "exprs": ["x - 1 = 0", "x + 1 \\neq 0", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 1 } { x + 1 }"}, {"id": "x - 1 = 0"}, {"id": "分式 $\\frac { x - 1 } { x + 1 }$ 的值为零"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "x + 1 \\neq 0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "\\frac { x - 1 } { x + 1 }", "target": "x - 1 = 0"}, {"rel": "被描述", "source": "\\frac { x - 1 } { x + 1 }", "target": "x + 1 \\neq 0"}, {"rel": "联立", "source": "x - 1 = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - 1 } { x + 1 }$ 的值为零", "target": "x - 1 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x - 1 = 0"}, {"rel": "联立", "source": "x + 1 \\neq 0", "target": "x = 1"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x + 1 \\neq 0"}]}}
|
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{"content": "The polynomial $3 x ^ { 2 } - 2 xy - y ^ { 2 }$ subtracted by the polynomial $m$ equals $- 5 x ^ { 2 } + xy - 2 y ^ { 2 }$. Find the value of $m$.", "answer": "8 x ^ { 2 } - 3 xy + y ^ { 2 }", "steps": "Because the polynomial $( 3 x ^ 2 - 2 xy - y ^ 2 ) - ( - 5 x ^ 2 + xy - 2 y ^ 2 ) = 3 x ^ 2 - 2 xy - y ^ 2 + 5 x ^ 2 - xy + 2 y ^ 2 = 8 x ^ 2 - 3 xy + y ^ 2$, we can solve for $m$ to be $8 x ^ 2 - 3 xy + y ^ 2$.", "expr_cands": ["3 x ^ { 2 } - 2 xy - y ^ { 2 }", "y", "x", "m", "- 5 x ^ { 2 } + xy - 2 y ^ { 2 }", "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )", "8 x ^ { 2 } - 3 xy + y ^ { 2 }", "m = 8 x ^ { 2 } - 3 xy + y ^ { 2 }"], "exprs": ["( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )", "8 x ^ { 2 } - 3 xy + y ^ { 2 }", "m = 8 x ^ { 2 } - 3 xy + y ^ { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { 2 } - 2 xy - y ^ { 2 }"}, {"id": "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )"}, {"id": "- 5 x ^ { 2 } + xy - 2 y ^ { 2 }"}, {"id": "m"}, {"id": "多项式 $3 x ^ { 2 } - 2 xy - y ^ { 2 }$ 减去多项式 $m$"}, {"id": "所得的差为 $- 5 x ^ { 2 } + xy - 2 y ^ { 2 }$"}, {"id": "8 x ^ { 2 } - 3 xy + y ^ { 2 }"}, {"id": "m = 8 x ^ { 2 } - 3 xy + y ^ { 2 }"}], "links": [{"rel": "被描述", "source": "3 x ^ { 2 } - 2 xy - y ^ { 2 }", "target": "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )"}, {"rel": "计算", "source": "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )", "target": "8 x ^ { 2 } - 3 xy + y ^ { 2 }"}, {"rel": "被描述", "source": "- 5 x ^ { 2 } + xy - 2 y ^ { 2 }", "target": "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )"}, {"rel": "被描述", "source": "m", "target": "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )"}, {"rel": "被描述", "source": "m", "target": "m = 8 x ^ { 2 } - 3 xy + y ^ { 2 }"}, {"rel": "限制性描述", "source": "多项式 $3 x ^ { 2 } - 2 xy - y ^ { 2 }$ 减去多项式 $m$", "target": "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )"}, {"rel": "限制性描述", "source": "多项式 $3 x ^ { 2 } - 2 xy - y ^ { 2 }$ 减去多项式 $m$", "target": "m = 8 x ^ { 2 } - 3 xy + y ^ { 2 }"}, {"rel": "限制性描述", "source": "所得的差为 $- 5 x ^ { 2 } + xy - 2 y ^ { 2 }$", "target": "( 3 x ^ { 2 } - 2 xy - y ^ { 2 } ) - ( - 5 x ^ { 2 } + xy - 2 y ^ { 2 } )"}, {"rel": "限制性描述", "source": "所得的差为 $- 5 x ^ { 2 } + xy - 2 y ^ { 2 }$", "target": "m = 8 x ^ { 2 } - 3 xy + y ^ { 2 }"}, {"rel": "被描述", "source": "8 x ^ { 2 } - 3 xy + y ^ { 2 }", "target": "m = 8 x ^ { 2 } - 3 xy + y ^ { 2 }"}]}}
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{"content": "Given the equation $a ( 2 x - 1 ) = 3 x - 2$ with a solution of $- 1$ for $x$, what is the value of $a$?", "answer": "\\frac { 5 } { 3 }", "steps": "Substituting $x = - 1$, we get $a ( - 2 - 1 ) = - 3 - 2$. Solving for $a$, we get $a = \\frac { 5 } { 3 }$.", "expr_cands": ["x", "a ( 2 x - 1 ) = 3 x - 2", "a", "- 1", "x = - 1", "a ( - 2 - 1 ) = - 3 - 2", "a = \\frac { 5 } { 3 }"], "exprs": ["x = - 1", "a ( - 2 - 1 ) = - 3 - 2", "a = \\frac { 5 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x"}, {"id": "x = - 1"}, {"id": "a ( 2 x - 1 ) = 3 x - 2"}, {"id": "- 1"}, {"id": "关于 $x$ 的方程 $a ( 2 x - 1 ) = 3 x - 2$ 解为 $- 1$"}, {"id": "a ( - 2 - 1 ) = - 3 - 2"}, {"id": "a = \\frac { 5 } { 3 }"}], "links": [{"rel": "被描述", "source": "x", "target": "x = - 1"}, {"rel": "代入", "source": "x = - 1", "target": "a ( - 2 - 1 ) = - 3 - 2"}, {"rel": "被描述", "source": "a ( 2 x - 1 ) = 3 x - 2", "target": "x = - 1"}, {"rel": "被代入", "source": "a ( 2 x - 1 ) = 3 x - 2", "target": "a ( - 2 - 1 ) = - 3 - 2"}, {"rel": "被描述", "source": "- 1", "target": "x = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $a ( 2 x - 1 ) = 3 x - 2$ 解为 $- 1$", "target": "x = - 1"}, {"rel": "等式方程求解", "source": "a ( - 2 - 1 ) = - 3 - 2", "target": "a = \\frac { 5 } { 3 }"}]}}
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{"content": "If the equation $( k - 1 ) { x } ^ { | k | } - 1 = 0$ is a linear equation in one variable $x$, then the value of $k$ is ____ ?", "answer": "- 1", "steps": "From the given condition, we have $| k | = 1$ and $k - 1 \\neq 0$. Solving for $k$, we get $k = - 1$.", "expr_cands": ["x", "( k - 1 ) { x } ^ { | k | } - 1 = 0", "k", "| k | = 1", "k = - 1", "k = 1", "k - 1 \\neq 0", "k \\neq 1"], "exprs": ["| k | = 1", "k - 1 \\neq 0", "k = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( k - 1 ) { x } ^ { | k | } - 1 = 0"}, {"id": "| k | = 1"}, {"id": "关于 $x$ 的方程 $( k - 1 ) { x } ^ { | k | } - 1 = 0$ 是一元一次方程"}, {"id": "k - 1 \\neq 0"}, {"id": "k = - 1"}], "links": [{"rel": "被描述", "source": "( k - 1 ) { x } ^ { | k | } - 1 = 0", "target": "| k | = 1"}, {"rel": "被描述", "source": "( k - 1 ) { x } ^ { | k | } - 1 = 0", "target": "k - 1 \\neq 0"}, {"rel": "联立", "source": "| k | = 1", "target": "k = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $( k - 1 ) { x } ^ { | k | } - 1 = 0$ 是一元一次方程", "target": "| k | = 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $( k - 1 ) { x } ^ { | k | } - 1 = 0$ 是一元一次方程", "target": "k - 1 \\neq 0"}, {"rel": "联立", "source": "k - 1 \\neq 0", "target": "k = - 1"}]}}
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{"content": "The coefficient and the sum of the exponents of the monomial $- 3 xy ^ 2 z ^ 3$ are ____?", "answer": "3", "steps": "Because the coefficient of the monomial $- 3 xy ^ 2 z ^ 3$ is $- 3$, the exponent is $6$, so the sum of the coefficient and exponent is $6 - 3 = 3$.", "expr_cands": ["- 3 xy ^ { 2 } z ^ { 3 }", "y", "z", "x", "- 3", "6", "6 - 3", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 3 xy ^ { 2 } z ^ { 3 }"}, {"id": "3"}, {"id": "单项式 $- 3 xy ^ { 2 } z ^ { 3 }$ 的系数与次数的和"}], "links": [{"rel": "被描述", "source": "- 3 xy ^ { 2 } z ^ { 3 }", "target": "3"}, {"rel": "限制性描述", "source": "单项式 $- 3 xy ^ { 2 } z ^ { 3 }$ 的系数与次数的和", "target": "3"}]}}
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{"content": "Given $8 { a } ^ { 3 } { b } ^ { m } \\div 4 { a } ^ { n } { b } ^ { 2 } = 2 { b } ^ { 2 }$, what is the value of $mn$?", "answer": "12", "steps": "$8 { a } ^ { 3 } { b } ^ { m } \\div 4 { a } ^ { n } { b } ^ { 2 } = 2 { a } ^ { 3 - n } { b } ^ { m - 2 } = 2 { b } ^ { 2 }$ , according to the problem, we have $3 - n = 0$ and $m - 2 = 2$, which gives us $m = 4$ and $n = 3$. Therefore, $mn = 4 * 3 = 12$.", "expr_cands": ["8 { a } ^ { 3 } { b } ^ { m } \\div 4 { a } ^ { n } { b } ^ { 2 } = 2 { b } ^ { 2 }", "a", "m", "b", "n", "mn", "3 - n = 0", "n = 3", "m - 2 = 2", "m = 4", "12"], "exprs": ["3 - n = 0", "m - 2 = 2", "n = 3", "m = 4", "12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "8 { a } ^ { 3 } { b } ^ { m } \\div 4 { a } ^ { n } { b } ^ { 2 } = 2 { b } ^ { 2 }"}, {"id": "3 - n = 0"}, {"id": "根据题意得 : $3 - n = 0$ , $m - 2 = 2$"}, {"id": "m - 2 = 2"}, {"id": "m = 4"}, {"id": "n = 3"}, {"id": "mn"}, {"id": "12"}], "links": [{"rel": "被描述", "source": "8 { a } ^ { 3 } { b } ^ { m } \\div 4 { a } ^ { n } { b } ^ { 2 } = 2 { b } ^ { 2 }", "target": "3 - n = 0"}, {"rel": "被描述", "source": "8 { a } ^ { 3 } { b } ^ { m } \\div 4 { a } ^ { n } { b } ^ { 2 } = 2 { b } ^ { 2 }", "target": "m - 2 = 2"}, {"rel": "等式方程求解", "source": "3 - n = 0", "target": "n = 3"}, {"rel": "限制性描述", "source": "根据题意得 : $3 - n = 0$ , $m - 2 = 2$", "target": "3 - n = 0"}, {"rel": "限制性描述", "source": "根据题意得 : $3 - n = 0$ , $m - 2 = 2$", "target": "m - 2 = 2"}, {"rel": "等式方程求解", "source": "m - 2 = 2", "target": "m = 4"}, {"rel": "代入", "source": "m = 4", "target": "12"}, {"rel": "代入", "source": "n = 3", "target": "12"}, {"rel": "被代入", "source": "mn", "target": "12"}]}}
|
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{"content": "If the line $y = - 2 x + 1$ is translated $2$ units to the left, the equation of the new line is ____?", "answer": "y = - 2 x - 3", "steps": "According to the principle of adding to the left and subtracting from the right, we know that the equation of the line $y = - 2 x + 1$ will shift 2 units to the left, and the new equation of the line will be $y = - 2 ( x + 2 ) + 1$, which simplifies to $y = - 2 x - 3$.", "expr_cands": ["y = - 2 x + 1", "y", "x", "2", "y = - 2 ( x + 2 ) + 1", "1 - 2 x = - 2 ( x + 2 ) + 1", "- 2 x - 3"], "exprs": ["y = - 2 ( x + 2 ) + 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "y = - 2 ( x + 2 ) + 1"}, {"id": "y = - 2 x + 1"}, {"id": "直线 $y = - 2 x + 1$ 向左平移 $2$ 个单位"}], "links": [{"rel": "被描述", "source": "2", "target": "y = - 2 ( x + 2 ) + 1"}, {"rel": "被描述", "source": "y = - 2 x + 1", "target": "y = - 2 ( x + 2 ) + 1"}, {"rel": "限制性描述", "source": "直线 $y = - 2 x + 1$ 向左平移 $2$ 个单位", "target": "y = - 2 ( x + 2 ) + 1"}]}}
|
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{"content": "$15$, the square root of a positive number $x$ is $2 a 3$ and $5 a$, then $a$ = ____?", "answer": "- 2", "steps": "According to the problem, we have $2 a - 3 + 5 - a = 0$. Solving for $a$, we get $a = - 2$.", "expr_cands": ["15", "x", "2 a 3", "a", "5 a", "2 a - 3 + 5 - a = 0", "a = - 2"], "exprs": ["2 a - 3 + 5 - a = 0", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "15"}, {"id": "2 a - 3 + 5 - a = 0"}, {"id": "2 a 3"}, {"id": "一个正数 $x$ 的平方根是 $2 a 3$ 与 $5 a$"}, {"id": "平方根互为相反数"}, {"id": "a = - 2"}], "links": [{"rel": "被描述", "source": "15", "target": "2 a - 3 + 5 - a = 0"}, {"rel": "等式方程求解", "source": "2 a - 3 + 5 - a = 0", "target": "a = - 2"}, {"rel": "被描述", "source": "2 a 3", "target": "2 a - 3 + 5 - a = 0"}, {"rel": "限制性描述", "source": "一个正数 $x$ 的平方根是 $2 a 3$ 与 $5 a$", "target": "2 a - 3 + 5 - a = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "2 a - 3 + 5 - a = 0"}]}}
|
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{"content": "The difference between $2 x$ and $5$ times $( x - 3 )$ is ____ ?", "answer": "- 3 x + 15", "steps": "$2 x$ multiplied by $5$ less than $( x - 3 )$ can be expressed as $2 x - 5 ( x - 3 ) = 2 x - 5 x + 15 = - 3 x + 15$.", "expr_cands": ["2 x", "x", "( x - 3 )", "5", "2 x - 5 ( x - 3 )", "- 3 x + 15"], "exprs": ["2 x - 5 ( x - 3 )", "- 3 x + 15"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x"}, {"id": "2 x - 5 ( x - 3 )"}, {"id": "( x - 3 )"}, {"id": "5"}, {"id": "$2 x$ 与 $( x - 3 )$ 的 $5$ 倍的差"}, {"id": "- 3 x + 15"}], "links": [{"rel": "被描述", "source": "2 x", "target": "2 x - 5 ( x - 3 )"}, {"rel": "计算", "source": "2 x - 5 ( x - 3 )", "target": "- 3 x + 15"}, {"rel": "被描述", "source": "( x - 3 )", "target": "2 x - 5 ( x - 3 )"}, {"rel": "被描述", "source": "5", "target": "2 x - 5 ( x - 3 )"}, {"rel": "限制性描述", "source": "$2 x$ 与 $( x - 3 )$ 的 $5$ 倍的差", "target": "2 x - 5 ( x - 3 )"}]}}
|
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{"content": "If the equation $\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ 2 - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ 2$ holds, then $a + b$ = ____?", "answer": "2", "steps": "According to the problem, we have $2 a + 1 = 1$ and $3 b - 4 = 2$. Solving for $a$ and $b$, we get $a = 0$ and $b = 2$. Therefore, $a + b = 0 + 2 = 2$.", "expr_cands": ["\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ { 2 } - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ { 2 }", "y", "b", "x", "a", "a + b", "2 a + 1 = 1", "a = 0", "3 b - 4 = 2", "b = 2", "2"], "exprs": ["2 a + 1 = 1", "3 b - 4 = 2", "a = 0", "b = 2", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ { 2 } - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ { 2 }"}, {"id": "2 a + 1 = 1"}, {"id": "式 $\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ { 2 } - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ { 2 }$ 成立"}, {"id": "3 b - 4 = 2"}, {"id": "a = 0"}, {"id": "b = 2"}, {"id": "a + b"}, {"id": "2"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ { 2 } - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ { 2 }", "target": "2 a + 1 = 1"}, {"rel": "被描述", "source": "\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ { 2 } - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ { 2 }", "target": "3 b - 4 = 2"}, {"rel": "等式方程求解", "source": "2 a + 1 = 1", "target": "a = 0"}, {"rel": "限制性描述", "source": "式 $\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ { 2 } - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ { 2 }$ 成立", "target": "2 a + 1 = 1"}, {"rel": "限制性描述", "source": "式 $\\frac { 1 } { 2 } x ^ { 2 a + 1 } y ^ { 2 } - \\frac { 1 } { 4 } xy ^ { 3 b - 4 } = \\frac { 1 } { 4 } xy ^ { 2 }$ 成立", "target": "3 b - 4 = 2"}, {"rel": "等式方程求解", "source": "3 b - 4 = 2", "target": "b = 2"}, {"rel": "代入", "source": "a = 0", "target": "2"}, {"rel": "代入", "source": "b = 2", "target": "2"}, {"rel": "被代入", "source": "a + b", "target": "2"}]}}
|
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{"content": "Given $y - 3 = kx$, and $x = 2$ when $y = 7$. What is the value of $y$ when $x = 4$?", "answer": "11", "steps": "From $y - 3 = kx$. Substituting $x = 2$, $y = 7$, we get $7 - 3 = 2 k$, $\\therefore k = 2$. Thus, $y = 2 x + 3$. When $x = 4$, $y = 2 \\times 4 + 3 = 11$.", "expr_cands": ["y - 3 = kx", "k", "y", "x", "x = 2", "y = 7", "x = 4", "2 k = 2 k", "7 - 3 = 2 k", "k = 2", "y = 2 x + 3", "y = 11"], "exprs": ["7 - 3 = 2 k", "k = 2", "y = 2 x + 3", "y = 11"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y - 3 = kx"}, {"id": "7 - 3 = 2 k"}, {"id": "x = 2"}, {"id": "y = 7"}, {"id": "k = 2"}, {"id": "y = 2 x + 3"}, {"id": "x = 4"}, {"id": "y = 11"}], "links": [{"rel": "被代入", "source": "y - 3 = kx", "target": "7 - 3 = 2 k"}, {"rel": "联立", "source": "y - 3 = kx", "target": "y = 2 x + 3"}, {"rel": "等式方程求解", "source": "7 - 3 = 2 k", "target": "k = 2"}, {"rel": "代入", "source": "x = 2", "target": "7 - 3 = 2 k"}, {"rel": "代入", "source": "y = 7", "target": "7 - 3 = 2 k"}, {"rel": "联立", "source": "k = 2", "target": "y = 2 x + 3"}, {"rel": "被代入", "source": "y = 2 x + 3", "target": "y = 11"}, {"rel": "代入", "source": "x = 4", "target": "y = 11"}]}}
|
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{"content": "If $x = 4$ is a solution to the equation $x - 2 a = 0$ with respect to $x$, then the value of $a$ is", "answer": "2", "steps": "According to the problem, substituting $x = 4$ into the given equation yields $4 - 2 a = 0$, which can be solved to obtain $a = 2$.", "expr_cands": ["x = 4", "x", "x - 2 a = 0", "a", "4 - 2 a = 0", "a = 2"], "exprs": ["4 - 2 a = 0", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 4"}, {"id": "4 - 2 a = 0"}, {"id": "x - 2 a = 0"}, {"id": "a = 2"}], "links": [{"rel": "代入", "source": "x = 4", "target": "4 - 2 a = 0"}, {"rel": "等式方程求解", "source": "4 - 2 a = 0", "target": "a = 2"}, {"rel": "被代入", "source": "x - 2 a = 0", "target": "4 - 2 a = 0"}]}}
|
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{"content": "The inequality $- 2 x > - 4$ has a positive integer solution of $x$ = ____ ?", "answer": "1", "steps": "Because $- 2 x > - 4$, therefore $x < 2$, therefore the positive integer solution is: $x = 1$.", "expr_cands": ["- 2 x > - 4", "x", "x < 2", "x = 1"], "exprs": ["x < 2", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 x > - 4"}, {"id": "x < 2"}, {"id": "x = 1"}, {"id": "不等式 $- 2 x > - 4$ 的正整数解为 $x$ ="}], "links": [{"rel": "不等式方程求解", "source": "- 2 x > - 4", "target": "x < 2"}, {"rel": "被描述", "source": "x < 2", "target": "x = 1"}, {"rel": "限制性描述", "source": "不等式 $- 2 x > - 4$ 的正整数解为 $x$ =", "target": "x = 1"}]}}
|
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{"content": "If $ab \\neq 0$ and $\\frac { | a | } { a } + \\frac { | b | } { b } = 0$, then the value of $2017 - \\frac { | ab | } { ab }$ is ____?", "answer": "2018", "steps": "\\because $ab \\neq 0$ and $\\frac { | a | } { a } + \\frac { | b | } { b } = 0$ , \\therefore $a$ and $b$ have opposite signs, \\therefore the original expression $= 2017 + 1 = 2018$.", "expr_cands": ["ab \\neq 0", "b", "a", "\\frac { | a | } { a } + \\frac { | b | } { b } = 0", "2017 - \\frac { | ab | } { ab }", "2017 + 1", "2018"], "exprs": ["2017 + 1", "2018"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ab \\neq 0"}, {"id": "2017 + 1"}, {"id": "\\frac { | a | } { a } + \\frac { | b | } { b } = 0"}, {"id": "2017 - \\frac { | ab | } { ab }"}, {"id": ", $ab \\neq 0$ 且 $\\frac { | a | } { a } + \\frac { | b | } { b } = 0$"}, {"id": ", $a$ , $b$ 一负一正"}, {"id": "2018"}], "links": [{"rel": "被描述", "source": "ab \\neq 0", "target": "2017 + 1"}, {"rel": "计算", "source": "2017 + 1", "target": "2018"}, {"rel": "被描述", "source": "\\frac { | a | } { a } + \\frac { | b | } { b } = 0", "target": "2017 + 1"}, {"rel": "被描述", "source": "2017 - \\frac { | ab | } { ab }", "target": "2017 + 1"}, {"rel": "限制性描述", "source": ", $ab \\neq 0$ 且 $\\frac { | a | } { a } + \\frac { | b | } { b } = 0$", "target": "2017 + 1"}, {"rel": "限制性描述", "source": ", $a$ , $b$ 一负一正", "target": "2017 + 1"}]}}
|
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{"content": "If the equation $\\sqrt { x } = k + 1$ has no real solutions, then the range of values for $k$ is ____?", "answer": "k < - 1", "steps": "$\\because$ The equation $\\sqrt { x } = k + 1$ has no real solutions, $\\therefore$ $k + 1 < 0$, which implies $k < - 1$.", "expr_cands": ["\\sqrt { x } = k + 1", "k", "x", "k + 1 < 0", "k < - 1"], "exprs": ["k + 1 < 0", "k < - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x } = k + 1"}, {"id": "k + 1 < 0"}, {"id": "方程 $\\sqrt { x } = k + 1$ 无实数解"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "$k$ 的取值范围"}, {"id": "k < - 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { x } = k + 1", "target": "k + 1 < 0"}, {"rel": "不等式方程求解", "source": "k + 1 < 0", "target": "k < - 1"}, {"rel": "限制性描述", "source": "方程 $\\sqrt { x } = k + 1$ 无实数解", "target": "k + 1 < 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "k + 1 < 0"}, {"rel": "限制性描述", "source": "$k$ 的取值范围", "target": "k + 1 < 0"}]}}
|
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{"content": "When $m$ = ____ ?, the solution of the equation $x = 10 - 4 x$ is the same as the solution of the equation $5 x + 2 m = 2$.", "answer": "- 4", "steps": "Since $x = 10 - 4 x$, therefore $x = 2$. Since the solution to the equation $x = 10 - 4 x$ is the same as the solution to the equation $5 x + 2 m = 2$, therefore substituting $x = 2$ into the equation $5 x + 2 m = 2$ yields $5 * 2 + 2 m = 2$. Therefore, $m = - 4$.", "expr_cands": ["m", "x = 10 - 4 x", "x", "5 x + 2 m = 2", "x = 2", "2 m - 20 x + 50 = 2", "5 * 2 + 2 m = 2", "m = - 4"], "exprs": ["x = 2", "m = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 10 - 4 x"}, {"id": "x = 2"}, {"id": "5 x + 2 m = 2"}, {"id": "m = - 4"}], "links": [{"rel": "等式方程求解", "source": "x = 10 - 4 x", "target": "x = 2"}, {"rel": "联立", "source": "x = 2", "target": "m = - 4"}, {"rel": "联立", "source": "5 x + 2 m = 2", "target": "m = - 4"}]}}
|
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{"content": "The coefficient sum of the monomial $- \\frac { 2 xy } { 3 }$ and $x ^ 3$ is ____?", "answer": "\\frac { 1 } { 3 }", "steps": "The coefficient sum of the monomial $- \\frac { 2 xy } { 3 }$ and $x ^ 3$ is $- \\frac { 2 } { 3 } + 1 = \\frac { 1 } { 3 }$.", "expr_cands": ["- \\frac { 2 xy } { 3 }", "x", "y", "x ^ { 3 }", "- \\frac { 2 } { 3 } + 1", "\\frac { 1 } { 3 }"], "exprs": ["\\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 2 xy } { 3 }"}, {"id": "\\frac { 1 } { 3 }"}, {"id": "x ^ { 3 }"}, {"id": "单项式 $- \\frac { 2 xy } { 3 }$ 与 $x ^ { 3 }$ 的系数之和"}, {"id": "单项式 $- \\frac { 2 xy } { 3 }$ 与 $x ^ { 3 }$ 的系数之和为 $- \\frac { 2 } { 3 } + 1 = \\frac { 1 } { 3 }$"}], "links": [{"rel": "被描述", "source": "- \\frac { 2 xy } { 3 }", "target": "\\frac { 1 } { 3 }"}, {"rel": "被描述", "source": "x ^ { 3 }", "target": "\\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { 2 xy } { 3 }$ 与 $x ^ { 3 }$ 的系数之和", "target": "\\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { 2 xy } { 3 }$ 与 $x ^ { 3 }$ 的系数之和为 $- \\frac { 2 } { 3 } + 1 = \\frac { 1 } { 3 }$", "target": "\\frac { 1 } { 3 }"}]}}
|
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{"content": "If the value of the algebraic expression $7 a - 5$ is the opposite of $3 - 5 a$, then the value of $a$ is ____?", "answer": "1", "steps": "From the given problem, we have $7 a - 5 + 3 - 5 a = 0$. Solving for $a$, we get $a = 1$.", "expr_cands": ["7 a - 5", "a", "3 - 5 a", "7 a - 5 + 3 - 5 a = 0", "a = 1"], "exprs": ["7 a - 5 + 3 - 5 a = 0", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "7 a - 5"}, {"id": "7 a - 5 + 3 - 5 a = 0"}, {"id": "3 - 5 a"}, {"id": "代数式 $7 a - 5$ 与 $3 - 5 a$ 的值互为相反数"}, {"id": "a = 1"}], "links": [{"rel": "被描述", "source": "7 a - 5", "target": "7 a - 5 + 3 - 5 a = 0"}, {"rel": "等式方程求解", "source": "7 a - 5 + 3 - 5 a = 0", "target": "a = 1"}, {"rel": "被描述", "source": "3 - 5 a", "target": "7 a - 5 + 3 - 5 a = 0"}, {"rel": "限制性描述", "source": "代数式 $7 a - 5$ 与 $3 - 5 a$ 的值互为相反数", "target": "7 a - 5 + 3 - 5 a = 0"}]}}
|
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{"content": "Given that the value of the algebraic expression $x - 2 y$ is $3$, what is the value of the algebraic expression $15 - x + 2 y$?", "answer": "12", "steps": "$\\because$ The value of the algebraic expression $x - 2 y$ is $3$, $\\therefore$ $x - 2 y = 3$, $\\therefore$ $15 - x + 2 y = 15 - ( x - 2 y ) = 15 - 3 = 12$.", "expr_cands": ["x - 2 y", "y", "x", "3", "15 - x + 2 y", "x - 2 y = 3", "15 - ( x - 2 y )", "12"], "exprs": ["x - 2 y = 3", "15 - ( x - 2 y )", "12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - 2 y"}, {"id": "x - 2 y = 3"}, {"id": "3"}, {"id": "代数式 $x - 2 y$ 的值是 $3$"}, {"id": "15 - ( x - 2 y )"}, {"id": "15 - x + 2 y"}, {"id": "12"}], "links": [{"rel": "被描述", "source": "x - 2 y", "target": "x - 2 y = 3"}, {"rel": "提取因式参考", "source": "x - 2 y", "target": "15 - ( x - 2 y )"}, {"rel": "代入", "source": "x - 2 y = 3", "target": "12"}, {"rel": "被描述", "source": "3", "target": "x - 2 y = 3"}, {"rel": "限制性描述", "source": "代数式 $x - 2 y$ 的值是 $3$", "target": "x - 2 y = 3"}, {"rel": "被代入", "source": "15 - ( x - 2 y )", "target": "12"}, {"rel": "提取因式", "source": "15 - x + 2 y", "target": "15 - ( x - 2 y )"}]}}
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{"content": "The algebraic expression $5 m + \\frac { 1 } { 4 }$ and $5 ( m - \\frac { 1 } { 4 } )$ are opposite, then $m$ = ____ ?", "answer": "\\frac { 1 } { 10 }", "steps": "From the given problem, we have $5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 }) = 0$. Therefore, $5 m + \\frac { 1 } { 4 } + 5 m - \\frac { 5 } { 4 } = 0$. This simplifies to $10 m - 1 = 0$, and solving for $m$ gives $m = \\frac { 1 } { 10 }$.", "expr_cands": ["5 m + \\frac { 1 } { 4 }", "m", "5 ( m - \\frac { 1 } { 4 } )", "5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 } ) = 0", "m = \\frac { 1 } { 10 }", "5 m + \\frac { 1 } { 4 } + 5 m - \\frac { 5 } { 4 } = 0", "10 m - 1 = 0"], "exprs": ["5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 } ) = 0", "m = \\frac { 1 } { 10 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 m + \\frac { 1 } { 4 }"}, {"id": "5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 } ) = 0"}, {"id": "5 ( m - \\frac { 1 } { 4 } )"}, {"id": "代数式 $5 m + \\frac { 1 } { 4 }$ 与 $5 ( m - \\frac { 1 } { 4 } )$ 互为相反数"}, {"id": "m = \\frac { 1 } { 10 }"}], "links": [{"rel": "被描述", "source": "5 m + \\frac { 1 } { 4 }", "target": "5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 } ) = 0"}, {"rel": "等式方程求解", "source": "5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 } ) = 0", "target": "m = \\frac { 1 } { 10 }"}, {"rel": "被描述", "source": "5 ( m - \\frac { 1 } { 4 } )", "target": "5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 } ) = 0"}, {"rel": "属性描述", "source": "代数式 $5 m + \\frac { 1 } { 4 }$ 与 $5 ( m - \\frac { 1 } { 4 } )$ 互为相反数", "target": "5 m + \\frac { 1 } { 4 } + 5 ( m - \\frac { 1 } { 4 } ) = 0"}]}}
|
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{"content": "Given $x + y = 2019$, $x - y = \\frac { 2020 } { 2019 }$, what is the value of ${ x } ^ { 2 } - { y } ^ { 2 }$?", "answer": "2020", "steps": "$\\because x + y = 2019$, $x - y = \\frac { 2020 } { 2019 }$, $\\therefore x ^ 2 - y ^ 2 = ( x + y ) ( x - y ) = 2019 * \\frac { 2020 } { 2019 } = 2020$.", "expr_cands": ["x + y = 2019", "y", "x", "x - y = \\frac { 2020 } { 2019 }", "{ x } ^ { 2 } - { y } ^ { 2 }", "( x + y ) ( x - y )", "2020"], "exprs": ["( x + y ) ( x - y )", "2020"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { 2 } - { y } ^ { 2 }"}, {"id": "( x + y ) ( x - y )"}, {"id": "x + y = 2019"}, {"id": "x - y = \\frac { 2020 } { 2019 }"}, {"id": "2020"}], "links": [{"rel": "提取因式", "source": "{ x } ^ { 2 } - { y } ^ { 2 }", "target": "( x + y ) ( x - y )"}, {"rel": "被代入", "source": "( x + y ) ( x - y )", "target": "2020"}, {"rel": "提取因式参考", "source": "x + y = 2019", "target": "( x + y ) ( x - y )"}, {"rel": "代入", "source": "x + y = 2019", "target": "2020"}, {"rel": "提取因式参考", "source": "x - y = \\frac { 2020 } { 2019 }", "target": "( x + y ) ( x - y )"}, {"rel": "代入", "source": "x - y = \\frac { 2020 } { 2019 }", "target": "2020"}]}}
|
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{"content": "The value of $x$ that makes the fraction $\\frac { | x | - \\sqrt { 2 }} {( x + 1 ) ( x - \\sqrt { 2 })}$ equal to $0$ is _____.", "answer": "- \\sqrt { 2 }", "steps": "According to the problem, we have $| x | - \\sqrt { 2 } = 0$ and $( x + 1 ) ( x - \\sqrt { 2 }) \\neq 0$. Solving for $x$, we get $x = - \\sqrt { 2 }$.", "expr_cands": ["\\frac { | x | - \\sqrt { 2 } } { ( x + 1 ) ( x - \\sqrt { 2 } ) }", "x", "0", "| x | - \\sqrt { 2 } = 0", "x = \\sqrt { 2 }", "x = - \\sqrt { 2 }", "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0", "( - 1 < x \\wedge x < \\sqrt { 2 })", "\\sqrt { 2 } < x", "x < - 1"], "exprs": ["| x | - \\sqrt { 2 } = 0", "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0", "x = - \\sqrt { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { | x | - \\sqrt { 2 } } { ( x + 1 ) ( x - \\sqrt { 2 } ) }"}, {"id": "| x | - \\sqrt { 2 } = 0"}, {"id": "x"}, {"id": "0"}, {"id": "使分式 $\\frac { | x | - \\sqrt { 2 } } { ( x + 1 ) ( x - \\sqrt { 2 } ) }$ 的值为 $0$ 的 $x$ 值"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0"}, {"id": "x = - \\sqrt { 2 }"}], "links": [{"rel": "被描述", "source": "\\frac { | x | - \\sqrt { 2 } } { ( x + 1 ) ( x - \\sqrt { 2 } ) }", "target": "| x | - \\sqrt { 2 } = 0"}, {"rel": "被描述", "source": "\\frac { | x | - \\sqrt { 2 } } { ( x + 1 ) ( x - \\sqrt { 2 } ) }", "target": "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0"}, {"rel": "联立", "source": "| x | - \\sqrt { 2 } = 0", "target": "x = - \\sqrt { 2 }"}, {"rel": "被描述", "source": "x", "target": "| x | - \\sqrt { 2 } = 0"}, {"rel": "被描述", "source": "x", "target": "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0"}, {"rel": "被描述", "source": "0", "target": "| x | - \\sqrt { 2 } = 0"}, {"rel": "被描述", "source": "0", "target": "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0"}, {"rel": "限制性描述", "source": "使分式 $\\frac { | x | - \\sqrt { 2 } } { ( x + 1 ) ( x - \\sqrt { 2 } ) }$ 的值为 $0$ 的 $x$ 值", "target": "| x | - \\sqrt { 2 } = 0"}, {"rel": "限制性描述", "source": "使分式 $\\frac { | x | - \\sqrt { 2 } } { ( x + 1 ) ( x - \\sqrt { 2 } ) }$ 的值为 $0$ 的 $x$ 值", "target": "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "| x | - \\sqrt { 2 } = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0"}, {"rel": "联立", "source": "( x + 1 ) ( x - \\sqrt { 2 } ) \\neq 0", "target": "x = - \\sqrt { 2 }"}]}}
|
||||
{"content": "If the value of the algebraic expression $\\frac { 2 - x } { 3 }$ is non-negative, then the condition that $x$ satisfies is ____?", "answer": "x \\le 2", "steps": "From the given condition, we have $\\frac { 2 - x } { 3 } \\geq 0$, which implies that $x \\leq 2$.", "expr_cands": ["\\frac { 2 - x } { 3 }", "x", "\\frac { 2 - x } { 3 } \\ge 0", "x \\le 2"], "exprs": ["\\frac { 2 - x } { 3 } \\ge 0", "x \\le 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 - x } { 3 }"}, {"id": "\\frac { 2 - x } { 3 } \\ge 0"}, {"id": "代数式 $\\frac { 2 - x } { 3 }$ 的值是非负数"}, {"id": "x \\le 2"}], "links": [{"rel": "被描述", "source": "\\frac { 2 - x } { 3 }", "target": "\\frac { 2 - x } { 3 } \\ge 0"}, {"rel": "不等式方程求解", "source": "\\frac { 2 - x } { 3 } \\ge 0", "target": "x \\le 2"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 2 - x } { 3 }$ 的值是非负数", "target": "\\frac { 2 - x } { 3 } \\ge 0"}]}}
|
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{"content": "What is the maximum integer value of $x$ such that the value of the algebraic expression $4 x - \\frac { 3 } { 2 }$ is not greater than the value of the expression $3 x + 5$?", "answer": "6", "steps": "From the given condition, we have $4 x - \\frac { 3 } { 2 } \\leq 3 x + 5$. Solving for $x$, we get $x \\leq 6.5$. Therefore, the largest integer value of $x$ is $6$.", "expr_cands": ["4 x - \\frac { 3 } { 2 }", "x", "3 x + 5", "4 x - \\frac { 3 } { 2 } \\le 3 x + 5", "x \\le \\frac { 13 } { 2 }", "x \\le 6.5", "6"], "exprs": ["4 x - \\frac { 3 } { 2 } \\le 3 x + 5", "x \\le 6.5", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x - \\frac { 3 } { 2 }"}, {"id": "4 x - \\frac { 3 } { 2 } \\le 3 x + 5"}, {"id": "3 x + 5"}, {"id": "使代数式 $4 x - \\frac { 3 } { 2 }$ 的值不大于 $3 x + 5$ 的值的 $x$ 的最大整数值"}, {"id": "x \\le 6.5"}, {"id": "6"}, {"id": "最大整数为 $6$"}], "links": [{"rel": "被描述", "source": "4 x - \\frac { 3 } { 2 }", "target": "4 x - \\frac { 3 } { 2 } \\le 3 x + 5"}, {"rel": "不等式方程求解", "source": "4 x - \\frac { 3 } { 2 } \\le 3 x + 5", "target": "x \\le 6.5"}, {"rel": "被描述", "source": "3 x + 5", "target": "4 x - \\frac { 3 } { 2 } \\le 3 x + 5"}, {"rel": "限制性描述", "source": "使代数式 $4 x - \\frac { 3 } { 2 }$ 的值不大于 $3 x + 5$ 的值的 $x$ 的最大整数值", "target": "4 x - \\frac { 3 } { 2 } \\le 3 x + 5"}, {"rel": "被描述", "source": "x \\le 6.5", "target": "6"}, {"rel": "限制性描述", "source": "最大整数为 $6$", "target": "6"}]}}
|
||||
{"content": "The range of values for the variable $x$ in the fraction $\\frac { x } { 2 x + 3 }$ is _____.", "answer": "x \\neq - \\frac { 3 } { 2 }", "steps": "From the given equation, it can be deduced that $2 x + 3 \\neq 0$. Solving for $x$, we get $x \\neq - \\frac { 3 } { 2 }$.", "expr_cands": ["\\frac { x } { 2 x + 3 }", "x", "2 x + 3 \\neq 0", "x \\neq - \\frac { 3 } { 2 }"], "exprs": ["2 x + 3 \\neq 0", "x \\neq - \\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { 2 x + 3 }"}, {"id": "2 x + 3 \\neq 0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq - \\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\frac { x } { 2 x + 3 }", "target": "2 x + 3 \\neq 0"}, {"rel": "不等式方程求解", "source": "2 x + 3 \\neq 0", "target": "x \\neq - \\frac { 3 } { 2 }"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "2 x + 3 \\neq 0"}]}}
|
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{"content": "If $3 k - 2 x ^ { 2 k - 1 } > 0$ is a one-variable linear inequality, then $k$ = ____ ?", "answer": "1", "steps": "According to the problem, we have $2 k - 1 = 1$. Solving for $k$, we get $k = 1$.", "expr_cands": ["3 k - 2 x ^ { 2 k - 1 } > 0", "x", "k", "2 k - 1 = 1", "k = 1"], "exprs": ["2 k - 1 = 1", "k = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 k - 2 x ^ { 2 k - 1 } > 0"}, {"id": "2 k - 1 = 1"}, {"id": "$3 k - 2 x ^ { 2 k - 1 } > 0$ 是关于 $x$ 的一元一次不等式"}, {"id": "k = 1"}], "links": [{"rel": "被描述", "source": "3 k - 2 x ^ { 2 k - 1 } > 0", "target": "2 k - 1 = 1"}, {"rel": "等式方程求解", "source": "2 k - 1 = 1", "target": "k = 1"}, {"rel": "限制性描述", "source": "$3 k - 2 x ^ { 2 k - 1 } > 0$ 是关于 $x$ 的一元一次不等式", "target": "2 k - 1 = 1"}]}}
|
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{"content": "If $| x + 2 | = 0$, then the value of $2 x$ is ____?", "answer": "- 4", "steps": "From $| x + 2 | = 0$, we get $x + 2 = 0$, which means $x = - 2$. Therefore, $2 x = - 4$.", "expr_cands": ["| x + 2 | = 0", "x", "2 x", "x = - 2", "x + 2 = 0", "- 4"], "exprs": ["x = - 2", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x + 2 | = 0"}, {"id": "x = - 2"}, {"id": "2 x"}, {"id": "- 4"}], "links": [{"rel": "等式方程求解", "source": "| x + 2 | = 0", "target": "x = - 2"}, {"rel": "代入", "source": "x = - 2", "target": "- 4"}, {"rel": "被代入", "source": "2 x", "target": "- 4"}]}}
|
||||
{"content": "The quadratic coefficient of the equation $x ^ 2 - 3 x + 2 = 0$ is ____?", "answer": "1", "steps": "The quadratic coefficient of the equation $x ^ 2 - 3 x + 2 = 0$ is $1$.", "expr_cands": ["x ^ { 2 } - 3 x + 2 = 0", "x", "x = 1", "x = 2", "1"], "exprs": ["1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 x + 2 = 0"}, {"id": "1"}, {"id": "方程 $x ^ { 2 } - 3 x + 2 = 0$ 的二次项系数"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 3 x + 2 = 0", "target": "1"}, {"rel": "限制性描述", "source": "方程 $x ^ { 2 } - 3 x + 2 = 0$ 的二次项系数", "target": "1"}]}}
|
||||
{"content": "If $a ^ { 2 } + b ^ { 2 } + 2 c ^ { 2 } + 2 ac - 2 bc = 0$, then $a + b$ = ____ ?", "answer": "0", "steps": "Since $a ^ { 2 } + b ^ { 2 } + 2 c ^ { 2 } + 2 ac - 2 bc = ( a + c ) ^ { 2 } + ( b - c ) ^ { 2 } = 0$, it follows that $a + c = 0$ and $b - c = 0$, which implies that $a = - c$ and $b = c$. Therefore, $a + b = - c + c = 0$.", "expr_cands": ["a ^ { 2 } + b ^ { 2 } + 2 c ^ { 2 } + 2 ac - 2 bc = 0", "b", "a", "c", "a + b", "( a + c ) ^ { 2 } + ( b - c ) ^ { 2 } = 0", "a + c = 0", "b - c = 0", "a = - c", "b = c", "0"], "exprs": ["( a + c ) ^ { 2 } + ( b - c ) ^ { 2 } = 0", "a + c = 0", "b - c = 0", "a = - c", "b = c", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } + b ^ { 2 } + 2 c ^ { 2 } + 2 ac - 2 bc = 0"}, {"id": "( a + c ) ^ { 2 } + ( b - c ) ^ { 2 } = 0"}, {"id": "a + c = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "b - c = 0"}, {"id": "a = - c"}, {"id": "b = c"}, {"id": "a + b"}, {"id": "0"}], "links": [{"rel": "提取因式", "source": "a ^ { 2 } + b ^ { 2 } + 2 c ^ { 2 } + 2 ac - 2 bc = 0", "target": "( a + c ) ^ { 2 } + ( b - c ) ^ { 2 } = 0"}, {"rel": "被描述", "source": "( a + c ) ^ { 2 } + ( b - c ) ^ { 2 } = 0", "target": "a + c = 0"}, {"rel": "被描述", "source": "( a + c ) ^ { 2 } + ( b - c ) ^ { 2 } = 0", "target": "b - c = 0"}, {"rel": "移项", "source": "a + c = 0", "target": "a = - c"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "a + c = 0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "b - c = 0"}, {"rel": "移项", "source": "b - c = 0", "target": "b = c"}, {"rel": "代入", "source": "a = - c", "target": "0"}, {"rel": "代入", "source": "b = c", "target": "0"}, {"rel": "被代入", "source": "a + b", "target": "0"}]}}
|
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{"content": "If $\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }$ and $2 { x } ^ { n - 1 } { y } ^ { 2 }$ can be combined into one term, then the value of $n ^ { - m } + ( m - n ) ^ { 2 }$ is ____?", "answer": "\\frac { 10 } { 9 }", "steps": "Because $\\frac { 1 } { 3 } x ^ 2 y ^ m$ and $2 x ^ { n - 1 } y ^ 2$ can be combined into one term, therefore $\\frac { 1 } { 3 } x ^ 2 y ^ m$ and $2 x ^ { n - 1 } y ^ 2$ are like terms, therefore $n - 1 = 2$, $m = 2$, therefore $n = 3$, $m = 2$, therefore $n ^ { - m } + ( m - n ) ^ 23 ^ { - 2 } + ( 2 - 3 ) ^ 2 = \\frac { 1 } { 9 } + 1 = \\frac { 10 } { 9 }$.", "expr_cands": ["\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }", "m", "y", "x", "2 x ^ { n - 1 } y ^ { 2 }", "n", "n ^ { - m } + ( m - n ) ^ { 2 }", "n - 1 = 2", "n = 3", "m = 2", "n ^ { - m } + ( m - n ) ^ { 2 } 3 ^ { - 2 } + ( 2 - 3 ) ^ { 2 } = \\frac { 10 } { 9 }", "n ^ { - m } + ( m - n ) ^ { 2 } 3 ^ { - 2 } + ( 2 - 3 ) ^ { 2 }", "\\frac { 10 } { 9 }"], "exprs": ["n - 1 = 2", "m = 2", "n = 3", "\\frac { 10 } { 9 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }"}, {"id": "n - 1 = 2"}, {"id": "2 x ^ { n - 1 } y ^ { 2 }"}, {"id": "$\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }$ 与 $2 x ^ { n - 1 } y ^ { 2 }$ 可以合并成一项"}, {"id": "n = 3"}, {"id": "m = 2"}, {"id": "n ^ { - m } + ( m - n ) ^ { 2 }"}, {"id": "\\frac { 10 } { 9 }"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }", "target": "n - 1 = 2"}, {"rel": "被描述", "source": "\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }", "target": "m = 2"}, {"rel": "等式方程求解", "source": "n - 1 = 2", "target": "n = 3"}, {"rel": "被描述", "source": "2 x ^ { n - 1 } y ^ { 2 }", "target": "n - 1 = 2"}, {"rel": "被描述", "source": "2 x ^ { n - 1 } y ^ { 2 }", "target": "m = 2"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }$ 与 $2 x ^ { n - 1 } y ^ { 2 }$ 可以合并成一项", "target": "n - 1 = 2"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 3 } { x } ^ { 2 } { y } ^ { m }$ 与 $2 x ^ { n - 1 } y ^ { 2 }$ 可以合并成一项", "target": "m = 2"}, {"rel": "代入", "source": "n = 3", "target": "\\frac { 10 } { 9 }"}, {"rel": "代入", "source": "m = 2", "target": "\\frac { 10 } { 9 }"}, {"rel": "被代入", "source": "n ^ { - m } + ( m - n ) ^ { 2 }", "target": "\\frac { 10 } { 9 }"}]}}
|
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{"content": "Given that the degree of the monomial $3 x ^ { a - 1 } y ^ 3$ is $6$, what is the value of $a$?", "answer": "4", "steps": "It is known from the problem that $a - 1 + 3 = 6$, which implies that $a = 4$.", "expr_cands": ["3 x ^ { a - 1 } y ^ { 3 }", "y", "x", "a", "6", "a - 1 + 3 = 6", "a = 4"], "exprs": ["a - 1 + 3 = 6", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { a - 1 } y ^ { 3 }"}, {"id": "a - 1 + 3 = 6"}, {"id": "6"}, {"id": "单项式 $3 x ^ { a - 1 } y ^ { 3 }$ 的次数是 $6$"}, {"id": "a = 4"}], "links": [{"rel": "被描述", "source": "3 x ^ { a - 1 } y ^ { 3 }", "target": "a - 1 + 3 = 6"}, {"rel": "等式方程求解", "source": "a - 1 + 3 = 6", "target": "a = 4"}, {"rel": "被描述", "source": "6", "target": "a - 1 + 3 = 6"}, {"rel": "限制性描述", "source": "单项式 $3 x ^ { a - 1 } y ^ { 3 }$ 的次数是 $6$", "target": "a - 1 + 3 = 6"}]}}
|
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{"content": "Given $a ^ { 2 } + 2 a - 3 = 0$, what is the value of the algebraic expression $2 a ^ { 2 } + 4 a - 3$?", "answer": "3", "steps": "From the given information, we know that $a ^ 2 + 2 a = 3$. Therefore, $2 a ^ 2 + 4 a - 3 = 2 ( a ^ 2 + 2 a ) - 3 = 2 * 3 - 3 = 3$.", "expr_cands": ["a ^ { 2 } + 2 a - 3 = 0", "a", "2 a ^ { 2 } + 4 a - 3", "a ^ { 2 } + 2 a = 3", "a = - 3", "a = 1", "2 ( { a } ^ { 2 } + 2 a ) - 3", "3"], "exprs": ["a ^ { 2 } + 2 a = 3", "2 ( { a } ^ { 2 } + 2 a ) - 3", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } + 2 a - 3 = 0"}, {"id": "a ^ { 2 } + 2 a = 3"}, {"id": "2 ( { a } ^ { 2 } + 2 a ) - 3"}, {"id": "2 a ^ { 2 } + 4 a - 3"}, {"id": "3"}], "links": [{"rel": "移项", "source": "a ^ { 2 } + 2 a - 3 = 0", "target": "a ^ { 2 } + 2 a = 3"}, {"rel": "提取因式参考", "source": "a ^ { 2 } + 2 a = 3", "target": "2 ( { a } ^ { 2 } + 2 a ) - 3"}, {"rel": "代入", "source": "a ^ { 2 } + 2 a = 3", "target": "3"}, {"rel": "被代入", "source": "2 ( { a } ^ { 2 } + 2 a ) - 3", "target": "3"}, {"rel": "提取因式", "source": "2 a ^ { 2 } + 4 a - 3", "target": "2 ( { a } ^ { 2 } + 2 a ) - 3"}]}}
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{"content": "The sum of the two roots of the equation $x ^ 2 - 6 x + 5 = 0$ is _____.", "answer": "6", "steps": "The sum of the two roots of the equation $x ^ 2 - 6 x + 5 = 0$ is 6.", "expr_cands": ["x ^ { 2 } - 6 x + 5 = 0", "x", "x = 1", "x = 5", "6"], "exprs": ["6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 6 x + 5 = 0"}, {"id": "6"}, {"id": "方程 $x ^ { 2 } - 6 x + 5 = 0$ 的两个根之和"}, {"id": "一元二次方程根与系数关系，两根之和"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 6 x + 5 = 0", "target": "6"}, {"rel": "限制性描述", "source": "方程 $x ^ { 2 } - 6 x + 5 = 0$ 的两个根之和", "target": "6"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "6"}]}}
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{"content": "When $x$ = ____ ?, the fraction $\\frac { 9 } { x - 3 }$ is undefined.", "answer": "3", "steps": "When the denominator $x - 3 = 0$, that is $x = 3$, the fraction $\\frac { 9 } { x - 3 }$ is undefined.", "expr_cands": ["x", "\\frac { 9 } { x - 3 }", "x - 3 = 0", "x = 3"], "exprs": ["x - 3 = 0", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 9 } { x - 3 }"}, {"id": "x - 3 = 0"}, {"id": "分式 $\\frac { 9 } { x - 3 }$ 无意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "\\frac { 9 } { x - 3 }", "target": "x - 3 = 0"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "限制性描述", "source": "分式 $\\frac { 9 } { x - 3 }$ 无意义", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 3 = 0"}]}}
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{"content": "The square root of a positive number is $- m - 3$ and $2 m + 1$. What is the positive number?", "answer": "25", "steps": "$\\because$ The square root of a positive number is $- m - 3$ and $2 m + 1$, $\\therefore$ $- m - 3 + 2 m + 1 = 0$, which gives $m = 2$, $\\therefore$ $2 m + 1 = 5$, $\\therefore$ the number is $5 ^ 2 = 25$.", "expr_cands": ["- m - 3", "m", "2 m + 1", "- m - 3 + 2 m + 1 = 0", "m = 2", "5", "5 ^ { 2 }", "25"], "exprs": ["- m - 3 + 2 m + 1 = 0", "m = 2", "5", "25"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- m - 3"}, {"id": "- m - 3 + 2 m + 1 = 0"}, {"id": "2 m + 1"}, {"id": "一个正数的平方根分别为 $- m - 3$ 和 $2 m + 1$"}, {"id": "平方根互为相反数"}, {"id": "m = 2"}, {"id": "5"}, {"id": "25"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "- m - 3", "target": "- m - 3 + 2 m + 1 = 0"}, {"rel": "等式方程求解", "source": "- m - 3 + 2 m + 1 = 0", "target": "m = 2"}, {"rel": "被描述", "source": "2 m + 1", "target": "- m - 3 + 2 m + 1 = 0"}, {"rel": "被代入", "source": "2 m + 1", "target": "5"}, {"rel": "限制性描述", "source": "一个正数的平方根分别为 $- m - 3$ 和 $2 m + 1$", "target": "- m - 3 + 2 m + 1 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "- m - 3 + 2 m + 1 = 0"}, {"rel": "代入", "source": "m = 2", "target": "5"}, {"rel": "被描述", "source": "5", "target": "25"}, {"rel": "限制性描述", "source": "平方", "target": "25"}]}}
|
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{"content": "$11$. The range of values for the independent variable that makes the function $y = \\frac { 3 } { \\sqrt { 1 - 2 x }}$ meaningful is ____?", "answer": "x < \\frac { 1 } { 2 }", "steps": "From the given condition, we have $1 - 2 x > 0$. Solving for $x$, we get $x < \\frac { 1 } { 2 }$.", "expr_cands": ["11", "y = \\frac { 3 } { \\sqrt { 1 - 2 x } }", "y", "x", "1 - 2 x > 0", "x < \\frac { 1 } { 2 }"], "exprs": ["1 - 2 x > 0", "x < \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 3 } { \\sqrt { 1 - 2 x } }"}, {"id": "1 - 2 x > 0"}, {"id": "使函数 $y = \\frac { 3 } { \\sqrt { 1 - 2 x } }$ 有意义的自变量的取值范围"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x < \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "y = \\frac { 3 } { \\sqrt { 1 - 2 x } }", "target": "1 - 2 x > 0"}, {"rel": "不等式方程求解", "source": "1 - 2 x > 0", "target": "x < \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "使函数 $y = \\frac { 3 } { \\sqrt { 1 - 2 x } }$ 有意义的自变量的取值范围", "target": "1 - 2 x > 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "1 - 2 x > 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "1 - 2 x > 0"}]}}
|
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{"content": "When $2 x + 1$ and $- 3 x + 2$ are opposite in sign, then $x ^ 2 - 2 x + 1$ = ____?", "answer": "4", "steps": "According to the problem, we have $2 x + 1 - 3 x + 2 = 0$. Moving and combining terms, we get $- x = - 3$. Solving for $x$, we get $x = 3$. Therefore, the original expression is equal to $9 - 6 + 1 = 4$.", "expr_cands": ["2 x + 1", "x", "- 3 x + 2", "x ^ { 2 } - 2 x + 1", "2 x + 1 - 3 x + 2 = 0", "x = 3", "- x = - 3", "9 - 6 + 1", "4"], "exprs": ["2 x + 1 - 3 x + 2 = 0", "x = 3", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 1"}, {"id": "2 x + 1 - 3 x + 2 = 0"}, {"id": "- 3 x + 2"}, {"id": "当 $2 x + 1$ 和 $- 3 x + 2$ 互为相反数时"}, {"id": "x = 3"}, {"id": "x ^ { 2 } - 2 x + 1"}, {"id": "4"}], "links": [{"rel": "被描述", "source": "2 x + 1", "target": "2 x + 1 - 3 x + 2 = 0"}, {"rel": "等式方程求解", "source": "2 x + 1 - 3 x + 2 = 0", "target": "x = 3"}, {"rel": "被描述", "source": "- 3 x + 2", "target": "2 x + 1 - 3 x + 2 = 0"}, {"rel": "限制性描述", "source": "当 $2 x + 1$ 和 $- 3 x + 2$ 互为相反数时", "target": "2 x + 1 - 3 x + 2 = 0"}, {"rel": "代入", "source": "x = 3", "target": "4"}, {"rel": "被代入", "source": "x ^ { 2 } - 2 x + 1", "target": "4"}]}}
|
||||
{"content": "Given, $\\sqrt { x } = 2$, what is $x ^ 2$?", "answer": "16", "steps": "Because the square root of x is equal to 2, therefore x is equal to 4, and thus x squared is equal to 4 squared, which is equal to 16.", "expr_cands": ["\\sqrt { x } = 2", "x", "x ^ { 2 }", "x = 4", "16"], "exprs": ["16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 }"}, {"id": "16"}, {"id": "\\sqrt { x } = 2"}], "links": [{"rel": "被代入", "source": "x ^ { 2 }", "target": "16"}, {"rel": "代入", "source": "\\sqrt { x } = 2", "target": "16"}]}}
|
||||
{"content": "Given that the value of the polynomial $x + 2 y$ is $6$, what is the value of the polynomial $3 x + 6 y + 1$?", "answer": "19", "steps": "From the given information, we have $x + 2 y = 6$. Therefore, $3 x + 6 y + 1 = 3 ( x + 2 y ) + 1 = 3 * 6 + 1 = 19$.", "expr_cands": ["x + 2 y", "y", "x", "6", "3 x + 6 y + 1", "x + 2 y = 6", "3 ( x + 2 y ) + 1", "19"], "exprs": ["x + 2 y = 6", "19"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 2 y"}, {"id": "x + 2 y = 6"}, {"id": "6"}, {"id": "多项式 $x + 2 y$ 的值是 $6$"}, {"id": "19"}, {"id": "3 x + 6 y + 1"}, {"id": "多项式 $3 x + 6 y + 1$ 的值"}], "links": [{"rel": "被描述", "source": "x + 2 y", "target": "x + 2 y = 6"}, {"rel": "被描述", "source": "x + 2 y = 6", "target": "19"}, {"rel": "被描述", "source": "6", "target": "x + 2 y = 6"}, {"rel": "限制性描述", "source": "多项式 $x + 2 y$ 的值是 $6$", "target": "x + 2 y = 6"}, {"rel": "被描述", "source": "3 x + 6 y + 1", "target": "19"}, {"rel": "限制性描述", "source": "多项式 $3 x + 6 y + 1$ 的值", "target": "19"}]}}
|
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{"content": "If the equation $\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }$ has a positive root, then $a$ = ____?", "answer": "4", "steps": "Going to the denominator, we get $x = 2 ( x - 4 ) + a$. Simplifying, we get $x + a - 8 = 0$. Since the equation $\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }$ has a repeated root, we have $x - 4 = 0$, which means $x = 4$. Therefore, $4 + a - 8 = 0$, and we get $a = 4$.", "expr_cands": ["\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }", "x", "a", "x = 2 ( x - 4 ) + a", "x + a - 8 = 0", "2 a + 2 x - 8 - 8 = 0", "\\frac { a + 2 x - 8 } { a + 2 x - 8 - 4 } = \\frac { a } { a + 2 x - 8 - 4 } + 2", "x - 4 = 0", "x = 4", "4 + a - 8 = 0", "a = 4"], "exprs": ["x = 2 ( x - 4 ) + a", "x - 4 = 0", "x + a - 8 = 0", "x = 4", "4 + a - 8 = 0", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }"}, {"id": "x = 2 ( x - 4 ) + a"}, {"id": "x + a - 8 = 0"}, {"id": "x - 4 = 0"}, {"id": "方程 $\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }$ 有增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 4"}, {"id": "4 + a - 8 = 0"}, {"id": "a = 4"}], "links": [{"rel": "同乘除", "source": "\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }", "target": "x = 2 ( x - 4 ) + a"}, {"rel": "被描述", "source": "\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }", "target": "x - 4 = 0"}, {"rel": "移项", "source": "x = 2 ( x - 4 ) + a", "target": "x + a - 8 = 0"}, {"rel": "被代入", "source": "x + a - 8 = 0", "target": "4 + a - 8 = 0"}, {"rel": "等式方程求解", "source": "x - 4 = 0", "target": "x = 4"}, {"rel": "限制性描述", "source": "方程 $\\frac { x } { x - 4 } = 2 + \\frac { a } { x - 4 }$ 有增根", "target": "x - 4 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 4 = 0"}, {"rel": "代入", "source": "x = 4", "target": "4 + a - 8 = 0"}, {"rel": "等式方程求解", "source": "4 + a - 8 = 0", "target": "a = 4"}]}}
|
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{"content": "If $( m + 1 ) x ^ { | m | } < 2019$ is a one-variable linear inequality, then $m$ = ____?", "answer": "1", "steps": "$\\because$ $( m + 1 ) x ^ { | m | } < 2019$ is a one-variable linear inequality in terms of $x$, $\\therefore$ $m + 1 \\neq 0$, $| m | = 1$, and solving yields: $m = 1$.", "expr_cands": ["( m + 1 ) x ^ { | m | } < 2019", "x", "m", "m + 1 \\neq 0", "m \\neq - 1", "| m | = 1", "m = - 1", "m = 1"], "exprs": ["m + 1 \\neq 0", "| m | = 1", "m \\neq - 1", "m = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m + 1 ) x ^ { | m | } < 2019"}, {"id": "m + 1 \\neq 0"}, {"id": "$( m + 1 ) x ^ { | m | } < 2019$ 是关于 $x$ 的一元一次不等式"}, {"id": "m \\neq - 1"}, {"id": "| m | = 1"}, {"id": "m = 1"}], "links": [{"rel": "被描述", "source": "( m + 1 ) x ^ { | m | } < 2019", "target": "m + 1 \\neq 0"}, {"rel": "被描述", "source": "( m + 1 ) x ^ { | m | } < 2019", "target": "| m | = 1"}, {"rel": "不等式方程求解", "source": "m + 1 \\neq 0", "target": "m \\neq - 1"}, {"rel": "限制性描述", "source": "$( m + 1 ) x ^ { | m | } < 2019$ 是关于 $x$ 的一元一次不等式", "target": "m + 1 \\neq 0"}, {"rel": "限制性描述", "source": "$( m + 1 ) x ^ { | m | } < 2019$ 是关于 $x$ 的一元一次不等式", "target": "| m | = 1"}, {"rel": "联立", "source": "m \\neq - 1", "target": "m = 1"}, {"rel": "联立", "source": "| m | = 1", "target": "m = 1"}]}}
|
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{"content": "If $( a + 1 ) x ^ { | a | } + 3 y = 1$ is a linear equation in two variables $x$ and $y$, then $a$ = ____ ?", "answer": "1", "steps": "The equation $( a + 1 ) x ^ { | a | } + 3 y = 1$ is a second-degree equation in two variables $x$ and $y$. It is known that $| a | = 1$ and $a + 1 \\neq 0$, and solving the equation yields $a = 1$.", "expr_cands": ["( a + 1 ) x ^ { | a | } + 3 y = 1", "a", "x", "y", "| a | = 1", "a = - 1", "a = 1", "a + 1 \\neq 0", "a \\neq - 1"], "exprs": ["| a | = 1", "a + 1 \\neq 0", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a + 1 ) x ^ { | a | } + 3 y = 1"}, {"id": "| a | = 1"}, {"id": "$( a + 1 ) x ^ { | a | } + 3 y = 1$ 是关于 $x$ , $y$ 的二元一次方程"}, {"id": "a + 1 \\neq 0"}, {"id": "a = 1"}], "links": [{"rel": "被描述", "source": "( a + 1 ) x ^ { | a | } + 3 y = 1", "target": "| a | = 1"}, {"rel": "被描述", "source": "( a + 1 ) x ^ { | a | } + 3 y = 1", "target": "a + 1 \\neq 0"}, {"rel": "联立", "source": "| a | = 1", "target": "a = 1"}, {"rel": "限制性描述", "source": "$( a + 1 ) x ^ { | a | } + 3 y = 1$ 是关于 $x$ , $y$ 的二元一次方程", "target": "| a | = 1"}, {"rel": "限制性描述", "source": "$( a + 1 ) x ^ { | a | } + 3 y = 1$ 是关于 $x$ , $y$ 的二元一次方程", "target": "a + 1 \\neq 0"}, {"rel": "联立", "source": "a + 1 \\neq 0", "target": "a = 1"}]}}
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||||
{"content": "The solution set of the inequality $\\frac { x - 1 } { 2 } < - 1$ is _____.", "answer": "x < - 1", "steps": "Going to the denominator, we get $x - 1 < - 2$. Moving terms, we get $x < - 2 + 1$. Combining like terms, we get $x < - 1$.", "expr_cands": ["\\frac { x - 1 } { 2 } < - 1", "x", "x - 1 < - 2", "x < - 1", "x < - 2 + 1"], "exprs": ["x - 1 < - 2", "x < - 2 + 1", "x < - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 1 } { 2 } < - 1"}, {"id": "x - 1 < - 2"}, {"id": "x < - 2 + 1"}, {"id": "x < - 1"}], "links": [{"rel": "同乘除", "source": "\\frac { x - 1 } { 2 } < - 1", "target": "x - 1 < - 2"}, {"rel": "移项", "source": "x - 1 < - 2", "target": "x < - 2 + 1"}, {"rel": "不等式方程求解", "source": "x - 1 < - 2", "target": "x < - 1"}]}}
|
||||
{"content": "What is the minimum value of $| x - 6 | + | x - 1 |$?", "answer": "5", "steps": "The minimum value of $| x - 6 | + | x - 1 |$ is $6 - 1 = 5$.", "expr_cands": ["| x - 6 | + | x - 1 |", "x", "6 - 1", "5"], "exprs": ["6 - 1", "5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x - 6 | + | x - 1 |"}, {"id": "6 - 1"}, {"id": "$| x - 6 | + | x - 1 |$ 的最小值"}, {"id": "绝对值恒大于等于0"}, {"id": "5"}], "links": [{"rel": "被描述", "source": "| x - 6 | + | x - 1 |", "target": "6 - 1"}, {"rel": "计算", "source": "6 - 1", "target": "5"}, {"rel": "限制性描述", "source": "$| x - 6 | + | x - 1 |$ 的最小值", "target": "6 - 1"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "6 - 1"}]}}
|
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{"content": "Translate $y = 2 x + 3$ to the right by $2$ units, the new equation of the line is ____? ", "answer": "y = 2 x - 1", "steps": "\\because $y = 2 x + 3$ is shifted $2$ units to the right to obtain $y = 2 ( x - 2 ) + 3$ , \\therefore the new equation after translation is: $y = 2 x - 1$.", "expr_cands": ["y = 2 x + 3", "y", "x", "2", "y = 2 ( x - 2 ) + 3", "2 x + 3 = 2 ( x - 2 ) + 3", "2 x - 1"], "exprs": ["y = 2 ( x - 2 ) + 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "y = 2 ( x - 2 ) + 3"}, {"id": "y = 2 x + 3"}, {"id": "将 $y = 2 x + 3$ 相右平移 $2$ 个单位"}], "links": [{"rel": "被描述", "source": "2", "target": "y = 2 ( x - 2 ) + 3"}, {"rel": "被描述", "source": "y = 2 x + 3", "target": "y = 2 ( x - 2 ) + 3"}, {"rel": "限制性描述", "source": "将 $y = 2 x + 3$ 相右平移 $2$ 个单位", "target": "y = 2 ( x - 2 ) + 3"}]}}
|
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{"content": "Given that $x = 2$ is a root of the quadratic equation $x ^ 2 - 4 x + c = 0$, what is the value of $c$?", "answer": "4", "steps": "Substituting $x = 2$ into $x ^ 2 - 4 x + c = 0$, we get $2 ^ 2 - 4 \\times 2 + c = 0$. Solving for $c$, we get $c = 4$.", "expr_cands": ["x = 2", "x", "x ^ { 2 } - 4 x + c = 0", "c", "c - 4 = 0", "2 ^ { 2 } - 4 * 2 + c = 0", "c = 4"], "exprs": ["2 ^ { 2 } - 4 * 2 + c = 0", "c = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 4 x + c = 0"}, {"id": "2 ^ { 2 } - 4 * 2 + c = 0"}, {"id": "x = 2"}, {"id": "c = 4"}], "links": [{"rel": "被代入", "source": "x ^ { 2 } - 4 x + c = 0", "target": "2 ^ { 2 } - 4 * 2 + c = 0"}, {"rel": "等式方程求解", "source": "2 ^ { 2 } - 4 * 2 + c = 0", "target": "c = 4"}, {"rel": "代入", "source": "x = 2", "target": "2 ^ { 2 } - 4 * 2 + c = 0"}]}}
|
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{"content": "$\\frac { a } { 2 } = \\frac { b } { 3 } \\neq 0$ , what is the value of the algebraic expression $\\frac { 5 a - b } { a ^ { 2 } - 4 b ^ { 2 } } \\times ( a - 2 b )$? ", "answer": "\\frac { 7 } { 8 }", "steps": "Let $\\frac { a } { 2 } = \\frac { b } { 3 } = k$, then $a = 2 k$, $b = 3 k$. The original expression is $\\frac { 5 a - b } {( a + 2 b ) ( a - 2 b )} \\times ( a - 2 b ) = \\frac { 5 a - b } { a + 2 b } = \\frac { 5 \\times 2 k - 3 k } { 2 k + 2 \\times 3 k } = \\frac { 7 } { 8 }$.", "expr_cands": ["\\frac { a } { 2 } = \\frac { b } { 3 } \\neq 0", "\\frac { 5 a - b } { a ^ { 2 } - 4 b ^ { 2 } } \\times ( a - 2 b )", "a", "b", "\\frac { a } { 2 } = k", "k", "a = 2 k", "b = 3 k", "\\frac { 5 a - b } { ( a + 2 b ) ( a - 2 b ) } \\times ( a - 2 b )", "\\frac { 7 } { 8 }"], "exprs": ["a = 2 k", "b = 3 k", "\\frac { 7 } { 8 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $\\frac { a } { 2 } = \\frac { b } { 3 } = k$"}, {"id": "a = 2 k"}, {"id": "b = 3 k"}, {"id": "\\frac { 5 a - b } { a ^ { 2 } - 4 b ^ { 2 } } \\times ( a - 2 b )"}, {"id": "\\frac { 7 } { 8 }"}], "links": [{"rel": "假设描述", "source": "设 $\\frac { a } { 2 } = \\frac { b } { 3 } = k$", "target": "a = 2 k"}, {"rel": "假设描述", "source": "设 $\\frac { a } { 2 } = \\frac { b } { 3 } = k$", "target": "b = 3 k"}, {"rel": "代入", "source": "a = 2 k", "target": "\\frac { 7 } { 8 }"}, {"rel": "代入", "source": "b = 3 k", "target": "\\frac { 7 } { 8 }"}, {"rel": "被代入", "source": "\\frac { 5 a - b } { a ^ { 2 } - 4 b ^ { 2 } } \\times ( a - 2 b )", "target": "\\frac { 7 } { 8 }"}]}}
|
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{"content": "When $x$ = ____ ?, the value of the algebraic expression $\\frac { x - 2 } { x }$ is 0. ", "answer": "2", "steps": "It is known from the problem that $x - 2 = 0$ and $x \\neq 0$. Solving for $x$, we get $x = 2$.", "expr_cands": ["x", "\\frac { x - 2 } { x }", "0", "x - 2 = 0", "x = 2", "x \\neq 0"], "exprs": ["x - 2 = 0", "x \\neq 0", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 2 } { x }"}, {"id": "x - 2 = 0"}, {"id": "0"}, {"id": "代数式 $\\frac { x - 2 } { x }$ 的值为 $0$"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "x = 2"}, {"id": "x \\neq 0"}], "links": [{"rel": "被描述", "source": "\\frac { x - 2 } { x }", "target": "x - 2 = 0"}, {"rel": "被描述", "source": "\\frac { x - 2 } { x }", "target": "x \\neq 0"}, {"rel": "联立", "source": "x - 2 = 0", "target": "x = 2"}, {"rel": "被描述", "source": "0", "target": "x - 2 = 0"}, {"rel": "被描述", "source": "0", "target": "x \\neq 0"}, {"rel": "限制性描述", "source": "代数式 $\\frac { x - 2 } { x }$ 的值为 $0$", "target": "x - 2 = 0"}, {"rel": "限制性描述", "source": "代数式 $\\frac { x - 2 } { x }$ 的值为 $0$", "target": "x \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x - 2 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x \\neq 0"}, {"rel": "联立", "source": "x \\neq 0", "target": "x = 2"}]}}
|
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{"content": "Given that $x = 2$ is a root of the quadratic equation $x ^ 2 - mx - 6 = 0$, what is the value of $m$?", "answer": "- 1", "steps": "$\\because x = 2$ is a root of the quadratic equation $x ^ 2 - mx - 6 = 0$, $\\therefore 2 ^ 2 - 2 m - 6 = 0$, $\\therefore m = - 1$.", "expr_cands": ["x = 2", "x", "x ^ { 2 } - mx - 6 = 0", "m", "- 2 m - 6 + 4 = 0", "2 ^ { 2 } - 2 m - 6 = 0", "m = - 1"], "exprs": ["2 ^ { 2 } - 2 m - 6 = 0", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - mx - 6 = 0"}, {"id": "2 ^ { 2 } - 2 m - 6 = 0"}, {"id": "x = 2"}, {"id": "$x = 2$ 是一元二次方程方程 $x ^ { 2 } - mx - 6 = 0$ 的一个根"}, {"id": "m = - 1"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - mx - 6 = 0", "target": "2 ^ { 2 } - 2 m - 6 = 0"}, {"rel": "等式方程求解", "source": "2 ^ { 2 } - 2 m - 6 = 0", "target": "m = - 1"}, {"rel": "被描述", "source": "x = 2", "target": "2 ^ { 2 } - 2 m - 6 = 0"}, {"rel": "限制性描述", "source": "$x = 2$ 是一元二次方程方程 $x ^ { 2 } - mx - 6 = 0$ 的一个根", "target": "2 ^ { 2 } - 2 m - 6 = 0"}]}}
|
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{"content": "Given that the equation $( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0$ is a quadratic equation, what is the value of $a$?", "answer": "- 4", "steps": "From the given condition, we can obtain $| a | - 2 = 2$ and $a - 4 \\neq 0$. Solving these equations, we get $a = - 4$.", "expr_cands": ["( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0", "a", "x", "| a | - 2 = 2", "a = - 4", "a = 4", "a - 4 \\neq 0", "a \\neq 4"], "exprs": ["| a | - 2 = 2", "a - 4 \\neq 0", "a = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0"}, {"id": "| a | - 2 = 2"}, {"id": "方程 $( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0$ 是一个一元二次方程"}, {"id": "a - 4 \\neq 0"}, {"id": "a = - 4"}], "links": [{"rel": "被描述", "source": "( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0", "target": "| a | - 2 = 2"}, {"rel": "被描述", "source": "( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0", "target": "a - 4 \\neq 0"}, {"rel": "联立", "source": "| a | - 2 = 2", "target": "a = - 4"}, {"rel": "限制性描述", "source": "方程 $( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0$ 是一个一元二次方程", "target": "| a | - 2 = 2"}, {"rel": "限制性描述", "source": "方程 $( a - 4 ) x ^ { | a | - 2 } + 2 x + a = 0$ 是一个一元二次方程", "target": "a - 4 \\neq 0"}, {"rel": "联立", "source": "a - 4 \\neq 0", "target": "a = - 4"}]}}
|
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{"content": "Given $\\frac { x - y } { x + y } = - \\frac { 1 } { 5 }$, what is the value of $\\frac { x } { y }$?", "answer": "\\frac { 2 } { 3 }", "steps": "$\\because \\frac { x - y } { x + y } = - \\frac { 1 } { 5 }$, $\\therefore 5 ( x - y ) = - x - y$, which means $x = \\frac { 2 } { 3 } y$. Substituting $x = \\frac { 2 } { 3 } y$ yields $\\frac { \\frac { 2 } { 3 } y } { y } = \\frac { 2 } { 3 }$.", "expr_cands": ["\\frac { x - y } { x + y } = - \\frac { 1 } { 5 }", "x", "y", "\\frac { x } { y }", "5 ( x - y ) = - x - y", "x = \\frac { 2 } { 3 } y", "\\frac { 2 } { 3 } y", "\\frac { \\frac { 2 } { 3 } y } { y }", "\\frac { 2 } { 3 }"], "exprs": ["5 ( x - y ) = - x - y", "x = \\frac { 2 } { 3 } y", "\\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - y } { x + y } = - \\frac { 1 } { 5 }"}, {"id": "5 ( x - y ) = - x - y"}, {"id": "x = \\frac { 2 } { 3 } y"}, {"id": "\\frac { x } { y }"}, {"id": "\\frac { 2 } { 3 }"}], "links": [{"rel": "同乘除", "source": "\\frac { x - y } { x + y } = - \\frac { 1 } { 5 }", "target": "5 ( x - y ) = - x - y"}, {"rel": "等式方程部分求解", "source": "5 ( x - y ) = - x - y", "target": "x = \\frac { 2 } { 3 } y"}, {"rel": "代入", "source": "x = \\frac { 2 } { 3 } y", "target": "\\frac { 2 } { 3 }"}, {"rel": "被代入", "source": "\\frac { x } { y }", "target": "\\frac { 2 } { 3 }"}]}}
|
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{"content": "If the value of the algebraic expression $1 - 3 a$ is $- 2$, then $a$ is ____?", "answer": "1", "steps": "According to the problem, we have $1 - 3 a = - 2$. By rearranging and combining terms, we get $- 3 a = - 3$. Solving for $a$, we get $a = 1$.", "expr_cands": ["1 - 3 a", "a", "- 2", "1 - 3 a = - 2", "a = 1", "- 3 a = - 3"], "exprs": ["1 - 3 a = - 2", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1 - 3 a"}, {"id": "1 - 3 a = - 2"}, {"id": "- 2"}, {"id": "代数式 $1 - 3 a$ 的值为 $- 2$"}, {"id": "a = 1"}], "links": [{"rel": "被描述", "source": "1 - 3 a", "target": "1 - 3 a = - 2"}, {"rel": "等式方程求解", "source": "1 - 3 a = - 2", "target": "a = 1"}, {"rel": "被描述", "source": "- 2", "target": "1 - 3 a = - 2"}, {"rel": "限制性描述", "source": "代数式 $1 - 3 a$ 的值为 $- 2$", "target": "1 - 3 a = - 2"}]}}
|
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{"content": "Given a one-variable linear equation in $x$, $2019 x + 3 = 2 x + b$, with a solution of $2 x - k - 3 = 0$, what is the solution to the one-variable linear equation in $y$, $2019 ( y + 1 ) + 3 = 2 ( y + 1 ) + b$?", "answer": "y = 1", "steps": "According to the problem, we get: $y + 1 = 2$, solving it, we get: $y = 1$.", "expr_cands": ["x", "2019 x + 3 = 2 x + b", "b", "2 x - k - 3 = 0", "k", "y", "2019 ( y + 1 ) + 3 = 2 ( y + 1 ) + b", "y + 1 = 2", "y = 1"], "exprs": ["y + 1 = 2", "y = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2019 ( y + 1 ) + 3 = 2 ( y + 1 ) + b"}, {"id": "y + 1 = 2"}, {"id": "2019 x + 3 = 2 x + b"}, {"id": "2 x - k - 3 = 0"}, {"id": "关于 $x$ 的一元一次方程 $2019 x + 3 = 2 x + b$ 的解为 $2 x - k - 3 = 0$"}, {"id": "关于 $y$ 的一元一次方程 $2019 ( y + 1 ) + 3 = 2 ( y + 1 ) + b$ 的解"}, {"id": "y = 1"}], "links": [{"rel": "被描述", "source": "2019 ( y + 1 ) + 3 = 2 ( y + 1 ) + b", "target": "y + 1 = 2"}, {"rel": "等式方程求解", "source": "y + 1 = 2", "target": "y = 1"}, {"rel": "被描述", "source": "2019 x + 3 = 2 x + b", "target": "y + 1 = 2"}, {"rel": "被描述", "source": "2 x - k - 3 = 0", "target": "y + 1 = 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元一次方程 $2019 x + 3 = 2 x + b$ 的解为 $2 x - k - 3 = 0$", "target": "y + 1 = 2"}, {"rel": "限制性描述", "source": "关于 $y$ 的一元一次方程 $2019 ( y + 1 ) + 3 = 2 ( y + 1 ) + b$ 的解", "target": "y + 1 = 2"}]}}
|
||||
{"content": "To make the algebraic expression $\\sqrt { 1 - 2 x }$ meaningful, what is the maximum value of $x$?", "answer": "\\frac { 1 } { 2 }", "steps": "$\\because$ The algebraic expression $\\sqrt { 1 - 2 x }$ is meaningful, $\\therefore$ $1 - 2 x \\ge 0$, which leads to $x \\le \\frac { 1 } { 2 }$. $\\therefore$ The maximum value of $x$ is $\\frac { 1 } { 2 }$.", "expr_cands": ["\\sqrt { 1 - 2 x }", "x", "1 - 2 x \\ge 0", "x \\le \\frac { 1 } { 2 }", "\\frac { 1 } { 2 }"], "exprs": ["1 - 2 x \\ge 0", "x \\le \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 1 - 2 x }"}, {"id": "1 - 2 x \\ge 0"}, {"id": "要使代数式 $\\sqrt { 1 - 2 x }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\le \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\sqrt { 1 - 2 x }", "target": "1 - 2 x \\ge 0"}, {"rel": "不等式方程求解", "source": "1 - 2 x \\ge 0", "target": "x \\le \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "要使代数式 $\\sqrt { 1 - 2 x }$ 有意义", "target": "1 - 2 x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "1 - 2 x \\ge 0"}]}}
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{"content": "The two square roots of a positive number are $m + 1$ and $m - 3$. Find the value of the positive number.", "answer": "4", "steps": "According to the problem, $( m + 1 ) + ( m - 3 ) = 0$. Solving for $m$, we get $m = 1$. Then, $m + 1 = 1 + 1 = 2$. Since $2 ^ 2 = 4$, the positive integer in question is $4$.", "expr_cands": ["m + 1", "m", "m - 3", "( m + 1 ) + ( m - 3 ) = 0", "m = 1", "2", "2 ^ { 2 }", "4"], "exprs": ["( m + 1 ) + ( m - 3 ) = 0", "m = 1", "2", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m + 1"}, {"id": "( m + 1 ) + ( m - 3 ) = 0"}, {"id": "m - 3"}, {"id": "一个正数的两个平方根为 $m + 1$ 和 $m - 3$"}, {"id": "平方根互为相反数"}, {"id": "m = 1"}, {"id": "2"}, {"id": "4"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "m + 1", "target": "( m + 1 ) + ( m - 3 ) = 0"}, {"rel": "被代入", "source": "m + 1", "target": "2"}, {"rel": "等式方程求解", "source": "( m + 1 ) + ( m - 3 ) = 0", "target": "m = 1"}, {"rel": "被描述", "source": "m - 3", "target": "( m + 1 ) + ( m - 3 ) = 0"}, {"rel": "限制性描述", "source": "一个正数的两个平方根为 $m + 1$ 和 $m - 3$", "target": "( m + 1 ) + ( m - 3 ) = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "( m + 1 ) + ( m - 3 ) = 0"}, {"rel": "代入", "source": "m = 1", "target": "2"}, {"rel": "被描述", "source": "2", "target": "4"}, {"rel": "限制性描述", "source": "平方", "target": "4"}]}}
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{"content": "If $m = 2016$, then $( m + 1 ) ^ 2 - m ( m + 1 )$ = ____?", "answer": "2017", "steps": "$( m + 1 ) ^ 2 - m ( m + 1 ) = ( m + 1 ) ( m + 1 - m ) = m + 1$, when $m = 2016$, the original expression equals $2016 + 1 = 2017$.", "expr_cands": ["m = 2016", "m", "( m + 1 ) ^ { 2 } - m ( m + 1 )", "m + 1", "2016 + 1", "2017"], "exprs": ["2017"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m = 2016"}, {"id": "2017"}, {"id": "( m + 1 ) ^ { 2 } - m ( m + 1 )"}], "links": [{"rel": "代入", "source": "m = 2016", "target": "2017"}, {"rel": "被代入", "source": "( m + 1 ) ^ { 2 } - m ( m + 1 )", "target": "2017"}]}}
|
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{"content": "The range of values for $x$ that make the function $y = \\sqrt { 2 - x }$ meaningful is ____ ?", "answer": "x \\le 2", "steps": "From the given condition, we have $2 - x \\geq 0$, which implies $x \\leq 2$.", "expr_cands": ["y = \\sqrt { 2 - x }", "y", "x", "2 - x \\ge 0", "x \\le 2"], "exprs": ["2 - x \\ge 0", "x \\le 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { 2 - x }"}, {"id": "2 - x \\ge 0"}, {"id": "函数 $y = \\sqrt { 2 - x }$ 有意义的 $x$ 的取值范围"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\le 2"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { 2 - x }", "target": "2 - x \\ge 0"}, {"rel": "不等式方程求解", "source": "2 - x \\ge 0", "target": "x \\le 2"}, {"rel": "限制性描述", "source": "函数 $y = \\sqrt { 2 - x }$ 有意义的 $x$ 的取值范围", "target": "2 - x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 - x \\ge 0"}]}}
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||||
{"content": "If the fractional equation $\\frac { x } { x - 3 } = \\frac { 3 } { x - 3 } + 2$ has a positive root, then the possible root is ____?", "answer": "x = 3", "steps": "$\\because$ The original equation has an extraneous root, $\\therefore$ the simplest common denominator is $x - 3 = 0$, which gives the solution $x = 3$. Thus, the extraneous root is $x = 3$.", "expr_cands": ["\\frac { x } { x - 3 } = \\frac { 3 } { x - 3 } + 2", "x", "x - 3 = 0", "x = 3"], "exprs": ["x - 3 = 0", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x - 3 } = \\frac { 3 } { x - 3 } + 2"}, {"id": "x - 3 = 0"}, {"id": "分式方程 $\\frac { x } { x - 3 } = \\frac { 3 } { x - 3 } + 2$ 有增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "\\frac { x } { x - 3 } = \\frac { 3 } { x - 3 } + 2", "target": "x - 3 = 0"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "限制性描述", "source": "分式方程 $\\frac { x } { x - 3 } = \\frac { 3 } { x - 3 } + 2$ 有增根", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 3 = 0"}]}}
|
||||
{"content": "Given the function $f ( x ) = \\frac { x } { 2 x + 3 }$, what is the range of values for the independent variable $x$?", "answer": "x \\neq - \\frac { 3 } { 2 }", "steps": "Because $2 x + 3 \\neq 0$, therefore $x \\neq - \\frac { 3 } { 2 }$.", "expr_cands": ["f ( x ) = \\frac { x } { 2 x + 3 }", "f", "x", "2 x + 3 \\neq 0", "x \\neq - \\frac { 3 } { 2 }"], "exprs": ["2 x + 3 \\neq 0", "x \\neq - \\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "f ( x ) = \\frac { x } { 2 x + 3 }"}, {"id": "2 x + 3 \\neq 0"}, {"id": "分式有意义，则分母不为0"}, {"id": "函数 $f ( x ) = \\frac { x } { 2 x + 3 }$"}, {"id": "x \\neq - \\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "f ( x ) = \\frac { x } { 2 x + 3 }", "target": "2 x + 3 \\neq 0"}, {"rel": "不等式方程求解", "source": "2 x + 3 \\neq 0", "target": "x \\neq - \\frac { 3 } { 2 }"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "2 x + 3 \\neq 0"}, {"rel": "限制性描述", "source": "函数 $f ( x ) = \\frac { x } { 2 x + 3 }$", "target": "2 x + 3 \\neq 0"}]}}
|
||||
{"content": "If the solution to the equation $x - 4 = a + 2 x$ is $x = - 2$, then $a$ = ____?", "answer": "- 2", "steps": "Substituting $x = - 2$ into the equation gives $- 2 - 4 = a - 4$, solving for $a$ gives $a = - 2$.", "expr_cands": ["x - 4 = a + 2 x", "a", "x", "x = - 2", "- 2 - 4 = a - 4", "a = - 2"], "exprs": ["- 2 - 4 = a - 4", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - 4 = a + 2 x"}, {"id": "- 2 - 4 = a - 4"}, {"id": "x = - 2"}, {"id": "a = - 2"}], "links": [{"rel": "被代入", "source": "x - 4 = a + 2 x", "target": "- 2 - 4 = a - 4"}, {"rel": "等式方程求解", "source": "- 2 - 4 = a - 4", "target": "a = - 2"}, {"rel": "代入", "source": "x = - 2", "target": "- 2 - 4 = a - 4"}]}}
|
||||
{"content": "Given $m$ and $n$ are opposite numbers, what is the value of $m + n - 3$?", "answer": "- 3", "steps": "$\\because$ $m$ and $n$ are opposite in sign, $\\therefore$ $m + n = 0$, $\\therefore$ $m + n - 3 = - 3$.", "expr_cands": ["m", "n", "m + n - 3", "m + n = 0", "- 3"], "exprs": ["m + n = 0", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m"}, {"id": "m + n = 0"}, {"id": "n"}, {"id": "$m$ , $n$ 互为相反数"}, {"id": "m + n - 3"}, {"id": "- 3"}], "links": [{"rel": "被描述", "source": "m", "target": "m + n = 0"}, {"rel": "代入", "source": "m + n = 0", "target": "- 3"}, {"rel": "被描述", "source": "n", "target": "m + n = 0"}, {"rel": "限制性描述", "source": "$m$ , $n$ 互为相反数", "target": "m + n = 0"}, {"rel": "被代入", "source": "m + n - 3", "target": "- 3"}]}}
|
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{"content": "The solution set of the inequality $3 x - 1 > 8$ is ____?", "answer": "x > 3", "steps": "$3 x > 8 + 1$ means three times x is greater than eight plus one.$3 x > 9$ means three times x is greater than nine.$x > 3$ means x is greater than three.", "expr_cands": ["3 x - 1 > 8", "x", "3 x > 8 + 1", "3 < x", "3 x > 9", "x > 3"], "exprs": ["x > 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 1 > 8"}, {"id": "x > 3"}], "links": [{"rel": "不等式方程求解", "source": "3 x - 1 > 8", "target": "x > 3"}]}}
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{"content": "On each branch of the hyperbola $y = \\frac { 2 - k } { x }$, $y$ decreases as $x$ increases. The range of values for $k$ is ____?", "answer": "k < 2", "steps": "From the given condition, we have $2 - k > 0$, which implies that $k < 2$.", "expr_cands": ["y = \\frac { 2 - k } { x }", "y", "k", "x", "2 - k > 0", "k < 2"], "exprs": ["2 - k > 0", "k < 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 2 - k } { x }"}, {"id": "2 - k > 0"}, {"id": "在双曲线 $y = \\frac { 2 - k } { x }$ 的每一支上"}, {"id": "$y$ 都随着 $x$ 的增大而减小"}, {"id": "k < 2"}], "links": [{"rel": "被描述", "source": "y = \\frac { 2 - k } { x }", "target": "2 - k > 0"}, {"rel": "不等式方程求解", "source": "2 - k > 0", "target": "k < 2"}, {"rel": "限制性描述", "source": "在双曲线 $y = \\frac { 2 - k } { x }$ 的每一支上", "target": "2 - k > 0"}, {"rel": "限制性描述", "source": "$y$ 都随着 $x$ 的增大而减小", "target": "2 - k > 0"}]}}
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{"content": "The solution set of the inequality $- \\frac { 1 } { 3 } x + 1 \\le - 5$ is ____ ?", "answer": "x \\ge 18", "steps": "Moving terms yields: $- \\frac { 1 } { 3 } x \\le - 5 - 1$, combining like terms gives: $- \\frac { 1 } { 3 } x \\le - 6$, multiplying both sides by $- 3$ gives: $x \\ge 18$. Therefore, the solution set for the inequality $- \\frac { 1 } { 3 } x + 1 \\le - 5$ is $x \\ge 18$.", "expr_cands": ["- \\frac { 1 } { 3 } x + 1 \\le - 5", "x", "- \\frac { 1 } { 3 } x \\le - 5 - 1", "18 \\le x", "- \\frac { 1 } { 3 } x \\le - 6", "1", "x \\ge 18"], "exprs": ["- \\frac { 1 } { 3 } x \\le - 5 - 1", "- \\frac { 1 } { 3 } x \\le - 6", "x \\ge 18"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 1 } { 3 } x + 1 \\le - 5"}, {"id": "- \\frac { 1 } { 3 } x \\le - 5 - 1"}, {"id": "- \\frac { 1 } { 3 } x \\le - 6"}, {"id": "x \\ge 18"}], "links": [{"rel": "移项", "source": "- \\frac { 1 } { 3 } x + 1 \\le - 5", "target": "- \\frac { 1 } { 3 } x \\le - 5 - 1"}, {"rel": "计算", "source": "- \\frac { 1 } { 3 } x \\le - 5 - 1", "target": "- \\frac { 1 } { 3 } x \\le - 6"}, {"rel": "不等式方程求解", "source": "- \\frac { 1 } { 3 } x \\le - 6", "target": "x \\ge 18"}]}}
|
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{"content": "If $y = x + 5 - b$ is a direct proportion function, then $b$ = ____?", "answer": "5", "steps": "According to the problem, we get: $5 - b = 0$, solving for $b$, we get: $b = 5$.", "expr_cands": ["y = x + 5 - b", "y", "b", "x", "5 - b = 0", "b = 5"], "exprs": ["5 - b = 0", "b = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x + 5 - b"}, {"id": "5 - b = 0"}, {"id": "$y = x + 5 - b$ 是正比例函数"}, {"id": "b = 5"}], "links": [{"rel": "被描述", "source": "y = x + 5 - b", "target": "5 - b = 0"}, {"rel": "等式方程求解", "source": "5 - b = 0", "target": "b = 5"}, {"rel": "限制性描述", "source": "$y = x + 5 - b$ 是正比例函数", "target": "5 - b = 0"}]}}
|
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{"content": "The equation in terms of $x$: $3 x ^ { m - 1 } - 2 m = 0$ is a linear equation in one variable. What is the value of $m$?", "answer": "2", "steps": "From the given information, we have $m - 1 = 1$, which implies $m = 2$ after solving for $m$. ", "expr_cands": ["x", "3 x ^ { m - 1 } - 2 m = 0", "m", "m - 1 = 1", "m = 2"], "exprs": ["m - 1 = 1", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { m - 1 } - 2 m = 0"}, {"id": "m - 1 = 1"}, {"id": "关于 $x$ 的方程 : $3 x ^ { m - 1 } - 2 m = 0$ 是一元一次方程"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "3 x ^ { m - 1 } - 2 m = 0", "target": "m - 1 = 1"}, {"rel": "等式方程求解", "source": "m - 1 = 1", "target": "m = 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 : $3 x ^ { m - 1 } - 2 m = 0$ 是一元一次方程", "target": "m - 1 = 1"}]}}
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{"content": "Given that the equation $2 x ^ 2 - x - 1 = 0$ has two roots $x _ 1$ and $x _ 2$, what is the value of $x _ 1 + x _ 2$?", "answer": "\\frac { 1 } { 2 }", "steps": "Since the equation $2 x ^ { 2 } - x - 1 = 0$ has two roots, denoted as $x _ { 1 }$ and $x _ { 2 }$, we have $x _ { 1 } + x _ { 2 } = - \\frac { b } { a }$. Therefore, $x _ { 1 } + x _ { 2 } = - \\frac { - 1 } { 2 } = \\frac { 1 } { 2 }$.", "expr_cands": ["2 x ^ { 2 } - x - 1 = 0", "x", "x _ { 1 }", "x _ { 2 }", "x _ { 1 } + x _ { 2 }", "x = - \\frac { 1 } { 2 }", "x = 1", "x _ { 1 } + x _ { 2 } = - \\frac { b } { a }", "b", "a", "x _ { 1 } + x _ { 2 } = \\frac { 1 } { 2 }", "\\frac { 1 } { 2 }"], "exprs": ["x _ { 1 } + x _ { 2 } = - \\frac { b } { a }", "\\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 2 } - x - 1 = 0"}, {"id": "x _ { 1 } + x _ { 2 } = - \\frac { b } { a }"}, {"id": "方程 $2 x ^ { 2 } - x - 1 = 0$ 的两根分别是 $x _ { 1 }$ 和 $x _ { 2 }$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "\\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "2 x ^ { 2 } - x - 1 = 0", "target": "x _ { 1 } + x _ { 2 } = - \\frac { b } { a }"}, {"rel": "被描述", "source": "2 x ^ { 2 } - x - 1 = 0", "target": "\\frac { 1 } { 2 }"}, {"rel": "被描述", "source": "x _ { 1 } + x _ { 2 } = - \\frac { b } { a }", "target": "\\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "方程 $2 x ^ { 2 } - x - 1 = 0$ 的两根分别是 $x _ { 1 }$ 和 $x _ { 2 }$", "target": "x _ { 1 } + x _ { 2 } = - \\frac { b } { a }"}, {"rel": "限制性描述", "source": "方程 $2 x ^ { 2 } - x - 1 = 0$ 的两根分别是 $x _ { 1 }$ 和 $x _ { 2 }$", "target": "\\frac { 1 } { 2 }"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + x _ { 2 } = - \\frac { b } { a }"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "\\frac { 1 } { 2 }"}]}}
|
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{"content": "The equation of the parabola $y = x ^ 2 + 1$ is shifted down by $2$ units and right by $3$ units. The new equation of the parabola is _____.", "answer": "y = ( x - 3 ) ^ { 2 } - 1", "steps": "The parabola $y = x ^ 2 + 1$ is shifted down by $2$ units, so its new equation is $y = x ^ 2 + 1 - 2 = x ^ 2 - 1$. Then it is shifted to the right by $3$ units, so its new equation is $y = ( x - 3 ) ^ 2 - 1$.", "expr_cands": ["y = x ^ { 2 } + 1", "y", "x", "2", "3", "y = x ^ { 2 } - 1", "y = ( x - 3 ) ^ { 2 } - 1", "x ^ { 2 } - 1 = ( x - 3 ) ^ { 2 } - 1", "x ^ { 2 } - 1"], "exprs": ["y = ( x - 3 ) ^ { 2 } - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "y = ( x - 3 ) ^ { 2 } - 1"}, {"id": "y = x ^ { 2 } + 1"}, {"id": "3"}, {"id": "将抛物线 $y = x ^ { 2 } + 1$ 向下平移 $2$ 个单位"}, {"id": "向右平移 $3$ 个单位"}], "links": [{"rel": "被描述", "source": "2", "target": "y = ( x - 3 ) ^ { 2 } - 1"}, {"rel": "被描述", "source": "y = x ^ { 2 } + 1", "target": "y = ( x - 3 ) ^ { 2 } - 1"}, {"rel": "被描述", "source": "3", "target": "y = ( x - 3 ) ^ { 2 } - 1"}, {"rel": "限制性描述", "source": "将抛物线 $y = x ^ { 2 } + 1$ 向下平移 $2$ 个单位", "target": "y = ( x - 3 ) ^ { 2 } - 1"}, {"rel": "限制性描述", "source": "向右平移 $3$ 个单位", "target": "y = ( x - 3 ) ^ { 2 } - 1"}]}}
|
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{"content": "If $- 2 a ^ { m } b ^ { 4 }$ and $5 a ^ { 3 } b ^ { 2 + n }$ can be combined into one term, then $m ^ { n }$ = ____ ?", "answer": "9", "steps": "Because $- 2 a ^ { m } b ^ { 4 }$ and $5 a ^ { 3 } b ^ { 2 + n }$ can be combined into one term, therefore $m = 3$, $4 = 2 + n$, therefore $m = 3$, $n = 2$, therefore $m ^ { n } = 3 ^ { 2 } = 9$.", "expr_cands": ["- 2 a ^ { m } b ^ { 4 }", "a", "b", "m", "5 a ^ { 3 } b ^ { 2 + n }", "n", "m ^ { n }", "m = 3", "4 = 2 + n", "n = 2", "9"], "exprs": ["m = 3", "4 = 2 + n", "n = 2", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 a ^ { m } b ^ { 4 }"}, {"id": "m = 3"}, {"id": "5 a ^ { 3 } b ^ { 2 + n }"}, {"id": "$- 2 a ^ { m } b ^ { 4 }$ 与 $5 a ^ { 3 } b ^ { 2 + n }$ 可以合并成一项"}, {"id": "4 = 2 + n"}, {"id": "n = 2"}, {"id": "m ^ { n }"}, {"id": "9"}], "links": [{"rel": "被描述", "source": "- 2 a ^ { m } b ^ { 4 }", "target": "m = 3"}, {"rel": "被描述", "source": "- 2 a ^ { m } b ^ { 4 }", "target": "4 = 2 + n"}, {"rel": "代入", "source": "m = 3", "target": "9"}, {"rel": "被描述", "source": "5 a ^ { 3 } b ^ { 2 + n }", "target": "m = 3"}, {"rel": "被描述", "source": "5 a ^ { 3 } b ^ { 2 + n }", "target": "4 = 2 + n"}, {"rel": "限制性描述", "source": "$- 2 a ^ { m } b ^ { 4 }$ 与 $5 a ^ { 3 } b ^ { 2 + n }$ 可以合并成一项", "target": "m = 3"}, {"rel": "限制性描述", "source": "$- 2 a ^ { m } b ^ { 4 }$ 与 $5 a ^ { 3 } b ^ { 2 + n }$ 可以合并成一项", "target": "4 = 2 + n"}, {"rel": "等式方程求解", "source": "4 = 2 + n", "target": "n = 2"}, {"rel": "代入", "source": "n = 2", "target": "9"}, {"rel": "被代入", "source": "m ^ { n }", "target": "9"}]}}
|
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{"content": "The coefficient of the $x ^ 2$ term in the product of the polynomials $( x ^ 4 - 2 x ^ 3 + x ^ 2 - 8 x + 1 )$ and $( x ^ 2 + 2 x - 3 )$ is _____.", "answer": "- 18", "steps": "$( { x } ^ { 4 } - 2 { x } ^ { 3 } + { x } ^ { 2 } - 8 x + 1 ) ( { x } ^ { 2 } + 2 x - 3 ) = x ^ { 6 } + 2 x ^ { 5 } - 3 x ^ { 4 } - 2 x ^ { 5 } - 4 x ^ { 4 } + 6 x ^ { 3 } + x ^ { 4 } + 2 x ^ { 3 } - 3 x ^ { 2 } - 8 x ^ { 3 } - 16 x ^ { 2 } + 24 x + x ^ { 2 } + 2 x - 3 = x ^ { 6 } - 6 x ^ { 4 } - 18 x ^ { 2 } + 26 x - 3$, therefore the coefficient of the $x ^ 2$ term in the product is $- 18$.", "expr_cands": ["( { x } ^ { 4 } - 2 { x } ^ { 3 } + { x } ^ { 2 } - 8 x + 1 ) ( { x } ^ { 2 } + 2 x - 3 )", "x", "x ^ { 2 }", "x ^ { 6 } - 6 x ^ { 4 } - 18 x ^ { 2 } + 26 x - 3", "- 18"], "exprs": ["x ^ { 6 } - 6 x ^ { 4 } - 18 x ^ { 2 } + 26 x - 3", "- 18"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( { x } ^ { 4 } - 2 { x } ^ { 3 } + { x } ^ { 2 } - 8 x + 1 ) ( { x } ^ { 2 } + 2 x - 3 )"}, {"id": "x ^ { 6 } - 6 x ^ { 4 } - 18 x ^ { 2 } + 26 x - 3"}, {"id": "- 18"}, {"id": "多项式 $( { x } ^ { 4 } - 2 { x } ^ { 3 } + { x } ^ { 2 } - 8 x + 1 ) ( { x } ^ { 2 } + 2 x - 3 )$ 的积中 $x ^ { 2 }$ 项的系数"}], "links": [{"rel": "展开", "source": "( { x } ^ { 4 } - 2 { x } ^ { 3 } + { x } ^ { 2 } - 8 x + 1 ) ( { x } ^ { 2 } + 2 x - 3 )", "target": "x ^ { 6 } - 6 x ^ { 4 } - 18 x ^ { 2 } + 26 x - 3"}, {"rel": "被描述", "source": "x ^ { 6 } - 6 x ^ { 4 } - 18 x ^ { 2 } + 26 x - 3", "target": "- 18"}, {"rel": "限制性描述", "source": "多项式 $( { x } ^ { 4 } - 2 { x } ^ { 3 } + { x } ^ { 2 } - 8 x + 1 ) ( { x } ^ { 2 } + 2 x - 3 )$ 的积中 $x ^ { 2 }$ 项的系数", "target": "- 18"}]}}
|
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{"content": "If $9 a ^ { x } b ^ { 7 }$ and $- 7 a ^ { 3 x - 4 } b ^ { 7 }$ are like terms, then $x$ = ____ ?", "answer": "2", "steps": "$9 a ^ { x } b ^ { 7 }$ and $- 7 a ^ { 3 x - 4 } b ^ { 7 }$ are like terms, so we have $x = 3 x - 4$, which gives $x = 2$ upon solving.", "expr_cands": ["9 a ^ { x } b ^ { 7 }", "b", "x", "a", "- 7 a ^ { 3 x - 4 } b ^ { 7 }", "x = 3 x - 4", "x = 2"], "exprs": ["x = 3 x - 4", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 7 a ^ { 3 x - 4 } b ^ { 7 }"}, {"id": "x = 3 x - 4"}, {"id": "9 a ^ { x } b ^ { 7 }"}, {"id": "$9 a ^ { x } b ^ { 7 }$ 与 $- 7 a ^ { 3 x - 4 } b ^ { 7 }$ 是同类项"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "- 7 a ^ { 3 x - 4 } b ^ { 7 }", "target": "x = 3 x - 4"}, {"rel": "等式方程求解", "source": "x = 3 x - 4", "target": "x = 2"}, {"rel": "被描述", "source": "9 a ^ { x } b ^ { 7 }", "target": "x = 3 x - 4"}, {"rel": "限制性描述", "source": "$9 a ^ { x } b ^ { 7 }$ 与 $- 7 a ^ { 3 x - 4 } b ^ { 7 }$ 是同类项", "target": "x = 3 x - 4"}]}}
|
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{"content": "The coefficient times the degree of the monomial $- \\frac { x ^ 2 y } { 3 }$ is ____?", "answer": "- 1", "steps": "The coefficient and degree of the monomial $- \\frac { x ^ 2 y } { 3 }$ are $- \\frac { 1 } { 3 }$ and $3$ respectively, so $- \\frac { 1 } { 3 } \\times 3 = - 1$.", "expr_cands": ["- \\frac { x ^ { 2 } y } { 3 }", "x", "y", "- \\frac { 1 } { 3 }", "3", "- \\frac { 1 } { 3 } * 3", "- 1"], "exprs": ["- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { x ^ { 2 } y } { 3 }"}, {"id": "- 1"}, {"id": "单项式 $- \\frac { x ^ { 2 } y } { 3 }$ 的系数与次数的积"}, {"id": "单项式 $- \\frac { x ^ { 2 } y } { 3 }$ 的系数与次数分别为 : $- \\frac { 1 } { 3 }$ , $3$"}], "links": [{"rel": "被描述", "source": "- \\frac { x ^ { 2 } y } { 3 }", "target": "- 1"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { x ^ { 2 } y } { 3 }$ 的系数与次数的积", "target": "- 1"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { x ^ { 2 } y } { 3 }$ 的系数与次数分别为 : $- \\frac { 1 } { 3 }$ , $3$", "target": "- 1"}]}}
|
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{"content": "The value of the algebraic expression $\\frac { m - 1 } { 3 } - 1$ is positive. What is the range of possible values for $m$?", "answer": "m > 4", "steps": "$\\because$ The value of the algebraic expression $\\frac { m - 1 } { 3 } - 1$ is positive. $\\therefore$ $\\frac { m - 1 } { 3 } - 1 > 0$. Solving for $m$, we get $m > 4$.", "expr_cands": ["\\frac { m - 1 } { 3 } - 1", "m", "\\frac { m - 1 } { 3 } - 1 > 0", "4 < m", "m > 4"], "exprs": ["\\frac { m - 1 } { 3 } - 1 > 0", "m > 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { m - 1 } { 3 } - 1"}, {"id": "\\frac { m - 1 } { 3 } - 1 > 0"}, {"id": "代数式 $\\frac { m - 1 } { 3 } - 1$ 的值为正数"}, {"id": "m > 4"}], "links": [{"rel": "被描述", "source": "\\frac { m - 1 } { 3 } - 1", "target": "\\frac { m - 1 } { 3 } - 1 > 0"}, {"rel": "不等式方程求解", "source": "\\frac { m - 1 } { 3 } - 1 > 0", "target": "m > 4"}, {"rel": "限制性描述", "source": "代数式 $\\frac { m - 1 } { 3 } - 1$ 的值为正数", "target": "\\frac { m - 1 } { 3 } - 1 > 0"}]}}
|
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{"content": "If the inequality $ax ^ { | a - 1 | } > 2$ is a linear inequality with one variable, then $a$ = ____ ?", "answer": "2", "steps": "According to the problem, we have $| a - 1 | = 1$ and $a \\neq 0$. Solving for $a$, we get $a = 2$.", "expr_cands": ["ax ^ { | a - 1 | } > 2", "x", "a", "| a - 1 | = 1", "a = 0", "a = 2", "a \\neq 0"], "exprs": ["| a - 1 | = 1", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ax ^ { | a - 1 | } > 2"}, {"id": "| a - 1 | = 1"}, {"id": "不等式 $ax ^ { | a - 1 | } > 2$ 是一元一次不等式"}, {"id": "a = 2"}, {"id": "a \\neq 0"}], "links": [{"rel": "被描述", "source": "ax ^ { | a - 1 | } > 2", "target": "| a - 1 | = 1"}, {"rel": "联立", "source": "| a - 1 | = 1", "target": "a = 2"}, {"rel": "限制性描述", "source": "不等式 $ax ^ { | a - 1 | } > 2$ 是一元一次不等式", "target": "| a - 1 | = 1"}, {"rel": "联立", "source": "a \\neq 0", "target": "a = 2"}]}}
|
||||
{"content": "Given $| a + 8 | + | b - 9 | = 0$, what is $a - b$?", "answer": "- 17", "steps": "Since $| a + 8 | + | b - 9 | = 0$, it follows that $a + 8 = 0$ and $b - 9 = 0$. Solving for $a$ and $b$, we get $a = - 8$ and $b = 9$. Therefore, $a - b = - 8 - 9 = - 17$.", "expr_cands": ["| a + 8 | + | b - 9 | = 0", "a", "b", "a - b", "a + 8 = 0", "a = - 8", "b - 9 = 0", "b = 9", "- 17"], "exprs": ["a + 8 = 0", "b - 9 = 0", "a = - 8", "b = 9", "- 17"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a + 8 | + | b - 9 | = 0"}, {"id": "a + 8 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "b - 9 = 0"}, {"id": "a = - 8"}, {"id": "b = 9"}, {"id": "a - b"}, {"id": "- 17"}], "links": [{"rel": "被描述", "source": "| a + 8 | + | b - 9 | = 0", "target": "a + 8 = 0"}, {"rel": "被描述", "source": "| a + 8 | + | b - 9 | = 0", "target": "b - 9 = 0"}, {"rel": "等式方程求解", "source": "a + 8 = 0", "target": "a = - 8"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a + 8 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b - 9 = 0"}, {"rel": "等式方程求解", "source": "b - 9 = 0", "target": "b = 9"}, {"rel": "代入", "source": "a = - 8", "target": "- 17"}, {"rel": "代入", "source": "b = 9", "target": "- 17"}, {"rel": "被代入", "source": "a - b", "target": "- 17"}]}}
|
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{"content": "If $3 x ^ { 3 } y ^ { m + 1 }$ and $- 5 x ^ { n - 2 } y ^ { 2 }$ are like terms, then the value of $m - n$ is ____?", "answer": "- 4", "steps": "$\\because$ $3 x ^ { 3 } y ^ { m + 1 }$ and $- 5 x ^ { n - 2 } y ^ { 2 }$ are like terms, $\\therefore$ $3 = n - 2$, $m + 1 = 2$, solving for $n$ and $m$, we get: $n = 5$, $m = 1$, thus $m - n = 1 - 5 = - 4$.", "expr_cands": ["3 x ^ { 3 } y ^ { m + 1 }", "y", "m", "x", "- 5 x ^ { n - 2 } y ^ { 2 }", "n", "m - n", "3 = n - 2", "n = 5", "m + 1 = 2", "m = 1", "- 4"], "exprs": ["3 = n - 2", "m + 1 = 2", "n = 5", "m = 1", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 5 x ^ { n - 2 } y ^ { 2 }"}, {"id": "3 = n - 2"}, {"id": "3 x ^ { 3 } y ^ { m + 1 }"}, {"id": "$3 x ^ { 3 } y ^ { m + 1 }$ 与 $- 5 x ^ { n - 2 } y ^ { 2 }$ 是同类项"}, {"id": "m + 1 = 2"}, {"id": "n = 5"}, {"id": "m = 1"}, {"id": "m - n"}, {"id": "- 4"}], "links": [{"rel": "被描述", "source": "- 5 x ^ { n - 2 } y ^ { 2 }", "target": "3 = n - 2"}, {"rel": "被描述", "source": "- 5 x ^ { n - 2 } y ^ { 2 }", "target": "m + 1 = 2"}, {"rel": "等式方程求解", "source": "3 = n - 2", "target": "n = 5"}, {"rel": "被描述", "source": "3 x ^ { 3 } y ^ { m + 1 }", "target": "3 = n - 2"}, {"rel": "被描述", "source": "3 x ^ { 3 } y ^ { m + 1 }", "target": "m + 1 = 2"}, {"rel": "限制性描述", "source": "$3 x ^ { 3 } y ^ { m + 1 }$ 与 $- 5 x ^ { n - 2 } y ^ { 2 }$ 是同类项", "target": "3 = n - 2"}, {"rel": "限制性描述", "source": "$3 x ^ { 3 } y ^ { m + 1 }$ 与 $- 5 x ^ { n - 2 } y ^ { 2 }$ 是同类项", "target": "m + 1 = 2"}, {"rel": "等式方程求解", "source": "m + 1 = 2", "target": "m = 1"}, {"rel": "代入", "source": "n = 5", "target": "- 4"}, {"rel": "代入", "source": "m = 1", "target": "- 4"}, {"rel": "被代入", "source": "m - n", "target": "- 4"}]}}
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{"content": "If the value of $\\frac { 3 x - 1 } { 5 }$ is the reciprocal of $- \\frac { 5 } { 3 }$, then the value of $x$ is ____?", "answer": "- \\frac { 2 } { 3 }", "steps": "$\\because$ $\\frac { 3 x - 1 } { 5 }$ is the reciprocal of $- \\frac { 5 } { 3 }$, $\\therefore$ $\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }$, solving for $x$ gives $x = - \\frac { 2 } { 3 }$.", "expr_cands": ["\\frac { 3 x - 1 } { 5 }", "x", "- \\frac { 5 } { 3 }", "\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }", "x = - \\frac { 2 } { 3 }"], "exprs": ["\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }", "x = - \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 x - 1 } { 5 }"}, {"id": "\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }"}, {"id": "- \\frac { 5 } { 3 }"}, {"id": "$\\frac { 3 x - 1 } { 5 }$ 的值与 $- \\frac { 5 } { 3 }$ 互为倒数"}, {"id": "x = - \\frac { 2 } { 3 }"}], "links": [{"rel": "被描述", "source": "\\frac { 3 x - 1 } { 5 }", "target": "\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }"}, {"rel": "等式方程求解", "source": "\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }", "target": "x = - \\frac { 2 } { 3 }"}, {"rel": "被描述", "source": "- \\frac { 5 } { 3 }", "target": "\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }"}, {"rel": "限制性描述", "source": "$\\frac { 3 x - 1 } { 5 }$ 的值与 $- \\frac { 5 } { 3 }$ 互为倒数", "target": "\\frac { 3 x - 1 } { 5 } = - \\frac { 3 } { 5 }"}]}}
|
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{"content": "If $x = - 1$ is a solution of the equation $2 ( x - b ) + a = 0$ with respect to $x$, then the value of $a - 2 b + 1$ is ____?", "answer": "3", "steps": "Substituting $x = - 1$ into the equation, we get $2 ( - 1 - b ) + a = 0$. Therefore, $a - 2 b = 2$. Thus, $a - 2 b + 1 = 2 + 1 = 3$.", "expr_cands": ["x = - 1", "x", "2 ( x - b ) + a = 0", "a", "b", "a - 2 b + 1", "2 ( - 1 - b ) + a = 0", "a - 2 b = 2", "3"], "exprs": ["2 ( - 1 - b ) + a = 0", "a - 2 b = 2", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = - 1"}, {"id": "2 ( - 1 - b ) + a = 0"}, {"id": "2 ( x - b ) + a = 0"}, {"id": "a - 2 b = 2"}, {"id": "a - 2 b + 1"}, {"id": "3"}], "links": [{"rel": "代入", "source": "x = - 1", "target": "2 ( - 1 - b ) + a = 0"}, {"rel": "移项", "source": "2 ( - 1 - b ) + a = 0", "target": "a - 2 b = 2"}, {"rel": "被代入", "source": "2 ( x - b ) + a = 0", "target": "2 ( - 1 - b ) + a = 0"}, {"rel": "代入", "source": "a - 2 b = 2", "target": "3"}, {"rel": "被代入", "source": "a - 2 b + 1", "target": "3"}]}}
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||||
{"content": "If the polynomial ${ x } ^ { 2 } + ( k - 1 ) x + 3$ does not contain a linear term in $x$, then $k$ = ____ ?", "answer": "1", "steps": "$\\because$ The polynomial ${ x } ^ { 2 } + ( k - 1 ) x + 3$ does not contain a linear term in $x$, $\\therefore$ $k - 1 = 0$, $\\therefore$ $k = 1$.", "expr_cands": ["{ x } ^ { 2 } + ( k - 1 ) x + 3", "k", "x", "k - 1 = 0", "k = 1"], "exprs": ["k - 1 = 0", "k = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { 2 } + ( k - 1 ) x + 3"}, {"id": "k - 1 = 0"}, {"id": "多项式 ${ x } ^ { 2 } + ( k - 1 ) x + 3$ 中不含有 $x$ 的一次项"}, {"id": "k = 1"}], "links": [{"rel": "被描述", "source": "{ x } ^ { 2 } + ( k - 1 ) x + 3", "target": "k - 1 = 0"}, {"rel": "等式方程求解", "source": "k - 1 = 0", "target": "k = 1"}, {"rel": "限制性描述", "source": "多项式 ${ x } ^ { 2 } + ( k - 1 ) x + 3$ 中不含有 $x$ 的一次项", "target": "k - 1 = 0"}]}}
|
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{"content": "If $ab = - 1$, then the value of $b$ when $a = \\sqrt { 25 }$ is", "answer": "- \\frac { 1 } { 5 }", "steps": "$a = \\sqrt { 25 } = 5$ , because $ab = - 1$ , therefore $b = - \\frac { 1 } { b } = - \\frac { 1 } { 5 }$ .", "expr_cands": ["ab = - 1", "a", "b", "a = \\sqrt { 25 }", "a = 5", "5 b = - 1", "b = - \\frac { 1 } { 5 }"], "exprs": ["a = 5", "b = - \\frac { 1 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a = \\sqrt { 25 }"}, {"id": "a = 5"}, {"id": "ab = - 1"}, {"id": "b = - \\frac { 1 } { 5 }"}], "links": [{"rel": "计算", "source": "a = \\sqrt { 25 }", "target": "a = 5"}, {"rel": "联立", "source": "a = 5", "target": "b = - \\frac { 1 } { 5 }"}, {"rel": "联立", "source": "ab = - 1", "target": "b = - \\frac { 1 } { 5 }"}]}}
|
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{"content": "The smallest integer solution that satisfies $x + 2019 > 0$ is ____ ?", "answer": "- 2018", "steps": "Since $x + 2019 > 0$, it follows that $x > - 2019$. Therefore, the smallest integer solution to the inequality is $- 2018$.", "expr_cands": ["x + 2019 > 0", "x", "- 2019 < x", "x > - 2019", "- 2018"], "exprs": ["x > - 2019", "- 2018"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 2019 > 0"}, {"id": "x > - 2019"}, {"id": "- 2018"}, {"id": "满足 $x + 2019 > 0$ 的最小整数解"}], "links": [{"rel": "不等式方程求解", "source": "x + 2019 > 0", "target": "x > - 2019"}, {"rel": "被描述", "source": "x > - 2019", "target": "- 2018"}, {"rel": "限制性描述", "source": "满足 $x + 2019 > 0$ 的最小整数解", "target": "- 2018"}]}}
|
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{"content": "If $a$ is the arithmetic square root of $2021$, then the arithmetic square root of $\\frac { 2021 } { 100 }$ is ____?", "answer": "\\frac { a } { 10 }", "steps": "$\\because$ $a$ is the arithmetic square root of $2021$, $\\therefore$ $a ^ 2 = 2021$. Thus, the arithmetic square root of $\\frac { 2021 } { 100 }$ is $\\sqrt { \\frac { 2021 } { 100 }} = \\sqrt { \\frac { a ^ 2 } { 100 }} = \\frac { a } { 10 }$.", "expr_cands": ["a", "2021", "\\frac { 2021 } { 100 }", "a ^ { 2 } = 2021", "a = \\sqrt { 2021 }", "a = - \\sqrt { 2021 }", "\\sqrt { \\frac { 2021 } { 100 } } = \\frac { a } { 10 }", "\\sqrt { \\frac { 2021 } { 100 } }", "\\frac { a } { 10 }"], "exprs": ["a ^ { 2 } = 2021", "\\frac { a } { 10 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a ^ { 2 } = 2021"}, {"id": "2021"}, {"id": "$a$ 是 $2021$ 是算术平方根"}, {"id": "\\frac { a } { 10 }"}, {"id": "\\frac { 2021 } { 100 }"}, {"id": "$\\frac { 2021 } { 100 }$ 的算术平方根"}], "links": [{"rel": "被描述", "source": "a", "target": "a ^ { 2 } = 2021"}, {"rel": "被描述", "source": "a ^ { 2 } = 2021", "target": "\\frac { a } { 10 }"}, {"rel": "被描述", "source": "2021", "target": "a ^ { 2 } = 2021"}, {"rel": "限制性描述", "source": "$a$ 是 $2021$ 是算术平方根", "target": "a ^ { 2 } = 2021"}, {"rel": "被描述", "source": "\\frac { 2021 } { 100 }", "target": "\\frac { a } { 10 }"}, {"rel": "限制性描述", "source": "$\\frac { 2021 } { 100 }$ 的算术平方根", "target": "\\frac { a } { 10 }"}]}}
|
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{"content": "Factorization: $ax ^ 2 - 10 ax + 25 a$ = ____ ?", "answer": "a ( x - 5 ) ^ { 2 }", "steps": "$ax ^ { 2 } - 10 ax + 25 a = a ( x ^ { 2 } - 10 x + 25 ) - -$ (Extracting common factor) = $a ( x - 5 ) ^ { 2 }$.$ - - (Completing the square formula)", "expr_cands": ["ax ^ { 2 } - 10 ax + 25 a", "x", "a", "a ( x ^ { 2 } - 10 x + 25 ) - -", "a ( x - 5 ) ^ { 2 }"], "exprs": ["a ( x - 5 ) ^ { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ax ^ { 2 } - 10 ax + 25 a"}, {"id": "a ( x - 5 ) ^ { 2 }"}], "links": [{"rel": "提取因式", "source": "ax ^ { 2 } - 10 ax + 25 a", "target": "a ( x - 5 ) ^ { 2 }"}]}}
|
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{"content": "When $x$ = ____ ?, the value of the fraction $\\frac { | x | - 2 } { x - 2 }$ is zero.", "answer": "- 2", "steps": "When $| x | - 2 = 0$ and $x - 2 \\neq 0$, i.e. $x = - 2$, the value of the fraction $\\frac { | x | - 2 } { x - 2 }$ is zero.", "expr_cands": ["x", "\\frac { | x | - 2 } { x - 2 }", "| x | - 2 = 0", "x = - 2", "x = 2", "x - 2 \\neq 0", "x \\neq 2"], "exprs": ["| x | - 2 = 0", "x - 2 \\neq 0", "x = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { | x | - 2 } { x - 2 }"}, {"id": "| x | - 2 = 0"}, {"id": "分式 $\\frac { | x | - 2 } { x - 2 }$ 值为零"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "x - 2 \\neq 0"}, {"id": "x = - 2"}], "links": [{"rel": "被描述", "source": "\\frac { | x | - 2 } { x - 2 }", "target": "| x | - 2 = 0"}, {"rel": "被描述", "source": "\\frac { | x | - 2 } { x - 2 }", "target": "x - 2 \\neq 0"}, {"rel": "联立", "source": "| x | - 2 = 0", "target": "x = - 2"}, {"rel": "限制性描述", "source": "分式 $\\frac { | x | - 2 } { x - 2 }$ 值为零", "target": "| x | - 2 = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { | x | - 2 } { x - 2 }$ 值为零", "target": "x - 2 \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "| x | - 2 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x - 2 \\neq 0"}, {"rel": "联立", "source": "x - 2 \\neq 0", "target": "x = - 2"}]}}
|
||||
{"content": "If the polynomial $( 2 x + m ) ( x - 1 )$ does not contain the linear term of $x$, then the value of $m$ is ____?", "answer": "2", "steps": "$$(2x+m)(x-1)=2x^2+(m-2)x-m$$From the fact that $( 2 x + m ) ( x - 1 )$ does not contain a linear term in $x$, we have $m - 2 = 0$. Solving for $m$, we get $m = 2$.", "expr_cands": ["( 2 x + m ) ( x - 1 )", "x", "m", "2 x ^ { 2 } + ( m - 2 ) x - m", "m - 2 = 0", "m = 2"], "exprs": ["2 x ^ { 2 } + ( m - 2 ) x - m", "m - 2 = 0", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 2 x + m ) ( x - 1 )"}, {"id": "2 x ^ { 2 } + ( m - 2 ) x - m"}, {"id": "x"}, {"id": "m - 2 = 0"}, {"id": "整式 $( 2 x + m ) ( x - 1 )$ 不含有 $x$ 的一次项"}, {"id": "m = 2"}], "links": [{"rel": "提取因式", "source": "( 2 x + m ) ( x - 1 )", "target": "2 x ^ { 2 } + ( m - 2 ) x - m"}, {"rel": "被描述", "source": "2 x ^ { 2 } + ( m - 2 ) x - m", "target": "m - 2 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "2 x ^ { 2 } + ( m - 2 ) x - m"}, {"rel": "等式方程求解", "source": "m - 2 = 0", "target": "m = 2"}, {"rel": "限制性描述", "source": "整式 $( 2 x + m ) ( x - 1 )$ 不含有 $x$ 的一次项", "target": "m - 2 = 0"}]}}
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||||
{"content": "The solution set of the inequality $- x \\le 3 x - 4$ is ____ ?", "answer": "x \\ge 1", "steps": "Moving terms, we get $- x - 3 x \\le - 4$. Combining like terms, we get $- 4 x \\le - 4$. Dividing by $- 4$ (or multiplying by $- 1$ and flipping the inequality), we get $x \\ge 1$.", "expr_cands": ["- x \\le 3 x - 4", "x", "- x - 3 x \\le - 4", "1 \\le x", "- 4 x \\le - 4", "1", "x \\ge 1"], "exprs": ["x \\ge 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- x \\le 3 x - 4"}, {"id": "x \\ge 1"}], "links": [{"rel": "不等式方程求解", "source": "- x \\le 3 x - 4", "target": "x \\ge 1"}]}}
|
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{"content": "When does $\\frac { x } { \\sqrt { x - 1 }}$ have a meaningful value in the real number range?", "answer": "x > 1", "steps": "According to the meaning of quadratic radicals and the denominator cannot be zero, we have $x - 1 > 0$, which implies $x > 1$.", "expr_cands": ["x", "\\frac { x } { \\sqrt { x - 1 } }", "0", "x - 1 > 0", "1 < x", "x > 1"], "exprs": ["x - 1 > 0", "x > 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { \\sqrt { x - 1 } }"}, {"id": "x - 1 > 0"}, {"id": "$\\frac { x } { \\sqrt { x - 1 } }$ 在实数范围内有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x > 1"}], "links": [{"rel": "被描述", "source": "\\frac { x } { \\sqrt { x - 1 } }", "target": "x - 1 > 0"}, {"rel": "不等式方程求解", "source": "x - 1 > 0", "target": "x > 1"}, {"rel": "限制性描述", "source": "$\\frac { x } { \\sqrt { x - 1 } }$ 在实数范围内有意义", "target": "x - 1 > 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 1 > 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 1 > 0"}]}}
|
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{"content": "Given $3 a = 2 b ( a \\neq 0 )$, what is $\\frac { b } { a }$?", "answer": "\\frac { 3 } { 2 }", "steps": "Because $3 a = 2 b$, therefore $b = \\frac { 3 a } { 2 }$. Because $a \\neq 0$, therefore $\\frac { b } { a } = \\frac { \\frac { 3 } { 2 } a } { a } = \\frac { 3 } { 2 }$.", "expr_cands": ["3 a = 2 b ( a \\neq 0 )", "a", "b", "\\frac { b } { a }", "3 a = 2 b", "b = \\frac { 3 a } { 2 }", "b = b", "a \\neq 0", "\\frac { 3 } { 2 }"], "exprs": ["b = \\frac { 3 a } { 2 }", "\\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a = 2 b ( a \\neq 0 )"}, {"id": "b = \\frac { 3 a } { 2 }"}, {"id": "\\frac { b } { a }"}, {"id": "\\frac { 3 } { 2 }"}], "links": [{"rel": "等式方程部分求解", "source": "3 a = 2 b ( a \\neq 0 )", "target": "b = \\frac { 3 a } { 2 }"}, {"rel": "代入", "source": "b = \\frac { 3 a } { 2 }", "target": "\\frac { 3 } { 2 }"}, {"rel": "被代入", "source": "\\frac { b } { a }", "target": "\\frac { 3 } { 2 }"}]}}
|
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{"content": "The equation $2 x + a = 2$ has a solution of $x = 2$, then $a$ = ____ ?", "answer": "- 2", "steps": "$\\because$ The solution to the equation $2 x + a = 2$ is $x = 2$, $\\therefore$ $2 \\times 2 + a = 2$, which gives us $a = - 2$.", "expr_cands": ["2 x + a = 2", "a", "x", "x = 2", "2 * 2 + a = 2", "a = - 2"], "exprs": ["2 * 2 + a = 2", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + a = 2"}, {"id": "2 * 2 + a = 2"}, {"id": "x = 2"}, {"id": "a = - 2"}], "links": [{"rel": "被代入", "source": "2 x + a = 2", "target": "2 * 2 + a = 2"}, {"rel": "等式方程求解", "source": "2 * 2 + a = 2", "target": "a = - 2"}, {"rel": "代入", "source": "x = 2", "target": "2 * 2 + a = 2"}]}}
|
||||
{"content": "If $\\sqrt { 2 x - 5 }$ does not exist, then the possible values of $x$ are _____.", "answer": "x < \\frac { 5 } { 2 }", "steps": "It is known from the problem that $2 x - 5 < 0$, which implies that $x < \\frac { 5 } { 2 }$.", "expr_cands": ["2 x - 5", "x", "2 x - 5 < 0", "x < \\frac { 5 } { 2 }"], "exprs": ["2 x - 5 < 0", "x < \\frac { 5 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 5"}, {"id": "2 x - 5 < 0"}, {"id": "$2 x - 5$ 没有平方根"}, {"id": "x < \\frac { 5 } { 2 }"}], "links": [{"rel": "被描述", "source": "2 x - 5", "target": "2 x - 5 < 0"}, {"rel": "不等式方程求解", "source": "2 x - 5 < 0", "target": "x < \\frac { 5 } { 2 }"}, {"rel": "限制性描述", "source": "$2 x - 5$ 没有平方根", "target": "2 x - 5 < 0"}]}}
|
||||
{"content": "Given the function $y = ax + a - 3$ is a proportional function, what is the value of $a$?", "answer": "3", "steps": "$\\because$ The function $y = ax + a - 3$ is a direct proportion function, $\\therefore$ $a - 3 = 0$ and $a \\neq 0$, solving for $a$, we get: $a = 3$.", "expr_cands": ["y = ax + a - 3", "y", "x", "a", "a - 3 = 0", "a = 3", "a \\neq 0"], "exprs": ["a - 3 = 0", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ax + a - 3"}, {"id": "a - 3 = 0"}, {"id": "函数 $y = ax + a - 3$ 是正比例函数"}, {"id": "a = 3"}], "links": [{"rel": "被描述", "source": "y = ax + a - 3", "target": "a - 3 = 0"}, {"rel": "等式方程求解", "source": "a - 3 = 0", "target": "a = 3"}, {"rel": "限制性描述", "source": "函数 $y = ax + a - 3$ 是正比例函数", "target": "a - 3 = 0"}]}}
|
||||
{"content": "If $2 x ^ { m } ^ + ^ { n } y ^ { 2 }$ and $- 5 x ^ { 4 } y ^ { n } ^ - ^ { m }$ can be combined, then the value of $mn$ is ____?", "answer": "3", "steps": "Because if $2 x ^ { m + n } y ^ { 2 }$ and $- 5 x ^ { 4 } y ^ { n - m }$ can be combined, therefore $2 x ^ { m + n } y ^ { 2 }$ and $- 5 x ^ { 4 } y ^ { n - m }$ are like terms. Therefore, $m + n = 4$, $2 = n - m$, and thus we can solve for $m = 1$ and $n = 3$. Therefore, $mn = 1 * 3 = 3$.", "expr_cands": ["2 x ^ { m } ^ + ^ { n } y ^ { 2 }", "m", "x", "- 5 x ^ { 4 } y ^ { n } ^ - ^ { m }", "y", "n", "mn", "2 x ^ { m + n } y ^ { 2 }", "- 5 x ^ { 4 } y ^ { n - m }", "m + n = 4", "2 = n - m", "m = 1", "n = 3", "3"], "exprs": ["m + n = 4", "2 = n - m", "m = 1", "n = 3", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { m + n } y ^ { 2 }"}, {"id": "m + n = 4"}, {"id": "- 5 x ^ { 4 } y ^ { n - m }"}, {"id": "$2 x ^ { m + n } y ^ { 2 }$ 与 $- 5 x ^ { 4 } y ^ { n - m }$ 能合并"}, {"id": ", $2 x ^ { m + n } y ^ { 2 }$ 与 $- 5 x ^ { 4 } y ^ { n - m }$ 是同类项"}, {"id": "2 = n - m"}, {"id": "m = 1"}, {"id": "n = 3"}, {"id": "mn"}, {"id": "3"}], "links": [{"rel": "被描述", "source": "2 x ^ { m + n } y ^ { 2 }", "target": "m + n = 4"}, {"rel": "被描述", "source": "2 x ^ { m + n } y ^ { 2 }", "target": "2 = n - m"}, {"rel": "联立", "source": "m + n = 4", "target": "m = 1"}, {"rel": "联立", "source": "m + n = 4", "target": "n = 3"}, {"rel": "被描述", "source": "- 5 x ^ { 4 } y ^ { n - m }", "target": "m + n = 4"}, {"rel": "被描述", "source": "- 5 x ^ { 4 } y ^ { n - m }", "target": "2 = n - m"}, {"rel": "限制性描述", "source": "$2 x ^ { m + n } y ^ { 2 }$ 与 $- 5 x ^ { 4 } y ^ { n - m }$ 能合并", "target": "m + n = 4"}, {"rel": "限制性描述", "source": "$2 x ^ { m + n } y ^ { 2 }$ 与 $- 5 x ^ { 4 } y ^ { n - m }$ 能合并", "target": "2 = n - m"}, {"rel": "限制性描述", "source": ", $2 x ^ { m + n } y ^ { 2 }$ 与 $- 5 x ^ { 4 } y ^ { n - m }$ 是同类项", "target": "m + n = 4"}, {"rel": "限制性描述", "source": ", $2 x ^ { m + n } y ^ { 2 }$ 与 $- 5 x ^ { 4 } y ^ { n - m }$ 是同类项", "target": "2 = n - m"}, {"rel": "联立", "source": "2 = n - m", "target": "m = 1"}, {"rel": "联立", "source": "2 = n - m", "target": "n = 3"}, {"rel": "代入", "source": "m = 1", "target": "3"}, {"rel": "代入", "source": "n = 3", "target": "3"}, {"rel": "被代入", "source": "mn", "target": "3"}]}}
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{"content": "If $\\frac { a } { b } = 20$, $\\frac { b } { c } = 10$, then the value of $\\frac { a + b } { b + c }$ is ____?", "answer": "\\frac { 210 } { 11 }", "steps": "Because $\\frac { a } { b } = 20$, $\\frac { b } { c } = 10$, we have $a = 20 b$, $b = 10 c$, and $a = 200 c$. Therefore, $\\frac { a + b } { b + c } = \\frac { 200 c + 10 c } { 10 c + c } = \\frac { 210 c } { 11 c } = \\frac { 210 } { 11 }$.", "expr_cands": ["\\frac { a } { b } = 20", "b", "a", "\\frac { b } { c } = 10", "c", "\\frac { a + b } { b + c }", "a = 20 b", "b = 10 c", "200 c", "\\frac { 210 } { 11 }"], "exprs": ["a = 20 b", "b = 10 c", "\\frac { 210 } { 11 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a } { b } = 20"}, {"id": "a = 20 b"}, {"id": "\\frac { b } { c } = 10"}, {"id": "b = 10 c"}, {"id": "\\frac { a + b } { b + c }"}, {"id": "\\frac { 210 } { 11 }"}], "links": [{"rel": "同乘除", "source": "\\frac { a } { b } = 20", "target": "a = 20 b"}, {"rel": "代入", "source": "a = 20 b", "target": "\\frac { 210 } { 11 }"}, {"rel": "同乘除", "source": "\\frac { b } { c } = 10", "target": "b = 10 c"}, {"rel": "代入", "source": "b = 10 c", "target": "\\frac { 210 } { 11 }"}, {"rel": "被代入", "source": "\\frac { a + b } { b + c }", "target": "\\frac { 210 } { 11 }"}]}}
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{"content": "If $2 a + 3 b - 6 = 0$, then the value of the polynomial $6 a + 9 b - 12$ is ____?", "answer": "6", "steps": "Since $2 a + 3 b - 6 = 0$, therefore $2 a + 3 b = 6$, therefore $6 a + 9 b - 12 = 3 ( 2 a + 3 b ) - 12 = 3 * 6 - 12 = 18 - 12 = 6$. That is, the value of the polynomial $6 a + 9 b - 12$ is $6$.", "expr_cands": ["2 a + 3 b - 6 = 0", "a", "b", "6 a + 9 b - 12", "2 a + 3 b = 6", "3 ( 2 a + 3 b ) - 12", "6"], "exprs": ["2 a + 3 b = 6", "3 ( 2 a + 3 b ) - 12", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a + 3 b - 6 = 0"}, {"id": "2 a + 3 b = 6"}, {"id": "6 a + 9 b - 12"}, {"id": "3 ( 2 a + 3 b ) - 12"}, {"id": "6"}], "links": [{"rel": "移项", "source": "2 a + 3 b - 6 = 0", "target": "2 a + 3 b = 6"}, {"rel": "提取因式参考", "source": "2 a + 3 b = 6", "target": "3 ( 2 a + 3 b ) - 12"}, {"rel": "代入", "source": "2 a + 3 b = 6", "target": "6"}, {"rel": "提取因式", "source": "6 a + 9 b - 12", "target": "3 ( 2 a + 3 b ) - 12"}, {"rel": "被代入", "source": "3 ( 2 a + 3 b ) - 12", "target": "6"}]}}
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{"content": "If the equation $x - 2 y + 3 z = 0$ holds and $y = 2$ when $x = 1$, then $z$ = ____ ?", "answer": "1", "steps": "Substituting $x = 1$ and $y = 2$ into the equation gives $1 - 4 + 3 z = 0$, which yields $z = 1$ as the solution.", "expr_cands": ["x - 2 y + 3 z = 0", "y", "x", "z", "x = 1", "y = 2", "1 - 4 + 3 z = 0", "z = 1"], "exprs": ["1 - 4 + 3 z = 0", "z = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - 2 y + 3 z = 0"}, {"id": "1 - 4 + 3 z = 0"}, {"id": "x = 1"}, {"id": "y = 2"}, {"id": "z = 1"}], "links": [{"rel": "被代入", "source": "x - 2 y + 3 z = 0", "target": "1 - 4 + 3 z = 0"}, {"rel": "等式方程求解", "source": "1 - 4 + 3 z = 0", "target": "z = 1"}, {"rel": "代入", "source": "x = 1", "target": "1 - 4 + 3 z = 0"}, {"rel": "代入", "source": "y = 2", "target": "1 - 4 + 3 z = 0"}]}}
|
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{"content": "The algebraic expression $\\frac { 1 } { x - 1 }$ is meaningful, and the condition that $x$ should satisfy is ____?", "answer": "x \\neq 1", "steps": "From the given condition, we have $x - 1 \\neq 0$, which implies that $x \\neq 1$.", "expr_cands": ["\\frac { 1 } { x - 1 }", "x", "x - 1 \\neq 0", "x \\neq 1"], "exprs": ["x - 1 \\neq 0", "x \\neq 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x - 1 }"}, {"id": "x - 1 \\neq 0"}, {"id": "代数式 $\\frac { 1 } { x - 1 }$ 有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 1"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { x - 1 }", "target": "x - 1 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 1 \\neq 0", "target": "x \\neq 1"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 1 } { x - 1 }$ 有意义", "target": "x - 1 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 1 \\neq 0"}]}}
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{"content": "$\\sqrt { a - 1 } + | b - 4 | = 0$ , then $\\sqrt { ab }$ = ____ ?", "answer": "2", "steps": "From the given information, we have $a - 1 = 0$ and $b - 4 = 0$. Solving for $a$ and $b$, we get $a = 1$ and $b = 4$. Therefore, $\\sqrt { ab } = 2$.", "expr_cands": ["\\sqrt { a - 1 } + | b - 4 | = 0", "b", "a", "\\sqrt { ab }", "a - 1 = 0", "a = 1", "b - 4 = 0", "b = 4", "2"], "exprs": ["a - 1 = 0", "b - 4 = 0", "a = 1", "b = 4", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { a - 1 } + | b - 4 | = 0"}, {"id": "a - 1 = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "b - 4 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "a = 1"}, {"id": "b = 4"}, {"id": "\\sqrt { ab }"}, {"id": "2"}], "links": [{"rel": "被描述", "source": "\\sqrt { a - 1 } + | b - 4 | = 0", "target": "a - 1 = 0"}, {"rel": "被描述", "source": "\\sqrt { a - 1 } + | b - 4 | = 0", "target": "b - 4 = 0"}, {"rel": "等式方程求解", "source": "a - 1 = 0", "target": "a = 1"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "a - 1 = 0"}, {"rel": "等式方程求解", "source": "b - 4 = 0", "target": "b = 4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b - 4 = 0"}, {"rel": "代入", "source": "a = 1", "target": "2"}, {"rel": "代入", "source": "b = 4", "target": "2"}, {"rel": "被代入", "source": "\\sqrt { ab }", "target": "2"}]}}
|
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{"content": "If the polynomial $- 2 mx ^ 2 - 5 x ^ 2 + x ^ 2 - 2 x + 9$ does not contain the term $x ^ 2$, then $m$ = ____?", "answer": "- 2", "steps": "$\\because$ The polynomial in $x$, $- 2 mx ^ 2 - 5 x ^ 2 + x ^ 2 - 2 x + 9$, does not contain the term $x ^ 2$, $\\therefore$ $- 2 m - 5 + 1 = 0$, solving for $m$, we get: $m = - 2$.", "expr_cands": ["x", "- 2 mx ^ { 2 } - 5 x ^ { 2 } + x ^ { 2 } - 2 x + 9", "m", "x ^ { 2 }", "- 2 m - 5 + 1 = 0", "m = - 2"], "exprs": ["- 2 m - 5 + 1 = 0", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 mx ^ { 2 } - 5 x ^ { 2 } + x ^ { 2 } - 2 x + 9"}, {"id": "- 2 m - 5 + 1 = 0"}, {"id": "关于 $x$ 的多项式 $- 2 mx ^ { 2 } - 5 x ^ { 2 } + x ^ { 2 } - 2 x + 9$ 中不含有 $x ^ { 2 }$ 项"}, {"id": "m = - 2"}], "links": [{"rel": "被描述", "source": "- 2 mx ^ { 2 } - 5 x ^ { 2 } + x ^ { 2 } - 2 x + 9", "target": "- 2 m - 5 + 1 = 0"}, {"rel": "等式方程求解", "source": "- 2 m - 5 + 1 = 0", "target": "m = - 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的多项式 $- 2 mx ^ { 2 } - 5 x ^ { 2 } + x ^ { 2 } - 2 x + 9$ 中不含有 $x ^ { 2 }$ 项", "target": "- 2 m - 5 + 1 = 0"}]}}
|
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{"content": "If the simplest quadratic radical $\\sqrt { 5 x + 2 }$ and $\\sqrt { 8 - x }$ are of the same type, then $x$ = ____?", "answer": "1", "steps": "From the concept of similar quadratic radicals, we have $5 x + 2 = 8 - x$, which can be solved to obtain $x = 1$.", "expr_cands": ["\\sqrt { 5 x + 2 }", "x", "\\sqrt { 8 - x }", "5 x + 2 = 8 - x", "x = 1"], "exprs": ["5 x + 2 = 8 - x", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 5 x + 2 }"}, {"id": "5 x + 2 = 8 - x"}, {"id": "\\sqrt { 8 - x }"}, {"id": "最简二次根式 $\\sqrt { 5 x + 2 }$ 与 $\\sqrt { 8 - x }$ 是同类二次根式"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { 5 x + 2 }", "target": "5 x + 2 = 8 - x"}, {"rel": "等式方程求解", "source": "5 x + 2 = 8 - x", "target": "x = 1"}, {"rel": "被描述", "source": "\\sqrt { 8 - x }", "target": "5 x + 2 = 8 - x"}, {"rel": "限制性描述", "source": "最简二次根式 $\\sqrt { 5 x + 2 }$ 与 $\\sqrt { 8 - x }$ 是同类二次根式", "target": "5 x + 2 = 8 - x"}]}}
|
||||
{"content": "Given $a ^ { 3 m + n } = 27$ and $a ^ { m } = 3$, what is the value of $n$?", "answer": "0", "steps": "$\\because a ^ { m } = 3$, $\\therefore a ^ { 3 m } = 3 ^ { 3 } = 27$, $\\because a ^ { 3 m + n } = 27$, $\\therefore a ^ { n } = 1$, solving for $n$ yields $n = 0$.", "expr_cands": ["a ^ { 3 m + n } = 27", "a", "m", "n", "a ^ { m } = 3", "a ^ { 3 m }", "27", "27 a ^ { n } = 27", "a ^ { n } = 1", "n = 0"], "exprs": ["27 a ^ { n } = 27", "a ^ { n } = 1", "n = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 3 m + n } = 27"}, {"id": "27 a ^ { n } = 27"}, {"id": "a ^ { m } = 3"}, {"id": "a ^ { n } = 1"}, {"id": "n = 0"}, {"id": "多项式零次方项，若底数不为0，则恒等于1"}], "links": [{"rel": "被代入", "source": "a ^ { 3 m + n } = 27", "target": "27 a ^ { n } = 27"}, {"rel": "同乘除", "source": "27 a ^ { n } = 27", "target": "a ^ { n } = 1"}, {"rel": "代入", "source": "a ^ { m } = 3", "target": "27 a ^ { n } = 27"}, {"rel": "被描述", "source": "a ^ { n } = 1", "target": "n = 0"}, {"rel": "属性描述", "source": "多项式零次方项，若底数不为0，则恒等于1", "target": "n = 0"}]}}
|
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{"content": "$\\sqrt { ( 5 - x ) ^ { 2 } } = x - 5$ , then ____ ?", "answer": "x \\ge 5", "steps": "Because the square root of the quantity $( 5 - x ) ^ 2$ is equal to $x - 5$, it follows that $5 - x$ is non-negative. Therefore, $x$ must be greater than or equal to $5$.", "expr_cands": ["\\sqrt { ( 5 - x ) ^ { 2 } } = x - 5", "x", "5 - x \\le 0", "5 \\le x", "x \\ge 5"], "exprs": ["5 - x \\le 0", "x \\ge 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { ( 5 - x ) ^ { 2 } } = x - 5"}, {"id": "5 - x \\le 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 5"}], "links": [{"rel": "被描述", "source": "\\sqrt { ( 5 - x ) ^ { 2 } } = x - 5", "target": "5 - x \\le 0"}, {"rel": "不等式方程求解", "source": "5 - x \\le 0", "target": "x \\ge 5"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "5 - x \\le 0"}]}}
|
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{"content": "If the equation $2 x + 1 = 5$ and the equation $7 - kx = x + 2 k$ about $x$ have the same solution, then the value of $k$ is ____?", "answer": "\\frac { 5 } { 4 }", "steps": "The solution to the equation $2 x + 1 = 5$ is $x = 2$. Since the equation $2 x + 1 = 5$ is equivalent to the equation $7 - kx = x + 2 k$ in terms of $x$, the solution to the equation $7 - kx = x + 2 k$ is also $x = 2$. Therefore, $7 - 2 k = 2 + 2 k$, and solving for $k$ yields $k = \\frac { 5 } { 4 }$.", "expr_cands": ["2 x + 1 = 5", "x", "7 - kx = x + 2 k", "k", "x = 2", "7 - 2 k = 2 k + 2", "2 k + 2 = 2 k + 2", "7 - 2 k = 2 + 2 k", "k = \\frac { 5 } { 4 }"], "exprs": ["x = 2", "7 - 2 k = 2 + 2 k", "k = \\frac { 5 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 1 = 5"}, {"id": "x = 2"}, {"id": "7 - kx = x + 2 k"}, {"id": "7 - 2 k = 2 + 2 k"}, {"id": "k = \\frac { 5 } { 4 }"}], "links": [{"rel": "等式方程求解", "source": "2 x + 1 = 5", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "7 - 2 k = 2 + 2 k"}, {"rel": "被代入", "source": "7 - kx = x + 2 k", "target": "7 - 2 k = 2 + 2 k"}, {"rel": "等式方程求解", "source": "7 - 2 k = 2 + 2 k", "target": "k = \\frac { 5 } { 4 }"}]}}
|
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{"content": "Given line segments $a = 2$, $b = 3$, $c = 4$, if $a$, $b$, $c$, $d$ are in proportion, then $d$ = ____?", "answer": "6", "steps": "$\\because$ Four line segments $a$, $b$, $c$, $d$ are in proportion, $\\therefore \\frac { a } { b } = \\frac { c } { d }$, $\\because a = 2$, $b = 3$, $c = 4$. $\\therefore \\frac { 2 } { 3 } = \\frac { 4 } { d }$, solving for $d$, we get: $d = 6$.", "expr_cands": ["a = 2", "a", "b = 3", "b", "c = 4", "c", "d", "\\frac { a } { b } = \\frac { c } { d }", "\\frac { 2 } { 3 } = \\frac { 4 } { d }", "d = 6"], "exprs": ["\\frac { a } { b } = \\frac { c } { d }", "\\frac { 2 } { 3 } = \\frac { 4 } { d }", "d = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "\\frac { a } { b } = \\frac { c } { d }"}, {"id": "b"}, {"id": "c"}, {"id": "d"}, {"id": "$a$ , $b$ , $c$ , $d$ 四条线段成比例"}, {"id": "a = 2"}, {"id": "\\frac { 2 } { 3 } = \\frac { 4 } { d }"}, {"id": "b = 3"}, {"id": "c = 4"}, {"id": "d = 6"}], "links": [{"rel": "被描述", "source": "a", "target": "\\frac { a } { b } = \\frac { c } { d }"}, {"rel": "被代入", "source": "\\frac { a } { b } = \\frac { c } { d }", "target": "\\frac { 2 } { 3 } = \\frac { 4 } { d }"}, {"rel": "被描述", "source": "b", "target": "\\frac { a } { b } = \\frac { c } { d }"}, {"rel": "被描述", "source": "c", "target": "\\frac { a } { b } = \\frac { c } { d }"}, {"rel": "被描述", "source": "d", "target": "\\frac { a } { b } = \\frac { c } { d }"}, {"rel": "限制性描述", "source": "$a$ , $b$ , $c$ , $d$ 四条线段成比例", "target": "\\frac { a } { b } = \\frac { c } { d }"}, {"rel": "代入", "source": "a = 2", "target": "\\frac { 2 } { 3 } = \\frac { 4 } { d }"}, {"rel": "等式方程求解", "source": "\\frac { 2 } { 3 } = \\frac { 4 } { d }", "target": "d = 6"}, {"rel": "代入", "source": "b = 3", "target": "\\frac { 2 } { 3 } = \\frac { 4 } { d }"}, {"rel": "代入", "source": "c = 4", "target": "\\frac { 2 } { 3 } = \\frac { 4 } { d }"}]}}
|
||||
{"content": "If the value of the fraction $\\frac { 2 } { x - 2 }$ is equal to the value of the fraction $\\frac { 4 } { x + 3 }$, then the value of $x$ is ____?", "answer": "7", "steps": "According to the problem, we have $\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }$. Multiplying both sides by $( x - 2 ) ( x + 3 )$, we get $2 ( x + 3 ) = 4 ( x - 2 )$. Solving for $x$, we get $x = 7$. Checking, we see that $x = 7$ is indeed a solution to the equation.", "expr_cands": ["\\frac { 2 } { x - 2 }", "x", "\\frac { 4 } { x + 3 }", "\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }", "x = 7", "2 x + 6 = 4 x - 8"], "exprs": ["\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }", "x = 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { x - 2 }"}, {"id": "\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }"}, {"id": "\\frac { 4 } { x + 3 }"}, {"id": "分式 $\\frac { 2 } { x - 2 }$ 与 $\\frac { 4 } { x + 3 }$ 的值相"}, {"id": "x = 7"}], "links": [{"rel": "被描述", "source": "\\frac { 2 } { x - 2 }", "target": "\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }"}, {"rel": "等式方程求解", "source": "\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }", "target": "x = 7"}, {"rel": "被描述", "source": "\\frac { 4 } { x + 3 }", "target": "\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }"}, {"rel": "限制性描述", "source": "分式 $\\frac { 2 } { x - 2 }$ 与 $\\frac { 4 } { x + 3 }$ 的值相", "target": "\\frac { 2 } { x - 2 } = \\frac { 4 } { x + 3 }"}]}}
|
||||
{"content": "If $y = ( m ^ 2 - 1 ) x ^ { m ^ 2 - m }$ is a quadratic function, then $m$ = ____?", "answer": "2", "steps": "From the given condition, we have $m ^ 2 - m = 2$, and $m ^ 2 - 1 \\neq 0$. Solving for $m$, we get $m = 2$.", "expr_cands": ["y = ( m ^ { 2 } - 1 ) x ^ { m ^ { 2 } - m }", "y", "x", "m", "m ^ { 2 } - m = 2", "m = - 1", "m = 2", "m ^ { 2 } - 1 \\neq 0", "( - 1 < m \\wedge m < 1 )", "1 < m", "m < - 1"], "exprs": ["m ^ { 2 } - m = 2", "m ^ { 2 } - 1 \\neq 0", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m ^ { 2 } - 1 ) x ^ { m ^ { 2 } - m }"}, {"id": "m ^ { 2 } - m = 2"}, {"id": "$y = ( m ^ { 2 } - 1 ) x ^ { m ^ { 2 } - m }$ 是二次函数"}, {"id": "m ^ { 2 } - 1 \\neq 0"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "y = ( m ^ { 2 } - 1 ) x ^ { m ^ { 2 } - m }", "target": "m ^ { 2 } - m = 2"}, {"rel": "被描述", "source": "y = ( m ^ { 2 } - 1 ) x ^ { m ^ { 2 } - m }", "target": "m ^ { 2 } - 1 \\neq 0"}, {"rel": "联立", "source": "m ^ { 2 } - m = 2", "target": "m = 2"}, {"rel": "限制性描述", "source": "$y = ( m ^ { 2 } - 1 ) x ^ { m ^ { 2 } - m }$ 是二次函数", "target": "m ^ { 2 } - m = 2"}, {"rel": "限制性描述", "source": "$y = ( m ^ { 2 } - 1 ) x ^ { m ^ { 2 } - m }$ 是二次函数", "target": "m ^ { 2 } - 1 \\neq 0"}, {"rel": "联立", "source": "m ^ { 2 } - 1 \\neq 0", "target": "m = 2"}]}}
|
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{"content": "If the solution set of the inequality $3 m - 2 x < 5$ with respect to $x$ is $x > 2$, then the value of the real number $m$ is ____?", "answer": "3", "steps": "Solve the inequality $3 m - 2 x < 5$ to get $x > \\frac { 3 m - 5 } { 2 }$. Since the solution set of this inequality is $x > 2$, we have $\\frac { 3 m - 5 } { 2 } = 2$. Therefore, $m = 3$.", "expr_cands": ["x", "3 m - 2 x < 5", "m", "x > 2", "x > \\frac { 3 m - 5 } { 2 }", "\\frac { 3 m - 5 } { 2 } = 2", "m = 3"], "exprs": ["x > \\frac { 3 m - 5 } { 2 }", "\\frac { 3 m - 5 } { 2 } = 2", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 m - 2 x < 5"}, {"id": "x > \\frac { 3 m - 5 } { 2 }"}, {"id": "x > 2"}, {"id": "\\frac { 3 m - 5 } { 2 } = 2"}, {"id": "m = 3"}], "links": [{"rel": "不等式方程部分求解", "source": "3 m - 2 x < 5", "target": "x > \\frac { 3 m - 5 } { 2 }"}, {"rel": "联立", "source": "x > \\frac { 3 m - 5 } { 2 }", "target": "\\frac { 3 m - 5 } { 2 } = 2"}, {"rel": "联立", "source": "x > 2", "target": "\\frac { 3 m - 5 } { 2 } = 2"}, {"rel": "等式方程求解", "source": "\\frac { 3 m - 5 } { 2 } = 2", "target": "m = 3"}]}}
|
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{"content": "Given $a = 99$, what is the value of $\\frac { a ^ 2 - 4 } { a - 2 }$?", "answer": "101", "steps": "Since $a = 99$, therefore $\\frac { a ^ 2 - 4 } { a - 2 } = \\frac {( a + 2 ) ( a - 2 )} { a - 2 } = a + 2$, the original expression is $a + 2 = 99 + 2 = 101$.", "expr_cands": ["a = 99", "a", "\\frac { a ^ { 2 } - 4 } { a - 2 }", "a + 2", "101"], "exprs": ["a + 2", "101"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a ^ { 2 } - 4 } { a - 2 }"}, {"id": "a + 2"}, {"id": "101"}, {"id": "a = 99"}], "links": [{"rel": "计算", "source": "\\frac { a ^ { 2 } - 4 } { a - 2 }", "target": "a + 2"}, {"rel": "被代入", "source": "a + 2", "target": "101"}, {"rel": "代入", "source": "a = 99", "target": "101"}]}}
|
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{"content": "The condition for the square root expression $\\sqrt { x - 5 }$ to be meaningful is ____ ?", "answer": "x \\ge 5", "steps": "From the given condition, we have $x - 5 \\ge 0$, which implies $x \\ge 5$.", "expr_cands": ["\\sqrt { x - 5 }", "x", "x - 5 \\ge 0", "5 \\le x", "x \\ge 5"], "exprs": ["x - 5 \\ge 0", "x \\ge 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x - 5 }"}, {"id": "x - 5 \\ge 0"}, {"id": "使二次根式 $\\sqrt { x - 5 }$ 有意义的条件"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 5"}], "links": [{"rel": "被描述", "source": "\\sqrt { x - 5 }", "target": "x - 5 \\ge 0"}, {"rel": "不等式方程求解", "source": "x - 5 \\ge 0", "target": "x \\ge 5"}, {"rel": "限制性描述", "source": "使二次根式 $\\sqrt { x - 5 }$ 有意义的条件", "target": "x - 5 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 5 \\ge 0"}]}}
|
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{"content": "If the value of the fraction $\\frac { 6 } { 2 x - 1 }$ is $2$, then the value of $x$ is ____?", "answer": "2", "steps": "Because the value of the fraction $\\frac { 6 } { 2 x - 1 }$ is $2$, we can obtain: $2 x - 1 = 3$, solving for $x$, we get: $x = 2$.", "expr_cands": ["\\frac { 6 } { 2 x - 1 }", "x", "2", "2 x - 1 = 3", "x = 2"], "exprs": ["2 x - 1 = 3", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 6 } { 2 x - 1 }"}, {"id": "2 x - 1 = 3"}, {"id": "2"}, {"id": "分式 $\\frac { 6 } { 2 x - 1 }$ 的值为 $2$"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "\\frac { 6 } { 2 x - 1 }", "target": "2 x - 1 = 3"}, {"rel": "等式方程求解", "source": "2 x - 1 = 3", "target": "x = 2"}, {"rel": "被描述", "source": "2", "target": "2 x - 1 = 3"}, {"rel": "限制性描述", "source": "分式 $\\frac { 6 } { 2 x - 1 }$ 的值为 $2$", "target": "2 x - 1 = 3"}]}}
|
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{"content": "The expression of the line obtained by translating the line $y = 3 x + 1$ down $5$ units is ____ ?", "answer": "y = 3 x - 4", "steps": "According to the principle of adding up and subtracting down, we know that the expression of the line $y = 3 x + 1$ after being translated down by $5$ units is $y = 3 x + 1 - 5$, which is $y = 3 x - 4$.", "expr_cands": ["y = 3 x + 1", "y", "x", "5", "y = 3 x + 1 - 5", "3 x + 1 = 3 x + 1 - 5", "3 x - 4"], "exprs": ["y = 3 x + 1 - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 3 x + 1"}, {"id": "y = 3 x + 1 - 5"}, {"id": "5"}, {"id": "直线 $y = 3 x + 1$ 向下平移 $5$ 个单位后得到直线的表达式是 : $y = 3 x + 1 - 5$"}], "links": [{"rel": "被描述", "source": "y = 3 x + 1", "target": "y = 3 x + 1 - 5"}, {"rel": "被描述", "source": "5", "target": "y = 3 x + 1 - 5"}, {"rel": "限制性描述", "source": "直线 $y = 3 x + 1$ 向下平移 $5$ 个单位后得到直线的表达式是 : $y = 3 x + 1 - 5$", "target": "y = 3 x + 1 - 5"}]}}
|
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{"content": "If $a + b = 2$, $ab = 2 \\sqrt { 2 }$, then the value of $a ^ 2 + b ^ 2$ is ____?", "answer": "4 - 4 \\sqrt { 2 }", "steps": "Because $a + b = 2$, $ab = 2 \\sqrt { 2 }$, we have $a ^ 2 + b ^ 2 = ( a + b ) ^ 2 - 2 ab = 2 ^ 2 - 2 * 2 \\sqrt { 2 } = 4 - 4 \\sqrt { 2 }$.", "expr_cands": ["a + b = 2", "b", "a", "ab = 2 \\sqrt { 2 }", "a ^ { 2 } + b ^ { 2 }", "( a + b ) ^ { 2 } - 2 ab", "4 - 4 \\sqrt { 2 }"], "exprs": ["( a + b ) ^ { 2 } - 2 ab", "4 - 4 \\sqrt { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } + b ^ { 2 }"}, {"id": "( a + b ) ^ { 2 } - 2 ab"}, {"id": "a + b = 2"}, {"id": "ab = 2 \\sqrt { 2 }"}, {"id": "4 - 4 \\sqrt { 2 }"}], "links": [{"rel": "提取因式", "source": "a ^ { 2 } + b ^ { 2 }", "target": "( a + b ) ^ { 2 } - 2 ab"}, {"rel": "被代入", "source": "( a + b ) ^ { 2 } - 2 ab", "target": "4 - 4 \\sqrt { 2 }"}, {"rel": "提取因式参考", "source": "a + b = 2", "target": "( a + b ) ^ { 2 } - 2 ab"}, {"rel": "代入", "source": "a + b = 2", "target": "4 - 4 \\sqrt { 2 }"}, {"rel": "提取因式参考", "source": "ab = 2 \\sqrt { 2 }", "target": "( a + b ) ^ { 2 } - 2 ab"}, {"rel": "代入", "source": "ab = 2 \\sqrt { 2 }", "target": "4 - 4 \\sqrt { 2 }"}]}}
|
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{"content": "Given the equation $ax + 8 = a - 2$ and $3 x - 4 = 4 x$ have the same solution, what is the value of $a$?", "answer": "2", "steps": "The equation $3 x - 4 = 4 x$ has a solution of $x = - 4$. Substituting $x = - 4$ into the equation $ax + 8 = a - 2$ yields $- 4 a + 8 = a - 2$, which can be solved to obtain $a = 2$.", "expr_cands": ["ax + 8 = a - 2", "x", "a", "3 x - 4 = 4 x", "x = - 4", "8 - 4 a = a - 2", "- 4 a + 8 = a - 2", "a = 2"], "exprs": ["x = - 4", "- 4 a + 8 = a - 2", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 4 = 4 x"}, {"id": "x = - 4"}, {"id": "ax + 8 = a - 2"}, {"id": "- 4 a + 8 = a - 2"}, {"id": "方程 $ax + 8 = a - 2$ 与 $3 x - 4 = 4 x$ 解相同"}, {"id": "a = 2"}], "links": [{"rel": "等式方程求解", "source": "3 x - 4 = 4 x", "target": "x = - 4"}, {"rel": "被描述", "source": "x = - 4", "target": "- 4 a + 8 = a - 2"}, {"rel": "被描述", "source": "ax + 8 = a - 2", "target": "- 4 a + 8 = a - 2"}, {"rel": "等式方程求解", "source": "- 4 a + 8 = a - 2", "target": "a = 2"}, {"rel": "限制性描述", "source": "方程 $ax + 8 = a - 2$ 与 $3 x - 4 = 4 x$ 解相同", "target": "- 4 a + 8 = a - 2"}]}}
|
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{"content": "The algebraic expression $3 x - 1$ and $3 ( x - \\frac { 5 } { 3 })$ are opposite in sign. What is the value of $x$?", "answer": "1", "steps": "According to the problem, we have $3 x - 1 + 3 ( x - \\frac { 5 } { 3 }) = 0$. Expanding the brackets, we get $3 x - 1 + 3 x - 5 = 0$. Rearranging, we have $3 x + 3 x = 5 + 1$. Combining like terms, we get $6 x = 6$. Dividing both sides by 6, we get $x = 1$.", "expr_cands": ["3 x - 1", "x", "3 ( x - \\frac { 5 } { 3 } )", "3 x - 1 + 3 ( x - \\frac { 5 } { 3 } ) = 0", "x = 1", "3 x - 1 + 3 x - 5 = 0", "3 x + 3 x = 5 + 1", "6 x = 6", "1"], "exprs": ["3 x - 1 + 3 ( x - \\frac { 5 } { 3 } ) = 0", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 1"}, {"id": "3 x - 1 + 3 ( x - \\frac { 5 } { 3 } ) = 0"}, {"id": "3 ( x - \\frac { 5 } { 3 } )"}, {"id": "代数式 $3 x - 1$ 与 $3 ( x - \\frac { 5 } { 3 } )$ 互为相反数"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "3 x - 1", "target": "3 x - 1 + 3 ( x - \\frac { 5 } { 3 } ) = 0"}, {"rel": "等式方程求解", "source": "3 x - 1 + 3 ( x - \\frac { 5 } { 3 } ) = 0", "target": "x = 1"}, {"rel": "被描述", "source": "3 ( x - \\frac { 5 } { 3 } )", "target": "3 x - 1 + 3 ( x - \\frac { 5 } { 3 } ) = 0"}, {"rel": "限制性描述", "source": "代数式 $3 x - 1$ 与 $3 ( x - \\frac { 5 } { 3 } )$ 互为相反数", "target": "3 x - 1 + 3 ( x - \\frac { 5 } { 3 } ) = 0"}]}}
|
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{"content": "To make the fraction $\\frac { 1 } { x + 2 }$ meaningful, the value of $x$ should satisfy:", "answer": "x \\neq - 2", "steps": "$\\because$ The fraction $\\frac { 1 } { x + 2 }$ is meaningful, $\\therefore$ $x + 2 \\neq 0$, $\\therefore$ $x \\neq - 2$, which means that the value of $x$ should satisfy: $x \\neq - 2$.", "expr_cands": ["\\frac { 1 } { x + 2 }", "x", "x + 2 \\neq 0", "x \\neq - 2"], "exprs": ["x + 2 \\neq 0", "x \\neq - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x + 2 }"}, {"id": "x + 2 \\neq 0"}, {"id": "要使分式 $\\frac { 1 } { x + 2 }$ 有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq - 2"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { x + 2 }", "target": "x + 2 \\neq 0"}, {"rel": "不等式方程求解", "source": "x + 2 \\neq 0", "target": "x \\neq - 2"}, {"rel": "限制性描述", "source": "要使分式 $\\frac { 1 } { x + 2 }$ 有意义", "target": "x + 2 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x + 2 \\neq 0"}]}}
|
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{"content": "Given $a = 2 + \\sqrt { 3 }$, what is the reciprocal of $a$?", "answer": "2 - \\sqrt { 3 }", "steps": "The reciprocal of $2 + \\sqrt { 3 }$ is $\\frac { 1 } { 2 + \\sqrt { 3 }} = \\frac { 2 - \\sqrt { 3 }} {( 2 + \\sqrt { 3 }) ( 2 - \\sqrt { 3 })} = \\frac { 2 - \\sqrt { 3 }} { 4 - 3 } = 2 - \\sqrt { 3 }$.", "expr_cands": ["a = 2 + \\sqrt { 3 }", "a", "2 + \\sqrt { 3 }", "\\frac { 1 } { 2 + \\sqrt { 3 } }", "2 - \\sqrt { 3 }"], "exprs": ["2 - \\sqrt { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a = 2 + \\sqrt { 3 }"}, {"id": "2 - \\sqrt { 3 }"}, {"id": "$a$ 的倒数"}], "links": [{"rel": "被描述", "source": "a = 2 + \\sqrt { 3 }", "target": "2 - \\sqrt { 3 }"}, {"rel": "限制性描述", "source": "$a$ 的倒数", "target": "2 - \\sqrt { 3 }"}]}}
|
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{"content": "If one root of the equation $x ^ 2 - kx - 12 = 0$ is $3$, then the value of $k$ is ____?", "answer": "- 1", "steps": "Substituting $x = 3$ into the equation $x ^ 2 - kx - 12 = 0$ yields $9 - 3 k - 12 = 0$, which can be solved to obtain $k = - 1$.", "expr_cands": ["x", "x ^ { 2 } - kx - 12 = 0", "k", "3", "x = 3", "- 3 k - 12 + 9 = 0", "9 - 3 k - 12 = 0", "k = - 1"], "exprs": ["x = 3", "9 - 3 k - 12 = 0", "k = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "x = 3"}, {"id": "x"}, {"id": "x ^ { 2 } - kx - 12 = 0"}, {"id": "关于 $x$ 的方程 $x ^ { 2 } - kx - 12 = 0$ 的一个根为 $3$"}, {"id": "9 - 3 k - 12 = 0"}, {"id": "k = - 1"}], "links": [{"rel": "被描述", "source": "3", "target": "x = 3"}, {"rel": "代入", "source": "x = 3", "target": "9 - 3 k - 12 = 0"}, {"rel": "被描述", "source": "x", "target": "x = 3"}, {"rel": "被描述", "source": "x ^ { 2 } - kx - 12 = 0", "target": "x = 3"}, {"rel": "被代入", "source": "x ^ { 2 } - kx - 12 = 0", "target": "9 - 3 k - 12 = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x ^ { 2 } - kx - 12 = 0$ 的一个根为 $3$", "target": "x = 3"}, {"rel": "等式方程求解", "source": "9 - 3 k - 12 = 0", "target": "k = - 1"}]}}
|
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{"content": "If the value of the fraction $\\frac { a ^ 2 - 16 } { 4 + a }$ is $0$, then the value of $a$ is ____?", "answer": "4", "steps": "From the given information, we can obtain that $a ^ 2 - 16 = 0$ and $a + 4 \\neq 0$, which implies that $a = 4$.", "expr_cands": ["\\frac { a ^ { 2 } - 16 } { 4 + a }", "a", "0", "a ^ { 2 } - 16 = 0", "a = - 4", "a = 4", "a + 4 \\neq 0", "a \\neq - 4", "x = 4", "x"], "exprs": ["a ^ { 2 } - 16 = 0", "a + 4 \\neq 0", "x = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a ^ { 2 } - 16 } { 4 + a }"}, {"id": "a ^ { 2 } - 16 = 0"}, {"id": "0"}, {"id": "分式 $\\frac { a ^ { 2 } - 16 } { 4 + a }$ 的值为 $0$"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "a + 4 \\neq 0"}, {"id": "x = 4"}], "links": [{"rel": "被描述", "source": "\\frac { a ^ { 2 } - 16 } { 4 + a }", "target": "a ^ { 2 } - 16 = 0"}, {"rel": "被描述", "source": "\\frac { a ^ { 2 } - 16 } { 4 + a }", "target": "a + 4 \\neq 0"}, {"rel": "联立", "source": "a ^ { 2 } - 16 = 0", "target": "x = 4"}, {"rel": "被描述", "source": "0", "target": "a ^ { 2 } - 16 = 0"}, {"rel": "被描述", "source": "0", "target": "a + 4 \\neq 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { a ^ { 2 } - 16 } { 4 + a }$ 的值为 $0$", "target": "a ^ { 2 } - 16 = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { a ^ { 2 } - 16 } { 4 + a }$ 的值为 $0$", "target": "a + 4 \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "a ^ { 2 } - 16 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "a + 4 \\neq 0"}, {"rel": "联立", "source": "a + 4 \\neq 0", "target": "x = 4"}]}}
|
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{"content": "The two square roots of a positive number $a$ are $2 b - 1$ and $b + 4$. Find the cube root of $a + b$.", "answer": "2", "steps": "Since the two square roots of a positive number $a$ are $2 b - 1$ and $b + 4$, therefore $2 b - 1 + b + 4 = 0$, therefore $b = - 1$. Therefore $b + 4 = - 1 + 4 = 3$, therefore $a = 9$. Therefore $a + b = 9 + ( - 1 ) = 8$. Since the cube root of $8$ is $2$, therefore the cube root of $a + b$ is $2$.", "expr_cands": ["a", "2 b - 1", "b", "b + 4", "a + b", "2 b - 1 + b + 4 = 0", "b = - 1", "3", "a = 9", "8", "2"], "exprs": ["2 b - 1 + b + 4 = 0", "a = 9", "b = - 1", "3", "8", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 b - 1"}, {"id": "2 b - 1 + b + 4 = 0"}, {"id": "b + 4"}, {"id": "一个正数 $a$ 的两个平方根是 $2 b - 1$ 和 $b + 4$"}, {"id": "平方根互为相反数"}, {"id": "b = - 1"}, {"id": "3"}, {"id": "a = 9"}, {"id": "a + b"}, {"id": "8"}, {"id": "2"}, {"id": "$a + b$ 的立方根"}, {"id": ", $8$ 的立方根为 $2$"}, {"id": ", $a + b$ 的立方根为 $2$"}], "links": [{"rel": "被描述", "source": "2 b - 1", "target": "2 b - 1 + b + 4 = 0"}, {"rel": "被描述", "source": "2 b - 1", "target": "a = 9"}, {"rel": "等式方程求解", "source": "2 b - 1 + b + 4 = 0", "target": "b = - 1"}, {"rel": "被描述", "source": "b + 4", "target": "2 b - 1 + b + 4 = 0"}, {"rel": "被代入", "source": "b + 4", "target": "3"}, {"rel": "被描述", "source": "b + 4", "target": "a = 9"}, {"rel": "限制性描述", "source": "一个正数 $a$ 的两个平方根是 $2 b - 1$ 和 $b + 4$", "target": "2 b - 1 + b + 4 = 0"}, {"rel": "限制性描述", "source": "一个正数 $a$ 的两个平方根是 $2 b - 1$ 和 $b + 4$", "target": "a = 9"}, {"rel": "限制性描述", "source": "平方根互为相反数", "target": "2 b - 1 + b + 4 = 0"}, {"rel": "代入", "source": "b = - 1", "target": "3"}, {"rel": "代入", "source": "b = - 1", "target": "8"}, {"rel": "代入", "source": "a = 9", "target": "8"}, {"rel": "被代入", "source": "a + b", "target": "8"}, {"rel": "被描述", "source": "a + b", "target": "2"}, {"rel": "被描述", "source": "8", "target": "2"}, {"rel": "限制性描述", "source": "$a + b$ 的立方根", "target": "2"}, {"rel": "限制性描述", "source": ", $8$ 的立方根为 $2$", "target": "2"}, {"rel": "限制性描述", "source": ", $a + b$ 的立方根为 $2$", "target": "2"}]}}
|
||||
{"content": "What is the smallest integer solution that satisfies the inequality $18 + 2 x > 0$?", "answer": "- 8", "steps": "$18 + 2 x > 0$, moving terms yields: $2 x > - 18$, therefore $x > - 9$. The smallest integer satisfying this inequality is $- 8$.", "expr_cands": ["18 + 2 x > 0", "x", "- 9 < x", "2 x > - 18", "x > - 9", "- 8"], "exprs": ["x > - 9", "- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "18 + 2 x > 0"}, {"id": "x > - 9"}, {"id": "- 8"}, {"id": "满足不等式 $18 + 2 x > 0$ 的最小整数解"}], "links": [{"rel": "不等式方程求解", "source": "18 + 2 x > 0", "target": "x > - 9"}, {"rel": "被描述", "source": "x > - 9", "target": "- 8"}, {"rel": "限制性描述", "source": "满足不等式 $18 + 2 x > 0$ 的最小整数解", "target": "- 8"}]}}
|
||||
{"content": "Solve the fractional equation in terms of $m$: $\\frac { 5 } { m - 3 } = - 1$ ____?", "answer": "m = - 2", "steps": "Going to the denominator, we get $- m + 3 = 5$. Solving for $m$, we get $m = - 2$. Upon checking, we find that $m = - 2$ is a solution to the fractional equation.", "expr_cands": ["m", "\\frac { 5 } { m - 3 } = - 1", "- m + 3 = 5", "m = - 2"], "exprs": ["m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 5 } { m - 3 } = - 1"}, {"id": "m = - 2"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 5 } { m - 3 } = - 1", "target": "m = - 2"}]}}
|
||||
{"content": "The range of values of $x$ that make $\\sqrt { x + 4 }$ meaningful is ____ ?", "answer": "x \\ge - 4", "steps": "From the given information, we have $x + 4 \\ge 0$, which implies $x \\ge - 4$.", "expr_cands": ["\\sqrt { x + 4 }", "x", "x + 4 \\ge 0", "- 4 \\le x", "x \\ge - 4"], "exprs": ["x + 4 \\ge 0", "x \\ge - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + 4 }"}, {"id": "x + 4 \\ge 0"}, {"id": "使得 $\\sqrt { x + 4 }$ 有意义的 $x$ 的取值范围"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge - 4"}], "links": [{"rel": "被描述", "source": "\\sqrt { x + 4 }", "target": "x + 4 \\ge 0"}, {"rel": "不等式方程求解", "source": "x + 4 \\ge 0", "target": "x \\ge - 4"}, {"rel": "限制性描述", "source": "使得 $\\sqrt { x + 4 }$ 有意义的 $x$ 的取值范围", "target": "x + 4 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x + 4 \\ge 0"}]}}
|
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{"content": "To make the expansion of $( x ^ 2 - x + 5 ) ( 2 x ^ 2 - ax - 4 )$ not contain the term $x ^ 2$, the value of $a$ is ____?", "answer": "- 6", "steps": "$( x ^ { 2 } - x + 5 ) ( 2 x ^ { 2 } - ax - 4 ) = 2 { x } ^ { 4 } - ( a + 2 ) { x } ^ { 3 } + ( 6 + a ) { x } ^ { 2 } + ( 4 - 5 a ) x - 20$ , from the expansion, we can see that there is no $x ^ { 2 }$ term, so we get $6 + a = 0$, which gives us $a = - 6$.", "expr_cands": ["( x ^ { 2 } - x + 5 ) ( 2 x ^ { 2 } - ax - 4 )", "a", "x", "x ^ { 2 }", "2 { x } ^ { 4 } - ( a + 2 ) { x } ^ { 3 } + ( 6 + a ) { x } ^ { 2 } + ( 4 - 5 a ) x - 20", "6 + a = 0", "a = - 6"], "exprs": ["2 { x } ^ { 4 } - ( a + 2 ) { x } ^ { 3 } + ( 6 + a ) { x } ^ { 2 } + ( 4 - 5 a ) x - 20", "6 + a = 0", "a = - 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x ^ { 2 } - x + 5 ) ( 2 x ^ { 2 } - ax - 4 )"}, {"id": "2 { x } ^ { 4 } - ( a + 2 ) { x } ^ { 3 } + ( 6 + a ) { x } ^ { 2 } + ( 4 - 5 a ) x - 20"}, {"id": "x ^ { 2 }"}, {"id": "6 + a = 0"}, {"id": "要使 $( x ^ { 2 } - x + 5 ) ( 2 x ^ { 2 } - ax - 4 )$ 展开式中不含 $x ^ { 2 }$ 项"}, {"id": "a = - 6"}], "links": [{"rel": "提取因式", "source": "( x ^ { 2 } - x + 5 ) ( 2 x ^ { 2 } - ax - 4 )", "target": "2 { x } ^ { 4 } - ( a + 2 ) { x } ^ { 3 } + ( 6 + a ) { x } ^ { 2 } + ( 4 - 5 a ) x - 20"}, {"rel": "被描述", "source": "2 { x } ^ { 4 } - ( a + 2 ) { x } ^ { 3 } + ( 6 + a ) { x } ^ { 2 } + ( 4 - 5 a ) x - 20", "target": "6 + a = 0"}, {"rel": "提取因式参考", "source": "x ^ { 2 }", "target": "2 { x } ^ { 4 } - ( a + 2 ) { x } ^ { 3 } + ( 6 + a ) { x } ^ { 2 } + ( 4 - 5 a ) x - 20"}, {"rel": "等式方程求解", "source": "6 + a = 0", "target": "a = - 6"}, {"rel": "限制性描述", "source": "要使 $( x ^ { 2 } - x + 5 ) ( 2 x ^ { 2 } - ax - 4 )$ 展开式中不含 $x ^ { 2 }$ 项", "target": "6 + a = 0"}]}}
|
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{"content": "If the fractional equation about $x$, $\\frac { 2 } { x - 1 } = 3 + \\frac { m } { 1 - x }$, has a root that makes the denominator equal to zero, then the value of $m$ is ____?", "answer": "- 2", "steps": "Solving the equation, we get $2 = 3 ( x - 1 ) - m$, which gives us $m = 3 x - 5$. If the denominator is zero, then $x - 1 = 0$, which means $x = 1$. Therefore, $m = 3 - 5 = - 2$.", "expr_cands": ["x", "\\frac { 2 } { x - 1 } = 3 + \\frac { m } { 1 - x }", "m", "2 = 3 ( x - 1 ) - m", "m = 3 x - 5", "x - 1 = 0", "x = 1", "m = - 2"], "exprs": ["2 = 3 ( x - 1 ) - m", "x - 1 = 0", "m = 3 x - 5", "x = 1", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { x - 1 } = 3 + \\frac { m } { 1 - x }"}, {"id": "2 = 3 ( x - 1 ) - m"}, {"id": "m = 3 x - 5"}, {"id": "x - 1 = 0"}, {"id": "关于 $x$ 的分式方程 $\\frac { 2 } { x - 1 } = 3 + \\frac { m } { 1 - x }$ 有使分母为零的根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 1"}, {"id": "m = - 2"}], "links": [{"rel": "同乘除", "source": "\\frac { 2 } { x - 1 } = 3 + \\frac { m } { 1 - x }", "target": "2 = 3 ( x - 1 ) - m"}, {"rel": "被描述", "source": "\\frac { 2 } { x - 1 } = 3 + \\frac { m } { 1 - x }", "target": "x - 1 = 0"}, {"rel": "等式方程部分求解", "source": "2 = 3 ( x - 1 ) - m", "target": "m = 3 x - 5"}, {"rel": "被代入", "source": "m = 3 x - 5", "target": "m = - 2"}, {"rel": "等式方程求解", "source": "x - 1 = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的分式方程 $\\frac { 2 } { x - 1 } = 3 + \\frac { m } { 1 - x }$ 有使分母为零的根", "target": "x - 1 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 1 = 0"}, {"rel": "代入", "source": "x = 1", "target": "m = - 2"}]}}
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{"content": "If $\\frac { 1 } { 3 } ( y + 1 )$ is the opposite of $3 - 2 y$, then $y$ equals", "answer": "2", "steps": "From the given information, we have $\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0$, which yields $y = 2$ as the solution.", "expr_cands": ["\\frac { 1 } { 3 } ( y + 1 )", "y", "3 - 2 y", "\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0", "y = 2"], "exprs": ["\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0", "y = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 3 } ( y + 1 )"}, {"id": "\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0"}, {"id": "3 - 2 y"}, {"id": "$\\frac { 1 } { 3 } ( y + 1 )$ 与 $3 - 2 y$ 互为相反数"}, {"id": "y = 2"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 3 } ( y + 1 )", "target": "\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0"}, {"rel": "等式方程求解", "source": "\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0", "target": "y = 2"}, {"rel": "被描述", "source": "3 - 2 y", "target": "\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 3 } ( y + 1 )$ 与 $3 - 2 y$ 互为相反数", "target": "\\frac { 1 } { 3 } ( y + 1 ) + 3 - 2 y = 0"}]}}
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{"content": "If the constant term in the expansion of $( x + m ) ( x ^ 2 + nx + 1 )$ is $- 2$ and there is no $x ^ 2$ term, then the coefficient of the linear term in the expansion is ____?", "answer": "- 3", "steps": "$( x + m ) ( x ^ { 2 } + nx + 1 ) = x ^ { 3 } + nx ^ { 2 } + x + mx ^ { 2 } + mnx + m = x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m$ Since the constant term in the expansion is $- 2$ and there is no $x ^ 2$ term, we have $m = - 2$, $m + n = 0$, and thus $n = 2$. Therefore, $mn + 1 = - 3$.", "expr_cands": ["( x + m ) ( x ^ { 2 } + nx + 1 )", "m", "n", "x", "- 2", "x ^ { 2 }", "x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m", "m = - 2", "m + n = 0", "n - 2 = 0", "n = 2", "mn + 1", "- 3"], "exprs": ["x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m", "m = - 2", "m + n = 0", "mn + 1", "n - 2 = 0", "n = 2", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + m ) ( x ^ { 2 } + nx + 1 )"}, {"id": "x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m"}, {"id": "m = - 2"}, {"id": "$( x + m ) ( x ^ { 2 } + nx + 1 )$ 的展开式中常数项为 $- 2$"}, {"id": "m + n = 0"}, {"id": "且不含 $x ^ { 2 }$ 项"}, {"id": "n - 2 = 0"}, {"id": "n = 2"}, {"id": "mn + 1"}, {"id": "展开式中的一次项系数"}, {"id": "- 3"}], "links": [{"rel": "展开", "source": "( x + m ) ( x ^ { 2 } + nx + 1 )", "target": "x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m"}, {"rel": "被描述", "source": "x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m", "target": "m = - 2"}, {"rel": "被描述", "source": "x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m", "target": "m + n = 0"}, {"rel": "被描述", "source": "x ^ { 3 } + ( m + n ) x ^ { 2 } + ( mn + 1 ) x + m", "target": "mn + 1"}, {"rel": "代入", "source": "m = - 2", "target": "n - 2 = 0"}, {"rel": "代入", "source": "m = - 2", "target": "- 3"}, {"rel": "限制性描述", "source": "$( x + m ) ( x ^ { 2 } + nx + 1 )$ 的展开式中常数项为 $- 2$", "target": "m = - 2"}, {"rel": "限制性描述", "source": "$( x + m ) ( x ^ { 2 } + nx + 1 )$ 的展开式中常数项为 $- 2$", "target": "m + n = 0"}, {"rel": "被代入", "source": "m + n = 0", "target": "n - 2 = 0"}, {"rel": "限制性描述", "source": "且不含 $x ^ { 2 }$ 项", "target": "m + n = 0"}, {"rel": "等式方程求解", "source": "n - 2 = 0", "target": "n = 2"}, {"rel": "代入", "source": "n = 2", "target": "- 3"}, {"rel": "被代入", "source": "mn + 1", "target": "- 3"}, {"rel": "限制性描述", "source": "展开式中的一次项系数", "target": "mn + 1"}]}}
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{"content": "Regarding the algebraic expression $( ax - 2 ) ({ x } ^ { 2 } + 3 x - 1 )$, if there is no term containing $x ^ { 2 }$ in the expanded expression, then $a$ = ____?", "answer": "\\frac { 2 } { 3 }", "steps": "$( ax - 2 ) ( x ^ { 2 } + 3 x - 1 ) = ax ^ { 3 } + 3 ax ^ { 2 } - ax - 2 x ^ { 2 } - 6 x + 2 = ax ^ { 3 } + ( 3 a - 2 ) x ^ { 2 } - ax - 6 x + 2$ From the given information, we know that $3 a - 2 = 0$, therefore $a = \\frac { 2 } { 3 }$.", "expr_cands": ["x", "( ax - 2 ) ( { x } ^ { 2 } + 3 x - 1 )", "a", "x ^ { 2 }", "( ax - 2 ) ( x ^ { 2 } + 3 x - 1 )", "ax ^ { 3 } + ( 3 a - 2 ) x ^ { 2 } - ax - 6 x + 2", "3 a - 2 = 0", "a = \\frac { 2 } { 3 }"], "exprs": ["ax ^ { 3 } + ( 3 a - 2 ) x ^ { 2 } - ax - 6 x + 2", "3 a - 2 = 0", "a = \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( ax - 2 ) ( x ^ { 2 } + 3 x - 1 )"}, {"id": "ax ^ { 3 } + ( 3 a - 2 ) x ^ { 2 } - ax - 6 x + 2"}, {"id": "x ^ { 2 }"}, {"id": "3 a - 2 = 0"}, {"id": "关于 $x$ 的代数式 $( ax - 2 ) ( { x } ^ { 2 } + 3 x - 1 )$ 的展开式中不含 $x ^ { 2 }$ 项"}, {"id": "a = \\frac { 2 } { 3 }"}], "links": [{"rel": "提取因式", "source": "( ax - 2 ) ( x ^ { 2 } + 3 x - 1 )", "target": "ax ^ { 3 } + ( 3 a - 2 ) x ^ { 2 } - ax - 6 x + 2"}, {"rel": "被描述", "source": "ax ^ { 3 } + ( 3 a - 2 ) x ^ { 2 } - ax - 6 x + 2", "target": "3 a - 2 = 0"}, {"rel": "提取因式参考", "source": "x ^ { 2 }", "target": "ax ^ { 3 } + ( 3 a - 2 ) x ^ { 2 } - ax - 6 x + 2"}, {"rel": "等式方程求解", "source": "3 a - 2 = 0", "target": "a = \\frac { 2 } { 3 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的代数式 $( ax - 2 ) ( { x } ^ { 2 } + 3 x - 1 )$ 的展开式中不含 $x ^ { 2 }$ 项", "target": "3 a - 2 = 0"}]}}
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{"content": "If the fraction $\\frac { 3 } { a + 3 }$ is undefined and the value of $\\frac { b + 4 } { b ^ 2 + 1 }$ is $0$, then $a + b$ = ____?", "answer": "- 7", "steps": "From the given information, we can obtain $a + 3 = 0$, which implies $a = - 3$, and $b + 4 = 0$, which implies $b = - 4$. When $b = - 4$, the denominator is not equal to zero, therefore $a + b = - 7$.", "expr_cands": ["\\frac { 3 } { a + 3 }", "a", "\\frac { b + 4 } { b ^ { 2 } + 1 }", "b", "0", "a + b", "a + 3 = 0", "a = - 3", "b + 4 = 0", "b = - 4", "- 7"], "exprs": ["a + 3 = 0", "b + 4 = 0", "a = - 3", "b = - 4", "- 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 } { a + 3 }"}, {"id": "a + 3 = 0"}, {"id": "分式 $\\frac { 3 } { a + 3 }$ 无意义"}, {"id": "分式有增根，则分母为0"}, {"id": "a = - 3"}, {"id": "\\frac { b + 4 } { b ^ { 2 } + 1 }"}, {"id": "b + 4 = 0"}, {"id": "0"}, {"id": "$\\frac { b + 4 } { b ^ { 2 } + 1 }$ 的值为 $0$"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "b = - 4"}, {"id": "a + b"}, {"id": "- 7"}], "links": [{"rel": "被描述", "source": "\\frac { 3 } { a + 3 }", "target": "a + 3 = 0"}, {"rel": "等式方程求解", "source": "a + 3 = 0", "target": "a = - 3"}, {"rel": "限制性描述", "source": "分式 $\\frac { 3 } { a + 3 }$ 无意义", "target": "a + 3 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "a + 3 = 0"}, {"rel": "代入", "source": "a = - 3", "target": "- 7"}, {"rel": "被描述", "source": "\\frac { b + 4 } { b ^ { 2 } + 1 }", "target": "b + 4 = 0"}, {"rel": "等式方程求解", "source": "b + 4 = 0", "target": "b = - 4"}, {"rel": "被描述", "source": "0", "target": "b + 4 = 0"}, {"rel": "限制性描述", "source": "$\\frac { b + 4 } { b ^ { 2 } + 1 }$ 的值为 $0$", "target": "b + 4 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "b + 4 = 0"}, {"rel": "代入", "source": "b = - 4", "target": "- 7"}, {"rel": "被代入", "source": "a + b", "target": "- 7"}]}}
|
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{"content": "If $a ^ { - 1 } + a = 2$, find the value of $a ^ 2 + a ^ { - 2 }$.", "answer": "2", "steps": "Since $a ^ { - 1 } + a = 2$, it follows that $a ^ 2 + a ^ { - 2 } = ( a ^ { - 1 } + a ) ^ 2 - 2 = 4 - 2 = 2$.", "expr_cands": ["a ^ { - 1 } + a = 2", "a", "a ^ { 2 } + a ^ { - 2 }", "a = 1", "( a ^ { - 1 } + a ) ^ { 2 } - 2", "2"], "exprs": ["( a ^ { - 1 } + a ) ^ { 2 } - 2", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } + a ^ { - 2 }"}, {"id": "( a ^ { - 1 } + a ) ^ { 2 } - 2"}, {"id": "a ^ { - 1 } + a = 2"}, {"id": "2"}], "links": [{"rel": "提取因式", "source": "a ^ { 2 } + a ^ { - 2 }", "target": "( a ^ { - 1 } + a ) ^ { 2 } - 2"}, {"rel": "被代入", "source": "( a ^ { - 1 } + a ) ^ { 2 } - 2", "target": "2"}, {"rel": "提取因式参考", "source": "a ^ { - 1 } + a = 2", "target": "( a ^ { - 1 } + a ) ^ { 2 } - 2"}, {"rel": "代入", "source": "a ^ { - 1 } + a = 2", "target": "2"}]}}
|
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{"content": "If $\\sqrt { a - 3 }$ exists, then the possible values of real number $a$ are _____.", "answer": "a \\ge 3", "steps": "According to the problem, we have $a - 3 \\ge 0$, which implies $a \\ge 3$.", "expr_cands": ["a - 3", "a", "a - 3 \\ge 0", "3 \\le a", "a \\ge 3"], "exprs": ["a - 3 \\ge 0", "a \\ge 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - 3"}, {"id": "a - 3 \\ge 0"}, {"id": "$a - 3$ 有平方根"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a \\ge 3"}], "links": [{"rel": "被描述", "source": "a - 3", "target": "a - 3 \\ge 0"}, {"rel": "不等式方程求解", "source": "a - 3 \\ge 0", "target": "a \\ge 3"}, {"rel": "限制性描述", "source": "$a - 3$ 有平方根", "target": "a - 3 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "a - 3 \\ge 0"}]}}
|
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{"content": "Given the algebraic expression $x - 2 y = 5$, what is the value of the algebraic expression $9 - 2 x + 4 y$?", "answer": "- 1", "steps": "When $x - 2 y = 5$, then $9 - 2 x + 4 y = 9 - 2 ( x - 2 y ) = 9 - 2 * 5 = 9 - 10 = - 1$.", "expr_cands": ["x - 2 y = 5", "y", "x", "9 - 2 x + 4 y", "9 - 2 ( x - 2 y )", "- 1"], "exprs": ["9 - 2 ( x - 2 y )", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "9 - 2 x + 4 y"}, {"id": "9 - 2 ( x - 2 y )"}, {"id": "x - 2 y = 5"}, {"id": "- 1"}], "links": [{"rel": "提取因式", "source": "9 - 2 x + 4 y", "target": "9 - 2 ( x - 2 y )"}, {"rel": "被代入", "source": "9 - 2 ( x - 2 y )", "target": "- 1"}, {"rel": "提取因式参考", "source": "x - 2 y = 5", "target": "9 - 2 ( x - 2 y )"}, {"rel": "代入", "source": "x - 2 y = 5", "target": "- 1"}]}}
|
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{"content": "If $\\sqrt { 1 - 3 a }$ and $\\sqrt { b - 27 }$ are opposite in sign, then the value of $ab$ is ____?", "answer": "9", "steps": "Because $\\sqrt { 1 - 3 a }$ and $\\sqrt { b - 27 }$ are opposite in sign, therefore $\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0$. Since $1 - 3 a \\ge 0$ and $b - 27 \\ge 0$, it follows that $1 - 3 a = 0$ and $b - 27 = 0$. Solving for $a$ and $b$, we get $a = \\frac { 1 } { 3 }$ and $b = 27$. Therefore, $ab = \\frac { 1 } { 3 } * 27 = 9$.", "expr_cands": ["\\sqrt { 1 - 3 a }", "a", "\\sqrt { b - 27 }", "b", "ab", "\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0", "1 - 3 a \\ge 0", "a \\le \\frac { 1 } { 3 }", "b - 27 \\ge 0", "27 \\le b", "1 - 3 a = 0", "a = \\frac { 1 } { 3 }", "b - 27 = 0", "b = 27", "9"], "exprs": ["\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0", "1 - 3 a = 0", "b - 27 = 0", "a = \\frac { 1 } { 3 }", "b = 27", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 1 - 3 a }"}, {"id": "\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0"}, {"id": "\\sqrt { b - 27 }"}, {"id": "$\\sqrt { 1 - 3 a }$ 与 $\\sqrt { b - 27 }$ 互为相反数"}, {"id": "1 - 3 a = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "b - 27 = 0"}, {"id": "a = \\frac { 1 } { 3 }"}, {"id": "b = 27"}, {"id": "ab"}, {"id": "9"}], "links": [{"rel": "被描述", "source": "\\sqrt { 1 - 3 a }", "target": "\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0"}, {"rel": "被描述", "source": "\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0", "target": "1 - 3 a = 0"}, {"rel": "被描述", "source": "\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0", "target": "b - 27 = 0"}, {"rel": "被描述", "source": "\\sqrt { b - 27 }", "target": "\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0"}, {"rel": "限制性描述", "source": "$\\sqrt { 1 - 3 a }$ 与 $\\sqrt { b - 27 }$ 互为相反数", "target": "\\sqrt { 1 - 3 a } + \\sqrt { b - 27 } = 0"}, {"rel": "等式方程求解", "source": "1 - 3 a = 0", "target": "a = \\frac { 1 } { 3 }"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "1 - 3 a = 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "b - 27 = 0"}, {"rel": "等式方程求解", "source": "b - 27 = 0", "target": "b = 27"}, {"rel": "代入", "source": "a = \\frac { 1 } { 3 }", "target": "9"}, {"rel": "代入", "source": "b = 27", "target": "9"}, {"rel": "被代入", "source": "ab", "target": "9"}]}}
|
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{"content": "If two simplest quadratic surds $\\sqrt { 5 a - 4 }$ and $\\sqrt { 2 a + 5 }$ can be combined, then the value of $a$ is ____?", "answer": "3", "steps": "$\\because$ the two simplest quadratic surds $\\sqrt { 5 a - 4 }$ and $\\sqrt { 2 a + 5 }$ can be combined, $\\therefore$ $5 a - 4 = 2 a + 5$, $\\therefore$ $a = 3$.", "expr_cands": ["\\sqrt { 5 a - 4 }", "a", "\\sqrt { 2 a + 5 }", "5 a - 4 = 2 a + 5", "a = 3"], "exprs": ["5 a - 4 = 2 a + 5", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 5 a - 4 }"}, {"id": "5 a - 4 = 2 a + 5"}, {"id": "\\sqrt { 2 a + 5 }"}, {"id": "两个最简二次根式 $\\sqrt { 5 a - 4 }$ 与 $\\sqrt { 2 a + 5 }$ 可以合并"}, {"id": "a = 3"}], "links": [{"rel": "被描述", "source": "\\sqrt { 5 a - 4 }", "target": "5 a - 4 = 2 a + 5"}, {"rel": "等式方程求解", "source": "5 a - 4 = 2 a + 5", "target": "a = 3"}, {"rel": "被描述", "source": "\\sqrt { 2 a + 5 }", "target": "5 a - 4 = 2 a + 5"}, {"rel": "限制性描述", "source": "两个最简二次根式 $\\sqrt { 5 a - 4 }$ 与 $\\sqrt { 2 a + 5 }$ 可以合并", "target": "5 a - 4 = 2 a + 5"}]}}
|
||||
{"content": "When $a = 2$, what is the value of the algebraic expression $\\frac { a ( a + 1 )} { 2 }$?", "answer": "3", "steps": "$\\frac { a ( a + 1 ) } { 2 } = \\frac { a ^ { 2 } + a } { 2 }$ , when $a = 2$, the original expression equals $\\frac { 2 ^ { 2 } + 2 } { 2 } = 3$.", "expr_cands": ["a = 2", "a", "\\frac { a ( a + 1 ) } { 2 }", "\\frac { a ^ { 2 } + a } { 2 }", "\\frac { 2 ^ { 2 } + 2 } { 2 }", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a = 2"}, {"id": "3"}, {"id": "\\frac { a ( a + 1 ) } { 2 }"}], "links": [{"rel": "代入", "source": "a = 2", "target": "3"}, {"rel": "被代入", "source": "\\frac { a ( a + 1 ) } { 2 }", "target": "3"}]}}
|
||||
{"content": "If one root of the equation $x ^ 2 + bx - 3 = 0$ with respect to $x$ is $- 1$, then the value of $b$ is ____?", "answer": "- 2", "steps": "$\\because$ One root of the equation $x ^ 2 + bx - 3 = 0$ with respect to $x$ is $- 1$, $\\therefore ( - 1 ) ^ 2 + b * ( - 1 ) - 3 = 0$, solving for $b = - 2$.", "expr_cands": ["x", "x ^ { 2 } + bx - 3 = 0", "b", "- 1", "( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0", "b = - 2"], "exprs": ["( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0", "b = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + bx - 3 = 0"}, {"id": "( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0"}, {"id": "x"}, {"id": "- 1"}, {"id": "关于 $x$ 的方程 $x ^ { 2 } + bx - 3 = 0$ 的一个根是 $- 1$"}, {"id": "b = - 2"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + bx - 3 = 0", "target": "( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0"}, {"rel": "等式方程求解", "source": "( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0", "target": "b = - 2"}, {"rel": "被描述", "source": "x", "target": "( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0"}, {"rel": "被描述", "source": "- 1", "target": "( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x ^ { 2 } + bx - 3 = 0$ 的一个根是 $- 1$", "target": "( - 1 ) ^ { 2 } + b * ( - 1 ) - 3 = 0"}]}}
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{"content": "If the solution set of the inequality $( a - 4 ) x > 1$ is $x < \\frac { 1 } { a - 4 }$, then ____?", "answer": "a < 4", "steps": "Since the solution set of the inequality $( a - 4 ) x > 1$ is $x < \\frac { 1 } { a - 4 }$, it can be known that the direction of the inequality has changed. It can be inferred that $a - 4 < 0$, so $a < 4$.", "expr_cands": ["( a - 4 ) x > 1", "a", "x", "x < \\frac { 1 } { a - 4 }", "a - 4 < 0", "a < 4"], "exprs": ["a - 4 < 0", "a < 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 4 ) x > 1"}, {"id": "a - 4 < 0"}, {"id": "x < \\frac { 1 } { a - 4 }"}, {"id": "不等式 $( a - 4 ) x > 1$ 的解集为 $x < \\frac { 1 } { a - 4 }$"}, {"id": "不等式两边都乘或除同一个负数，不等号的方向改变"}, {"id": "a < 4"}], "links": [{"rel": "被描述", "source": "( a - 4 ) x > 1", "target": "a - 4 < 0"}, {"rel": "不等式方程求解", "source": "a - 4 < 0", "target": "a < 4"}, {"rel": "被描述", "source": "x < \\frac { 1 } { a - 4 }", "target": "a - 4 < 0"}, {"rel": "限制性描述", "source": "不等式 $( a - 4 ) x > 1$ 的解集为 $x < \\frac { 1 } { a - 4 }$", "target": "a - 4 < 0"}, {"rel": "属性描述", "source": "不等式两边都乘或除同一个负数，不等号的方向改变", "target": "a - 4 < 0"}]}}
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{"content": "Given that $nx ^ { | n - 1 | } + 5 = 0$ is a linear equation in one variable, what is the value of $n$?", "answer": "2", "steps": "$\\because nx ^ { | n - 1 | } + 5 = 0$ is a linear equation in one variable, $\\therefore n - 1 = 1$, and $n \\neq 0$, solving for $n$, we get: $n = 2$.", "expr_cands": ["nx ^ { | n - 1 | } + 5 = 0", "x", "n", "n - 1 = 1", "n = 2", "n \\neq 0"], "exprs": ["n - 1 = 1", "n \\neq 0", "n = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "nx ^ { | n - 1 | } + 5 = 0"}, {"id": "n - 1 = 1"}, {"id": "$nx ^ { | n - 1 | } + 5 = 0$ 为一元一次方程"}, {"id": "n = 2"}, {"id": "n \\neq 0"}], "links": [{"rel": "被描述", "source": "nx ^ { | n - 1 | } + 5 = 0", "target": "n - 1 = 1"}, {"rel": "被描述", "source": "nx ^ { | n - 1 | } + 5 = 0", "target": "n \\neq 0"}, {"rel": "等式方程求解", "source": "n - 1 = 1", "target": "n = 2"}, {"rel": "限制性描述", "source": "$nx ^ { | n - 1 | } + 5 = 0$ 为一元一次方程", "target": "n - 1 = 1"}, {"rel": "限制性描述", "source": "$nx ^ { | n - 1 | } + 5 = 0$ 为一元一次方程", "target": "n \\neq 0"}]}}
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{"content": "If $x = 3$ is a solution of the equation $kx - 8 = k$ with respect to $x$, then the value of $k$ is ____?", "answer": "4", "steps": "Substituting $x = 3$ into the equation $kx - 8 = k$ yields $3 k - 8 = k$. Solving for $k$ gives $k = 4$.", "expr_cands": ["x = 3", "x", "kx - 8 = k", "k", "3 k - 8 = k", "k = 4"], "exprs": ["3 k - 8 = k", "k = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "kx - 8 = k"}, {"id": "3 k - 8 = k"}, {"id": "x = 3"}, {"id": "k = 4"}], "links": [{"rel": "被代入", "source": "kx - 8 = k", "target": "3 k - 8 = k"}, {"rel": "等式方程求解", "source": "3 k - 8 = k", "target": "k = 4"}, {"rel": "代入", "source": "x = 3", "target": "3 k - 8 = k"}]}}
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{"content": "If the solution to the equation $x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1$ with respect to $x$ is a positive integer, then the sum of all integers $a$ that satisfy the condition is ____?", "answer": "- 7", "steps": "$x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1$ Taking the denominator, we get $6 x - 4 + ax = x + 4 - 6$ Moving terms and combining like terms, we get $( 5 + a ) x = 2$ , $x = \\frac { 2 } { 5 + a }$ , From the given condition, we have $a = - 3$ , $- 4$ , so the sum of all integers $a$ that satisfy the condition is $- 3 - 4 = - 7$.", "expr_cands": ["x", "x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1", "a", "6 x - 4 + ax = x + 4 - 6", "( 5 + a ) x = 2", "x = \\frac { 2 } { 5 + a }", "a = - 3", "- 4", "- 3 - 4", "- 7"], "exprs": ["x = \\frac { 2 } { 5 + a }", "a = - 3", "- 4", "- 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1"}, {"id": "x = \\frac { 2 } { 5 + a }"}, {"id": "a"}, {"id": "a = - 3"}, {"id": "关于 $x$ 的方程 $x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1$ 的解是正整数"}, {"id": "- 4"}, {"id": "- 7"}, {"id": "符合条件的所有整数 $a$ 的和是 $- 3 - 4 = - 7$"}], "links": [{"rel": "等式方程部分求解", "source": "x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1", "target": "x = \\frac { 2 } { 5 + a }"}, {"rel": "被描述", "source": "x = \\frac { 2 } { 5 + a }", "target": "a = - 3"}, {"rel": "被描述", "source": "x = \\frac { 2 } { 5 + a }", "target": "- 4"}, {"rel": "被描述", "source": "a", "target": "a = - 3"}, {"rel": "被描述", "source": "a", "target": "- 4"}, {"rel": "被描述", "source": "a = - 3", "target": "- 7"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1$ 的解是正整数", "target": "a = - 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x - \\frac { 4 - ax } { 6 } = \\frac { x + 4 } { 6 } - 1$ 的解是正整数", "target": "- 4"}, {"rel": "被描述", "source": "- 4", "target": "- 7"}, {"rel": "限制性描述", "source": "符合条件的所有整数 $a$ 的和是 $- 3 - 4 = - 7$", "target": "- 7"}]}}
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{"content": "If $a$, $b$ are opposite numbers, $c$, $d$ are reciprocal numbers, and $m$ is the largest negative integer, then the value of $- 2 | - m | + cd - \\frac { a + b } { m }$ is ____?", "answer": "- 1", "steps": "Since $a$ and $b$ are opposite numbers, and $c$ and $d$ are reciprocals, and $m$ is the largest negative integer, therefore $a + b = 0$, $cd = 1$, $m = - 1$. Therefore, $- 2 | - m | + cd - \\frac { a + b } { m } = - 2 * | - 1 | + 1 - \\frac { 0 } { - 1 } = - 2 * 1 + 1 - 0 = - 2 + 1 - 0 = - 1$.", "expr_cands": ["a", "b", "c", "d", "m", "- 2 | - m | + cd - \\frac { a + b } { m }", "a + b = 0", "cd = 1", "m = - 1", "- 1"], "exprs": ["a + b = 0", "cd = 1", "m = - 1", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "c"}, {"id": "cd = 1"}, {"id": "d"}, {"id": "$c$ , $d$ 互为倒数"}, {"id": "m"}, {"id": "m = - 1"}, {"id": "$m$ 是最大的负整数"}, {"id": "- 2 | - m | + cd - \\frac { a + b } { m }"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "- 1"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "c", "target": "cd = 1"}, {"rel": "代入", "source": "cd = 1", "target": "- 1"}, {"rel": "被描述", "source": "d", "target": "cd = 1"}, {"rel": "限制性描述", "source": "$c$ , $d$ 互为倒数", "target": "cd = 1"}, {"rel": "被描述", "source": "m", "target": "m = - 1"}, {"rel": "代入", "source": "m = - 1", "target": "- 1"}, {"rel": "限制性描述", "source": "$m$ 是最大的负整数", "target": "m = - 1"}, {"rel": "被代入", "source": "- 2 | - m | + cd - \\frac { a + b } { m }", "target": "- 1"}]}}
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{"content": "If the linear function $y = ( 3 a - 2 ) x + 6$ increases as $x$ increases, then the range of possible values for $a$ is _____.", "answer": "a > \\frac { 2 } { 3 }", "steps": "According to the problem, we have $3 a - 2 > 0$, which leads to $a > \\frac { 2 } { 3 }$.", "expr_cands": ["y = ( 3 a - 2 ) x + 6", "a", "y", "x", "3 a - 2 > 0", "\\frac { 2 } { 3 } < a", "a > \\frac { 2 } { 3 }"], "exprs": ["3 a - 2 > 0", "a > \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( 3 a - 2 ) x + 6"}, {"id": "3 a - 2 > 0"}, {"id": "一次函数 $y = ( 3 a - 2 ) x + 6$ 随着 $x$ 的增大而增大"}, {"id": "$a$ 的取值范围"}, {"id": "a > \\frac { 2 } { 3 }"}], "links": [{"rel": "被描述", "source": "y = ( 3 a - 2 ) x + 6", "target": "3 a - 2 > 0"}, {"rel": "不等式方程求解", "source": "3 a - 2 > 0", "target": "a > \\frac { 2 } { 3 }"}, {"rel": "限制性描述", "source": "一次函数 $y = ( 3 a - 2 ) x + 6$ 随着 $x$ 的增大而增大", "target": "3 a - 2 > 0"}, {"rel": "限制性描述", "source": "$a$ 的取值范围", "target": "3 a - 2 > 0"}]}}
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{"content": "In the equation $5 x - 2 y + z = 3$, if $x = 1$ and $y = 2$, then $z$ = ____ ?", "answer": "2", "steps": "To find the value of the third unknown variable in a system of three linear equations, substitute the given values of the first two variables and solve for the third. For example, if $x = 1$ and $y = 2$, substituting into the equation $5 x - 2 y + z = 3$ yields $z = 2$.", "expr_cands": ["5 x - 2 y + z = 3", "z", "x", "y", "x = 1", "y = 2", "z + 1 = 3", "z = 2"], "exprs": ["z + 1 = 3", "z = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x - 2 y + z = 3"}, {"id": "z + 1 = 3"}, {"id": "x = 1"}, {"id": "y = 2"}, {"id": "z = 2"}], "links": [{"rel": "被代入", "source": "5 x - 2 y + z = 3", "target": "z + 1 = 3"}, {"rel": "等式方程求解", "source": "z + 1 = 3", "target": "z = 2"}, {"rel": "代入", "source": "x = 1", "target": "z + 1 = 3"}, {"rel": "代入", "source": "y = 2", "target": "z + 1 = 3"}]}}
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{"content": "If the solution set of the inequality $3 x + a > 2$ is $x > 1$, then $a$ = ____ ?", "answer": "- 1", "steps": "$\\because$ $3 x + a > 2$, $\\therefore$ $x > \\frac { 2 - a } { 3 }$. $\\because$ the solution set of the inequality $3 x + a > 2$ is $x > 1$, $\\therefore$ $\\frac { 2 - a } { 3 } = 1$, which yields $a = - 1$.", "expr_cands": ["3 x + a > 2", "a", "x", "x > 1", "x > \\frac { 2 - a } { 3 }", "\\frac { 2 - a } { 3 } = 1", "a = - 1"], "exprs": ["x > \\frac { 2 - a } { 3 }", "\\frac { 2 - a } { 3 } = 1", "a = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + a > 2"}, {"id": "x > \\frac { 2 - a } { 3 }"}, {"id": "x > 1"}, {"id": "\\frac { 2 - a } { 3 } = 1"}, {"id": "不等式 $3 x + a > 2$ 的解集是 $x > 1$"}, {"id": "a = - 1"}], "links": [{"rel": "不等式方程部分求解", "source": "3 x + a > 2", "target": "x > \\frac { 2 - a } { 3 }"}, {"rel": "被描述", "source": "x > \\frac { 2 - a } { 3 }", "target": "\\frac { 2 - a } { 3 } = 1"}, {"rel": "被描述", "source": "x > 1", "target": "\\frac { 2 - a } { 3 } = 1"}, {"rel": "等式方程求解", "source": "\\frac { 2 - a } { 3 } = 1", "target": "a = - 1"}, {"rel": "限制性描述", "source": "不等式 $3 x + a > 2$ 的解集是 $x > 1$", "target": "\\frac { 2 - a } { 3 } = 1"}]}}
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{"content": "Given a quadratic equation in one variable $x$, $5 x ^ 2 + kx - 6 = 0$, with one root being $2$. Find the other root.", "answer": "- \\frac { 3 } { 5 }", "steps": "Let the other root of the equation be $t$. According to the problem, we have $2 * t = \\frac { - 6 } { 5 }$, solving which we get $t = - \\frac { 3 } { 5 }$.", "expr_cands": ["x", "5 x ^ { 2 } + kx - 6 = 0", "k", "2", "t", "2 * t = \\frac { - 6 } { 5 }", "t = - \\frac { 3 } { 5 }"], "exprs": ["2 * t = \\frac { - 6 } { 5 }", "t = - \\frac { 3 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x ^ { 2 } + kx - 6 = 0"}, {"id": "2 * t = \\frac { - 6 } { 5 }"}, {"id": "2"}, {"id": "t"}, {"id": "x"}, {"id": "关于 $x$ 的一元二次方程 $5 x ^ { 2 } + kx - 6 = 0$ 的一个根是 $2$"}, {"id": "设方程的另一个根为 $t$"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "t = - \\frac { 3 } { 5 }"}], "links": [{"rel": "被描述", "source": "5 x ^ { 2 } + kx - 6 = 0", "target": "2 * t = \\frac { - 6 } { 5 }"}, {"rel": "等式方程求解", "source": "2 * t = \\frac { - 6 } { 5 }", "target": "t = - \\frac { 3 } { 5 }"}, {"rel": "被描述", "source": "2", "target": "2 * t = \\frac { - 6 } { 5 }"}, {"rel": "被描述", "source": "t", "target": "2 * t = \\frac { - 6 } { 5 }"}, {"rel": "被描述", "source": "x", "target": "2 * t = \\frac { - 6 } { 5 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $5 x ^ { 2 } + kx - 6 = 0$ 的一个根是 $2$", "target": "2 * t = \\frac { - 6 } { 5 }"}, {"rel": "限制性描述", "source": "设方程的另一个根为 $t$", "target": "2 * t = \\frac { - 6 } { 5 }"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "2 * t = \\frac { - 6 } { 5 }"}]}}
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{"content": "$\\sqrt { ( x - 3 ) ^ { 2 } } = x - 3$ , the range of $x$ is: ____ ?", "answer": "x \\ge 3", "steps": "From the given information, we have $x - 3 \\ge 0$, which implies $x \\ge 3$.", "expr_cands": ["\\sqrt { ( x - 3 ) ^ { 2 } } = x - 3", "x", "x - 3 \\ge 0", "3 \\le x", "x \\ge 3"], "exprs": ["x - 3 \\ge 0", "x \\ge 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { ( x - 3 ) ^ { 2 } } = x - 3"}, {"id": "x - 3 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 3"}], "links": [{"rel": "被描述", "source": "\\sqrt { ( x - 3 ) ^ { 2 } } = x - 3", "target": "x - 3 \\ge 0"}, {"rel": "不等式方程求解", "source": "x - 3 \\ge 0", "target": "x \\ge 3"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 3 \\ge 0"}]}}
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{"content": "Given a quadratic equation $x ^ 2 - 6 x + a = 0$ with one root being $2$, the other root is ____?", "answer": "4", "steps": "The other root of the equation is $x$. According to the given condition, we have $x + 2 = 6$. Solving for $x$, we get $x = 4$.", "expr_cands": ["x ^ { 2 } - 6 x + a = 0", "a", "x", "2", "x + 2 = 6", "x = 4"], "exprs": ["x + 2 = 6", "x = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 6 x + a = 0"}, {"id": "x + 2 = 6"}, {"id": "2"}, {"id": "x"}, {"id": "一元二次方程 $x ^ { 2 } - 6 x + a = 0$ 有一个根为 $2$"}, {"id": "x = 4"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 6 x + a = 0", "target": "x + 2 = 6"}, {"rel": "等式方程求解", "source": "x + 2 = 6", "target": "x = 4"}, {"rel": "被描述", "source": "2", "target": "x + 2 = 6"}, {"rel": "被描述", "source": "x", "target": "x + 2 = 6"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - 6 x + a = 0$ 有一个根为 $2$", "target": "x + 2 = 6"}]}}
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{"content": "The condition for the expression $\\sqrt { 2 x - 1 }$ to be meaningful is ____ ?", "answer": "x \\ge \\frac { 1 } { 2 }", "steps": "Because $\\sqrt { 2 x - 1 }$ is meaningful, therefore $2 x - 1 \\ge 0$, which leads to $x \\ge \\frac { 1 } { 2 }$ after solving.", "expr_cands": ["\\sqrt { 2 x - 1 }", "x", "2 x - 1 \\ge 0", "\\frac { 1 } { 2 } \\le x", "x \\ge \\frac { 1 } { 2 }"], "exprs": ["2 x - 1 \\ge 0", "x \\ge \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 2 x - 1 }"}, {"id": "2 x - 1 \\ge 0"}, {"id": "式子 $\\sqrt { 2 x - 1 }$ 有意义的条件"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\sqrt { 2 x - 1 }", "target": "2 x - 1 \\ge 0"}, {"rel": "不等式方程求解", "source": "2 x - 1 \\ge 0", "target": "x \\ge \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "式子 $\\sqrt { 2 x - 1 }$ 有意义的条件", "target": "2 x - 1 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 x - 1 \\ge 0"}]}}
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{"content": "If the polynomial in $x$, $3 { x } ^ 2 - k { x } ^ 2 - 2 { x } ^ 3 + x - 1$, does not contain a quadratic term, then the value of $k$ is ____?", "answer": "3", "steps": "Original expression = $( 3 - k ) x ^ 2 - 2 x ^ 3 + x - 1$. Since the polynomial $3 x ^ 2 - kx ^ 2 - 2 x ^ 3 + x - 1$ does not contain a quadratic term with respect to $x$, we have $3 - k = 0$. Solving for $k$, we get $k = 3$.", "expr_cands": ["x", "3 { x } ^ { 2 } - k { x } ^ { 2 } - 2 { x } ^ { 3 } + x - 1", "k", "( 3 - k ) x ^ { 2 } - 2 x ^ { 3 } + x - 1", "3 - k = 0", "k = 3"], "exprs": ["( 3 - k ) x ^ { 2 } - 2 x ^ { 3 } + x - 1", "3 - k = 0", "k = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 { x } ^ { 2 } - k { x } ^ { 2 } - 2 { x } ^ { 3 } + x - 1"}, {"id": "( 3 - k ) x ^ { 2 } - 2 x ^ { 3 } + x - 1"}, {"id": "x"}, {"id": "3 - k = 0"}, {"id": "关于 $x$ 的多项式 $3 { x } ^ { 2 } - k { x } ^ { 2 } - 2 { x } ^ { 3 } + x - 1$ 不含二次项"}, {"id": "k = 3"}], "links": [{"rel": "提取因式", "source": "3 { x } ^ { 2 } - k { x } ^ { 2 } - 2 { x } ^ { 3 } + x - 1", "target": "( 3 - k ) x ^ { 2 } - 2 x ^ { 3 } + x - 1"}, {"rel": "被描述", "source": "( 3 - k ) x ^ { 2 } - 2 x ^ { 3 } + x - 1", "target": "3 - k = 0"}, {"rel": "提取因式参考", "source": "x", "target": "( 3 - k ) x ^ { 2 } - 2 x ^ { 3 } + x - 1"}, {"rel": "等式方程求解", "source": "3 - k = 0", "target": "k = 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的多项式 $3 { x } ^ { 2 } - k { x } ^ { 2 } - 2 { x } ^ { 3 } + x - 1$ 不含二次项", "target": "3 - k = 0"}]}}
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{"content": "If the result of $( x + m ) ( 2 x - 1 )$ does not contain a linear term in $x$, then the value of the constant $m$ is ____?", "answer": "\\frac { 1 } { 2 }", "steps": "$( x + m ) ( 2 x - 1 ) = 2 { x } ^ { 2 } - x + 2 mx - m = 2 { x } ^ { 2 } + ( 2 m - 1 ) x - m$ Because there is no linear term in $x$ in the result, therefore $2 m - 1 = 0$, which gives $m = \\frac { 1 } { 2 }$.", "expr_cands": ["( x + m ) ( 2 x - 1 )", "m", "x", "2 { x } ^ { 2 } + ( 2 m - 1 ) x - m", "2 m - 1 = 0", "m = \\frac { 1 } { 2 }"], "exprs": ["2 { x } ^ { 2 } + ( 2 m - 1 ) x - m", "2 m - 1 = 0", "m = \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + m ) ( 2 x - 1 )"}, {"id": "2 { x } ^ { 2 } + ( 2 m - 1 ) x - m"}, {"id": "x"}, {"id": "2 m - 1 = 0"}, {"id": "$( x + m ) ( 2 x - 1 )$ 的结果中不含 $x$ 的一次项"}, {"id": "m = \\frac { 1 } { 2 }"}], "links": [{"rel": "提取因式", "source": "( x + m ) ( 2 x - 1 )", "target": "2 { x } ^ { 2 } + ( 2 m - 1 ) x - m"}, {"rel": "被描述", "source": "2 { x } ^ { 2 } + ( 2 m - 1 ) x - m", "target": "2 m - 1 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "2 { x } ^ { 2 } + ( 2 m - 1 ) x - m"}, {"rel": "等式方程求解", "source": "2 m - 1 = 0", "target": "m = \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "$( x + m ) ( 2 x - 1 )$ 的结果中不含 $x$ 的一次项", "target": "2 m - 1 = 0"}]}}
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{"content": "For the inequality $- \\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }$ involving $x$, if we ignore the - in front of $- \\frac { 1 + x } { 2 }$, the obtained solution set is $x \\le 1$. What is the correct solution set for this inequality?", "answer": "x \\le 13", "steps": "According to the problem, we have the solution set of $\\frac { 1 + x } { 2 } \\leq a - \\frac { 1 + 2 x } { 3 }$ as $x \\leq 1$. Solving the inequality $\\frac { 1 + x } { 2 } \\leq a - \\frac { 1 + 2 x } { 3 }$ gives us $x \\leq \\frac { 6 a - 5 } { 7 }$. Therefore, $\\frac { 6 a - 5 } { 7 } = 1$, and thus $a = 2$. Solving the inequality $- \\frac { 1 + x } { 2 } \\leq 2 - \\frac { 1 + 2 x } { 3 }$ gives us $x \\leq 13$.", "expr_cands": ["x", "- \\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }", "a", "- \\frac { 1 + x } { 2 }", "x \\le 1", "\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }", "x \\le \\frac { 6 a - 5 } { 7 }", "\\frac { 6 a - 5 } { 7 } = 1", "a = 2", "- \\frac { 1 + x } { 2 } \\le 2 - \\frac { 1 + 2 x } { 3 }", "x \\le 13"], "exprs": ["\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }", "x \\le \\frac { 6 a - 5 } { 7 }", "\\frac { 6 a - 5 } { 7 } = 1", "a = 2", "- \\frac { 1 + x } { 2 } \\le 2 - \\frac { 1 + 2 x } { 3 }", "x \\le 13"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }"}, {"id": "\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }"}, {"id": "- \\frac { 1 + x } { 2 }"}, {"id": "解关于 $x$ 的不等式 $- \\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }$ 时"}, {"id": "忽略了 $- \\frac { 1 + x } { 2 }$ 前面的 \" - \""}, {"id": "x \\le \\frac { 6 a - 5 } { 7 }"}, {"id": "x \\le 1"}, {"id": "\\frac { 6 a - 5 } { 7 } = 1"}, {"id": "导致求出的解集为 $x \\le 1$"}, {"id": "a = 2"}, {"id": "- \\frac { 1 + x } { 2 } \\le 2 - \\frac { 1 + 2 x } { 3 }"}, {"id": "x \\le 13"}], "links": [{"rel": "被描述", "source": "- \\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }", "target": "\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }"}, {"rel": "被代入", "source": "- \\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }", "target": "- \\frac { 1 + x } { 2 } \\le 2 - \\frac { 1 + 2 x } { 3 }"}, {"rel": "不等式方程部分求解", "source": "\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }", "target": "x \\le \\frac { 6 a - 5 } { 7 }"}, {"rel": "被描述", "source": "- \\frac { 1 + x } { 2 }", "target": "\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }"}, {"rel": "限制性描述", "source": "解关于 $x$ 的不等式 $- \\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }$ 时", "target": "\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }"}, {"rel": "限制性描述", "source": "忽略了 $- \\frac { 1 + x } { 2 }$ 前面的 \" - \"", "target": "\\frac { 1 + x } { 2 } \\le a - \\frac { 1 + 2 x } { 3 }"}, {"rel": "被描述", "source": "x \\le \\frac { 6 a - 5 } { 7 }", "target": "\\frac { 6 a - 5 } { 7 } = 1"}, {"rel": "被描述", "source": "x \\le 1", "target": "\\frac { 6 a - 5 } { 7 } = 1"}, {"rel": "等式方程求解", "source": "\\frac { 6 a - 5 } { 7 } = 1", "target": "a = 2"}, {"rel": "限制性描述", "source": "导致求出的解集为 $x \\le 1$", "target": "\\frac { 6 a - 5 } { 7 } = 1"}, {"rel": "代入", "source": "a = 2", "target": "- \\frac { 1 + x } { 2 } \\le 2 - \\frac { 1 + 2 x } { 3 }"}, {"rel": "不等式方程求解", "source": "- \\frac { 1 + x } { 2 } \\le 2 - \\frac { 1 + 2 x } { 3 }", "target": "x \\le 13"}]}}
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{"content": "Given that the solution to the equation $3 x - 2 m = 4$ is $x = m$, the value of $m$ is ____?", "answer": "4", "steps": "According to the problem, we have $3 m - 2 m = 4$, which means $m = 4$.", "expr_cands": ["x", "3 x - 2 m = 4", "m", "x = m", "3 m - 2 m = 4", "m = 4"], "exprs": ["3 m - 2 m = 4", "m = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x"}, {"id": "3 m - 2 m = 4"}, {"id": "3 x - 2 m = 4"}, {"id": "x = m"}, {"id": "关于 $x$ 的方程 $3 x - 2 m = 4$ 的解是 $x = m$"}, {"id": "m = 4"}], "links": [{"rel": "被描述", "source": "x", "target": "3 m - 2 m = 4"}, {"rel": "等式方程求解", "source": "3 m - 2 m = 4", "target": "m = 4"}, {"rel": "被描述", "source": "3 x - 2 m = 4", "target": "3 m - 2 m = 4"}, {"rel": "被描述", "source": "x = m", "target": "3 m - 2 m = 4"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $3 x - 2 m = 4$ 的解是 $x = m$", "target": "3 m - 2 m = 4"}]}}
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{"content": "If $m + n = 10$ and $m - n = 2$, then $m ^ { 2 } - n ^ { 2 }$ = ____ ?", "answer": "20", "steps": "Because $m + n = 10$ and $m - n = 2$, therefore $m ^ 2 - n ^ 2 = ( m + n ) ( m - n ) = 10 * 2 = 20$.", "expr_cands": ["m + n = 10", "n", "m", "m - n = 2", "m ^ { 2 } - n ^ { 2 }", "( m + n ) ( m - n )", "20"], "exprs": ["( m + n ) ( m - n )", "20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m ^ { 2 } - n ^ { 2 }"}, {"id": "( m + n ) ( m - n )"}, {"id": "m + n = 10"}, {"id": "m - n = 2"}, {"id": "20"}], "links": [{"rel": "提取因式", "source": "m ^ { 2 } - n ^ { 2 }", "target": "( m + n ) ( m - n )"}, {"rel": "被代入", "source": "( m + n ) ( m - n )", "target": "20"}, {"rel": "提取因式参考", "source": "m + n = 10", "target": "( m + n ) ( m - n )"}, {"rel": "代入", "source": "m + n = 10", "target": "20"}, {"rel": "提取因式参考", "source": "m - n = 2", "target": "( m + n ) ( m - n )"}, {"rel": "代入", "source": "m - n = 2", "target": "20"}]}}
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{"content": "If $| a | = 8$, $| b | = 5$, $ab < 0$, and $a + b > 0$, then what is the value of $a - b$?", "answer": "13", "steps": "$\\because | a | = 8$, $| b | = 5$, and $ab < 0$, $a + b > 0$, $\\therefore a = 8$, $b = - 5$, so $a - b = 13$.", "expr_cands": ["| a | = 8", "a", "| b | = 5", "b", "ab < 0", "a + b > 0", "a - b", "a = - 8", "a = 8", "b = - 5", "b = 5", "13"], "exprs": ["a = 8", "b = - 5", "13"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a | = 8"}, {"id": "a = 8"}, {"id": "旦 $a + b > 0$"}, {"id": "且 $ab < 0$ , $a + b > 0$"}, {"id": "绝对值恒大于等于0"}, {"id": "| b | = 5"}, {"id": "b = - 5"}, {"id": "a - b"}, {"id": "13"}], "links": [{"rel": "被描述", "source": "| a | = 8", "target": "a = 8"}, {"rel": "代入", "source": "a = 8", "target": "13"}, {"rel": "限制性描述", "source": "旦 $a + b > 0$", "target": "a = 8"}, {"rel": "限制性描述", "source": "旦 $a + b > 0$", "target": "b = - 5"}, {"rel": "限制性描述", "source": "且 $ab < 0$ , $a + b > 0$", "target": "a = 8"}, {"rel": "限制性描述", "source": "且 $ab < 0$ , $a + b > 0$", "target": "b = - 5"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a = 8"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b = - 5"}, {"rel": "被描述", "source": "| b | = 5", "target": "b = - 5"}, {"rel": "代入", "source": "b = - 5", "target": "13"}, {"rel": "被代入", "source": "a - b", "target": "13"}]}}
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{"content": "If $y + 1$ is directly proportional to $x - 2$, and $x = - 1$ when $y = 5$, then the functional relationship between $y$ and $x$ is ____?", "answer": "y = - 2 x + 3", "steps": "Because $y + 1$ is directly proportional to $x - 2$, we can assume that $y + 1 = k ( x - 2 )$, where $k \\neq 0$. Since $x = - 1$ when $y = 5$, we have $6 = - 3 k$, which gives us $k = - 2$. Therefore, $y = - 2 x + 3$.", "expr_cands": ["y + 1", "y", "x - 2", "x", "x = - 1", "y = 5", "y + 1 = k ( x - 2 ) ( k \\neq 0 )", "k", "6 = - 3 k", "k = - 2", "y = - 2 x + 3"], "exprs": ["y + 1 = k ( x - 2 ) ( k \\neq 0 )", "6 = - 3 k", "k = - 2", "y = - 2 x + 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $y + 1 = k ( x - 2 ) ( k \\neq 0 )$"}, {"id": "y + 1 = k ( x - 2 ) ( k \\neq 0 )"}, {"id": "x = - 1"}, {"id": "6 = - 3 k"}, {"id": "y = 5"}, {"id": "k = - 2"}, {"id": "y = - 2 x + 3"}], "links": [{"rel": "假设描述", "source": "设 $y + 1 = k ( x - 2 ) ( k \\neq 0 )$", "target": "y + 1 = k ( x - 2 ) ( k \\neq 0 )"}, {"rel": "被代入", "source": "y + 1 = k ( x - 2 ) ( k \\neq 0 )", "target": "6 = - 3 k"}, {"rel": "联立", "source": "y + 1 = k ( x - 2 ) ( k \\neq 0 )", "target": "y = - 2 x + 3"}, {"rel": "代入", "source": "x = - 1", "target": "6 = - 3 k"}, {"rel": "等式方程求解", "source": "6 = - 3 k", "target": "k = - 2"}, {"rel": "代入", "source": "y = 5", "target": "6 = - 3 k"}, {"rel": "联立", "source": "k = - 2", "target": "y = - 2 x + 3"}]}}
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{"content": "If the equation $( a - 3 ) x = 5$ has a solution with respect to $x$, then the possible values of $a$ are _____.", "answer": "a \\neq 3", "steps": "$\\because$ The equation $( a - 3 ) x = 5$ has a solution for $x$, $\\therefore$ $a - 3 \\neq 0$. Solving for $a$, we get $a \\neq 3$.", "expr_cands": ["x", "( a - 3 ) x = 5", "a", "a - 3 \\neq 0", "a \\neq 3"], "exprs": ["a - 3 \\neq 0", "a \\neq 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 3 ) x = 5"}, {"id": "a - 3 \\neq 0"}, {"id": "关于 $x$ 的方程 $( a - 3 ) x = 5$ 有解"}, {"id": "a \\neq 3"}], "links": [{"rel": "被描述", "source": "( a - 3 ) x = 5", "target": "a - 3 \\neq 0"}, {"rel": "不等式方程求解", "source": "a - 3 \\neq 0", "target": "a \\neq 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $( a - 3 ) x = 5$ 有解", "target": "a - 3 \\neq 0"}]}}
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{"content": "Given $y = 5 + \\sqrt { 3 - x } - \\sqrt { x - 3 }$, what is $y ^ { x }$?", "answer": "125", "steps": "$\\because y = 5 + \\sqrt { 3 - x } - \\sqrt { x - 3 }$ , $\\therefore$ by the property of non-negative numbers, we have $x = 3$ , $\\therefore$ $y = 5$ , so $y ^ { x } = 5 ^ { 3 } = 125$.", "expr_cands": ["y = 5 + \\sqrt { 3 - x } - \\sqrt { x - 3 }", "x", "y", "y ^ { x }", "x = 3", "y = 5", "125"], "exprs": ["x = 3", "y = 5", "125"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 5 + \\sqrt { 3 - x } - \\sqrt { x - 3 }"}, {"id": "x = 3"}, {"id": "非负数的性质得 $x = 3$"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "y = 5"}, {"id": "y ^ { x }"}, {"id": "125"}], "links": [{"rel": "被描述", "source": "y = 5 + \\sqrt { 3 - x } - \\sqrt { x - 3 }", "target": "x = 3"}, {"rel": "被代入", "source": "y = 5 + \\sqrt { 3 - x } - \\sqrt { x - 3 }", "target": "y = 5"}, {"rel": "代入", "source": "x = 3", "target": "y = 5"}, {"rel": "代入", "source": "x = 3", "target": "125"}, {"rel": "限制性描述", "source": "非负数的性质得 $x = 3$", "target": "x = 3"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x = 3"}, {"rel": "代入", "source": "y = 5", "target": "125"}, {"rel": "被代入", "source": "y ^ { x }", "target": "125"}]}}
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{"content": "If $x = 2$ is a solution of the equation $ax - 3 = 5$, then $a$ = ____ ?", "answer": "4", "steps": "From the given information, we can derive that $2 a - 3 = 5$. Solving for $a$, we get $a = 4$.", "expr_cands": ["x = 2", "x", "ax - 3 = 5", "a", "2 a - 3 = 5", "a = 4"], "exprs": ["2 a - 3 = 5", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 2"}, {"id": "2 a - 3 = 5"}, {"id": "ax - 3 = 5"}, {"id": "a = 4"}], "links": [{"rel": "代入", "source": "x = 2", "target": "2 a - 3 = 5"}, {"rel": "等式方程求解", "source": "2 a - 3 = 5", "target": "a = 4"}, {"rel": "被代入", "source": "ax - 3 = 5", "target": "2 a - 3 = 5"}]}}
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{"content": "If three times $x$ plus five equals nine times $x$ minus seven, what is the value of $x$?", "answer": "2", "steps": "According to the problem, we have $3 x + 5 = 9 x - 7$. Moving terms and combining like terms, we get $6 x = 12$. Solving for $x$, we get $x = 2$.", "expr_cands": ["x", "3", "5", "9", "7", "3 x + 5 = 9 x - 7", "x = 2", "6 x = 12"], "exprs": ["3 x + 5 = 9 x - 7", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "3 x + 5 = 9 x - 7"}, {"id": "x"}, {"id": "5"}, {"id": "9"}, {"id": "7"}, {"id": "$x$ 的 $3$ 倍与 $5$ 的和等于 $x$ 的 $9$ 倍与 $7$ 的差"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "3", "target": "3 x + 5 = 9 x - 7"}, {"rel": "等式方程求解", "source": "3 x + 5 = 9 x - 7", "target": "x = 2"}, {"rel": "被描述", "source": "x", "target": "3 x + 5 = 9 x - 7"}, {"rel": "被描述", "source": "5", "target": "3 x + 5 = 9 x - 7"}, {"rel": "被描述", "source": "9", "target": "3 x + 5 = 9 x - 7"}, {"rel": "被描述", "source": "7", "target": "3 x + 5 = 9 x - 7"}, {"rel": "限制性描述", "source": "$x$ 的 $3$ 倍与 $5$ 的和等于 $x$ 的 $9$ 倍与 $7$ 的差", "target": "3 x + 5 = 9 x - 7"}]}}
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{"content": "If the solution to the equation $2 x + 1 = 3 k$ in terms of $x$ is negative, then the range of values for $k$ is ____?", "answer": "k < \\frac { 1 } { 3 }", "steps": "Solve $2 x + 1 = 3 k$, we get $x = \\frac { 3 k - 1 } { 2 }$. According to the problem, we have $\\frac { 3 k - 1 } { 2 } < 0$, which leads to $k < \\frac { 1 } { 3 }$.", "expr_cands": ["x", "2 x + 1 = 3 k", "k", "x = \\frac { 3 k - 1 } { 2 }", "\\frac { 3 k - 1 } { 2 } < 0", "k < \\frac { 1 } { 3 }"], "exprs": ["x = \\frac { 3 k - 1 } { 2 }", "\\frac { 3 k - 1 } { 2 } < 0", "k < \\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 1 = 3 k"}, {"id": "x = \\frac { 3 k - 1 } { 2 }"}, {"id": "\\frac { 3 k - 1 } { 2 } < 0"}, {"id": "关于 $x$ 的方程 $2 x + 1 = 3 k$ 的解是负数"}, {"id": "k < \\frac { 1 } { 3 }"}], "links": [{"rel": "等式方程部分求解", "source": "2 x + 1 = 3 k", "target": "x = \\frac { 3 k - 1 } { 2 }"}, {"rel": "被描述", "source": "x = \\frac { 3 k - 1 } { 2 }", "target": "\\frac { 3 k - 1 } { 2 } < 0"}, {"rel": "不等式方程求解", "source": "\\frac { 3 k - 1 } { 2 } < 0", "target": "k < \\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $2 x + 1 = 3 k$ 的解是负数", "target": "\\frac { 3 k - 1 } { 2 } < 0"}]}}
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{"content": "The line $y = x - 1$is translated upward by $3$ units to obtain the line $l$. The function expression corresponding to the line $l$ is ____?", "answer": "y = x + 2", "steps": "The line $y = x - 1$ is translated upwards by $3$ units to obtain the line $l$. Therefore, the function expression for line $l$ is $y = x - 1 + 3$, which simplifies to $y = x + 2$.", "expr_cands": ["y = x - 1", "y", "x", "3", "l", "y = x - 1 + 3", "x - 1 = x - 1 + 3", "x + 2"], "exprs": ["x + 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x - 1"}, {"id": "x + 2"}, {"id": "3"}, {"id": "l"}, {"id": "将直线 $y = x - 1$ 向上平移 $3$ 个单位得到直线 $l$"}, {"id": "直线 $l$ 对应的函数表达式"}], "links": [{"rel": "被描述", "source": "y = x - 1", "target": "x + 2"}, {"rel": "被描述", "source": "3", "target": "x + 2"}, {"rel": "被描述", "source": "l", "target": "x + 2"}, {"rel": "限制性描述", "source": "将直线 $y = x - 1$ 向上平移 $3$ 个单位得到直线 $l$", "target": "x + 2"}, {"rel": "限制性描述", "source": "直线 $l$ 对应的函数表达式", "target": "x + 2"}]}}
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{"content": "When the rational number $a < 0$, what is the value of $- a - | a |$?", "answer": "0", "steps": "Since $a < 0$, it follows that $- a - | a | = - a - ( - a ) = - a + a = 0$.", "expr_cands": ["a < 0", "a", "- a - | a |", "- a - ( - a )", "0"], "exprs": ["- a - ( - a )", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a < 0"}, {"id": "- a - ( - a )"}, {"id": "- a - | a |"}, {"id": "绝对值恒大于等于0"}, {"id": "当有理数 $a < 0$ 时"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "a < 0", "target": "- a - ( - a )"}, {"rel": "计算", "source": "- a - ( - a )", "target": "0"}, {"rel": "被描述", "source": "- a - | a |", "target": "- a - ( - a )"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "- a - ( - a )"}, {"rel": "限制性描述", "source": "当有理数 $a < 0$ 时", "target": "- a - ( - a )"}]}}
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{"content": "The positive integer solution of the inequality $3 x - 4 < x$ is ____?", "answer": "1", "steps": "$3 x - 4 < x$, $3 x - x < 4$, $2 x < 4$, $x < 2$, so the positive integer solution to the inequality $3 x - 4 < x$ is $1$.", "expr_cands": ["3 x - 4 < x", "x", "x < 2", "3 x - x < 4", "2 x < 4", "1"], "exprs": ["x < 2", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x - 4 < x"}, {"id": "x < 2"}, {"id": "1"}, {"id": "不等式 $3 x - 4 < x$ 的正整数解"}], "links": [{"rel": "不等式方程求解", "source": "3 x - 4 < x", "target": "x < 2"}, {"rel": "被描述", "source": "x < 2", "target": "1"}, {"rel": "限制性描述", "source": "不等式 $3 x - 4 < x$ 的正整数解", "target": "1"}]}}
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{"content": "The value of $a$ for which one root of the quadratic equation $( a - 3 ) x ^ 2 - 2 x + a ^ 2 - 9 = 0$ is $0$ is _____.", "answer": "- 3", "steps": "According to the problem, we have $a ^ 2 - 9 = 0$ and $a - 3 \\neq 0$. Solving for $a$, we get $a = - 3$.", "expr_cands": ["( a - 3 ) x ^ { 2 } - 2 x + a ^ { 2 } - 9 = 0", "a", "x", "0", "a ^ { 2 } - 9 = 0", "a = - 3", "a = 3", "a - 3 \\neq 0", "a \\neq 3"], "exprs": ["a ^ { 2 } - 9 = 0", "a - 3 \\neq 0", "a = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 3 ) x ^ { 2 } - 2 x + a ^ { 2 } - 9 = 0"}, {"id": "a ^ { 2 } - 9 = 0"}, {"id": "x"}, {"id": "0"}, {"id": "一元二次方程 $( a - 3 ) x ^ { 2 } - 2 x + a ^ { 2 } - 9 = 0$ 的一个根是 $0$"}, {"id": "a - 3 \\neq 0"}, {"id": "a = - 3"}], "links": [{"rel": "被描述", "source": "( a - 3 ) x ^ { 2 } - 2 x + a ^ { 2 } - 9 = 0", "target": "a ^ { 2 } - 9 = 0"}, {"rel": "被描述", "source": "( a - 3 ) x ^ { 2 } - 2 x + a ^ { 2 } - 9 = 0", "target": "a - 3 \\neq 0"}, {"rel": "联立", "source": "a ^ { 2 } - 9 = 0", "target": "a = - 3"}, {"rel": "被描述", "source": "x", "target": "a ^ { 2 } - 9 = 0"}, {"rel": "被描述", "source": "0", "target": "a ^ { 2 } - 9 = 0"}, {"rel": "限制性描述", "source": "一元二次方程 $( a - 3 ) x ^ { 2 } - 2 x + a ^ { 2 } - 9 = 0$ 的一个根是 $0$", "target": "a ^ { 2 } - 9 = 0"}, {"rel": "限制性描述", "source": "一元二次方程 $( a - 3 ) x ^ { 2 } - 2 x + a ^ { 2 } - 9 = 0$ 的一个根是 $0$", "target": "a - 3 \\neq 0"}, {"rel": "联立", "source": "a - 3 \\neq 0", "target": "a = - 3"}]}}
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{"content": "If $| x - 3 | + ( y + 2 ) ^ { 2 } = 0$, then the value of $xy$ is ____?", "answer": "- 6", "steps": "From the given information, we have $x - 3 = 0$ and $y + 2 = 0$, which implies $x = 3$ and $y = - 2$. Therefore, $xy = 3 * ( - 2 ) = - 6$.", "expr_cands": ["| x - 3 | + ( y + 2 ) ^ { 2 } = 0", "y", "x", "xy", "x - 3 = 0", "x = 3", "y + 2 = 0", "y = - 2", "- 6"], "exprs": ["x - 3 = 0", "y + 2 = 0", "x = 3", "y = - 2", "- 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x - 3 | + ( y + 2 ) ^ { 2 } = 0"}, {"id": "x - 3 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "y + 2 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "x = 3"}, {"id": "y = - 2"}, {"id": "xy"}, {"id": "- 6"}], "links": [{"rel": "被描述", "source": "| x - 3 | + ( y + 2 ) ^ { 2 } = 0", "target": "x - 3 = 0"}, {"rel": "被描述", "source": "| x - 3 | + ( y + 2 ) ^ { 2 } = 0", "target": "y + 2 = 0"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x - 3 = 0"}, {"rel": "等式方程求解", "source": "y + 2 = 0", "target": "y = - 2"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "y + 2 = 0"}, {"rel": "代入", "source": "x = 3", "target": "- 6"}, {"rel": "代入", "source": "y = - 2", "target": "- 6"}, {"rel": "被代入", "source": "xy", "target": "- 6"}]}}
|
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{"content": "The value of the fraction $\\frac { 3 x + 9 } { x - 2 }$ is zero, then $x$ = ____ ?", "answer": "- 3", "steps": "$\\because$ The value of the fraction $\\frac { 3 x + 9 } { x - 2 }$ is zero, $\\therefore$ $3 x + 9 = 0$, solving for $x$ gives $x = - 3$.", "expr_cands": ["\\frac { 3 x + 9 } { x - 2 }", "x", "3 x + 9 = 0", "x = - 3"], "exprs": ["3 x + 9 = 0", "x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 x + 9 } { x - 2 }"}, {"id": "3 x + 9 = 0"}, {"id": "分式 $\\frac { 3 x + 9 } { x - 2 }$ 的值为零"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "x = - 3"}], "links": [{"rel": "被描述", "source": "\\frac { 3 x + 9 } { x - 2 }", "target": "3 x + 9 = 0"}, {"rel": "等式方程求解", "source": "3 x + 9 = 0", "target": "x = - 3"}, {"rel": "限制性描述", "source": "分式 $\\frac { 3 x + 9 } { x - 2 }$ 的值为零", "target": "3 x + 9 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "3 x + 9 = 0"}]}}
|
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{"content": "If $a - 2 b = 7$, then the value of $6 - 2 a + 4 b$ is ____?", "answer": "- 8", "steps": "Since $a - 2 b = 7$, therefore $6 - 2 a + 4 b = 6 - 2 ( a - 2 b ) = 6 - 2 { * } 7 = - 8$.", "expr_cands": ["a - 2 b = 7", "b", "a", "6 - 2 a + 4 b", "6 - 2 ( a - 2 b )", "- 8"], "exprs": ["6 - 2 ( a - 2 b )", "- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "6 - 2 a + 4 b"}, {"id": "6 - 2 ( a - 2 b )"}, {"id": "a - 2 b = 7"}, {"id": "- 8"}], "links": [{"rel": "提取因式", "source": "6 - 2 a + 4 b", "target": "6 - 2 ( a - 2 b )"}, {"rel": "被代入", "source": "6 - 2 ( a - 2 b )", "target": "- 8"}, {"rel": "提取因式参考", "source": "a - 2 b = 7", "target": "6 - 2 ( a - 2 b )"}, {"rel": "代入", "source": "a - 2 b = 7", "target": "- 8"}]}}
|
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{"content": "If $4 : 3 = 5 : x$, then the value of $x$ is ____?", "answer": "\\frac { 15 } { 4 }", "steps": "By the property of proportion, we have $4 x = 3 * 5$, solving for $x$ gives $x = \\frac { 15 } { 4 }$.", "expr_cands": ["4 : 3 = 5 : x", "x", "4 x = 3 * 5", "x = \\frac { 15 } { 4 }"], "exprs": ["x = \\frac { 15 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 : 3 = 5 : x"}, {"id": "x = \\frac { 15 } { 4 }"}], "links": [{"rel": "等式方程求解", "source": "4 : 3 = 5 : x", "target": "x = \\frac { 15 } { 4 }"}]}}
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{"content": "If $m$ is a solution to the equation $ax ^ 2 + bx + 5 = 0$ with respect to $x$, then $am ^ 2 + bm - 7$ = ____ ?", "answer": "- 12", "steps": "Substituting $x = m$ into $ax ^ { 2 } + bx + 5 = 0$, we get $am ^ { 2 } + bm + 5 = 0$. Therefore, $am ^ { 2 } + bm = - 5$, so $am ^ { 2 } + bm - 7 = - 5 - 7 = - 12$.", "expr_cands": ["m", "x", "ax ^ { 2 } + bx + 5 = 0", "a", "b", "am ^ { 2 } + bm - 7", "x = m", "a m ^ { 2 } + b m + 5 = 0", "am ^ { 2 } + bm + 5 = 0", "am ^ { 2 } + bm = - 5", "- 12"], "exprs": ["x = m", "am ^ { 2 } + bm = - 5", "am ^ { 2 } + bm + 5 = 0", "- 12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m"}, {"id": "x = m"}, {"id": "x"}, {"id": "ax ^ { 2 } + bx + 5 = 0"}, {"id": "$m$ 是关于 $x$ 的方程 $ax ^ { 2 } + bx + 5 = 0$ 的一个解"}, {"id": "am ^ { 2 } + bm + 5 = 0"}, {"id": "am ^ { 2 } + bm = - 5"}, {"id": "am ^ { 2 } + bm - 7"}, {"id": "- 12"}], "links": [{"rel": "被描述", "source": "m", "target": "x = m"}, {"rel": "代入", "source": "x = m", "target": "am ^ { 2 } + bm + 5 = 0"}, {"rel": "被描述", "source": "x", "target": "x = m"}, {"rel": "被描述", "source": "ax ^ { 2 } + bx + 5 = 0", "target": "x = m"}, {"rel": "被代入", "source": "ax ^ { 2 } + bx + 5 = 0", "target": "am ^ { 2 } + bm + 5 = 0"}, {"rel": "移项", "source": "ax ^ { 2 } + bx + 5 = 0", "target": "am ^ { 2 } + bm = - 5"}, {"rel": "限制性描述", "source": "$m$ 是关于 $x$ 的方程 $ax ^ { 2 } + bx + 5 = 0$ 的一个解", "target": "x = m"}, {"rel": "代入", "source": "am ^ { 2 } + bm = - 5", "target": "- 12"}, {"rel": "被代入", "source": "am ^ { 2 } + bm - 7", "target": "- 12"}]}}
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||||
{"content": "Given a quadratic equation in one variable $x$, $x ^ 2 - 4 x - k = 0$, with one root being $3$, what is the other root?", "answer": "1", "steps": "If the equation has another root of $a$, then according to the relationship between roots and coefficients, we have $a + 3 = 4$, and solving for $a$ gives $a = 1$.", "expr_cands": ["x", "x ^ { 2 } - 4 x - k = 0", "k", "3", "a", "a + 3 = 4", "a = 1"], "exprs": ["a", "a + 3 = 4", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程的另一个根为 $a$"}, {"id": "a"}, {"id": "3"}, {"id": "a + 3 = 4"}, {"id": "x"}, {"id": "x ^ { 2 } - 4 x - k = 0"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 4 x - k = 0$ 的一个根为 $3$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "a = 1"}], "links": [{"rel": "假设描述", "source": "设方程的另一个根为 $a$", "target": "a"}, {"rel": "限制性描述", "source": "设方程的另一个根为 $a$", "target": "a + 3 = 4"}, {"rel": "被描述", "source": "a", "target": "a + 3 = 4"}, {"rel": "被描述", "source": "3", "target": "a + 3 = 4"}, {"rel": "等式方程求解", "source": "a + 3 = 4", "target": "a = 1"}, {"rel": "被描述", "source": "x", "target": "a + 3 = 4"}, {"rel": "被描述", "source": "x ^ { 2 } - 4 x - k = 0", "target": "a + 3 = 4"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 4 x - k = 0$ 的一个根为 $3$", "target": "a + 3 = 4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "a + 3 = 4"}]}}
|
||||
{"content": "$13$, What is the positive integer solution of the inequality $- 3 x + 6 > 0$?", "answer": "1", "steps": "The solution set of $- 3 x + 6 > 0$ is $x < 2$; within the solution set of $x < 2$, the positive integer is $1$.", "expr_cands": ["13", "- 3 x + 6 > 0", "x", "x < 2", "1"], "exprs": ["x < 2", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 3 x + 6 > 0"}, {"id": "x < 2"}, {"id": "1"}, {"id": "不等式 $- 3 x + 6 > 0$ 的正整数解"}], "links": [{"rel": "不等式方程求解", "source": "- 3 x + 6 > 0", "target": "x < 2"}, {"rel": "被描述", "source": "x < 2", "target": "1"}, {"rel": "限制性描述", "source": "不等式 $- 3 x + 6 > 0$ 的正整数解", "target": "1"}]}}
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{"content": "In the linear equation $x - 2 y = 3$, when $x = 1$, $y$ = ____ ?", "answer": "- 1", "steps": "Substituting $x = 1$ into the equation yields $1 - 2 y = 3$, which can be solved to obtain $y = - 1$.", "expr_cands": ["x - 2 y = 3", "y", "x", "x = 1", "1 - 2 y = 3", "y = - 1"], "exprs": ["1 - 2 y = 3", "y = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - 2 y = 3"}, {"id": "1 - 2 y = 3"}, {"id": "x = 1"}, {"id": "y = - 1"}], "links": [{"rel": "被代入", "source": "x - 2 y = 3", "target": "1 - 2 y = 3"}, {"rel": "等式方程求解", "source": "1 - 2 y = 3", "target": "y = - 1"}, {"rel": "代入", "source": "x = 1", "target": "1 - 2 y = 3"}]}}
|
||||
{"content": "Given that $x = 6$ is a solution to the equation $3 x + m = - 1$ in terms of $x$, the value of $m$ is:", "answer": "- 19", "steps": "$\\because x = 6$ is a solution to the equation $3 x + m = - 1$ in terms of $x$, $\\therefore$ substituting $x = 6$ yields $18 + m = - 1$, which can be solved to obtain $m = - 19$.", "expr_cands": ["x = 6", "x", "3 x + m = - 1", "m", "m + 18 = - 1", "18 + m = - 1", "m = - 19"], "exprs": ["18 + m = - 1", "m = - 19"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + m = - 1"}, {"id": "18 + m = - 1"}, {"id": "x = 6"}, {"id": "m = - 19"}], "links": [{"rel": "被代入", "source": "3 x + m = - 1", "target": "18 + m = - 1"}, {"rel": "等式方程求解", "source": "18 + m = - 1", "target": "m = - 19"}, {"rel": "代入", "source": "x = 6", "target": "18 + m = - 1"}]}}
|
||||
{"content": "If $x = 0$ is a root of the quadratic equation $( m - 1 ) x ^ 2 - mx + m ^ 2 - 1 = 0$, what is the value of $m$?", "answer": "- 1", "steps": "Substituting $x = 0$ into the equation $( m - 1 ) x ^ 2 - mx + m ^ 2 - 1 = 0$, we get $m ^ 2 - 1 = 0$, and since $m - 1 \\neq 0$, we can solve for $m$ to get $m = - 1$.", "expr_cands": ["x = 0", "x", "( m - 1 ) x ^ { 2 } - mx + m ^ { 2 } - 1 = 0", "m", "m ^ { 2 } - 1 = 0", "m = - 1", "m = 1", "m - 1 \\neq 0", "m \\neq 1"], "exprs": ["m ^ { 2 } - 1 = 0", "m - 1 \\neq 0", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 1 ) x ^ { 2 } - mx + m ^ { 2 } - 1 = 0"}, {"id": "m ^ { 2 } - 1 = 0"}, {"id": "x = 0"}, {"id": "m - 1 \\neq 0"}, {"id": "$x = 0$ 是二次方程 $( m - 1 ) x ^ { 2 } - mx + m ^ { 2 } - 1 = 0$ 的一个根"}, {"id": "m = - 1"}], "links": [{"rel": "被代入", "source": "( m - 1 ) x ^ { 2 } - mx + m ^ { 2 } - 1 = 0", "target": "m ^ { 2 } - 1 = 0"}, {"rel": "被描述", "source": "( m - 1 ) x ^ { 2 } - mx + m ^ { 2 } - 1 = 0", "target": "m - 1 \\neq 0"}, {"rel": "联立", "source": "m ^ { 2 } - 1 = 0", "target": "m = - 1"}, {"rel": "代入", "source": "x = 0", "target": "m ^ { 2 } - 1 = 0"}, {"rel": "联立", "source": "m - 1 \\neq 0", "target": "m = - 1"}, {"rel": "限制性描述", "source": "$x = 0$ 是二次方程 $( m - 1 ) x ^ { 2 } - mx + m ^ { 2 } - 1 = 0$ 的一个根", "target": "m - 1 \\neq 0"}]}}
|
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{"content": "When $x$ = ____ ?, the fraction $\\frac { x } { x + 2 }$ is undefined.", "answer": "- 2", "steps": "From the given information, we have $x + 2 = 0$, which implies that $x = - 2$.", "expr_cands": ["x", "\\frac { x } { x + 2 }", "x + 2 = 0", "x = - 2"], "exprs": ["x + 2 = 0", "x = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x + 2 }"}, {"id": "x + 2 = 0"}, {"id": "分式 $\\frac { x } { x + 2 }$ 没有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x = - 2"}], "links": [{"rel": "被描述", "source": "\\frac { x } { x + 2 }", "target": "x + 2 = 0"}, {"rel": "等式方程求解", "source": "x + 2 = 0", "target": "x = - 2"}, {"rel": "限制性描述", "source": "分式 $\\frac { x } { x + 2 }$ 没有意义", "target": "x + 2 = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x + 2 = 0"}]}}
|
||||
{"content": "When $x = - 2$, what is the value of the algebraic expression $( x - 2 ) ^ 2$?", "answer": "16", "steps": "When $x = - 2$, the value of the algebraic expression $( x - 2 ) ^ 2$ is: $( - 2 - 2 ) ^ 2 = 16$.", "expr_cands": ["x = - 2", "x", "( x - 2 ) ^ { 2 }", "( - 2 - 2 ) ^ { 2 }", "16"], "exprs": ["( - 2 - 2 ) ^ { 2 }", "16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = - 2"}, {"id": "( - 2 - 2 ) ^ { 2 }"}, {"id": "( x - 2 ) ^ { 2 }"}, {"id": "16"}], "links": [{"rel": "代入", "source": "x = - 2", "target": "( - 2 - 2 ) ^ { 2 }"}, {"rel": "计算", "source": "( - 2 - 2 ) ^ { 2 }", "target": "16"}, {"rel": "被代入", "source": "( x - 2 ) ^ { 2 }", "target": "( - 2 - 2 ) ^ { 2 }"}]}}
|
||||
{"content": "The algebraic expression $2 x + 3$ is the opposite of $5$, then $x$ equals", "answer": "- 4", "steps": "$\\because$ The algebraic expression $2 x + 3$ is the opposite of $5$, $\\therefore$ $2 x + 3 = - 5$. Solving for $x$, we get $x = - 4$.", "expr_cands": ["2 x + 3", "x", "5", "2 x + 3 = - 5", "x = - 4"], "exprs": ["2 x + 3 = - 5", "x = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5"}, {"id": "2 x + 3 = - 5"}, {"id": "2 x + 3"}, {"id": "代数式 $2 x + 3$ 与 $5$ 互为相反数"}, {"id": "x = - 4"}], "links": [{"rel": "被描述", "source": "5", "target": "2 x + 3 = - 5"}, {"rel": "等式方程求解", "source": "2 x + 3 = - 5", "target": "x = - 4"}, {"rel": "被描述", "source": "2 x + 3", "target": "2 x + 3 = - 5"}, {"rel": "限制性描述", "source": "代数式 $2 x + 3$ 与 $5$ 互为相反数", "target": "2 x + 3 = - 5"}]}}
|
||||
{"content": "What is the smallest integer solution of the inequality $2 x - 5 > 0$?", "answer": "3", "steps": "The solution set of the inequality is $x > 2.5$, so the smallest integer solution of the inequality $2 x - 5 > 0$ is $3$.", "expr_cands": ["2 x - 5 > 0", "x", "x > 2.5", "\\frac { 5 } { 2 } < x", "3"], "exprs": ["x > 2.5", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 5 > 0"}, {"id": "x > 2.5"}, {"id": "3"}, {"id": "不等式 $2 x - 5 > 0$ 的最小整数解"}, {"id": "不等式的解集是 $x > 2.5$"}], "links": [{"rel": "不等式方程求解", "source": "2 x - 5 > 0", "target": "x > 2.5"}, {"rel": "被描述", "source": "x > 2.5", "target": "3"}, {"rel": "限制性描述", "source": "不等式 $2 x - 5 > 0$ 的最小整数解", "target": "3"}, {"rel": "限制性描述", "source": "不等式的解集是 $x > 2.5$", "target": "3"}]}}
|
||||
{"content": "When $x$ = ____ ?, the value of the algebraic expression $2 x - 8$ is equal to the value of $1 + 3 x$.", "answer": "- 9", "steps": "According to the problem, we have $2 x - 8 = 1 + 3 x$. By rearranging and combining terms, we get $- x = 9$, which means $x = - 9$.", "expr_cands": ["x", "2 x - 8", "1 + 3 x", "2 x - 8 = 1 + 3 x", "x = - 9", "- x = 9"], "exprs": ["2 x - 8 = 1 + 3 x", "x = - 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 8"}, {"id": "2 x - 8 = 1 + 3 x"}, {"id": "1 + 3 x"}, {"id": "代数式 $2 x - 8$ 的值与 $1 + 3 x$ 的值相"}, {"id": "x = - 9"}], "links": [{"rel": "被描述", "source": "2 x - 8", "target": "2 x - 8 = 1 + 3 x"}, {"rel": "等式方程求解", "source": "2 x - 8 = 1 + 3 x", "target": "x = - 9"}, {"rel": "被描述", "source": "1 + 3 x", "target": "2 x - 8 = 1 + 3 x"}, {"rel": "限制性描述", "source": "代数式 $2 x - 8$ 的值与 $1 + 3 x$ 的值相", "target": "2 x - 8 = 1 + 3 x"}]}}
|
||||
{"content": "The constant term of the quadratic equation $3 x ^ 2 - 2 x - 8 = 0$ with respect to $x$ is _____.", "answer": "- 8", "steps": "The constant term of the quadratic equation in one variable $x$, $3 x ^ 2 - 2 x - 8 = 0$, is $- 8$.", "expr_cands": ["x", "3 x ^ { 2 } - 2 x - 8 = 0", "x = - \\frac { 4 } { 3 }", "x = 2", "- 8"], "exprs": ["- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { 2 } - 2 x - 8 = 0"}, {"id": "- 8"}, {"id": "关于 $x$ 的一元二次方程 $3 x ^ { 2 } - 2 x - 8 = 0$ 的常数项"}], "links": [{"rel": "被描述", "source": "3 x ^ { 2 } - 2 x - 8 = 0", "target": "- 8"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $3 x ^ { 2 } - 2 x - 8 = 0$ 的常数项", "target": "- 8"}]}}
|
||||
{"content": "In $y = \\frac { x } { x - 3 }$, the range of values for $x$ is ____?", "answer": "x \\neq 3", "steps": "According to the problem, we have $x - 3 \\neq 0$, which implies $x \\neq 3$.", "expr_cands": ["y = \\frac { x } { x - 3 }", "x", "y", "x - 3 \\neq 0", "x \\neq 3"], "exprs": ["x - 3 \\neq 0", "x \\neq 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { x } { x - 3 }"}, {"id": "x - 3 \\neq 0"}, {"id": "在 $y = \\frac { x } { x - 3 }$ 中"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 3"}], "links": [{"rel": "被描述", "source": "y = \\frac { x } { x - 3 }", "target": "x - 3 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 3 \\neq 0", "target": "x \\neq 3"}, {"rel": "限制性描述", "source": "在 $y = \\frac { x } { x - 3 }$ 中", "target": "x - 3 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 3 \\neq 0"}]}}
|
||||
{"content": "If the simplest quadratic radical $\\sqrt { a ^ { 2 } + 3 a }$ and $\\sqrt { a + 15 }$ are of the same type, then $a$ = ____?", "answer": "- 5", "steps": "From the given condition, we have $a ^ 2 + 3 a = a + 15$. Rearranging, we get $a ^ 2 + 2 a - 15 = 0$. Solving this quadratic equation, we get $a = 3$ or $a = - 5$. When $a = 3$, the square roots $\\sqrt { a ^ 2 + 3 a }$ and $\\sqrt { a + 15 }$ are not in their simplest form. Therefore, we have $a = - 5$.", "expr_cands": ["\\sqrt { a ^ { 2 } + 3 a }", "a", "\\sqrt { a + 15 }", "a ^ { 2 } + 3 a = a + 15", "a = - 5", "a = 3", "a ^ { 2 } + 2 a - 15 = 0"], "exprs": ["a ^ { 2 } + 3 a = a + 15", "a ^ { 2 } + 2 a - 15 = 0", "a = 3", "a = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { a ^ { 2 } + 3 a }"}, {"id": "a ^ { 2 } + 3 a = a + 15"}, {"id": "\\sqrt { a + 15 }"}, {"id": "最简二次根式 $\\sqrt { a ^ { 2 } + 3 a }$ 与 $\\sqrt { a + 15 }$ 是同类二次根式"}, {"id": "a ^ { 2 } + 2 a - 15 = 0"}, {"id": "a = 3"}, {"id": "当 $a = 3$ 时"}, {"id": "二次根式 $\\sqrt { a ^ { 2 } + 3 a }$ 与 $\\sqrt { a + 15 }$ 不是最简二次根式"}, {"id": "a = - 5"}], "links": [{"rel": "被描述", "source": "\\sqrt { a ^ { 2 } + 3 a }", "target": "a ^ { 2 } + 3 a = a + 15"}, {"rel": "移项", "source": "a ^ { 2 } + 3 a = a + 15", "target": "a ^ { 2 } + 2 a - 15 = 0"}, {"rel": "被描述", "source": "a ^ { 2 } + 3 a = a + 15", "target": "a = 3"}, {"rel": "等式方程求解", "source": "a ^ { 2 } + 3 a = a + 15", "target": "a = - 5"}, {"rel": "被描述", "source": "\\sqrt { a + 15 }", "target": "a ^ { 2 } + 3 a = a + 15"}, {"rel": "限制性描述", "source": "最简二次根式 $\\sqrt { a ^ { 2 } + 3 a }$ 与 $\\sqrt { a + 15 }$ 是同类二次根式", "target": "a ^ { 2 } + 3 a = a + 15"}, {"rel": "限制性描述", "source": "当 $a = 3$ 时", "target": "a = 3"}, {"rel": "限制性描述", "source": "二次根式 $\\sqrt { a ^ { 2 } + 3 a }$ 与 $\\sqrt { a + 15 }$ 不是最简二次根式", "target": "a = 3"}]}}
|
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{"content": "Given that $( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5$ is a one-variable linear equation in $x$, the solution to this equation is ____?", "answer": "x = - \\frac { 1 } { 12 }", "steps": "According to the problem, we have $| m - 1 | = 1$ and $m - 2 \\neq 0$. Solving for $m$, we get $m = 0$. Therefore, the equation becomes $- 2 x + \\frac { 1 } { 3 } = 0.5$, and solving for $x$, we get $x = - \\frac { 1 } { 12 }$.", "expr_cands": ["( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5", "m", "x", "| m - 1 | = 1", "m = 0", "m = 2", "m - 2 \\neq 0", "m \\neq 2", "- 2 x + \\frac { 1 } { 3 } = 0.5", "x = - 0.0833333333333333", "x = - \\frac { 1 } { 12 }"], "exprs": ["| m - 1 | = 1", "m - 2 \\neq 0", "m = 0", "- 2 x + \\frac { 1 } { 3 } = 0.5", "x = - \\frac { 1 } { 12 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5"}, {"id": "| m - 1 | = 1"}, {"id": "$( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5$ 是关于 $x$ 的一元一次方程"}, {"id": "m - 2 \\neq 0"}, {"id": "m = 0"}, {"id": "- 2 x + \\frac { 1 } { 3 } = 0.5"}, {"id": "x = - \\frac { 1 } { 12 }"}], "links": [{"rel": "被描述", "source": "( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5", "target": "| m - 1 | = 1"}, {"rel": "被描述", "source": "( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5", "target": "m - 2 \\neq 0"}, {"rel": "被代入", "source": "( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5", "target": "- 2 x + \\frac { 1 } { 3 } = 0.5"}, {"rel": "联立", "source": "| m - 1 | = 1", "target": "m = 0"}, {"rel": "限制性描述", "source": "$( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5$ 是关于 $x$ 的一元一次方程", "target": "| m - 1 | = 1"}, {"rel": "限制性描述", "source": "$( m - 2 ) { x } ^ { | m - 1 | } + \\frac { 1 } { 3 } = 0.5$ 是关于 $x$ 的一元一次方程", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "m - 2 \\neq 0", "target": "m = 0"}, {"rel": "代入", "source": "m = 0", "target": "- 2 x + \\frac { 1 } { 3 } = 0.5"}, {"rel": "等式方程求解", "source": "- 2 x + \\frac { 1 } { 3 } = 0.5", "target": "x = - \\frac { 1 } { 12 }"}]}}
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{"content": "The equation $x - a = 2$ has a non-negative solution for $x$, what is the range of possible values for $a$?", "answer": "a \\ge - 2", "steps": "$x - a = 2$, $x = a + 2$. Since $x$ is non-negative, therefore $a + 2 \\ge 0$, $a \\ge - 2$.", "expr_cands": ["x", "x - a = 2", "a", "x = a + 2", "a + 2 \\ge 0", "- 2 \\le a", "a \\ge - 2"], "exprs": ["x = a + 2", "a + 2 \\ge 0", "a \\ge - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - a = 2"}, {"id": "x = a + 2"}, {"id": "a + 2 \\ge 0"}, {"id": "关于 $x$ 的方程 $x - a = 2$ 的解是非负数"}, {"id": "a \\ge - 2"}], "links": [{"rel": "等式方程部分求解", "source": "x - a = 2", "target": "x = a + 2"}, {"rel": "被描述", "source": "x = a + 2", "target": "a + 2 \\ge 0"}, {"rel": "不等式方程求解", "source": "a + 2 \\ge 0", "target": "a \\ge - 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x - a = 2$ 的解是非负数", "target": "a + 2 \\ge 0"}]}}
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{"content": "Given $2 ^ { m } * 8 ^ { 2 } = 4 ^ { 4 }$, what is the value of $m$?", "answer": "2", "steps": "$2 ^ { m } * 8 ^ { 2 } = 4 ^ { 4 }$ , $2 ^ { m } * 2 ^ { 6 } = 2 ^ { 8 }$ , $2 ^ { m + 6 } = 2 ^ { 8 }$ , then $m + 6 = 8$ , solving for $m$ gives $m = 2$.", "expr_cands": ["2 ^ { m } * 8 ^ { 2 } = 4 ^ { 4 }", "m", "m = 2", "2 ^ { m } * 2 ^ { 6 }", "256", "2 ^ { m + 6 } = 2 ^ { 8 }", "m + 6 = 8"], "exprs": ["m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ^ { m } * 8 ^ { 2 } = 4 ^ { 4 }"}, {"id": "m = 2"}], "links": [{"rel": "等式方程求解", "source": "2 ^ { m } * 8 ^ { 2 } = 4 ^ { 4 }", "target": "m = 2"}]}}
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{"content": "When ____ ?, $| 3 - x | = x - 3$.", "answer": "x \\ge 3", "steps": "From the given condition, we can obtain $3 - x \\le 0$, which implies $x \\ge 3$.", "expr_cands": ["| 3 - x | = x - 3", "x", "3 - x \\le 0", "3 \\le x", "x \\ge 3"], "exprs": ["3 - x \\le 0", "x \\ge 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| 3 - x | = x - 3"}, {"id": "3 - x \\le 0"}, {"id": "绝对值恒大于等于0"}, {"id": "x \\ge 3"}], "links": [{"rel": "被描述", "source": "| 3 - x | = x - 3", "target": "3 - x \\le 0"}, {"rel": "不等式方程求解", "source": "3 - x \\le 0", "target": "x \\ge 3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "3 - x \\le 0"}]}}
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{"content": "If the polynomial $( mx ^ { 2 } + x - 3 ) ( x ^ { 2 } - 2 x + 1 )$ does not contain the term $x ^ { 2 }$ after the operation, then $m$ = ____ ?", "answer": "5", "steps": "$( mx ^ { 2 } + x - 3 ) ( x ^ { 2 } - 2 x + 1 ) = mx ^ { 4 } - 2 mx ^ { 3 } + mx ^ { 2 } + x ^ { 3 } - 2 x ^ { 2 } + x - 3 x ^ { 2 } + 6 x - 3 = mx ^ { 4 } + ( 1 - 2 m ) x ^ { 3 } + ( m - 2 - 3 ) x ^ { 2 } + 7 x - 3$ \\because There is no $x ^ { 2 }$ in the polynomial, \\therefore $m - 2 - 3 = 0$ , \\therefore $m = 5$.", "expr_cands": ["( mx ^ { 2 } + x - 3 ) ( x ^ { 2 } - 2 x + 1 )", "m", "x", "x ^ { 2 }", "mx ^ { 4 } + ( 1 - 2 m ) x ^ { 3 } + ( m - 2 - 3 ) x ^ { 2 } + 7 x - 3", "m - 2 - 3 = 0", "m = 5"], "exprs": ["mx ^ { 4 } + ( 1 - 2 m ) x ^ { 3 } + ( m - 2 - 3 ) x ^ { 2 } + 7 x - 3", "m - 2 - 3 = 0", "m = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( mx ^ { 2 } + x - 3 ) ( x ^ { 2 } - 2 x + 1 )"}, {"id": "mx ^ { 4 } + ( 1 - 2 m ) x ^ { 3 } + ( m - 2 - 3 ) x ^ { 2 } + 7 x - 3"}, {"id": "x"}, {"id": "m - 2 - 3 = 0"}, {"id": "整式 $( mx ^ { 2 } + x - 3 ) ( x ^ { 2 } - 2 x + 1 )$ 运算后不含 $x ^ { 2 }$ 项"}, {"id": "m = 5"}], "links": [{"rel": "提取因式", "source": "( mx ^ { 2 } + x - 3 ) ( x ^ { 2 } - 2 x + 1 )", "target": "mx ^ { 4 } + ( 1 - 2 m ) x ^ { 3 } + ( m - 2 - 3 ) x ^ { 2 } + 7 x - 3"}, {"rel": "被描述", "source": "mx ^ { 4 } + ( 1 - 2 m ) x ^ { 3 } + ( m - 2 - 3 ) x ^ { 2 } + 7 x - 3", "target": "m - 2 - 3 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "mx ^ { 4 } + ( 1 - 2 m ) x ^ { 3 } + ( m - 2 - 3 ) x ^ { 2 } + 7 x - 3"}, {"rel": "等式方程求解", "source": "m - 2 - 3 = 0", "target": "m = 5"}, {"rel": "限制性描述", "source": "整式 $( mx ^ { 2 } + x - 3 ) ( x ^ { 2 } - 2 x + 1 )$ 运算后不含 $x ^ { 2 }$ 项", "target": "m - 2 - 3 = 0"}]}}
|
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{"content": "If the solution to the equation $mx - 5 = 2 x - 1$ for $x$ is $x = 2$, then the solution to the equation $2 y - m = 2$ for $y$ is ____?", "answer": "3", "steps": "From the given information, we have $2 m - 5 = 2 * 2 - 1$, which gives us $m = 4$. Substituting $m = 4$ into $2 y - m = 2$, we get $2 y - 4 = 2$, which gives us $y = 3$.", "expr_cands": ["x", "mx - 5 = 2 x - 1", "m", "x = 2", "y", "2 y - m = 2", "2 m - 5 = 2 * 2 - 1", "m = 4", "2 y - 4 = 2", "y = 3"], "exprs": ["2 m - 5 = 2 * 2 - 1", "m = 4", "2 y - 4 = 2", "y = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x"}, {"id": "2 m - 5 = 2 * 2 - 1"}, {"id": "mx - 5 = 2 x - 1"}, {"id": "x = 2"}, {"id": "关于 $x$ 的方程 $mx - 5 = 2 x - 1$ 的解是 $x = 2$"}, {"id": "m = 4"}, {"id": "2 y - m = 2"}, {"id": "2 y - 4 = 2"}, {"id": "y = 3"}], "links": [{"rel": "被描述", "source": "x", "target": "2 m - 5 = 2 * 2 - 1"}, {"rel": "等式方程求解", "source": "2 m - 5 = 2 * 2 - 1", "target": "m = 4"}, {"rel": "被描述", "source": "mx - 5 = 2 x - 1", "target": "2 m - 5 = 2 * 2 - 1"}, {"rel": "被描述", "source": "x = 2", "target": "2 m - 5 = 2 * 2 - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $mx - 5 = 2 x - 1$ 的解是 $x = 2$", "target": "2 m - 5 = 2 * 2 - 1"}, {"rel": "代入", "source": "m = 4", "target": "2 y - 4 = 2"}, {"rel": "被代入", "source": "2 y - m = 2", "target": "2 y - 4 = 2"}, {"rel": "等式方程求解", "source": "2 y - 4 = 2", "target": "y = 3"}]}}
|
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{"content": "If the product of $( x + 2 ) ( x + p )$ does not contain a linear term in $x$, then $p$ = ____?", "answer": "- 2", "steps": "$( x + 2 ) ( x + p ) = x ^ { 2 } + ( p + 2 ) x + 2 p$ , from the linear term in the product, we get $p + 2 = 0$ , which means $p = - 2$ .", "expr_cands": ["( x + 2 ) ( x + p )", "p", "x", "x ^ { 2 } + ( p + 2 ) x + 2 p", "p + 2 = 0", "p = - 2"], "exprs": ["x ^ { 2 } + ( p + 2 ) x + 2 p", "p + 2 = 0", "p = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + 2 ) ( x + p )"}, {"id": "x ^ { 2 } + ( p + 2 ) x + 2 p"}, {"id": "x"}, {"id": "p + 2 = 0"}, {"id": "$( x + 2 ) ( x + p )$ 的乘积不含 $x$ 的一次项"}, {"id": "p = - 2"}], "links": [{"rel": "提取因式", "source": "( x + 2 ) ( x + p )", "target": "x ^ { 2 } + ( p + 2 ) x + 2 p"}, {"rel": "被描述", "source": "x ^ { 2 } + ( p + 2 ) x + 2 p", "target": "p + 2 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "x ^ { 2 } + ( p + 2 ) x + 2 p"}, {"rel": "等式方程求解", "source": "p + 2 = 0", "target": "p = - 2"}, {"rel": "限制性描述", "source": "$( x + 2 ) ( x + p )$ 的乘积不含 $x$ 的一次项", "target": "p + 2 = 0"}]}}
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{"content": "The coefficient of the linear term in the equation $3 x ( x + 1 ) = 0$ is ____ ?", "answer": "3", "steps": "The equation can be rearranged as: $3 x ^ 2 + 3 x = 0$, and the coefficient of the linear term is $3$.", "expr_cands": ["3 x ( x + 1 ) = 0", "x", "3 x ^ { 2 } + 3 x = 0", "x = - 1", "x = 0", "3"], "exprs": ["3 x ^ { 2 } + 3 x = 0", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ( x + 1 ) = 0"}, {"id": "3 x ^ { 2 } + 3 x = 0"}, {"id": "3"}, {"id": "方程 $3 x ( x + 1 ) = 0$ 的一次项系数"}], "links": [{"rel": "展开", "source": "3 x ( x + 1 ) = 0", "target": "3 x ^ { 2 } + 3 x = 0"}, {"rel": "被描述", "source": "3 x ^ { 2 } + 3 x = 0", "target": "3"}, {"rel": "限制性描述", "source": "方程 $3 x ( x + 1 ) = 0$ 的一次项系数", "target": "3"}]}}
|
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{"content": "If $x = - 5$ is a solution to the equation $x + 2 y = 3$, then $y$ = ____ ?", "answer": "4", "steps": "Substituting $x = - 5$ into $x + 2 y = 3$ yields $- 5 + 2 y = 3$, which can be solved to obtain $y = 4$.", "expr_cands": ["x = - 5", "x", "x + 2 y = 3", "y", "2 y - 5 = 3", "- 5 + 2 y = 3", "y = 4"], "exprs": ["- 5 + 2 y = 3", "y = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 2 y = 3"}, {"id": "- 5 + 2 y = 3"}, {"id": "x = - 5"}, {"id": "y = 4"}], "links": [{"rel": "被代入", "source": "x + 2 y = 3", "target": "- 5 + 2 y = 3"}, {"rel": "等式方程求解", "source": "- 5 + 2 y = 3", "target": "y = 4"}, {"rel": "代入", "source": "x = - 5", "target": "- 5 + 2 y = 3"}]}}
|
||||
{"content": "Given the equation $ax - 6 = 12 + a$ has a solution of $x = - 2$, what is the value of $a$?", "answer": "- 6", "steps": "Substituting $x = - 2$ into $ax - 6 = 12 + a$, we get $- 2 a - 6 = 12 + a$. Therefore, $a = - 6$.", "expr_cands": ["x", "ax - 6 = 12 + a", "a", "x = - 2", "- 2 a - 6 = 12 + a", "a = - 6"], "exprs": ["- 2 a - 6 = 12 + a", "a = - 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ax - 6 = 12 + a"}, {"id": "- 2 a - 6 = 12 + a"}, {"id": "x = - 2"}, {"id": "a = - 6"}], "links": [{"rel": "被代入", "source": "ax - 6 = 12 + a", "target": "- 2 a - 6 = 12 + a"}, {"rel": "等式方程求解", "source": "- 2 a - 6 = 12 + a", "target": "a = - 6"}, {"rel": "代入", "source": "x = - 2", "target": "- 2 a - 6 = 12 + a"}]}}
|
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{"content": "The solution set of the inequality $3 x + 12 > 0$ is:", "answer": "x > - 4", "steps": "Since 3x is greater than negative 12, it follows that x is greater than negative 4.", "expr_cands": ["3 x + 12 > 0", "x", "3 x > - 12", "- 4 < x", "x > - 4"], "exprs": ["x > - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + 12 > 0"}, {"id": "x > - 4"}], "links": [{"rel": "不等式方程求解", "source": "3 x + 12 > 0", "target": "x > - 4"}]}}
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{"content": "Let $S = ( 2 + 1 ) ( 2 ^ { 2 } + 1 ) ( 2 ^ { 4 } + 1 ) ( 2 ^ { 8 } + 1 ) ( 2 ^ { 16 } + 1 )$, then $S + 1$ = ____ ?", "answer": "2 ^ { 32 }", "steps": "$S = ( 2 + 1 ) ( 2 ^ { 2 } + 1 ) ( 2 ^ { 4 } + 1 ) ( 2 ^ { 8 } + 1 ) ( 2 ^ { 16 } + 1 ) = ( 2 - 1 ) * ( 2 + 1 ) * ( 2 ^ { 2 } + 1 ) * ( 2 ^ { 4 } + 1 ) * ( 2 ^ { 8 } + 1 ) * ( 2 ^ { 16 } + 1 ) = ( 2 ^ { 2 } - 1 ) * ( 2 ^ { 2 } + 1 ) * ( 2 ^ { 4 } + 1 ) * ( 2 ^ { 8 } + 1 ) * ( 2 ^ { 16 } + 1 ) = ( 2 ^ { 4 } - 1 ) ( 2 ^ { 4 } + 1 ) * ( 2 ^ { 8 } + 1 ) * ( 2 ^ { 16 } + 1 ) = ( 2 ^ { 8 } - 1 ) ( 2 ^ { 8 } + 1 ) * ( 2 ^ { 16 } + 1 ) = ( 2 ^ { 16 } - 1 ) * ( 2 ^ { 16 } + 1 ) = 2 ^ { 32 } - 1$ , so $S + 1 = 2 ^ { 32 }$.", "expr_cands": ["S = ( 2 + 1 ) ( 2 ^ { 2 } + 1 ) ( 2 ^ { 4 } + 1 ) ( 2 ^ { 8 } + 1 ) ( 2 ^ { 16 } + 1 )", "S", "S + 1", "S = 2 ^ { 32 } - 1", "4294967296"], "exprs": ["S = 2 ^ { 32 } - 1", "4294967296"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "S = ( 2 + 1 ) ( 2 ^ { 2 } + 1 ) ( 2 ^ { 4 } + 1 ) ( 2 ^ { 8 } + 1 ) ( 2 ^ { 16 } + 1 )"}, {"id": "S = 2 ^ { 32 } - 1"}, {"id": "S + 1"}, {"id": "4294967296"}], "links": [{"rel": "计算", "source": "S = ( 2 + 1 ) ( 2 ^ { 2 } + 1 ) ( 2 ^ { 4 } + 1 ) ( 2 ^ { 8 } + 1 ) ( 2 ^ { 16 } + 1 )", "target": "S = 2 ^ { 32 } - 1"}, {"rel": "代入", "source": "S = 2 ^ { 32 } - 1", "target": "4294967296"}, {"rel": "被代入", "source": "S + 1", "target": "4294967296"}]}}
|
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{"content": "If $\\sqrt { 12 } + \\sqrt { y } = \\sqrt { 27 }$, then the value of $y$ is ____?", "answer": "3", "steps": "Because $\\sqrt { 12 } + \\sqrt { y } = \\sqrt { 27 }$, we have $\\sqrt { y } = \\sqrt { 27 } - \\sqrt { 12 } = 3 \\sqrt { 3 } - 2 \\sqrt { 3 } = \\sqrt { 3 }$. Therefore, $y = 3$.", "expr_cands": ["\\sqrt { 12 } + \\sqrt { y } = \\sqrt { 27 }", "y", "y = 3", "\\sqrt { y } = \\sqrt { 3 }"], "exprs": ["y = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 12 } + \\sqrt { y } = \\sqrt { 27 }"}, {"id": "y = 3"}], "links": [{"rel": "等式方程求解", "source": "\\sqrt { 12 } + \\sqrt { y } = \\sqrt { 27 }", "target": "y = 3"}]}}
|
||||
{"content": "If $2 x - 1 = 8$, then what is the value of $2 ( 2 x + 1 ) - 3$?", "answer": "17", "steps": "Because $2 x - 1 = 8$, therefore $2 x = 9$, therefore $2 ( 2 x + 1 ) - 3 = 2 * ( 9 + 1 ) - 3 = 20 - 3 = 17$.", "expr_cands": ["2 x - 1 = 8", "x", "2 ( 2 x + 1 ) - 3", "x = \\frac { 9 } { 2 }", "2 x = 9", "17"], "exprs": ["2 x = 9", "17"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 1 = 8"}, {"id": "2 x = 9"}, {"id": "2 ( 2 x + 1 ) - 3"}, {"id": "17"}], "links": [{"rel": "移项", "source": "2 x - 1 = 8", "target": "2 x = 9"}, {"rel": "代入", "source": "2 x = 9", "target": "17"}, {"rel": "被代入", "source": "2 ( 2 x + 1 ) - 3", "target": "17"}]}}
|
||||
{"content": "The positive integer solution of the inequality $- \\frac { 1 } { 2 } x + 1 > 0$ is ____ ?", "answer": "1", "steps": "Since $- \\frac { 1 } { 2 } x + 1 > 0$, therefore $- \\frac { 1 } { 2 } x > - 1$, therefore $x < 2$. Thus, the positive integer solution of the inequality is $1$.", "expr_cands": ["- \\frac { 1 } { 2 } x + 1 > 0", "x", "x < 2", "- \\frac { 1 } { 2 } x > - 1", "1"], "exprs": ["x < 2", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 1 } { 2 } x + 1 > 0"}, {"id": "x < 2"}, {"id": "1"}, {"id": "不等式的正整数解是 $1$"}, {"id": "不等式 $- \\frac { 1 } { 2 } x + 1 > 0$ 的正整数解"}], "links": [{"rel": "不等式方程求解", "source": "- \\frac { 1 } { 2 } x + 1 > 0", "target": "x < 2"}, {"rel": "被描述", "source": "x < 2", "target": "1"}, {"rel": "限制性描述", "source": "不等式的正整数解是 $1$", "target": "1"}, {"rel": "限制性描述", "source": "不等式 $- \\frac { 1 } { 2 } x + 1 > 0$ 的正整数解", "target": "1"}]}}
|
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{"content": "Given that the value of the algebraic expression $x ^ 2 - \\frac { 3 } { 2 } x$ is $10$, find the value of $2 x ^ 2 - 3 x + 2000$.", "answer": "2020", "steps": "Because the value of $x ^ 2 - \\frac { 3 } { 2 } x$ is $10$, therefore $2 x ^ 2 - 3 x = 20$. The original expression is equal to $20 + 2000 = 2020$.", "expr_cands": ["x ^ { 2 } - \\frac { 3 } { 2 } x", "x", "10", "2 x ^ { 2 } - 3 x + 2000", "2 x ^ { 2 } - 3 x = 20", "x = - \\frac { 5 } { 2 }", "x = 4", "20 + 2000", "2020"], "exprs": ["2 x ^ { 2 } - 3 x = 20", "20 + 2000", "2020"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - \\frac { 3 } { 2 } x"}, {"id": "2 x ^ { 2 } - 3 x = 20"}, {"id": "10"}, {"id": "代数式 $x ^ { 2 } - \\frac { 3 } { 2 } x$ 的值为 $10$"}, {"id": "2 x ^ { 2 } - 3 x + 2000"}, {"id": "20 + 2000"}, {"id": "2020"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - \\frac { 3 } { 2 } x", "target": "2 x ^ { 2 } - 3 x = 20"}, {"rel": "代入", "source": "2 x ^ { 2 } - 3 x = 20", "target": "20 + 2000"}, {"rel": "被描述", "source": "10", "target": "2 x ^ { 2 } - 3 x = 20"}, {"rel": "限制性描述", "source": "代数式 $x ^ { 2 } - \\frac { 3 } { 2 } x$ 的值为 $10$", "target": "2 x ^ { 2 } - 3 x = 20"}, {"rel": "被代入", "source": "2 x ^ { 2 } - 3 x + 2000", "target": "20 + 2000"}, {"rel": "计算", "source": "20 + 2000", "target": "2020"}]}}
|
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{"content": "The degree of the monomial $- \\frac { 2 } { 5 } x ^ 2 yz$ is ____?", "answer": "4", "steps": "The degree of the monomial $- \\frac { 2 } { 5 } x ^ 2 yz$ is $2 + 1 + 1 = 4$.", "expr_cands": ["- \\frac { 2 } { 5 } { x } ^ { 2 } yz", "y", "x", "z", "- \\frac { 2 } { 5 } x ^ { 2 } yz", "2 + 1 + 1", "4"], "exprs": ["4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 2 } { 5 } x ^ { 2 } yz"}, {"id": "4"}, {"id": "单项式 $- \\frac { 2 } { 5 } { x } ^ { 2 } yz$ 的次数"}], "links": [{"rel": "被描述", "source": "- \\frac { 2 } { 5 } x ^ { 2 } yz", "target": "4"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { 2 } { 5 } { x } ^ { 2 } yz$ 的次数", "target": "4"}]}}
|
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{"content": "The smallest integer solution that makes $\\frac { - 6 x - 2 } { 3 } \\leq \\frac { 3 x } { 2 } + \\frac { 1 } { 2 }$ true is ____ ?", "answer": "0", "steps": "Solve the inequality. Multiplying both sides by $6$ yields $- 12 x - 4 \\le 9 x + 3$. Rearranging gives $- 12 x - 9 x \\le 4 + 3$, which simplifies to $- 21 x \\le 7$. Therefore, $x \\ge - \\frac { 1 } { 3 }$. The smallest integer satisfying this inequality is $0$.", "expr_cands": ["\\frac { - 6 x - 2 } { 3 } \\le \\frac { 3 x } { 2 } + \\frac { 1 } { 2 }", "x", "6", "- 12 x - 4 \\le 9 x + 3", "- \\frac { 1 } { 3 } \\le x", "- 12 x - 9 x \\le 4 + 3", "- 21 x \\le 7", "x \\ge - \\frac { 1 } { 3 }", "0"], "exprs": ["x \\ge - \\frac { 1 } { 3 }", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { - 6 x - 2 } { 3 } \\le \\frac { 3 x } { 2 } + \\frac { 1 } { 2 }"}, {"id": "x \\ge - \\frac { 1 } { 3 }"}, {"id": "0"}, {"id": "使 $\\frac { - 6 x - 2 } { 3 } \\le \\frac { 3 x } { 2 } + \\frac { 1 } { 2 }$ 成立的最小整数解"}], "links": [{"rel": "不等式方程求解", "source": "\\frac { - 6 x - 2 } { 3 } \\le \\frac { 3 x } { 2 } + \\frac { 1 } { 2 }", "target": "x \\ge - \\frac { 1 } { 3 }"}, {"rel": "被描述", "source": "x \\ge - \\frac { 1 } { 3 }", "target": "0"}, {"rel": "限制性描述", "source": "使 $\\frac { - 6 x - 2 } { 3 } \\le \\frac { 3 x } { 2 } + \\frac { 1 } { 2 }$ 成立的最小整数解", "target": "0"}]}}
|
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{"content": "In $( k + 8 ) ^ { 3 } = - 27$, what is the value of $k$?", "answer": "- 11", "steps": "$k + 8 = - 3$, solving for k, we get $k = - 11$.", "expr_cands": ["( k + 8 ) ^ { 3 } = - 27", "k", "k + 8 = - 3", "k = - 11"], "exprs": ["k = - 11"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( k + 8 ) ^ { 3 } = - 27"}, {"id": "k = - 11"}], "links": [{"rel": "等式方程求解", "source": "( k + 8 ) ^ { 3 } = - 27", "target": "k = - 11"}]}}
|
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{"content": "If $x ^ 2 - x + m$ is a perfect square, then the value of $m$ is:", "answer": "\\frac { 1 } { 4 }", "steps": "$x ^ { 2 } - x + m = x ^ { 2 } - 2 \\times \\frac { 1 } { 2 } x + m$, because $x ^ { 2 } - x + m$ is a perfect square, therefore $m = ( \\frac { 1 } { 2 }) ^ { 2 } = \\frac { 1 } { 4 }$.", "expr_cands": ["x ^ { 2 } - x + m", "m", "x", "x ^ { 2 } - 2 \\times \\frac { 1 } { 2 } x + m", "m = \\frac { 1 } { 4 }"], "exprs": ["m = \\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - x + m"}, {"id": "m = \\frac { 1 } { 4 }"}, {"id": "$x ^ { 2 } - x + m$ 完全平方式"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - x + m", "target": "m = \\frac { 1 } { 4 }"}, {"rel": "限制性描述", "source": "$x ^ { 2 } - x + m$ 完全平方式", "target": "m = \\frac { 1 } { 4 }"}]}}
|
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{"content": "If the monomials $2 xy$ and $5 x ^ { 2 m - 3 } y$ are like terms, what is the value of $m$?", "answer": "2", "steps": "Because the monomials $2 xy$ and $5 x ^ { 2 m - 3 } y$ are like terms, we have $2 m - 3 = 1$, which implies $m = 2$.", "expr_cands": ["2 xy", "y", "x", "5 x ^ { 2 m - 3 } y", "m", "2 m - 3 = 1", "m = 2"], "exprs": ["2 m - 3 = 1", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x ^ { 2 m - 3 } y"}, {"id": "2 m - 3 = 1"}, {"id": "2 xy"}, {"id": "单项式 $2 xy$ 与 $5 x ^ { 2 m - 3 } y$ 是同类项"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "5 x ^ { 2 m - 3 } y", "target": "2 m - 3 = 1"}, {"rel": "等式方程求解", "source": "2 m - 3 = 1", "target": "m = 2"}, {"rel": "被描述", "source": "2 xy", "target": "2 m - 3 = 1"}, {"rel": "限制性描述", "source": "单项式 $2 xy$ 与 $5 x ^ { 2 m - 3 } y$ 是同类项", "target": "2 m - 3 = 1"}]}}
|
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{"content": "Given: The two square roots of a positive number are $- 5$ and $a + 1$. Find the value of $a$.", "answer": "4", "steps": "According to the problem, we have $- 5 + a + 1 = 0$, and solving for $a$ gives $a = 4$.", "expr_cands": ["- 5", "a + 1", "a", "- 5 + a + 1 = 0", "a = 4"], "exprs": ["- 5 + a + 1 = 0", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 5"}, {"id": "- 5 + a + 1 = 0"}, {"id": "a + 1"}, {"id": ": 一个正数的两个平方根分别是 $- 5$ 和 $a + 1$"}, {"id": "平方根互为相反数"}, {"id": "a = 4"}], "links": [{"rel": "被描述", "source": "- 5", "target": "- 5 + a + 1 = 0"}, {"rel": "等式方程求解", "source": "- 5 + a + 1 = 0", "target": "a = 4"}, {"rel": "被描述", "source": "a + 1", "target": "- 5 + a + 1 = 0"}, {"rel": "限制性描述", "source": ": 一个正数的两个平方根分别是 $- 5$ 和 $a + 1$", "target": "- 5 + a + 1 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "- 5 + a + 1 = 0"}]}}
|
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{"content": "Regarding the fractional equation in $x$, $\\frac { x - m } { x - 3 } = \\frac { 1 } { 3 - x }$, there is a positive root. What is the positive root?", "answer": "3", "steps": "Because the equation has a repeated root, $\\therefore$ the repeated root is $x - 3 = 0$ or $3 - x = 0$. That is, $x = 3$.", "expr_cands": ["x", "\\frac { x - m } { x - 3 } = \\frac { 1 } { 3 - x }", "m", "x - 3 = 0", "x = 3", "3 - x = 0"], "exprs": ["x - 3 = 0", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - m } { x - 3 } = \\frac { 1 } { 3 - x }"}, {"id": "x - 3 = 0"}, {"id": "x"}, {"id": "关于 $x$ 的分式方程 $\\frac { x - m } { x - 3 } = \\frac { 1 } { 3 - x }$ 有增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "\\frac { x - m } { x - 3 } = \\frac { 1 } { 3 - x }", "target": "x - 3 = 0"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "被描述", "source": "x", "target": "x - 3 = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的分式方程 $\\frac { x - m } { x - 3 } = \\frac { 1 } { 3 - x }$ 有增根", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 3 = 0"}]}}
|
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{"content": "In the same Cartesian coordinate system, if the line $y = 2 x + 3$ is parallel to the line $y = kx - 1$, then the value of $k$ is ____?", "answer": "2", "steps": "$\\because$ The line $y = 2 x + 3$ is parallel to the line $y = kx - 1$, $\\therefore$ $k = 2$.", "expr_cands": ["y = 2 x + 3", "y", "x", "y = kx - 1", "k", "2 x + 3 = kx - 1", "k = 2"], "exprs": ["k = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 x + 3"}, {"id": "k = 2"}, {"id": "y = kx - 1"}, {"id": "在同一直角坐标系中"}, {"id": "直线 $y = 2 x + 3$ 与直线 $y = kx - 1$ 平行"}, {"id": "$k$ 值"}], "links": [{"rel": "被描述", "source": "y = 2 x + 3", "target": "k = 2"}, {"rel": "被描述", "source": "y = kx - 1", "target": "k = 2"}, {"rel": "限制性描述", "source": "在同一直角坐标系中", "target": "k = 2"}, {"rel": "限制性描述", "source": "直线 $y = 2 x + 3$ 与直线 $y = kx - 1$ 平行", "target": "k = 2"}, {"rel": "限制性描述", "source": "$k$ 值", "target": "k = 2"}]}}
|
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{"content": "Given $5 x + 3 = 8 x - 3$ and $\\frac { 5 x + a } { 6 } = \\frac { 7 } { 3 }$, if the solutions of these two equations are opposite, then $a$ = ____?", "answer": "24", "steps": "Solve the equation $5 x + 3 = 8 x - 3$ to get $x = 2$. Then, according to the problem, we have $\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }$. Solving for $a$, we get $a = 24$.", "expr_cands": ["5 x + 3 = 8 x - 3", "x", "\\frac { 5 x + a } { 6 } = \\frac { 7 } { 3 }", "a", "x = 2", "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }", "a = 24"], "exprs": ["x = 2", "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }", "a = 24"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x + 3 = 8 x - 3"}, {"id": "x = 2"}, {"id": "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }"}, {"id": "\\frac { 5 x + a } { 6 } = \\frac { 7 } { 3 }"}, {"id": "$5 x + 3 = 8 x - 3$ 和 $\\frac { 5 x + a } { 6 } = \\frac { 7 } { 3 }$ 这两个方程的解是互为相反数"}, {"id": "a = 24"}], "links": [{"rel": "等式方程求解", "source": "5 x + 3 = 8 x - 3", "target": "x = 2"}, {"rel": "被描述", "source": "5 x + 3 = 8 x - 3", "target": "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }"}, {"rel": "被描述", "source": "x = 2", "target": "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }"}, {"rel": "等式方程求解", "source": "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }", "target": "a = 24"}, {"rel": "被描述", "source": "\\frac { 5 x + a } { 6 } = \\frac { 7 } { 3 }", "target": "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }"}, {"rel": "限制性描述", "source": "$5 x + 3 = 8 x - 3$ 和 $\\frac { 5 x + a } { 6 } = \\frac { 7 } { 3 }$ 这两个方程的解是互为相反数", "target": "\\frac { 5 * ( - 2 ) + a } { 6 } = \\frac { 7 } { 3 }"}]}}
|
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{"content": "In $a ^ { 2 } + ( 2 k - 6 ) ab + b ^ { 2 } + 9$, if there is no $ab$ term, then $k$ = ____ ?", "answer": "3", "steps": "$\\because$ The polynomial $a ^ 2 + ( 2 k - 6 ) ab + b ^ 2 + 9$ does not contain a term with $ab$, $\\therefore$ $2 k - 6 = 0$, which solves to $k = 3$.", "expr_cands": ["a ^ { 2 } + ( 2 k - 6 ) ab + b ^ { 2 } + 9", "k", "b", "a", "ab", "2 k - 6 = 0", "k = 3"], "exprs": ["2 k - 6 = 0", "k = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } + ( 2 k - 6 ) ab + b ^ { 2 } + 9"}, {"id": "2 k - 6 = 0"}, {"id": "多项式 $a ^ { 2 } + ( 2 k - 6 ) ab + b ^ { 2 } + 9$ 不含 $ab$ 的项"}, {"id": "k = 3"}], "links": [{"rel": "被描述", "source": "a ^ { 2 } + ( 2 k - 6 ) ab + b ^ { 2 } + 9", "target": "2 k - 6 = 0"}, {"rel": "等式方程求解", "source": "2 k - 6 = 0", "target": "k = 3"}, {"rel": "限制性描述", "source": "多项式 $a ^ { 2 } + ( 2 k - 6 ) ab + b ^ { 2 } + 9$ 不含 $ab$ 的项", "target": "2 k - 6 = 0"}]}}
|
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{"content": "If $\\frac { x } { 2 } = \\frac { y } { 3 }$, find $\\frac { x + y } { y }$.", "answer": "\\frac { 5 } { 3 }", "steps": "Because $\\frac { x } { 2 } = \\frac { y } { 3 }$, therefore $\\frac { x } { y } = \\frac { 2 } { 3 }$; let $x = 2 k$, then $y = 3 k$; therefore $\\frac { x + y } { y } = \\frac { 2 k + 3 k } { 3 k } = \\frac { 5 } { 3 }$.", "expr_cands": ["\\frac { x } { 2 } = \\frac { y } { 3 }", "y", "x", "\\frac { x + y } { y }", "\\frac { x } { y } = \\frac { 2 } { 3 }", "x = 2 k", "k", "y = 3 k", "\\frac { 5 } { 3 }"], "exprs": ["x = 2 k", "y = 3 k", "\\frac { 5 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $x = 2 k$"}, {"id": "x = 2 k"}, {"id": "y = 3 k"}, {"id": "\\frac { x } { 2 } = \\frac { y } { 3 }"}, {"id": "\\frac { x + y } { y }"}, {"id": "\\frac { 5 } { 3 }"}], "links": [{"rel": "假设描述", "source": "设 $x = 2 k$", "target": "x = 2 k"}, {"rel": "联立", "source": "x = 2 k", "target": "y = 3 k"}, {"rel": "代入", "source": "x = 2 k", "target": "\\frac { 5 } { 3 }"}, {"rel": "代入", "source": "y = 3 k", "target": "\\frac { 5 } { 3 }"}, {"rel": "联立", "source": "\\frac { x } { 2 } = \\frac { y } { 3 }", "target": "y = 3 k"}, {"rel": "被代入", "source": "\\frac { x + y } { y }", "target": "\\frac { 5 } { 3 }"}]}}
|
||||
{"content": "The range of the independent variable for the function $y = \\frac { 1 } { x + 2019 }$ is ____ ?", "answer": "x \\neq - 2019", "steps": "From the given condition, we can deduce that $x + 2019 \\neq 0$. Solving for $x$, we get $x \\neq - 2019$.", "expr_cands": ["y = \\frac { 1 } { x + 2019 }", "x", "y", "x + 2019 \\neq 0", "x \\neq - 2019"], "exprs": ["x + 2019 \\neq 0", "x \\neq - 2019"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 1 } { x + 2019 }"}, {"id": "x + 2019 \\neq 0"}, {"id": "函数 $y = \\frac { 1 } { x + 2019 }$ 自变量的取值范围"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq - 2019"}], "links": [{"rel": "被描述", "source": "y = \\frac { 1 } { x + 2019 }", "target": "x + 2019 \\neq 0"}, {"rel": "不等式方程求解", "source": "x + 2019 \\neq 0", "target": "x \\neq - 2019"}, {"rel": "限制性描述", "source": "函数 $y = \\frac { 1 } { x + 2019 }$ 自变量的取值范围", "target": "x + 2019 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x + 2019 \\neq 0"}]}}
|
||||
{"content": "In the linear function $y = ( m + 1 ) x + 5$, as $x$ increases, the value of $y$ decreases. What is the range of values for $m$?", "answer": "m < - 1", "steps": "$\\because$ The function $y = ( m + 1 ) x + 5$ is a linear function, and $y$ decreases as $x$ increases. $\\therefore$ $m + 1 < 0$, which implies $m < - 1$.", "expr_cands": ["y = ( m + 1 ) x + 5", "y", "x", "m", "m + 1 < 0", "m < - 1"], "exprs": ["m + 1 < 0", "m < - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m + 1 ) x + 5"}, {"id": "m + 1 < 0"}, {"id": "一次函数 $y = ( m + 1 ) x + 5$ 中"}, {"id": "$y$ 值随 $x$ 的增大而减少"}, {"id": "m < - 1"}], "links": [{"rel": "被描述", "source": "y = ( m + 1 ) x + 5", "target": "m + 1 < 0"}, {"rel": "不等式方程求解", "source": "m + 1 < 0", "target": "m < - 1"}, {"rel": "限制性描述", "source": "一次函数 $y = ( m + 1 ) x + 5$ 中", "target": "m + 1 < 0"}, {"rel": "限制性描述", "source": "$y$ 值随 $x$ 的增大而减少", "target": "m + 1 < 0"}]}}
|
||||
{"content": "To make $\\sqrt { 2 x - 6 }$ meaningful in the real number range, the range of values for $x$ is ____?", "answer": "x \\ge 3", "steps": "Since $\\sqrt { 2 x - 6 }$ is defined in the real number range, it follows that $2 x - 6 \\ge 0$, which implies that $x \\ge 3$.", "expr_cands": ["\\sqrt { 2 x - 6 }", "x", "2 x - 6 \\ge 0", "3 \\le x", "x \\ge 3"], "exprs": ["2 x - 6 \\ge 0", "x \\ge 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 2 x - 6 }"}, {"id": "2 x - 6 \\ge 0"}, {"id": "要使 $\\sqrt { 2 x - 6 }$ 在实数范围内有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 3"}], "links": [{"rel": "被描述", "source": "\\sqrt { 2 x - 6 }", "target": "2 x - 6 \\ge 0"}, {"rel": "不等式方程求解", "source": "2 x - 6 \\ge 0", "target": "x \\ge 3"}, {"rel": "限制性描述", "source": "要使 $\\sqrt { 2 x - 6 }$ 在实数范围内有意义", "target": "2 x - 6 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 x - 6 \\ge 0"}]}}
|
||||
{"content": "Given that $x = 1$ is a solution to the equation $x + 2 m = 7$, then $m$ = ____ ?", "answer": "3", "steps": "Because $x = 1$ is a solution of the equation $x + 2 m = 7$, therefore $1 + 2 m = 7$. Solving for $m$, we get $m = 3$.", "expr_cands": ["x = 1", "x", "x + 2 m = 7", "m", "2 m + 1 = 7", "1 + 2 m = 7", "m = 3"], "exprs": ["1 + 2 m = 7", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 2 m = 7"}, {"id": "1 + 2 m = 7"}, {"id": "x = 1"}, {"id": "m = 3"}], "links": [{"rel": "被代入", "source": "x + 2 m = 7", "target": "1 + 2 m = 7"}, {"rel": "等式方程求解", "source": "1 + 2 m = 7", "target": "m = 3"}, {"rel": "代入", "source": "x = 1", "target": "1 + 2 m = 7"}]}}
|
||||
{"content": "The polynomial $x ^ { 2 } - 2 kxy - 5 y ^ { 2 } + \\frac { 1 } { 3 } xy - 6$ after combining like terms does not contain the term $xy$. What is the value of $k$?", "answer": "\\frac { 1 } { 6 }", "steps": "$\\because$ The polynomial $x ^ 2 - 2 kxy - 5 y ^ 2 + \\frac { 1 } { 3 } xy - 6$ after combining like terms does not contain the term $xy$, $\\therefore$ $- 2 k + \\frac { 1 } { 3 } = 0$, solving for $k$ gives: $k = \\frac { 1 } { 6 }$.", "expr_cands": ["x ^ { 2 } - 2 kxy - 5 y ^ { 2 } + \\frac { 1 } { 3 } xy - 6", "k", "y", "x", "xy", "- 2 k + \\frac { 1 } { 3 } = 0", "k = \\frac { 1 } { 6 }"], "exprs": ["- 2 k + \\frac { 1 } { 3 } = 0", "k = \\frac { 1 } { 6 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 2 kxy - 5 y ^ { 2 } + \\frac { 1 } { 3 } xy - 6"}, {"id": "- 2 k + \\frac { 1 } { 3 } = 0"}, {"id": "多项式 $x ^ { 2 } - 2 kxy - 5 y ^ { 2 } + \\frac { 1 } { 3 } xy - 6$ 合并同类项后不含 $xy$ 项"}, {"id": "k = \\frac { 1 } { 6 }"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 2 kxy - 5 y ^ { 2 } + \\frac { 1 } { 3 } xy - 6", "target": "- 2 k + \\frac { 1 } { 3 } = 0"}, {"rel": "等式方程求解", "source": "- 2 k + \\frac { 1 } { 3 } = 0", "target": "k = \\frac { 1 } { 6 }"}, {"rel": "限制性描述", "source": "多项式 $x ^ { 2 } - 2 kxy - 5 y ^ { 2 } + \\frac { 1 } { 3 } xy - 6$ 合并同类项后不含 $xy$ 项", "target": "- 2 k + \\frac { 1 } { 3 } = 0"}]}}
|
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{"content": "When $m = - 2$, the fractional equation $\\frac { x } { x - 2 } - 3 = \\frac { m } { 2 - x }$ has a positive root. What is the positive root of the equation?", "answer": "2", "steps": "$\\because$ When $m = - 2$, the fractional equation $\\frac { x } { x - 2 } - 3 = \\frac { m } { 2 - x }$ has a root that increases with $x$. $\\therefore$ $x - 2 = 0$, which gives $x = 2$.", "expr_cands": ["m = - 2", "m", "\\frac { x } { x - 2 } - 3 = \\frac { m } { 2 - x }", "x", "\\frac { x } { x - 2 } - 3 = - \\frac { 2 } { 2 - x }", "x - 2 = 0", "x = 2"], "exprs": ["x - 2 = 0", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x - 2 } - 3 = \\frac { m } { 2 - x }"}, {"id": "x - 2 = 0"}, {"id": "分式方程 $\\frac { x } { x - 2 } - 3 = \\frac { m } { 2 - x }$ 有增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "\\frac { x } { x - 2 } - 3 = \\frac { m } { 2 - x }", "target": "x - 2 = 0"}, {"rel": "等式方程求解", "source": "x - 2 = 0", "target": "x = 2"}, {"rel": "限制性描述", "source": "分式方程 $\\frac { x } { x - 2 } - 3 = \\frac { m } { 2 - x }$ 有增根", "target": "x - 2 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 2 = 0"}]}}
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{"content": "The equation $4 x ^ { 2 } - kx + 6 = 0$ has a root of $2$, then $k$ = ____ ?", "answer": "11", "steps": "Substituting $x = 2$ into the equation $4 x ^ 2 - kx + 6 = 0$, we get $16 - 2 k + 6 = 0$, so $k = 11$.", "expr_cands": ["4 x ^ { 2 } - kx + 6 = 0", "k", "x", "2", "x = 2", "22 - 2 k = 0", "16 - 2 k + 6 = 0", "k = 11"], "exprs": ["x = 2", "16 - 2 k + 6 = 0", "k = 11"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "x = 2"}, {"id": "4 x ^ { 2 } - kx + 6 = 0"}, {"id": "x"}, {"id": "方程 $4 x ^ { 2 } - kx + 6 = 0$ 的一个根是 $2$"}, {"id": "16 - 2 k + 6 = 0"}, {"id": "k = 11"}], "links": [{"rel": "被描述", "source": "2", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "16 - 2 k + 6 = 0"}, {"rel": "被描述", "source": "4 x ^ { 2 } - kx + 6 = 0", "target": "x = 2"}, {"rel": "被代入", "source": "4 x ^ { 2 } - kx + 6 = 0", "target": "16 - 2 k + 6 = 0"}, {"rel": "被描述", "source": "x", "target": "x = 2"}, {"rel": "限制性描述", "source": "方程 $4 x ^ { 2 } - kx + 6 = 0$ 的一个根是 $2$", "target": "x = 2"}, {"rel": "等式方程求解", "source": "16 - 2 k + 6 = 0", "target": "k = 11"}]}}
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{"content": "If the result of $( 4 x ^ 2 + 2 x ) ( x + a )$ does not contain a term of $x ^ 2$, then the value of $a$ is ____?", "answer": "- \\frac { 1 } { 2 }", "steps": "Original expression = $4 x ^ { 3 } + ( 4 a + 2 ) x ^ { 2 } + 2 ax$. Since there is no term containing $x ^ { 2 }$ in the result, we get $4 a + 2 = 0$, which gives $a = - \\frac { 1 } { 2 }$.", "expr_cands": ["( 4 x ^ { 2 } + 2 x ) ( x + a )", "x", "a", "x ^ { 2 }", "4 x ^ { 3 } + ( 4 a + 2 ) x ^ { 2 } + 2 ax", "4 a + 2 = 0", "a = - \\frac { 1 } { 2 }"], "exprs": ["4 x ^ { 3 } + ( 4 a + 2 ) x ^ { 2 } + 2 ax", "4 a + 2 = 0", "a = - \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 4 x ^ { 2 } + 2 x ) ( x + a )"}, {"id": "4 x ^ { 3 } + ( 4 a + 2 ) x ^ { 2 } + 2 ax"}, {"id": "x ^ { 2 }"}, {"id": "4 a + 2 = 0"}, {"id": "$( 4 x ^ { 2 } + 2 x ) ( x + a )$ 的运算结果中不含 $x ^ { 2 }$ 的项"}, {"id": "a = - \\frac { 1 } { 2 }"}], "links": [{"rel": "提取因式", "source": "( 4 x ^ { 2 } + 2 x ) ( x + a )", "target": "4 x ^ { 3 } + ( 4 a + 2 ) x ^ { 2 } + 2 ax"}, {"rel": "被描述", "source": "4 x ^ { 3 } + ( 4 a + 2 ) x ^ { 2 } + 2 ax", "target": "4 a + 2 = 0"}, {"rel": "提取因式参考", "source": "x ^ { 2 }", "target": "4 x ^ { 3 } + ( 4 a + 2 ) x ^ { 2 } + 2 ax"}, {"rel": "等式方程求解", "source": "4 a + 2 = 0", "target": "a = - \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "$( 4 x ^ { 2 } + 2 x ) ( x + a )$ 的运算结果中不含 $x ^ { 2 }$ 的项", "target": "4 a + 2 = 0"}]}}
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{"content": "If the one-variable linear equation $2 x + 3 = 7$ and $4 x + 3 a = 20$ are the same solution equation, then the value of $a$ is ____?", "answer": "4", "steps": "$2 x + 3 = 7$, therefore $2 x = 7 - 3$, therefore $2 x = 4$, which gives us $x = 2$. Since $2 x + 3 = 7$ and $4 x + 3 a = 20$ are equations with the same solution, $x = 2$ is also a solution to the equation $4 x + 3 a = 20$. Therefore, $4 * 2 + 3 a = 20$, which simplifies to $8 + 3 a = 20$. Solving for $a$, we get $a = 4$.", "expr_cands": ["x", "2 x + 3 = 7", "4 x + 3 a = 20", "a", "x = 2", "2 x = 7 - 3", "2 x", "4", "3 a + 8 = 20", "4 * 2 + 3 a = 20", "a = 4", "8 + 3 a = 20", "3 a = 12"], "exprs": ["x = 2", "4 * 2 + 3 a = 20", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 3 = 7"}, {"id": "x = 2"}, {"id": "4 x + 3 a = 20"}, {"id": "4 * 2 + 3 a = 20"}, {"id": "关于 $x$ 的一元一次方程 $2 x + 3 = 7$ 与 $4 x + 3 a = 20$ 是同解方程"}, {"id": "根据 $2 x + 3 = 7$ 和 $4 x + 3 a = 20$ 是同解方程"}, {"id": ", $x = 2$ 是方程 $4 x + 3 a = 20$ 的解"}, {"id": "a = 4"}], "links": [{"rel": "等式方程求解", "source": "2 x + 3 = 7", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "4 * 2 + 3 a = 20"}, {"rel": "被代入", "source": "4 x + 3 a = 20", "target": "4 * 2 + 3 a = 20"}, {"rel": "等式方程求解", "source": "4 * 2 + 3 a = 20", "target": "a = 4"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元一次方程 $2 x + 3 = 7$ 与 $4 x + 3 a = 20$ 是同解方程", "target": "4 * 2 + 3 a = 20"}, {"rel": "限制性描述", "source": "根据 $2 x + 3 = 7$ 和 $4 x + 3 a = 20$ 是同解方程", "target": "4 * 2 + 3 a = 20"}, {"rel": "限制性描述", "source": ", $x = 2$ 是方程 $4 x + 3 a = 20$ 的解", "target": "4 * 2 + 3 a = 20"}]}}
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{"content": "If $| n + 2 | + | m - 7 | = 0$, then $n - m$ is equal to ____?", "answer": "- 9", "steps": "Since $| n + 2 | + | m - 7 | = 0$, it follows that $n = - 2$ and $m = 7$. Therefore, $n - m = - 2 - 7 = - 9$.", "expr_cands": ["| n + 2 | + | m - 7 | = 0", "m", "n", "n - m", "n = - 2", "m = 7", "- 9"], "exprs": ["n = - 2", "m = 7", "- 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| n + 2 | + | m - 7 | = 0"}, {"id": "n = - 2"}, {"id": "绝对值恒大于等于0"}, {"id": "m = 7"}, {"id": "n - m"}, {"id": "- 9"}], "links": [{"rel": "被描述", "source": "| n + 2 | + | m - 7 | = 0", "target": "n = - 2"}, {"rel": "被描述", "source": "| n + 2 | + | m - 7 | = 0", "target": "m = 7"}, {"rel": "代入", "source": "n = - 2", "target": "- 9"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "n = - 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m = 7"}, {"rel": "代入", "source": "m = 7", "target": "- 9"}, {"rel": "被代入", "source": "n - m", "target": "- 9"}]}}
|
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{"content": "If the solution to the equation $2 x + a - 4 = 0$ with respect to $x$ is $x = - 2$, then the value of $a$ is ____?", "answer": "8", "steps": "Substituting $x = - 2$ into the equation, we get $- 4 + a - 4 = 0$, which gives us $a = 8$.", "expr_cands": ["x", "2 x + a - 4 = 0", "a", "x = - 2", "- 4 + a - 4 = 0", "a = 8"], "exprs": ["- 4 + a - 4 = 0", "a = 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + a - 4 = 0"}, {"id": "- 4 + a - 4 = 0"}, {"id": "x = - 2"}, {"id": "a = 8"}], "links": [{"rel": "被代入", "source": "2 x + a - 4 = 0", "target": "- 4 + a - 4 = 0"}, {"rel": "等式方程求解", "source": "- 4 + a - 4 = 0", "target": "a = 8"}, {"rel": "代入", "source": "x = - 2", "target": "- 4 + a - 4 = 0"}]}}
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{"content": "If $( a ^ 4 ) ^ 3 \\div ( a ^ 2 ) ^ 5 = 64$, and $a < 0$, then $a$ = ____ ?", "answer": "- 8", "steps": "Since $( a ^ 4 ) ^ 3 \\div ( a ^ 2 ) ^ 5 = 64$, it follows that $a ^ { 12 } \\div a ^ { 10 } = 64$. Therefore, $a ^ 2 = 64$. Also, since $a < 0$, it follows that $a = - 8$.", "expr_cands": ["( a ^ { 4 } ) ^ { 3 } \\div ( a ^ { 2 } ) ^ { 5 } = 64", "a", "a < 0", "a = - 8", "a = 8", "a ^ { 12 } \\div a ^ { 10 } = 64", "a ^ { 2 } = 64"], "exprs": ["a = - 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a ^ { 4 } ) ^ { 3 } \\div ( a ^ { 2 } ) ^ { 5 } = 64"}, {"id": "a = - 8"}, {"id": "a < 0"}], "links": [{"rel": "联立", "source": "( a ^ { 4 } ) ^ { 3 } \\div ( a ^ { 2 } ) ^ { 5 } = 64", "target": "a = - 8"}, {"rel": "联立", "source": "a < 0", "target": "a = - 8"}]}}
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{"content": "If the fraction $\\frac { x } { 3 x - 6 }$ is meaningful, then the condition that $x$ needs to satisfy is ____?", "answer": "x \\neq 2", "steps": "The fraction $\\frac { x } { 3 x - 6 }$ is defined only if $3 x - 6 \\neq 0$, which implies $x \\neq 2$.", "expr_cands": ["\\frac { x } { 3 x - 6 }", "x", "3 x - 6 \\neq 0", "x \\neq 2"], "exprs": ["3 x - 6 \\neq 0", "x \\neq 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { 3 x - 6 }"}, {"id": "3 x - 6 \\neq 0"}, {"id": "分式 $\\frac { x } { 3 x - 6 }$ 有意义"}, {"id": "$x$ 需要满足的条件"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 2"}], "links": [{"rel": "被描述", "source": "\\frac { x } { 3 x - 6 }", "target": "3 x - 6 \\neq 0"}, {"rel": "不等式方程求解", "source": "3 x - 6 \\neq 0", "target": "x \\neq 2"}, {"rel": "限制性描述", "source": "分式 $\\frac { x } { 3 x - 6 }$ 有意义", "target": "3 x - 6 \\neq 0"}, {"rel": "限制性描述", "source": "$x$ 需要满足的条件", "target": "3 x - 6 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "3 x - 6 \\neq 0"}]}}
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{"content": "If $- 2 a ^ { m + 1 } b ^ { 3 }$ and $5 a ^ { 3 } b ^ { 2 n - 3 }$ can be combined into one term, then the value of $mn$ is ____?", "answer": "6", "steps": "According to the given problem, $- 2 a ^ { m + 1 } b ^ 3$ and $5 a ^ 3 b ^ { 2 n - 3 }$ are like terms. Therefore, we have $m + 1 = 3$ and $2 n - 3 = 3$. Solving for $m$ and $n$, we get $m = 2$ and $n = 3$. Thus, $mn = 2 \\times 3 = 6$.", "expr_cands": ["- 2 a ^ { m + 1 } b ^ { 3 }", "b", "a", "m", "5 a ^ { 3 } b ^ { 2 n - 3 }", "n", "mn", "m + 1 = 3", "m = 2", "2 n - 3 = 3", "n = 3", "6"], "exprs": ["m + 1 = 3", "2 n - 3 = 3", "m = 2", "n = 3", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 a ^ { m + 1 } b ^ { 3 }"}, {"id": "m + 1 = 3"}, {"id": "5 a ^ { 3 } b ^ { 2 n - 3 }"}, {"id": "$- 2 a ^ { m + 1 } b ^ { 3 }$ 与 $5 a ^ { 3 } b ^ { 2 n - 3 }$ 可以合并成一项"}, {"id": "$- 2 a ^ { m + 1 } b ^ { 3 }$ 与 $5 a ^ { 3 } b ^ { 2 n - 3 }$ 是同类项"}, {"id": "2 n - 3 = 3"}, {"id": "m = 2"}, {"id": "n = 3"}, {"id": "mn"}, {"id": "6"}], "links": [{"rel": "被描述", "source": "- 2 a ^ { m + 1 } b ^ { 3 }", "target": "m + 1 = 3"}, {"rel": "被描述", "source": "- 2 a ^ { m + 1 } b ^ { 3 }", "target": "2 n - 3 = 3"}, {"rel": "等式方程求解", "source": "m + 1 = 3", "target": "m = 2"}, {"rel": "被描述", "source": "5 a ^ { 3 } b ^ { 2 n - 3 }", "target": "m + 1 = 3"}, {"rel": "被描述", "source": "5 a ^ { 3 } b ^ { 2 n - 3 }", "target": "2 n - 3 = 3"}, {"rel": "限制性描述", "source": "$- 2 a ^ { m + 1 } b ^ { 3 }$ 与 $5 a ^ { 3 } b ^ { 2 n - 3 }$ 可以合并成一项", "target": "m + 1 = 3"}, {"rel": "限制性描述", "source": "$- 2 a ^ { m + 1 } b ^ { 3 }$ 与 $5 a ^ { 3 } b ^ { 2 n - 3 }$ 可以合并成一项", "target": "2 n - 3 = 3"}, {"rel": "限制性描述", "source": "$- 2 a ^ { m + 1 } b ^ { 3 }$ 与 $5 a ^ { 3 } b ^ { 2 n - 3 }$ 是同类项", "target": "m + 1 = 3"}, {"rel": "限制性描述", "source": "$- 2 a ^ { m + 1 } b ^ { 3 }$ 与 $5 a ^ { 3 } b ^ { 2 n - 3 }$ 是同类项", "target": "2 n - 3 = 3"}, {"rel": "等式方程求解", "source": "2 n - 3 = 3", "target": "n = 3"}, {"rel": "代入", "source": "m = 2", "target": "6"}, {"rel": "代入", "source": "n = 3", "target": "6"}, {"rel": "被代入", "source": "mn", "target": "6"}]}}
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{"content": "Given that the degree of the monomial $8 x ^ { 2 } y ^ { 3 m - 1 }$ is $4$, what is the value of $m$?", "answer": "1", "steps": "$\\because$ The degree of the monomial $8 x ^ { 2 } y ^ { 3 m - 1 }$ is $4$, $\\therefore$ $2 + 3 m - 1 = 4$, $\\therefore$ $m = 1$.", "expr_cands": ["8 x ^ { 2 } y ^ { 3 m - 1 }", "m", "y", "x", "4", "2 + 3 m - 1 = 4", "m = 1"], "exprs": ["2 + 3 m - 1 = 4", "m = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "8 x ^ { 2 } y ^ { 3 m - 1 }"}, {"id": "2 + 3 m - 1 = 4"}, {"id": "4"}, {"id": "单项式 $8 x ^ { 2 } y ^ { 3 m - 1 }$ 的次数是 $4$"}, {"id": "m = 1"}], "links": [{"rel": "被描述", "source": "8 x ^ { 2 } y ^ { 3 m - 1 }", "target": "2 + 3 m - 1 = 4"}, {"rel": "等式方程求解", "source": "2 + 3 m - 1 = 4", "target": "m = 1"}, {"rel": "被描述", "source": "4", "target": "2 + 3 m - 1 = 4"}, {"rel": "限制性描述", "source": "单项式 $8 x ^ { 2 } y ^ { 3 m - 1 }$ 的次数是 $4$", "target": "2 + 3 m - 1 = 4"}]}}
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{"content": "The function $y = x + m - 1$ is a proportional function, then $m$ = ____?", "answer": "1", "steps": "Since $y = x + m - 1$ is a direct proportion function, therefore $m - 1 = 0$. Solving for $m$, we get $m = 1$.", "expr_cands": ["y = x + m - 1", "y", "m", "x", "m - 1 = 0", "m = 1"], "exprs": ["m - 1 = 0", "m = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x + m - 1"}, {"id": "m - 1 = 0"}, {"id": "函数 $y = x + m - 1$ 是正比例函数"}, {"id": "m = 1"}], "links": [{"rel": "被描述", "source": "y = x + m - 1", "target": "m - 1 = 0"}, {"rel": "等式方程求解", "source": "m - 1 = 0", "target": "m = 1"}, {"rel": "限制性描述", "source": "函数 $y = x + m - 1$ 是正比例函数", "target": "m - 1 = 0"}]}}
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{"content": "The meaningful condition for the algebraic expression $\\frac { 1 } { \\sqrt { x + 2 }}$ is ____?", "answer": "x > - 2", "steps": "From the given condition, we have $x + 2 > 0$, which implies that $x > - 2$.", "expr_cands": ["\\frac { 1 } { \\sqrt { x + 2 } }", "x", "x + 2 > 0", "- 2 < x", "x > - 2"], "exprs": ["x + 2 > 0", "x > - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { \\sqrt { x + 2 } }"}, {"id": "x + 2 > 0"}, {"id": "代数式 $\\frac { 1 } { \\sqrt { x + 2 } }$ 有意义的条件"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x > - 2"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { \\sqrt { x + 2 } }", "target": "x + 2 > 0"}, {"rel": "不等式方程求解", "source": "x + 2 > 0", "target": "x > - 2"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 1 } { \\sqrt { x + 2 } }$ 有意义的条件", "target": "x + 2 > 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x + 2 > 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x + 2 > 0"}]}}
|
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{"content": "If $a - 5 = - b$, then ${ ( a + b ) } ^ { 2 } - 4 ( a + b )$ is ____?", "answer": "5", "steps": "Since $a - 5 = - b$, it follows that $a + b = 5$. Therefore, $( a + b ) ^ 2 - 4 ( a + b ) = ( a + b ) ( a + b - 4 ) = 5 * ( 5 - 4 ) = 5$.", "expr_cands": ["a - 5 = - b", "b", "a", "{ ( a + b ) } ^ { 2 } - 4 ( a + b )", "a + b = 5", "( a + b ) ^ { 2 } - 4 ( a + b )", "5"], "exprs": ["a + b = 5", "5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - 5 = - b"}, {"id": "a + b = 5"}, {"id": "{ ( a + b ) } ^ { 2 } - 4 ( a + b )"}, {"id": "5"}], "links": [{"rel": "移项", "source": "a - 5 = - b", "target": "a + b = 5"}, {"rel": "代入", "source": "a + b = 5", "target": "5"}, {"rel": "被代入", "source": "{ ( a + b ) } ^ { 2 } - 4 ( a + b )", "target": "5"}]}}
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{"content": "If the fraction $\\frac { 1 } { x - 2 }$ is undefined, then the value of the real number $x$ is ____?", "answer": "2", "steps": "According to the problem, we have $x - 2 = 0$, which means $x = 2$.", "expr_cands": ["\\frac { 1 } { x - 2 }", "x", "x - 2 = 0", "x = 2"], "exprs": ["x - 2 = 0", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x - 2 }"}, {"id": "x - 2 = 0"}, {"id": "分式 $\\frac { 1 } { x - 2 }$ 无意义"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { x - 2 }", "target": "x - 2 = 0"}, {"rel": "等式方程求解", "source": "x - 2 = 0", "target": "x = 2"}, {"rel": "限制性描述", "source": "分式 $\\frac { 1 } { x - 2 }$ 无意义", "target": "x - 2 = 0"}]}}
|
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{"content": "Given $a + 3 b = 7$, what is $2 a + 6 b - 8$?", "answer": "6", "steps": "Because $a + 3 b = 7$, therefore the original expression is equal to $2 ( a + 3 b ) - 8 = 14 - 8 = 6$.", "expr_cands": ["a + 3 b = 7", "b", "a", "2 a + 6 b - 8", "2 ( a + 3 b ) - 8", "6"], "exprs": ["2 ( a + 3 b ) - 8", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a + 6 b - 8"}, {"id": "2 ( a + 3 b ) - 8"}, {"id": "a + 3 b = 7"}, {"id": "6"}], "links": [{"rel": "提取因式", "source": "2 a + 6 b - 8", "target": "2 ( a + 3 b ) - 8"}, {"rel": "被代入", "source": "2 ( a + 3 b ) - 8", "target": "6"}, {"rel": "提取因式参考", "source": "a + 3 b = 7", "target": "2 ( a + 3 b ) - 8"}, {"rel": "代入", "source": "a + 3 b = 7", "target": "6"}]}}
|
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{"content": "If $a$ and $b$ are opposite numbers, $c$ and $d$ are reciprocal numbers, $| m | = 2$, then $\\frac { a + b } { m } + \\sqrt { m ^ 2 - cd }$ = ____?", "answer": "\\sqrt { 3 }", "steps": "From the given conditions, we have $a + b = 0$, $cd = 1$, and $m = 2$ or $- 2$. Therefore, the original expression is equal to $0 + \\sqrt { 4 - 1 } = \\sqrt { 3 }$.", "expr_cands": ["a", "b", "c", "d", "| m | = 2", "m", "\\frac { a + b } { m } + \\sqrt { { m } ^ { 2 } - cd }", "a + b = 0", "cd = 1", "m = 2", "- 2", "0 + \\sqrt { 4 - 1 }", "\\sqrt { 3 }"], "exprs": ["a + b = 0", "cd = 1", "0 + \\sqrt { 4 - 1 }", "\\sqrt { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ 与 $b$ 互为相反数"}, {"id": "c"}, {"id": "cd = 1"}, {"id": "d"}, {"id": "$c$ 与 $d$ 互为倒数"}, {"id": "| m | = 2"}, {"id": "0 + \\sqrt { 4 - 1 }"}, {"id": "\\frac { a + b } { m } + \\sqrt { { m } ^ { 2 } - cd }"}, {"id": "\\sqrt { 3 }"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "0 + \\sqrt { 4 - 1 }"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ 与 $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "c", "target": "cd = 1"}, {"rel": "代入", "source": "cd = 1", "target": "0 + \\sqrt { 4 - 1 }"}, {"rel": "被描述", "source": "d", "target": "cd = 1"}, {"rel": "限制性描述", "source": "$c$ 与 $d$ 互为倒数", "target": "cd = 1"}, {"rel": "代入", "source": "| m | = 2", "target": "0 + \\sqrt { 4 - 1 }"}, {"rel": "计算", "source": "0 + \\sqrt { 4 - 1 }", "target": "\\sqrt { 3 }"}, {"rel": "被代入", "source": "\\frac { a + b } { m } + \\sqrt { { m } ^ { 2 } - cd }", "target": "0 + \\sqrt { 4 - 1 }"}]}}
|
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{"content": "If $x - y = 6$, $y - z = - 2$, what is the value of $z - x$?", "answer": "- 4", "steps": "$\\because x - y = 6$, $y - z = - 2$, $\\therefore$ adding the two equations gives: $x - z = 4$, $\\therefore$ $z - x = - 4$.", "expr_cands": ["x - y = 6", "y", "x", "y - z = - 2", "z", "z - x", "x - z = 4", "- 4"], "exprs": ["x - z = 4", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - y = 6"}, {"id": "x - z = 4"}, {"id": "y - z = - 2"}, {"id": "z - x"}, {"id": "- 4"}], "links": [{"rel": "联立", "source": "x - y = 6", "target": "x - z = 4"}, {"rel": "代入", "source": "x - z = 4", "target": "- 4"}, {"rel": "联立", "source": "y - z = - 2", "target": "x - z = 4"}, {"rel": "被代入", "source": "z - x", "target": "- 4"}]}}
|
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{"content": "Given $a - 2 = b + c$, what is the value of the algebraic expression $a ( a - b - c ) - b ( a - b - c ) - c ( a - b - c )$?", "answer": "4", "steps": "From $a - 2 = b + c$, we get $a - b - c = 2$. Therefore, the original expression is $( a - b - c ) ( a - b - c ) = 4$.", "expr_cands": ["a - 2 = b + c", "b", "a", "c", "a ( a - b - c ) - b ( a - b - c ) - c ( a - b - c )", "a - b - c = 2", "( a - b - c ) ( a - b - c )", "4"], "exprs": ["a - b - c = 2", "( a - b - c ) ( a - b - c )", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - 2 = b + c"}, {"id": "a - b - c = 2"}, {"id": "a ( a - b - c ) - b ( a - b - c ) - c ( a - b - c )"}, {"id": "( a - b - c ) ( a - b - c )"}, {"id": "4"}], "links": [{"rel": "移项", "source": "a - 2 = b + c", "target": "a - b - c = 2"}, {"rel": "提取因式参考", "source": "a - b - c = 2", "target": "( a - b - c ) ( a - b - c )"}, {"rel": "代入", "source": "a - b - c = 2", "target": "4"}, {"rel": "提取因式", "source": "a ( a - b - c ) - b ( a - b - c ) - c ( a - b - c )", "target": "( a - b - c ) ( a - b - c )"}, {"rel": "被代入", "source": "( a - b - c ) ( a - b - c )", "target": "4"}]}}
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{"content": "If the value of the expression $5 x - 8$ is the opposite of $3 x$, then the value of $x$ is ____?", "answer": "1", "steps": "According to the problem, we have $5 x - 8 + 3 x = 0$. By rearranging and combining like terms, we get $8 x = 8$. Solving for $x$, we get $x = 1$.", "expr_cands": ["5 x - 8", "x", "3 x", "5 x - 8 + 3 x = 0", "x = 1", "8 x = 8"], "exprs": ["5 x - 8 + 3 x = 0", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x - 8"}, {"id": "5 x - 8 + 3 x = 0"}, {"id": "3 x"}, {"id": "式子 $5 x - 8$ 的值与 $3 x$ 互为相反数"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "5 x - 8", "target": "5 x - 8 + 3 x = 0"}, {"rel": "等式方程求解", "source": "5 x - 8 + 3 x = 0", "target": "x = 1"}, {"rel": "被描述", "source": "3 x", "target": "5 x - 8 + 3 x = 0"}, {"rel": "限制性描述", "source": "式子 $5 x - 8$ 的值与 $3 x$ 互为相反数", "target": "5 x - 8 + 3 x = 0"}]}}
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{"content": "The equation about $x$, $x ^ 2 + x + a ^ 2 - 1 = 0$, has one root $x _ 1 = 0$. Find the other root $x _ 2$.", "answer": "- 1", "steps": "According to the problem, we have $x _ 1 + x _ 2 = - 1$, which means $0 + x _ 2 = - 1$, so $x _ 2 = - 1$.", "expr_cands": ["x", "x ^ { 2 } + x + a ^ { 2 } - 1 = 0", "a", "x _ { 1 } = 0", "x _ { 1 }", "x _ { 2 }", "x _ { 1 } + x _ { 2 } = - 1", "0 + x _ { 2 } = - 1", "x_{2} = - 1", "x _ { 2 } = - 1"], "exprs": ["x _ { 1 } + x _ { 2 } = - 1", "0 + x _ { 2 } = - 1", "x _ { 2 } = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + x + a ^ { 2 } - 1 = 0"}, {"id": "x _ { 1 } + x _ { 2 } = - 1"}, {"id": "x _ { 1 }"}, {"id": "x _ { 2 }"}, {"id": "关于 $x$ 的方程 $x ^ { 2 } + x + a ^ { 2 } - 1 = 0$ 的一个根是 $x _ { 1 } = 0$"}, {"id": "另一个根 $x _ { 2 }$ ="}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "x _ { 1 } = 0"}, {"id": "0 + x _ { 2 } = - 1"}, {"id": "x _ { 2 } = - 1"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + x + a ^ { 2 } - 1 = 0", "target": "x _ { 1 } + x _ { 2 } = - 1"}, {"rel": "被代入", "source": "x _ { 1 } + x _ { 2 } = - 1", "target": "0 + x _ { 2 } = - 1"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } + x _ { 2 } = - 1"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } + x _ { 2 } = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x ^ { 2 } + x + a ^ { 2 } - 1 = 0$ 的一个根是 $x _ { 1 } = 0$", "target": "x _ { 1 } + x _ { 2 } = - 1"}, {"rel": "限制性描述", "source": "另一个根 $x _ { 2 }$ =", "target": "x _ { 1 } + x _ { 2 } = - 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + x _ { 2 } = - 1"}, {"rel": "代入", "source": "x _ { 1 } = 0", "target": "0 + x _ { 2 } = - 1"}, {"rel": "等式方程求解", "source": "0 + x _ { 2 } = - 1", "target": "x _ { 2 } = - 1"}]}}
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{"content": "The solution set of the inequality $ax < 2$ is $x > - 1$. What is the value of $a$?", "answer": "- 2", "steps": "The solution set of $ax < 2$ is $x > - 1$. Therefore, we have $x > \\frac { 2 } { a }$. Substituting $\\frac { 2 } { a } = - 1$, we get $a = - 2$.", "expr_cands": ["ax < 2", "x", "a", "x > - 1", "x > \\frac { 2 } { a }", "\\frac { 2 } { a } = - 1", "a = - 2"], "exprs": ["x > \\frac { 2 } { a }", "\\frac { 2 } { a } = - 1", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ax < 2"}, {"id": "x > \\frac { 2 } { a }"}, {"id": "x > - 1"}, {"id": "\\frac { 2 } { a } = - 1"}, {"id": "不等式 $ax < 2$ 的解集是 $x > - 1$"}, {"id": "a = - 2"}], "links": [{"rel": "不等式方程部分求解", "source": "ax < 2", "target": "x > \\frac { 2 } { a }"}, {"rel": "被描述", "source": "x > \\frac { 2 } { a }", "target": "\\frac { 2 } { a } = - 1"}, {"rel": "被描述", "source": "x > - 1", "target": "\\frac { 2 } { a } = - 1"}, {"rel": "等式方程求解", "source": "\\frac { 2 } { a } = - 1", "target": "a = - 2"}, {"rel": "限制性描述", "source": "不等式 $ax < 2$ 的解集是 $x > - 1$", "target": "\\frac { 2 } { a } = - 1"}]}}
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{"content": "If $x$ is a real number and $y = \\sqrt { x - 2 } + \\sqrt { 2 - x } - 1$, then $x + y$ = ____ ?", "answer": "1", "steps": "From $y = \\sqrt { x - 2 } + \\sqrt { 2 - x } - 1$, we obtain $x - 2 \\ge 0$, $2 - x \\ge 0$. Solving for $x$, we get $x = 2$. When $x = 2$, $y = - 1$, so $x + y = 2 + ( - 1 ) = 1$.", "expr_cands": ["x", "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } - 1", "y", "x + y", "x - 2 \\ge 0", "2 \\le x", "2 - x \\ge 0", "x \\le 2", "x = 2", "y = - 1", "1"], "exprs": ["x - 2 \\ge 0", "2 - x \\ge 0", "x = 2", "y = - 1", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } - 1"}, {"id": "x - 2 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "2 - x \\ge 0"}, {"id": "x = 2"}, {"id": "y = - 1"}, {"id": "x + y"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } - 1", "target": "x - 2 \\ge 0"}, {"rel": "被描述", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } - 1", "target": "2 - x \\ge 0"}, {"rel": "被代入", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } - 1", "target": "y = - 1"}, {"rel": "联立", "source": "x - 2 \\ge 0", "target": "x = 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 2 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 - x \\ge 0"}, {"rel": "联立", "source": "2 - x \\ge 0", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "y = - 1"}, {"rel": "代入", "source": "x = 2", "target": "1"}, {"rel": "代入", "source": "y = - 1", "target": "1"}, {"rel": "被代入", "source": "x + y", "target": "1"}]}}
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{"content": "If the degree of the monomial $\\frac { 2 } { 3 } x ^ my ^ 2$ is the same as that of $- x ^ 4 y$, then the value of $m$ is ____?", "answer": "3", "steps": "From the given information, we have $m + 2 = 4 + 1$, which can be solved to obtain $m = 3$.", "expr_cands": ["\\frac { 2 } { 3 } x ^ { m } y ^ { 2 }", "m", "y", "x", "- x ^ { 4 } y", "m + 2 = 4 + 1", "m = 3"], "exprs": ["m + 2 = 4 + 1", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { 3 } x ^ { m } y ^ { 2 }"}, {"id": "m + 2 = 4 + 1"}, {"id": "- x ^ { 4 } y"}, {"id": "单项式 $\\frac { 2 } { 3 } x ^ { m } y ^ { 2 }$ 与 $- x ^ { 4 } y$ 的次数相同"}, {"id": "m = 3"}], "links": [{"rel": "被描述", "source": "\\frac { 2 } { 3 } x ^ { m } y ^ { 2 }", "target": "m + 2 = 4 + 1"}, {"rel": "等式方程求解", "source": "m + 2 = 4 + 1", "target": "m = 3"}, {"rel": "被描述", "source": "- x ^ { 4 } y", "target": "m + 2 = 4 + 1"}, {"rel": "限制性描述", "source": "单项式 $\\frac { 2 } { 3 } x ^ { m } y ^ { 2 }$ 与 $- x ^ { 4 } y$ 的次数相同", "target": "m + 2 = 4 + 1"}]}}
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{"content": "If $( x + m ) ( x + n ) = x ^ 2 - 7 x + mn$, then the value of $- m - n$ is ____?", "answer": "7", "steps": "Because $( x + m ) ( x + n ) = x ^ 2 + ( m + n ) x + mn = x ^ 2 - 7 x + mn$, therefore $m + n = - 7$, therefore $- m - n = 7$.", "expr_cands": ["( x + m ) ( x + n ) = x ^ { 2 } - 7 x + { mn }", "m", "x", "n", "- m - n", "( x + m ) ( x + n ) = x ^ { 2 } - 7 x + mn", "m + n = - 7", "7"], "exprs": ["m + n = - 7", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + m ) ( x + n ) = x ^ { 2 } - 7 x + { mn }"}, {"id": "m + n = - 7"}, {"id": "- m - n"}, {"id": "7"}], "links": [{"rel": "移项", "source": "( x + m ) ( x + n ) = x ^ { 2 } - 7 x + { mn }", "target": "m + n = - 7"}, {"rel": "代入", "source": "m + n = - 7", "target": "7"}, {"rel": "被代入", "source": "- m - n", "target": "7"}]}}
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{"content": "If $a$ and $b$ are real numbers and satisfy $| a - 2 | + \\sqrt { - b ^ 2 } = 0$, then $a + b$ = ____?", "answer": "2", "steps": "From the given information, we have $a - 2 = 0$ and $- { b } ^ { 2 } = 0$. Solving for $a$ and $b$, we get $a = 2$ and $b = 0$. Therefore, $a + b = 2 + 0 = 2$.", "expr_cands": ["a", "b", "| a - 2 | + \\sqrt { - { b } ^ { 2 } } = 0", "a + b", "a - 2 = 0", "a = 2", "- { b } ^ { 2 } = 0", "b = 0", "2"], "exprs": ["a - 2 = 0", "- { b } ^ { 2 } = 0", "a = 2", "b = 0", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 2 | + \\sqrt { - { b } ^ { 2 } } = 0"}, {"id": "a - 2 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "- { b } ^ { 2 } = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a = 2"}, {"id": "b = 0"}, {"id": "a + b"}, {"id": "2"}], "links": [{"rel": "被描述", "source": "| a - 2 | + \\sqrt { - { b } ^ { 2 } } = 0", "target": "a - 2 = 0"}, {"rel": "被描述", "source": "| a - 2 | + \\sqrt { - { b } ^ { 2 } } = 0", "target": "- { b } ^ { 2 } = 0"}, {"rel": "等式方程求解", "source": "a - 2 = 0", "target": "a = 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a - 2 = 0"}, {"rel": "等式方程求解", "source": "- { b } ^ { 2 } = 0", "target": "b = 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "- { b } ^ { 2 } = 0"}, {"rel": "代入", "source": "a = 2", "target": "2"}, {"rel": "代入", "source": "b = 0", "target": "2"}, {"rel": "被代入", "source": "a + b", "target": "2"}]}}
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{"content": "Given a linear function $y = ( m - 3 ) x - 2$, where $y$ decreases as $x$ increases, the possible range of values for $m$ is ____?", "answer": "m < 3", "steps": "$\\because$ The linear function $y = ( m - 3 ) x - 2$ decreases as $x$ increases, $\\therefore$ $m - 3 < 0$, which implies $m < 3$.", "expr_cands": ["y = ( m - 3 ) x - 2", "x", "m", "y", "m - 3 < 0", "m < 3"], "exprs": ["m - 3 < 0", "m < 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m - 3 ) x - 2"}, {"id": "m - 3 < 0"}, {"id": "一次函数 $y = ( m - 3 ) x - 2$"}, {"id": "其中 $y$ 随 $x$ 的增大而减小"}, {"id": "$m$ 的取值范围"}, {"id": "m < 3"}], "links": [{"rel": "被描述", "source": "y = ( m - 3 ) x - 2", "target": "m - 3 < 0"}, {"rel": "不等式方程求解", "source": "m - 3 < 0", "target": "m < 3"}, {"rel": "限制性描述", "source": "一次函数 $y = ( m - 3 ) x - 2$", "target": "m - 3 < 0"}, {"rel": "限制性描述", "source": "其中 $y$ 随 $x$ 的增大而减小", "target": "m - 3 < 0"}, {"rel": "限制性描述", "source": "$m$ 的取值范围", "target": "m - 3 < 0"}]}}
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{"content": "When $x = - 2$, the value of the algebraic expression $x ( 2 - m ) + 4$ is $18$. What is the value of this algebraic expression when $x = 3$?", "answer": "- 17", "steps": "Substituting $x = - 2$ into $x ( 2 - m ) + 4 = 18$, we get $- 4 + 2 m + 4 = 18$. Solving for $m$, we get $m = 9$. Therefore, the algebraic expression is $- 7 x + 4$. Substituting $x = 3$ into the expression, we get $- 7 \\times 3 + 4 = - 17$.", "expr_cands": ["x = - 2", "x", "x ( 2 - m ) + 4", "m", "18", "x = 3", "x ( 2 - m ) + 4 = 18", "2 m = 18", "- 4 + 2 m + 4 = 18", "m = 9", "- 7 x + 4", "- 7 * 3 + 4", "- 17"], "exprs": ["x ( 2 - m ) + 4 = 18", "- 4 + 2 m + 4 = 18", "m = 9", "- 17"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ( 2 - m ) + 4"}, {"id": "x ( 2 - m ) + 4 = 18"}, {"id": "18"}, {"id": "代数式 $x ( 2 - m ) + 4$ 的值等于 $18$"}, {"id": "当 $x = - 2$ 时"}, {"id": "- 4 + 2 m + 4 = 18"}, {"id": "x = - 2"}, {"id": "m = 9"}, {"id": "- 17"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "x ( 2 - m ) + 4", "target": "x ( 2 - m ) + 4 = 18"}, {"rel": "被代入", "source": "x ( 2 - m ) + 4", "target": "- 17"}, {"rel": "被代入", "source": "x ( 2 - m ) + 4 = 18", "target": "- 4 + 2 m + 4 = 18"}, {"rel": "被描述", "source": "18", "target": "x ( 2 - m ) + 4 = 18"}, {"rel": "限制性描述", "source": "代数式 $x ( 2 - m ) + 4$ 的值等于 $18$", "target": "x ( 2 - m ) + 4 = 18"}, {"rel": "限制性描述", "source": "当 $x = - 2$ 时", "target": "x ( 2 - m ) + 4 = 18"}, {"rel": "等式方程求解", "source": "- 4 + 2 m + 4 = 18", "target": "m = 9"}, {"rel": "代入", "source": "x = - 2", "target": "- 4 + 2 m + 4 = 18"}, {"rel": "代入", "source": "m = 9", "target": "- 17"}, {"rel": "代入", "source": "x = 3", "target": "- 17"}]}}
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{"content": "The degree of the monomial $2 ^ { 5 } a ^ { 2 } b ^ { 5 }$ is ____ ?", "answer": "7", "steps": "The degree of the monomial $2 ^ { 5 } a ^ { 2 } b ^ { 5 }$ is $2 + 5 = 7$.", "expr_cands": ["2 ^ { 5 } a ^ { 2 } b ^ { 5 }", "b", "a", "2 + 5", "7"], "exprs": ["7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ^ { 5 } a ^ { 2 } b ^ { 5 }"}, {"id": "7"}, {"id": "单项式 $2 ^ { 5 } a ^ { 2 } b ^ { 5 }$ 的次数"}], "links": [{"rel": "被描述", "source": "2 ^ { 5 } a ^ { 2 } b ^ { 5 }", "target": "7"}, {"rel": "限制性描述", "source": "单项式 $2 ^ { 5 } a ^ { 2 } b ^ { 5 }$ 的次数", "target": "7"}]}}
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{"content": "Given $( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5$ is a one-variable linear inequality in $x$, then $a$ = ____?", "answer": "- 2", "steps": "According to the problem, $| a | - 1 = 1$, and $a - 2 \\neq 0$. Solving for $| a | = 2$ and $a \\neq 2$, we get $a = - 2$.", "expr_cands": ["( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5", "a", "x", "| a | - 1 = 1", "a = - 2", "a = 2", "a - 2 \\neq 0", "a \\neq 2", "| a | = 2"], "exprs": ["| a | - 1 = 1", "a \\neq 2", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5"}, {"id": "| a | - 1 = 1"}, {"id": "$( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5$ 是关于 $x$ 的一元一次不等式"}, {"id": "a \\neq 2"}, {"id": "a = - 2"}], "links": [{"rel": "被描述", "source": "( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5", "target": "| a | - 1 = 1"}, {"rel": "被描述", "source": "( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5", "target": "a \\neq 2"}, {"rel": "联立", "source": "| a | - 1 = 1", "target": "a = - 2"}, {"rel": "限制性描述", "source": "$( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5$ 是关于 $x$ 的一元一次不等式", "target": "| a | - 1 = 1"}, {"rel": "限制性描述", "source": "$( a - 2 ) { x } ^ { | a | - 1 } + 3 > 5$ 是关于 $x$ 的一元一次不等式", "target": "a \\neq 2"}, {"rel": "联立", "source": "a \\neq 2", "target": "a = - 2"}]}}
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{"content": "Given that the equation $\\frac { 32 } { 99 } x - m = 4 x - 2$ has a solution of $x = - 3$, what is the solution to the equation $\\frac { 32 } { 99 } ( x + 5 ) - m = 4 ( x + 5 ) - 2$?", "answer": "- 8", "steps": "$\\because$ The equation $\\frac { 32 } { 99 } x - m = 4 x - 2$ has a solution of $x = - 3$ with respect to $x$, $\\therefore$ the solution to the equation $\\frac { 32 } { 99 } ( x + 5 ) - m = 4 ( x + 5 ) - 2$ is $x + 5 = - 3$, which means $x = - 8$.", "expr_cands": ["x", "\\frac { 32 } { 99 } x - m = 4 x - 2", "m", "x = - 3", "\\frac { 32 } { 99 } ( x + 5 ) - m = 4 ( x + 5 ) - 2", "\\frac { 64 } { 99 } - m = 6", "x + 5 = - 3", "x = - 8"], "exprs": ["x + 5 = - 3", "x = - 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 32 } { 99 } ( x + 5 ) - m = 4 ( x + 5 ) - 2"}, {"id": "x + 5 = - 3"}, {"id": "\\frac { 32 } { 99 } x - m = 4 x - 2"}, {"id": "x = - 3"}, {"id": "关于 $x$ 的方程 $\\frac { 32 } { 99 } x - m = 4 x - 2$ 的解为 $x = - 3$"}, {"id": "关于 $x$ 的方程 $\\frac { 32 } { 99 } ( x + 5 ) - m = 4 ( x + 5 ) - 2$ 的解"}, {"id": "x = - 8"}], "links": [{"rel": "被描述", "source": "\\frac { 32 } { 99 } ( x + 5 ) - m = 4 ( x + 5 ) - 2", "target": "x + 5 = - 3"}, {"rel": "等式方程求解", "source": "x + 5 = - 3", "target": "x = - 8"}, {"rel": "被描述", "source": "\\frac { 32 } { 99 } x - m = 4 x - 2", "target": "x + 5 = - 3"}, {"rel": "被描述", "source": "x = - 3", "target": "x + 5 = - 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $\\frac { 32 } { 99 } x - m = 4 x - 2$ 的解为 $x = - 3$", "target": "x + 5 = - 3"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $\\frac { 32 } { 99 } ( x + 5 ) - m = 4 ( x + 5 ) - 2$ 的解", "target": "x + 5 = - 3"}]}}
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{"content": "Given that the equation $x ^ 2 - 4 x - 1 = 0$ has two roots $x _ 1$ and $x _ 2$, what is $( 1 - x _ 1 ) ( 1 - x _ 2 )$?", "answer": "- 4", "steps": "According to the problem, we have $x _ 1 + x _ 2 = 4$ and $x _ 1 x _ 2 = - 1$. Therefore, $( 1 - x _ 1 ) ( 1 - x _ 2 ) = 1 - ( x _ 1 + x _ 2 ) + x _ 1 x _ 2 = 1 - 4 + ( - 1 ) = - 4$.", "expr_cands": ["x ^ { 2 } - 4 x - 1 = 0", "x", "x _ { 1 }", "x _ { 2 }", "( 1 - x _ { 1 } ) ( 1 - x _ { 2 } )", "x _ { 1 } + x _ { 2 } = 4", "x _ { 1 } x _ { 2 } = - 1", "1 - ( x _ { 1 } + x _ { 2 } ) + x _ { 1 } x _ { 2 }", "- 4"], "exprs": ["x _ { 1 } + x _ { 2 } = 4", "x _ { 1 } x _ { 2 } = - 1", "1 - ( x _ { 1 } + x _ { 2 } ) + x _ { 1 } x _ { 2 }", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 4 x - 1 = 0"}, {"id": "x _ { 1 } + x _ { 2 } = 4"}, {"id": "x _ { 1 }"}, {"id": "x _ { 2 }"}, {"id": "方程 $x ^ { 2 } - 4 x - 1 = 0$ 的两根为 $x _ { 1 }$ , $x _ { 2 }$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "x _ { 1 } x _ { 2 } = - 1"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "( 1 - x _ { 1 } ) ( 1 - x _ { 2 } )"}, {"id": "1 - ( x _ { 1 } + x _ { 2 } ) + x _ { 1 } x _ { 2 }"}, {"id": "- 4"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 4 x - 1 = 0", "target": "x _ { 1 } + x _ { 2 } = 4"}, {"rel": "被描述", "source": "x ^ { 2 } - 4 x - 1 = 0", "target": "x _ { 1 } x _ { 2 } = - 1"}, {"rel": "提取因式参考", "source": "x _ { 1 } + x _ { 2 } = 4", "target": "1 - ( x _ { 1 } + x _ { 2 } ) + x _ { 1 } x _ { 2 }"}, {"rel": "代入", "source": "x _ { 1 } + x _ { 2 } = 4", "target": "- 4"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } + x _ { 2 } = 4"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } x _ { 2 } = - 1"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } + x _ { 2 } = 4"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } x _ { 2 } = - 1"}, {"rel": "限制性描述", "source": "方程 $x ^ { 2 } - 4 x - 1 = 0$ 的两根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "x _ { 1 } + x _ { 2 } = 4"}, {"rel": "限制性描述", "source": "方程 $x ^ { 2 } - 4 x - 1 = 0$ 的两根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "x _ { 1 } x _ { 2 } = - 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + x _ { 2 } = 4"}, {"rel": "提取因式参考", "source": "x _ { 1 } x _ { 2 } = - 1", "target": "1 - ( x _ { 1 } + x _ { 2 } ) + x _ { 1 } x _ { 2 }"}, {"rel": "代入", "source": "x _ { 1 } x _ { 2 } = - 1", "target": "- 4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "x _ { 1 } x _ { 2 } = - 1"}, {"rel": "提取因式", "source": "( 1 - x _ { 1 } ) ( 1 - x _ { 2 } )", "target": "1 - ( x _ { 1 } + x _ { 2 } ) + x _ { 1 } x _ { 2 }"}, {"rel": "被代入", "source": "1 - ( x _ { 1 } + x _ { 2 } ) + x _ { 1 } x _ { 2 }", "target": "- 4"}]}}
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{"content": "The inequality $\\frac { x - 1 } { 2 } - \\frac { x - 2 } { 4 } > 1$ after eliminating the denominators is ____ ?", "answer": "2 ( x - 1 ) - x + 2 > 4", "steps": "The inequality $\\frac { x - 1 } { 2 } - \\frac { x - 2 } { 4 } > 1$ is given. Multiplying both sides of the inequality by $4$ yields $2 ( x - 1 ) - x + 2 > 4$.", "expr_cands": ["\\frac { x - 1 } { 2 } - \\frac { x - 2 } { 4 } > 1", "x", "4 < x", "4", "2 ( x - 1 ) - x + 2 > 4"], "exprs": ["2 ( x - 1 ) - x + 2 > 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 1 } { 2 } - \\frac { x - 2 } { 4 } > 1"}, {"id": "2 ( x - 1 ) - x + 2 > 4"}], "links": [{"rel": "同乘除", "source": "\\frac { x - 1 } { 2 } - \\frac { x - 2 } { 4 } > 1", "target": "2 ( x - 1 ) - x + 2 > 4"}]}}
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{"content": "The sum of positive integer solutions that satisfy the inequality $2 m - 1 < \\frac { 3 m + 2 } { 2 }$ is ____?", "answer": "6", "steps": "To eliminate the denominator, we get $4 m - 2 < 3 m + 2$. Moving terms around, we get $4 m - 3 m < 2 + 2$, which simplifies to $m < 4$. Therefore, the sum of positive integer solutions is $1 + 2 + 3 = 6$.", "expr_cands": ["2 m - 1 < \\frac { 3 m + 2 } { 2 }", "m", "4 m - 2 < 3 m + 2", "m < 4", "4 m - 3 m < 2 + 2", "1 + 2 + 3", "6"], "exprs": ["m < 4", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 m - 1 < \\frac { 3 m + 2 } { 2 }"}, {"id": "m < 4"}, {"id": "6"}, {"id": "故正整数解和为 : $1 + 2 + 3 = 6$"}], "links": [{"rel": "不等式方程求解", "source": "2 m - 1 < \\frac { 3 m + 2 } { 2 }", "target": "m < 4"}, {"rel": "被描述", "source": "m < 4", "target": "6"}, {"rel": "限制性描述", "source": "故正整数解和为 : $1 + 2 + 3 = 6$", "target": "6"}]}}
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{"content": "The domain of the function $y = \\frac { 2 } { x - 1 }$ is _____.", "answer": "x \\neq 1", "steps": "According to the problem, we have $x - 1 \\neq 0$, which implies that $x \\neq 1$.", "expr_cands": ["y = \\frac { 2 } { x - 1 }", "x", "y", "x - 1 \\neq 0", "x \\neq 1"], "exprs": ["x - 1 \\neq 0", "x \\neq 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 2 } { x - 1 }"}, {"id": "x - 1 \\neq 0"}, {"id": "函数 $y = \\frac { 2 } { x - 1 }$ 的定义域"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 1"}], "links": [{"rel": "被描述", "source": "y = \\frac { 2 } { x - 1 }", "target": "x - 1 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 1 \\neq 0", "target": "x \\neq 1"}, {"rel": "限制性描述", "source": "函数 $y = \\frac { 2 } { x - 1 }$ 的定义域", "target": "x - 1 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 1 \\neq 0"}]}}
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{"content": "If the solution to the equation $3 x - 2 a = 12$ is $x = - 2$, then the value of $a$ is ____?", "answer": "- 9", "steps": "Substituting $x = - 2$ into the equation gives: $- 6 - 2 a = 12$, solving for $a$ gives: $a = - 9$.", "expr_cands": ["{ 3 } x - 2 a = 12", "a", "x", "x = - 2", "- 6 - 2 a = 12", "a = - 9"], "exprs": ["- 6 - 2 a = 12", "a = - 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ 3 } x - 2 a = 12"}, {"id": "- 6 - 2 a = 12"}, {"id": "x = - 2"}, {"id": "a = - 9"}], "links": [{"rel": "被代入", "source": "{ 3 } x - 2 a = 12", "target": "- 6 - 2 a = 12"}, {"rel": "等式方程求解", "source": "- 6 - 2 a = 12", "target": "a = - 9"}, {"rel": "代入", "source": "x = - 2", "target": "- 6 - 2 a = 12"}]}}
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{"content": "When $x$ is equal to what, the value of the algebraic expression $x ^ 2 + 4 x$ is the opposite of the value of the algebraic expression $1 - 2 x$?", "answer": "- 1", "steps": "According to the problem, we have $x ^ 2 + 4 x + 1 - 2 x = 0$, which simplifies to $x ^ 2 + 2 x + 1 = 0$. Therefore, $( x + 1 ) ^ 2 = 0$, and we have $x = - 1$.", "expr_cands": ["x", "x ^ { 2 } + 4 x", "1 - 2 x", "x ^ { 2 } + 4 x + 1 - 2 x = 0", "x = - 1", "x ^ { 2 } + 2 x + 1 = 0", "( x + 1 ) ^ { 2 } = 0"], "exprs": ["x ^ { 2 } + 4 x + 1 - 2 x = 0", "x = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + 4 x"}, {"id": "x ^ { 2 } + 4 x + 1 - 2 x = 0"}, {"id": "1 - 2 x"}, {"id": "代数式 $x ^ { 2 } + 4 x$ 的值与代数式 $1 - 2 x$ 的值互为相反数"}, {"id": "x = - 1"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + 4 x", "target": "x ^ { 2 } + 4 x + 1 - 2 x = 0"}, {"rel": "等式方程求解", "source": "x ^ { 2 } + 4 x + 1 - 2 x = 0", "target": "x = - 1"}, {"rel": "被描述", "source": "1 - 2 x", "target": "x ^ { 2 } + 4 x + 1 - 2 x = 0"}, {"rel": "限制性描述", "source": "代数式 $x ^ { 2 } + 4 x$ 的值与代数式 $1 - 2 x$ 的值互为相反数", "target": "x ^ { 2 } + 4 x + 1 - 2 x = 0"}]}}
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{"content": "$N$ is a non-zero integer, and the smallest natural number $x$ that satisfies $180 x = N ^ 2$ is _____.", "answer": "5", "steps": "$\\sqrt { 180 } = 6 \\sqrt { 5 }$ , then the smallest natural number $x$ that makes $180 x = N ^ { 2 }$ true is $x = 5$.", "expr_cands": ["N", "0", "180 x = N ^ { 2 }", "x", "\\sqrt { 180 }", "6 \\sqrt { 5 }", "x = 5"], "exprs": ["6 \\sqrt { 5 }", "x = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 180 }"}, {"id": "6 \\sqrt { 5 }"}, {"id": "180 x = N ^ { 2 }"}, {"id": "x = 5"}, {"id": "使 $180 x = N ^ { 2 }$ 成立的最小自然数"}], "links": [{"rel": "计算", "source": "\\sqrt { 180 }", "target": "6 \\sqrt { 5 }"}, {"rel": "被描述", "source": "6 \\sqrt { 5 }", "target": "x = 5"}, {"rel": "被描述", "source": "180 x = N ^ { 2 }", "target": "x = 5"}, {"rel": "限制性描述", "source": "使 $180 x = N ^ { 2 }$ 成立的最小自然数", "target": "x = 5"}]}}
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{"content": "If $m$ satisfies the equation $| 2019 - m | = 2019 + | m |$, then $| m - 2020 |$ is equal to ____?", "answer": "- m + 2020", "steps": "Because $| 2019 - m | = 2019 + | m |$, therefore $m \\le 0$, therefore $m - 2020 < 0$, therefore $| m - 2020 | = - m + 2020$.", "expr_cands": ["m", "| 2019 - m | = 2019 + | m |", "| m - 2020 |", "m \\le 0", "m - 2020 < 0", "m < 2020", "| m - 2020 | = - m + 2020", "- m + 2020"], "exprs": ["m \\le 0", "- m + 2020"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| 2019 - m | = 2019 + | m |"}, {"id": "m \\le 0"}, {"id": "m"}, {"id": "绝对值恒大于等于0"}, {"id": "$m$ 满足方程 $| 2019 - m | = 2019 + | m |$"}, {"id": "| m - 2020 |"}, {"id": "- m + 2020"}], "links": [{"rel": "被描述", "source": "| 2019 - m | = 2019 + | m |", "target": "m \\le 0"}, {"rel": "被描述", "source": "m \\le 0", "target": "- m + 2020"}, {"rel": "被描述", "source": "m", "target": "m \\le 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m \\le 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "- m + 2020"}, {"rel": "限制性描述", "source": "$m$ 满足方程 $| 2019 - m | = 2019 + | m |$", "target": "m \\le 0"}, {"rel": "被描述", "source": "| m - 2020 |", "target": "- m + 2020"}]}}
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{"content": "Given that $y$ is an inverse proportion function of $x$, when $x = 3$, $y = 2$, the functional relationship between $y$ and $x$ is ____?", "answer": "y = \\frac { 6 } { x }", "steps": "The analytical expression of the inverse proportion function is $y = \\frac { k } { x } ( k \\neq 0 )$. Since $x = 3$ and $y = 2$, we have $2 = \\frac { k } { 3 }$, which gives $k = 6$. Therefore, the analytical expression of the inverse proportion function is $y = \\frac { 6 } { x }$.", "expr_cands": ["y", "x", "x = 3", "y = 2", "y = \\frac { k } { x } ( k \\neq 0 )", "k", "2 = \\frac { k } { 3 }", "k = 6", "y = \\frac { 6 } { x }"], "exprs": ["y = \\frac { k } { x } ( k \\neq 0 )", "2 = \\frac { k } { 3 }", "k = 6", "y = \\frac { 6 } { x }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设反比例函数的解析式为 $y = \\frac { k } { x } ( k \\neq 0 )$"}, {"id": "y = \\frac { k } { x } ( k \\neq 0 )"}, {"id": "x = 3"}, {"id": "2 = \\frac { k } { 3 }"}, {"id": "y = 2"}, {"id": "k = 6"}, {"id": "y = \\frac { 6 } { x }"}], "links": [{"rel": "假设描述", "source": "设反比例函数的解析式为 $y = \\frac { k } { x } ( k \\neq 0 )$", "target": "y = \\frac { k } { x } ( k \\neq 0 )"}, {"rel": "被代入", "source": "y = \\frac { k } { x } ( k \\neq 0 )", "target": "2 = \\frac { k } { 3 }"}, {"rel": "被代入", "source": "y = \\frac { k } { x } ( k \\neq 0 )", "target": "y = \\frac { 6 } { x }"}, {"rel": "代入", "source": "x = 3", "target": "2 = \\frac { k } { 3 }"}, {"rel": "等式方程求解", "source": "2 = \\frac { k } { 3 }", "target": "k = 6"}, {"rel": "代入", "source": "y = 2", "target": "2 = \\frac { k } { 3 }"}, {"rel": "代入", "source": "k = 6", "target": "y = \\frac { 6 } { x }"}]}}
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{"content": "If the solution to the fractional equation $\\frac { ax } { x + 2 } = 2$ is $x = 2$, then the value of $a$ is ____?", "answer": "4", "steps": "Substituting $x = 2$ into the original equation yields $\\frac { 2 a } { 2 + 2 } = 2$, which gives the solution $a = 4$.", "expr_cands": ["\\frac { ax } { x + 2 } = 2", "a", "x", "x = 2", "\\frac { 2 a } { 2 + 2 } = 2", "a = 4"], "exprs": ["\\frac { 2 a } { 2 + 2 } = 2", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { ax } { x + 2 } = 2"}, {"id": "\\frac { 2 a } { 2 + 2 } = 2"}, {"id": "x = 2"}, {"id": "a = 4"}], "links": [{"rel": "被代入", "source": "\\frac { ax } { x + 2 } = 2", "target": "\\frac { 2 a } { 2 + 2 } = 2"}, {"rel": "等式方程求解", "source": "\\frac { 2 a } { 2 + 2 } = 2", "target": "a = 4"}, {"rel": "代入", "source": "x = 2", "target": "\\frac { 2 a } { 2 + 2 } = 2"}]}}
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{"content": "If the parabola $y = x ^ { 2 } - 2 x + m$ has a common point with the $x$-axis, then the range of values for $m$ is ____?", "answer": "m \\le 1", "steps": "$\\because$ The parabola $y = x ^ 2 - 2 x + m$ has a common point with the $x$-axis, $\\therefore ( - 2 ) ^ 2 - 4 \\cdot 1 \\cdot m \\ge 0$, which yields $m \\le 1$.", "expr_cands": ["y = x ^ { 2 } - 2 x + m", "m", "y", "x", "( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0", "m \\le 1"], "exprs": ["( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0", "m \\le 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 } - 2 x + m"}, {"id": "( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0"}, {"id": "x"}, {"id": "抛物线 $y = x ^ { 2 } - 2 x + m$ 与 $x$ 轴有公共点"}, {"id": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0"}, {"id": "m \\le 1"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 } - 2 x + m", "target": "( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0"}, {"rel": "不等式方程求解", "source": "( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0", "target": "m \\le 1"}, {"rel": "被描述", "source": "x", "target": "( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0"}, {"rel": "限制性描述", "source": "抛物线 $y = x ^ { 2 } - 2 x + m$ 与 $x$ 轴有公共点", "target": "( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0", "target": "( - 2 ) ^ { 2 } - 4 * 1 * m \\ge 0"}]}}
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{"content": "The coefficient of $x ^ 2$ in the result of $( x + 1 ) ( 2 x ^ 2 + ax + 1 )$ is $- 1$. What is the value of $a$?", "answer": "- 3", "steps": "$( x + 1 ) ( 2 x ^ 2 + ax + 1 ) = 2 x ^ 3 + ax ^ 2 + x + 2 x ^ 2 + ax + 1 = 2 x ^ 3 + ( a + 2 ) x ^ 2 + ( 1 + a ) x + 1$; since the coefficient of $x ^ 2$ in the result is $- 1$, therefore $a + 2 = - 1$, which solves to $a = - 3$.", "expr_cands": ["( x + 1 ) ( 2 x ^ { 2 } + ax + 1 )", "a", "x", "x ^ { 2 }", "- 1", "2 x ^ { 3 } + ( a + 2 ) x ^ { 2 } + ( 1 + a ) x + 1", "a + 2 = - 1", "a = - 3"], "exprs": ["2 x ^ { 3 } + ( a + 2 ) x ^ { 2 } + ( 1 + a ) x + 1", "a + 2 = - 1", "a = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + 1 ) ( 2 x ^ { 2 } + ax + 1 )"}, {"id": "2 x ^ { 3 } + ( a + 2 ) x ^ { 2 } + ( 1 + a ) x + 1"}, {"id": "x"}, {"id": "a + 2 = - 1"}, {"id": "- 1"}, {"id": "在 $( x + 1 ) ( 2 x ^ { 2 } + ax + 1 )$ 的运算结果中 $x ^ { 2 }$ 的系数是 $- 1$"}, {"id": "a = - 3"}], "links": [{"rel": "提取因式", "source": "( x + 1 ) ( 2 x ^ { 2 } + ax + 1 )", "target": "2 x ^ { 3 } + ( a + 2 ) x ^ { 2 } + ( 1 + a ) x + 1"}, {"rel": "被描述", "source": "2 x ^ { 3 } + ( a + 2 ) x ^ { 2 } + ( 1 + a ) x + 1", "target": "a + 2 = - 1"}, {"rel": "提取因式参考", "source": "x", "target": "2 x ^ { 3 } + ( a + 2 ) x ^ { 2 } + ( 1 + a ) x + 1"}, {"rel": "等式方程求解", "source": "a + 2 = - 1", "target": "a = - 3"}, {"rel": "被描述", "source": "- 1", "target": "a + 2 = - 1"}, {"rel": "限制性描述", "source": "在 $( x + 1 ) ( 2 x ^ { 2 } + ax + 1 )$ 的运算结果中 $x ^ { 2 }$ 的系数是 $- 1$", "target": "a + 2 = - 1"}]}}
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{"content": "Given $a ^ { 2 } - 2 b - 3 = 0$, find the value of the polynomial $4 a ^ { 2 } - 8 b + 5$ is ____?", "answer": "17", "steps": "Since $a ^ 2 - 2 b - 3 = 0$, it follows that $a ^ 2 - 2 b = 3$. Therefore, $4 a ^ 2 - 8 b + 5 = 4 * ( a ^ 2 - 2 b ) + 5 = 17$.", "expr_cands": ["a ^ { 2 } - 2 b - 3 = 0", "b", "a", "4 a ^ { 2 } - 8 b + 5", "a ^ { 2 } - 2 b = 3", "4 * ( a ^ { 2 } - 2 b ) + 5", "17"], "exprs": ["a ^ { 2 } - 2 b = 3", "4 * ( a ^ { 2 } - 2 b ) + 5", "17"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } - 2 b - 3 = 0"}, {"id": "a ^ { 2 } - 2 b = 3"}, {"id": "4 a ^ { 2 } - 8 b + 5"}, {"id": "4 * ( a ^ { 2 } - 2 b ) + 5"}, {"id": "17"}], "links": [{"rel": "移项", "source": "a ^ { 2 } - 2 b - 3 = 0", "target": "a ^ { 2 } - 2 b = 3"}, {"rel": "提取因式参考", "source": "a ^ { 2 } - 2 b = 3", "target": "4 * ( a ^ { 2 } - 2 b ) + 5"}, {"rel": "代入", "source": "a ^ { 2 } - 2 b = 3", "target": "17"}, {"rel": "提取因式", "source": "4 a ^ { 2 } - 8 b + 5", "target": "4 * ( a ^ { 2 } - 2 b ) + 5"}, {"rel": "被代入", "source": "4 * ( a ^ { 2 } - 2 b ) + 5", "target": "17"}]}}
|
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{"content": "The equation $2 { x } ^ { 2 } - 4 x + m = 0$ has one root as $- 1$, then the other root is ____?", "answer": "3", "steps": "Let the other root of the equation $2 x ^ 2 - 4 x + m = 0$ be $a$. Since the equation has a root of $- 1$, we have $a + ( - 1 ) = 2$. Solving for $a$, we get $a = 3$.", "expr_cands": ["2 { x } ^ { 2 } - 4 x + m = 0", "m", "x", "- 1", "2 x ^ { 2 } - 4 x + m = 0", "a", "a + ( - 1 ) = 2", "a = 3"], "exprs": ["a", "a + ( - 1 ) = 2", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程 $2 x ^ { 2 } - 4 x + m = 0$ 的另一个根是 $a$"}, {"id": "a"}, {"id": "- 1"}, {"id": "a + ( - 1 ) = 2"}, {"id": "2 x ^ { 2 } - 4 x + m = 0"}, {"id": "方程 $2 { x } ^ { 2 } - 4 x + m = 0$ 有一个根为 $- 1$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "a = 3"}], "links": [{"rel": "假设描述", "source": "设方程 $2 x ^ { 2 } - 4 x + m = 0$ 的另一个根是 $a$", "target": "a"}, {"rel": "假设描述", "source": "设方程 $2 x ^ { 2 } - 4 x + m = 0$ 的另一个根是 $a$", "target": "a + ( - 1 ) = 2"}, {"rel": "被描述", "source": "a", "target": "a + ( - 1 ) = 2"}, {"rel": "被描述", "source": "- 1", "target": "a + ( - 1 ) = 2"}, {"rel": "等式方程求解", "source": "a + ( - 1 ) = 2", "target": "a = 3"}, {"rel": "被描述", "source": "2 x ^ { 2 } - 4 x + m = 0", "target": "a + ( - 1 ) = 2"}, {"rel": "限制性描述", "source": "方程 $2 { x } ^ { 2 } - 4 x + m = 0$ 有一个根为 $- 1$", "target": "a + ( - 1 ) = 2"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "a + ( - 1 ) = 2"}]}}
|
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{"content": "Given the polynomial ${ x } ^ { 2 } - 2 kxy - 3 ( { x } ^ { 2 } - 12 xy + x )$ does not contain any product terms of $x$ and $y$. Find the value of $k$.", "answer": "18", "steps": "Original expression = $x ^ { 2 } - 2 kxy - 3 x ^ { 2 } + 36 xy - 3 x = - 2 x ^ { 2 } + ( 36 - 2 k ) xy - 3 x$ . Since the result does not contain a product term of $x$ and $y$, we obtain $36 - 2 k = 0$, and solve for $k = 18$.", "expr_cands": ["{ x } ^ { 2 } - 2 kxy - 3 ( { x } ^ { 2 } - 12 xy + x )", "x", "y", "k", "x ^ { 2 } - 2 kxy - 3 x ^ { 2 } + 36 xy - 3 x", "- 2 x ^ { 2 } + ( 36 - 2 k ) xy - 3 x", "36 - 2 k = 0", "k = 18"], "exprs": ["x ^ { 2 } - 2 kxy - 3 x ^ { 2 } + 36 xy - 3 x", "36 - 2 k = 0", "k = 18"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { 2 } - 2 kxy - 3 ( { x } ^ { 2 } - 12 xy + x )"}, {"id": "x ^ { 2 } - 2 kxy - 3 x ^ { 2 } + 36 xy - 3 x"}, {"id": "36 - 2 k = 0"}, {"id": "多项式 ${ x } ^ { 2 } - 2 kxy - 3 ( { x } ^ { 2 } - 12 xy + x )$ 不含 $x$ , $y$ 的乘积项"}, {"id": "k = 18"}], "links": [{"rel": "展开", "source": "{ x } ^ { 2 } - 2 kxy - 3 ( { x } ^ { 2 } - 12 xy + x )", "target": "x ^ { 2 } - 2 kxy - 3 x ^ { 2 } + 36 xy - 3 x"}, {"rel": "被描述", "source": "x ^ { 2 } - 2 kxy - 3 x ^ { 2 } + 36 xy - 3 x", "target": "36 - 2 k = 0"}, {"rel": "等式方程求解", "source": "36 - 2 k = 0", "target": "k = 18"}, {"rel": "限制性描述", "source": "多项式 ${ x } ^ { 2 } - 2 kxy - 3 ( { x } ^ { 2 } - 12 xy + x )$ 不含 $x$ , $y$ 的乘积项", "target": "36 - 2 k = 0"}]}}
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{"content": "The equation $ax + 2 bx = 3$ has a solution of $x = 1$. What is the solution to the equation $a ( y - 1 ) + 2 b ( y - 1 ) = 3$?", "answer": "y = 2", "steps": "From the given information, we have $a + 2 b = 3$ and $a ( y - 1 ) + 2 b ( y - 1 ) = 3$. Simplifying the second equation, we get $( a + 2 b ) y - ( a + 2 b ) = 3$, which can be further simplified to $3 y - 3 = 3$. Therefore, $y = 2$.", "expr_cands": ["ax + 2 bx = 3", "a", "x", "b", "x = 1", "a ( y - 1 ) + 2 b ( y - 1 ) = 3", "y", "a + 2 b = 3", "( a + 2 b ) y - ( a + 2 b ) = 3", "3 y - 3 = 3", "y = 2"], "exprs": ["a + 2 b = 3", "( a + 2 b ) y - ( a + 2 b ) = 3", "3 y - 3 = 3", "y = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ax + 2 bx = 3"}, {"id": "a + 2 b = 3"}, {"id": "x = 1"}, {"id": "a ( y - 1 ) + 2 b ( y - 1 ) = 3"}, {"id": "( a + 2 b ) y - ( a + 2 b ) = 3"}, {"id": "3 y - 3 = 3"}, {"id": "y = 2"}], "links": [{"rel": "被代入", "source": "ax + 2 bx = 3", "target": "a + 2 b = 3"}, {"rel": "提取因式参考", "source": "a + 2 b = 3", "target": "( a + 2 b ) y - ( a + 2 b ) = 3"}, {"rel": "代入", "source": "a + 2 b = 3", "target": "3 y - 3 = 3"}, {"rel": "代入", "source": "x = 1", "target": "a + 2 b = 3"}, {"rel": "提取因式", "source": "a ( y - 1 ) + 2 b ( y - 1 ) = 3", "target": "( a + 2 b ) y - ( a + 2 b ) = 3"}, {"rel": "被代入", "source": "( a + 2 b ) y - ( a + 2 b ) = 3", "target": "3 y - 3 = 3"}, {"rel": "等式方程求解", "source": "3 y - 3 = 3", "target": "y = 2"}]}}
|
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{"content": "Given that the opposite of $a + 3$ is ____?", "answer": "- a - 3", "steps": "The opposite of $a + 3$ is $- ( a + 3 ) = - a - 3$.", "expr_cands": ["a + 3", "a", "- ( a + 3 )", "- a - 3"], "exprs": ["- ( a + 3 )", "- a - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a + 3"}, {"id": "- ( a + 3 )"}, {"id": "$a + 3$ 的相反数"}, {"id": "- a - 3"}], "links": [{"rel": "被描述", "source": "a + 3", "target": "- ( a + 3 )"}, {"rel": "计算", "source": "- ( a + 3 )", "target": "- a - 3"}, {"rel": "限制性描述", "source": "$a + 3$ 的相反数", "target": "- ( a + 3 )"}]}}
|
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{"content": "Given that $m$ and $n$ are the roots of the equation ${ x } ^ { 2 } - 2 x - 1 = 0$, what is the value of $2 mn - m - n$?", "answer": "- 4", "steps": "According to the relationship between the root and the coefficient, we have: $m + n = 2$, $mn = - 1$, $2 mn - m - n = 2 mn - ( m + n ) = - 2 - 2 = - 4$.", "expr_cands": ["m", "n", "{ x } ^ { 2 } - 2 x - 1 = 0", "x", "2 mn - m - n", "m + n = 2", "mn = - 1", "- 4"], "exprs": ["m + n = 2", "mn = - 1", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { 2 } - 2 x - 1 = 0"}, {"id": "m + n = 2"}, {"id": "m"}, {"id": "n"}, {"id": "$m$ , $n$ 是方程 ${ x } ^ { 2 } - 2 x - 1 = 0$ 的两根"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "mn = - 1"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "2 mn - m - n"}, {"id": "- 4"}], "links": [{"rel": "被描述", "source": "{ x } ^ { 2 } - 2 x - 1 = 0", "target": "m + n = 2"}, {"rel": "被描述", "source": "{ x } ^ { 2 } - 2 x - 1 = 0", "target": "mn = - 1"}, {"rel": "代入", "source": "m + n = 2", "target": "- 4"}, {"rel": "被描述", "source": "m", "target": "m + n = 2"}, {"rel": "被描述", "source": "m", "target": "mn = - 1"}, {"rel": "被描述", "source": "n", "target": "m + n = 2"}, {"rel": "被描述", "source": "n", "target": "mn = - 1"}, {"rel": "限制性描述", "source": "$m$ , $n$ 是方程 ${ x } ^ { 2 } - 2 x - 1 = 0$ 的两根", "target": "m + n = 2"}, {"rel": "限制性描述", "source": "$m$ , $n$ 是方程 ${ x } ^ { 2 } - 2 x - 1 = 0$ 的两根", "target": "mn = - 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "m + n = 2"}, {"rel": "代入", "source": "mn = - 1", "target": "- 4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "mn = - 1"}, {"rel": "被代入", "source": "2 mn - m - n", "target": "- 4"}]}}
|
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{"content": "If $2 ^ m = 2$, $2 ^ n = 4$, then $2 ^ { m - n }$ is equal to ____?", "answer": "\\frac { 1 } { 2 }", "steps": "Because $2 ^ m = 2$ and $2 ^ n = 4$, therefore $2 ^ { m - n } = 2 ^ m \\div 2 ^ n = 2 \\div 4 = \\frac { 1 } { 2 }$.", "expr_cands": ["2 ^ { m } = 2", "m", "2 ^ { n } = 4", "n", "2 ^ { m - n }", "m = 1", "n = 2", "\\frac { 1 } { 2 }"], "exprs": ["\\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ^ { m - n }"}, {"id": "\\frac { 1 } { 2 }"}, {"id": "2 ^ { m } = 2"}, {"id": "2 ^ { n } = 4"}], "links": [{"rel": "被代入", "source": "2 ^ { m - n }", "target": "\\frac { 1 } { 2 }"}, {"rel": "代入", "source": "2 ^ { m } = 2", "target": "\\frac { 1 } { 2 }"}, {"rel": "代入", "source": "2 ^ { n } = 4", "target": "\\frac { 1 } { 2 }"}]}}
|
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{"content": "Given $a ^ { m } = 4$ and $a ^ { n } = \\frac { 1 } { 2 }$, what is $a ^ { 2 m - 2 n }$?", "answer": "64", "steps": "Since $a ^ { m } = 4$ and $a ^ { n } = \\frac { 1 } { 2 }$, therefore $a ^ { 2 m - 2 n } = ( a ^ { m } ) ^ { 2 } \\div ( a ^ { n } ) ^ { 2 } = 4 ^ { 2 } \\div ( \\frac { 1 } { 2 } ) ^ { 2 } = 16 \\div \\frac { 1 } { 4 } = 64$.", "expr_cands": ["a ^ { m } = 4", "m", "a", "a ^ { n } = \\frac { 1 } { 2 }", "n", "a ^ { 2 m - 2 n }", "64"], "exprs": ["64"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 m - 2 n }"}, {"id": "64"}, {"id": "a ^ { m } = 4"}, {"id": "a ^ { n } = \\frac { 1 } { 2 }"}], "links": [{"rel": "被代入", "source": "a ^ { 2 m - 2 n }", "target": "64"}, {"rel": "代入", "source": "a ^ { m } = 4", "target": "64"}, {"rel": "代入", "source": "a ^ { n } = \\frac { 1 } { 2 }", "target": "64"}]}}
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{"content": "The solution to the equation $\\frac { x + 1 } { 2 } + \\frac { x + 3 } { 6 } = 3$ is ____ ?", "answer": "x = 3", "steps": "Going to the denominator, we get $3 ( x + 1 ) + ( x + 3 ) = 18$. Expanding the brackets, we get $3 x + 3 + x + 3 = 18$. Rearranging, we get $3 x + x = 18 - 3 - 3$. Combining like terms, we get $4 x = 12$. Dividing by the coefficient, we get $x = 3$.", "expr_cands": ["\\frac { x + 1 } { 2 } + \\frac { x + 3 } { 6 } = 3", "x", "3 ( x + 1 ) + ( x + 3 ) = 18", "x = 3", "3 x + 3 + x + 3 = 18", "3 x + x = 18 - 3 - 3", "4 x = 12", "1"], "exprs": ["x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x + 1 } { 2 } + \\frac { x + 3 } { 6 } = 3"}, {"id": "x = 3"}], "links": [{"rel": "等式方程求解", "source": "\\frac { x + 1 } { 2 } + \\frac { x + 3 } { 6 } = 3", "target": "x = 3"}]}}
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{"content": "Given $5 x ^ { 1 + m } y ^ { 4 }$ and $x ^ { 3 } y ^ { 4 }$ are similar terms, what is the value of $m$?", "answer": "2", "steps": "$\\because$ $5 x ^ { 1 + m } y ^ { 4 }$ and $x ^ { 3 } y ^ { 4 }$ are similar terms, $\\therefore$ $1 + m = 3$, which solves for $m = 2$.", "expr_cands": ["5 x ^ { 1 + m } y ^ { 4 }", "m", "y", "x", "x ^ { 3 } y ^ { 4 }", "1 + m = 3", "m = 2"], "exprs": ["1 + m = 3", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x ^ { 1 + m } y ^ { 4 }"}, {"id": "1 + m = 3"}, {"id": "x ^ { 3 } y ^ { 4 }"}, {"id": "$5 x ^ { 1 + m } y ^ { 4 }$ 与 $x ^ { 3 } y ^ { 4 }$ 是同类项"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "5 x ^ { 1 + m } y ^ { 4 }", "target": "1 + m = 3"}, {"rel": "等式方程求解", "source": "1 + m = 3", "target": "m = 2"}, {"rel": "被描述", "source": "x ^ { 3 } y ^ { 4 }", "target": "1 + m = 3"}, {"rel": "限制性描述", "source": "$5 x ^ { 1 + m } y ^ { 4 }$ 与 $x ^ { 3 } y ^ { 4 }$ 是同类项", "target": "1 + m = 3"}]}}
|
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{"content": "Given that $x = - 2$, $y = 1$ is a solution to the system of equations $3 x + 5 y - k = 1$, what is the value of $2 k - 1$?", "answer": "- 5", "steps": "Substituting $x = - 2$ and $y = 1$ into the linear equation $3 x + 5 y - k = 1$, we get $- 6 + 5 - k = 1$. Solving for $k$, we get $k = - 2$. Therefore, $2 k - 1 = - 4 - 1 = - 5$.", "expr_cands": ["x = - 2", "x", "y = 1", "y", "3 x + 5 y - k = 1", "k", "2 k - 1", "- k - 1 = 1", "- 6 + 5 - k = 1", "k = - 2", "- 5"], "exprs": ["- 6 + 5 - k = 1", "k = - 2", "- 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + 5 y - k = 1"}, {"id": "- 6 + 5 - k = 1"}, {"id": "x = - 2"}, {"id": "y = 1"}, {"id": "k = - 2"}, {"id": "2 k - 1"}, {"id": "- 5"}], "links": [{"rel": "被代入", "source": "3 x + 5 y - k = 1", "target": "- 6 + 5 - k = 1"}, {"rel": "等式方程求解", "source": "- 6 + 5 - k = 1", "target": "k = - 2"}, {"rel": "代入", "source": "x = - 2", "target": "- 6 + 5 - k = 1"}, {"rel": "代入", "source": "y = 1", "target": "- 6 + 5 - k = 1"}, {"rel": "代入", "source": "k = - 2", "target": "- 5"}, {"rel": "被代入", "source": "2 k - 1", "target": "- 5"}]}}
|
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{"content": "Given a quadratic equation in one variable $x$, $x ^ 2 - 3 x + k + 1 = 0$, and the product of its two roots is $- 4$. The value of $k$ is ____?", "answer": "- 5", "steps": "$\\because$ For the quadratic equation in one variable $x$, $x ^ 2 - 3 x + k + 1 = 0$, the product of its two roots is $- 4$, $\\therefore$ $k + 1 = - 4$, $\\therefore$ $k = - 5$.", "expr_cands": ["x", "x ^ { 2 } - 3 x + k + 1 = 0", "k", "- 4", "k + 1 = - 4", "k = - 5"], "exprs": ["k + 1 = - 4", "k = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 x + k + 1 = 0"}, {"id": "k + 1 = - 4"}, {"id": "- 4"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 3 x + k + 1 = 0$"}, {"id": "它的两根之积为 $- 4$"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "k = - 5"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 3 x + k + 1 = 0", "target": "k + 1 = - 4"}, {"rel": "等式方程求解", "source": "k + 1 = - 4", "target": "k = - 5"}, {"rel": "被描述", "source": "- 4", "target": "k + 1 = - 4"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 3 x + k + 1 = 0$", "target": "k + 1 = - 4"}, {"rel": "限制性描述", "source": "它的两根之积为 $- 4$", "target": "k + 1 = - 4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "k + 1 = - 4"}]}}
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{"content": "Suppose $P = x ^ { 2 } - 3 xy$ , $Q = 3 xy - 9 y ^ { 2 }$ , if $P = Q$ , then the value of $\\frac { x } { y }$ is ____ ?", "answer": "3", "steps": "Since $P = x ^ 2 - 3 xy$ and $Q = 3 xy - 9 y ^ 2$ and $P = Q$, we have $x ^ 2 - 3 xy = 3 xy - 9 y ^ 2$. Simplifying this equation, we get $x ^ 2 - 6 xy + 9 y ^ 2 = 0$. Factoring this equation, we get $( x - 3 y ) ^ 2 = 0$. Therefore, $x = 3 y$. Thus, $\\frac { x } { y } = \\frac { 3 y } { y } = 3$.", "expr_cands": ["P = x ^ { 2 } - 3 xy", "x", "y", "P", "Q = 3 xy - 9 y ^ { 2 }", "Q", "P = Q", "\\frac { x } { y }", "x ^ { 2 } - 3 x y = 3 x y - 9 y ^ { 2 }", "x ^ { 2 } - 3 xy = 3 xy - 9 y ^ { 2 }", "x ^ { 2 } - 6 xy + 9 y ^ { 2 } = 0", "{ ( x - 3 y ) } ^ { 2 } = 0", "x = 3 y", "3"], "exprs": ["x ^ { 2 } - 3 xy = 3 xy - 9 y ^ { 2 }", "x ^ { 2 } - 6 xy + 9 y ^ { 2 } = 0", "{ ( x - 3 y ) } ^ { 2 } = 0", "x = 3 y", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "Q = 3 xy - 9 y ^ { 2 }"}, {"id": "x ^ { 2 } - 3 xy = 3 xy - 9 y ^ { 2 }"}, {"id": "P = Q"}, {"id": "P = x ^ { 2 } - 3 xy"}, {"id": "x ^ { 2 } - 6 xy + 9 y ^ { 2 } = 0"}, {"id": "{ ( x - 3 y ) } ^ { 2 } = 0"}, {"id": "x = 3 y"}, {"id": "\\frac { x } { y }"}, {"id": "3"}], "links": [{"rel": "代入", "source": "Q = 3 xy - 9 y ^ { 2 }", "target": "x ^ { 2 } - 3 xy = 3 xy - 9 y ^ { 2 }"}, {"rel": "移项", "source": "x ^ { 2 } - 3 xy = 3 xy - 9 y ^ { 2 }", "target": "x ^ { 2 } - 6 xy + 9 y ^ { 2 } = 0"}, {"rel": "被代入", "source": "P = Q", "target": "x ^ { 2 } - 3 xy = 3 xy - 9 y ^ { 2 }"}, {"rel": "代入", "source": "P = x ^ { 2 } - 3 xy", "target": "x ^ { 2 } - 3 xy = 3 xy - 9 y ^ { 2 }"}, {"rel": "提取因式", "source": "x ^ { 2 } - 6 xy + 9 y ^ { 2 } = 0", "target": "{ ( x - 3 y ) } ^ { 2 } = 0"}, {"rel": "等式方程部分求解", "source": "{ ( x - 3 y ) } ^ { 2 } = 0", "target": "x = 3 y"}, {"rel": "代入", "source": "x = 3 y", "target": "3"}, {"rel": "被代入", "source": "\\frac { x } { y }", "target": "3"}]}}
|
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{"content": "If the polynomial $5 x ^ { 2 } + 2 x - 2$ multiplied by the polynomial $ax + 1$ does not contain the term $x ^ { 2 }$, then the constant $a$ = ____ ?", "answer": "- \\frac { 5 } { 2 }", "steps": "According to the problem, we have $( 5 x ^ 2 + 2 x - 2 ) ( ax + 1 ) = 5 ax ^ 3 + ( 5 + 2 a ) x ^ 2 + 2 x - 2 ax - 2$. Since the result does not contain the term $x ^ 2$, we get $5 + 2 a = 0$, which gives us $a = - \\frac { 5 } { 2 }$.", "expr_cands": ["5 x ^ { 2 } + 2 x - 2", "x", "ax + 1", "a", "x ^ { 2 }", "( 5 x ^ { 2 } + 2 x - 2 ) ( ax + 1 )", "5 ax ^ { 3 } + ( 5 + 2 a ) x ^ { 2 } + 2 x - 2 ax - 2", "5 + 2 a = 0", "a = - \\frac { 5 } { 2 }"], "exprs": ["( 5 x ^ { 2 } + 2 x - 2 ) ( ax + 1 )", "5 ax ^ { 3 } + ( 5 + 2 a ) x ^ { 2 } + 2 x - 2 ax - 2", "5 + 2 a = 0", "a = - \\frac { 5 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x ^ { 2 } + 2 x - 2"}, {"id": "( 5 x ^ { 2 } + 2 x - 2 ) ( ax + 1 )"}, {"id": "ax + 1"}, {"id": "多项式 $5 x ^ { 2 } + 2 x - 2$ 与多项式 $ax + 1$ 的乘积中"}, {"id": "5 ax ^ { 3 } + ( 5 + 2 a ) x ^ { 2 } + 2 x - 2 ax - 2"}, {"id": "x ^ { 2 }"}, {"id": "5 + 2 a = 0"}, {"id": "不含 $x ^ { 2 }$ 项"}, {"id": "a = - \\frac { 5 } { 2 }"}], "links": [{"rel": "被描述", "source": "5 x ^ { 2 } + 2 x - 2", "target": "( 5 x ^ { 2 } + 2 x - 2 ) ( ax + 1 )"}, {"rel": "提取因式", "source": "( 5 x ^ { 2 } + 2 x - 2 ) ( ax + 1 )", "target": "5 ax ^ { 3 } + ( 5 + 2 a ) x ^ { 2 } + 2 x - 2 ax - 2"}, {"rel": "被描述", "source": "ax + 1", "target": "( 5 x ^ { 2 } + 2 x - 2 ) ( ax + 1 )"}, {"rel": "限制性描述", "source": "多项式 $5 x ^ { 2 } + 2 x - 2$ 与多项式 $ax + 1$ 的乘积中", "target": "( 5 x ^ { 2 } + 2 x - 2 ) ( ax + 1 )"}, {"rel": "限制性描述", "source": "多项式 $5 x ^ { 2 } + 2 x - 2$ 与多项式 $ax + 1$ 的乘积中", "target": "5 + 2 a = 0"}, {"rel": "被描述", "source": "5 ax ^ { 3 } + ( 5 + 2 a ) x ^ { 2 } + 2 x - 2 ax - 2", "target": "5 + 2 a = 0"}, {"rel": "提取因式参考", "source": "x ^ { 2 }", "target": "5 ax ^ { 3 } + ( 5 + 2 a ) x ^ { 2 } + 2 x - 2 ax - 2"}, {"rel": "等式方程求解", "source": "5 + 2 a = 0", "target": "a = - \\frac { 5 } { 2 }"}, {"rel": "限制性描述", "source": "不含 $x ^ { 2 }$ 项", "target": "5 + 2 a = 0"}]}}
|
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{"content": "Given $a : b : c = 3 : 5 : 7$ and $2 a - b + c = 16$, find $\\frac { a + c } { b }$.", "answer": "2", "steps": "According to the problem, let $a = 3 k$, $b = 5 k$, $c = 7 k$ ($k \\neq 0$). Then we have $6 k - 5 k + 7 k = 16$, which gives us $k = 2$. Therefore, $a + c = 10 k = 20$ and $b = 5 k = 10$. Hence, $\\frac { a + c } { b } = \\frac { 20 } { 10 } = 2$.", "expr_cands": ["a : b : c = 3 : 5 : 7", "a", "c", "b", "2 a - b + c = 16", "\\frac { a + c } { b }", "a = 3 k", "k", "b = 5 k", "c = 7 k ( k \\neq 0 )", "6 k - 5 k + 7 k = 16", "k = 2", "a + c", "20", "b = 10", "2"], "exprs": ["a = 3 k", "b = 5 k", "c = 7 k ( k \\neq 0 )", "6 k - 5 k + 7 k = 16", "k = 2", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $a = 3 k$ , $b = 5 k$ , $c = 7 k ( k \\neq 0 )$"}, {"id": "a = 3 k"}, {"id": "b = 5 k"}, {"id": "c = 7 k ( k \\neq 0 )"}, {"id": "2 a - b + c = 16"}, {"id": "6 k - 5 k + 7 k = 16"}, {"id": "k = 2"}, {"id": "\\frac { a + c } { b }"}, {"id": "2"}], "links": [{"rel": "假设描述", "source": "设 $a = 3 k$ , $b = 5 k$ , $c = 7 k ( k \\neq 0 )$", "target": "a = 3 k"}, {"rel": "假设描述", "source": "设 $a = 3 k$ , $b = 5 k$ , $c = 7 k ( k \\neq 0 )$", "target": "b = 5 k"}, {"rel": "假设描述", "source": "设 $a = 3 k$ , $b = 5 k$ , $c = 7 k ( k \\neq 0 )$", "target": "c = 7 k ( k \\neq 0 )"}, {"rel": "代入", "source": "a = 3 k", "target": "6 k - 5 k + 7 k = 16"}, {"rel": "代入", "source": "a = 3 k", "target": "2"}, {"rel": "代入", "source": "b = 5 k", "target": "6 k - 5 k + 7 k = 16"}, {"rel": "代入", "source": "b = 5 k", "target": "2"}, {"rel": "代入", "source": "c = 7 k ( k \\neq 0 )", "target": "6 k - 5 k + 7 k = 16"}, {"rel": "代入", "source": "c = 7 k ( k \\neq 0 )", "target": "2"}, {"rel": "被代入", "source": "2 a - b + c = 16", "target": "6 k - 5 k + 7 k = 16"}, {"rel": "等式方程求解", "source": "6 k - 5 k + 7 k = 16", "target": "k = 2"}, {"rel": "代入", "source": "k = 2", "target": "2"}, {"rel": "被代入", "source": "\\frac { a + c } { b }", "target": "2"}]}}
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{"content": "If $- 5 { x } ^ { 2 - m } { y } ^ { 4 }$ and $2 { x } ^ { 3 } { y } ^ { n - 1 }$ are like terms, then $m + n$ = ____ ?", "answer": "4", "steps": "According to the problem, we have $2 - m = 3$ and $n - 1 = 4$. Solving for $m$ and $n$, we get $m = - 1$ and $n = 5$. Therefore, $m + n = 4$.", "expr_cands": ["- 5 { x } ^ { 2 - m } { y } ^ { 4 }", "x", "m", "y", "2 { x } ^ { 3 } { y } ^ { n - 1 }", "n", "m + n", "2 - m = 3", "m = - 1", "n - 1 = 4", "n = 5", "4"], "exprs": ["2 - m = 3", "n - 1 = 4", "m = - 1", "n = 5", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 5 { x } ^ { 2 - m } { y } ^ { 4 }"}, {"id": "2 - m = 3"}, {"id": "2 { x } ^ { 3 } { y } ^ { n - 1 }"}, {"id": "$- 5 { x } ^ { 2 - m } { y } ^ { 4 }$ 与 $2 { x } ^ { 3 } { y } ^ { n - 1 }$ 为同类项"}, {"id": "n - 1 = 4"}, {"id": "m = - 1"}, {"id": "n = 5"}, {"id": "m + n"}, {"id": "4"}], "links": [{"rel": "被描述", "source": "- 5 { x } ^ { 2 - m } { y } ^ { 4 }", "target": "2 - m = 3"}, {"rel": "被描述", "source": "- 5 { x } ^ { 2 - m } { y } ^ { 4 }", "target": "n - 1 = 4"}, {"rel": "等式方程求解", "source": "2 - m = 3", "target": "m = - 1"}, {"rel": "被描述", "source": "2 { x } ^ { 3 } { y } ^ { n - 1 }", "target": "2 - m = 3"}, {"rel": "被描述", "source": "2 { x } ^ { 3 } { y } ^ { n - 1 }", "target": "n - 1 = 4"}, {"rel": "限制性描述", "source": "$- 5 { x } ^ { 2 - m } { y } ^ { 4 }$ 与 $2 { x } ^ { 3 } { y } ^ { n - 1 }$ 为同类项", "target": "2 - m = 3"}, {"rel": "限制性描述", "source": "$- 5 { x } ^ { 2 - m } { y } ^ { 4 }$ 与 $2 { x } ^ { 3 } { y } ^ { n - 1 }$ 为同类项", "target": "n - 1 = 4"}, {"rel": "等式方程求解", "source": "n - 1 = 4", "target": "n = 5"}, {"rel": "代入", "source": "m = - 1", "target": "4"}, {"rel": "代入", "source": "n = 5", "target": "4"}, {"rel": "被代入", "source": "m + n", "target": "4"}]}}
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{"content": "Given that the value of the fraction $\\frac { x - 3 } { x + 1 }$ is $0$, what is the value of $x$?", "answer": "3", "steps": "$x - 3 = 0$ , and $x + 1 \\neq 0$ , $x = 3$.", "expr_cands": ["\\frac { x - 3 } { x + 1 }", "x", "0", "x - 3 = 0", "x = 3", "x + 1 \\neq 0", "x \\neq - 1"], "exprs": ["x - 3 = 0", "x + 1 \\neq 0", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 3 } { x + 1 }"}, {"id": "x - 3 = 0"}, {"id": "0"}, {"id": "分式 $\\frac { x - 3 } { x + 1 }$ 的值为 $0$"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "x + 1 \\neq 0"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "\\frac { x - 3 } { x + 1 }", "target": "x - 3 = 0"}, {"rel": "被描述", "source": "\\frac { x - 3 } { x + 1 }", "target": "x + 1 \\neq 0"}, {"rel": "联立", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "被描述", "source": "0", "target": "x - 3 = 0"}, {"rel": "被描述", "source": "0", "target": "x + 1 \\neq 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - 3 } { x + 1 }$ 的值为 $0$", "target": "x - 3 = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - 3 } { x + 1 }$ 的值为 $0$", "target": "x + 1 \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x + 1 \\neq 0"}, {"rel": "联立", "source": "x + 1 \\neq 0", "target": "x = 3"}]}}
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{"content": "The solution to the equation $( x + 3 ) ( 2 x - 5 ) - ( 2 x + 1 ) ( x - 8 ) = 41$ is ____?", "answer": "x = 3", "steps": "$2 { x } ^ { 2 } - 5 x + 6 x - 15 - ( 2 { x } ^ { 2 } - 16 x + x - 8 ) = 41$ The equation above is solved as follows:$2 { x } ^ { 2 } - 5 x + 6 x - 15 - 2 { x } ^ { 2 } + 16 x - x + 8 = 41$Combining like terms, we get:$16 x - 7 = 41$Solving for x, we get:$16 x = 48$$x = 3$", "expr_cands": ["( x + 3 ) ( 2 x - 5 ) - ( 2 x + 1 ) ( x - 8 ) = 41", "x", "2 { x } ^ { 2 } - 5 x + 6 x - 15 - ( 2 { x } ^ { 2 } - 16 x + x - 8 ) = 41", "x = 3", "2 { x } ^ { 2 } - 5 x + 6 x - 15 - 2 { x } ^ { 2 } + 16 x - x + 8 = 41", "16 x - 7 = 41", "16 x = 48"], "exprs": ["x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + 3 ) ( 2 x - 5 ) - ( 2 x + 1 ) ( x - 8 ) = 41"}, {"id": "x = 3"}], "links": [{"rel": "等式方程求解", "source": "( x + 3 ) ( 2 x - 5 ) - ( 2 x + 1 ) ( x - 8 ) = 41", "target": "x = 3"}]}}
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{"content": "If $a ^ { m } = 6$, $a ^ { n } = 3$, calculate $a ^ { 2 m - n }$.", "answer": "12", "steps": "Since $a ^ { m } = 6$, therefore $a ^ { 2 m } = 36$, therefore $a ^ { 2 m - n } = a ^ { 2 m } \\div a ^ { n } = 12$.", "expr_cands": ["a ^ { m } = 6", "a", "m", "a ^ { n } = 3", "n", "a ^ { 2 m - n }", "a ^ { 2 m }", "36", "a ^ { 2 m } \\div a ^ { n } = 12", "a ^ { 2 m } \\div a ^ { n }", "12"], "exprs": ["12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { m } = 6"}, {"id": "12"}, {"id": "a ^ { n } = 3"}, {"id": "a ^ { 2 m - n }"}], "links": [{"rel": "代入", "source": "a ^ { m } = 6", "target": "12"}, {"rel": "代入", "source": "a ^ { n } = 3", "target": "12"}, {"rel": "被代入", "source": "a ^ { 2 m - n }", "target": "12"}]}}
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{"content": "If $\\sqrt { 4 - 4 a + a ^ 2 } = 2 - a$, then the relationship between $a$ and $2$ is ____?", "answer": "a \\le 2", "steps": "$\\sqrt { 4 - 4 a + a ^ { 2 } } = 2 - a$ means the square root of 4 minus 4 times a plus a squared equals 2 minus a. It follows that 2 minus a must be greater than or equal to 0 and a must be less than or equal to 2.", "expr_cands": ["\\sqrt { 4 - 4 a + a ^ { 2 } } = 2 - a", "a", "2", "2 - a \\ge 0", "a \\le 2"], "exprs": ["2 - a \\ge 0", "a \\le 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 4 - 4 a + a ^ { 2 } } = 2 - a"}, {"id": "2 - a \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a \\le 2"}], "links": [{"rel": "被描述", "source": "\\sqrt { 4 - 4 a + a ^ { 2 } } = 2 - a", "target": "2 - a \\ge 0"}, {"rel": "不等式方程求解", "source": "2 - a \\ge 0", "target": "a \\le 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 - a \\ge 0"}]}}
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{"content": "What is the result of $( - 1 ) ^ { 2 n } + ( - 1 ) ^ { 2 n + 1 }$? (where n is a positive integer)", "answer": "0", "steps": "$\\because$ $n$ is a positive integer, $\\therefore$ $2 n$ is even, $2 n + 1$ is odd, $\\therefore$ $( - 1 ) ^ { 2 n + 1 } + ( - 1 ) ^ { 2 n } = - 1 + 1 = 0$.", "expr_cands": ["( - 1 ) ^ { 2 n } + ( - 1 ) ^ { 2 n + 1 }", "n", "2 n", "2 n + 1", "( - 1 ) ^ { 2 n + 1 } + ( - 1 ) ^ { 2 n }", "0"], "exprs": ["0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( - 1 ) ^ { 2 n + 1 } + ( - 1 ) ^ { 2 n }"}, {"id": "0"}, {"id": ", $n$ 为正整数"}, {"id": ", $2 n$ 为偶数"}, {"id": "$2 n + 1$ 为奇数"}], "links": [{"rel": "被描述", "source": "( - 1 ) ^ { 2 n + 1 } + ( - 1 ) ^ { 2 n }", "target": "0"}, {"rel": "限制性描述", "source": ", $n$ 为正整数", "target": "0"}, {"rel": "限制性描述", "source": ", $2 n$ 为偶数", "target": "0"}, {"rel": "限制性描述", "source": "$2 n + 1$ 为奇数", "target": "0"}]}}
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{"content": "If real numbers $m$ and $n$ satisfy $| m - 2 | + \\sqrt { n - 4 } = 0$, then $m ^ { 2 } + n$ is equal to ____?", "answer": "8", "steps": "Because $| m - 2 | + \\sqrt { n - 4 } = 0$, therefore $m - 2 = 0$, $n - 4 = 0$, therefore $m = 2$, $n = 4$, therefore $m ^ { 2 } + n = 2 ^ { 2 } + 4 = 8$.", "expr_cands": ["m", "n", "| m - 2 | + \\sqrt { n - 4 } = 0", "m ^ { 2 } + n", "m - 2 = 0", "m = 2", "n - 4 = 0", "n = 4", "8"], "exprs": ["m - 2 = 0", "n - 4 = 0", "m = 2", "n = 4", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| m - 2 | + \\sqrt { n - 4 } = 0"}, {"id": "m - 2 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "n - 4 = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "m = 2"}, {"id": "n = 4"}, {"id": "m ^ { 2 } + n"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "| m - 2 | + \\sqrt { n - 4 } = 0", "target": "m - 2 = 0"}, {"rel": "被描述", "source": "| m - 2 | + \\sqrt { n - 4 } = 0", "target": "n - 4 = 0"}, {"rel": "等式方程求解", "source": "m - 2 = 0", "target": "m = 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m - 2 = 0"}, {"rel": "等式方程求解", "source": "n - 4 = 0", "target": "n = 4"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "n - 4 = 0"}, {"rel": "代入", "source": "m = 2", "target": "8"}, {"rel": "代入", "source": "n = 4", "target": "8"}, {"rel": "被代入", "source": "m ^ { 2 } + n", "target": "8"}]}}
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{"content": "Given a linear function $y = kx - 4$, when $x = 2$, $y = - 3$. What is the value of $y$ when $x = - 2$?", "answer": "- 5", "steps": "$\\because$ The linear function $y = kx - 4$ has a value of $- 3$ when $x = 2$. $\\therefore$ $2 k - 4 = - 3$, which solves for $k = \\frac { 1 } { 2 }$. $\\therefore$ The analytical expression for the linear function is $y = \\frac { 1 } { 2 } x - 4$. $\\therefore$ When $x = - 2$, $y = \\frac { 1 } { 2 } * ( - 2 ) - 4 = - 1 - 4 = - 5$.", "expr_cands": ["y = kx - 4", "k", "y", "x", "x = 2", "y = - 3", "x = - 2", "2 k - 4 = - 3", "k = \\frac { 1 } { 2 }", "y = \\frac { 1 } { 2 } x - 4", "y = - 5"], "exprs": ["2 k - 4 = - 3", "k = \\frac { 1 } { 2 }", "y = \\frac { 1 } { 2 } x - 4", "y = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = kx - 4"}, {"id": "2 k - 4 = - 3"}, {"id": "x = 2"}, {"id": "y = - 3"}, {"id": "k = \\frac { 1 } { 2 }"}, {"id": "y = \\frac { 1 } { 2 } x - 4"}, {"id": "x = - 2"}, {"id": "y = - 5"}], "links": [{"rel": "被代入", "source": "y = kx - 4", "target": "2 k - 4 = - 3"}, {"rel": "被代入", "source": "y = kx - 4", "target": "y = \\frac { 1 } { 2 } x - 4"}, {"rel": "等式方程求解", "source": "2 k - 4 = - 3", "target": "k = \\frac { 1 } { 2 }"}, {"rel": "代入", "source": "x = 2", "target": "2 k - 4 = - 3"}, {"rel": "代入", "source": "y = - 3", "target": "2 k - 4 = - 3"}, {"rel": "代入", "source": "k = \\frac { 1 } { 2 }", "target": "y = \\frac { 1 } { 2 } x - 4"}, {"rel": "被代入", "source": "y = \\frac { 1 } { 2 } x - 4", "target": "y = - 5"}, {"rel": "代入", "source": "x = - 2", "target": "y = - 5"}]}}
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{"content": "The sum of all positive integer solutions to the inequality $2 x - 3 < 5$ is _____.", "answer": "6", "steps": "Moving the terms, we get $2 x < 8$. Dividing both sides by $2$, we get $x < 4$. Therefore, the positive integer solutions to the inequality $2 x - 3 < 5$ are $1$, $2$, and $3$. Thus, the sum of all positive integer solutions is $1 + 2 + 3 = 6$.", "expr_cands": ["2 x - 3 < 5", "x", "2 x < 8", "x < 4", "1", "2", "3", "1 + 2 + 3", "6"], "exprs": ["x < 4", "1", "2", "3", "1 + 2 + 3", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 3 < 5"}, {"id": "x < 4"}, {"id": "1"}, {"id": "不等式 $2 x - 3 < 5$ 的正整数解为 $1$ , $2$ , $3$"}, {"id": "2"}, {"id": "3"}, {"id": "1 + 2 + 3"}, {"id": "不等式 $2 x - 3 < 5$ 的所有正整数解的和"}, {"id": "6"}], "links": [{"rel": "不等式方程求解", "source": "2 x - 3 < 5", "target": "x < 4"}, {"rel": "被描述", "source": "x < 4", "target": "1"}, {"rel": "被描述", "source": "x < 4", "target": "2"}, {"rel": "被描述", "source": "x < 4", "target": "3"}, {"rel": "被描述", "source": "1", "target": "1 + 2 + 3"}, {"rel": "限制性描述", "source": "不等式 $2 x - 3 < 5$ 的正整数解为 $1$ , $2$ , $3$", "target": "1"}, {"rel": "限制性描述", "source": "不等式 $2 x - 3 < 5$ 的正整数解为 $1$ , $2$ , $3$", "target": "2"}, {"rel": "限制性描述", "source": "不等式 $2 x - 3 < 5$ 的正整数解为 $1$ , $2$ , $3$", "target": "3"}, {"rel": "被描述", "source": "2", "target": "1 + 2 + 3"}, {"rel": "被描述", "source": "3", "target": "1 + 2 + 3"}, {"rel": "计算", "source": "1 + 2 + 3", "target": "6"}, {"rel": "限制性描述", "source": "不等式 $2 x - 3 < 5$ 的所有正整数解的和", "target": "1 + 2 + 3"}]}}
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{"content": "If the value of $2 x ^ { 2 } - 3 x$ is $- 1$, then the value of $\\frac { 3 } { 2 } x - x ^ { 2 } + 3$ is ____?", "answer": "\\frac { 7 } { 2 }", "steps": "Since $2 x ^ 2 - 3 x = - 1$, we can rearrange the equation to get $x ^ 2 - \\frac { 3 } { 2 } x = - \\frac { 1 } { 2 }$. Therefore, $- \\left ( x ^ 2 - \\frac { 3 } { 2 } x \\right ) + 3 = \\frac { 1 } { 2 } + 3 = \\frac { 7 } { 2 }$.", "expr_cands": ["2 x ^ { 2 } - 3 x", "x", "- 1", "\\frac { 3 } { 2 } x - x ^ { 2 } + 3", "2 x ^ { 2 } - 3 x = - 1", "x = \\frac { 1 } { 2 }", "x = 1", "x ^ { 2 } - \\frac { 3 } { 2 } x = - \\frac { 1 } { 2 }", "- ( x ^ { 2 } - \\frac { 3 } { 2 } x ) + 3", "\\frac { 7 } { 2 }"], "exprs": ["2 x ^ { 2 } - 3 x = - 1", "x ^ { 2 } - \\frac { 3 } { 2 } x = - \\frac { 1 } { 2 }", "- ( x ^ { 2 } - \\frac { 3 } { 2 } x ) + 3", "\\frac { 7 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 2 } - 3 x"}, {"id": "2 x ^ { 2 } - 3 x = - 1"}, {"id": "- 1"}, {"id": "$2 x ^ { 2 } - 3 x$ 的值为 $- 1$"}, {"id": "x ^ { 2 } - \\frac { 3 } { 2 } x = - \\frac { 1 } { 2 }"}, {"id": "\\frac { 3 } { 2 } x - x ^ { 2 } + 3"}, {"id": "- ( x ^ { 2 } - \\frac { 3 } { 2 } x ) + 3"}, {"id": "\\frac { 7 } { 2 }"}], "links": [{"rel": "被描述", "source": "2 x ^ { 2 } - 3 x", "target": "2 x ^ { 2 } - 3 x = - 1"}, {"rel": "同乘除", "source": "2 x ^ { 2 } - 3 x = - 1", "target": "x ^ { 2 } - \\frac { 3 } { 2 } x = - \\frac { 1 } { 2 }"}, {"rel": "被描述", "source": "- 1", "target": "2 x ^ { 2 } - 3 x = - 1"}, {"rel": "限制性描述", "source": "$2 x ^ { 2 } - 3 x$ 的值为 $- 1$", "target": "2 x ^ { 2 } - 3 x = - 1"}, {"rel": "提取因式参考", "source": "x ^ { 2 } - \\frac { 3 } { 2 } x = - \\frac { 1 } { 2 }", "target": "- ( x ^ { 2 } - \\frac { 3 } { 2 } x ) + 3"}, {"rel": "代入", "source": "x ^ { 2 } - \\frac { 3 } { 2 } x = - \\frac { 1 } { 2 }", "target": "\\frac { 7 } { 2 }"}, {"rel": "提取因式", "source": "\\frac { 3 } { 2 } x - x ^ { 2 } + 3", "target": "- ( x ^ { 2 } - \\frac { 3 } { 2 } x ) + 3"}, {"rel": "被代入", "source": "- ( x ^ { 2 } - \\frac { 3 } { 2 } x ) + 3", "target": "\\frac { 7 } { 2 }"}]}}
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{"content": "$- a ^ { 6 } \\div ( - a ) ^ { 2 }$ equals ____ ?", "answer": "- a ^ { 4 }", "steps": "$- a ^ { 6 } \\div ( - a ) ^ { 2 }$ equals $- a ^ { 6 } \\div a ^ { 2 } = - a ^ { 4 }$.", "expr_cands": ["- a ^ { 6 } \\div ( - a ) ^ { 2 }", "a", "- a ^ { 6 } \\div a ^ { 2 }", "- a ^ { 4 }"], "exprs": ["- a ^ { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- a ^ { 6 } \\div ( - a ) ^ { 2 }"}, {"id": "- a ^ { 4 }"}], "links": [{"rel": "计算", "source": "- a ^ { 6 } \\div ( - a ) ^ { 2 }", "target": "- a ^ { 4 }"}]}}
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{"content": "If $a = 2 + \\sqrt { 7 }$, then the value of $a ^ { 2 } - 4 a + 5$ is ____?", "answer": "8", "steps": "Since $a = 2 + \\sqrt { 7 }$, it follows that $a - 2 = \\sqrt { 7 }$. Therefore, $a ^ 2 - 4 a + 5 = ( a - 2 ) ^ 2 + 1 = ( \\sqrt { 7 }) ^ 2 + 1 = 8$.", "expr_cands": ["a = 2 + \\sqrt { 7 }", "a", "a ^ { 2 } - 4 a + 5", "a - 2", "\\sqrt { 7 }", "8"], "exprs": ["8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } - 4 a + 5"}, {"id": "8"}, {"id": "a = 2 + \\sqrt { 7 }"}], "links": [{"rel": "被代入", "source": "a ^ { 2 } - 4 a + 5", "target": "8"}, {"rel": "代入", "source": "a = 2 + \\sqrt { 7 }", "target": "8"}]}}
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{"content": "If $2 a = 12$, then $6 a$ = ____ ?", "answer": "36", "steps": "Because $2 a = 12$, therefore $a = 6$, therefore $6 a = 36$.", "expr_cands": ["2 a = 12", "a", "6 a", "a = 6", "36"], "exprs": ["a = 6", "36"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a = 12"}, {"id": "a = 6"}, {"id": "6 a"}, {"id": "36"}], "links": [{"rel": "等式方程求解", "source": "2 a = 12", "target": "a = 6"}, {"rel": "代入", "source": "a = 6", "target": "36"}, {"rel": "被代入", "source": "6 a", "target": "36"}]}}
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{"content": "If the value of the algebraic expression $- 2 { x } ^ 2 - 5 y + n { x } ^ 2 - 1$ is independent of the value of $x$, then the value of $n$ is ____?", "answer": "2", "steps": "= $( - 2 + n ) x ^ { 2 } - 5 y - 1$ Since the value of the algebraic expression is independent of the value of the variable $x$, we have $- 2 + n = 0$, which implies $n = 2$.", "expr_cands": ["x", "- 2 { x } ^ { 2 } - 5 y + n { x } ^ { 2 } - 1", "y", "n", "( - 2 + n ) x ^ { 2 } - 5 y - 1", "- 2 + n = 0", "n = 2"], "exprs": ["( - 2 + n ) x ^ { 2 } - 5 y - 1", "- 2 + n = 0", "n = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 { x } ^ { 2 } - 5 y + n { x } ^ { 2 } - 1"}, {"id": "( - 2 + n ) x ^ { 2 } - 5 y - 1"}, {"id": "x"}, {"id": "- 2 + n = 0"}, {"id": "关于 $x$ 的代数式 $- 2 { x } ^ { 2 } - 5 y + n { x } ^ { 2 } - 1$ 的值与 $x$ 的取值无关"}, {"id": "n = 2"}], "links": [{"rel": "提取因式", "source": "- 2 { x } ^ { 2 } - 5 y + n { x } ^ { 2 } - 1", "target": "( - 2 + n ) x ^ { 2 } - 5 y - 1"}, {"rel": "被描述", "source": "( - 2 + n ) x ^ { 2 } - 5 y - 1", "target": "- 2 + n = 0"}, {"rel": "提取因式参考", "source": "x", "target": "( - 2 + n ) x ^ { 2 } - 5 y - 1"}, {"rel": "等式方程求解", "source": "- 2 + n = 0", "target": "n = 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的代数式 $- 2 { x } ^ { 2 } - 5 y + n { x } ^ { 2 } - 1$ 的值与 $x$ 的取值无关", "target": "- 2 + n = 0"}]}}
|
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{"content": "If real numbers $a$ and $b$ satisfy $a - 2 b = 4$ and $ab = 2$, then $a ^ 2 + 4 b ^ 2$ = ____?", "answer": "24", "steps": "$\\because$ Real numbers $a$ and $b$ satisfy $a - 2 b = 4$ and $ab = 2$. $\\therefore$ $a ^ 2 + 4 b ^ 2 = ( a - 2 b ) ^ 2 + 4 ab = 4 ^ 2 + 4 * 2 = 24$.", "expr_cands": ["a", "b", "a - 2 b = 4", "ab = 2", "a ^ { 2 } + 4 b ^ { 2 }", "( a - 2 b ) ^ { 2 } + 4 ab", "24"], "exprs": ["( a - 2 b ) ^ { 2 } + 4 ab", "24"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } + 4 b ^ { 2 }"}, {"id": "( a - 2 b ) ^ { 2 } + 4 ab"}, {"id": "a - 2 b = 4"}, {"id": "ab = 2"}, {"id": "24"}], "links": [{"rel": "提取因式", "source": "a ^ { 2 } + 4 b ^ { 2 }", "target": "( a - 2 b ) ^ { 2 } + 4 ab"}, {"rel": "被代入", "source": "( a - 2 b ) ^ { 2 } + 4 ab", "target": "24"}, {"rel": "提取因式参考", "source": "a - 2 b = 4", "target": "( a - 2 b ) ^ { 2 } + 4 ab"}, {"rel": "代入", "source": "a - 2 b = 4", "target": "24"}, {"rel": "提取因式参考", "source": "ab = 2", "target": "( a - 2 b ) ^ { 2 } + 4 ab"}, {"rel": "代入", "source": "ab = 2", "target": "24"}]}}
|
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{"content": "Given $y = \\sqrt { 2 - x } + \\sqrt { x - 2 } + 5$, find the value of $\\frac { x } { y }$.", "answer": "\\frac { 2 } { 5 }", "steps": "From the given conditions, we have $x - 2 \\ge 0$ and $2 - x \\ge 0$, which implies $x \\ge 2$ and $x \\le 2$. Therefore, $x = 2$ and $y = 5$. Thus, $\\frac { x } { y } = \\frac { 2 } { 5 }$.", "expr_cands": ["y = \\sqrt { 2 - x } + \\sqrt { x - 2 } + 5", "x", "y", "\\frac { x } { y }", "x - 2 \\ge 0", "2 \\le x", "2 - x \\ge 0", "x \\le 2", "x \\ge 2", "x = 2", "y = 5", "\\frac { 2 } { 5 }"], "exprs": ["x - 2 \\ge 0", "2 - x \\ge 0", "x = 2", "y = 5", "\\frac { 2 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { 2 - x } + \\sqrt { x - 2 } + 5"}, {"id": "x - 2 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "2 - x \\ge 0"}, {"id": "x = 2"}, {"id": "y = 5"}, {"id": "\\frac { x } { y }"}, {"id": "\\frac { 2 } { 5 }"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { 2 - x } + \\sqrt { x - 2 } + 5", "target": "x - 2 \\ge 0"}, {"rel": "被描述", "source": "y = \\sqrt { 2 - x } + \\sqrt { x - 2 } + 5", "target": "2 - x \\ge 0"}, {"rel": "被代入", "source": "y = \\sqrt { 2 - x } + \\sqrt { x - 2 } + 5", "target": "y = 5"}, {"rel": "联立", "source": "x - 2 \\ge 0", "target": "x = 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 2 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 - x \\ge 0"}, {"rel": "联立", "source": "2 - x \\ge 0", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "y = 5"}, {"rel": "代入", "source": "x = 2", "target": "\\frac { 2 } { 5 }"}, {"rel": "代入", "source": "y = 5", "target": "\\frac { 2 } { 5 }"}, {"rel": "被代入", "source": "\\frac { x } { y }", "target": "\\frac { 2 } { 5 }"}]}}
|
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{"content": "$2 x - 5 y - 3 = 0$ , then ${ { 4 } ^ { x } } \\div { 32 } ^ { y }$ = ____ ?", "answer": "8", "steps": "Since $2 x - 5 y - 3 = 0$, therefore $2 x - 5 y = 3$, therefore the original expression is equal to ${ 2 } ^ { 2 x } \\div { 2 } ^ { 5 y } = { 2 } ^ { 2 x - 5 y } = { 2 } ^ { 3 } = 8$.", "expr_cands": ["2 x - 5 y - 3 = 0", "x", "y", "{ { 4 } ^ { x } } \\div { 32 } ^ { y }", "2 x - 5 y = 3", "{ 2 } ^ { 2 x - 5 y }", "8"], "exprs": ["2 x - 5 y = 3", "{ 2 } ^ { 2 x - 5 y }", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 5 y - 3 = 0"}, {"id": "2 x - 5 y = 3"}, {"id": "{ { 4 } ^ { x } } \\div { 32 } ^ { y }"}, {"id": "{ 2 } ^ { 2 x - 5 y }"}, {"id": "8"}], "links": [{"rel": "移项", "source": "2 x - 5 y - 3 = 0", "target": "2 x - 5 y = 3"}, {"rel": "代入", "source": "2 x - 5 y = 3", "target": "8"}, {"rel": "计算", "source": "{ { 4 } ^ { x } } \\div { 32 } ^ { y }", "target": "{ 2 } ^ { 2 x - 5 y }"}, {"rel": "被代入", "source": "{ 2 } ^ { 2 x - 5 y }", "target": "8"}]}}
|
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{"content": "If the value of $5 x - 5$ is the opposite of the value of $2 x - 9$, then $x$ equals ____?", "answer": "2", "steps": "According to the problem, we have $5 x - 5 + 2 x - 9 = 0$. By rearranging and combining like terms, we get $7 x = 14$. Solving for $x$, we get $x = 2$.", "expr_cands": ["5 x - 5", "x", "2 x - 9", "5 x - 5 + 2 x - 9 = 0", "x = 2", "7 x = 14"], "exprs": ["5 x - 5 + 2 x - 9 = 0", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x - 5"}, {"id": "5 x - 5 + 2 x - 9 = 0"}, {"id": "2 x - 9"}, {"id": "$5 x - 5$ 的值与 $2 x - 9$ 的值互为相反数"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "5 x - 5", "target": "5 x - 5 + 2 x - 9 = 0"}, {"rel": "等式方程求解", "source": "5 x - 5 + 2 x - 9 = 0", "target": "x = 2"}, {"rel": "被描述", "source": "2 x - 9", "target": "5 x - 5 + 2 x - 9 = 0"}, {"rel": "限制性描述", "source": "$5 x - 5$ 的值与 $2 x - 9$ 的值互为相反数", "target": "5 x - 5 + 2 x - 9 = 0"}]}}
|
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{"content": "If the solutions of the equations $4 x - 2 m = 3 x + 2$ and $x = 2 x - 3 m$ with respect to $x$ are the same, then $m$ = ____?", "answer": "2", "steps": "Solve the equation $4 x - 2 m = 3 x + 2$, we get $x = 2 + 2 m$. Solve the equation $2 x - 3 m = x$, we get $x = 3 m$. Since the solutions of the two equations are the same, we have $3 m = 2 + 2 m$. Solving for $m$, we get $m = 2$.", "expr_cands": ["x", "4 x - 2 m = 3 x + 2", "m", "x = 2 x - 3 m", "x = 2 + 2 m", "2 x - 3 m = x", "( - 1 ) 3 m + 4 m + 4 = 2 m + 2", "x = 3 m", "2 m + 2 = 3 m", "3 m = 2 + 2 m", "m = 2"], "exprs": ["x = 2 + 2 m", "x = 3 m", "3 m = 2 + 2 m", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x - 2 m = 3 x + 2"}, {"id": "x = 2 + 2 m"}, {"id": "x = 2 x - 3 m"}, {"id": "x = 3 m"}, {"id": "3 m = 2 + 2 m"}, {"id": "关于 $x$ 的方程 $4 x - 2 m = 3 x + 2$ 和 $x = 2 x - 3 m$ 的解相同"}, {"id": "m = 2"}], "links": [{"rel": "等式方程部分求解", "source": "4 x - 2 m = 3 x + 2", "target": "x = 2 + 2 m"}, {"rel": "被描述", "source": "x = 2 + 2 m", "target": "3 m = 2 + 2 m"}, {"rel": "等式方程部分求解", "source": "x = 2 x - 3 m", "target": "x = 3 m"}, {"rel": "被描述", "source": "x = 3 m", "target": "3 m = 2 + 2 m"}, {"rel": "等式方程求解", "source": "3 m = 2 + 2 m", "target": "m = 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $4 x - 2 m = 3 x + 2$ 和 $x = 2 x - 3 m$ 的解相同", "target": "3 m = 2 + 2 m"}]}}
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{"content": "For the proportional function $y = mx ^ { m ^ { 2 } - 3 }$, the value of $y$ decreases as the value of $x$ increases. What is the value of $m$?", "answer": "- 2", "steps": "$\\because$ The value of $y$ decreases as the value of $x$ increases, $\\therefore$ $m < 0$. $\\because$ The function is a direct proportion function $y = mx ^ { m ^ 2 - 3 }$, $\\therefore$ $m ^ 2 - 3 = 1$, $\\therefore$ $m = - 2$.", "expr_cands": ["y = mx ^ { m ^ { 2 } - 3 }", "x", "m", "y", "m < 0", "m ^ { 2 } - 3 = 1", "m = - 2", "m = 2"], "exprs": ["m < 0", "m ^ { 2 } - 3 = 1", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = mx ^ { m ^ { 2 } - 3 }"}, {"id": "m < 0"}, {"id": "对于正比例函数 $y = mx ^ { m ^ { 2 } - 3 }$ , $y$ 的值随 $x$ 的值增大而减小"}, {"id": ", $y$ 的值随 $x$ 的值增大而减小"}, {"id": "m ^ { 2 } - 3 = 1"}, {"id": "m = - 2"}], "links": [{"rel": "被描述", "source": "y = mx ^ { m ^ { 2 } - 3 }", "target": "m < 0"}, {"rel": "被描述", "source": "y = mx ^ { m ^ { 2 } - 3 }", "target": "m ^ { 2 } - 3 = 1"}, {"rel": "联立", "source": "m < 0", "target": "m = - 2"}, {"rel": "限制性描述", "source": "对于正比例函数 $y = mx ^ { m ^ { 2 } - 3 }$ , $y$ 的值随 $x$ 的值增大而减小", "target": "m < 0"}, {"rel": "限制性描述", "source": "对于正比例函数 $y = mx ^ { m ^ { 2 } - 3 }$ , $y$ 的值随 $x$ 的值增大而减小", "target": "m ^ { 2 } - 3 = 1"}, {"rel": "限制性描述", "source": ", $y$ 的值随 $x$ 的值增大而减小", "target": "m < 0"}, {"rel": "限制性描述", "source": ", $y$ 的值随 $x$ 的值增大而减小", "target": "m ^ { 2 } - 3 = 1"}, {"rel": "联立", "source": "m ^ { 2 } - 3 = 1", "target": "m = - 2"}]}}
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{"content": "If $a = 2009 x + 2007$, $b = 2009 x + 2008$, $c = 2009 x + 2009$, then the value of $a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ab - bc - ca$ is ____?", "answer": "3", "steps": "Since $a = 2009 x + 2007$, $b = 2009 x + 2008$, and $c = 2009 x + 2009$, we have $a - b = - 1$, $b - c = - 1$, and $c - a = 2$. Therefore, $a ^ 2 + b ^ 2 + c ^ 2 - ab - bc - ca = \\frac { 1 } { 2 } ( 2 a ^ 2 + 2 b ^ 2 + 2 c ^ 2 - 2 ab - 2 bc - 2 ca ) = \\frac { 1 } { 2 } [( a - b ) ^ 2 + ( b - c ) ^ 2 + ( c - a ) ^ 2 ] = \\frac { 1 } { 2 } ( 1 + 1 + 4 ) = 3$.", "expr_cands": ["a = 2009 x + 2007", "x", "a", "b = 2009 x + 2008", "b", "c = 2009 x + 2009", "c", "a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ab - bc - ca", "a - b", "- 1", "b - c", "c - a", "2", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ab - bc - ca"}, {"id": "3"}, {"id": "a = 2009 x + 2007"}, {"id": "b = 2009 x + 2008"}, {"id": "c = 2009 x + 2009"}], "links": [{"rel": "被代入", "source": "a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ab - bc - ca", "target": "3"}, {"rel": "代入", "source": "a = 2009 x + 2007", "target": "3"}, {"rel": "代入", "source": "b = 2009 x + 2008", "target": "3"}, {"rel": "代入", "source": "c = 2009 x + 2009", "target": "3"}]}}
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{"content": "Given a quadratic equation in one variable $x$, $x ^ 2 + 6 x + 5 = 0$ has two roots $x _ 1$ and $x _ 2$. Find the value of $x _ 1 x _ 2 + x _ 1 + x _ 2$.", "answer": "- 1", "steps": "$\\because$ The quadratic equation in one variable $x$, $x ^ 2 + 6 x + 5 = 0$, has two roots $x _ 1$ and $x _ 2$. $\\therefore$ $x _ 1 x _ 2 = 5$, $x _ 1 + x _ 2 = - 6$, $\\therefore$ $x _ 1 x _ 2 + x _ 1 + x _ 2 = - 1$.", "expr_cands": ["x", "x ^ { 2 } + 6 x + 5 = 0", "x _ { 1 }", "x _ { 2 }", "x _ { 1 } x _ { 2 } + x _ { 1 } + x _ { 2 }", "x = - 5", "x = - 1", "x _ { 1 } x _ { 2 } = 5", "x _ { 1 } + x _ { 2 } = - 6", "- 1"], "exprs": ["x _ { 1 } x _ { 2 } = 5", "x _ { 1 } + x _ { 2 } = - 6", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x"}, {"id": "x _ { 1 } x _ { 2 } = 5"}, {"id": "x ^ { 2 } + 6 x + 5 = 0"}, {"id": "x _ { 1 }"}, {"id": "x _ { 2 }"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } + 6 x + 5 = 0$ 有两个根为 $x _ { 1 }$ 和 $x _ { 2 }$"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "x _ { 1 } + x _ { 2 } = - 6"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "x _ { 1 } x _ { 2 } + x _ { 1 } + x _ { 2 }"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "x", "target": "x _ { 1 } x _ { 2 } = 5"}, {"rel": "被描述", "source": "x", "target": "x _ { 1 } + x _ { 2 } = - 6"}, {"rel": "代入", "source": "x _ { 1 } x _ { 2 } = 5", "target": "- 1"}, {"rel": "被描述", "source": "x ^ { 2 } + 6 x + 5 = 0", "target": "x _ { 1 } x _ { 2 } = 5"}, {"rel": "被描述", "source": "x ^ { 2 } + 6 x + 5 = 0", "target": "x _ { 1 } + x _ { 2 } = - 6"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } x _ { 2 } = 5"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } + x _ { 2 } = - 6"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } x _ { 2 } = 5"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } + x _ { 2 } = - 6"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } + 6 x + 5 = 0$ 有两个根为 $x _ { 1 }$ 和 $x _ { 2 }$", "target": "x _ { 1 } x _ { 2 } = 5"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } + 6 x + 5 = 0$ 有两个根为 $x _ { 1 }$ 和 $x _ { 2 }$", "target": "x _ { 1 } + x _ { 2 } = - 6"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "x _ { 1 } x _ { 2 } = 5"}, {"rel": "代入", "source": "x _ { 1 } + x _ { 2 } = - 6", "target": "- 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + x _ { 2 } = - 6"}, {"rel": "被代入", "source": "x _ { 1 } x _ { 2 } + x _ { 1 } + x _ { 2 }", "target": "- 1"}]}}
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{"content": "The opposite of the solution to the equation $- 4 x - 1 = 0$ is ____ ?", "answer": "\\frac { 1 } { 4 }", "steps": "$- 4 x - 1 = 0$, rearranging gives $- 4 x = 1$, solving gives $x = - \\frac { 1 } { 4 }$, therefore the opposite of $x$ is $\\frac { 1 } { 4 }$.", "expr_cands": ["- 4 x - 1 = 0", "x", "x = - \\frac { 1 } { 4 }", "- 4 x = 1", "\\frac { 1 } { 4 }"], "exprs": ["x = - \\frac { 1 } { 4 }", "\\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 4 x - 1 = 0"}, {"id": "x = - \\frac { 1 } { 4 }"}, {"id": "\\frac { 1 } { 4 }"}, {"id": "方程 $- 4 x - 1 = 0$ 的解的相反数"}, {"id": "$x$ 的相反数为 $\\frac { 1 } { 4 }$"}], "links": [{"rel": "等式方程求解", "source": "- 4 x - 1 = 0", "target": "x = - \\frac { 1 } { 4 }"}, {"rel": "被描述", "source": "x = - \\frac { 1 } { 4 }", "target": "\\frac { 1 } { 4 }"}, {"rel": "限制性描述", "source": "方程 $- 4 x - 1 = 0$ 的解的相反数", "target": "\\frac { 1 } { 4 }"}, {"rel": "限制性描述", "source": "$x$ 的相反数为 $\\frac { 1 } { 4 }$", "target": "\\frac { 1 } { 4 }"}]}}
|
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{"content": "The equation of the parabola obtained by shifting the parabola $y = x ^ 2 + 1$ 2 units to the left and 3 units down is _____.", "answer": "y = x ^ { 2 } + 4 x + 2", "steps": "The equation of the parabola $y = x ^ 2 + 1$ shifted 2 units to the left is $y = ( x + 2 ) ^ 2 + 1$. Shifting it down 3 units gives $y = ( x + 2 ) ^ 2 + 1 - 3$, which simplifies to $y = x ^ 2 + 4 x + 2$.", "expr_cands": ["y = x ^ { 2 } + 1", "x", "y", "2", "3", "y = ( x + 2 ) ^ { 2 } + 1", "x ^ { 2 } + 1 = ( x + 2 ) ^ { 2 } + 1", "y = ( x + 2 ) ^ { 2 } + 1 - 3", "( x + 2 ) ^ { 2 } + 1 = ( x + 2 ) ^ { 2 } + 1 - 3", "x ^ { 2 } + 4 x + 2"], "exprs": ["y = ( x + 2 ) ^ { 2 } + 1 - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 } + 1"}, {"id": "y = ( x + 2 ) ^ { 2 } + 1 - 3"}, {"id": "2"}, {"id": "3"}, {"id": "将抛物线 $y = x ^ { 2 } + 1$ 向左平移 $2$ 个单位"}, {"id": "再向下平移 $3$ 个单位"}, {"id": "所得到的抛物线的关系式"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 } + 1", "target": "y = ( x + 2 ) ^ { 2 } + 1 - 3"}, {"rel": "被描述", "source": "2", "target": "y = ( x + 2 ) ^ { 2 } + 1 - 3"}, {"rel": "被描述", "source": "3", "target": "y = ( x + 2 ) ^ { 2 } + 1 - 3"}, {"rel": "限制性描述", "source": "将抛物线 $y = x ^ { 2 } + 1$ 向左平移 $2$ 个单位", "target": "y = ( x + 2 ) ^ { 2 } + 1 - 3"}, {"rel": "限制性描述", "source": "再向下平移 $3$ 个单位", "target": "y = ( x + 2 ) ^ { 2 } + 1 - 3"}, {"rel": "限制性描述", "source": "所得到的抛物线的关系式", "target": "y = ( x + 2 ) ^ { 2 } + 1 - 3"}]}}
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||||
{"content": "Given the equation $4 x - 3 m = 22$ and $x + 2 m = 0$ with the same solution for $x$, what is the value of $m$?", "answer": "- 2", "steps": "Solve the equation $4 x - 3 m = 22$ to get $x = \\frac { 3 m + 22 } { 4 }$. Solve the equation $x + 2 m = 0$ to get $x = - 2 m$. Since the solutions of the two equations are the same, we have $\\frac { 3 m + 22 } { 4 } = - 2 m$. Solving for $m$, we get $m = - 2$.", "expr_cands": ["x", "4 x - 3 m = 22", "m", "x + 2 m = 0", "x = \\frac { 3 m + 22 } { 4 }", "\\frac { 11 m } { 4 } + \\frac { 11 } { 2 } = 0", "x = - 2 m", "\\frac { 3 m } { 4 } + \\frac { 11 } { 2 } = - 2 m", "\\frac { 3 m + 22 } { 4 } = - 2 m", "m = - 2"], "exprs": ["x = \\frac { 3 m + 22 } { 4 }", "x = - 2 m", "\\frac { 3 m + 22 } { 4 } = - 2 m", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x - 3 m = 22"}, {"id": "x = \\frac { 3 m + 22 } { 4 }"}, {"id": "x + 2 m = 0"}, {"id": "x = - 2 m"}, {"id": "\\frac { 3 m + 22 } { 4 } = - 2 m"}, {"id": "关于 $x$ 的方程 $4 x - 3 m = 22$ 与 $x + 2 m = 0$"}, {"id": "它们有相同的解"}, {"id": "m = - 2"}], "links": [{"rel": "等式方程部分求解", "source": "4 x - 3 m = 22", "target": "x = \\frac { 3 m + 22 } { 4 }"}, {"rel": "被描述", "source": "x = \\frac { 3 m + 22 } { 4 }", "target": "\\frac { 3 m + 22 } { 4 } = - 2 m"}, {"rel": "等式方程部分求解", "source": "x + 2 m = 0", "target": "x = - 2 m"}, {"rel": "被描述", "source": "x = - 2 m", "target": "\\frac { 3 m + 22 } { 4 } = - 2 m"}, {"rel": "等式方程求解", "source": "\\frac { 3 m + 22 } { 4 } = - 2 m", "target": "m = - 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $4 x - 3 m = 22$ 与 $x + 2 m = 0$", "target": "\\frac { 3 m + 22 } { 4 } = - 2 m"}, {"rel": "限制性描述", "source": "它们有相同的解", "target": "\\frac { 3 m + 22 } { 4 } = - 2 m"}]}}
|
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{"content": "If $a$ and $b$ are opposite numbers, and $c$ and $d$ are reciprocal numbers, then $2017 ( a + b ) ^ { 5 } + 2018 ( cd ) ^ { 5 }$ = ____ ?", "answer": "2018", "steps": "Since $a$ and $b$ are opposite numbers, and $c$ and $d$ are reciprocal numbers, therefore $a + b = 0$, $cd = 1$. Thus, $2017 ( a + b ) ^ 5 + 2018 ( cd ) ^ 5 = 2017 * 0 ^ 5 + 2018 * 1 ^ 5 = 0 + 2018 = 2018$.", "expr_cands": ["a", "b", "c", "d", "2017 ( a + b ) ^ { 5 } + 2018 ( cd ) ^ { 5 }", "a + b = 0", "cd = 1", "2018"], "exprs": ["a + b = 0", "cd = 1", "2018"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "c"}, {"id": "cd = 1"}, {"id": "d"}, {"id": "$c$ , $d$ 互为倒数"}, {"id": "2017 ( a + b ) ^ { 5 } + 2018 ( cd ) ^ { 5 }"}, {"id": "2018"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "2018"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "c", "target": "cd = 1"}, {"rel": "代入", "source": "cd = 1", "target": "2018"}, {"rel": "被描述", "source": "d", "target": "cd = 1"}, {"rel": "限制性描述", "source": "$c$ , $d$ 互为倒数", "target": "cd = 1"}, {"rel": "被代入", "source": "2017 ( a + b ) ^ { 5 } + 2018 ( cd ) ^ { 5 }", "target": "2018"}]}}
|
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{"content": "If the difference between $7 x ^ { m } y ^ { 3 }$ and $- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }$ is still a monomial, then $n ^ { m }$ = ____ ?", "answer": "9", "steps": "Since $7 x ^ my ^ 3$ and $- \\frac { 1 } { 2 } x ^ 2 y ^ n$ have a difference that is still a monomial, therefore $m = 2$, $n = 3$, and thus $n ^ m = 3 ^ 2 = 9$.", "expr_cands": ["7 x ^ { m } y ^ { 3 }", "m", "y", "x", "- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }", "n", "n ^ { m }", "m = 2", "n = 3", "9"], "exprs": ["m = 2", "n = 3", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "7 x ^ { m } y ^ { 3 }"}, {"id": "m = 2"}, {"id": "- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }"}, {"id": "$7 x ^ { m } y ^ { 3 }$ 与 $- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }$ 的差仍然是单项式"}, {"id": "n = 3"}, {"id": "n ^ { m }"}, {"id": "9"}], "links": [{"rel": "被描述", "source": "7 x ^ { m } y ^ { 3 }", "target": "m = 2"}, {"rel": "被描述", "source": "7 x ^ { m } y ^ { 3 }", "target": "n = 3"}, {"rel": "代入", "source": "m = 2", "target": "9"}, {"rel": "被描述", "source": "- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }", "target": "m = 2"}, {"rel": "被描述", "source": "- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }", "target": "n = 3"}, {"rel": "限制性描述", "source": "$7 x ^ { m } y ^ { 3 }$ 与 $- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }$ 的差仍然是单项式", "target": "m = 2"}, {"rel": "限制性描述", "source": "$7 x ^ { m } y ^ { 3 }$ 与 $- \\frac { 1 } { 2 } x ^ { 2 } y ^ { n }$ 的差仍然是单项式", "target": "n = 3"}, {"rel": "代入", "source": "n = 3", "target": "9"}, {"rel": "被代入", "source": "n ^ { m }", "target": "9"}]}}
|
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{"content": "The algebraic expression $\\frac { 1 } { \\sqrt { 3 x - 1 } }$ is meaningful in the range of real numbers, then the range of values that $x$ can take.", "answer": "x > \\frac { 1 } { 3 }", "steps": "From the given condition, we have $3 x - 1 > 0$, which implies $x > \\frac { 1 } { 3 }$.", "expr_cands": ["\\frac { 1 } { \\sqrt { 3 x - 1 } }", "x", "3 x - 1 > 0", "\\frac { 1 } { 3 } < x", "x > \\frac { 1 } { 3 }"], "exprs": ["3 x - 1 > 0", "x > \\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { \\sqrt { 3 x - 1 } }"}, {"id": "3 x - 1 > 0"}, {"id": "代数式 $\\frac { 1 } { \\sqrt { 3 x - 1 } }$ 在实数范围内有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x > \\frac { 1 } { 3 }"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { \\sqrt { 3 x - 1 } }", "target": "3 x - 1 > 0"}, {"rel": "不等式方程求解", "source": "3 x - 1 > 0", "target": "x > \\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 1 } { \\sqrt { 3 x - 1 } }$ 在实数范围内有意义", "target": "3 x - 1 > 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "3 x - 1 > 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "3 x - 1 > 0"}]}}
|
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{"content": "If the fractional equation $\\frac { x - a } { x - 1 } - \\frac { 3 } { x - 1 } = 0$ has no solution for $x$, then the value of the real number $a$ is ____?", "answer": "- 2", "steps": "To eliminate the denominator, we get $x - a - 3 = 0$. Since the fractional equation has no solution, we obtain $x - 1 = 0$, which means $x = 1$. Substituting $x = 1$ into the equation, we get $1 - a - 3 = 0$, and solving for $a$, we get $a = - 2$.", "expr_cands": ["x", "\\frac { x - a } { x - 1 } - \\frac { 3 } { x - 1 } = 0", "a", "x - a - 3 = 0", "x - 1 = 0", "x = 1", "1 - a - 3 = 0", "a = - 2"], "exprs": ["x - a - 3 = 0", "x - 1 = 0", "x = 1", "1 - a - 3 = 0", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - a } { x - 1 } - \\frac { 3 } { x - 1 } = 0"}, {"id": "x - a - 3 = 0"}, {"id": "x - 1 = 0"}, {"id": "有关 $x$ 的分式方程 $\\frac { x - a } { x - 1 } - \\frac { 3 } { x - 1 } = 0$ 无解"}, {"id": "分式方程无解，则分母为0"}, {"id": "x = 1"}, {"id": "1 - a - 3 = 0"}, {"id": "a = - 2"}], "links": [{"rel": "同乘除", "source": "\\frac { x - a } { x - 1 } - \\frac { 3 } { x - 1 } = 0", "target": "x - a - 3 = 0"}, {"rel": "被描述", "source": "\\frac { x - a } { x - 1 } - \\frac { 3 } { x - 1 } = 0", "target": "x - 1 = 0"}, {"rel": "被代入", "source": "x - a - 3 = 0", "target": "1 - a - 3 = 0"}, {"rel": "等式方程求解", "source": "x - 1 = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "有关 $x$ 的分式方程 $\\frac { x - a } { x - 1 } - \\frac { 3 } { x - 1 } = 0$ 无解", "target": "x - 1 = 0"}, {"rel": "属性描述", "source": "分式方程无解，则分母为0", "target": "x - 1 = 0"}, {"rel": "代入", "source": "x = 1", "target": "1 - a - 3 = 0"}, {"rel": "等式方程求解", "source": "1 - a - 3 = 0", "target": "a = - 2"}]}}
|
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{"content": "If $\\frac { x } { 10 } = \\frac { y } { 8 } = \\frac { z } { 9 }$, then $\\frac { x + y + z } { y + z }$ = ____ ?", "answer": "\\frac { 27 } { 17 }", "steps": "Let $\\frac { x } { 10 } = \\frac { y } { 8 } = \\frac { z } { 9 } = k$, then $x = 10 k$, $y = 8 k$, $z = 9 k$. Therefore, $\\frac { x + y + z } { y + z } = \\frac { 10 k + 8 k + 9 k } { 8 k + 9 k } = \\frac { 27 k } { 17 k } = \\frac { 27 } { 17 }$.", "expr_cands": ["\\frac { x } { 10 } = \\frac { y } { 8 } = \\frac { z } { 9 }", "\\frac { x + y + z } { y + z }", "x", "z", "y", "\\frac { x } { 10 } = k", "k", "x = 10 k", "y = 8 k", "z = 9 k", "\\frac { 27 } { 17 }"], "exprs": ["x = 10 k", "y = 8 k", "z = 9 k", "\\frac { 27 } { 17 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $\\frac { x } { 10 } = \\frac { y } { 8 } = \\frac { z } { 9 } = k$"}, {"id": "x = 10 k"}, {"id": "y = 8 k"}, {"id": "z = 9 k"}, {"id": "\\frac { x + y + z } { y + z }"}, {"id": "\\frac { 27 } { 17 }"}], "links": [{"rel": "假设描述", "source": "设 $\\frac { x } { 10 } = \\frac { y } { 8 } = \\frac { z } { 9 } = k$", "target": "x = 10 k"}, {"rel": "假设描述", "source": "设 $\\frac { x } { 10 } = \\frac { y } { 8 } = \\frac { z } { 9 } = k$", "target": "y = 8 k"}, {"rel": "假设描述", "source": "设 $\\frac { x } { 10 } = \\frac { y } { 8 } = \\frac { z } { 9 } = k$", "target": "z = 9 k"}, {"rel": "代入", "source": "x = 10 k", "target": "\\frac { 27 } { 17 }"}, {"rel": "代入", "source": "y = 8 k", "target": "\\frac { 27 } { 17 }"}, {"rel": "代入", "source": "z = 9 k", "target": "\\frac { 27 } { 17 }"}, {"rel": "被代入", "source": "\\frac { x + y + z } { y + z }", "target": "\\frac { 27 } { 17 }"}]}}
|
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{"content": "The simplest quadratic radical $\\sqrt { 16 - 3 m }$ can be combined with $\\sqrt { 4 m - 5 }$. What is the value of $m$?", "answer": "3", "steps": "$\\because$ The simplest quadratic radicals $\\sqrt { 16 - 3 m }$ and $\\sqrt { 4 m - 5 }$ can be combined, $\\therefore$ they are of the same type of quadratic radicals. $\\therefore$ $16 - 3 m = 4 m - 5$, solving for $m$ gives $m = 3$.", "expr_cands": ["\\sqrt { 16 - 3 m }", "m", "\\sqrt { 4 m - 5 }", "16 - 3 m = 4 m - 5", "m = 3"], "exprs": ["16 - 3 m = 4 m - 5", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 16 - 3 m }"}, {"id": "16 - 3 m = 4 m - 5"}, {"id": "\\sqrt { 4 m - 5 }"}, {"id": "最简二次根式 $\\sqrt { 16 - 3 m }$ 与 $\\sqrt { 4 m - 5 }$ 可以合并"}, {"id": "最简二次根式 $\\sqrt { 16 - 3 m }$ 与 $\\sqrt { 4 m - 5 }$ 是同类二次根式"}, {"id": "m = 3"}], "links": [{"rel": "被描述", "source": "\\sqrt { 16 - 3 m }", "target": "16 - 3 m = 4 m - 5"}, {"rel": "等式方程求解", "source": "16 - 3 m = 4 m - 5", "target": "m = 3"}, {"rel": "被描述", "source": "\\sqrt { 4 m - 5 }", "target": "16 - 3 m = 4 m - 5"}, {"rel": "限制性描述", "source": "最简二次根式 $\\sqrt { 16 - 3 m }$ 与 $\\sqrt { 4 m - 5 }$ 可以合并", "target": "16 - 3 m = 4 m - 5"}, {"rel": "限制性描述", "source": "最简二次根式 $\\sqrt { 16 - 3 m }$ 与 $\\sqrt { 4 m - 5 }$ 是同类二次根式", "target": "16 - 3 m = 4 m - 5"}]}}
|
||||
{"content": "If $| 5 - 10 x | = 10 x - 5$, then the possible values of $x$ are _____.", "answer": "x \\ge \\frac { 1 } { 2 }", "steps": "Because $| 5 - 10 x | = 10 x - 5$, therefore $10 x - 5 \\ge 0$, which implies that $x \\ge \\frac { 1 } { 2 }$ after solving for $x$.", "expr_cands": ["| 5 - 10 x | = 10 x - 5", "x", "10 x - 5 \\ge 0", "\\frac { 1 } { 2 } \\le x", "x \\ge \\frac { 1 } { 2 }"], "exprs": ["10 x - 5 \\ge 0", "x \\ge \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| 5 - 10 x | = 10 x - 5"}, {"id": "10 x - 5 \\ge 0"}, {"id": "绝对值恒大于等于0"}, {"id": "x \\ge \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "| 5 - 10 x | = 10 x - 5", "target": "10 x - 5 \\ge 0"}, {"rel": "不等式方程求解", "source": "10 x - 5 \\ge 0", "target": "x \\ge \\frac { 1 } { 2 }"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "10 x - 5 \\ge 0"}]}}
|
||||
{"content": "If the value of the fraction $\\frac { x } { x ^ 2 + 2 }$ is positive, then the range of values for the real number $x$ is ____?", "answer": "x > 0", "steps": "Due to $x ^ { 2 } + 2 > 0$, and since $\\frac { x } { x ^ { 2 } + 2 } > 0$, therefore $x > 0$.", "expr_cands": ["\\frac { x } { x ^ { 2 } + 2 }", "x", "x ^ { 2 } + 2 > 0", "\\frac { x } { x ^ { 2 } + 2 } > 0", "0 < x", "x > 0"], "exprs": ["\\frac { x } { x ^ { 2 } + 2 } > 0", "x > 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x ^ { 2 } + 2 }"}, {"id": "\\frac { x } { x ^ { 2 } + 2 } > 0"}, {"id": "分式 $\\frac { x } { x ^ { 2 } + 2 }$ 的值为正"}, {"id": "分式为正数，则分子分母同号"}, {"id": "x > 0"}, {"id": "多项式偶次方项恒大于等于0"}], "links": [{"rel": "被描述", "source": "\\frac { x } { x ^ { 2 } + 2 }", "target": "\\frac { x } { x ^ { 2 } + 2 } > 0"}, {"rel": "被描述", "source": "\\frac { x } { x ^ { 2 } + 2 } > 0", "target": "x > 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x } { x ^ { 2 } + 2 }$ 的值为正", "target": "\\frac { x } { x ^ { 2 } + 2 } > 0"}, {"rel": "属性描述", "source": "分式为正数，则分子分母同号", "target": "\\frac { x } { x ^ { 2 } + 2 } > 0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "x > 0"}]}}
|
||||
{"content": "If $y = \\sqrt { x - 2 } + \\sqrt { 2 - x }$, then the value of ${ ( x + 2018 ) } ^ { y }$ is ____ ?", "answer": "1", "steps": "\\because $x - 2 \\ge 0$ , $2 - x \\ge 0$ , \\therefore $x = 2$ , so $y = 0$ , \\therefore the original expression = ${ ( 2 + 2018 ) } ^ { 0 } = 1$ .", "expr_cands": ["y = \\sqrt { x - 2 } + \\sqrt { 2 - x }", "x", "y", "{ ( x + 2018 ) } ^ { y }", "x - 2 \\ge 0", "2 \\le x", "2 - x \\ge 0", "x \\le 2", "x = 2", "y = 0", "{ ( 2 + 2018 ) } ^ { 0 }", "1"], "exprs": ["x - 2 \\ge 0", "2 - x \\ge 0", "x = 2", "y = 0", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x }"}, {"id": "x - 2 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "2 - x \\ge 0"}, {"id": "x = 2"}, {"id": "y = 0"}, {"id": "{ ( x + 2018 ) } ^ { y }"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x }", "target": "x - 2 \\ge 0"}, {"rel": "被描述", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x }", "target": "2 - x \\ge 0"}, {"rel": "被代入", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x }", "target": "y = 0"}, {"rel": "联立", "source": "x - 2 \\ge 0", "target": "x = 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 2 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 - x \\ge 0"}, {"rel": "联立", "source": "2 - x \\ge 0", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "y = 0"}, {"rel": "代入", "source": "x = 2", "target": "1"}, {"rel": "代入", "source": "y = 0", "target": "1"}, {"rel": "被代入", "source": "{ ( x + 2018 ) } ^ { y }", "target": "1"}]}}
|
||||
{"content": "If we want the fraction $\\frac { 5 } { - x + 5 }$ to be meaningful, the range of values for $x$ is ____?", "answer": "x \\neq 5", "steps": "$\\because$ The fraction $\\frac { 5 } { - x + 5 }$ is meaningful, $\\therefore$ $- x + 5 \\neq 0$, $\\therefore$ $x \\neq 5$.", "expr_cands": ["\\frac { 5 } { - x + 5 }", "x", "- x + 5 \\neq 0", "x \\neq 5"], "exprs": ["- x + 5 \\neq 0", "x \\neq 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 5 } { - x + 5 }"}, {"id": "- x + 5 \\neq 0"}, {"id": "要使分式 $\\frac { 5 } { - x + 5 }$ 有意义"}, {"id": "$x$ 的取值范围"}, {"id": "分式 $\\frac { 5 } { - x + 5 }$ 有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 5"}], "links": [{"rel": "被描述", "source": "\\frac { 5 } { - x + 5 }", "target": "- x + 5 \\neq 0"}, {"rel": "不等式方程求解", "source": "- x + 5 \\neq 0", "target": "x \\neq 5"}, {"rel": "限制性描述", "source": "要使分式 $\\frac { 5 } { - x + 5 }$ 有意义", "target": "- x + 5 \\neq 0"}, {"rel": "限制性描述", "source": "$x$ 的取值范围", "target": "- x + 5 \\neq 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { 5 } { - x + 5 }$ 有意义", "target": "- x + 5 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "- x + 5 \\neq 0"}]}}
|
||||
{"content": "If $( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0$ is a linear equation in one variable, then $m$ = ____ ?", "answer": "2", "steps": "From the given condition, we have $| m | - 1 = 1$ and $m + 2 \\neq 0$. Solving for $m$, we get $m = 2$.", "expr_cands": ["( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0", "m", "x", "| m | - 1 = 1", "m = - 2", "m = 2", "m + 2 \\neq 0", "m \\neq - 2"], "exprs": ["| m | - 1 = 1", "m + 2 \\neq 0", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0"}, {"id": "| m | - 1 = 1"}, {"id": "$( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0$ 是一元一次方程"}, {"id": "m + 2 \\neq 0"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0", "target": "| m | - 1 = 1"}, {"rel": "被描述", "source": "( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0", "target": "m + 2 \\neq 0"}, {"rel": "联立", "source": "| m | - 1 = 1", "target": "m = 2"}, {"rel": "限制性描述", "source": "$( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0$ 是一元一次方程", "target": "| m | - 1 = 1"}, {"rel": "限制性描述", "source": "$( m + 2 ) { x } ^ { | m | - 1 } + 8 = 0$ 是一元一次方程", "target": "m + 2 \\neq 0"}, {"rel": "联立", "source": "m + 2 \\neq 0", "target": "m = 2"}]}}
|
||||
{"content": "If $x = 3 - 2 a$ is a solution to the inequality $2 ( x - 3 ) < x - 1$, then the possible values of $a$ are _____.", "answer": "a > - 1", "steps": "From the given problem, we can derive that $2 ( 3 - 2 a - 3 ) < 3 - 2 a - 1$, $- 4 a < 2 - 2 a$, $- 2 a < 2$, and $a > - 1$.", "expr_cands": ["x = 3 - 2 a", "a", "x", "2 ( x - 3 ) < x - 1", "2 ( 3 - 2 a - 3 ) < 3 - 2 a - 1", "- 1 < a", "- 4 a < 2 - 2 a", "- 2 a < 2", "a > - 1"], "exprs": ["a > - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 3 - 2 a"}, {"id": "a > - 1"}, {"id": "2 ( x - 3 ) < x - 1"}], "links": [{"rel": "联立", "source": "x = 3 - 2 a", "target": "a > - 1"}, {"rel": "联立", "source": "2 ( x - 3 ) < x - 1", "target": "a > - 1"}]}}
|
||||
{"content": "The equation of the line obtained by translating the line $y = \\frac { 1 } { 3 } x + 1$ downward by $3$ units is _____.", "answer": "y = \\frac { 1 } { 3 } x - 2", "steps": "The equation of the line $y = \\frac { 1 } { 3 } x + 1$ after being translated downward by $3$ units is $y = \\frac { 1 } { 3 } x + 1 - 3$, which simplifies to $y = \\frac { 1 } { 3 } x - 2$.", "expr_cands": ["y = \\frac { 1 } { 3 } x + 1", "y", "x", "3", "y = \\frac { 1 } { 3 } x + 1 - 3", "\\frac { x } { 3 } + 1 = \\frac { 1 } { 3 } x + 1 - 3", "\\frac { 1 } { 3 } x - 2"], "exprs": ["y = \\frac { 1 } { 3 } x + 1 - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 1 } { 3 } x + 1"}, {"id": "y = \\frac { 1 } { 3 } x + 1 - 3"}, {"id": "3"}, {"id": "将直线 $y = \\frac { 1 } { 3 } x + 1$ 向下平移 $3$ 个单位所得直线的解析式"}], "links": [{"rel": "被描述", "source": "y = \\frac { 1 } { 3 } x + 1", "target": "y = \\frac { 1 } { 3 } x + 1 - 3"}, {"rel": "被描述", "source": "3", "target": "y = \\frac { 1 } { 3 } x + 1 - 3"}, {"rel": "限制性描述", "source": "将直线 $y = \\frac { 1 } { 3 } x + 1$ 向下平移 $3$ 个单位所得直线的解析式", "target": "y = \\frac { 1 } { 3 } x + 1 - 3"}]}}
|
||||
{"content": "If $x = 2$ is a root of the quadratic equation $x ^ 2 - mx - 10 = 0$, then $m$ is equal to ____?", "answer": "- 3", "steps": "Substituting $x = 2$ into $x ^ 2 - mx - 10 = 0$, we get $4 - 2 m - 10 = 0$, so $m = - 3$.", "expr_cands": ["x = 2", "x", "x ^ { 2 } - mx - 10 = 0", "m", "- 2 m - 10 + 4 = 0", "4 - 2 m - 10 = 0", "m = - 3"], "exprs": ["4 - 2 m - 10 = 0", "m = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - mx - 10 = 0"}, {"id": "4 - 2 m - 10 = 0"}, {"id": "x = 2"}, {"id": "m = - 3"}], "links": [{"rel": "被代入", "source": "x ^ { 2 } - mx - 10 = 0", "target": "4 - 2 m - 10 = 0"}, {"rel": "等式方程求解", "source": "4 - 2 m - 10 = 0", "target": "m = - 3"}, {"rel": "代入", "source": "x = 2", "target": "4 - 2 m - 10 = 0"}]}}
|
||||
{"content": "If the function $y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }$ is an inverse proportion function, then the value of $n$ is ____?", "answer": "1", "steps": "According to the problem, we have $n ^ 2 - 5 n + 3 = - 1$ and $n - 4 \\neq 0$. Solving for $n$, we get $n = 1$.", "expr_cands": ["y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }", "y", "n", "x", "n ^ { 2 } - 5 n + 3 = - 1", "n = 1", "n = 4", "n - 4 \\neq 0", "n \\neq 4"], "exprs": ["n ^ { 2 } - 5 n + 3 = - 1", "n - 4 \\neq 0", "n = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }"}, {"id": "n ^ { 2 } - 5 n + 3 = - 1"}, {"id": "函数 $y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }$ 是反比例函数"}, {"id": "n - 4 \\neq 0"}, {"id": "n = 1"}], "links": [{"rel": "被描述", "source": "y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }", "target": "n ^ { 2 } - 5 n + 3 = - 1"}, {"rel": "被描述", "source": "y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }", "target": "n - 4 \\neq 0"}, {"rel": "联立", "source": "n ^ { 2 } - 5 n + 3 = - 1", "target": "n = 1"}, {"rel": "限制性描述", "source": "函数 $y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }$ 是反比例函数", "target": "n ^ { 2 } - 5 n + 3 = - 1"}, {"rel": "限制性描述", "source": "函数 $y = ( n - 4 ) x ^ { { n } ^ { 2 } - 5 n + 3 }$ 是反比例函数", "target": "n - 4 \\neq 0"}, {"rel": "联立", "source": "n - 4 \\neq 0", "target": "n = 1"}]}}
|
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{"content": "If the inverse proportion function $y = \\frac { k - 1 } { x }$ is in the first and third quadrants, then the range of values for $k$ is ____?", "answer": "k > 1", "steps": "According to the problem, we have $k - 1 > 0$, which implies $k > 1$ after solving.", "expr_cands": ["y = \\frac { k - 1 } { x }", "y", "k", "x", "k - 1 > 0", "1 < k", "k > 1"], "exprs": ["k - 1 > 0", "k > 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { k - 1 } { x }"}, {"id": "k - 1 > 0"}, {"id": "反比例函数 $y = \\frac { k - 1 } { x }$ 在第一"}, {"id": "三象限"}, {"id": "k > 1"}], "links": [{"rel": "被描述", "source": "y = \\frac { k - 1 } { x }", "target": "k - 1 > 0"}, {"rel": "不等式方程求解", "source": "k - 1 > 0", "target": "k > 1"}, {"rel": "限制性描述", "source": "反比例函数 $y = \\frac { k - 1 } { x }$ 在第一", "target": "k - 1 > 0"}, {"rel": "限制性描述", "source": "三象限", "target": "k - 1 > 0"}]}}
|
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{"content": "In the quadratic equation $y = x + 5$, if $x = - 3$, then $y$ = ____ ?", "answer": "2", "steps": "Substituting $x = - 3$ into $y = x + 5$, we get $y = - 3 + 5$, which is equal to $2$. Therefore, the value of $y$ is $2$.", "expr_cands": ["y = x + 5", "y", "x", "x = - 3", "y = 2", "y = - 3 + 5", "2"], "exprs": ["y = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x + 5"}, {"id": "y = 2"}, {"id": "x = - 3"}], "links": [{"rel": "被代入", "source": "y = x + 5", "target": "y = 2"}, {"rel": "代入", "source": "x = - 3", "target": "y = 2"}]}}
|
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{"content": "If the fractional equation $1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }$ has a positive root, then $k$ = ____?", "answer": "- 1", "steps": "$1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }$ , $x - 2 + k = - 1$ , $x = 1 - k$ , since the fractional equation $1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }$ has a proper root, therefore $x - 2 = 0$, which gives $x = 2$. Hence, $2 = 1 - k$, and we get $k = - 1$. ", "expr_cands": ["1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }", "k", "x", "x - 2 + k = - 1", "x = 1 - k", "\\frac { 1 } { k + 1 } = \\frac { 1 } { k + 1 }", "x - 2 = 0", "x = 2", "2 = 1 - k", "k = - 1"], "exprs": ["x - 2 + k = - 1", "x - 2 = 0", "x = 1 - k", "x = 2", "2 = 1 - k", "k = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }"}, {"id": "x - 2 + k = - 1"}, {"id": "x = 1 - k"}, {"id": "x - 2 = 0"}, {"id": "分式方程 $1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }$ 有增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 2"}, {"id": "2 = 1 - k"}, {"id": "k = - 1"}], "links": [{"rel": "同乘除", "source": "1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }", "target": "x - 2 + k = - 1"}, {"rel": "被描述", "source": "1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }", "target": "x - 2 = 0"}, {"rel": "等式方程部分求解", "source": "x - 2 + k = - 1", "target": "x = 1 - k"}, {"rel": "被代入", "source": "x = 1 - k", "target": "2 = 1 - k"}, {"rel": "等式方程求解", "source": "x - 2 = 0", "target": "x = 2"}, {"rel": "限制性描述", "source": "分式方程 $1 + \\frac { k } { x - 2 } = \\frac { 1 } { 2 - x }$ 有增根", "target": "x - 2 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 2 = 0"}, {"rel": "代入", "source": "x = 2", "target": "2 = 1 - k"}, {"rel": "等式方程求解", "source": "2 = 1 - k", "target": "k = - 1"}]}}
|
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{"content": "$mx ^ { n } y$ is a monomial in terms of $x$ and $y$, with a coefficient of $3$ and a degree of $4$. Then $m + n$ = ____?", "answer": "6", "steps": "From the given information, we know that $m = 3$, $n + 1 = 4$, so $m = 3$ and $n = 3$. Therefore, $m + n = 6$.", "expr_cands": ["mx ^ { n } y", "n", "y", "m", "x", "3", "4", "m + n", "m = 3", "n + 1 = 4", "n = 3", "6"], "exprs": ["m = 3", "n + 1 = 4", "n = 3", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "m = 3"}, {"id": "mx ^ { n } y"}, {"id": "y"}, {"id": "x"}, {"id": "$mx ^ { n } y$ 是关于 $x$ , $y$ 的一个单项式"}, {"id": "且系数是 $3$"}, {"id": "n + 1 = 4"}, {"id": "4"}, {"id": "次数是 $4$"}, {"id": "n = 3"}, {"id": "m + n"}, {"id": "6"}], "links": [{"rel": "被描述", "source": "3", "target": "m = 3"}, {"rel": "代入", "source": "m = 3", "target": "6"}, {"rel": "被描述", "source": "mx ^ { n } y", "target": "m = 3"}, {"rel": "被描述", "source": "mx ^ { n } y", "target": "n + 1 = 4"}, {"rel": "被描述", "source": "y", "target": "m = 3"}, {"rel": "被描述", "source": "y", "target": "n + 1 = 4"}, {"rel": "被描述", "source": "x", "target": "m = 3"}, {"rel": "被描述", "source": "x", "target": "n + 1 = 4"}, {"rel": "限制性描述", "source": "$mx ^ { n } y$ 是关于 $x$ , $y$ 的一个单项式", "target": "m = 3"}, {"rel": "限制性描述", "source": "$mx ^ { n } y$ 是关于 $x$ , $y$ 的一个单项式", "target": "n + 1 = 4"}, {"rel": "限制性描述", "source": "且系数是 $3$", "target": "m = 3"}, {"rel": "等式方程求解", "source": "n + 1 = 4", "target": "n = 3"}, {"rel": "被描述", "source": "4", "target": "n + 1 = 4"}, {"rel": "限制性描述", "source": "次数是 $4$", "target": "n + 1 = 4"}, {"rel": "代入", "source": "n = 3", "target": "6"}, {"rel": "被代入", "source": "m + n", "target": "6"}]}}
|
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{"content": "If the monomial $0.2 { a } ^ { n } { b } ^ { m + 2 }$ and the monomial $4 { a } ^ { 4 } { b } ^ { n }$ sum up to a monomial, then $( - m ) ^ { n }$ = ____ ?", "answer": "16", "steps": "From the given information, we can conclude that the two expressions can be combined, which means that they are like terms. Therefore, we have $n = 4$ and $m + 2 = n$. Thus, $m = 2$. Therefore, $( - m ) ^ n = ( - 2 ) ^ 4 = 16$.", "expr_cands": ["0.2 { a } ^ { n } { b } ^ { m + 2 }", "a", "n", "b", "m", "4 { a } ^ { 4 } { b } ^ { n }", "( - m ) ^ { n }", "n = 4", "m + 2 = n", "m + 2 = 4", "m = 2", "16"], "exprs": ["n = 4", "m + 2 = n", "m + 2 = 4", "m = 2", "16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "0.2 { a } ^ { n } { b } ^ { m + 2 }"}, {"id": "n = 4"}, {"id": "4 { a } ^ { 4 } { b } ^ { n }"}, {"id": "单项式 $0.2 { a } ^ { n } { b } ^ { m + 2 }$ 与单项式 $4 { a } ^ { 4 } { b } ^ { n }$ 的和仍为单项式"}, {"id": "两者可以合并说明两式为同类项"}, {"id": "m + 2 = n"}, {"id": "m + 2 = 4"}, {"id": "m = 2"}, {"id": "( - m ) ^ { n }"}, {"id": "16"}], "links": [{"rel": "被描述", "source": "0.2 { a } ^ { n } { b } ^ { m + 2 }", "target": "n = 4"}, {"rel": "被描述", "source": "0.2 { a } ^ { n } { b } ^ { m + 2 }", "target": "m + 2 = n"}, {"rel": "代入", "source": "n = 4", "target": "m + 2 = 4"}, {"rel": "代入", "source": "n = 4", "target": "16"}, {"rel": "被描述", "source": "4 { a } ^ { 4 } { b } ^ { n }", "target": "n = 4"}, {"rel": "被描述", "source": "4 { a } ^ { 4 } { b } ^ { n }", "target": "m + 2 = n"}, {"rel": "限制性描述", "source": "单项式 $0.2 { a } ^ { n } { b } ^ { m + 2 }$ 与单项式 $4 { a } ^ { 4 } { b } ^ { n }$ 的和仍为单项式", "target": "n = 4"}, {"rel": "限制性描述", "source": "单项式 $0.2 { a } ^ { n } { b } ^ { m + 2 }$ 与单项式 $4 { a } ^ { 4 } { b } ^ { n }$ 的和仍为单项式", "target": "m + 2 = n"}, {"rel": "限制性描述", "source": "两者可以合并说明两式为同类项", "target": "n = 4"}, {"rel": "限制性描述", "source": "两者可以合并说明两式为同类项", "target": "m + 2 = n"}, {"rel": "被代入", "source": "m + 2 = n", "target": "m + 2 = 4"}, {"rel": "等式方程求解", "source": "m + 2 = 4", "target": "m = 2"}, {"rel": "代入", "source": "m = 2", "target": "16"}, {"rel": "被代入", "source": "( - m ) ^ { n }", "target": "16"}]}}
|
||||
{"content": "If $x = 1$ is a real root of the quadratic equation $x ^ 2 + ( k + 1 ) x + 2 = 0$ in terms of $x$, then the other real root is ____?", "answer": "2", "steps": "The quadratic equation $x ^ 2 + ( k + 1 ) x + 2 = 0$ has two real roots, which are $1$ and $m$. The product of the roots, $x _ 1 \\times x _ 2$, is equal to $\\frac { c } { a }$. Since $1 \\times m = 2$, we have $m = 2$.", "expr_cands": ["x = 1", "x", "x ^ { 2 } + ( k + 1 ) x + 2 = 0", "k", "1", "m", "x _ { 1 } \\times x _ { 2 } = \\frac { c } { a }", "c", "x _ { 1 }", "x _ { 2 }", "a", "1 * m = 2", "m = 2"], "exprs": ["m", "1 * m = 2", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设一元二次方程 $x ^ { 2 } + ( k + 1 ) x + 2 = 0$ 的两个实数根分别为 $1$ 和 $m$ , $x _ { 1 } \\times x _ { 2 } = \\frac { c } { a }$ , $1 * m = 2$"}, {"id": "m"}, {"id": "x = 1"}, {"id": "1 * m = 2"}, {"id": "x ^ { 2 } + ( k + 1 ) x + 2 = 0"}, {"id": "$x = 1$ 是关于 $x$ 的一元二次方程 $x ^ { 2 } + ( k + 1 ) x + 2 = 0$ 的一个实数根"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "m = 2"}], "links": [{"rel": "假设描述", "source": "设一元二次方程 $x ^ { 2 } + ( k + 1 ) x + 2 = 0$ 的两个实数根分别为 $1$ 和 $m$ , $x _ { 1 } \\times x _ { 2 } = \\frac { c } { a }$ , $1 * m = 2$", "target": "m"}, {"rel": "限制性描述", "source": "设一元二次方程 $x ^ { 2 } + ( k + 1 ) x + 2 = 0$ 的两个实数根分别为 $1$ 和 $m$ , $x _ { 1 } \\times x _ { 2 } = \\frac { c } { a }$ , $1 * m = 2$", "target": "1 * m = 2"}, {"rel": "被描述", "source": "m", "target": "1 * m = 2"}, {"rel": "被描述", "source": "x = 1", "target": "1 * m = 2"}, {"rel": "等式方程求解", "source": "1 * m = 2", "target": "m = 2"}, {"rel": "被描述", "source": "x ^ { 2 } + ( k + 1 ) x + 2 = 0", "target": "1 * m = 2"}, {"rel": "限制性描述", "source": "$x = 1$ 是关于 $x$ 的一元二次方程 $x ^ { 2 } + ( k + 1 ) x + 2 = 0$ 的一个实数根", "target": "1 * m = 2"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "1 * m = 2"}]}}
|
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{"content": "If $m$ and $n$ are opposite numbers, then $m + ( - 5 ) + n$ = ____?", "answer": "- 5", "steps": "$\\because$ $m$ and $n$ are opposite numbers, $\\therefore$ $m + n = 0$, $m + ( - 5 ) + n = m + n - 5 = 0 - 5 = - 5$.", "expr_cands": ["m", "n", "m + ( - 5 ) + n", "m + n = 0", "m + n - 5", "- 5"], "exprs": ["m + n = 0", "- 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m"}, {"id": "m + n = 0"}, {"id": "n"}, {"id": "$m$ , $n$ 互为相反数"}, {"id": ", $m$ , $n$ 互为相反数"}, {"id": "m + ( - 5 ) + n"}, {"id": "- 5"}], "links": [{"rel": "被描述", "source": "m", "target": "m + n = 0"}, {"rel": "代入", "source": "m + n = 0", "target": "- 5"}, {"rel": "被描述", "source": "n", "target": "m + n = 0"}, {"rel": "限制性描述", "source": "$m$ , $n$ 互为相反数", "target": "m + n = 0"}, {"rel": "限制性描述", "source": ", $m$ , $n$ 互为相反数", "target": "m + n = 0"}, {"rel": "被代入", "source": "m + ( - 5 ) + n", "target": "- 5"}]}}
|
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{"content": "When $a$ = ____ ?, $y = x ^ { 2 a - 1 }$ is a direct proportion function.", "answer": "1", "steps": "Since $y = x ^ { 2 a - 1 }$ is a direct proportion function, therefore $2 a - 1 = 1$, solving for $a$ gives $a = 1$.", "expr_cands": ["a", "y = x ^ { 2 a - 1 }", "y", "x", "2 a - 1 = 1", "a = 1"], "exprs": ["2 a - 1 = 1", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 a - 1 }"}, {"id": "2 a - 1 = 1"}, {"id": "$y = x ^ { 2 a - 1 }$ 是正比例函数"}, {"id": "a = 1"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 a - 1 }", "target": "2 a - 1 = 1"}, {"rel": "等式方程求解", "source": "2 a - 1 = 1", "target": "a = 1"}, {"rel": "限制性描述", "source": "$y = x ^ { 2 a - 1 }$ 是正比例函数", "target": "2 a - 1 = 1"}]}}
|
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{"content": "If $y = 2 x ^ { m - 5 }$ is an inverse proportion function, then $m$ = ____?", "answer": "4", "steps": "Since $y = 2 x ^ { m - 5 }$ is an inverse proportion function, therefore $m - 5 = - 1$, which solves to $m = 4$.", "expr_cands": ["y = 2 x ^ { m - 5 }", "x", "y", "m", "m - 5 = - 1", "m = 4"], "exprs": ["m - 5 = - 1", "m = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 x ^ { m - 5 }"}, {"id": "m - 5 = - 1"}, {"id": "$y = 2 x ^ { m - 5 }$ 为反比例函数"}, {"id": "m = 4"}], "links": [{"rel": "被描述", "source": "y = 2 x ^ { m - 5 }", "target": "m - 5 = - 1"}, {"rel": "等式方程求解", "source": "m - 5 = - 1", "target": "m = 4"}, {"rel": "限制性描述", "source": "$y = 2 x ^ { m - 5 }$ 为反比例函数", "target": "m - 5 = - 1"}]}}
|
||||
{"content": "The roots of the equation $( x - 1 ) ^ 3 - 8 = 0$ are _____.", "answer": "x = 3", "steps": "Move terms, we get $( x - 1 ) ^ 3 = 8$, take the cube root, we get $x - 1 = 2$ $\\therefore$ $x = 3$.", "expr_cands": ["( x - 1 ) ^ { 3 } - 8 = 0", "x", "( x - 1 ) ^ { 3 } = 8", "x = 3", "x - 1 = 2"], "exprs": ["x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x - 1 ) ^ { 3 } - 8 = 0"}, {"id": "x = 3"}], "links": [{"rel": "等式方程求解", "source": "( x - 1 ) ^ { 3 } - 8 = 0", "target": "x = 3"}]}}
|
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{"content": "In the real number system, what is the value of the algebraic expression $| | \\sqrt { - ( x + 5 ) ^ { 2 } } - 2 | - 3 |$?", "answer": "1", "steps": "From the fact that the radicand of the quadratic radical is greater than or equal to $0$, we know that $- ( x + 5 ) ^ 2 = 0$. Therefore, the original expression is $|| 0 - 2 | - 3 | = | 2 - 3 | = | - 1 | = 1$.", "expr_cands": ["| | \\sqrt { - ( x + 5 ) ^ { 2 } } - 2 | - 3 |", "x", "0", "- ( x + 5 ) ^ { 2 } = 0", "x = - 5", "| | 0 - 2 | - 3 |", "1"], "exprs": ["- ( x + 5 ) ^ { 2 } = 0", "x = - 5", "| | 0 - 2 | - 3 |", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| | \\sqrt { - ( x + 5 ) ^ { 2 } } - 2 | - 3 |"}, {"id": "- ( x + 5 ) ^ { 2 } = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x = - 5"}, {"id": "| | 0 - 2 | - 3 |"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "| | \\sqrt { - ( x + 5 ) ^ { 2 } } - 2 | - 3 |", "target": "- ( x + 5 ) ^ { 2 } = 0"}, {"rel": "被代入", "source": "| | \\sqrt { - ( x + 5 ) ^ { 2 } } - 2 | - 3 |", "target": "| | 0 - 2 | - 3 |"}, {"rel": "等式方程求解", "source": "- ( x + 5 ) ^ { 2 } = 0", "target": "x = - 5"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "- ( x + 5 ) ^ { 2 } = 0"}, {"rel": "代入", "source": "x = - 5", "target": "| | 0 - 2 | - 3 |"}, {"rel": "计算", "source": "| | 0 - 2 | - 3 |", "target": "1"}]}}
|
||||
{"content": "If $x = 1$ is a real root of the quadratic equation $x ^ 2 + 2 x + k = 0$, then the value of $k$ is ____?", "answer": "- 3", "steps": "Substituting $x = 1$ into the equation $x ^ 2 + 2 x + k = 0$ yields $1 + 2 + k = 0$, which gives the solution $k = - 3$.", "expr_cands": ["x = 1", "x", "x ^ { 2 } + 2 x + k = 0", "k", "k + 3 = 0", "1 + 2 + k = 0", "k = - 3"], "exprs": ["1 + 2 + k = 0", "k = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + 2 x + k = 0"}, {"id": "1 + 2 + k = 0"}, {"id": "x = 1"}, {"id": "k = - 3"}], "links": [{"rel": "被代入", "source": "x ^ { 2 } + 2 x + k = 0", "target": "1 + 2 + k = 0"}, {"rel": "等式方程求解", "source": "1 + 2 + k = 0", "target": "k = - 3"}, {"rel": "代入", "source": "x = 1", "target": "1 + 2 + k = 0"}]}}
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{"content": "Given a linear function $y = ( a - 2 ) x + 3$ where the function value $y$ decreases as the independent variable $x$ increases, the range of real numbers for $a$ is ____?", "answer": "a < 2", "steps": "$\\because$ For a linear function in terms of $x$, $y = ( a - 2 ) x + 3$, $y$ decreases as $x$ increases. $\\therefore$ $a - 2 < 0$, which implies $a < 2$.", "expr_cands": ["y = ( a - 2 ) x + 3", "y", "a", "x", "a - 2 < 0", "a < 2"], "exprs": ["a - 2 < 0", "a < 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( a - 2 ) x + 3"}, {"id": "a - 2 < 0"}, {"id": "y"}, {"id": "x"}, {"id": "一次函数 $y = ( a - 2 ) x + 3$ 的函数值 $y$ 随着自变量 $x$ 的值增大而减小"}, {"id": "a < 2"}], "links": [{"rel": "被描述", "source": "y = ( a - 2 ) x + 3", "target": "a - 2 < 0"}, {"rel": "不等式方程求解", "source": "a - 2 < 0", "target": "a < 2"}, {"rel": "被描述", "source": "y", "target": "a - 2 < 0"}, {"rel": "被描述", "source": "x", "target": "a - 2 < 0"}, {"rel": "限制性描述", "source": "一次函数 $y = ( a - 2 ) x + 3$ 的函数值 $y$ 随着自变量 $x$ 的值增大而减小", "target": "a - 2 < 0"}]}}
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{"content": "The equation of the line obtained by translating the line $y = 3 x$ $2$ units to the right along the $x$-axis is _____.", "answer": "y = 3 x - 6", "steps": "Translate the above math content in English, you should keep the content wrapped in $unchanged.The straight line$y = 3x$is translated$2$units to the right along the$x$- axis , and the resulting function is$y = 3(x-2)$, which can also be written as$y = 3x - 6$.", "expr_cands": ["y = 3 x", "y", "x", "2", "y = 3 ( x - 2 )", "3 x = 3 ( x - 2 )", "3 x - 6"], "exprs": ["y = 3 ( x - 2 )"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "y = 3 ( x - 2 )"}, {"id": "y = 3 x"}, {"id": "x"}, {"id": "把直线 $y = 3 x$ 沿 $x$ 轴向右平移 $2$ 个单位"}, {"id": "到的直线解析式"}], "links": [{"rel": "被描述", "source": "2", "target": "y = 3 ( x - 2 )"}, {"rel": "被描述", "source": "y = 3 x", "target": "y = 3 ( x - 2 )"}, {"rel": "被描述", "source": "x", "target": "y = 3 ( x - 2 )"}, {"rel": "限制性描述", "source": "把直线 $y = 3 x$ 沿 $x$ 轴向右平移 $2$ 个单位", "target": "y = 3 ( x - 2 )"}, {"rel": "限制性描述", "source": "到的直线解析式", "target": "y = 3 ( x - 2 )"}]}}
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{"content": "$( a + 8 ) ^ { 2 } + | b - 5 | + \\sqrt { 3 - c } = 0$, then a + b + c = ____ ?", "answer": "0", "steps": "From the given conditions, we have $a + 8 = 0$, $b - 5 = 0$, and $3 - c = 0$. Solving for $a$, $b$, and $c$, we get $a = - 8$, $b = 5$, and $c = 3$. Therefore, $a + b + c = 0$.", "expr_cands": ["( a + 8 ) ^ { 2 } + | b - 5 | + \\sqrt { 3 - c } = 0", "c", "a", "b", "a + b + c", "a + 8 = 0", "a = - 8", "b - 5 = 0", "b = 5", "3 - c = 0", "c = 3", "0"], "exprs": ["a + 8 = 0", "b - 5 = 0", "3 - c = 0", "a = - 8", "b = 5", "c = 3", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a + 8 ) ^ { 2 } + | b - 5 | + \\sqrt { 3 - c } = 0"}, {"id": "a + 8 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "b - 5 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "3 - c = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a = - 8"}, {"id": "b = 5"}, {"id": "c = 3"}, {"id": "a + b + c"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "( a + 8 ) ^ { 2 } + | b - 5 | + \\sqrt { 3 - c } = 0", "target": "a + 8 = 0"}, {"rel": "被描述", "source": "( a + 8 ) ^ { 2 } + | b - 5 | + \\sqrt { 3 - c } = 0", "target": "b - 5 = 0"}, {"rel": "被描述", "source": "( a + 8 ) ^ { 2 } + | b - 5 | + \\sqrt { 3 - c } = 0", "target": "3 - c = 0"}, {"rel": "等式方程求解", "source": "a + 8 = 0", "target": "a = - 8"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "a + 8 = 0"}, {"rel": "等式方程求解", "source": "b - 5 = 0", "target": "b = 5"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b - 5 = 0"}, {"rel": "等式方程求解", "source": "3 - c = 0", "target": "c = 3"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "3 - c = 0"}, {"rel": "代入", "source": "a = - 8", "target": "0"}, {"rel": "代入", "source": "b = 5", "target": "0"}, {"rel": "代入", "source": "c = 3", "target": "0"}, {"rel": "被代入", "source": "a + b + c", "target": "0"}]}}
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{"content": "If $\\sqrt { m }$ and $\\sqrt { 18 }$ are of the same type of quadratic surd, then the smallest positive integer value of $m$ is ____?", "answer": "2", "steps": "$\\sqrt { 18 } = 3 \\sqrt { 2 }$ , because $\\sqrt { m }$ and $\\sqrt { 18 }$ are of the same type of quadratic radical, therefore the smallest positive integer value of $m$ is $2$.", "expr_cands": ["\\sqrt { m }", "m", "\\sqrt { 18 }", "3 \\sqrt { 2 }", "2"], "exprs": ["3 \\sqrt { 2 }", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 18 }"}, {"id": "3 \\sqrt { 2 }"}, {"id": "\\sqrt { m }"}, {"id": "2"}, {"id": "m"}, {"id": "$\\sqrt { m }$ 与 $\\sqrt { 18 }$ 是同类二次根式"}, {"id": "$m$ 的最小正整数值"}], "links": [{"rel": "计算", "source": "\\sqrt { 18 }", "target": "3 \\sqrt { 2 }"}, {"rel": "被描述", "source": "3 \\sqrt { 2 }", "target": "2"}, {"rel": "被描述", "source": "\\sqrt { m }", "target": "2"}, {"rel": "被描述", "source": "m", "target": "2"}, {"rel": "限制性描述", "source": "$\\sqrt { m }$ 与 $\\sqrt { 18 }$ 是同类二次根式", "target": "2"}, {"rel": "限制性描述", "source": "$m$ 的最小正整数值", "target": "2"}]}}
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{"content": "If $a$ and $b$ are opposite numbers ($a \\neq 0$), and $x$ and $y$ are reciprocal, then the value of the algebraic expression $\\frac { a + b } { 2 } - xy - \\frac { a } { b }$ is ____?", "answer": "0", "steps": "According to the problem, we have $a + b = 0$, $xy = 1$, and $\\frac { a } { b } = - 1$. Therefore, the original expression is equal to $0 - 1 + 1 = 0$.", "expr_cands": ["a", "b", "x", "y", "\\frac { a + b } { 2 } - xy - \\frac { a } { b }", "a + b = 0", "xy = 1", "\\frac { a } { b } = - 1", "0 - 1 + 1", "0"], "exprs": ["a + b = 0", "xy = 1", "\\frac { a } { b } = - 1", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数 a \\neq 0 ) , $x$ , $y$ 互为倒数"}, {"id": "x"}, {"id": "xy = 1"}, {"id": "y"}, {"id": "\\frac { a } { b } = - 1"}, {"id": "\\frac { a + b } { 2 } - xy - \\frac { a } { b }"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "被描述", "source": "a", "target": "\\frac { a } { b } = - 1"}, {"rel": "代入", "source": "a + b = 0", "target": "0"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "被描述", "source": "b", "target": "\\frac { a } { b } = - 1"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数 a \\neq 0 ) , $x$ , $y$ 互为倒数", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数 a \\neq 0 ) , $x$ , $y$ 互为倒数", "target": "xy = 1"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数 a \\neq 0 ) , $x$ , $y$ 互为倒数", "target": "\\frac { a } { b } = - 1"}, {"rel": "被描述", "source": "x", "target": "xy = 1"}, {"rel": "代入", "source": "xy = 1", "target": "0"}, {"rel": "被描述", "source": "y", "target": "xy = 1"}, {"rel": "代入", "source": "\\frac { a } { b } = - 1", "target": "0"}, {"rel": "被代入", "source": "\\frac { a + b } { 2 } - xy - \\frac { a } { b }", "target": "0"}]}}
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{"content": "The two roots of the quadratic equation $x ^ 2 + 4 x - 5 = 0$ are denoted by $a$ and $b$. Find the value of $a ^ 2 + b ^ 2$.", "answer": "26", "steps": "Because the equation $x ^ 2 + 4 x - 5 = 0$ has two roots, denoted as $a$ and $b$, therefore $a + b = - 4$ and $ab = - 5$. Then, we have $a ^ 2 + b ^ 2 = ( a + b ) ^ 2 - 2 ab = 16 + 10 = 26$.", "expr_cands": ["x ^ { 2 } + 4 x - 5 = 0", "x", "a", "b", "a ^ { 2 } + b ^ { 2 }", "x = - 5", "x = 1", "a + b = - 4", "ab = - 5", "( a + b ) ^ { 2 } - 2 ab", "26"], "exprs": ["a + b = - 4", "ab = - 5", "( a + b ) ^ { 2 } - 2 ab", "26"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + 4 x - 5 = 0"}, {"id": "a + b = - 4"}, {"id": "a"}, {"id": "b"}, {"id": "一元二次方程 $x ^ { 2 } + 4 x - 5 = 0$ 的两根分别为 $a$ 和 $b$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "ab = - 5"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "a ^ { 2 } + b ^ { 2 }"}, {"id": "( a + b ) ^ { 2 } - 2 ab"}, {"id": "26"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + 4 x - 5 = 0", "target": "a + b = - 4"}, {"rel": "被描述", "source": "x ^ { 2 } + 4 x - 5 = 0", "target": "ab = - 5"}, {"rel": "提取因式参考", "source": "a + b = - 4", "target": "( a + b ) ^ { 2 } - 2 ab"}, {"rel": "代入", "source": "a + b = - 4", "target": "26"}, {"rel": "被描述", "source": "a", "target": "a + b = - 4"}, {"rel": "被描述", "source": "a", "target": "ab = - 5"}, {"rel": "被描述", "source": "b", "target": "a + b = - 4"}, {"rel": "被描述", "source": "b", "target": "ab = - 5"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } + 4 x - 5 = 0$ 的两根分别为 $a$ 和 $b$", "target": "a + b = - 4"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } + 4 x - 5 = 0$ 的两根分别为 $a$ 和 $b$", "target": "ab = - 5"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "a + b = - 4"}, {"rel": "提取因式参考", "source": "ab = - 5", "target": "( a + b ) ^ { 2 } - 2 ab"}, {"rel": "代入", "source": "ab = - 5", "target": "26"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "ab = - 5"}, {"rel": "提取因式", "source": "a ^ { 2 } + b ^ { 2 }", "target": "( a + b ) ^ { 2 } - 2 ab"}, {"rel": "被代入", "source": "( a + b ) ^ { 2 } - 2 ab", "target": "26"}]}}
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{"content": "Given $( 5 - 3 x + mx ^ 2 - 6 x ^ 3 ) ( 1 - 2 x )$ does not contain a term with $x ^ 3$, what is the value of $m$?", "answer": "- 3", "steps": "$\\because$ $( 5 - 3 x + mx ^ 2 - 6 x ^ 3 ) ( 1 - 2 x ) = 5 - 13 x + ( m + 6 ) x ^ 2 + ( - 6 - 2 m ) x ^ 3 + 12 x ^ 4$. $\\therefore$ Since there is no term containing $x ^ 3$ in the result, $- 2 m - 6 = 0$, which yields $m = - 3$.", "expr_cands": ["( 5 - 3 x + mx ^ { 2 } - 6 x ^ { 3 } ) ( 1 - 2 x )", "m", "x", "x ^ { 3 }", "5 - 13 x + ( m + 6 ) x ^ { 2 } + ( - 6 - 2 m ) x ^ { 3 } + 12 x ^ { 4 }", "- 2 m - 6 = 0", "m = - 3"], "exprs": ["5 - 13 x + ( m + 6 ) x ^ { 2 } + ( - 6 - 2 m ) x ^ { 3 } + 12 x ^ { 4 }", "- 2 m - 6 = 0", "m = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 5 - 3 x + mx ^ { 2 } - 6 x ^ { 3 } ) ( 1 - 2 x )"}, {"id": "5 - 13 x + ( m + 6 ) x ^ { 2 } + ( - 6 - 2 m ) x ^ { 3 } + 12 x ^ { 4 }"}, {"id": "- 2 m - 6 = 0"}, {"id": "$( 5 - 3 x + mx ^ { 2 } - 6 x ^ { 3 } ) ( 1 - 2 x )$ 的计算结果中不含 $x ^ { 3 }$ 的项"}, {"id": "m = - 3"}], "links": [{"rel": "展开", "source": "( 5 - 3 x + mx ^ { 2 } - 6 x ^ { 3 } ) ( 1 - 2 x )", "target": "5 - 13 x + ( m + 6 ) x ^ { 2 } + ( - 6 - 2 m ) x ^ { 3 } + 12 x ^ { 4 }"}, {"rel": "被描述", "source": "5 - 13 x + ( m + 6 ) x ^ { 2 } + ( - 6 - 2 m ) x ^ { 3 } + 12 x ^ { 4 }", "target": "- 2 m - 6 = 0"}, {"rel": "等式方程求解", "source": "- 2 m - 6 = 0", "target": "m = - 3"}, {"rel": "限制性描述", "source": "$( 5 - 3 x + mx ^ { 2 } - 6 x ^ { 3 } ) ( 1 - 2 x )$ 的计算结果中不含 $x ^ { 3 }$ 的项", "target": "- 2 m - 6 = 0"}]}}
|
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{"content": "If the value of the algebraic expression $x + 2 y$ is $3$, then the value of the algebraic expression $2 x + 4 y + 1$ is [ ].", "answer": "7", "steps": "Since $x + 2 y = 3$, it follows that $2 x + 4 y + 1 = 2 ( x + 2 y ) + 1 = 2 * 3 + 1 = 6 + 1 = 7$.", "expr_cands": ["x + 2 y", "x", "y", "3", "2 x + 4 y + 1", "x + 2 y = 3", "2 ( x + 2 y ) + 1", "7"], "exprs": ["x + 2 y = 3", "2 ( x + 2 y ) + 1", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 2 y"}, {"id": "x + 2 y = 3"}, {"id": "3"}, {"id": "代数式 $x + 2 y$ 的值是 $3$"}, {"id": "2 ( x + 2 y ) + 1"}, {"id": "2 x + 4 y + 1"}, {"id": "7"}], "links": [{"rel": "被描述", "source": "x + 2 y", "target": "x + 2 y = 3"}, {"rel": "提取因式参考", "source": "x + 2 y", "target": "2 ( x + 2 y ) + 1"}, {"rel": "代入", "source": "x + 2 y = 3", "target": "7"}, {"rel": "被描述", "source": "3", "target": "x + 2 y = 3"}, {"rel": "限制性描述", "source": "代数式 $x + 2 y$ 的值是 $3$", "target": "x + 2 y = 3"}, {"rel": "被代入", "source": "2 ( x + 2 y ) + 1", "target": "7"}, {"rel": "提取因式", "source": "2 x + 4 y + 1", "target": "2 ( x + 2 y ) + 1"}]}}
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{"content": "The degree of the monomial $5 a ^ 3 bc ^ 4$ is ____?", "answer": "8", "steps": "The degree of the monomial $5 a ^ { 3 } bc ^ { 4 }$ is $3 + 1 + 4 = 8$.", "expr_cands": ["5 a ^ { 3 } bc ^ { 4 }", "b", "c", "a", "3 + 1 + 4", "8"], "exprs": ["3 + 1 + 4", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 a ^ { 3 } bc ^ { 4 }"}, {"id": "3 + 1 + 4"}, {"id": "单项式 $5 a ^ { 3 } bc ^ { 4 }$ 的次数"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "5 a ^ { 3 } bc ^ { 4 }", "target": "3 + 1 + 4"}, {"rel": "计算", "source": "3 + 1 + 4", "target": "8"}, {"rel": "限制性描述", "source": "单项式 $5 a ^ { 3 } bc ^ { 4 }$ 的次数", "target": "3 + 1 + 4"}]}}
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{"content": "The maximum integer solution of the inequality $- x \\ge 2 x + 3$ is ____ ?", "answer": "- 1", "steps": "Moving the terms, we get $- x - 2 x \\ge 3$, which simplifies to $- 3 x \\ge 3$. Solving for $x$, we get $x \\le - 1$. Therefore, the largest integer solution to the inequality $- x \\ge 2 x + 3$ is $- 1$.", "expr_cands": ["- x \\ge 2 x + 3", "x", "- x - 2 x \\ge 3", "x \\le - 1", "- 3 x \\ge 3", "- 1"], "exprs": ["x \\le - 1", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- x \\ge 2 x + 3"}, {"id": "x \\le - 1"}, {"id": "- 1"}, {"id": "一元一次不等式 $- x \\ge 2 x + 3$ 的最大整数解"}], "links": [{"rel": "不等式方程部分求解", "source": "- x \\ge 2 x + 3", "target": "x \\le - 1"}, {"rel": "被描述", "source": "x \\le - 1", "target": "- 1"}, {"rel": "限制性描述", "source": "一元一次不等式 $- x \\ge 2 x + 3$ 的最大整数解", "target": "- 1"}]}}
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{"content": "The monomial $- \\frac { 1 } { 4 } x ^ { a + b } y ^ { | a | - 1 }$ and $( a - 2 ) x ^ { 3 } y$ are not equal to $0$, and their sum is a monomial. What is the value of $a - b$?", "answer": "- 7", "steps": "From the given information, we have $a + b = 3$, $| a | - 1 = 1$, and $a - 2 \\neq 0$. Solving for $a$ and $b$, we get $a = - 2$ and $b = 5$. Therefore, $a - b = - 2 - 5 = - 7$.", "expr_cands": ["- \\frac { 1 } { 4 } x ^ { a + b } y ^ { | a | - 1 }", "b", "y", "x", "a", "( a - 2 ) x ^ { 3 } y", "0", "a - b", "a + b = 3", "| a | - 1 = 1", "a = - 2", "a = 2", "a - 2 \\neq 0", "a \\neq 2", "b = 5", "- 7"], "exprs": ["a + b = 3", "| a | - 1 = 1", "a - 2 \\neq 0", "a = - 2", "b = 5", "- 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 1 } { 4 } x ^ { a + b } y ^ { | a | - 1 }"}, {"id": "a + b = 3"}, {"id": "( a - 2 ) x ^ { 3 } y"}, {"id": "且和是单项式"}, {"id": "| a | - 1 = 1"}, {"id": "a - 2 \\neq 0"}, {"id": "单项式 $- \\frac { 1 } { 4 } x ^ { a + b } y ^ { | a | - 1 }$ 与 $( a - 2 ) x ^ { 3 } y$ 不为 $0$"}, {"id": "a = - 2"}, {"id": "b = 5"}, {"id": "a - b"}, {"id": "- 7"}], "links": [{"rel": "被描述", "source": "- \\frac { 1 } { 4 } x ^ { a + b } y ^ { | a | - 1 }", "target": "a + b = 3"}, {"rel": "被描述", "source": "- \\frac { 1 } { 4 } x ^ { a + b } y ^ { | a | - 1 }", "target": "| a | - 1 = 1"}, {"rel": "联立", "source": "a + b = 3", "target": "b = 5"}, {"rel": "被描述", "source": "( a - 2 ) x ^ { 3 } y", "target": "a + b = 3"}, {"rel": "被描述", "source": "( a - 2 ) x ^ { 3 } y", "target": "| a | - 1 = 1"}, {"rel": "被描述", "source": "( a - 2 ) x ^ { 3 } y", "target": "a - 2 \\neq 0"}, {"rel": "限制性描述", "source": "且和是单项式", "target": "a + b = 3"}, {"rel": "限制性描述", "source": "且和是单项式", "target": "| a | - 1 = 1"}, {"rel": "联立", "source": "| a | - 1 = 1", "target": "a = - 2"}, {"rel": "联立", "source": "a - 2 \\neq 0", "target": "a = - 2"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { 1 } { 4 } x ^ { a + b } y ^ { | a | - 1 }$ 与 $( a - 2 ) x ^ { 3 } y$ 不为 $0$", "target": "a - 2 \\neq 0"}, {"rel": "联立", "source": "a = - 2", "target": "b = 5"}, {"rel": "代入", "source": "a = - 2", "target": "- 7"}, {"rel": "代入", "source": "b = 5", "target": "- 7"}, {"rel": "被代入", "source": "a - b", "target": "- 7"}]}}
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{"content": "Given that the value of the algebraic expression $2 x ^ 2 - 5 x + 9$ is $7$, what is the value of $x ^ 2 - \\frac { 5 } { 2 } x + 9$?", "answer": "8", "steps": "Since $2 x ^ { 2 } - 5 x + 9 = 7$, therefore $2 x ^ { 2 } - 5 x = - 2$. Thus, the original expression is equal to $\\frac { 1 } { 2 } ( 2 x ^ { 2 } - 5 x ) + 9 = \\frac { 1 } { 2 } * ( - 2 ) + 9 = - 1 + 9 = 8$.", "expr_cands": ["2 x ^ { 2 } - 5 x + 9", "x", "7", "x ^ { 2 } - \\frac { 5 } { 2 } x + 9", "2 x ^ { 2 } - 5 x + 9 = 7", "x = \\frac { 1 } { 2 }", "x = 2", "2 x ^ { 2 } - 5 x = - 2", "\\frac { 1 } { 2 } ( 2 x ^ { 2 } - 5 x ) + 9", "8"], "exprs": ["2 x ^ { 2 } - 5 x + 9 = 7", "2 x ^ { 2 } - 5 x = - 2", "\\frac { 1 } { 2 } ( 2 x ^ { 2 } - 5 x ) + 9", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 2 } - 5 x + 9"}, {"id": "2 x ^ { 2 } - 5 x + 9 = 7"}, {"id": "7"}, {"id": "代数式 $2 x ^ { 2 } - 5 x + 9$ 的值为 $7$"}, {"id": "2 x ^ { 2 } - 5 x = - 2"}, {"id": "x ^ { 2 } - \\frac { 5 } { 2 } x + 9"}, {"id": "\\frac { 1 } { 2 } ( 2 x ^ { 2 } - 5 x ) + 9"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "2 x ^ { 2 } - 5 x + 9", "target": "2 x ^ { 2 } - 5 x + 9 = 7"}, {"rel": "移项", "source": "2 x ^ { 2 } - 5 x + 9 = 7", "target": "2 x ^ { 2 } - 5 x = - 2"}, {"rel": "被描述", "source": "7", "target": "2 x ^ { 2 } - 5 x + 9 = 7"}, {"rel": "限制性描述", "source": "代数式 $2 x ^ { 2 } - 5 x + 9$ 的值为 $7$", "target": "2 x ^ { 2 } - 5 x + 9 = 7"}, {"rel": "提取因式参考", "source": "2 x ^ { 2 } - 5 x = - 2", "target": "\\frac { 1 } { 2 } ( 2 x ^ { 2 } - 5 x ) + 9"}, {"rel": "代入", "source": "2 x ^ { 2 } - 5 x = - 2", "target": "8"}, {"rel": "提取因式", "source": "x ^ { 2 } - \\frac { 5 } { 2 } x + 9", "target": "\\frac { 1 } { 2 } ( 2 x ^ { 2 } - 5 x ) + 9"}, {"rel": "被代入", "source": "\\frac { 1 } { 2 } ( 2 x ^ { 2 } - 5 x ) + 9", "target": "8"}]}}
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{"content": "If the equation $\\frac { x + 3 } { x - 1 } = \\frac { 1 - m } { 1 - x }$ has a positive root, then $m$ = ____ ?", "answer": "5", "steps": "To eliminate the denominator, we get $x + 3 = m - 1$. Since the fractional equation has an extraneous root, we have $x - 1 = 0$, which means $x = 1$. Substituting $x = 1$ into the polynomial equation, we get $m = 5$.", "expr_cands": ["x", "\\frac { x + 3 } { x - 1 } = \\frac { 1 - m } { 1 - x }", "m", "x + 3 = m - 1", "x - 1 = 0", "x = 1", "m = 5"], "exprs": ["x + 3 = m - 1", "x - 1 = 0", "x = 1", "m = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x + 3 } { x - 1 } = \\frac { 1 - m } { 1 - x }"}, {"id": "x + 3 = m - 1"}, {"id": "x - 1 = 0"}, {"id": "关于 $x$ 的方程 $\\frac { x + 3 } { x - 1 } = \\frac { 1 - m } { 1 - x }$ 有増根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 1"}, {"id": "m = 5"}], "links": [{"rel": "同乘除", "source": "\\frac { x + 3 } { x - 1 } = \\frac { 1 - m } { 1 - x }", "target": "x + 3 = m - 1"}, {"rel": "被描述", "source": "\\frac { x + 3 } { x - 1 } = \\frac { 1 - m } { 1 - x }", "target": "x - 1 = 0"}, {"rel": "联立", "source": "x + 3 = m - 1", "target": "m = 5"}, {"rel": "等式方程求解", "source": "x - 1 = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $\\frac { x + 3 } { x - 1 } = \\frac { 1 - m } { 1 - x }$ 有増根", "target": "x - 1 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 1 = 0"}, {"rel": "联立", "source": "x = 1", "target": "m = 5"}]}}
|
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{"content": "If $p$ and $q$ are prime numbers and $5 p + 7 q = 29$, then $p ^ q + q ^ p - q$ = ____?", "answer": "15", "steps": "Since $p$ and $q$ are prime numbers, $5 p$ and $7 q$ must be either both odd or one odd and one even. Since $5 p + 7 q = 29$ is odd, $5 p$ and $7 q$ must be one odd and one even. Therefore, one of $p$ and $q$ must be $2$. When $p = 2$, $q = \\frac { 19 } { 7 }$ (which is discarded). When $q = 2$, $p = 3$, and $3$ is a prime number. Therefore, $p ^ q + q ^ p - q = 15$.", "expr_cands": ["p", "q", "5 p + 7 q = 29", "p ^ { q } + q ^ { p } - q", "5 p", "7 q", "2", "p = 2", "q = \\frac { 19 } { 7 }", "q = 2", "p = 3", "3", "15"], "exprs": ["q = 2", "p = 3", "15"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "p"}, {"id": "q = 2"}, {"id": "q"}, {"id": "5 p + 7 q = 29"}, {"id": "$p$ 和 $q$ 为质数"}, {"id": ", $5 p$ , $7 q$ 必有一个为奇数或两个都为奇数"}, {"id": ", $p$ , $q$ 中必有一数为 $2$"}, {"id": ", $5 p + 7 q = 29$ 为奇数"}, {"id": ", $5 p$ , $7 q$ 必为一奇一偶"}, {"id": "当 $p = 2$ 时"}, {"id": "$q = \\frac { 19 } { 7 }$ ( 舍去 )"}, {"id": "p = 3"}, {"id": "p ^ { q } + q ^ { p } - q"}, {"id": "15"}], "links": [{"rel": "被描述", "source": "p", "target": "q = 2"}, {"rel": "联立", "source": "q = 2", "target": "p = 3"}, {"rel": "代入", "source": "q = 2", "target": "15"}, {"rel": "被描述", "source": "q", "target": "q = 2"}, {"rel": "被描述", "source": "5 p + 7 q = 29", "target": "q = 2"}, {"rel": "联立", "source": "5 p + 7 q = 29", "target": "p = 3"}, {"rel": "限制性描述", "source": "$p$ 和 $q$ 为质数", "target": "q = 2"}, {"rel": "限制性描述", "source": ", $5 p$ , $7 q$ 必有一个为奇数或两个都为奇数", "target": "q = 2"}, {"rel": "限制性描述", "source": ", $p$ , $q$ 中必有一数为 $2$", "target": "q = 2"}, {"rel": "限制性描述", "source": ", $5 p + 7 q = 29$ 为奇数", "target": "q = 2"}, {"rel": "限制性描述", "source": ", $5 p$ , $7 q$ 必为一奇一偶", "target": "q = 2"}, {"rel": "限制性描述", "source": "当 $p = 2$ 时", "target": "q = 2"}, {"rel": "限制性描述", "source": "$q = \\frac { 19 } { 7 }$ ( 舍去 )", "target": "q = 2"}, {"rel": "代入", "source": "p = 3", "target": "15"}, {"rel": "被代入", "source": "p ^ { q } + q ^ { p } - q", "target": "15"}]}}
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{"content": "If $y + 2$ is directly proportional to $2 x$, and when $x = 1$, $y = 4$, then the functional relationship between $y$ and $x$ is ____?", "answer": "y = 6 x - 2", "steps": "\\because $y + 2$ is directly proportional to $2 x$, \\therefore let $y + 2 = 2 kx$, substituting $x = 1$ and $y = 4$ into $y + 2 = 2 kx$ gives $4 + 2 = 2 k$, \\therefore $k = 3$, \\therefore $y = 6 x - 2$.", "expr_cands": ["y + 2", "y", "2 x", "x", "x = 1", "y = 4", "y + 2 = 2 kx", "k", "2 k = 2 k", "4 + 2 = 2 k", "k = 3", "y = 6 x - 2"], "exprs": ["y + 2 = 2 kx", "4 + 2 = 2 k", "k = 3", "y = 6 x - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $y + 2 = 2 kx$"}, {"id": "y + 2 = 2 kx"}, {"id": "4 + 2 = 2 k"}, {"id": "x = 1"}, {"id": "y = 4"}, {"id": "k = 3"}, {"id": "y = 6 x - 2"}], "links": [{"rel": "假设描述", "source": "设 $y + 2 = 2 kx$", "target": "y + 2 = 2 kx"}, {"rel": "被代入", "source": "y + 2 = 2 kx", "target": "4 + 2 = 2 k"}, {"rel": "联立", "source": "y + 2 = 2 kx", "target": "y = 6 x - 2"}, {"rel": "等式方程求解", "source": "4 + 2 = 2 k", "target": "k = 3"}, {"rel": "代入", "source": "x = 1", "target": "4 + 2 = 2 k"}, {"rel": "代入", "source": "y = 4", "target": "4 + 2 = 2 k"}, {"rel": "联立", "source": "k = 3", "target": "y = 6 x - 2"}]}}
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{"content": "In the quadratic function $y = ax ^ { 2 }$, when $x = 1$, $y = 2$. What is the value of $a$?", "answer": "2", "steps": "Substituting $x = 1$ and $y = 2$ into $y = ax ^ 2$, we get $2 = a$. Solving for $a$, we get $a = 2$.", "expr_cands": ["y = ax ^ { 2 }", "a", "y", "x", "x = 1", "y = 2", "2 = a", "a = 2"], "exprs": ["a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ax ^ { 2 }"}, {"id": "a = 2"}, {"id": "x = 1"}, {"id": "y = 2"}], "links": [{"rel": "被代入", "source": "y = ax ^ { 2 }", "target": "a = 2"}, {"rel": "代入", "source": "x = 1", "target": "a = 2"}, {"rel": "代入", "source": "y = 2", "target": "a = 2"}]}}
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{"content": "If the algebraic expression $x ^ 2 - 2 x - 3$ is transformed into the form $( x - m ) ^ 2 + k$, where $m$ and $k$ are constants, then $m + k$ = ____?", "answer": "- 3", "steps": "Since $x ^ 2 - 2 x - 3 = x ^ 2 - 2 x + 1 - 4 = ( x - 1 ) ^ 2 - 4$, we have $m = 1$ and $k = - 4$. Therefore, $m + k = - 3$.", "expr_cands": ["x ^ { 2 } - 2 x - 3", "x", "( x - m ) ^ { 2 } + k", "k", "m", "m + k", "( x - 1 ) ^ { 2 } - 4", "m = 1", "k = - 4", "- 3"], "exprs": ["( x - 1 ) ^ { 2 } - 4", "m = 1", "k = - 4", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 2 x - 3"}, {"id": "( x - 1 ) ^ { 2 } - 4"}, {"id": "( x - m ) ^ { 2 } + k"}, {"id": "m = 1"}, {"id": "把代数式 $x ^ { 2 } - 2 x - 3$ 化为 $( x - m ) ^ { 2 } + k$ 的形式"}, {"id": "k = - 4"}, {"id": "m + k"}, {"id": "- 3"}], "links": [{"rel": "提取因式", "source": "x ^ { 2 } - 2 x - 3", "target": "( x - 1 ) ^ { 2 } - 4"}, {"rel": "被描述", "source": "( x - 1 ) ^ { 2 } - 4", "target": "m = 1"}, {"rel": "被描述", "source": "( x - 1 ) ^ { 2 } - 4", "target": "k = - 4"}, {"rel": "被描述", "source": "( x - m ) ^ { 2 } + k", "target": "m = 1"}, {"rel": "被描述", "source": "( x - m ) ^ { 2 } + k", "target": "k = - 4"}, {"rel": "代入", "source": "m = 1", "target": "- 3"}, {"rel": "限制性描述", "source": "把代数式 $x ^ { 2 } - 2 x - 3$ 化为 $( x - m ) ^ { 2 } + k$ 的形式", "target": "m = 1"}, {"rel": "限制性描述", "source": "把代数式 $x ^ { 2 } - 2 x - 3$ 化为 $( x - m ) ^ { 2 } + k$ 的形式", "target": "k = - 4"}, {"rel": "代入", "source": "k = - 4", "target": "- 3"}, {"rel": "被代入", "source": "m + k", "target": "- 3"}]}}
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{"content": "When $y$ = ____ ?, the value of $1 - \\frac { 2 y - 5 } { 6 }$ is equal to the value of $\\frac { 3 - y } { 6 }$.", "answer": "8", "steps": "According to the problem, we have $1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }$. Simplifying this expression by getting rid of the denominators, we get $6 - ( 2 y - 5 ) = 3 - y$. Expanding the brackets, we get $6 - 2 y + 5 = 3 - y$. Rearranging and combining like terms, we get $y = 8$.", "expr_cands": ["y", "1 - \\frac { 2 y - 5 } { 6 }", "\\frac { 3 - y } { 6 }", "1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }", "y = 8", "6 - ( 2 y - 5 ) = 3 - y", "6 - 2 y + 5 = 3 - y"], "exprs": ["1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }", "y = 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1 - \\frac { 2 y - 5 } { 6 }"}, {"id": "1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }"}, {"id": "\\frac { 3 - y } { 6 }"}, {"id": "$1 - \\frac { 2 y - 5 } { 6 }$ 与 $\\frac { 3 - y } { 6 }$ 的值相"}, {"id": "y = 8"}], "links": [{"rel": "被描述", "source": "1 - \\frac { 2 y - 5 } { 6 }", "target": "1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }"}, {"rel": "等式方程求解", "source": "1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }", "target": "y = 8"}, {"rel": "被描述", "source": "\\frac { 3 - y } { 6 }", "target": "1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }"}, {"rel": "限制性描述", "source": "$1 - \\frac { 2 y - 5 } { 6 }$ 与 $\\frac { 3 - y } { 6 }$ 的值相", "target": "1 - \\frac { 2 y - 5 } { 6 } = \\frac { 3 - y } { 6 }"}]}}
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||||
{"content": "If $x \\ge 5$, then the minimum value of $x$ is $a$, and if $x \\le - 7$, then the maximum value of $x$ is $b$. What is the value of $ab$?", "answer": "- 35", "steps": "Because the minimum value of $x$ for $x \\ge 5$ is $a$, $a = 5$; the maximum value of $x$ for $x \\le - 7$ is $b$, so $b = - 7$; therefore, $ab = 5 * ( - 7 ) = - 35$.", "expr_cands": ["x \\ge 5", "x", "a", "x \\le - 7", "b", "ab", "a = 5", "b = - 7", "- 35"], "exprs": ["a = 5", "b = - 7", "- 35"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x \\ge 5"}, {"id": "a = 5"}, {"id": "a"}, {"id": "因为 $x \\ge 5$ 的最小值是 $a$ , $a = 5$ , $x \\le - 7$ 的最大值是 $b$"}, {"id": "x \\le - 7"}, {"id": "b = - 7"}, {"id": "b"}, {"id": "ab"}, {"id": "- 35"}], "links": [{"rel": "被描述", "source": "x \\ge 5", "target": "a = 5"}, {"rel": "代入", "source": "a = 5", "target": "- 35"}, {"rel": "被描述", "source": "a", "target": "a = 5"}, {"rel": "限制性描述", "source": "因为 $x \\ge 5$ 的最小值是 $a$ , $a = 5$ , $x \\le - 7$ 的最大值是 $b$", "target": "a = 5"}, {"rel": "限制性描述", "source": "因为 $x \\ge 5$ 的最小值是 $a$ , $a = 5$ , $x \\le - 7$ 的最大值是 $b$", "target": "b = - 7"}, {"rel": "被描述", "source": "x \\le - 7", "target": "b = - 7"}, {"rel": "代入", "source": "b = - 7", "target": "- 35"}, {"rel": "被描述", "source": "b", "target": "b = - 7"}, {"rel": "被代入", "source": "ab", "target": "- 35"}]}}
|
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{"content": "When $x$ = ____ ?, the value of the algebraic expression $x - \\frac { x - 2 } { 5 }$ is equal to $- 2$.", "answer": "- 3", "steps": "$x - \\frac { x - 2 } { 5 } = - 2$. To eliminate the denominator, we get $5 x - x + 2 = - 10$. Rearranging and combining like terms, we get $4 x = - 12$. Dividing by the coefficient of $x$, we get $x = - 3$.", "expr_cands": ["x", "x - \\frac { x - 2 } { 5 }", "- 2", "x - \\frac { x - 2 } { 5 } = - 2", "x = - 3", "5 x - x + 2 = - 10", "4 x = - 12", "1"], "exprs": ["x - \\frac { x - 2 } { 5 } = - 2", "x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - \\frac { x - 2 } { 5 }"}, {"id": "x - \\frac { x - 2 } { 5 } = - 2"}, {"id": "- 2"}, {"id": "代数式 $x - \\frac { x - 2 } { 5 }$ 的值等于 $- 2$"}, {"id": "x = - 3"}], "links": [{"rel": "被描述", "source": "x - \\frac { x - 2 } { 5 }", "target": "x - \\frac { x - 2 } { 5 } = - 2"}, {"rel": "等式方程求解", "source": "x - \\frac { x - 2 } { 5 } = - 2", "target": "x = - 3"}, {"rel": "被描述", "source": "- 2", "target": "x - \\frac { x - 2 } { 5 } = - 2"}, {"rel": "限制性描述", "source": "代数式 $x - \\frac { x - 2 } { 5 }$ 的值等于 $- 2$", "target": "x - \\frac { x - 2 } { 5 } = - 2"}]}}
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||||
{"content": "If $x = 4$ is a root of the equation $x ^ 2 - 3 x + c = 0$, then $c$ = ____ ?", "answer": "- 1", "steps": "Assuming the other root of the equation is $a$, since $x = 4$ is a root of the equation $x ^ 2 - 3 x + c = 0$, we have $a + 4 = 3$. Solving for $a$, we get $a = - 1$.", "expr_cands": ["x = 4", "x", "x ^ { 2 } - 3 x + c = 0", "c", "a", "c + 4 = 0", "a + 4 = 3", "a = - 1"], "exprs": ["a", "a + 4 = 3", "a = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程的另一个根为 $a$"}, {"id": "a"}, {"id": "x = 4"}, {"id": "a + 4 = 3"}, {"id": "x ^ { 2 } - 3 x + c = 0"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "a = - 1"}], "links": [{"rel": "假设描述", "source": "设方程的另一个根为 $a$", "target": "a"}, {"rel": "被描述", "source": "a", "target": "a + 4 = 3"}, {"rel": "被描述", "source": "x = 4", "target": "a + 4 = 3"}, {"rel": "等式方程求解", "source": "a + 4 = 3", "target": "a = - 1"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 x + c = 0", "target": "a + 4 = 3"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "a + 4 = 3"}]}}
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||||
{"content": "If the two non-equal square roots of a positive number are $2 a - 1$ and $a - 2$, what is the positive number?", "answer": "1", "steps": "According to the problem, we have $2 a - 1 + a - 2 = 0$. Solving for $a$, we get $a = 1$. Therefore, the number is $( 2 a - 1 ) ^ 2 = 1 ^ 2 = 1$.", "expr_cands": ["2 a - 1", "a", "a - 2", "2 a - 1 + a - 2 = 0", "a = 1", "( 2 a - 1 ) ^ { 2 }", "1"], "exprs": ["2 a - 1 + a - 2 = 0", "( 2 a - 1 ) ^ { 2 }", "a = 1", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a - 1"}, {"id": "2 a - 1 + a - 2 = 0"}, {"id": "a - 2"}, {"id": "一个正数的两个不相等的平方根是 $2 a - 1$ 和 $a - 2$"}, {"id": "平方根互为相反数"}, {"id": "a = 1"}, {"id": "( 2 a - 1 ) ^ { 2 }"}, {"id": "平方"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "2 a - 1", "target": "2 a - 1 + a - 2 = 0"}, {"rel": "被描述", "source": "2 a - 1", "target": "( 2 a - 1 ) ^ { 2 }"}, {"rel": "等式方程求解", "source": "2 a - 1 + a - 2 = 0", "target": "a = 1"}, {"rel": "被描述", "source": "a - 2", "target": "2 a - 1 + a - 2 = 0"}, {"rel": "限制性描述", "source": "一个正数的两个不相等的平方根是 $2 a - 1$ 和 $a - 2$", "target": "2 a - 1 + a - 2 = 0"}, {"rel": "限制性描述", "source": "平方根互为相反数", "target": "2 a - 1 + a - 2 = 0"}, {"rel": "代入", "source": "a = 1", "target": "1"}, {"rel": "被代入", "source": "( 2 a - 1 ) ^ { 2 }", "target": "1"}, {"rel": "限制性描述", "source": "平方", "target": "( 2 a - 1 ) ^ { 2 }"}]}}
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||||
{"content": "Given that the value of $a - 2 b$ is $- 2$, what is the value of $( a - 2 b ) ^ 2 + 2 ( a - 2 b )$?", "answer": "0", "steps": "Substituting $a - 2 b = - 2$ into $( a - 2 b ) ^ 2 + 2 ( a - 2 b )$ yields: $( - 2 ) ^ 2 + 2 * ( - 2 ) = 4 - 4 = 0$.", "expr_cands": ["a - 2 b", "b", "a", "- 2", "( a - 2 b ) ^ { 2 } + 2 ( a - 2 b )", "a - 2 b = - 2", "( - 2 ) ^ { 2 } + 2 * ( - 2 )", "0"], "exprs": ["a - 2 b = - 2", "( - 2 ) ^ { 2 } + 2 * ( - 2 )", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - 2 b"}, {"id": "a - 2 b = - 2"}, {"id": "- 2"}, {"id": "$a - 2 b$ 的值是 $- 2$"}, {"id": "( a - 2 b ) ^ { 2 } + 2 ( a - 2 b )"}, {"id": "( - 2 ) ^ { 2 } + 2 * ( - 2 )"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "a - 2 b", "target": "a - 2 b = - 2"}, {"rel": "代入", "source": "a - 2 b = - 2", "target": "( - 2 ) ^ { 2 } + 2 * ( - 2 )"}, {"rel": "被描述", "source": "- 2", "target": "a - 2 b = - 2"}, {"rel": "限制性描述", "source": "$a - 2 b$ 的值是 $- 2$", "target": "a - 2 b = - 2"}, {"rel": "被代入", "source": "( a - 2 b ) ^ { 2 } + 2 ( a - 2 b )", "target": "( - 2 ) ^ { 2 } + 2 * ( - 2 )"}, {"rel": "计算", "source": "( - 2 ) ^ { 2 } + 2 * ( - 2 )", "target": "0"}]}}
|
||||
{"content": "Given $a ^ 2 - b ^ 2 = 5$, $a + b = - 2$, what is the value of the algebraic expression $a - b$?", "answer": "- 2.5", "steps": "Since $a ^ 2 - b ^ 2 = 5$ and $a + b = - 2$, therefore $a - b = ( a ^ 2 - b ^ 2 ) / ( a + b ) = 5 / ( - 2 ) = - 2.5$.", "expr_cands": ["a ^ { 2 } - b ^ { 2 } = 5", "b", "a", "a + b = - 2", "a - b", "( a ^ { 2 } - b ^ { 2 } ) \\div ( a + b )", "- \\frac { 5 } { 2 }"], "exprs": ["( a ^ { 2 } - b ^ { 2 } ) \\div ( a + b )", "- \\frac { 5 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - b"}, {"id": "( a ^ { 2 } - b ^ { 2 } ) \\div ( a + b )"}, {"id": "a ^ { 2 } - b ^ { 2 } = 5"}, {"id": "a + b = - 2"}, {"id": "- \\frac { 5 } { 2 }"}], "links": [{"rel": "提取因式", "source": "a - b", "target": "( a ^ { 2 } - b ^ { 2 } ) \\div ( a + b )"}, {"rel": "被代入", "source": "( a ^ { 2 } - b ^ { 2 } ) \\div ( a + b )", "target": "- \\frac { 5 } { 2 }"}, {"rel": "提取因式参考", "source": "a ^ { 2 } - b ^ { 2 } = 5", "target": "( a ^ { 2 } - b ^ { 2 } ) \\div ( a + b )"}, {"rel": "代入", "source": "a ^ { 2 } - b ^ { 2 } = 5", "target": "- \\frac { 5 } { 2 }"}, {"rel": "提取因式参考", "source": "a + b = - 2", "target": "( a ^ { 2 } - b ^ { 2 } ) \\div ( a + b )"}, {"rel": "代入", "source": "a + b = - 2", "target": "- \\frac { 5 } { 2 }"}]}}
|
||||
{"content": "Given that $3 a + 1$ and $5$ are two square roots of a positive number $b$, what is the value of $a + b$?", "answer": "23", "steps": "From the given problem, we have $3 a + 1 + 5 = 0$, which gives us $a = - 2$. Therefore, $3 a + 1 = - 5$, and $b = 5 ^ 2 = 25$. Thus, $a + b = - 2 + 25 = 23$.", "expr_cands": ["3 a + 1", "a", "5", "b", "a + b", "3 a + 1 + 5 = 0", "a = - 2", "- 5", "b = 25", "23"], "exprs": ["3 a + 1 + 5 = 0", "a = - 2", "- 5", "b = 25", "23"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a + 1"}, {"id": "3 a + 1 + 5 = 0"}, {"id": "5"}, {"id": "b"}, {"id": "$3 a + 1$ 和 $5$ 是正数 $b$ 的两个平方根"}, {"id": "平方根互为相反数"}, {"id": "a = - 2"}, {"id": "- 5"}, {"id": "b = 25"}, {"id": "平方"}, {"id": "a + b"}, {"id": "23"}], "links": [{"rel": "被描述", "source": "3 a + 1", "target": "3 a + 1 + 5 = 0"}, {"rel": "被代入", "source": "3 a + 1", "target": "- 5"}, {"rel": "等式方程求解", "source": "3 a + 1 + 5 = 0", "target": "a = - 2"}, {"rel": "被描述", "source": "5", "target": "3 a + 1 + 5 = 0"}, {"rel": "被描述", "source": "b", "target": "3 a + 1 + 5 = 0"}, {"rel": "被描述", "source": "b", "target": "b = 25"}, {"rel": "限制性描述", "source": "$3 a + 1$ 和 $5$ 是正数 $b$ 的两个平方根", "target": "3 a + 1 + 5 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "3 a + 1 + 5 = 0"}, {"rel": "代入", "source": "a = - 2", "target": "- 5"}, {"rel": "代入", "source": "a = - 2", "target": "23"}, {"rel": "被描述", "source": "- 5", "target": "b = 25"}, {"rel": "代入", "source": "b = 25", "target": "23"}, {"rel": "限制性描述", "source": "平方", "target": "b = 25"}, {"rel": "被代入", "source": "a + b", "target": "23"}]}}
|
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{"content": "The solution to the equation $8 - 4 x = 64 + 4 x$ is ____ ?", "answer": "x = - 7", "steps": "Moving terms and combining, we get $- 8 x = 56$, which gives us the solution $x = - 7$.", "expr_cands": ["8 - 4 x = 64 + 4 x", "x", "- 8 x = 56", "x = - 7"], "exprs": ["- 8 x = 56", "x = - 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "8 - 4 x = 64 + 4 x"}, {"id": "- 8 x = 56"}, {"id": "x = - 7"}], "links": [{"rel": "移项", "source": "8 - 4 x = 64 + 4 x", "target": "- 8 x = 56"}, {"rel": "等式方程求解", "source": "- 8 x = 56", "target": "x = - 7"}]}}
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{"content": "In the real number system, if $\\sqrt { x + 1 }$ is defined, then the range of values for $x$ is ____?", "answer": "x \\ge - 1", "steps": "Since the square root of $x + 1$ is defined, we have $x + 1 \\geq 0$. Solving for $x$, we get $x \\geq - 1$.", "expr_cands": ["\\sqrt { x + 1 }", "x", "x + 1 \\ge 0", "- 1 \\le x", "x \\ge - 1"], "exprs": ["x + 1 \\ge 0", "x \\ge - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + 1 }"}, {"id": "x + 1 \\ge 0"}, {"id": "在实数范围内"}, {"id": "$\\sqrt { x + 1 }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge - 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { x + 1 }", "target": "x + 1 \\ge 0"}, {"rel": "不等式方程求解", "source": "x + 1 \\ge 0", "target": "x \\ge - 1"}, {"rel": "限制性描述", "source": "在实数范围内", "target": "x + 1 \\ge 0"}, {"rel": "限制性描述", "source": "$\\sqrt { x + 1 }$ 有意义", "target": "x + 1 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x + 1 \\ge 0"}]}}
|
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{"content": "The function of the line obtained by translating the line $y = 2 x - 3$ $8$ units upward along the $y$-axis is _____.", "answer": "y = 2 x + 5", "steps": "According to the principle of adding up and subtracting down, it can be known that the analytical expression of the straight line $y = 2 x - 3$ after being translated up by 8 units is $y = 2 x - 3 + 8$, that is, $y = 2 x + 5$.", "expr_cands": ["y = 2 x - 3", "x", "y", "8", "y = 2 x - 3 + 8", "2 x - 3 = 2 x - 3 + 8", "2 x + 5"], "exprs": ["y = 2 x - 3 + 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 x - 3"}, {"id": "y = 2 x - 3 + 8"}, {"id": "8"}, {"id": "直线 $y = 2 x - 3$ 沿 $y$ 轴向上平移 $8$ 个单位长度得到的直线函数关系式"}], "links": [{"rel": "被描述", "source": "y = 2 x - 3", "target": "y = 2 x - 3 + 8"}, {"rel": "被描述", "source": "8", "target": "y = 2 x - 3 + 8"}, {"rel": "限制性描述", "source": "直线 $y = 2 x - 3$ 沿 $y$ 轴向上平移 $8$ 个单位长度得到的直线函数关系式", "target": "y = 2 x - 3 + 8"}]}}
|
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{"content": "Given that $x _ 1$ and $x _ 2$ are the two real roots of the equation $2 x ^ 2 - 3 x - 1 = 0$, what is the value of $x _ 1 x _ 2 - x _ 1 - x _ 2$?", "answer": "- 2", "steps": "From the relationship between the root and the coefficient, we obtain $x _ 1 + x _ 2 = \\frac { 3 } { 2 }$, $x _ 1 \\times x _ 2 = - \\frac { 1 } { 2 }$, and $x _ 1 x _ 2 - x _ 1 - x _ 2 = x _ 1 x _ 2 - ( x _ 1 + x _ 2 ) = - \\frac { 1 } { 2 } - \\frac { 3 } { 2 } = - 2$.", "expr_cands": ["x _ { 1 }", "x _ { 2 }", "2 x ^ { 2 } - 3 x - 1 = 0", "x", "x _ { 1 } x _ { 2 } - x _ { 1 } - x _ { 2 }", "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }", "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }", "- 2"], "exprs": ["x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }", "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x _ { 1 }"}, {"id": "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }"}, {"id": "x _ { 2 }"}, {"id": "2 x ^ { 2 } - 3 x - 1 = 0"}, {"id": "$x _ { 1 }$ , $x _ { 2 }$ 是方程 $2 x ^ { 2 } - 3 x - 1 = 0$ 的两个实数根"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "x _ { 1 } x _ { 2 } - x _ { 1 } - x _ { 2 }"}, {"id": "- 2"}], "links": [{"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }"}, {"rel": "代入", "source": "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }", "target": "- 2"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }"}, {"rel": "被描述", "source": "2 x ^ { 2 } - 3 x - 1 = 0", "target": "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }"}, {"rel": "被描述", "source": "2 x ^ { 2 } - 3 x - 1 = 0", "target": "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "$x _ { 1 }$ , $x _ { 2 }$ 是方程 $2 x ^ { 2 } - 3 x - 1 = 0$ 的两个实数根", "target": "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }"}, {"rel": "限制性描述", "source": "$x _ { 1 }$ , $x _ { 2 }$ 是方程 $2 x ^ { 2 } - 3 x - 1 = 0$ 的两个实数根", "target": "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + x _ { 2 } = \\frac { 3 } { 2 }"}, {"rel": "代入", "source": "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }", "target": "- 2"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "x _ { 1 } \\times x _ { 2 } = - \\frac { 1 } { 2 }"}, {"rel": "被代入", "source": "x _ { 1 } x _ { 2 } - x _ { 1 } - x _ { 2 }", "target": "- 2"}]}}
|
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{"content": "If the equation $3 x = \\frac { 5 } { 2 } x - 4$ and $\\frac { 1 } { 2 } x - 2 ax = \\frac { a } { 4 } x + 5$ have the same solution for $x$, then $a$ = ____?", "answer": "\\frac { 1 } { 2 }", "steps": "Solving $3 x = \\frac { 5 } { 2 } x - 4$ gives $x = - 8$. Substituting $x = - 8$ into $\\frac { 1 } { 2 } x - 2 ax = \\frac { a } { 4 } x + 5$ yields $- 4 + 16 a = - 2 a + 5$. Solving for $a$ gives $a = \\frac { 1 } { 2 }$.", "expr_cands": ["x", "3 x = \\frac { 5 } { 2 } x - 4", "\\frac { 1 } { 2 } x - 2 ax = \\frac { a } { 4 } x + 5", "a", "x = - 8", "16 a - 4 = 5 - 2 a", "- 4 + 16 a = - 2 a + 5", "a = \\frac { 1 } { 2 }"], "exprs": ["x = - 8", "- 4 + 16 a = - 2 a + 5", "a = \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x = \\frac { 5 } { 2 } x - 4"}, {"id": "x = - 8"}, {"id": "\\frac { 1 } { 2 } x - 2 ax = \\frac { a } { 4 } x + 5"}, {"id": "- 4 + 16 a = - 2 a + 5"}, {"id": "a = \\frac { 1 } { 2 }"}], "links": [{"rel": "等式方程求解", "source": "3 x = \\frac { 5 } { 2 } x - 4", "target": "x = - 8"}, {"rel": "代入", "source": "x = - 8", "target": "- 4 + 16 a = - 2 a + 5"}, {"rel": "被代入", "source": "\\frac { 1 } { 2 } x - 2 ax = \\frac { a } { 4 } x + 5", "target": "- 4 + 16 a = - 2 a + 5"}, {"rel": "等式方程求解", "source": "- 4 + 16 a = - 2 a + 5", "target": "a = \\frac { 1 } { 2 }"}]}}
|
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{"content": "If the fraction $\\frac { 3 } { x - 5 }$ is meaningful, then $x$ should satisfy ____?", "answer": "x \\neq 5", "steps": "To make the fraction $\\frac { 3 } { x - 5 }$ meaningful, we need $x - 5 \\neq 0$, which means $x \\neq 5$.", "expr_cands": ["\\frac { 3 } { x - 5 }", "x", "x - 5 \\neq 0", "x \\neq 5"], "exprs": ["x - 5 \\neq 0", "x \\neq 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 } { x - 5 }"}, {"id": "x - 5 \\neq 0"}, {"id": "分式 $\\frac { 3 } { x - 5 }$ 有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 5"}], "links": [{"rel": "被描述", "source": "\\frac { 3 } { x - 5 }", "target": "x - 5 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 5 \\neq 0", "target": "x \\neq 5"}, {"rel": "限制性描述", "source": "分式 $\\frac { 3 } { x - 5 }$ 有意义", "target": "x - 5 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 5 \\neq 0"}]}}
|
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{"content": "The solution set of the inequality $26 - 8 x < 0$ is ____ ?", "answer": "x > \\frac { 13 } { 4 }", "steps": "Moving terms, we get $- 8 x < - 26$. Dividing both sides by $- 8$ to make the coefficient of $x$ equal to $1$, we get $x > \\frac { 13 } { 4 }$.", "expr_cands": ["26 - 8 x < 0", "x", "- 8 x < - 26", "\\frac { 13 } { 4 } < x", "1", "x > \\frac { 13 } { 4 }"], "exprs": ["x > \\frac { 13 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "26 - 8 x < 0"}, {"id": "x > \\frac { 13 } { 4 }"}], "links": [{"rel": "不等式方程求解", "source": "26 - 8 x < 0", "target": "x > \\frac { 13 } { 4 }"}]}}
|
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{"content": "If the parabola $y = 2 { x } ^ { 2 }$ is translated $3$ units to the right and $5$ units up, the resulting parabola is _____.", "answer": "y = 2 { ( x - 3 ) } ^ { 2 } + 5", "steps": "The expression of the parabola obtained by translating the parabola $y = 2 x ^ 2$ $3$ units to the right and $5$ units up is $y = 2 ( x - 3 ) ^ 2 + 5$.", "expr_cands": ["y = 2 { x } ^ { 2 }", "y", "x", "3", "5", "y = 2 x ^ { 2 }", "y = 2 ( x - 3 ) ^ { 2 } + 5", "2 x ^ { 2 } = 2 ( x - 3 ) ^ { 2 } + 5", "2 x ^ { 2 }", "y = 2 { ( x - 3 ) } ^ { 2 } + 5"], "exprs": ["y = 2 ( x - 3 ) ^ { 2 } + 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 { x } ^ { 2 }"}, {"id": "y = 2 ( x - 3 ) ^ { 2 } + 5"}, {"id": "3"}, {"id": "5"}, {"id": "将抛物线 $y = 2 { x } ^ { 2 }$ 向右平移 $3$ 个单位"}, {"id": "再向上平移 $5$ 个单位"}], "links": [{"rel": "被描述", "source": "y = 2 { x } ^ { 2 }", "target": "y = 2 ( x - 3 ) ^ { 2 } + 5"}, {"rel": "被描述", "source": "3", "target": "y = 2 ( x - 3 ) ^ { 2 } + 5"}, {"rel": "被描述", "source": "5", "target": "y = 2 ( x - 3 ) ^ { 2 } + 5"}, {"rel": "限制性描述", "source": "将抛物线 $y = 2 { x } ^ { 2 }$ 向右平移 $3$ 个单位", "target": "y = 2 ( x - 3 ) ^ { 2 } + 5"}, {"rel": "限制性描述", "source": "再向上平移 $5$ 个单位", "target": "y = 2 ( x - 3 ) ^ { 2 } + 5"}]}}
|
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{"content": "If $x = 0$, the value of $( x - 6 ) ^ 2 + ( x - 6 ) + m$ is $0$, then $m$ = ____ ?", "answer": "- 30", "steps": "According to the problem, we have $( - 6 ) ^ { 2 } - 6 + m = 0$, and solving for $m$ gives $m = - 30$.", "expr_cands": ["x = 0", "x", "( x - 6 ) ^ { 2 } + ( x - 6 ) + m", "m", "0", "( - 6 ) ^ { 2 } - 6 + m = 0", "m = - 30"], "exprs": ["( - 6 ) ^ { 2 } - 6 + m = 0", "m = - 30"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 0"}, {"id": "( - 6 ) ^ { 2 } - 6 + m = 0"}, {"id": "( x - 6 ) ^ { 2 } + ( x - 6 ) + m"}, {"id": "0"}, {"id": "$x = 0$ 时"}, {"id": "$( x - 6 ) ^ { 2 } + ( x - 6 ) + m$ 的计算结果是 $0$"}, {"id": "m = - 30"}], "links": [{"rel": "被描述", "source": "x = 0", "target": "( - 6 ) ^ { 2 } - 6 + m = 0"}, {"rel": "等式方程求解", "source": "( - 6 ) ^ { 2 } - 6 + m = 0", "target": "m = - 30"}, {"rel": "被描述", "source": "( x - 6 ) ^ { 2 } + ( x - 6 ) + m", "target": "( - 6 ) ^ { 2 } - 6 + m = 0"}, {"rel": "被描述", "source": "0", "target": "( - 6 ) ^ { 2 } - 6 + m = 0"}, {"rel": "限制性描述", "source": "$x = 0$ 时", "target": "( - 6 ) ^ { 2 } - 6 + m = 0"}, {"rel": "限制性描述", "source": "$( x - 6 ) ^ { 2 } + ( x - 6 ) + m$ 的计算结果是 $0$", "target": "( - 6 ) ^ { 2 } - 6 + m = 0"}]}}
|
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{"content": "Given $\\sqrt { x + 2 } = 2$, what is $( x + 2 ) ^ { 2 }$ equal to?", "answer": "16", "steps": "Because the square root of x plus 2 is equal to 2, therefore x plus 2 is equal to 4, therefore the quantity x plus 2 squared is equal to 16.", "expr_cands": ["\\sqrt { x + 2 } = 2", "x", "( x + 2 ) ^ { 2 }", "x = 2", "x + 2 = 4", "16"], "exprs": ["x = 2", "16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + 2 } = 2"}, {"id": "x = 2"}, {"id": "( x + 2 ) ^ { 2 }"}, {"id": "16"}], "links": [{"rel": "等式方程求解", "source": "\\sqrt { x + 2 } = 2", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "16"}, {"rel": "被代入", "source": "( x + 2 ) ^ { 2 }", "target": "16"}]}}
|
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{"content": "If $m < 0$ and $mn < 0$, what is the value of $| n - m + 1 | - | m - n - 5 |$?", "answer": "- 4", "steps": "According to the problem, we have $m < 0$ and $mn < 0$, so $n > 0$. It follows that $n - m + 1 > 0$ and $m - n - 5 < 0$. Therefore, the original expression is equal to $n - m + 1 + m - n - 5 = - 4$.", "expr_cands": ["m < 0", "m", "mn < 0", "n", "| n - m + 1 | - | m - n - 5 |", "n > 0", "n - m + 1 > 0", "m - n - 5 < 0", "n - m + 1 + m - n - 5", "- 4"], "exprs": ["n > 0", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m < 0"}, {"id": "n > 0"}, {"id": "mn < 0"}, {"id": "- 4"}, {"id": "| n - m + 1 | - | m - n - 5 |"}, {"id": "$| n - m + 1 | - | m - n - 5 |$ 的值"}, {"id": "绝对值恒大于等于0"}], "links": [{"rel": "联立", "source": "m < 0", "target": "n > 0"}, {"rel": "被描述", "source": "m < 0", "target": "- 4"}, {"rel": "被描述", "source": "n > 0", "target": "- 4"}, {"rel": "联立", "source": "mn < 0", "target": "n > 0"}, {"rel": "被描述", "source": "| n - m + 1 | - | m - n - 5 |", "target": "- 4"}, {"rel": "限制性描述", "source": "$| n - m + 1 | - | m - n - 5 |$ 的值", "target": "- 4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "- 4"}]}}
|
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{"content": "Given that the value of the polynomial $x + x + 1$ is $8$, what is the value of $4 x + 4 x + 9$?", "answer": "37", "steps": "Due to the fact that the value of $x + x + 1$ is $8$, i.e. $x + x + 1 = 8$, we have $x + x = 7$. Therefore, $4 x + 4 x = 4 \\times 7 = 28$. Hence, $4 x + 4 x + 9 = 37$.", "expr_cands": ["x + x + 1", "x", "8", "4 x + 4 x + 9", "x + x + 1 = 8", "x = \\frac { 7 } { 2 }", "x + x = 7", "4 x + 4 x = 28", "37"], "exprs": ["x + x + 1 = 8", "x = \\frac { 7 } { 2 }", "37"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + x + 1"}, {"id": "x + x + 1 = 8"}, {"id": "8"}, {"id": "整式 $x + x + 1$ 的值是 $8$"}, {"id": "x = \\frac { 7 } { 2 }"}, {"id": "4 x + 4 x + 9"}, {"id": "37"}], "links": [{"rel": "被描述", "source": "x + x + 1", "target": "x + x + 1 = 8"}, {"rel": "等式方程求解", "source": "x + x + 1 = 8", "target": "x = \\frac { 7 } { 2 }"}, {"rel": "被描述", "source": "8", "target": "x + x + 1 = 8"}, {"rel": "限制性描述", "source": "整式 $x + x + 1$ 的值是 $8$", "target": "x + x + 1 = 8"}, {"rel": "代入", "source": "x = \\frac { 7 } { 2 }", "target": "37"}, {"rel": "被代入", "source": "4 x + 4 x + 9", "target": "37"}]}}
|
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{"content": "In the algebraic expression $y = \\sqrt { x - 1 }$, what is the range of values for $x$?", "answer": "x \\ge 1", "steps": "From the given condition, we obtain $x - 1 \\ge 0$, which implies $x \\ge 1$ as a solution.", "expr_cands": ["y = \\sqrt { x - 1 }", "y", "x", "x - 1 \\ge 0", "1 \\le x", "x \\ge 1"], "exprs": ["x - 1 \\ge 0", "x \\ge 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 1 }"}, {"id": "x - 1 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 1"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 1 }", "target": "x - 1 \\ge 0"}, {"rel": "不等式方程求解", "source": "x - 1 \\ge 0", "target": "x \\ge 1"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 1 \\ge 0"}]}}
|
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{"content": "If ${ ( a - 2 ) } ^ { 0 } = 1$, then the range of possible values for the real number $a$ is:", "answer": "a \\neq { 2 }", "steps": "Since ${ ( a - 2 ) } ^ { 0 } = 1$, it follows that $a - 2 \\neq 0$. Solving for $a$, we get $a \\neq 2$.", "expr_cands": ["{ ( a - 2 ) } ^ { 0 } = 1", "a", "{ ( a - 2 ) } ^ { 0 }", "1", "a - 2 \\neq 0", "a \\neq 2"], "exprs": ["a - 2 \\neq 0", "a \\neq 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ ( a - 2 ) } ^ { 0 } = 1"}, {"id": "a - 2 \\neq 0"}, {"id": "${ ( a - 2 ) } ^ { 0 } = 1$ 实数 $a$ 的取值范围"}, {"id": "多项式零次方项，若底数不为0，则恒等于1"}, {"id": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0"}, {"id": "a \\neq 2"}], "links": [{"rel": "被描述", "source": "{ ( a - 2 ) } ^ { 0 } = 1", "target": "a - 2 \\neq 0"}, {"rel": "不等式方程求解", "source": "a - 2 \\neq 0", "target": "a \\neq 2"}, {"rel": "限制性描述", "source": "${ ( a - 2 ) } ^ { 0 } = 1$ 实数 $a$ 的取值范围", "target": "a - 2 \\neq 0"}, {"rel": "属性描述", "source": "多项式零次方项，若底数不为0，则恒等于1", "target": "a - 2 \\neq 0"}, {"rel": "属性描述", "source": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0", "target": "a - 2 \\neq 0"}]}}
|
||||
{"content": "If $y = \\frac { 9 } { 2 }$, then the result of $\\frac { 2 x + 6 } { 12 y } \\div \\frac { x + 3 } { 12 y ^ 2 }$ is ____?", "answer": "9", "steps": "$\\frac { 2 x + 6 } { 12 y } \\div \\frac { x + 3 } { 12 y ^ { 2 } } = \\frac { 2 ( x + 3 ) } { 12 y } * \\frac { 12 y ^ { 2 } } { x + 3 } = 2 y$ , when $y = \\frac { 9 } { 2 }$ , the original expression $= 2 * \\frac { 9 } { 2 } = 9$.", "expr_cands": ["y = \\frac { 9 } { 2 }", "y", "\\frac { 2 x + 6 } { 12 y } \\div \\frac { x + 3 } { 12 y ^ { 2 } }", "x", "2 y", "2 * \\frac { 9 } { 2 }", "9"], "exprs": ["9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 9 } { 2 }"}, {"id": "9"}, {"id": "\\frac { 2 x + 6 } { 12 y } \\div \\frac { x + 3 } { 12 y ^ { 2 } }"}], "links": [{"rel": "代入", "source": "y = \\frac { 9 } { 2 }", "target": "9"}, {"rel": "被代入", "source": "\\frac { 2 x + 6 } { 12 y } \\div \\frac { x + 3 } { 12 y ^ { 2 } }", "target": "9"}]}}
|
||||
{"content": "If $x = 1$, then $| x - 4 |$ = ____ ?", "answer": "3", "steps": "Since $x = 1$, it follows that $| x - 4 | = | 1 - 4 | = 3$.", "expr_cands": ["x = 1", "x", "| x - 4 |", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x - 4 |"}, {"id": "3"}, {"id": "x = 1"}, {"id": "绝对值恒大于等于0"}], "links": [{"rel": "被描述", "source": "| x - 4 |", "target": "3"}, {"rel": "被描述", "source": "x = 1", "target": "3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "3"}]}}
|
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{"content": "If $\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ and $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ can be combined into one term, then the value of ${ m } ^ { n }$ is ____?", "answer": "8", "steps": "$\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ and $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ can be combined into one term, that is: $\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ and $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ are like terms, $\\therefore m = 2$, $n = 3$, ${ m } ^ { n } = { 2 } ^ { 3 } = 8$, that is: the value of ${ m } ^ { n }$ is $8$.", "expr_cands": ["\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }", "b", "m", "a", "c", "- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }", "n", "{ m } ^ { n }", "m = 2", "n = 3", "8"], "exprs": ["m = 2", "n = 3", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }"}, {"id": "m = 2"}, {"id": "- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }"}, {"id": "$\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ 与 $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ 可以合并成一项"}, {"id": "即 : $\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ 与 $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ 是同类项"}, {"id": "n = 3"}, {"id": "{ m } ^ { n }"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }", "target": "m = 2"}, {"rel": "被描述", "source": "\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }", "target": "n = 3"}, {"rel": "代入", "source": "m = 2", "target": "8"}, {"rel": "被描述", "source": "- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }", "target": "m = 2"}, {"rel": "被描述", "source": "- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }", "target": "n = 3"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ 与 $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ 可以合并成一项", "target": "m = 2"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ 与 $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ 可以合并成一项", "target": "n = 3"}, {"rel": "限制性描述", "source": "即 : $\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ 与 $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ 是同类项", "target": "m = 2"}, {"rel": "限制性描述", "source": "即 : $\\frac { 1 } { 3 } { a } ^ { m } { b } ^ { 3 } { c } ^ { 4 }$ 与 $- 3 { a } ^ { 2 } { b } ^ { n } { c } ^ { 4 }$ 是同类项", "target": "n = 3"}, {"rel": "代入", "source": "n = 3", "target": "8"}, {"rel": "被代入", "source": "{ m } ^ { n }", "target": "8"}]}}
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{"content": "If $| a - 2 | = 2 - a$ and $| a | = 3$, then $a$ = ____ ?", "answer": "- 3", "steps": "Since $| a - 2 | = 2 - a$ and $| a | = 3$, it follows that $2 - a > 0$, which implies that $a < 2$. Therefore, $a = - 3$.", "expr_cands": ["| a - 2 | = 2 - a", "a", "| a | = 3", "a = - 3", "a = 3", "2 - a > 0", "a < 2"], "exprs": ["2 - a > 0", "a < 2", "a = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 2 | = 2 - a"}, {"id": "2 - a > 0"}, {"id": "绝对值恒大于等于0"}, {"id": "a < 2"}, {"id": "a = - 3"}, {"id": "| a | = 3"}], "links": [{"rel": "被描述", "source": "| a - 2 | = 2 - a", "target": "2 - a > 0"}, {"rel": "不等式方程求解", "source": "2 - a > 0", "target": "a < 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "2 - a > 0"}, {"rel": "联立", "source": "a < 2", "target": "a = - 3"}, {"rel": "联立", "source": "| a | = 3", "target": "a = - 3"}]}}
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{"content": "Given that the equation $x + k = 1$ has a solution of $x = 5$ for $x$, then $- | k + 2 |$ = ____ ?", "answer": "- 2", "steps": "$\\because$ The equation about $x$, $x + k = 1$, has a solution of $x = 5$. $\\therefore$ $5 + k = 1$, which gives us $k = - 4$. $\\therefore$ $- | k + 2 | = - | - 4 + 2 | = - | - 2 | = - 2$.", "expr_cands": ["x", "x + k = 1", "k", "x = 5", "- | k + 2 |", "5 + k = 1", "k = - 4", "- 2"], "exprs": ["5 + k = 1", "k = - 4", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + k = 1"}, {"id": "5 + k = 1"}, {"id": "x = 5"}, {"id": "k = - 4"}, {"id": "- | k + 2 |"}, {"id": "- 2"}], "links": [{"rel": "被代入", "source": "x + k = 1", "target": "5 + k = 1"}, {"rel": "等式方程求解", "source": "5 + k = 1", "target": "k = - 4"}, {"rel": "代入", "source": "x = 5", "target": "5 + k = 1"}, {"rel": "代入", "source": "k = - 4", "target": "- 2"}, {"rel": "被代入", "source": "- | k + 2 |", "target": "- 2"}]}}
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{"content": "The parabola $y = 3 x ^ { 2 }$ is shifted $1$ unit to the right and $2$ units down, resulting in the parabola _____.", "answer": "y = 3 ( x - 1 ) ^ { 2 } - 2", "steps": "According to the rule add up and subtract down, subtract left and add right, it can be known that the parabola $y = 3 x ^ 2$ is shifted one unit to the right and two units down, and the resulting parabola is $y = 3 ( x - 1 ) ^ 2 - 2$.", "expr_cands": ["y = 3 x ^ { 2 }", "y", "x", "1", "2", "y = 3 ( x - 1 ) ^ { 2 } - 2", "3 x ^ { 2 } = 3 ( x - 1 ) ^ { 2 } - 2"], "exprs": ["y = 3 ( x - 1 ) ^ { 2 } - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 3 x ^ { 2 }"}, {"id": "y = 3 ( x - 1 ) ^ { 2 } - 2"}, {"id": "1"}, {"id": "2"}, {"id": "抛物线 $y = 3 x ^ { 2 }$ 向右平移 $1$ 个单位"}, {"id": "再向下平移 $2$ 个单位"}, {"id": "所得到的抛物线"}], "links": [{"rel": "被描述", "source": "y = 3 x ^ { 2 }", "target": "y = 3 ( x - 1 ) ^ { 2 } - 2"}, {"rel": "被描述", "source": "1", "target": "y = 3 ( x - 1 ) ^ { 2 } - 2"}, {"rel": "被描述", "source": "2", "target": "y = 3 ( x - 1 ) ^ { 2 } - 2"}, {"rel": "限制性描述", "source": "抛物线 $y = 3 x ^ { 2 }$ 向右平移 $1$ 个单位", "target": "y = 3 ( x - 1 ) ^ { 2 } - 2"}, {"rel": "限制性描述", "source": "再向下平移 $2$ 个单位", "target": "y = 3 ( x - 1 ) ^ { 2 } - 2"}, {"rel": "限制性描述", "source": "所得到的抛物线", "target": "y = 3 ( x - 1 ) ^ { 2 } - 2"}]}}
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{"content": "Given $\\frac { a } { b } = \\frac { 2 } { 3 }$, what is the value of $\\frac { a } { a + b }$?", "answer": "\\frac { 2 } { 5 }", "steps": "Because $\\frac { a } { b } = \\frac { 2 } { 3 }$, let $a = 2 k$, $b = 3 k$, then $\\frac { a } { a + b } = \\frac { 2 k } { 2 k + 3 k } = \\frac { 2 k } { 5 k } = \\frac { 2 } { 5 }$.", "expr_cands": ["\\frac { a } { b } = \\frac { 2 } { 3 }", "b", "a", "\\frac { a } { a + b }", "a = 2 k", "k", "b = 3 k", "\\frac { 2 } { 5 }"], "exprs": ["a = 2 k", "b = 3 k", "\\frac { 2 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $a = 2 k$ , $b = 3 k$"}, {"id": "a = 2 k"}, {"id": "b = 3 k"}, {"id": "\\frac { a } { a + b }"}, {"id": "\\frac { 2 } { 5 }"}], "links": [{"rel": "假设描述", "source": "设 $a = 2 k$ , $b = 3 k$", "target": "a = 2 k"}, {"rel": "假设描述", "source": "设 $a = 2 k$ , $b = 3 k$", "target": "b = 3 k"}, {"rel": "代入", "source": "a = 2 k", "target": "\\frac { 2 } { 5 }"}, {"rel": "代入", "source": "b = 3 k", "target": "\\frac { 2 } { 5 }"}, {"rel": "被代入", "source": "\\frac { a } { a + b }", "target": "\\frac { 2 } { 5 }"}]}}
|
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{"content": "If the value of the expression $\\frac { 2 x - 1 } { 7 }$ is non-negative, then the range of possible values for $x$ is ____?", "answer": "x \\ge \\frac { 1 } { 2 }", "steps": "According to the problem, we have $\\frac { 2 x - 1 } { 7 } \\ge 0$. Simplifying, we get $2 x \\ge 1$, which means $x \\ge \\frac { 1 } { 2 }$.", "expr_cands": ["\\frac { 2 x - 1 } { 7 }", "x", "\\frac { 2 x - 1 } { 7 } \\ge 0", "\\frac { 1 } { 2 } \\le x", "2 x \\ge 1", "x \\ge \\frac { 1 } { 2 }"], "exprs": ["\\frac { 2 x - 1 } { 7 } \\ge 0", "x \\ge \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 x - 1 } { 7 }"}, {"id": "\\frac { 2 x - 1 } { 7 } \\ge 0"}, {"id": "式子 $\\frac { 2 x - 1 } { 7 }$ 的值是非负数"}, {"id": "x \\ge \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\frac { 2 x - 1 } { 7 }", "target": "\\frac { 2 x - 1 } { 7 } \\ge 0"}, {"rel": "不等式方程求解", "source": "\\frac { 2 x - 1 } { 7 } \\ge 0", "target": "x \\ge \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "式子 $\\frac { 2 x - 1 } { 7 }$ 的值是非负数", "target": "\\frac { 2 x - 1 } { 7 } \\ge 0"}]}}
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{"content": "If the square roots of a positive number are $a - 6$ and $3 a - 6$, then the number is ____?", "answer": "9", "steps": "According to the problem, we have $a - 6 + 3 a - 6 = 0$, which means $a = 3$. Therefore, the positive number is $( 3 - 6 ) ^ 2 = 9$.", "expr_cands": ["a - 6", "a", "3 a - 6", "a - 6 + 3 a - 6 = 0", "a = 3", "( 3 - 6 ) ^ { 2 }", "9"], "exprs": ["a - 6 + 3 a - 6 = 0", "a = 3", "( 3 - 6 ) ^ { 2 }", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - 6"}, {"id": "a - 6 + 3 a - 6 = 0"}, {"id": "3 a - 6"}, {"id": "一正数的平方根分别是 $a - 6$ 和 $3 a - 6$"}, {"id": "平方根互为相反数"}, {"id": "a = 3"}, {"id": "( 3 - 6 ) ^ { 2 }"}, {"id": "这个正数"}, {"id": "9"}], "links": [{"rel": "被描述", "source": "a - 6", "target": "a - 6 + 3 a - 6 = 0"}, {"rel": "被描述", "source": "a - 6", "target": "( 3 - 6 ) ^ { 2 }"}, {"rel": "等式方程求解", "source": "a - 6 + 3 a - 6 = 0", "target": "a = 3"}, {"rel": "被描述", "source": "3 a - 6", "target": "a - 6 + 3 a - 6 = 0"}, {"rel": "被描述", "source": "3 a - 6", "target": "( 3 - 6 ) ^ { 2 }"}, {"rel": "限制性描述", "source": "一正数的平方根分别是 $a - 6$ 和 $3 a - 6$", "target": "a - 6 + 3 a - 6 = 0"}, {"rel": "限制性描述", "source": "一正数的平方根分别是 $a - 6$ 和 $3 a - 6$", "target": "( 3 - 6 ) ^ { 2 }"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "a - 6 + 3 a - 6 = 0"}, {"rel": "被描述", "source": "a = 3", "target": "( 3 - 6 ) ^ { 2 }"}, {"rel": "计算", "source": "( 3 - 6 ) ^ { 2 }", "target": "9"}, {"rel": "限制性描述", "source": "这个正数", "target": "( 3 - 6 ) ^ { 2 }"}]}}
|
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{"content": "What is the smallest integer that satisfies the inequality $8 + 2 x > 0$?", "answer": "- 3", "steps": "Moving the term yields: $2 x > - 8$. Solving for $x$, we get: $x > - 4$. Therefore, the smallest integer is $- 3$.", "expr_cands": ["8 + 2 x > 0", "x", "2 x > - 8", "- 4 < x", "x > - 4", "- 3"], "exprs": ["x > - 4", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "8 + 2 x > 0"}, {"id": "x > - 4"}, {"id": "- 3"}, {"id": "满足不等式 $8 + 2 x > 0$ 的最小整数"}], "links": [{"rel": "不等式方程求解", "source": "8 + 2 x > 0", "target": "x > - 4"}, {"rel": "被描述", "source": "x > - 4", "target": "- 3"}, {"rel": "限制性描述", "source": "满足不等式 $8 + 2 x > 0$ 的最小整数", "target": "- 3"}]}}
|
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{"content": "Given $\\frac { c } { a - 2 b } = 3$, find the value of the algebraic expression $\\frac { 2 c } { a - 2 b } - \\frac { a - 2 b } { c } - \\frac { 5 } { 3 }$.", "answer": "4", "steps": "Because $\\frac { c } { a - 2 b } = 3$, therefore $\\frac { a - 2 b } { c } = \\frac { 1 } { 3 }$, then the original expression $= 2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 } = 4$.", "expr_cands": ["\\frac { c } { a - 2 b } = 3", "b", "c", "a", "\\frac { 2 c } { a - 2 b } - \\frac { a - 2 b } { c } - \\frac { 5 } { 3 }", "\\frac { a - 2 b } { c } = \\frac { 1 } { 3 }", "2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 }", "4"], "exprs": ["\\frac { a - 2 b } { c } = \\frac { 1 } { 3 }", "2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 }", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { c } { a - 2 b } = 3"}, {"id": "\\frac { a - 2 b } { c } = \\frac { 1 } { 3 }"}, {"id": "2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 }"}, {"id": "\\frac { 2 c } { a - 2 b } - \\frac { a - 2 b } { c } - \\frac { 5 } { 3 }"}, {"id": "4"}], "links": [{"rel": "同乘除", "source": "\\frac { c } { a - 2 b } = 3", "target": "\\frac { a - 2 b } { c } = \\frac { 1 } { 3 }"}, {"rel": "代入", "source": "\\frac { c } { a - 2 b } = 3", "target": "2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 }"}, {"rel": "代入", "source": "\\frac { a - 2 b } { c } = \\frac { 1 } { 3 }", "target": "2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 }"}, {"rel": "计算", "source": "2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 }", "target": "4"}, {"rel": "被代入", "source": "\\frac { 2 c } { a - 2 b } - \\frac { a - 2 b } { c } - \\frac { 5 } { 3 }", "target": "2 * 3 - \\frac { 1 } { 3 } - \\frac { 5 } { 3 }"}]}}
|
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{"content": "When $k$ = ____ ?, the polynomial $x ^ { 2 } - kxy - ( 3 y ^ { 2 } - 3 xy - 8 )$ in terms of $x$ and $y$ does not contain the term $xy$.", "answer": "3", "steps": "$x ^ { 2 } - kxy - ( 3 y ^ { 2 } - 3 xy - 8 ) = x ^ { 2 } - kxy - 3 y ^ { 2 } + 3 xy + 8 = x ^ { 2 } - 3 y ^ { 2 } + ( 3 - k ) xy + 8$ Let $3 - k = 0$, therefore $k = 3$.", "expr_cands": ["k", "x", "y", "x ^ { 2 } - kxy - ( 3 y ^ { 2 } - 3 xy - 8 )", "xy", "x ^ { 2 } - 3 y ^ { 2 } + ( 3 - k ) xy + 8", "3 - k = 0", "k = 3"], "exprs": ["x ^ { 2 } - 3 y ^ { 2 } + ( 3 - k ) xy + 8", "3 - k = 0", "k = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - kxy - ( 3 y ^ { 2 } - 3 xy - 8 )"}, {"id": "x ^ { 2 } - 3 y ^ { 2 } + ( 3 - k ) xy + 8"}, {"id": "xy"}, {"id": "3 - k = 0"}, {"id": "关于 $x$ , $y$ 的多项式 $x ^ { 2 } - kxy - ( 3 y ^ { 2 } - 3 xy - 8 )$ 中不含 $xy$ 项"}, {"id": "k = 3"}], "links": [{"rel": "提取因式", "source": "x ^ { 2 } - kxy - ( 3 y ^ { 2 } - 3 xy - 8 )", "target": "x ^ { 2 } - 3 y ^ { 2 } + ( 3 - k ) xy + 8"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 y ^ { 2 } + ( 3 - k ) xy + 8", "target": "3 - k = 0"}, {"rel": "提取因式参考", "source": "xy", "target": "x ^ { 2 } - 3 y ^ { 2 } + ( 3 - k ) xy + 8"}, {"rel": "等式方程求解", "source": "3 - k = 0", "target": "k = 3"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的多项式 $x ^ { 2 } - kxy - ( 3 y ^ { 2 } - 3 xy - 8 )$ 中不含 $xy$ 项", "target": "3 - k = 0"}]}}
|
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{"content": "The equation $3 x + 6 = 0$ and $5 x + m = 20$ have the same solution for $x$. What is the value of $m$?", "answer": "30", "steps": "The solution to the equation $3 x + 6 = 0$ is $x = - 2$. Substituting $x = - 2$ into the equation $5 x + m = 20$ yields $- 10 + m = 20$, which can be solved to obtain $m = 30$.", "expr_cands": ["x", "3 x + 6 = 0", "5 x + m = 20", "m", "x = - 2", "m - 10 = 20", "- 10 + m = 20", "m = 30"], "exprs": ["x = - 2", "- 10 + m = 20", "m = 30"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + 6 = 0"}, {"id": "x = - 2"}, {"id": "5 x + m = 20"}, {"id": "- 10 + m = 20"}, {"id": "m = 30"}], "links": [{"rel": "等式方程求解", "source": "3 x + 6 = 0", "target": "x = - 2"}, {"rel": "代入", "source": "x = - 2", "target": "- 10 + m = 20"}, {"rel": "被代入", "source": "5 x + m = 20", "target": "- 10 + m = 20"}, {"rel": "等式方程求解", "source": "- 10 + m = 20", "target": "m = 30"}]}}
|
||||
{"content": "If the value of the algebraic expression $\\frac { 6 - 2 x } { 3 }$ is negative, then the range of possible values for $x$ is ____?", "answer": "x > 3", "steps": "According to the problem, we have $\\frac { 6 - 2 x } { 3 } < 0$, and solving it gives us $x > 3$.", "expr_cands": ["\\frac { 6 - 2 x } { 3 }", "x", "\\frac { 6 - 2 x } { 3 } < 0", "3 < x", "x > 3"], "exprs": ["\\frac { 6 - 2 x } { 3 } < 0", "x > 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 6 - 2 x } { 3 }"}, {"id": "\\frac { 6 - 2 x } { 3 } < 0"}, {"id": "代数式 $\\frac { 6 - 2 x } { 3 }$ 的值是负数"}, {"id": "x > 3"}], "links": [{"rel": "被描述", "source": "\\frac { 6 - 2 x } { 3 }", "target": "\\frac { 6 - 2 x } { 3 } < 0"}, {"rel": "不等式方程求解", "source": "\\frac { 6 - 2 x } { 3 } < 0", "target": "x > 3"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 6 - 2 x } { 3 }$ 的值是负数", "target": "\\frac { 6 - 2 x } { 3 } < 0"}]}}
|
||||
{"content": "If the monomial $- \\frac { 4 } { 7 } \\pi xy ^ { 3 }$ is a like term with $- 8 x ^ { m } y ^ { n }$, then $m ^ { n }$ = ____ ?", "answer": "1", "steps": "$\\because$ The monomial $- \\frac { 4 } { 7 } \\pi xy ^ 3$ and $- 8 x ^ my ^ n$ are like terms, $\\therefore$ $m = 1$, $n = 3$, $\\therefore$ $m ^ n = 1 ^ 3 = 1$.", "expr_cands": ["- \\frac { 4 } { 7 } \\pi xy ^ { 3 }", "y", "x", "- 8 x ^ { m } y ^ { n }", "m", "n", "m ^ { n }", "m = 1", "n = 3", "1"], "exprs": ["m = 1", "n = 3", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 4 } { 7 } \\pi xy ^ { 3 }"}, {"id": "m = 1"}, {"id": "- 8 x ^ { m } y ^ { n }"}, {"id": "单项式 $- \\frac { 4 } { 7 } \\pi xy ^ { 3 }$ 与 $- 8 x ^ { m } y ^ { n }$ 是同类项"}, {"id": "n = 3"}, {"id": "m ^ { n }"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "- \\frac { 4 } { 7 } \\pi xy ^ { 3 }", "target": "m = 1"}, {"rel": "被描述", "source": "- \\frac { 4 } { 7 } \\pi xy ^ { 3 }", "target": "n = 3"}, {"rel": "代入", "source": "m = 1", "target": "1"}, {"rel": "被描述", "source": "- 8 x ^ { m } y ^ { n }", "target": "m = 1"}, {"rel": "被描述", "source": "- 8 x ^ { m } y ^ { n }", "target": "n = 3"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { 4 } { 7 } \\pi xy ^ { 3 }$ 与 $- 8 x ^ { m } y ^ { n }$ 是同类项", "target": "m = 1"}, {"rel": "限制性描述", "source": "单项式 $- \\frac { 4 } { 7 } \\pi xy ^ { 3 }$ 与 $- 8 x ^ { m } y ^ { n }$ 是同类项", "target": "n = 3"}, {"rel": "代入", "source": "n = 3", "target": "1"}, {"rel": "被代入", "source": "m ^ { n }", "target": "1"}]}}
|
||||
{"content": "What is the inequality that corresponds to the statement The sum of twice $a$ and $- 5$ is positive?", "answer": "2 a - 5 > 0", "steps": "The sum of twice $a$ and negative five is positive. The inequality written for this statement is $2 a - 5 > 0$.", "expr_cands": ["a", "2", "- 5", "2 a - 5 > 0", "\\frac { 5 } { 2 } < a"], "exprs": ["2 a - 5 > 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2"}, {"id": "2 a - 5 > 0"}, {"id": "a"}, {"id": "- 5"}, {"id": "根据 \" $a$ 的 $2$ 倍与 $- 5$ 的和是正数 \" 列出不等式"}], "links": [{"rel": "被描述", "source": "2", "target": "2 a - 5 > 0"}, {"rel": "被描述", "source": "a", "target": "2 a - 5 > 0"}, {"rel": "被描述", "source": "- 5", "target": "2 a - 5 > 0"}, {"rel": "限制性描述", "source": "根据 \" $a$ 的 $2$ 倍与 $- 5$ 的和是正数 \" 列出不等式", "target": "2 a - 5 > 0"}]}}
|
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{"content": "Given $2 ^ { x } \\cdot 8 ^ { x + 1 } = 2 ^ { 2 x + 5 }$, what is the value of $x$?", "answer": "1", "steps": "Because $2 ^ { x } \\cdot 8 ^ { x + 1 } = 2 ^ { 2 x + 5 }$, therefore $2 ^ { x } \\cdot ( 2 ^ { 3 } ) ^ { x + 1 } = 2 ^ { 2 x + 5 }$, therefore $2 ^ { x } \\cdot 2 ^ { 3 x + 3 } = 2 ^ { 2 x + 5 }$, therefore $2 ^ { 4 x + 3 } = 2 ^ { 2 x + 5 }$, therefore $4 x + 3 = 2 x + 5$, therefore $x = 1$.", "expr_cands": ["2 ^ { x } \\cdot 8 ^ { x + 1 } = 2 ^ { 2 x + 5 }", "x", "x = 1", "2 ^ { x } \\cdot ( 2 ^ { 3 } ) ^ { x + 1 } = 2 ^ { 2 x + 5 }", "2 ^ { x } \\cdot 2 ^ { 3 x + 3 } = 2 ^ { 2 x + 5 }", "2 ^ { 4 x + 3 } = 2 ^ { 2 x + 5 }", "4 x + 3 = 2 x + 5"], "exprs": ["x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ^ { x } \\cdot 8 ^ { x + 1 } = 2 ^ { 2 x + 5 }"}, {"id": "x = 1"}], "links": [{"rel": "等式方程求解", "source": "2 ^ { x } \\cdot 8 ^ { x + 1 } = 2 ^ { 2 x + 5 }", "target": "x = 1"}]}}
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{"content": "Given $2 a - b = 2$, what is the value of the algebraic expression $4 a ^ 2 - b ^ 2 - 4 b$?", "answer": "4", "steps": "$4 a ^ { 2 } - b ^ { 2 } - 4 b = 4 a ^ { 2 } - ( b ^ { 2 } + 4 b + 4 ) + 4 = ( 2 a ) ^ { 2 } - ( b + 2 ) ^ { 2 } + 4 = [ 2 a + ( b + 2 ) ] [ 2 a - ( b + 2 ) ] + 4 = ( 2 a + b + 2 ) ( 2 a - b - 2 ) + 4$ When $2 a - b = 2$, the original expression equals $0 + 4 = 4$. ", "expr_cands": ["2 a - b = 2", "b", "a", "4 { a } ^ { 2 } - { b } ^ { 2 } - 4 b", "4 a ^ { 2 } - b ^ { 2 } - 4 b", "( 2 a + b + 2 ) ( 2 a - b - 2 ) + 4", "0 + 4", "4"], "exprs": ["( 2 a + b + 2 ) ( 2 a - b - 2 ) + 4", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 a ^ { 2 } - b ^ { 2 } - 4 b"}, {"id": "( 2 a + b + 2 ) ( 2 a - b - 2 ) + 4"}, {"id": "2 a - b = 2"}, {"id": "4"}], "links": [{"rel": "提取因式", "source": "4 a ^ { 2 } - b ^ { 2 } - 4 b", "target": "( 2 a + b + 2 ) ( 2 a - b - 2 ) + 4"}, {"rel": "被代入", "source": "( 2 a + b + 2 ) ( 2 a - b - 2 ) + 4", "target": "4"}, {"rel": "提取因式参考", "source": "2 a - b = 2", "target": "( 2 a + b + 2 ) ( 2 a - b - 2 ) + 4"}, {"rel": "代入", "source": "2 a - b = 2", "target": "4"}]}}
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{"content": "$x = - 1$ is a root of the equation $x ^ 2 - mx + 6 = 0$. Find $m$ = ____?", "answer": "- 7", "steps": "Substituting $x = - 1$ into $x ^ 2 - mx + 6 = 0$, we get $( - 1 ) ^ { 2 } + m + 6 = 0$, which yields $m = - 7$.", "expr_cands": ["x = - 1", "x", "x ^ { 2 } - mx + 6 = 0", "m", "m + 7 = 0", "( - 1 ) ^ { 2 } + m + 6 = 0", "m = - 7"], "exprs": ["( - 1 ) ^ { 2 } + m + 6 = 0", "m = - 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - mx + 6 = 0"}, {"id": "( - 1 ) ^ { 2 } + m + 6 = 0"}, {"id": "x = - 1"}, {"id": "m = - 7"}], "links": [{"rel": "被代入", "source": "x ^ { 2 } - mx + 6 = 0", "target": "( - 1 ) ^ { 2 } + m + 6 = 0"}, {"rel": "等式方程求解", "source": "( - 1 ) ^ { 2 } + m + 6 = 0", "target": "m = - 7"}, {"rel": "代入", "source": "x = - 1", "target": "( - 1 ) ^ { 2 } + m + 6 = 0"}]}}
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{"content": "If $| m - 3 | + ( n + 2 ) ^ { 2 } = 0$, then the value of $mn$ is ____?", "answer": "- 6", "steps": "$\\because | m - 3 | + ( n + 2 ) ^ { 2 } = 0$ , $\\therefore m - 3 = 0$ , $n + 2 = 0$ , solving for $m = 3$ , $n = - 2$ , thus $mn = - 6$ .", "expr_cands": ["| m - 3 | + ( n + 2 ) ^ { 2 } = 0", "m", "n", "mn", "m - 3 = 0", "m = 3", "n + 2 = 0", "n = - 2", "- 6"], "exprs": ["m - 3 = 0", "n + 2 = 0", "m = 3", "n = - 2", "- 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| m - 3 | + ( n + 2 ) ^ { 2 } = 0"}, {"id": "m - 3 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "n + 2 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "m = 3"}, {"id": "n = - 2"}, {"id": "mn"}, {"id": "- 6"}], "links": [{"rel": "被描述", "source": "| m - 3 | + ( n + 2 ) ^ { 2 } = 0", "target": "m - 3 = 0"}, {"rel": "被描述", "source": "| m - 3 | + ( n + 2 ) ^ { 2 } = 0", "target": "n + 2 = 0"}, {"rel": "等式方程求解", "source": "m - 3 = 0", "target": "m = 3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m - 3 = 0"}, {"rel": "等式方程求解", "source": "n + 2 = 0", "target": "n = - 2"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "n + 2 = 0"}, {"rel": "代入", "source": "m = 3", "target": "- 6"}, {"rel": "代入", "source": "n = - 2", "target": "- 6"}, {"rel": "被代入", "source": "mn", "target": "- 6"}]}}
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{"content": "The polynomial $x ^ { 2 } - 3 kxy - 3 y ^ { 2 } + 3 xy - 8$ does not contain the term $xy$. What is the value of $k$?", "answer": "1", "steps": "Original expression = $x ^ { 2 } + ( - 3 k + 3 ) xy - 3 y ^ { 2 } - 8$. Since there is no $xy$ term, we have $- 3 k + 3 = 0$, which gives $k = 1$.", "expr_cands": ["x ^ { 2 } - 3 kxy - 3 y ^ { 2 } + 3 xy - 8", "y", "k", "x", "xy", "x ^ { 2 } + ( - 3 k + 3 ) xy - 3 y ^ { 2 } - 8", "- 3 k + 3 = 0", "k = 1"], "exprs": ["x ^ { 2 } + ( - 3 k + 3 ) xy - 3 y ^ { 2 } - 8", "- 3 k + 3 = 0", "k = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 kxy - 3 y ^ { 2 } + 3 xy - 8"}, {"id": "x ^ { 2 } + ( - 3 k + 3 ) xy - 3 y ^ { 2 } - 8"}, {"id": "xy"}, {"id": "- 3 k + 3 = 0"}, {"id": "多项式 $x ^ { 2 } - 3 kxy - 3 y ^ { 2 } + 3 xy - 8$ 中不含 $xy$ 项"}, {"id": "k = 1"}], "links": [{"rel": "提取因式", "source": "x ^ { 2 } - 3 kxy - 3 y ^ { 2 } + 3 xy - 8", "target": "x ^ { 2 } + ( - 3 k + 3 ) xy - 3 y ^ { 2 } - 8"}, {"rel": "被描述", "source": "x ^ { 2 } + ( - 3 k + 3 ) xy - 3 y ^ { 2 } - 8", "target": "- 3 k + 3 = 0"}, {"rel": "提取因式参考", "source": "xy", "target": "x ^ { 2 } + ( - 3 k + 3 ) xy - 3 y ^ { 2 } - 8"}, {"rel": "等式方程求解", "source": "- 3 k + 3 = 0", "target": "k = 1"}, {"rel": "限制性描述", "source": "多项式 $x ^ { 2 } - 3 kxy - 3 y ^ { 2 } + 3 xy - 8$ 中不含 $xy$ 项", "target": "- 3 k + 3 = 0"}]}}
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{"content": "If $a$ and $b$ are opposite numbers and $c$ and $d$ are reciprocal numbers, then $5 ( a + b ) - 6 cd$ = ____?", "answer": "- 6", "steps": "$\\because$ $a$ and $b$ are opposite numbers, $c$ and $d$ are reciprocal numbers, $\\therefore$ $a + b = 0$, $cd = 1$, $\\therefore$ the original expression $= 0 - 1 * 6 = - 6$.", "expr_cands": ["a", "b", "c", "d", "5 ( a + b ) - 6 cd", "a + b = 0", "cd = 1", "0 - 1 * 6", "- 6"], "exprs": ["a + b = 0", "cd = 1", "- 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ 与 $b$ 互为相反数"}, {"id": "c"}, {"id": "cd = 1"}, {"id": "d"}, {"id": "$c$ 与 $d$ 互为倒数"}, {"id": "5 ( a + b ) - 6 cd"}, {"id": "- 6"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "- 6"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ 与 $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "c", "target": "cd = 1"}, {"rel": "代入", "source": "cd = 1", "target": "- 6"}, {"rel": "被描述", "source": "d", "target": "cd = 1"}, {"rel": "限制性描述", "source": "$c$ 与 $d$ 互为倒数", "target": "cd = 1"}, {"rel": "被代入", "source": "5 ( a + b ) - 6 cd", "target": "- 6"}]}}
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{"content": "If $x ^ { m - 3 } \\cdot x ^ { 3 m } = x$, what is the value of $\\frac { 1 } { 2 } m ^ { 2 } - m + 1$?", "answer": "\\frac { 1 } { 2 }", "steps": "Since $x ^ { m - 3 } \\times x ^ { 3 m } = x$, we have $x ^ { 4 m - 3 } = x$. Therefore, $4 m - 3 = 1$, which gives us $m = 1$. Thus, $\\frac { 1 } { 2 } m ^ 2 - m + 1 = \\frac { 1 } { 2 } * 1 ^ 2 - 1 + 1 = \\frac { 1 } { 2 }$.", "expr_cands": ["x ^ { m - 3 } \\cdot x ^ { 3 m } = x", "m", "x", "\\frac { 1 } { 2 } m ^ { 2 } - m + 1", "x ^ { m - 3 } \\times x ^ { 3 m } = x", "x ^ { 4 m - 3 } = x", "4 m - 3 = 1", "m = 1", "\\frac { 1 } { 2 }"], "exprs": ["4 m - 3 = 1", "m = 1", "\\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { m - 3 } \\cdot x ^ { 3 m } = x"}, {"id": "4 m - 3 = 1"}, {"id": "m = 1"}, {"id": "\\frac { 1 } { 2 } m ^ { 2 } - m + 1"}, {"id": "\\frac { 1 } { 2 }"}], "links": [{"rel": "同取对数", "source": "x ^ { m - 3 } \\cdot x ^ { 3 m } = x", "target": "4 m - 3 = 1"}, {"rel": "等式方程求解", "source": "4 m - 3 = 1", "target": "m = 1"}, {"rel": "代入", "source": "m = 1", "target": "\\frac { 1 } { 2 }"}, {"rel": "被代入", "source": "\\frac { 1 } { 2 } m ^ { 2 } - m + 1", "target": "\\frac { 1 } { 2 }"}]}}
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{"content": "Given $a ^ 2 - 6 a + 9$ is the opposite of $| b - 1 |$, what is the value of $\\frac { a - b } { a + b }$?", "answer": "\\frac { 1 } { 2 }", "steps": "$\\because a ^ 2 - 6 a + 9$ is the opposite of $| b - 1 |$, $\\therefore a ^ 2 - 6 a + 9 + | b - 1 | = 0$, which means $( a - 3 ) ^ 2 + | b - 1 | = 0$. $\\therefore a - 3 = 0$, $b - 1 = 0$, $\\therefore a = 3$, $b = 1$. $\\therefore$ the original expression is $\\frac { 3 - 1 } { 3 + 1 } = \\frac { 2 } { 4 } = \\frac { 1 } { 2 }$.", "expr_cands": ["a ^ { 2 } - 6 a + 9", "a", "| b - 1 |", "b", "\\frac { a - b } { a + b }", "a ^ { 2 } - 6 a + 9 + | b - 1 | = 0", "( a - 3 ) ^ { 2 } + | b - 1 | = 0", "a - 3 = 0", "a = 3", "b - 1 = 0", "b = 1", "\\frac { 3 - 1 } { 3 + 1 }", "\\frac { 1 } { 2 }"], "exprs": ["a ^ { 2 } - 6 a + 9 + | b - 1 | = 0", "( a - 3 ) ^ { 2 } + | b - 1 | = 0", "a - 3 = 0", "b - 1 = 0", "a = 3", "b = 1", "\\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } - 6 a + 9"}, {"id": "a ^ { 2 } - 6 a + 9 + | b - 1 | = 0"}, {"id": "| b - 1 |"}, {"id": "$a ^ { 2 } - 6 a + 9$ 与 $| b - 1 |$ 互为相反数"}, {"id": "( a - 3 ) ^ { 2 } + | b - 1 | = 0"}, {"id": "a - 3 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "b - 1 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "a = 3"}, {"id": "b = 1"}, {"id": "\\frac { a - b } { a + b }"}, {"id": "\\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "a ^ { 2 } - 6 a + 9", "target": "a ^ { 2 } - 6 a + 9 + | b - 1 | = 0"}, {"rel": "提取因式", "source": "a ^ { 2 } - 6 a + 9 + | b - 1 | = 0", "target": "( a - 3 ) ^ { 2 } + | b - 1 | = 0"}, {"rel": "被描述", "source": "| b - 1 |", "target": "a ^ { 2 } - 6 a + 9 + | b - 1 | = 0"}, {"rel": "限制性描述", "source": "$a ^ { 2 } - 6 a + 9$ 与 $| b - 1 |$ 互为相反数", "target": "a ^ { 2 } - 6 a + 9 + | b - 1 | = 0"}, {"rel": "被描述", "source": "( a - 3 ) ^ { 2 } + | b - 1 | = 0", "target": "a - 3 = 0"}, {"rel": "被描述", "source": "( a - 3 ) ^ { 2 } + | b - 1 | = 0", "target": "b - 1 = 0"}, {"rel": "等式方程求解", "source": "a - 3 = 0", "target": "a = 3"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "a - 3 = 0"}, {"rel": "等式方程求解", "source": "b - 1 = 0", "target": "b = 1"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b - 1 = 0"}, {"rel": "代入", "source": "a = 3", "target": "\\frac { 1 } { 2 }"}, {"rel": "代入", "source": "b = 1", "target": "\\frac { 1 } { 2 }"}, {"rel": "被代入", "source": "\\frac { a - b } { a + b }", "target": "\\frac { 1 } { 2 }"}]}}
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{"content": "Given the equation $2 x + 3 = x + k$ has a solution of $x = 6$, then $k$ = ____ ?", "answer": "9", "steps": "Substituting $x = 6$ into the equation $2 x + 3 = x + k$ yields $2 \\times 6 + 3 = 6 + k$, which can be solved to obtain $k = 9$.", "expr_cands": ["x", "2 x + 3 = x + k", "k", "x = 6", "15 = k + 6", "2 * 6 + 3 = 6 + k", "k = 9"], "exprs": ["2 * 6 + 3 = 6 + k", "k = 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 3 = x + k"}, {"id": "2 * 6 + 3 = 6 + k"}, {"id": "x = 6"}, {"id": "k = 9"}], "links": [{"rel": "被代入", "source": "2 x + 3 = x + k", "target": "2 * 6 + 3 = 6 + k"}, {"rel": "等式方程求解", "source": "2 * 6 + 3 = 6 + k", "target": "k = 9"}, {"rel": "代入", "source": "x = 6", "target": "2 * 6 + 3 = 6 + k"}]}}
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{"content": "In the function equation $y = - \\frac { 1 } { 3 } x + 2$, if $x = 3$, then $y$ = ____ ?", "answer": "1", "steps": "Substituting $x = 3$ into $y = - \\frac { 1 } { 3 } x + 2$, we get $y = - \\frac { 1 } { 3 } * 3 + 2 = 1$.", "expr_cands": ["y = - \\frac { 1 } { 3 } x + 2", "y", "x", "x = 3", "y = 1"], "exprs": ["y = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - \\frac { 1 } { 3 } x + 2"}, {"id": "y = 1"}, {"id": "x = 3"}], "links": [{"rel": "被代入", "source": "y = - \\frac { 1 } { 3 } x + 2", "target": "y = 1"}, {"rel": "代入", "source": "x = 3", "target": "y = 1"}]}}
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{"content": "The value of $m$ for the quadratic equation in one variable $x ^ 2 - 7 x + 2 m = 0$ with one root being 2.5 times the other is ____?", "answer": "4", "steps": "Assuming the other root is $x _ 1$, then one of the roots is $2.5 x _ 1$. Since the quadratic equation in terms of $x$ is $x ^ 2 - 7 x + 2 m = 0$, we have $x _ 1 + 2.5 x _ 1 = 7$ and $x _ 1 \\cdot 2.5 x _ 1 = 2 m$. Solving for $x _ 1$ and $m$, we get $x _ 1 = 2$ and $m = 4$.", "expr_cands": ["x", "x ^ { 2 } - 7 x + 2 m = 0", "m", "2.5", "x _ { 1 }", "2.5 x _ { 1 }", "x _ { 1 } + 2.5 x _ { 1 } = 7", "x_{1} = 2.0", "x _ { 1 } * 2 x _ { 1 } = 2 m", "x _ { 1 } = 2", "m = 4"], "exprs": ["x _ { 1 }", "2.5 x _ { 1 }", "x _ { 1 } + 2.5 x _ { 1 } = 7", "x _ { 1 } * 2 x _ { 1 } = 2 m", "x _ { 1 } = 2", "m = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设另一个根为 $x _ { 1 }$"}, {"id": "x _ { 1 }"}, {"id": "2.5"}, {"id": "2.5 x _ { 1 }"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 7 x + 2 m = 0$ 的一个根是另一个 $2.5$ 倍"}, {"id": "x ^ { 2 } - 7 x + 2 m = 0"}, {"id": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"id": "其中一个根为 $2.5 x _ { 1 }$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "x _ { 1 } = 2"}, {"id": "m = 4"}], "links": [{"rel": "假设描述", "source": "设另一个根为 $x _ { 1 }$", "target": "x _ { 1 }"}, {"rel": "假设描述", "source": "设另一个根为 $x _ { 1 }$", "target": "2.5 x _ { 1 }"}, {"rel": "假设描述", "source": "设另一个根为 $x _ { 1 }$", "target": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"rel": "假设描述", "source": "设另一个根为 $x _ { 1 }$", "target": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "2.5 x _ { 1 }"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"rel": "被描述", "source": "2.5", "target": "2.5 x _ { 1 }"}, {"rel": "被描述", "source": "2.5 x _ { 1 }", "target": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"rel": "被描述", "source": "2.5 x _ { 1 }", "target": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 7 x + 2 m = 0$ 的一个根是另一个 $2.5$ 倍", "target": "2.5 x _ { 1 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 7 x + 2 m = 0$ 的一个根是另一个 $2.5$ 倍", "target": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 7 x + 2 m = 0$ 的一个根是另一个 $2.5$ 倍", "target": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"rel": "被描述", "source": "x ^ { 2 } - 7 x + 2 m = 0", "target": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"rel": "被描述", "source": "x ^ { 2 } - 7 x + 2 m = 0", "target": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"rel": "等式方程求解", "source": "x _ { 1 } + 2.5 x _ { 1 } = 7", "target": "x _ { 1 } = 2"}, {"rel": "限制性描述", "source": "其中一个根为 $2.5 x _ { 1 }$", "target": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"rel": "限制性描述", "source": "其中一个根为 $2.5 x _ { 1 }$", "target": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + 2.5 x _ { 1 } = 7"}, {"rel": "联立", "source": "x _ { 1 } * 2 x _ { 1 } = 2 m", "target": "m = 4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "x _ { 1 } * 2 x _ { 1 } = 2 m"}, {"rel": "联立", "source": "x _ { 1 } = 2", "target": "m = 4"}]}}
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{"content": "Given a parabola $y = ( m - 1 ) ^ 2 + 4$ whose vertex is the highest point of the parabola, what is the range of possible values for $m$?", "answer": "m < 1", "steps": "$\\because$ The vertex of the parabola $y = ( m - 1 ) x ^ 2 + 4$ is the highest point of the parabola. $\\therefore$ The parabola opens downwards. $\\therefore$ $m - 1 < 0$. $\\therefore$ $m < 1$.", "expr_cands": ["y = ( m - 1 ) ^ { 2 } + 4", "m", "y", "y = ( m - 1 ) x ^ { 2 } + 4", "x", "m - 1 < 0", "m < 1"], "exprs": ["m - 1 < 0", "m < 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m - 1 ) x ^ { 2 } + 4"}, {"id": "m - 1 < 0"}, {"id": "抛物线 $y = ( m - 1 ) x ^ { 2 } + 4$ 的顶点是此抛物线的最高点"}, {"id": "m < 1"}], "links": [{"rel": "被描述", "source": "y = ( m - 1 ) x ^ { 2 } + 4", "target": "m - 1 < 0"}, {"rel": "不等式方程求解", "source": "m - 1 < 0", "target": "m < 1"}, {"rel": "限制性描述", "source": "抛物线 $y = ( m - 1 ) x ^ { 2 } + 4$ 的顶点是此抛物线的最高点", "target": "m - 1 < 0"}]}}
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{"content": "If $x - 2 y + 3 = 0$, then the value of the algebraic expression $1 - 2 x + 4 y$ is ____?", "answer": "7", "steps": "Since $x - 2 y + 3 = 0$, therefore $x - 2 y = - 3$. When $x - 2 y = - 3$, $1 - 2 x + 4 y = 1 - 2 ( x - 2 y ) = 1 - 2 * ( - 3 ) = 1 + 6 = 7$.", "expr_cands": ["x - 2 y + 3 = 0", "y", "x", "1 - 2 x + 4 y", "x - 2 y = - 3", "1 - 2 ( x - 2 y )", "7"], "exprs": ["x - 2 y = - 3", "1 - 2 ( x - 2 y )", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - 2 y + 3 = 0"}, {"id": "x - 2 y = - 3"}, {"id": "1 - 2 x + 4 y"}, {"id": "1 - 2 ( x - 2 y )"}, {"id": "7"}], "links": [{"rel": "移项", "source": "x - 2 y + 3 = 0", "target": "x - 2 y = - 3"}, {"rel": "提取因式参考", "source": "x - 2 y = - 3", "target": "1 - 2 ( x - 2 y )"}, {"rel": "代入", "source": "x - 2 y = - 3", "target": "7"}, {"rel": "提取因式", "source": "1 - 2 x + 4 y", "target": "1 - 2 ( x - 2 y )"}, {"rel": "被代入", "source": "1 - 2 ( x - 2 y )", "target": "7"}]}}
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{"content": "The condition for the fraction $\\frac { x ^ 2 + 1 } { 1 - 3 x }$ to be negative is ____?", "answer": "x > \\frac { 1 } { 3 }", "steps": "$\\because$ The value of the fraction $\\frac { x ^ 2 + 1 } { 1 - 3 x }$ is negative, $x ^ 2 + 1 > 0$, $\\therefore$ $1 - 3 x < 0$, which leads to $x > \\frac { 1 } { 3 }$.", "expr_cands": ["\\frac { x ^ { 2 } + 1 } { 1 - 3 x }", "x", "x ^ { 2 } + 1 > 0", "1 - 3 x < 0", "\\frac { 1 } { 3 } < x", "x > \\frac { 1 } { 3 }"], "exprs": ["x ^ { 2 } + 1 > 0", "1 - 3 x < 0", "x > \\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x ^ { 2 } + 1 } { 1 - 3 x }"}, {"id": "x ^ { 2 } + 1 > 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "1 - 3 x < 0"}, {"id": "使分式 $\\frac { x ^ { 2 } + 1 } { 1 - 3 x }$ 的值为负的条件"}, {"id": "分式 $\\frac { x ^ { 2 } + 1 } { 1 - 3 x }$ 的值为负"}, {"id": "分式有意义，则分母不为0"}, {"id": "分式为负数，则分子分母异号"}, {"id": "x > \\frac { 1 } { 3 }"}], "links": [{"rel": "被描述", "source": "\\frac { x ^ { 2 } + 1 } { 1 - 3 x }", "target": "x ^ { 2 } + 1 > 0"}, {"rel": "被描述", "source": "\\frac { x ^ { 2 } + 1 } { 1 - 3 x }", "target": "1 - 3 x < 0"}, {"rel": "被描述", "source": "x ^ { 2 } + 1 > 0", "target": "1 - 3 x < 0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "x ^ { 2 } + 1 > 0"}, {"rel": "不等式方程求解", "source": "1 - 3 x < 0", "target": "x > \\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "使分式 $\\frac { x ^ { 2 } + 1 } { 1 - 3 x }$ 的值为负的条件", "target": "1 - 3 x < 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x ^ { 2 } + 1 } { 1 - 3 x }$ 的值为负", "target": "1 - 3 x < 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "1 - 3 x < 0"}, {"rel": "属性描述", "source": "分式为负数，则分子分母异号", "target": "1 - 3 x < 0"}]}}
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{"content": "If $\\sqrt {( x - 3 )} ^ 2 = 3 - x$, then the possible values of $x$ are ____?", "answer": "x \\le 3", "steps": "Because the square root of $( x - 3 ) ^ 2$ is equal to $3 - x$, it follows that $3 - x$ must be greater than or equal to zero. Solving for $x$, we get $x \\leq 3$.", "expr_cands": ["\\sqrt { ( x - 3 ) } ^ { 2 } = 3 - x", "x", "x = 3", "3 - x \\ge 0", "x \\le 3"], "exprs": ["3 - x \\ge 0", "x \\le 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { ( x - 3 ) } ^ { 2 } = 3 - x"}, {"id": "3 - x \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\le 3"}], "links": [{"rel": "被描述", "source": "\\sqrt { ( x - 3 ) } ^ { 2 } = 3 - x", "target": "3 - x \\ge 0"}, {"rel": "不等式方程求解", "source": "3 - x \\ge 0", "target": "x \\le 3"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "3 - x \\ge 0"}]}}
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{"content": "If $( 5 x + 3 )$ and $( - 2 x + 9 )$ are opposite numbers, then $x$ = ____?", "answer": "- 4", "steps": "According to the problem, we have $( 5 x + 3 ) + ( - 2 x + 9 ) = 0$. Removing the parentheses, we get $5 x + 3 - 2 x + 9 = 0$. Combining like terms and moving them to one side, we get $3 x = - 12$. Solving for $x$, we get $x = - 4$.", "expr_cands": ["( 5 x + 3 )", "x", "( - 2 x + 9 )", "( 5 x + 3 ) + ( - 2 x + 9 ) = 0", "x = - 4", "5 x + 3 - 2 x + 9 = 0", "3 x = - 12"], "exprs": ["( 5 x + 3 ) + ( - 2 x + 9 ) = 0", "x = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 5 x + 3 )"}, {"id": "( 5 x + 3 ) + ( - 2 x + 9 ) = 0"}, {"id": "( - 2 x + 9 )"}, {"id": "$( 5 x + 3 )$ 与 $( - 2 x + 9 )$ 互为相反数"}, {"id": "x = - 4"}], "links": [{"rel": "被描述", "source": "( 5 x + 3 )", "target": "( 5 x + 3 ) + ( - 2 x + 9 ) = 0"}, {"rel": "等式方程求解", "source": "( 5 x + 3 ) + ( - 2 x + 9 ) = 0", "target": "x = - 4"}, {"rel": "被描述", "source": "( - 2 x + 9 )", "target": "( 5 x + 3 ) + ( - 2 x + 9 ) = 0"}, {"rel": "限制性描述", "source": "$( 5 x + 3 )$ 与 $( - 2 x + 9 )$ 互为相反数", "target": "( 5 x + 3 ) + ( - 2 x + 9 ) = 0"}]}}
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{"content": "Given that $x = 3$ is a solution to the equation $4 x + 3 a = 6$ in terms of $x$, the value of $a$ is ____?", "answer": "- 2", "steps": "Substituting $x = 3$ into the equation gives: $12 + 3 a = 6$, solving for $a$ gives: $a = - 2$.", "expr_cands": ["x = 3", "x", "4 x + 3 a = 6", "a", "12 + 3 a = 6", "a = - 2"], "exprs": ["12 + 3 a = 6", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 3"}, {"id": "12 + 3 a = 6"}, {"id": "4 x + 3 a = 6"}, {"id": "a = - 2"}], "links": [{"rel": "代入", "source": "x = 3", "target": "12 + 3 a = 6"}, {"rel": "等式方程求解", "source": "12 + 3 a = 6", "target": "a = - 2"}, {"rel": "被代入", "source": "4 x + 3 a = 6", "target": "12 + 3 a = 6"}]}}
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{"content": "If $\\sqrt {( x - 3 )} ^ 2 = 3 - x$, then the possible values of $x$ are ____?", "answer": "7", "steps": "Because the square root of $( x - 3 ) ^ 2$ is equal to $3 - x$, it follows that $3 - x$ must be greater than or equal to zero. Solving for $x$, we get $x$ is less than or equal to 3.", "expr_cands": ["\\frac { a } { x - 4 } + 1 = \\frac { 1 - 2 x } { 4 - x }", "a", "x", "a + x - 4 = 2 x - 1", "x - 4 = 0", "x = 4", "a + 4 - 4 = 2 * 4 - 1", "a = 7"], "exprs": ["a + x - 4 = 2 x - 1", "x - 4 = 0", "x = 4", "a + 4 - 4 = 2 * 4 - 1", "a = 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a } { x - 4 } + 1 = \\frac { 1 - 2 x } { 4 - x }"}, {"id": "a + x - 4 = 2 x - 1"}, {"id": "x - 4 = 0"}, {"id": "用去分母法解方程 $\\frac { a } { x - 4 } + 1 = \\frac { 1 - 2 x } { 4 - x }$ 时会产生增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = 4"}, {"id": "a + 4 - 4 = 2 * 4 - 1"}, {"id": "a = 7"}], "links": [{"rel": "同乘除", "source": "\\frac { a } { x - 4 } + 1 = \\frac { 1 - 2 x } { 4 - x }", "target": "a + x - 4 = 2 x - 1"}, {"rel": "被描述", "source": "\\frac { a } { x - 4 } + 1 = \\frac { 1 - 2 x } { 4 - x }", "target": "x - 4 = 0"}, {"rel": "被代入", "source": "a + x - 4 = 2 x - 1", "target": "a + 4 - 4 = 2 * 4 - 1"}, {"rel": "等式方程求解", "source": "x - 4 = 0", "target": "x = 4"}, {"rel": "限制性描述", "source": "用去分母法解方程 $\\frac { a } { x - 4 } + 1 = \\frac { 1 - 2 x } { 4 - x }$ 时会产生增根", "target": "x - 4 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x - 4 = 0"}, {"rel": "代入", "source": "x = 4", "target": "a + 4 - 4 = 2 * 4 - 1"}, {"rel": "等式方程求解", "source": "a + 4 - 4 = 2 * 4 - 1", "target": "a = 7"}]}}
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{"content": "If $( 5 x + 3 )$ and $( - 2 x + 9 )$ are opposite numbers, then $x$ = ____?", "answer": "2", "steps": "According to the problem, we have $( 5 x + 3 ) + ( - 2 x + 9 ) = 0$. Removing the parentheses, we get $5 x + 3 - 2 x + 9 = 0$. Combining like terms and moving them to one side, we get $3 x = - 12$. Solving for $x$, we get $x = - 4$.", "expr_cands": ["y = ( a + 1 ) x ^ { a ^ { 2 } - a - 3 }", "a", "y", "x", "a + 1 \\neq 0", "a \\neq - 1", "{ a } ^ { 2 } - a - 3 = - 1", "a = - 1", "a = 2"], "exprs": ["a + 1 \\neq 0", "{ a } ^ { 2 } - a - 3 = - 1", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( a + 1 ) x ^ { a ^ { 2 } - a - 3 }"}, {"id": "a + 1 \\neq 0"}, {"id": "$y = ( a + 1 ) x ^ { a ^ { 2 } - a - 3 }$ 是反比例函数"}, {"id": "{ a } ^ { 2 } - a - 3 = - 1"}, {"id": "a = 2"}], "links": [{"rel": "被描述", "source": "y = ( a + 1 ) x ^ { a ^ { 2 } - a - 3 }", "target": "a + 1 \\neq 0"}, {"rel": "被描述", "source": "y = ( a + 1 ) x ^ { a ^ { 2 } - a - 3 }", "target": "{ a } ^ { 2 } - a - 3 = - 1"}, {"rel": "联立", "source": "a + 1 \\neq 0", "target": "a = 2"}, {"rel": "限制性描述", "source": "$y = ( a + 1 ) x ^ { a ^ { 2 } - a - 3 }$ 是反比例函数", "target": "a + 1 \\neq 0"}, {"rel": "限制性描述", "source": "$y = ( a + 1 ) x ^ { a ^ { 2 } - a - 3 }$ 是反比例函数", "target": "{ a } ^ { 2 } - a - 3 = - 1"}, {"rel": "联立", "source": "{ a } ^ { 2 } - a - 3 = - 1", "target": "a = 2"}]}}
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{"content": "Given that $x = 3$ is a solution to the equation $4 x + 3 a = 6$ in terms of $x$, the value of $a$ is ____?", "answer": "\\frac { 3 } { 10 }", "steps": "Substituting $x = 3$ into the equation gives: $12 + 3 a = 6$, solving for $a$ gives: $a = - 2$.", "expr_cands": ["m = 3", "m", "\\frac { { m } ^ { 2 } - 7 m } { { m } ^ { 2 } - 49 }", "\\frac { m ( m - 7 ) } { ( m + 7 ) ( m - 7 ) }", "\\frac { m } { m + 7 }", "\\frac { 3 } { 3 + 7 }", "\\frac { 3 } { 10 }"], "exprs": ["\\frac { 3 } { 10 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m = 3"}, {"id": "\\frac { 3 } { 10 }"}, {"id": "\\frac { { m } ^ { 2 } - 7 m } { { m } ^ { 2 } - 49 }"}], "links": [{"rel": "代入", "source": "m = 3", "target": "\\frac { 3 } { 10 }"}, {"rel": "被代入", "source": "\\frac { { m } ^ { 2 } - 7 m } { { m } ^ { 2 } - 49 }", "target": "\\frac { 3 } { 10 }"}]}}
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{"content": "If the equation $\\frac { a } { x - 4 } + 1 = \\frac { 1 - 2 x } { 4 - x }$ is solved using the method of eliminating denominators and extraneous roots are produced, then $a$ = ____ ?", "answer": "3", "steps": "To eliminate the denominator, we get $a + x - 4 = 2 x - 1$. Since the fractional equation has an extraneous root, we obtain $x - 4 = 0$, which means $x = 4$. Substituting $x = 4$ into the polynomial equation, we get $a + 4 - 4 = 2 * 4 - 1$, and solve for $a$, which is $a = 7$.", "expr_cands": ["x", "x ^ { 2 } - 3 x + 2 = 0", "x _ { 1 }", "x _ { 2 }", "x _ { 1 } + x _ { 2 }", "x = 1", "x = 2", "x _ { 1 } + x _ { 2 } = 3", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 x + 2 = 0"}, {"id": "3"}, {"id": "x _ { 1 }"}, {"id": "x _ { 2 }"}, {"id": "x _ { 1 } + x _ { 2 }"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 3 x + 2 = 0$ 两实数根为 $x _ { 1 }$ , $x _ { 2 }$"}, {"id": "一元二次方程根与系数关系，两根之和"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 3 x + 2 = 0", "target": "3"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "3"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "3"}, {"rel": "被描述", "source": "x _ { 1 } + x _ { 2 }", "target": "3"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 3 x + 2 = 0$ 两实数根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "3"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "3"}]}}
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{"content": "If $y = ( a + 1 ) x ^ { a ^ 2 - a - 3 }$ is an inverse proportion function, then the value of $a$ is ____?", "answer": "2", "steps": "From the given condition, we can deduce that $a + 1 \\neq 0$ and ${ a } ^ { 2 } - a - 3 = - 1$. Solving for $a$, we get $a = 2$.", "expr_cands": ["2 x + 7 > 3 x + 4", "x", "x < 3", "2 x - 3 x > 4 - 7", "- x > - 3", "1", "2", "1 * 2"], "exprs": ["x < 3", "1", "2", "1 * 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 7 > 3 x + 4"}, {"id": "x < 3"}, {"id": "1"}, {"id": "不等式 $2 x + 7 > 3 x + 4$ 的所有正整数解为 $1$ , $2$"}, {"id": "2"}, {"id": "1 * 2"}, {"id": "不等式 $2 x + 7 > 3 x + 4$ 的所有正整数解的乘积"}], "links": [{"rel": "不等式方程求解", "source": "2 x + 7 > 3 x + 4", "target": "x < 3"}, {"rel": "被描述", "source": "x < 3", "target": "1"}, {"rel": "被描述", "source": "x < 3", "target": "2"}, {"rel": "被描述", "source": "1", "target": "1 * 2"}, {"rel": "限制性描述", "source": "不等式 $2 x + 7 > 3 x + 4$ 的所有正整数解为 $1$ , $2$", "target": "1"}, {"rel": "限制性描述", "source": "不等式 $2 x + 7 > 3 x + 4$ 的所有正整数解为 $1$ , $2$", "target": "2"}, {"rel": "被描述", "source": "2", "target": "1 * 2"}, {"rel": "限制性描述", "source": "不等式 $2 x + 7 > 3 x + 4$ 的所有正整数解的乘积", "target": "1 * 2"}]}}
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{"content": "If $m = 3$, then the value of $\\frac {{ m } ^ { 2 } - 7 m } {{ m } ^ { 2 } - 49 }$ is ____?", "answer": "2", "steps": "Original expression = $\\frac { m ( m - 7 ) } { ( m + 7 ) ( m - 7 ) } = \\frac { m } { m + 7 }$. Substituting $m = 3$, we get the original expression = $\\frac { 3 } { 3 + 7 } = \\frac { 3 } { 10 }$.", "expr_cands": ["3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1", "x", "m", "y", "| m | + 2 = 4", "m = - 2", "m = 2", "m + 2 \\neq 0", "m \\neq - 2"], "exprs": ["| m | + 2 = 4", "m + 2 \\neq 0", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1"}, {"id": "| m | + 2 = 4"}, {"id": "多项式 $3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1$ 是一个四次三项式"}, {"id": "m + 2 \\neq 0"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1", "target": "| m | + 2 = 4"}, {"rel": "被描述", "source": "3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1", "target": "m + 2 \\neq 0"}, {"rel": "联立", "source": "| m | + 2 = 4", "target": "m = 2"}, {"rel": "限制性描述", "source": "多项式 $3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1$ 是一个四次三项式", "target": "| m | + 2 = 4"}, {"rel": "限制性描述", "source": "多项式 $3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1$ 是一个四次三项式", "target": "m + 2 \\neq 0"}, {"rel": "联立", "source": "m + 2 \\neq 0", "target": "m = 2"}]}}
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{"content": "Given a quadratic equation in one variable $x$, $x ^ 2 - 3 x + 2 = 0$, with two real roots $x _ 1$ and $x _ 2$, what is the value of $x _ 1 + x _ 2$?", "answer": "y = 2 { ( x - 3 ) } ^ { 2 }", "steps": "$\\because$ The quadratic equation in one variable $x$, $x ^ 2 - 3 x + 2 = 0$, has two real roots $x _ 1$ and $x _ 2$. $\\therefore$ $x _ 1 + x _ 2 = - ( - 3 ) = 3$.", "expr_cands": ["y = 2 { x } ^ { 2 }", "y", "x", "3", "y = 2 { ( x - 3 ) } ^ { 2 }", "2 x ^ { 2 } = 2 { ( x - 3 ) } ^ { 2 }", "2 x ^ { 2 }"], "exprs": ["y = 2 { ( x - 3 ) } ^ { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "y = 2 { ( x - 3 ) } ^ { 2 }"}, {"id": "y = 2 { x } ^ { 2 }"}, {"id": "将抛物线 $y = 2 { x } ^ { 2 }$ 向右平移 $3$ 个单位"}], "links": [{"rel": "被描述", "source": "3", "target": "y = 2 { ( x - 3 ) } ^ { 2 }"}, {"rel": "被描述", "source": "y = 2 { x } ^ { 2 }", "target": "y = 2 { ( x - 3 ) } ^ { 2 }"}, {"rel": "限制性描述", "source": "将抛物线 $y = 2 { x } ^ { 2 }$ 向右平移 $3$ 个单位", "target": "y = 2 { ( x - 3 ) } ^ { 2 }"}]}}
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{"content": "The product of all positive integer solutions to the inequality $2 x + 7 > 3 x + 4$ is _____.", "answer": "9", "steps": "$2 x + 7 > 3 x + 4$ is rearranged to $2 x - 3 x > 4 - 7$. Combining like terms gives $- x > - 3$. Dividing both sides by $- 1$ gives $x < 3$. Therefore, the positive integer solutions to the inequality $2 x + 7 > 3 x + 4$ are $1$ and $2$, and their product is $1 * 2 = 2$.", "expr_cands": ["x", "\\frac { 2 kx + a } { 3 } = 1 - \\frac { x - bk } { 6 }", "b", "a", "k", "x = 1", "2 a + b", "\\frac { 2 k + a } { 3 } = 1 - \\frac { 1 - bk } { 6 }", "( 4 - b ) k = 5 - 2 a", "4 - b = 0", "b = 4", "5 - 2 a = 0", "a = \\frac { 5 } { 2 }", "9"], "exprs": ["( 4 - b ) k = 5 - 2 a", "4 - b = 0", "5 - 2 a = 0", "b = 4", "a = \\frac { 5 } { 2 }", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 kx + a } { 3 } = 1 - \\frac { x - bk } { 6 }"}, {"id": "( 4 - b ) k = 5 - 2 a"}, {"id": "x = 1"}, {"id": "4 - b = 0"}, {"id": "它的解总是 $x = 1$"}, {"id": "无论 $k$ 为何值"}, {"id": "b = 4"}, {"id": "5 - 2 a = 0"}, {"id": "a = \\frac { 5 } { 2 }"}, {"id": "2 a + b"}, {"id": "9"}], "links": [{"rel": "联立", "source": "\\frac { 2 kx + a } { 3 } = 1 - \\frac { x - bk } { 6 }", "target": "( 4 - b ) k = 5 - 2 a"}, {"rel": "被描述", "source": "( 4 - b ) k = 5 - 2 a", "target": "4 - b = 0"}, {"rel": "被描述", "source": "( 4 - b ) k = 5 - 2 a", "target": "5 - 2 a = 0"}, {"rel": "联立", "source": "x = 1", "target": "( 4 - b ) k = 5 - 2 a"}, {"rel": "等式方程求解", "source": "4 - b = 0", "target": "b = 4"}, {"rel": "限制性描述", "source": "它的解总是 $x = 1$", "target": "4 - b = 0"}, {"rel": "限制性描述", "source": "它的解总是 $x = 1$", "target": "5 - 2 a = 0"}, {"rel": "限制性描述", "source": "无论 $k$ 为何值", "target": "4 - b = 0"}, {"rel": "限制性描述", "source": "无论 $k$ 为何值", "target": "5 - 2 a = 0"}, {"rel": "代入", "source": "b = 4", "target": "9"}, {"rel": "等式方程求解", "source": "5 - 2 a = 0", "target": "a = \\frac { 5 } { 2 }"}, {"rel": "代入", "source": "a = \\frac { 5 } { 2 }", "target": "9"}, {"rel": "被代入", "source": "2 a + b", "target": "9"}]}}
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{"content": "The polynomial $3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1$ is a quartic trinomial. What is the value of $m$?", "answer": "- 2", "steps": "$\\because$ The polynomial $3 x ^ { | m | } y ^ { 2 } - ( m + 2 ) x + 1$ is a quartic trinomial, $\\therefore$ $| m | + 2 = 4$, $m + 2 \\neq 0$, solving for $m$, we get: $m = 2$.", "expr_cands": ["a", "b", "c", "6 a - 2 b + 4 c", "a = - 1", "b = 0", "c = 1", "- 2"], "exprs": ["a = - 1", "b = 0", "c = 1", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a = - 1"}, {"id": "$a$ 是最大的负整数"}, {"id": "b"}, {"id": "b = 0"}, {"id": "$b$ 是最小的正整数"}, {"id": "c"}, {"id": "c = 1"}, {"id": "$c$ 为绝对值最小的数"}, {"id": "6 a - 2 b + 4 c"}, {"id": "- 2"}], "links": [{"rel": "被描述", "source": "a", "target": "a = - 1"}, {"rel": "代入", "source": "a = - 1", "target": "- 2"}, {"rel": "限制性描述", "source": "$a$ 是最大的负整数", "target": "a = - 1"}, {"rel": "被描述", "source": "b", "target": "b = 0"}, {"rel": "代入", "source": "b = 0", "target": "- 2"}, {"rel": "限制性描述", "source": "$b$ 是最小的正整数", "target": "b = 0"}, {"rel": "被描述", "source": "c", "target": "c = 1"}, {"rel": "代入", "source": "c = 1", "target": "- 2"}, {"rel": "限制性描述", "source": "$c$ 为绝对值最小的数", "target": "c = 1"}, {"rel": "被代入", "source": "6 a - 2 b + 4 c", "target": "- 2"}]}}
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{"content": "What is the equation of the parabola obtained by shifting the graph of$y = 2 { x } ^ { 2 }$ three units to the right?", "answer": "6", "steps": "The equation of the parabola $y = 2 { x } ^ { 2 }$ after shifting it $3$ units to the right is: $y = 2 { ( x - 3 ) } ^ { 2 }$.", "expr_cands": ["( ax - b ) ( 3 x + 4 ) = bx ^ { 2 } + cx + 72", "x", "b", "c", "a", "a + b + c", "3 a = b", "4 a - 3 b = c", "- 4 b = 72", "b = - 18", "a = - 6", "c = 30", "6"], "exprs": ["3 a = b", "4 a - 3 b = c", "- 4 b = 72", "b = - 18", "a = - 6", "c = 30", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( ax - b ) ( 3 x + 4 ) = bx ^ { 2 } + cx + 72"}, {"id": "3 a = b"}, {"id": "4 a - 3 b = c"}, {"id": "- 4 b = 72"}, {"id": "b = - 18"}, {"id": "a = - 6"}, {"id": "c = 30"}, {"id": "a + b + c"}, {"id": "6"}], "links": [{"rel": "移项", "source": "( ax - b ) ( 3 x + 4 ) = bx ^ { 2 } + cx + 72", "target": "3 a = b"}, {"rel": "移项", "source": "( ax - b ) ( 3 x + 4 ) = bx ^ { 2 } + cx + 72", "target": "4 a - 3 b = c"}, {"rel": "移项", "source": "( ax - b ) ( 3 x + 4 ) = bx ^ { 2 } + cx + 72", "target": "- 4 b = 72"}, {"rel": "联立", "source": "3 a = b", "target": "a = - 6"}, {"rel": "被代入", "source": "4 a - 3 b = c", "target": "c = 30"}, {"rel": "等式方程求解", "source": "- 4 b = 72", "target": "b = - 18"}, {"rel": "联立", "source": "b = - 18", "target": "a = - 6"}, {"rel": "代入", "source": "b = - 18", "target": "c = 30"}, {"rel": "代入", "source": "b = - 18", "target": "6"}, {"rel": "代入", "source": "a = - 6", "target": "c = 30"}, {"rel": "代入", "source": "a = - 6", "target": "6"}, {"rel": "代入", "source": "c = 30", "target": "6"}, {"rel": "被代入", "source": "a + b + c", "target": "6"}]}}
|
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{"content": "If the equation $\\frac { 2 kx + a } { 3 } = 1 - \\frac { x - bk } { 6 }$ about $x$ has a solution of $x = 1$ regardless of the value of $k$, then the algebraic expression $2 a + b$ is _____.", "answer": "m < 3", "steps": "Substituting $x = 1$ into the equation gives $\\frac { 2 k + a } { 3 } = 1 - \\frac { 1 - bk } { 6 }$, simplifying gives $( 4 - b ) k = 5 - 2 a$. Since the equation holds for any value of $k$, we have $4 - b = 0$ and $5 - 2 a = 0$. Solving gives $b = 4$ and $a = \\frac { 5 } { 2 }$. Therefore, $2 a + b = 2 * \\frac { 5 } { 2 } + 4 = 9$.", "expr_cands": ["y = ( m - 3 ) x + 5", "x", "m", "y", "m - 3 < 0", "m < 3"], "exprs": ["m - 3 < 0", "m < 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m - 3 ) x + 5"}, {"id": "m - 3 < 0"}, {"id": "一次函数 $y = ( m - 3 ) x + 5$ 的函数值 $y$ 随着 $x$ 的增大而减小"}, {"id": "$m$ 的取值范围"}, {"id": "m < 3"}], "links": [{"rel": "被描述", "source": "y = ( m - 3 ) x + 5", "target": "m - 3 < 0"}, {"rel": "不等式方程求解", "source": "m - 3 < 0", "target": "m < 3"}, {"rel": "限制性描述", "source": "一次函数 $y = ( m - 3 ) x + 5$ 的函数值 $y$ 随着 $x$ 的增大而减小", "target": "m - 3 < 0"}, {"rel": "限制性描述", "source": "$m$ 的取值范围", "target": "m - 3 < 0"}]}}
|
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{"content": "$a$ is the largest negative integer, $b$ is the smallest positive integer, and $c$ is the number with the smallest absolute value. What is the value of $6 a - 2 b + 4 c$?", "answer": "17", "steps": "From the given information, we know that $a = - 1$, $b = 0$, and $c = 1$. Therefore, $6 a - 2 b + 4 c = - 6 - 0 + 4 = - 2$.", "expr_cands": ["a + b = 5", "a", "b", "b - c = 12", "c", "a + 2 b - c", "a + b + ( b - c )", "17"], "exprs": ["a + b + ( b - c )", "17"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a + 2 b - c"}, {"id": "a + b + ( b - c )"}, {"id": "a + b = 5"}, {"id": "b - c = 12"}, {"id": "17"}], "links": [{"rel": "提取因式", "source": "a + 2 b - c", "target": "a + b + ( b - c )"}, {"rel": "被代入", "source": "a + b + ( b - c )", "target": "17"}, {"rel": "提取因式参考", "source": "a + b = 5", "target": "a + b + ( b - c )"}, {"rel": "代入", "source": "a + b = 5", "target": "17"}, {"rel": "提取因式参考", "source": "b - c = 12", "target": "a + b + ( b - c )"}, {"rel": "代入", "source": "b - c = 12", "target": "17"}]}}
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{"content": "If $( ax - b ) ( 3 x + 4 ) = bx ^ 2 + cx + 72$, then the value of $a + b + c$ is ____?", "answer": "0", "steps": "Because $( ax - b ) ( 3 x + 4 ) = 3 ax ^ 2 + ( 4 a - 3 b ) x - 4 b = bx ^ 2 + cx + 72$, therefore $3 a = b$, $4 a - 3 b = c$, $- 4 b = 72$. Solving for $a = - 6$, $b = - 18$, $c = 30$, then $a + b + c = - 6 - 18 + 30 = 6$.", "expr_cands": ["x = 1", "x", "\\sqrt { x + k } = x", "k", "\\sqrt { 1 + k } = 1", "k = 0", "1 + k = 1"], "exprs": ["\\sqrt { 1 + k } = 1", "k = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 1"}, {"id": "\\sqrt { 1 + k } = 1"}, {"id": "\\sqrt { x + k } = x"}, {"id": "k = 0"}], "links": [{"rel": "代入", "source": "x = 1", "target": "\\sqrt { 1 + k } = 1"}, {"rel": "等式方程求解", "source": "\\sqrt { 1 + k } = 1", "target": "k = 0"}, {"rel": "被代入", "source": "\\sqrt { x + k } = x", "target": "\\sqrt { 1 + k } = 1"}]}}
|
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{"content": "The range of $m$ for the linear function $y = ( m - 3 ) x + 5$ where the value of $y$ decreases as $x$ increases is _____.", "answer": "- 2", "steps": "According to the problem, we have $m - 3 < 0$, which implies $m < 3$.", "expr_cands": ["y = ( k - 2 ) { x } ^ { { k } ^ { 2 } - 5 }", "x", "k", "y", "k ^ { 2 } - 5 = - 1", "k = - 2", "k = 2", "k - 2 \\neq 0", "k \\neq 2", "m = - 2", "m"], "exprs": ["k ^ { 2 } - 5 = - 1", "k - 2 \\neq 0", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( k - 2 ) { x } ^ { { k } ^ { 2 } - 5 }"}, {"id": "k ^ { 2 } - 5 = - 1"}, {"id": "函数 $y = ( k - 2 ) { x } ^ { { k } ^ { 2 } - 5 }$ 是反比例函数"}, {"id": "k - 2 \\neq 0"}, {"id": "m = - 2"}], "links": [{"rel": "被描述", "source": "y = ( k - 2 ) { x } ^ { { k } ^ { 2 } - 5 }", "target": "k ^ { 2 } - 5 = - 1"}, {"rel": "被描述", "source": "y = ( k - 2 ) { x } ^ { { k } ^ { 2 } - 5 }", "target": "k - 2 \\neq 0"}, {"rel": "联立", "source": "k ^ { 2 } - 5 = - 1", "target": "m = - 2"}, {"rel": "限制性描述", "source": "函数 $y = ( k - 2 ) { x } ^ { { k } ^ { 2 } - 5 }$ 是反比例函数", "target": "k ^ { 2 } - 5 = - 1"}, {"rel": "限制性描述", "source": "函数 $y = ( k - 2 ) { x } ^ { { k } ^ { 2 } - 5 }$ 是反比例函数", "target": "k - 2 \\neq 0"}, {"rel": "联立", "source": "k - 2 \\neq 0", "target": "m = - 2"}]}}
|
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{"content": "Given $a + b = 5$, $b - c = 12$, what is the value of $a + 2 b - c$?", "answer": "- \\sqrt { 3 }", "steps": "Since $a + b = 5$ and $b - c = 12$, therefore $a + 2 b - c = a + b + ( b - c ) = 5 + 12 = 17$.", "expr_cands": ["x + 2 \\sqrt { 3 }", "x", "\\sqrt { 3 }", "x + 2 \\sqrt { 3 } = \\sqrt { 3 }", "x = - \\sqrt { 3 }"], "exprs": ["x + 2 \\sqrt { 3 } = \\sqrt { 3 }", "x = - \\sqrt { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 2 \\sqrt { 3 }"}, {"id": "x + 2 \\sqrt { 3 } = \\sqrt { 3 }"}, {"id": "\\sqrt { 3 }"}, {"id": "代数式 $x + 2 \\sqrt { 3 }$ 的值为 $\\sqrt { 3 }$"}, {"id": "x = - \\sqrt { 3 }"}], "links": [{"rel": "被描述", "source": "x + 2 \\sqrt { 3 }", "target": "x + 2 \\sqrt { 3 } = \\sqrt { 3 }"}, {"rel": "等式方程求解", "source": "x + 2 \\sqrt { 3 } = \\sqrt { 3 }", "target": "x = - \\sqrt { 3 }"}, {"rel": "被描述", "source": "\\sqrt { 3 }", "target": "x + 2 \\sqrt { 3 } = \\sqrt { 3 }"}, {"rel": "限制性描述", "source": "代数式 $x + 2 \\sqrt { 3 }$ 的值为 $\\sqrt { 3 }$", "target": "x + 2 \\sqrt { 3 } = \\sqrt { 3 }"}]}}
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{"content": "If $x = 1$ is a real root of the equation $\\sqrt { x + k } = x$ with respect to $x$, then $k$ = ____?", "answer": "x = 0", "steps": "Substituting $x = 1$ into the equation, we get $\\sqrt { 1 + k } = 1$. Squaring both sides, we get $1 + k = 1$, which gives us $k = 0$ upon solving. Upon checking, we see that $k = 0$ satisfies the given equation.", "expr_cands": ["( x + 2 ) ^ { 2 } - ( x - 2 ) ( x + 2 ) = 8", "x", "{ x } ^ { 2 } + 4 x + 4 - { x } ^ { 2 } + 4 = 8", "x = 0", "4 x = 0"], "exprs": ["x = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + 2 ) ^ { 2 } - ( x - 2 ) ( x + 2 ) = 8"}, {"id": "x = 0"}], "links": [{"rel": "等式方程求解", "source": "( x + 2 ) ^ { 2 } - ( x - 2 ) ( x + 2 ) = 8", "target": "x = 0"}]}}
|
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{"content": "If the function $y = ( k - 2 ) x ^ { k ^ 2 - 5 }$ is an inverse proportion, then $k$ = ____?", "answer": "m \\neq - 1", "steps": "Since the analytical expression of the inverse proportion function is $y = ( k - 2 ) x ^ { k ^ 2 - 5 }$, therefore $k ^ 2 - 5 = - 1$, and $k - 2 \\neq 0$. Solving for $m$, we get $m = - 2$.", "expr_cands": ["x", "mx - m = - x - 1", "m", "m + 1 \\neq 0", "m \\neq - 1"], "exprs": ["m + 1 \\neq 0", "m \\neq - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "mx - m = - x - 1"}, {"id": "m + 1 \\neq 0"}, {"id": "关于 $x$ 的方程 $mx - m = - x - 1$ 有解"}, {"id": "m \\neq - 1"}], "links": [{"rel": "被描述", "source": "mx - m = - x - 1", "target": "m + 1 \\neq 0"}, {"rel": "不等式方程求解", "source": "m + 1 \\neq 0", "target": "m \\neq - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $mx - m = - x - 1$ 有解", "target": "m + 1 \\neq 0"}]}}
|
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{"content": "If the value of the algebraic expression $x + 2 \\sqrt { 3 }$ is $\\sqrt { 3 }$, then $x$ is ____ ?", "answer": "- 2", "steps": "$\\because$ The value of the algebraic expression $x + 2 \\sqrt { 3 }$ is $\\sqrt { 3 }$ , $\\therefore$ $x + 2 \\sqrt { 3 } = \\sqrt { 3 }$ , $\\therefore$ $x = - \\sqrt { 3 }$.", "expr_cands": ["x \\ge 3", "x", "a", "x \\le - 5", "b", "a + b", "x = 3", "a = 3", "x = - 5", "b = - 5", "- 2"], "exprs": ["a = 3", "b = - 5", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a = 3"}, {"id": "x \\ge 3"}, {"id": "满足 $x \\ge 3$ 的 $x$ 的最小值是 $a$"}, {"id": "最小值为 $x = 3$"}, {"id": "即 $a = 3$"}, {"id": "b"}, {"id": "b = - 5"}, {"id": "x \\le - 5"}, {"id": "满足 $x \\le - 5$ 的 $x$ 的最大值是 $b$"}, {"id": "最大值为 $x = - 5$"}, {"id": "即 $b = - 5$"}, {"id": "a + b"}, {"id": "- 2"}], "links": [{"rel": "被描述", "source": "a", "target": "a = 3"}, {"rel": "代入", "source": "a = 3", "target": "- 2"}, {"rel": "被描述", "source": "x \\ge 3", "target": "a = 3"}, {"rel": "限制性描述", "source": "满足 $x \\ge 3$ 的 $x$ 的最小值是 $a$", "target": "a = 3"}, {"rel": "限制性描述", "source": "最小值为 $x = 3$", "target": "a = 3"}, {"rel": "限制性描述", "source": "即 $a = 3$", "target": "a = 3"}, {"rel": "被描述", "source": "b", "target": "b = - 5"}, {"rel": "代入", "source": "b = - 5", "target": "- 2"}, {"rel": "被描述", "source": "x \\le - 5", "target": "b = - 5"}, {"rel": "限制性描述", "source": "满足 $x \\le - 5$ 的 $x$ 的最大值是 $b$", "target": "b = - 5"}, {"rel": "限制性描述", "source": "最大值为 $x = - 5$", "target": "b = - 5"}, {"rel": "限制性描述", "source": "即 $b = - 5$", "target": "b = - 5"}, {"rel": "被代入", "source": "a + b", "target": "- 2"}]}}
|
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{"content": "The solution to the equation $( x + 2 ) ^ 2 - ( x - 2 ) ( x + 2 ) = 8$ is ____ ?", "answer": "1", "steps": "${ x } ^ { 2 } + 4 x + 4 - { x } ^ { 2 } + 4 = 8$ is an equation that needs to be solved for x. Simplifying the equation, we get 4x + 8 = 8. Solving for x, we get 4x = 0 which means x = 0.", "expr_cands": ["2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0", "x", "m", "y", "n", "n ^ { m }", "n - 3 = 1", "n = 4", "2 m + 1 = 1", "m = 0", "1"], "exprs": ["n - 3 = 1", "2 m + 1 = 1", "n = 4", "m = 0", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0"}, {"id": "n - 3 = 1"}, {"id": "$2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0$ 是关于 $x$ , $y$ 的二元一次方程"}, {"id": "2 m + 1 = 1"}, {"id": "n = 4"}, {"id": "m = 0"}, {"id": "n ^ { m }"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0", "target": "n - 3 = 1"}, {"rel": "被描述", "source": "2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0", "target": "2 m + 1 = 1"}, {"rel": "等式方程求解", "source": "n - 3 = 1", "target": "n = 4"}, {"rel": "限制性描述", "source": "$2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0$ 是关于 $x$ , $y$ 的二元一次方程", "target": "n - 3 = 1"}, {"rel": "限制性描述", "source": "$2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0$ 是关于 $x$ , $y$ 的二元一次方程", "target": "2 m + 1 = 1"}, {"rel": "等式方程求解", "source": "2 m + 1 = 1", "target": "m = 0"}, {"rel": "代入", "source": "n = 4", "target": "1"}, {"rel": "代入", "source": "m = 0", "target": "1"}, {"rel": "被代入", "source": "n ^ { m }", "target": "1"}]}}
|
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{"content": "The equation $mx - m = - x - 1$ has a solution for $x$, then the value of $m$ is ____?", "answer": "- 2", "steps": "$mx - m = - x - 1$ has a solution, so we have $m + 1 \\neq 0$. Solving for $m$, we get $m \\neq - 1$.", "expr_cands": ["( k - 2 ) x ^ { 3 } + ( | k | - 2 ) x ^ { 2 } - 5", "k", "x", "| k | - 2 = 0", "k = - 2", "k = 2", "k - 2 \\neq 0", "k \\neq 2"], "exprs": ["| k | - 2 = 0", "k - 2 \\neq 0", "k = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( k - 2 ) x ^ { 3 } + ( | k | - 2 ) x ^ { 2 } - 5"}, {"id": "| k | - 2 = 0"}, {"id": "$( k - 2 ) x ^ { 3 } + ( | k | - 2 ) x ^ { 2 } - 5$ 是关于 $x$ 的三次二项式"}, {"id": "k - 2 \\neq 0"}, {"id": "k = - 2"}], "links": [{"rel": "被描述", "source": "( k - 2 ) x ^ { 3 } + ( | k | - 2 ) x ^ { 2 } - 5", "target": "| k | - 2 = 0"}, {"rel": "被描述", "source": "( k - 2 ) x ^ { 3 } + ( | k | - 2 ) x ^ { 2 } - 5", "target": "k - 2 \\neq 0"}, {"rel": "联立", "source": "| k | - 2 = 0", "target": "k = - 2"}, {"rel": "限制性描述", "source": "$( k - 2 ) x ^ { 3 } + ( | k | - 2 ) x ^ { 2 } - 5$ 是关于 $x$ 的三次二项式", "target": "| k | - 2 = 0"}, {"rel": "限制性描述", "source": "$( k - 2 ) x ^ { 3 } + ( | k | - 2 ) x ^ { 2 } - 5$ 是关于 $x$ 的三次二项式", "target": "k - 2 \\neq 0"}, {"rel": "联立", "source": "k - 2 \\neq 0", "target": "k = - 2"}]}}
|
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{"content": "The minimum value of $x$ satisfying $x \\ge 3$ is $a$, and the maximum value of $x$ satisfying $x \\le - 5$ is $b$. Find the value of $a + b$.", "answer": "2", "steps": "$\\because x \\ge 3$, $\\therefore$ the minimum value is $x = 3$, which means $a = 3$. $\\because x \\le - 5$, $\\therefore$ the maximum value is $x = - 5$, which means $b = - 5$. Therefore, $a + b = 3 - 5 = - 2$.", "expr_cands": ["a", "b", "4 a + 4 b + 2", "a + b = 0", "4 ( a + b ) + 2", "2"], "exprs": ["a + b = 0", "4 ( a + b ) + 2", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ 与 $b$ 互为相反数"}, {"id": "4 a + 4 b + 2"}, {"id": "4 ( a + b ) + 2"}, {"id": "2"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "提取因式参考", "source": "a + b = 0", "target": "4 ( a + b ) + 2"}, {"rel": "代入", "source": "a + b = 0", "target": "2"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ 与 $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "提取因式", "source": "4 a + 4 b + 2", "target": "4 ( a + b ) + 2"}, {"rel": "被代入", "source": "4 ( a + b ) + 2", "target": "2"}]}}
|
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{"content": "Given that $2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0$ is a linear equation in $x$ and $y$, then $n ^ m$ = ____?", "answer": "4", "steps": "$\\because$ The equation $2 x ^ { n - 3 } - \\frac { 1 } { 3 } y ^ { 2 m + 1 } = 0$ is a two-variable linear equation in $x$ and $y$. $\\therefore$ We have $n - 3 = 1$ and $2 m + 1 = 1$, which implies $n = 4$ and $m = 0$. Thus, $n ^ { m } = 1$.", "expr_cands": ["2 ( x - 1 ) + 4 < 3 ( x + 1 ) + 2", "x", "2 x - { mx } = 4", "m", "- 3 < x", "x > - 3", "x = - 2", "2 x - mx = 4", "2 m - 4 = 4", "- 4 + 2 m = 4", "m = 4"], "exprs": ["x > - 3", "x = - 2", "- 4 + 2 m = 4", "m = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ( x - 1 ) + 4 < 3 ( x + 1 ) + 2"}, {"id": "x > - 3"}, {"id": "x = - 2"}, {"id": "不等式 $2 ( x - 1 ) + 4 < 3 ( x + 1 ) + 2$ 的最小整数解是关于 $x$ 的方程 $2 x - { mx } = 4$ 的解"}, {"id": "2 x - mx = 4"}, {"id": "- 4 + 2 m = 4"}, {"id": "m = 4"}], "links": [{"rel": "不等式方程求解", "source": "2 ( x - 1 ) + 4 < 3 ( x + 1 ) + 2", "target": "x > - 3"}, {"rel": "被描述", "source": "x > - 3", "target": "x = - 2"}, {"rel": "代入", "source": "x = - 2", "target": "- 4 + 2 m = 4"}, {"rel": "限制性描述", "source": "不等式 $2 ( x - 1 ) + 4 < 3 ( x + 1 ) + 2$ 的最小整数解是关于 $x$ 的方程 $2 x - { mx } = 4$ 的解", "target": "x = - 2"}, {"rel": "被代入", "source": "2 x - mx = 4", "target": "- 4 + 2 m = 4"}, {"rel": "等式方程求解", "source": "- 4 + 2 m = 4", "target": "m = 4"}]}}
|
||||
{"content": "If $( k - 2 ) x ^ 3 + ( | k | - 2 ) x ^ 2 - 5$ is a cubic binomial in terms of $x$, then the value of $k$ is ____?", "answer": "2", "steps": "From the given information, we have $| k | - 2 = 0$ and $k - 2 \\neq 0$. Solving for $k$, we get $k = - 2$.", "expr_cands": ["3 { x } ^ { m - 1 } + 3 y = 11", "x", "m", "y", "m - 1 = 1", "m = 2"], "exprs": ["m - 1 = 1", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 { x } ^ { m - 1 } + 3 y = 11"}, {"id": "m - 1 = 1"}, {"id": "$3 { x } ^ { m - 1 } + 3 y = 11$ 是关于 $x$ , $y$ 的二元一次方程"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "3 { x } ^ { m - 1 } + 3 y = 11", "target": "m - 1 = 1"}, {"rel": "等式方程求解", "source": "m - 1 = 1", "target": "m = 2"}, {"rel": "限制性描述", "source": "$3 { x } ^ { m - 1 } + 3 y = 11$ 是关于 $x$ , $y$ 的二元一次方程", "target": "m - 1 = 1"}]}}
|
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{"content": "If $a$ and $b$ are opposite numbers, then $4 a + 4 b + 2$ = ____?", "answer": "x < 2", "steps": "According to the problem, we have $a + b = 0$ and $4 a + 4 b + 2 = 4 ( a + b ) + 2 = 4 * 0 + 2 = 2$.", "expr_cands": ["m < - 5", "m", "( m + 3 ) x - 2 m - 6 > 0", "x", "m + 3 < - 2", "( m + 3 ) x > 2 ( m + 3 )", "x < 2"], "exprs": ["m + 3 < - 2", "( m + 3 ) x > 2 ( m + 3 )", "x < 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m < - 5"}, {"id": "m + 3 < - 2"}, {"id": "( m + 3 ) x - 2 m - 6 > 0"}, {"id": "( m + 3 ) x > 2 ( m + 3 )"}, {"id": "x < 2"}, {"id": "不等式两边都乘或除同一个负数，不等号的方向改变"}], "links": [{"rel": "移项", "source": "m < - 5", "target": "m + 3 < - 2"}, {"rel": "被描述", "source": "m + 3 < - 2", "target": "x < 2"}, {"rel": "移项", "source": "( m + 3 ) x - 2 m - 6 > 0", "target": "( m + 3 ) x > 2 ( m + 3 )"}, {"rel": "被描述", "source": "( m + 3 ) x > 2 ( m + 3 )", "target": "x < 2"}, {"rel": "属性描述", "source": "不等式两边都乘或除同一个负数，不等号的方向改变", "target": "x < 2"}]}}
|
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{"content": "Given that the minimum integer solution of the inequality $2 ( x - 1 ) + 4 < 3 ( x + 1 ) + 2$ is a solution of the equation $2 x - mx = 4$, then $m$ = ____ ?", "answer": "x \\ge \\frac { 5 } { 2 }", "steps": "$2 ( x - 1 ) + 4 < 3 ( x + 1 ) + 2$, so $x > - 3$. Therefore, the smallest integer solution to the inequality is $x = - 2$. Substituting $x = - 2$ into $2 x - mx = 4$, we get $- 4 + 2 m = 4$, which gives $m = 4$.", "expr_cands": ["\\sqrt { 2 x - 5 }", "x", "2 x - 5 \\ge 0", "\\frac { 5 } { 2 } \\le x", "x \\ge \\frac { 5 } { 2 }"], "exprs": ["2 x - 5 \\ge 0", "x \\ge \\frac { 5 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 2 x - 5 }"}, {"id": "2 x - 5 \\ge 0"}, {"id": "要是式子 $\\sqrt { 2 x - 5 }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge \\frac { 5 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\sqrt { 2 x - 5 }", "target": "2 x - 5 \\ge 0"}, {"rel": "不等式方程求解", "source": "2 x - 5 \\ge 0", "target": "x \\ge \\frac { 5 } { 2 }"}, {"rel": "限制性描述", "source": "要是式子 $\\sqrt { 2 x - 5 }$ 有意义", "target": "2 x - 5 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 x - 5 \\ge 0"}]}}
|
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{"content": "Given that $3 { x } ^ { m - 1 } + 3 y = 11$ is a two-variable linear equation, what is the value of $m$?", "answer": "- 11", "steps": "According to the definition of a quadratic equation, we know that $m - 1 = 1$. Therefore, $m = 2$.", "expr_cands": ["y ^ { 2 } - 2 y + 3", "y", "6", "- 2 y ^ { 2 } + 4 y - 5", "y ^ { 2 } - 2 y + 3 = 6", "y = - 1", "y = 3", "y ^ { 2 } - 2 y = 3", "- 2 ( y ^ { 2 } - 2 y ) - 5", "- 11"], "exprs": ["y ^ { 2 } - 2 y + 3 = 6", "y ^ { 2 } - 2 y = 3", "- 2 ( y ^ { 2 } - 2 y ) - 5", "- 11"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y ^ { 2 } - 2 y + 3"}, {"id": "y ^ { 2 } - 2 y + 3 = 6"}, {"id": "6"}, {"id": "代数式 $y ^ { 2 } - 2 y + 3$ 的值为 $6$"}, {"id": "y ^ { 2 } - 2 y = 3"}, {"id": "- 2 y ^ { 2 } + 4 y - 5"}, {"id": "- 2 ( y ^ { 2 } - 2 y ) - 5"}, {"id": "- 11"}], "links": [{"rel": "被描述", "source": "y ^ { 2 } - 2 y + 3", "target": "y ^ { 2 } - 2 y + 3 = 6"}, {"rel": "移项", "source": "y ^ { 2 } - 2 y + 3 = 6", "target": "y ^ { 2 } - 2 y = 3"}, {"rel": "被描述", "source": "6", "target": "y ^ { 2 } - 2 y + 3 = 6"}, {"rel": "限制性描述", "source": "代数式 $y ^ { 2 } - 2 y + 3$ 的值为 $6$", "target": "y ^ { 2 } - 2 y + 3 = 6"}, {"rel": "提取因式参考", "source": "y ^ { 2 } - 2 y = 3", "target": "- 2 ( y ^ { 2 } - 2 y ) - 5"}, {"rel": "代入", "source": "y ^ { 2 } - 2 y = 3", "target": "- 11"}, {"rel": "提取因式", "source": "- 2 y ^ { 2 } + 4 y - 5", "target": "- 2 ( y ^ { 2 } - 2 y ) - 5"}, {"rel": "被代入", "source": "- 2 ( y ^ { 2 } - 2 y ) - 5", "target": "- 11"}]}}
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{"content": "If $m < - 5$, what is the solution set of the inequality $( m + 3 ) x - 2 m - 6 > 0$?", "answer": "- \\frac { 7 } { 6 }", "steps": "Since $m < - 5$, it follows that $m + 3 < - 2$. Therefore, $( m + 3 ) x > 2 ( m + 3 )$, which implies that $x < 2$.", "expr_cands": ["x", "2 m + 5 x > 1", "m", "2 - 3 x < 0", "\\frac { 2 } { 3 } < x", "x > \\frac { 2 } { 3 }", "x > \\frac { 1 - 2 m } { 5 }", "\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }", "m = - \\frac { 7 } { 6 }"], "exprs": ["x > \\frac { 2 } { 3 }", "x > \\frac { 1 - 2 m } { 5 }", "\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }", "m = - \\frac { 7 } { 6 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 - 3 x < 0"}, {"id": "x > \\frac { 2 } { 3 }"}, {"id": "2 m + 5 x > 1"}, {"id": "x > \\frac { 1 - 2 m } { 5 }"}, {"id": "\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }"}, {"id": "关于 $x$ 的不等式 $2 m + 5 x > 1$ 与不等式 $2 - 3 x < 0$ 的解集相同"}, {"id": "m = - \\frac { 7 } { 6 }"}], "links": [{"rel": "不等式方程求解", "source": "2 - 3 x < 0", "target": "x > \\frac { 2 } { 3 }"}, {"rel": "被描述", "source": "x > \\frac { 2 } { 3 }", "target": "\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }"}, {"rel": "不等式方程部分求解", "source": "2 m + 5 x > 1", "target": "x > \\frac { 1 - 2 m } { 5 }"}, {"rel": "被描述", "source": "x > \\frac { 1 - 2 m } { 5 }", "target": "\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }"}, {"rel": "等式方程求解", "source": "\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }", "target": "m = - \\frac { 7 } { 6 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的不等式 $2 m + 5 x > 1$ 与不等式 $2 - 3 x < 0$ 的解集相同", "target": "\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }"}]}}
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{"content": "If the expression $\\sqrt { 2 x - 5 }$ is meaningful, what is the range of values for the variable $x$?", "answer": "- 6", "steps": "From the given condition, we have $2 x - 5 \\geq 0$. Solving for $x$, we get $x \\geq \\frac { 5 } { 2 }$.", "expr_cands": ["x > - 4", "x", "- 3", "- 2", "- 1", "- 3 - 2 - 1", "- 6"], "exprs": ["- 3", "- 2", "- 1", "- 3 - 2 - 1", "- 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x > - 4"}, {"id": "- 3"}, {"id": "$x$ 是负整数"}, {"id": "- 2"}, {"id": "- 1"}, {"id": "- 3 - 2 - 1"}, {"id": "$x$ 可取的负整数的和"}, {"id": "- 6"}], "links": [{"rel": "被描述", "source": "x > - 4", "target": "- 3"}, {"rel": "被描述", "source": "x > - 4", "target": "- 2"}, {"rel": "被描述", "source": "x > - 4", "target": "- 1"}, {"rel": "被描述", "source": "- 3", "target": "- 3 - 2 - 1"}, {"rel": "限制性描述", "source": "$x$ 是负整数", "target": "- 3"}, {"rel": "限制性描述", "source": "$x$ 是负整数", "target": "- 2"}, {"rel": "限制性描述", "source": "$x$ 是负整数", "target": "- 1"}, {"rel": "被描述", "source": "- 2", "target": "- 3 - 2 - 1"}, {"rel": "被描述", "source": "- 1", "target": "- 3 - 2 - 1"}, {"rel": "计算", "source": "- 3 - 2 - 1", "target": "- 6"}, {"rel": "限制性描述", "source": "$x$ 可取的负整数的和", "target": "- 3 - 2 - 1"}]}}
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{"content": "If the value of the algebraic expression $y ^ 2 - 2 y + 3$ is $6$, then the value of the algebraic expression $- 2 y ^ 2 + 4 y - 5$ is ____?", "answer": "x = - 3", "steps": "According to the problem, we have $y ^ 2 - 2 y + 3 = 6$, which means $y ^ 2 - 2 y = 3$. Therefore, $- 2 y ^ 2 + 4 y - 5 = - 2 ( y ^ 2 - 2 y ) - 5 = - 2 * 3 - 5 = - 11$.", "expr_cands": ["5 x + 1 = 2 x - 8", "x", "x = - 3", "5 x - 2 x = - 8 - 1", "3 x = - 9", "1"], "exprs": ["x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x + 1 = 2 x - 8"}, {"id": "x = - 3"}], "links": [{"rel": "等式方程求解", "source": "5 x + 1 = 2 x - 8", "target": "x = - 3"}]}}
|
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{"content": "Given the inequality $2 m + 5 x > 1$ and the inequality $2 - 3 x < 0$ have the same solution set, what is the value of $m$?", "answer": "- \\frac { 1 } { 2 }", "steps": "The solution set of the inequality $2 - 3 x < 0$ is $x > \\frac { 2 } { 3 }$; the solution set of the inequality $2 m + 5 x > 1$ is $x > \\frac { 1 - 2 m } { 5 }$. Since their solution sets are the same, we have $\\frac { 2 } { 3 } = \\frac { 1 - 2 m } { 5 }$, so $m = - \\frac { 7 } { 6 }$.", "expr_cands": ["2 x + 1 = 0", "x", "x = - \\frac { 1 } { 2 }", "2 x = - 1", "1"], "exprs": ["x = - \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 1 = 0"}, {"id": "x = - \\frac { 1 } { 2 }"}], "links": [{"rel": "等式方程求解", "source": "2 x + 1 = 0", "target": "x = - \\frac { 1 } { 2 }"}]}}
|
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{"content": "Given $x > - 4$, what is the sum of negative integers that $x$ can take?", "answer": "- 6", "steps": "Since $x$ is a negative integer, it can take on the values of $- 3$, $- 2$, or $- 1$. Therefore, the sum of the negative integer solutions is $- 3 - 2 - 1 = - 6$.", "expr_cands": ["x", "y", "2 mx ^ { 2 } - 2 x + y", "m", "- 6 x ^ { 2 } + 2 x - 3 y", "mx ^ { 2 } - 2 x + y", "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )", "( m + 6 ) x ^ { 2 } - 4 x + 4 y", "m + 6 = 0", "m = - 6"], "exprs": ["mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )", "( m + 6 ) x ^ { 2 } - 4 x + 4 y", "m + 6 = 0", "m = - 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 6 x ^ { 2 } + 2 x - 3 y"}, {"id": "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )"}, {"id": "x"}, {"id": "y"}, {"id": "2 mx ^ { 2 } - 2 x + y"}, {"id": "关于 $x$ , $y$ 的两个多项式 $2 mx ^ { 2 } - 2 x + y$ 与 $- 6 x ^ { 2 } + 2 x - 3 y$ 的差中不含二次项"}, {"id": "( m + 6 ) x ^ { 2 } - 4 x + 4 y"}, {"id": "m + 6 = 0"}, {"id": "m = - 6"}], "links": [{"rel": "被描述", "source": "- 6 x ^ { 2 } + 2 x - 3 y", "target": "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )"}, {"rel": "提取因式", "source": "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )", "target": "( m + 6 ) x ^ { 2 } - 4 x + 4 y"}, {"rel": "被描述", "source": "x", "target": "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )"}, {"rel": "被描述", "source": "y", "target": "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )"}, {"rel": "被描述", "source": "2 mx ^ { 2 } - 2 x + y", "target": "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的两个多项式 $2 mx ^ { 2 } - 2 x + y$ 与 $- 6 x ^ { 2 } + 2 x - 3 y$ 的差中不含二次项", "target": "mx ^ { 2 } - 2 x + y - ( - 6 x ^ { 2 } + 2 x - 3 y )"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的两个多项式 $2 mx ^ { 2 } - 2 x + y$ 与 $- 6 x ^ { 2 } + 2 x - 3 y$ 的差中不含二次项", "target": "m + 6 = 0"}, {"rel": "被描述", "source": "( m + 6 ) x ^ { 2 } - 4 x + 4 y", "target": "m + 6 = 0"}, {"rel": "等式方程求解", "source": "m + 6 = 0", "target": "m = - 6"}]}}
|
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{"content": "The solution to the equation $5 x + 1 = 2 x - 8$ is ____ ?", "answer": "1", "steps": "$5 x + 1 = 2 x - 8$, moving terms yields $5 x - 2 x = - 8 - 1$, combining like terms gives $3 x = - 9$, dividing both sides by the coefficient of $x$ gives $x = - 3$.", "expr_cands": ["a", "1 - 2 a", "a - 2", "1 - 2 a = a - 2", "a = 1", "- 3 a = - 3"], "exprs": ["1 - 2 a = a - 2", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1 - 2 a"}, {"id": "1 - 2 a = a - 2"}, {"id": "a - 2"}, {"id": "代数式 $1 - 2 a$ 与 $a - 2$ 的值相"}, {"id": "a = 1"}], "links": [{"rel": "被描述", "source": "1 - 2 a", "target": "1 - 2 a = a - 2"}, {"rel": "等式方程求解", "source": "1 - 2 a = a - 2", "target": "a = 1"}, {"rel": "被描述", "source": "a - 2", "target": "1 - 2 a = a - 2"}, {"rel": "限制性描述", "source": "代数式 $1 - 2 a$ 与 $a - 2$ 的值相", "target": "1 - 2 a = a - 2"}]}}
|
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{"content": "The solution to the equation $2 x + 1 = 0$ is ____ ?", "answer": "0", "steps": "$2 x + 1 = 0$, moving terms yields: $2 x = - 1$, dividing both sides by $2$ gives: $x = - \\frac { 1 } { 2 }$.", "expr_cands": ["a", "b", "c", "a + b + c = 0", "abc > 0", "\\frac { a } { | a | } + \\frac { b } { | b | } + \\frac { | c | } { c } + \\frac { abc } { | abc | }", "1 - 1 - 1 + 1", "0"], "exprs": ["1 - 1 - 1 + 1", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "1 - 1 - 1 + 1"}, {"id": "b"}, {"id": "c"}, {"id": "a + b + c = 0"}, {"id": "abc > 0"}, {"id": "\\frac { a } { | a | } + \\frac { b } { | b | } + \\frac { | c | } { c } + \\frac { abc } { | abc | }"}, {"id": "$a$ , $b$ , $c$ 是非零有理数"}, {"id": "$\\frac { a } { | a | } + \\frac { b } { | b | } + \\frac { | c | } { c } + \\frac { abc } { | abc | }$ 的值"}, {"id": "且 $a$ , $b$ , $c$ 是非零有理数"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "a", "target": "1 - 1 - 1 + 1"}, {"rel": "计算", "source": "1 - 1 - 1 + 1", "target": "0"}, {"rel": "被描述", "source": "b", "target": "1 - 1 - 1 + 1"}, {"rel": "被描述", "source": "c", "target": "1 - 1 - 1 + 1"}, {"rel": "被描述", "source": "a + b + c = 0", "target": "1 - 1 - 1 + 1"}, {"rel": "被描述", "source": "abc > 0", "target": "1 - 1 - 1 + 1"}, {"rel": "被描述", "source": "\\frac { a } { | a | } + \\frac { b } { | b | } + \\frac { | c | } { c } + \\frac { abc } { | abc | }", "target": "1 - 1 - 1 + 1"}, {"rel": "限制性描述", "source": "$a$ , $b$ , $c$ 是非零有理数", "target": "1 - 1 - 1 + 1"}, {"rel": "限制性描述", "source": "$\\frac { a } { | a | } + \\frac { b } { | b | } + \\frac { | c | } { c } + \\frac { abc } { | abc | }$ 的值", "target": "1 - 1 - 1 + 1"}, {"rel": "限制性描述", "source": "且 $a$ , $b$ , $c$ 是非零有理数", "target": "1 - 1 - 1 + 1"}]}}
|
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{"content": "Regarding the two polynomials in $x$ and $y$, $2 mx ^ 2 - 2 x + y$ and $- 6 x ^ 2 + 2 x - 3 y$, if their difference does not contain any quadratic term in $x$ and $y$, then $m$ = ____?", "answer": "y = 3 x + 7", "steps": "$\\because$ Two polynomials in $x$ and $y$, $mx ^ 2 - 2 x + y$ and $- 6 x ^ 2 + 2 x - 3 y$, have a difference that does not contain a quadratic term in $x$. $\\therefore$ $mx ^ 2 - 2 x + y - ( - 6 x ^ 2 + 2 x - 3 y ) = mx ^ 2 - 2 x + y + 6 x ^ 2 - 2 x + 3 y = ( m + 6 ) x ^ 2 - 4 x + 4 y$. Thus, $m + 6 = 0$, and we solve to get $m = - 6$.", "expr_cands": ["y = 3 x", "y", "x", "5", "4", "y = 3 ( x + 4 ) - 5", "3 x = 3 ( x + 4 ) - 5", "3 x + 7"], "exprs": ["y = 3 ( x + 4 ) - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 3 x"}, {"id": "y = 3 ( x + 4 ) - 5"}, {"id": "5"}, {"id": "4"}, {"id": "将直线 $y = 3 x$ 向下平移 $5$ 个单位"}, {"id": "再向左平移 $4$ 个单位"}, {"id": "所得直线的解析式为 $y = 3 ( x + 4 ) - 5$"}], "links": [{"rel": "被描述", "source": "y = 3 x", "target": "y = 3 ( x + 4 ) - 5"}, {"rel": "被描述", "source": "5", "target": "y = 3 ( x + 4 ) - 5"}, {"rel": "被描述", "source": "4", "target": "y = 3 ( x + 4 ) - 5"}, {"rel": "限制性描述", "source": "将直线 $y = 3 x$ 向下平移 $5$ 个单位", "target": "y = 3 ( x + 4 ) - 5"}, {"rel": "限制性描述", "source": "再向左平移 $4$ 个单位", "target": "y = 3 ( x + 4 ) - 5"}, {"rel": "限制性描述", "source": "所得直线的解析式为 $y = 3 ( x + 4 ) - 5$", "target": "y = 3 ( x + 4 ) - 5"}]}}
|
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{"content": "When $a$ = ____ ?, the algebraic expressions $1 - 2 a$ and $a - 2$ have the same value.", "answer": "- \\frac { 1 } { 2 }", "steps": "According to the problem, we have $1 - 2 a = a - 2$. By rearranging and combining like terms, we get $- 3 a = - 3$. Solving for $a$, we get $a = 1$.", "expr_cands": ["y = - 2 x", "x", "y", "y = 1", "- 2 x = 1", "x = - \\frac { 1 } { 2 }"], "exprs": ["- 2 x = 1", "x = - \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - 2 x"}, {"id": "- 2 x = 1"}, {"id": "y = 1"}, {"id": "x = - \\frac { 1 } { 2 }"}], "links": [{"rel": "被代入", "source": "y = - 2 x", "target": "- 2 x = 1"}, {"rel": "等式方程求解", "source": "- 2 x = 1", "target": "x = - \\frac { 1 } { 2 }"}, {"rel": "代入", "source": "y = 1", "target": "- 2 x = 1"}]}}
|
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{"content": "If $a$, $b$, $c$ are non-zero rational numbers, $a + b + c = 0$, and $abc > 0$, then the value of $\\frac { a } { | a | } + \\frac { b } { | b | } + \\frac { | c | } { c } + \\frac { abc } { | abc | }$ is ____?", "answer": "x < 4", "steps": "Since $a + b + c = 0$ and $a$, $b$, $c$ are non-zero rational numbers, therefore two of $a$, $b$, $c$ are negative. Thus, the original expression is equal to $1 - 1 - 1 + 1 = 0$.", "expr_cands": ["2 x - 3", "x", "5", "2 x - 3 < 5", "x < 4", "2 x < 8"], "exprs": ["2 x - 3 < 5", "x < 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 3"}, {"id": "2 x - 3 < 5"}, {"id": "5"}, {"id": "代数式 $2 x - 3$ 小于 $5$"}, {"id": "x < 4"}], "links": [{"rel": "被描述", "source": "2 x - 3", "target": "2 x - 3 < 5"}, {"rel": "不等式方程求解", "source": "2 x - 3 < 5", "target": "x < 4"}, {"rel": "被描述", "source": "5", "target": "2 x - 3 < 5"}, {"rel": "限制性描述", "source": "代数式 $2 x - 3$ 小于 $5$", "target": "2 x - 3 < 5"}]}}
|
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{"content": "The line $y = 3 x$ is translated down $5$ units and left $4$ units, resulting in the line _____.", "answer": "m \\neq 2", "steps": "The line $y = 3 x$ is translated down $5$ units and left $4$ units, resulting in the equation $y = 3 ( x + 4 ) - 5$, which simplifies to $y = 3 x + 7$.", "expr_cands": ["( m - 2 ) x - 1 = 0", "m", "x", "m - 2 = 0", "m = 2", "m \\neq 2"], "exprs": ["m \\neq 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 2 ) x - 1 = 0"}, {"id": "m \\neq 2"}, {"id": ", $( m - 2 ) x - 1 = 0$ 是关于 $x$ 的一元一次方程"}], "links": [{"rel": "被描述", "source": "( m - 2 ) x - 1 = 0", "target": "m \\neq 2"}, {"rel": "限制性描述", "source": ", $( m - 2 ) x - 1 = 0$ 是关于 $x$ 的一元一次方程", "target": "m \\neq 2"}]}}
|
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{"content": "Given the function $y = - 2 x$, when $x$ = ____ ?, $y = 1$.", "answer": "0", "steps": "When $y = 1$, $- 2 x = 1$, solving for $x$ gives $x = - \\frac { 1 } { 2 }$.", "expr_cands": ["y = x ^ { 2 } + mx + 4", "x", "m", "y", "- \\frac { m } { - 2 } = 0", "m = 0"], "exprs": ["- \\frac { m } { - 2 } = 0", "m = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 } + mx + 4"}, {"id": "- \\frac { m } { - 2 } = 0"}, {"id": "y"}, {"id": "因为二次函数 $y = x ^ { 2 } + mx + 4$ 的对称轴是 $y$ 轴"}, {"id": "二次函数y=a^x+bx+c 的对称轴为 - \\frac {b} {2a}"}, {"id": "m = 0"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 } + mx + 4", "target": "- \\frac { m } { - 2 } = 0"}, {"rel": "等式方程求解", "source": "- \\frac { m } { - 2 } = 0", "target": "m = 0"}, {"rel": "被描述", "source": "y", "target": "- \\frac { m } { - 2 } = 0"}, {"rel": "限制性描述", "source": "因为二次函数 $y = x ^ { 2 } + mx + 4$ 的对称轴是 $y$ 轴", "target": "- \\frac { m } { - 2 } = 0"}, {"rel": "属性描述", "source": "二次函数y=a^x+bx+c 的对称轴为 - \\frac {b} {2a}", "target": "- \\frac { m } { - 2 } = 0"}]}}
|
||||
{"content": "The algebraic expression $2 x - 3$ is less than $5$, then the range of values for $x$ is ____?", "answer": "- 4", "steps": "According to the problem, we have $2 x - 3 < 5$, which simplifies to $2 x < 8$. Therefore, $x < 4$.", "expr_cands": ["y = kx + b", "x", "k", "y", "b", "0.5", "2", "x = a", "a", "y = ak + b", "x = a - 0.5", "a = a - 0.5", "y + 2 = ( a - 0.5 ) k + b", "a k + b + 2 = ( a - 0.5 ) k + b", "2 = - 0.5 k", "k = - 4.0", "k = - 4"], "exprs": ["x = a", "y = ak + b", "x = a - 0.5", "y + 2 = ( a - 0.5 ) k + b", "2 = - 0.5 k", "k = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $x = a$ 时"}, {"id": "x = a"}, {"id": "y = kx + b"}, {"id": "y = ak + b"}, {"id": "0.5"}, {"id": "x = a - 0.5"}, {"id": "当 $x$ 的值每减小 $0.5$ 时"}, {"id": "$y$ 的值就增加 $2$"}, {"id": "一次函数 $y = kx + b$"}, {"id": "y + 2 = ( a - 0.5 ) k + b"}, {"id": "2 = - 0.5 k"}, {"id": "k = - 4"}], "links": [{"rel": "假设描述", "source": "设 $x = a$ 时", "target": "x = a"}, {"rel": "代入", "source": "x = a", "target": "y = ak + b"}, {"rel": "被描述", "source": "x = a", "target": "x = a - 0.5"}, {"rel": "被代入", "source": "y = kx + b", "target": "y = ak + b"}, {"rel": "被描述", "source": "y = kx + b", "target": "y + 2 = ( a - 0.5 ) k + b"}, {"rel": "联立", "source": "y = kx + b", "target": "2 = - 0.5 k"}, {"rel": "被描述", "source": "0.5", "target": "x = a - 0.5"}, {"rel": "被描述", "source": "x = a - 0.5", "target": "y + 2 = ( a - 0.5 ) k + b"}, {"rel": "限制性描述", "source": "当 $x$ 的值每减小 $0.5$ 时", "target": "x = a - 0.5"}, {"rel": "限制性描述", "source": "当 $x$ 的值每减小 $0.5$ 时", "target": "y + 2 = ( a - 0.5 ) k + b"}, {"rel": "限制性描述", "source": "$y$ 的值就增加 $2$", "target": "x = a - 0.5"}, {"rel": "限制性描述", "source": "$y$ 的值就增加 $2$", "target": "y + 2 = ( a - 0.5 ) k + b"}, {"rel": "限制性描述", "source": "一次函数 $y = kx + b$", "target": "x = a - 0.5"}, {"rel": "限制性描述", "source": "一次函数 $y = kx + b$", "target": "y + 2 = ( a - 0.5 ) k + b"}, {"rel": "联立", "source": "y + 2 = ( a - 0.5 ) k + b", "target": "2 = - 0.5 k"}, {"rel": "等式方程求解", "source": "2 = - 0.5 k", "target": "k = - 4"}]}}
|
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{"content": "Given: $( m - 2 ) x - 1 = 0$ is a linear equation in one variable $x$, then $m$ ____?", "answer": "1", "steps": "$\\because$ $( m - 2 ) x - 1 = 0$ is a linear equation in one variable $x$, $\\therefore$ $m - 2 = 0$. $\\therefore$ $m \\neq 2$.", "expr_cands": ["\\sqrt { x } + \\sqrt { - x }", "x", "\\sqrt { x + 1 }", "x \\ge 0", "- x \\ge 0", "x \\le 0", "x = 0", "1"], "exprs": ["x \\ge 0", "- x \\ge 0", "x = 0", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x } + \\sqrt { - x }"}, {"id": "x \\ge 0"}, {"id": "$\\sqrt { x } + \\sqrt { - x }$ 有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "- x \\ge 0"}, {"id": "x = 0"}, {"id": "\\sqrt { x + 1 }"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "\\sqrt { x } + \\sqrt { - x }", "target": "x \\ge 0"}, {"rel": "被描述", "source": "\\sqrt { x } + \\sqrt { - x }", "target": "- x \\ge 0"}, {"rel": "联立", "source": "x \\ge 0", "target": "x = 0"}, {"rel": "限制性描述", "source": "$\\sqrt { x } + \\sqrt { - x }$ 有意义", "target": "x \\ge 0"}, {"rel": "限制性描述", "source": "$\\sqrt { x } + \\sqrt { - x }$ 有意义", "target": "- x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "- x \\ge 0"}, {"rel": "联立", "source": "- x \\ge 0", "target": "x = 0"}, {"rel": "代入", "source": "x = 0", "target": "1"}, {"rel": "被代入", "source": "\\sqrt { x + 1 }", "target": "1"}]}}
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{"content": "The symmetric axis of the quadratic function $y = x ^ 2 + mx + 4$ is the $y$-axis. What is the value of $m$?", "answer": "- 1", "steps": "Because the axis of symmetry of the quadratic function $y = x ^ 2 + mx + 4$ is the $y$-axis, we have $- \\frac { m } { - 2 } = 0$, which implies $m = 0$.", "expr_cands": ["| x - 4 |", "x", "| 3 - y |", "y", "- x + y", "| x - 4 | + | 3 - y | = 0", "x - 4 = 0", "x = 4", "3 - y = 0", "y = 3", "- 1"], "exprs": ["| x - 4 | + | 3 - y | = 0", "x - 4 = 0", "3 - y = 0", "x = 4", "y = 3", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x - 4 |"}, {"id": "| x - 4 | + | 3 - y | = 0"}, {"id": "| 3 - y |"}, {"id": "$| x - 4 |$ 与 $| 3 - y |$ 互为相反数"}, {"id": "x - 4 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "3 - y = 0"}, {"id": "x = 4"}, {"id": "y = 3"}, {"id": "- x + y"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "| x - 4 |", "target": "| x - 4 | + | 3 - y | = 0"}, {"rel": "被描述", "source": "| x - 4 | + | 3 - y | = 0", "target": "x - 4 = 0"}, {"rel": "被描述", "source": "| x - 4 | + | 3 - y | = 0", "target": "3 - y = 0"}, {"rel": "被描述", "source": "| 3 - y |", "target": "| x - 4 | + | 3 - y | = 0"}, {"rel": "限制性描述", "source": "$| x - 4 |$ 与 $| 3 - y |$ 互为相反数", "target": "| x - 4 | + | 3 - y | = 0"}, {"rel": "等式方程求解", "source": "x - 4 = 0", "target": "x = 4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x - 4 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "3 - y = 0"}, {"rel": "等式方程求解", "source": "3 - y = 0", "target": "y = 3"}, {"rel": "代入", "source": "x = 4", "target": "- 1"}, {"rel": "代入", "source": "y = 3", "target": "- 1"}, {"rel": "被代入", "source": "- x + y", "target": "- 1"}]}}
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{"content": "Given a linear function $y = kx + b$, when $x$ decreases by $0.5$, $y$ increases by $2$. What is the value of $k$?", "answer": "x > 1", "steps": "When $x = a$, $y = ak + b$. Therefore, when $x = a - 0.5$, $y + 2 = ( a - 0.5 ) k + b$. Solving for $k$, we get $k = - 4$ since $2 = - 0.5 k$.", "expr_cands": ["4 x - y = 6", "y", "x", "x - \\frac { 1 } { 2 } y < 2", "y = 4 x - 6", "x - \\frac { 1 } { 2 } ( 4 x - 6 ) < 2", "1 < x", "x > 1"], "exprs": ["y = 4 x - 6", "x - \\frac { 1 } { 2 } ( 4 x - 6 ) < 2", "x > 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x - y = 6"}, {"id": "y = 4 x - 6"}, {"id": "x - \\frac { 1 } { 2 } y < 2"}, {"id": "x - \\frac { 1 } { 2 } ( 4 x - 6 ) < 2"}, {"id": "x > 1"}], "links": [{"rel": "移项", "source": "4 x - y = 6", "target": "y = 4 x - 6"}, {"rel": "代入", "source": "y = 4 x - 6", "target": "x - \\frac { 1 } { 2 } ( 4 x - 6 ) < 2"}, {"rel": "被代入", "source": "x - \\frac { 1 } { 2 } y < 2", "target": "x - \\frac { 1 } { 2 } ( 4 x - 6 ) < 2"}, {"rel": "不等式方程求解", "source": "x - \\frac { 1 } { 2 } ( 4 x - 6 ) < 2", "target": "x > 1"}]}}
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{"content": "If $\\sqrt { x } + \\sqrt { - x }$ is defined, then $\\sqrt { x + 1 }$ = ____?", "answer": "\\frac { 7 } { 3 }", "steps": "From the given condition, we have $x \\ge 0$ and $- x \\ge 0$, which implies $x = 0$. Therefore, $\\sqrt { x + 1 } = \\sqrt { 1 } = 1$.", "expr_cands": ["\\frac { - 2 a { x } ^ { 2 } } { 3 }", "x", "a", "- \\frac { 2 } { 3 }", "1 + 2", "3", "- \\frac { 2 } { 3 } + 3", "\\frac { 7 } { 3 }"], "exprs": ["- \\frac { 2 } { 3 }", "1 + 2", "3", "- \\frac { 2 } { 3 } + 3", "\\frac { 7 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { - 2 a { x } ^ { 2 } } { 3 }"}, {"id": "- \\frac { 2 } { 3 }"}, {"id": "项式 $\\frac { - 2 a { x } ^ { 2 } } { 3 }$ 的系数是 $- \\frac { 2 } { 3 }$"}, {"id": "1 + 2"}, {"id": "次数是 $1 + 2 = 3$"}, {"id": "3"}, {"id": "- \\frac { 2 } { 3 } + 3"}, {"id": "单项式 $\\frac { - 2 a { x } ^ { 2 } } { 3 }$ 的系数和次数之和"}, {"id": "\\frac { 7 } { 3 }"}], "links": [{"rel": "被描述", "source": "\\frac { - 2 a { x } ^ { 2 } } { 3 }", "target": "- \\frac { 2 } { 3 }"}, {"rel": "被描述", "source": "\\frac { - 2 a { x } ^ { 2 } } { 3 }", "target": "1 + 2"}, {"rel": "被描述", "source": "\\frac { - 2 a { x } ^ { 2 } } { 3 }", "target": "- \\frac { 2 } { 3 } + 3"}, {"rel": "被描述", "source": "- \\frac { 2 } { 3 }", "target": "- \\frac { 2 } { 3 } + 3"}, {"rel": "限制性描述", "source": "项式 $\\frac { - 2 a { x } ^ { 2 } } { 3 }$ 的系数是 $- \\frac { 2 } { 3 }$", "target": "- \\frac { 2 } { 3 }"}, {"rel": "计算", "source": "1 + 2", "target": "3"}, {"rel": "限制性描述", "source": "次数是 $1 + 2 = 3$", "target": "1 + 2"}, {"rel": "被描述", "source": "3", "target": "- \\frac { 2 } { 3 } + 3"}, {"rel": "计算", "source": "- \\frac { 2 } { 3 } + 3", "target": "\\frac { 7 } { 3 }"}, {"rel": "限制性描述", "source": "单项式 $\\frac { - 2 a { x } ^ { 2 } } { 3 }$ 的系数和次数之和", "target": "- \\frac { 2 } { 3 } + 3"}]}}
|
||||
{"content": "If $| x - 4 |$ and $| 3 - y |$ are opposite in sign, then $- x + y$ = ____ ?", "answer": "4", "steps": "$\\because$ $| x - 4 | + | 3 - y | = 0$, $\\therefore$ $x - 4 = 0$, $3 - y = 0$, so $x = 4$, $y = 3$, $- x + y = - 4 + 3 = - 1$.", "expr_cands": ["2 a - b = 2", "a", "b", "b - 2 ( a - 3 )", "b - 2 a + 6", "b - 2 a = - 2", "- 2 + 6", "4"], "exprs": ["b - 2 a + 6", "b - 2 a = - 2", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "b - 2 ( a - 3 )"}, {"id": "b - 2 a + 6"}, {"id": "2 a - b = 2"}, {"id": "b - 2 a = - 2"}, {"id": "4"}], "links": [{"rel": "展开", "source": "b - 2 ( a - 3 )", "target": "b - 2 a + 6"}, {"rel": "被代入", "source": "b - 2 a + 6", "target": "4"}, {"rel": "同乘除", "source": "2 a - b = 2", "target": "b - 2 a = - 2"}, {"rel": "代入", "source": "b - 2 a = - 2", "target": "4"}]}}
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{"content": "Given $4 x - y = 6$, $x - \\frac { 1 } { 2 } y < 2$, the range of $x$ is ____?", "answer": "- 3", "steps": "$\\because$ $4 x - y = 6$ , $\\therefore$ $y = 4 x - 6$ , $\\because$ $x - \\frac { 1 } { 2 } y < 2$ , $\\therefore$ $x - \\frac { 1 } { 2 } ( 4 x - 6 ) < 2$ , solving gives: $x > 1$ , which means the range of $x$ is $x > 1$.", "expr_cands": ["x", "y", "( a - 1 ) xy ^ { ( a + 1 ) ^ { 2 } } + a", "a", "a - 1 \\neq 0", "a \\neq 1", "a + 1 = 2", "a = 1", "a + 1 = - 2", "a = - 3"], "exprs": ["a - 1 \\neq 0", "a + 1 = - 2", "a = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 1 ) xy ^ { ( a + 1 ) ^ { 2 } } + a"}, {"id": "a - 1 \\neq 0"}, {"id": "关于 $x$ , $y$ 的整式 $( a - 1 ) xy ^ { ( a + 1 ) ^ { 2 } } + a$ 是五次多项式"}, {"id": "a + 1 = - 2"}, {"id": "a = - 3"}], "links": [{"rel": "被描述", "source": "( a - 1 ) xy ^ { ( a + 1 ) ^ { 2 } } + a", "target": "a - 1 \\neq 0"}, {"rel": "被描述", "source": "( a - 1 ) xy ^ { ( a + 1 ) ^ { 2 } } + a", "target": "a + 1 = - 2"}, {"rel": "被描述", "source": "a - 1 \\neq 0", "target": "a + 1 = - 2"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的整式 $( a - 1 ) xy ^ { ( a + 1 ) ^ { 2 } } + a$ 是五次多项式", "target": "a - 1 \\neq 0"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的整式 $( a - 1 ) xy ^ { ( a + 1 ) ^ { 2 } } + a$ 是五次多项式", "target": "a + 1 = - 2"}, {"rel": "等式方程求解", "source": "a + 1 = - 2", "target": "a = - 3"}]}}
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{"content": "The coefficient and degree sum of the monomial $\\frac { - 2 a { x } ^ { 2 }} { 3 }$ is ____?", "answer": "121", "steps": "The coefficient of the term $\\frac { - 2 a { x } ^ 2 } { 3 }$ is $- \\frac { 2 } { 3 }$; the degree is $1 + 2 = 3$; therefore, their sum is $- \\frac { 2 } { 3 } + 3 = \\frac { 7 } { 3 }$.", "expr_cands": ["m", "5 a + 1", "a", "a - 13", "5 a + 1 + a - 13 = 0", "a = 2", "11", "m = 121"], "exprs": ["5 a + 1 + a - 13 = 0", "a = 2", "11", "m = 121"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 a + 1"}, {"id": "5 a + 1 + a - 13 = 0"}, {"id": "a - 13"}, {"id": "一个正数 $m$ 的两个平方根是 $5 a + 1$ 和 $a - 13$"}, {"id": "平方根互为相反数"}, {"id": "a = 2"}, {"id": "11"}, {"id": "m = 121"}, {"id": "m"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "5 a + 1", "target": "5 a + 1 + a - 13 = 0"}, {"rel": "被代入", "source": "5 a + 1", "target": "11"}, {"rel": "等式方程求解", "source": "5 a + 1 + a - 13 = 0", "target": "a = 2"}, {"rel": "被描述", "source": "a - 13", "target": "5 a + 1 + a - 13 = 0"}, {"rel": "限制性描述", "source": "一个正数 $m$ 的两个平方根是 $5 a + 1$ 和 $a - 13$", "target": "5 a + 1 + a - 13 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "5 a + 1 + a - 13 = 0"}, {"rel": "代入", "source": "a = 2", "target": "11"}, {"rel": "被描述", "source": "11", "target": "m = 121"}, {"rel": "被描述", "source": "m", "target": "m = 121"}, {"rel": "限制性描述", "source": "平方", "target": "m = 121"}]}}
|
||||
{"content": "If $2 a - b = 2$, then $b - 2 ( a - 3 )$ = ____ ?", "answer": "1", "steps": "$b - 2 ( a - 3 ) = b - 2 a + 6$ , because $2 a - b = 2$ , therefore $b - 2 a = - 2$ , therefore the original expression equals $- 2 + 6 = 4$.", "expr_cands": ["x = 1", "x", "kx - 1 = 0", "k", "k - 1 = 0", "k = 1"], "exprs": ["k - 1 = 0", "k = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 1"}, {"id": "k - 1 = 0"}, {"id": "kx - 1 = 0"}, {"id": "k = 1"}], "links": [{"rel": "代入", "source": "x = 1", "target": "k - 1 = 0"}, {"rel": "等式方程求解", "source": "k - 1 = 0", "target": "k = 1"}, {"rel": "被代入", "source": "kx - 1 = 0", "target": "k - 1 = 0"}]}}
|
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{"content": "Regarding the polynomial in $x$ and $y$ given by $( a - 1 ) xy ^ {( a + 1 ) ^ 2 } + a$, if it is a fifth degree polynomial, then $a$ equals ____?", "answer": "1", "steps": "Because the polynomial $( a - 1 ) xy ^ {( a + 1 ) ^ 2 } + a$ in terms of $x$ and $y$ is a quintic polynomial, we know that $a - 1 \\neq 0$. Thus, $a + 1 = 2$ or $a + 1 = - 2$, which implies that $a = - 3$.", "expr_cands": ["16", "m", "- 27", "n", "m + n", "m = 4", "n = - 3", "1"], "exprs": ["m = 4", "n = - 3", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "16"}, {"id": "m = 4"}, {"id": "m"}, {"id": "$16$ 的算术平方根是 $m$ , $- 27$ 的立方根是 $n$"}, {"id": "- 27"}, {"id": "n = - 3"}, {"id": "n"}, {"id": "m + n"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "16", "target": "m = 4"}, {"rel": "代入", "source": "m = 4", "target": "1"}, {"rel": "被描述", "source": "m", "target": "m = 4"}, {"rel": "限制性描述", "source": "$16$ 的算术平方根是 $m$ , $- 27$ 的立方根是 $n$", "target": "m = 4"}, {"rel": "限制性描述", "source": "$16$ 的算术平方根是 $m$ , $- 27$ 的立方根是 $n$", "target": "n = - 3"}, {"rel": "被描述", "source": "- 27", "target": "n = - 3"}, {"rel": "代入", "source": "n = - 3", "target": "1"}, {"rel": "被描述", "source": "n", "target": "n = - 3"}, {"rel": "被代入", "source": "m + n", "target": "1"}]}}
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{"content": "Given that two square roots of a positive number $m$ are $5 a + 1$ and $a - 13$, find the value of $m$.", "answer": "\\frac { 1 } { 3 }", "steps": "Since $5 a + 1$ and $a - 13$ are two square roots of a positive number $m$, we have $5 a + 1 + a - 13 = 0$. Therefore, $a = 2$, $5 a + 1 = 11$, and $m = 11 ^ 2 = 121$.", "expr_cands": ["x", "2 x + a = 3", "a", "2 - \\frac { 3 x + 2 } { 3 } = 0", "x = \\frac { 4 } { 3 }", "a + \\frac { 8 } { 3 } = 3", "2 * \\frac { 4 } { 3 } + a = 3", "a = \\frac { 1 } { 3 }"], "exprs": ["x = \\frac { 4 } { 3 }", "2 * \\frac { 4 } { 3 } + a = 3", "a = \\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 - \\frac { 3 x + 2 } { 3 } = 0"}, {"id": "x = \\frac { 4 } { 3 }"}, {"id": "2 x + a = 3"}, {"id": "2 * \\frac { 4 } { 3 } + a = 3"}, {"id": "a = \\frac { 1 } { 3 }"}], "links": [{"rel": "等式方程求解", "source": "2 - \\frac { 3 x + 2 } { 3 } = 0", "target": "x = \\frac { 4 } { 3 }"}, {"rel": "代入", "source": "x = \\frac { 4 } { 3 }", "target": "2 * \\frac { 4 } { 3 } + a = 3"}, {"rel": "被代入", "source": "2 x + a = 3", "target": "2 * \\frac { 4 } { 3 } + a = 3"}, {"rel": "等式方程求解", "source": "2 * \\frac { 4 } { 3 } + a = 3", "target": "a = \\frac { 1 } { 3 }"}]}}
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{"content": "$x = 1$ is a solution to the equation $kx - 1 = 0$, then $k$ = ____ ?", "answer": "3", "steps": "Substituting $x = 1$ into the equation yields $k - 1 = 0$, which implies $k = 1$.", "expr_cands": ["ab = 3", "b", "a", "a + b = \\frac { 1 } { 3 }", "ab - ( 3 a - b ) - 4 b + 1", "ab - 3 ( a + b ) + 1", "3"], "exprs": ["ab - 3 ( a + b ) + 1", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ab - ( 3 a - b ) - 4 b + 1"}, {"id": "ab - 3 ( a + b ) + 1"}, {"id": "a + b = \\frac { 1 } { 3 }"}, {"id": "3"}, {"id": "ab = 3"}], "links": [{"rel": "提取因式", "source": "ab - ( 3 a - b ) - 4 b + 1", "target": "ab - 3 ( a + b ) + 1"}, {"rel": "被代入", "source": "ab - 3 ( a + b ) + 1", "target": "3"}, {"rel": "提取因式参考", "source": "a + b = \\frac { 1 } { 3 }", "target": "ab - 3 ( a + b ) + 1"}, {"rel": "代入", "source": "a + b = \\frac { 1 } { 3 }", "target": "3"}, {"rel": "代入", "source": "ab = 3", "target": "3"}]}}
|
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{"content": "If the arithmetic square root of $16$ is $m$, and the cube root of $- 27$ is $n$, then the value of $m + n$ is ____?", "answer": "8", "steps": "$\\because$ The arithmetic square root of $16$ is $m$, and the cube root of $- 27$ is $n$. $\\therefore$ $m = 4$ and $n = - 3$. $\\therefore$ $m + n = 4 + ( - 3 ) = 1$.", "expr_cands": ["y = x ^ { 2 } + bx + 3", "b", "y", "x", "x < - 2", "x > - 2", "x = 1", "x = - 2", "- \\frac { b } { 2 } = - 2", "b = 4", "y = x ^ { 2 } + 4 x + 3", "y = 8"], "exprs": ["- \\frac { b } { 2 } = - 2", "b = 4", "y = x ^ { 2 } + 4 x + 3", "y = 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 } + bx + 3"}, {"id": "- \\frac { b } { 2 } = - 2"}, {"id": "二次函数 $y = x ^ { 2 } + bx + 3$ 满足当 $x < - 2$ 时"}, {"id": "$y$ 随 $x$ 的增大而减小"}, {"id": "当 $x > - 2$ 时"}, {"id": "$y$ 随 $x$ 的增大而增大"}, {"id": "二次函数y=a^x+bx+c 的对称轴为 - \\frac {b} {2a}"}, {"id": "b = 4"}, {"id": "y = x ^ { 2 } + 4 x + 3"}, {"id": "x = 1"}, {"id": "y = 8"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 } + bx + 3", "target": "- \\frac { b } { 2 } = - 2"}, {"rel": "被代入", "source": "y = x ^ { 2 } + bx + 3", "target": "y = x ^ { 2 } + 4 x + 3"}, {"rel": "等式方程求解", "source": "- \\frac { b } { 2 } = - 2", "target": "b = 4"}, {"rel": "限制性描述", "source": "二次函数 $y = x ^ { 2 } + bx + 3$ 满足当 $x < - 2$ 时", "target": "- \\frac { b } { 2 } = - 2"}, {"rel": "限制性描述", "source": "$y$ 随 $x$ 的增大而减小", "target": "- \\frac { b } { 2 } = - 2"}, {"rel": "限制性描述", "source": "当 $x > - 2$ 时", "target": "- \\frac { b } { 2 } = - 2"}, {"rel": "限制性描述", "source": "$y$ 随 $x$ 的增大而增大", "target": "- \\frac { b } { 2 } = - 2"}, {"rel": "属性描述", "source": "二次函数y=a^x+bx+c 的对称轴为 - \\frac {b} {2a}", "target": "- \\frac { b } { 2 } = - 2"}, {"rel": "代入", "source": "b = 4", "target": "y = x ^ { 2 } + 4 x + 3"}, {"rel": "被代入", "source": "y = x ^ { 2 } + 4 x + 3", "target": "y = 8"}, {"rel": "代入", "source": "x = 1", "target": "y = 8"}]}}
|
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{"content": "If the equation $2 x + a = 3$ and the equation $2 - \\frac { 3 x + 2 } { 3 } = 0$ are the same solution equation about $x$, then the value of $a$ is ____?", "answer": "7", "steps": "Solve the equation $2 - \\frac { 3 x + 2 } { 3 } = 0$ to get $x = \\frac { 4 } { 3 }$. Substitute $x = \\frac { 4 } { 3 }$ into the equation $2 x + a = 3$ to get $2 * \\frac { 4 } { 3 } + a = 3$. Solving for $a$, we get $a = \\frac { 1 } { 3 }$.", "expr_cands": ["{ ( { a } ^ { 2 } ) } ^ { 3 } \\cdot { a } ^ { m } \\div { a } ^ { 4 } = { a } ^ { 9 }", "a", "m", "6 + m - 4 = 9", "m = 7"], "exprs": ["6 + m - 4 = 9", "m = 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ ( { a } ^ { 2 } ) } ^ { 3 } \\cdot { a } ^ { m } \\div { a } ^ { 4 } = { a } ^ { 9 }"}, {"id": "6 + m - 4 = 9"}, {"id": "m = 7"}], "links": [{"rel": "同取对数", "source": "{ ( { a } ^ { 2 } ) } ^ { 3 } \\cdot { a } ^ { m } \\div { a } ^ { 4 } = { a } ^ { 9 }", "target": "6 + m - 4 = 9"}, {"rel": "等式方程求解", "source": "6 + m - 4 = 9", "target": "m = 7"}]}}
|
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{"content": "If $ab = 3$, $a + b = \\frac { 1 } { 3 }$, then the value of $ab - ( 3 a - b ) - 4 b + 1$ is ____?", "answer": "x = 1", "steps": "\\because $ab = 3$, $a + b = \\frac { 1 } { 3 }$, \\therefore the original expression $= ab - 3 a + b - 4 b + 1 = ab - 3 ( a + b ) + 1 = 3 - 1 + 1 = 3$.", "expr_cands": ["x", "\\frac { 3 } { x - 1 }", "x - 1 = 0", "x = 1"], "exprs": ["x - 1 = 0", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 } { x - 1 }"}, {"id": "x - 1 = 0"}, {"id": "分式 $\\frac { 3 } { x - 1 }$ 无意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "\\frac { 3 } { x - 1 }", "target": "x - 1 = 0"}, {"rel": "等式方程求解", "source": "x - 1 = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "分式 $\\frac { 3 } { x - 1 }$ 无意义", "target": "x - 1 = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 1 = 0"}]}}
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{"content": "The quadratic function $y = x ^ { 2 } + bx + 3$ satisfies that when $x < - 2$, $y$ decreases as $x$ increases, and when $x > - 2$, $y$ increases as $x$ increases. What is the value of $y$ when $x = 1$?", "answer": "4 - \\sqrt { 7 }", "steps": "$\\because$ The quadratic function $y = x ^ 2 + bx + 3$ decreases as $x$ increases when $x < - 2$, and increases as $x$ increases when $x > - 2$. $\\therefore$ The axis of symmetry is $x = - 2$. $\\therefore$ $- \\frac { b } { 2 } = - 2$, $\\therefore$ $b = 4$. $\\therefore$ The quadratic function $y = x ^ 2 + 4 x + 3$ has a value of $8$ when $x = 1$.", "expr_cands": ["( 4 + \\sqrt { 7 } ) \\times a = b", "b", "a", "( 4 + \\sqrt { 7 } ) * ( 4 - \\sqrt { 7 } )", "9", "a = 4 - \\sqrt { 7 }"], "exprs": ["a = 4 - \\sqrt { 7 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 4 + \\sqrt { 7 } ) \\times a = b"}, {"id": "a = 4 - \\sqrt { 7 }"}, {"id": "$b$ 是整数"}, {"id": "$a$ 的值可能"}], "links": [{"rel": "被描述", "source": "( 4 + \\sqrt { 7 } ) \\times a = b", "target": "a = 4 - \\sqrt { 7 }"}, {"rel": "限制性描述", "source": "$b$ 是整数", "target": "a = 4 - \\sqrt { 7 }"}, {"rel": "限制性描述", "source": "$a$ 的值可能", "target": "a = 4 - \\sqrt { 7 }"}]}}
|
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{"content": "If ${ ( { a } ^ { 2 } ) } ^ { 3 } \\cdot { a } ^ { m } \\div { a } ^ { 4 } = { a } ^ { 9 }$, then $m$ = ____ ?", "answer": "15", "steps": "Because $({ a ^ 2 }) ^ 3 \\cdot { a ^ m } \\div { a ^ 4 } = { a ^ { 6 + m - 4 }} = { a ^ 9 }$, therefore $6 + m - 4 = 9$, which solves for $m = 7$.", "expr_cands": ["a + b = 4", "a", "b", "ab = 3", "( a + 2 ) ( b + 2 )", "ab + 2 ( a + b ) + 4", "15"], "exprs": ["ab + 2 ( a + b ) + 4", "15"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a + 2 ) ( b + 2 )"}, {"id": "ab + 2 ( a + b ) + 4"}, {"id": "a + b = 4"}, {"id": "ab = 3"}, {"id": "15"}], "links": [{"rel": "提取因式", "source": "( a + 2 ) ( b + 2 )", "target": "ab + 2 ( a + b ) + 4"}, {"rel": "被代入", "source": "ab + 2 ( a + b ) + 4", "target": "15"}, {"rel": "提取因式参考", "source": "a + b = 4", "target": "ab + 2 ( a + b ) + 4"}, {"rel": "代入", "source": "a + b = 4", "target": "15"}, {"rel": "提取因式参考", "source": "ab = 3", "target": "ab + 2 ( a + b ) + 4"}, {"rel": "代入", "source": "ab = 3", "target": "15"}]}}
|
||||
{"content": "When $x$ satisfies ____ ?, the fraction $\\frac { 3 } { x - 1 }$ is undefined.", "answer": "x = - 2", "steps": "$\\because$ The fraction $\\frac { 3 } { x - 1 }$ is undefined, $\\therefore$ $x - 1 = 0$, $\\therefore$ $x = 1$.", "expr_cands": ["| m - 2 | + { ( n - 2 ) } ^ { 2 } = 0", "m", "n", "x", "2 m + x = n", "m - 2 = 0", "m = 2", "n - 2 = 0", "n = 2", "x = - 2"], "exprs": ["m - 2 = 0", "n - 2 = 0", "m = 2", "n = 2", "x = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| m - 2 | + { ( n - 2 ) } ^ { 2 } = 0"}, {"id": "m - 2 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "n - 2 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "m = 2"}, {"id": "n = 2"}, {"id": "2 m + x = n"}, {"id": "x = - 2"}], "links": [{"rel": "被描述", "source": "| m - 2 | + { ( n - 2 ) } ^ { 2 } = 0", "target": "m - 2 = 0"}, {"rel": "被描述", "source": "| m - 2 | + { ( n - 2 ) } ^ { 2 } = 0", "target": "n - 2 = 0"}, {"rel": "等式方程求解", "source": "m - 2 = 0", "target": "m = 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "m - 2 = 0"}, {"rel": "等式方程求解", "source": "n - 2 = 0", "target": "n = 2"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "n - 2 = 0"}, {"rel": "联立", "source": "m = 2", "target": "x = - 2"}, {"rel": "联立", "source": "n = 2", "target": "x = - 2"}, {"rel": "联立", "source": "2 m + x = n", "target": "x = - 2"}]}}
|
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{"content": "Given $( 4 + \\sqrt { 7 }) \\times a = b$, if $b$ is an integer, then the possible value(s) of $a$ is/are ____? ", "answer": "- 3", "steps": "$\\because$ $( 4 + \\sqrt { 7 }) * ( 4 - \\sqrt { 7 }) = 16 - 7 = 9$, which satisfies the condition, $\\therefore$ $a = 4 - \\sqrt { 7 }$.", "expr_cands": ["a", "b", "c", "d", "2 ( a + b ) - 3 ( - cd ) ^ { 100 }", "a + b = 0", "cd = 1", "0 - 3", "- 3"], "exprs": ["a + b = 0", "cd = 1", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "c"}, {"id": "cd = 1"}, {"id": "d"}, {"id": "$c$ , $d$ 互为倒数"}, {"id": "2 ( a + b ) - 3 ( - cd ) ^ { 100 }"}, {"id": "- 3"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "- 3"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "c", "target": "cd = 1"}, {"rel": "代入", "source": "cd = 1", "target": "- 3"}, {"rel": "被描述", "source": "d", "target": "cd = 1"}, {"rel": "限制性描述", "source": "$c$ , $d$ 互为倒数", "target": "cd = 1"}, {"rel": "被代入", "source": "2 ( a + b ) - 3 ( - cd ) ^ { 100 }", "target": "- 3"}]}}
|
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{"content": "Given $a + b = 4$ and $ab = 3$, what is the value of the algebraic expression $( a + 2 ) ( b + 2 )$?", "answer": "\\frac { 3 } { 2 }", "steps": "Since $a + b = 4$ and $ab = 3$, it follows that $( a + 2 ) ( b + 2 ) = ab + 2 ( a + b ) + 4 = 3 + 2 * 4 + 4 = 15$.", "expr_cands": ["m", "n", "x ^ { 2 } + 2 x - 3 = 0", "x", "\\frac { m ^ { 2 } n - mn ^ { 2 } } { m ^ { 2 } - n ^ { 2 } }", "x = - 3", "x = 1", "m + n = - 2", "mn = - 3", "\\frac { mn ( m - n ) } { ( m + n ) ( m - n ) }", "\\frac { 3 } { 2 }"], "exprs": ["m + n = - 2", "mn = - 3", "\\frac { mn ( m - n ) } { ( m + n ) ( m - n ) }", "\\frac { 3 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m"}, {"id": "m + n = - 2"}, {"id": "n"}, {"id": "x ^ { 2 } + 2 x - 3 = 0"}, {"id": ", $m$ , $n$ 是一元二次方程 $x ^ { 2 } + 2 x - 3 = 0$ 的两根"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "mn = - 3"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "\\frac { m ^ { 2 } n - mn ^ { 2 } } { m ^ { 2 } - n ^ { 2 } }"}, {"id": "\\frac { mn ( m - n ) } { ( m + n ) ( m - n ) }"}, {"id": "\\frac { 3 } { 2 }"}], "links": [{"rel": "被描述", "source": "m", "target": "m + n = - 2"}, {"rel": "被描述", "source": "m", "target": "mn = - 3"}, {"rel": "提取因式参考", "source": "m + n = - 2", "target": "\\frac { mn ( m - n ) } { ( m + n ) ( m - n ) }"}, {"rel": "代入", "source": "m + n = - 2", "target": "\\frac { 3 } { 2 }"}, {"rel": "被描述", "source": "n", "target": "m + n = - 2"}, {"rel": "被描述", "source": "n", "target": "mn = - 3"}, {"rel": "被描述", "source": "x ^ { 2 } + 2 x - 3 = 0", "target": "m + n = - 2"}, {"rel": "被描述", "source": "x ^ { 2 } + 2 x - 3 = 0", "target": "mn = - 3"}, {"rel": "限制性描述", "source": ", $m$ , $n$ 是一元二次方程 $x ^ { 2 } + 2 x - 3 = 0$ 的两根", "target": "m + n = - 2"}, {"rel": "限制性描述", "source": ", $m$ , $n$ 是一元二次方程 $x ^ { 2 } + 2 x - 3 = 0$ 的两根", "target": "mn = - 3"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "m + n = - 2"}, {"rel": "提取因式参考", "source": "mn = - 3", "target": "\\frac { mn ( m - n ) } { ( m + n ) ( m - n ) }"}, {"rel": "代入", "source": "mn = - 3", "target": "\\frac { 3 } { 2 }"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "mn = - 3"}, {"rel": "提取因式", "source": "\\frac { m ^ { 2 } n - mn ^ { 2 } } { m ^ { 2 } - n ^ { 2 } }", "target": "\\frac { mn ( m - n ) } { ( m + n ) ( m - n ) }"}, {"rel": "被代入", "source": "\\frac { mn ( m - n ) } { ( m + n ) ( m - n ) }", "target": "\\frac { 3 } { 2 }"}]}}
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{"content": "Given $| m - 2 | + { ( n - 2 ) } ^ { 2 } = 0$, what is the solution to the equation $2 m + x = n$ in terms of $x$?", "answer": "0", "steps": "According to the problem, we have $m - 2 = 0$ and $n - 2 = 0$. Solving for $m$ and $n$, we get $m = 2$ and $n = 2$. Substituting these values into the equation, we get $x = - 2$.", "expr_cands": ["x", "- 1", "| - x - 1 |", "x = - 1", "0"], "exprs": ["x = - 1", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 1"}, {"id": "x = - 1"}, {"id": "x"}, {"id": "$x$ 的倒数是 $- 1$"}, {"id": "| - x - 1 |"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "- 1", "target": "x = - 1"}, {"rel": "代入", "source": "x = - 1", "target": "0"}, {"rel": "被描述", "source": "x", "target": "x = - 1"}, {"rel": "限制性描述", "source": "$x$ 的倒数是 $- 1$", "target": "x = - 1"}, {"rel": "被代入", "source": "| - x - 1 |", "target": "0"}]}}
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{"content": "Given $a$ and $b$ are opposite numbers, and $c$ and $d$ are reciprocal numbers, what is the value of $2 ( a + b ) - 3 ( - cd ) ^ { 100 }$?", "answer": "x \\ge 1", "steps": "According to the problem, we have $a + b = 0$ and $cd = 1$. Therefore, the original expression is equal to $0 - 3 = - 3$.", "expr_cands": ["y = \\sqrt { x - 1 }", "y", "x", "x - 1 \\ge 0", "1 \\le x", "x \\ge 1"], "exprs": ["x - 1 \\ge 0", "x \\ge 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 1 }"}, {"id": "x - 1 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 1"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 1 }", "target": "x - 1 \\ge 0"}, {"rel": "不等式方程求解", "source": "x - 1 \\ge 0", "target": "x \\ge 1"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 1 \\ge 0"}]}}
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{"content": "Let $m$ and $n$ be the two roots of the quadratic equation $x ^ 2 + 2 x - 3 = 0$. Find the value of $\\frac { m ^ 2 n - mn ^ 2 } { m ^ 2 - n ^ 2 }$.", "answer": "2", "steps": "Since $m$ and $n$ are the two roots of the quadratic equation $x ^ 2 + 2 x - 3 = 0$, we have $m + n = - 2$ and $mn = - 3$. Therefore, $\\frac { m ^ 2 n - mn ^ 2 } { m ^ 2 - n ^ 2 } = \\frac { mn ( m - n )} {( m + n ) ( m - n )} = \\frac { mn } { m + n } = \\frac { - 3 } { - 2 } = \\frac { 3 } { 2 }$.", "expr_cands": ["x", "- k - x + 6 > 0", "k", "1", "2", "3", "x < 6 - k", "6 - k = 4", "k = 2"], "exprs": ["x < 6 - k", "6 - k = 4", "k = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- k - x + 6 > 0"}, {"id": "x < 6 - k"}, {"id": "1"}, {"id": "6 - k = 4"}, {"id": "2"}, {"id": "3"}, {"id": "关于 $x$ 的不等式 $- k - x + 6 > 0$ 的正整数解为 $1$ , $2$ , $3$"}, {"id": "k = 2"}], "links": [{"rel": "不等式方程部分求解", "source": "- k - x + 6 > 0", "target": "x < 6 - k"}, {"rel": "被描述", "source": "x < 6 - k", "target": "6 - k = 4"}, {"rel": "被描述", "source": "1", "target": "6 - k = 4"}, {"rel": "等式方程求解", "source": "6 - k = 4", "target": "k = 2"}, {"rel": "被描述", "source": "2", "target": "6 - k = 4"}, {"rel": "被描述", "source": "3", "target": "6 - k = 4"}, {"rel": "限制性描述", "source": "关于 $x$ 的不等式 $- k - x + 6 > 0$ 的正整数解为 $1$ , $2$ , $3$", "target": "6 - k = 4"}]}}
|
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{"content": "If the reciprocal of $x$ is $- 1$, then $| - x - 1 |$ = ____?", "answer": "3", "steps": "The reciprocal of $x$ is $- 1$, and $x = - 1$. $| - x - 1 | = | 1 - 1 | = 0$.", "expr_cands": ["y = 2 { x } ^ { m - 2 } + 3", "m", "y", "x", "m - 2 = 1", "m = 3"], "exprs": ["m - 2 = 1", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 { x } ^ { m - 2 } + 3"}, {"id": "m - 2 = 1"}, {"id": "$y = 2 { x } ^ { m - 2 } + 3$ 是一次函数"}, {"id": "m = 3"}], "links": [{"rel": "被描述", "source": "y = 2 { x } ^ { m - 2 } + 3", "target": "m - 2 = 1"}, {"rel": "等式方程求解", "source": "m - 2 = 1", "target": "m = 3"}, {"rel": "限制性描述", "source": "$y = 2 { x } ^ { m - 2 } + 3$ 是一次函数", "target": "m - 2 = 1"}]}}
|
||||
{"content": "The domain of the function $y = \\sqrt { x - 1 }$ is ____ ?", "answer": "0", "steps": "From the given information, we have $x - 1 \\ge 0$ and $x \\ge 1$.", "expr_cands": ["y = ( m - 2 ) x ^ { | m - 1 | } + m - 4", "m", "y", "x", "| m - 1 | = 1", "m = 0", "m = 2", "m - 2 \\neq 0", "m \\neq 2"], "exprs": ["| m - 1 | = 1", "m - 2 \\neq 0", "m = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m - 2 ) x ^ { | m - 1 | } + m - 4"}, {"id": "| m - 1 | = 1"}, {"id": "$y = ( m - 2 ) x ^ { | m - 1 | } + m - 4$ 为一次函数"}, {"id": "m - 2 \\neq 0"}, {"id": "m = 0"}], "links": [{"rel": "被描述", "source": "y = ( m - 2 ) x ^ { | m - 1 | } + m - 4", "target": "| m - 1 | = 1"}, {"rel": "被描述", "source": "y = ( m - 2 ) x ^ { | m - 1 | } + m - 4", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "| m - 1 | = 1", "target": "m = 0"}, {"rel": "限制性描述", "source": "$y = ( m - 2 ) x ^ { | m - 1 | } + m - 4$ 为一次函数", "target": "| m - 1 | = 1"}, {"rel": "限制性描述", "source": "$y = ( m - 2 ) x ^ { | m - 1 | } + m - 4$ 为一次函数", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "m - 2 \\neq 0", "target": "m = 0"}]}}
|
||||
{"content": "If the positive integer solutions of the inequality $- k - x + 6 > 0$ with respect to $x$ are $1$, $2$, and $3$, then the value of the positive integer $k$ is ____?", "answer": "0", "steps": "The inequality $- k - x + 6 > 0$ is solved as $x < 6 - k$. Since the positive integer solutions of the inequality are $1$, $2$, and $3$, and $k$ is a positive integer, we have $6 - k = 4$, which implies $k = 2$.", "expr_cands": ["6 x ^ { m + 1 } + 3 = 0", "m", "x", "m + 1 = 1", "m = 0"], "exprs": ["m + 1 = 1", "m = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "6 x ^ { m + 1 } + 3 = 0"}, {"id": "m + 1 = 1"}, {"id": "$6 x ^ { m + 1 } + 3 = 0$ 是一元一次方程"}, {"id": "m = 0"}], "links": [{"rel": "被描述", "source": "6 x ^ { m + 1 } + 3 = 0", "target": "m + 1 = 1"}, {"rel": "等式方程求解", "source": "m + 1 = 1", "target": "m = 0"}, {"rel": "限制性描述", "source": "$6 x ^ { m + 1 } + 3 = 0$ 是一元一次方程", "target": "m + 1 = 1"}]}}
|
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{"content": "Given $y = 2 { x } ^ { m - 2 } + 3$ is a linear function, what is the value of $m$?", "answer": "- 6", "steps": "From the given condition, we have $m - 2 = 1$, solving for $m$ gives $m = 3$.", "expr_cands": ["x ^ { 2 } - 5 x + m", "m", "x", "( x + 1 )", "x + a", "a", "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m", "a + 1 = - 5", "a = - 6", "a = m", "m = - 6"], "exprs": ["x + a", "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m", "a + 1 = - 5", "a = m", "a = - 6", "m = - 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "题意设另一个多项式为 $x + a$"}, {"id": "x + a"}, {"id": "( x + 1 )"}, {"id": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m"}, {"id": "x ^ { 2 } - 5 x + m"}, {"id": "将 $x ^ { 2 } - 5 x + m$ 分解因式后有一个因式是 $( x + 1 )$"}, {"id": "a + 1 = - 5"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "a = m"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "a = - 6"}, {"id": "m = - 6"}], "links": [{"rel": "假设描述", "source": "题意设另一个多项式为 $x + a$", "target": "x + a"}, {"rel": "限制性描述", "source": "题意设另一个多项式为 $x + a$", "target": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m"}, {"rel": "被描述", "source": "x + a", "target": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m"}, {"rel": "被描述", "source": "( x + 1 )", "target": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m"}, {"rel": "被描述", "source": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m", "target": "a + 1 = - 5"}, {"rel": "被描述", "source": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m", "target": "a = m"}, {"rel": "被描述", "source": "x ^ { 2 } - 5 x + m", "target": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m"}, {"rel": "限制性描述", "source": "将 $x ^ { 2 } - 5 x + m$ 分解因式后有一个因式是 $( x + 1 )$", "target": "( x + 1 ) ( x + a ) = x ^ { 2 } - 5 x + m"}, {"rel": "等式方程求解", "source": "a + 1 = - 5", "target": "a = - 6"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "a + 1 = - 5"}, {"rel": "被代入", "source": "a = m", "target": "m = - 6"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "a = m"}, {"rel": "代入", "source": "a = - 6", "target": "m = - 6"}]}}
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{"content": "$y = ( m - 2 ) x ^ { | m - 1 | } + m - 4$ is a linear function, then $m$ = ____ ?", "answer": "1", "steps": "From the given condition, we have $| m - 1 | = 1$ and $m - 2 \\neq 0$. Therefore, $m = 2$ or $m = 0$, but $m \\neq 2$. Hence, $m = 0$.", "expr_cands": ["x = 12", "x", "| \\frac { 1 } { 4 } x - 2 | = b", "b", "| \\frac { 1 } { 4 } * 12 - 2 | = b", "b = 1"], "exprs": ["| \\frac { 1 } { 4 } * 12 - 2 | = b", "b = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 12"}, {"id": "| \\frac { 1 } { 4 } * 12 - 2 | = b"}, {"id": "| \\frac { 1 } { 4 } x - 2 | = b"}, {"id": "b = 1"}], "links": [{"rel": "代入", "source": "x = 12", "target": "| \\frac { 1 } { 4 } * 12 - 2 | = b"}, {"rel": "等式方程求解", "source": "| \\frac { 1 } { 4 } * 12 - 2 | = b", "target": "b = 1"}, {"rel": "被代入", "source": "| \\frac { 1 } { 4 } x - 2 | = b", "target": "| \\frac { 1 } { 4 } * 12 - 2 | = b"}]}}
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{"content": "If $6 x ^ { m + 1 } + 3 = 0$ is a linear equation, then $m$ = ____ ?", "answer": "- 2", "steps": "Since $6 x ^ { m + 1 } + 3 = 0$ is a linear equation with one variable, therefore $m + 1 = 1$, which solves to $m = 0$.", "expr_cands": ["x = 1", "x", "x ^ { 2 } + ax + 2 b = 0", "a", "b", "2 a + 4 b", "1 + a + 2 b = 0", "a + 2 b = - 1", "2 ( a + 2 b )", "- 2"], "exprs": ["1 + a + 2 b = 0", "a + 2 b = - 1", "2 ( a + 2 b )", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 1"}, {"id": "1 + a + 2 b = 0"}, {"id": "x ^ { 2 } + ax + 2 b = 0"}, {"id": "a + 2 b = - 1"}, {"id": "2 a + 4 b"}, {"id": "2 ( a + 2 b )"}, {"id": "- 2"}], "links": [{"rel": "代入", "source": "x = 1", "target": "1 + a + 2 b = 0"}, {"rel": "移项", "source": "1 + a + 2 b = 0", "target": "a + 2 b = - 1"}, {"rel": "被代入", "source": "x ^ { 2 } + ax + 2 b = 0", "target": "1 + a + 2 b = 0"}, {"rel": "提取因式参考", "source": "a + 2 b = - 1", "target": "2 ( a + 2 b )"}, {"rel": "代入", "source": "a + 2 b = - 1", "target": "- 2"}, {"rel": "提取因式", "source": "2 a + 4 b", "target": "2 ( a + 2 b )"}, {"rel": "被代入", "source": "2 ( a + 2 b )", "target": "- 2"}]}}
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{"content": "If $x ^ 2 - 5 x + m$ can be factored and one of the factors is $( x + 1 )$, then the value of $m$ is ____?", "answer": "35", "steps": "Let the other polynomial be $x + a$. Then $( x + 1 ) ( x + a ) = x ^ 2 + ( a + 1 ) x + a = x ^ 2 - 5 x + m$. Therefore, $a + 1 = - 5$ and $a = m$, which gives $m = - 6$.", "expr_cands": ["- x + 2 y = 5", "y", "x", "4 ( x - 2 y ) ^ { 2 } - 3 ( 2 y - x ) - 50", "x - 2 y = - 5", "2 y - x = 5", "4 * ( - 5 ) ^ { 2 } - 3 * 5 - 50", "35"], "exprs": ["x - 2 y = - 5", "2 y - x = 5", "35"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- x + 2 y = 5"}, {"id": "x - 2 y = - 5"}, {"id": "2 y - x = 5"}, {"id": "4 ( x - 2 y ) ^ { 2 } - 3 ( 2 y - x ) - 50"}, {"id": "35"}], "links": [{"rel": "同乘除", "source": "- x + 2 y = 5", "target": "x - 2 y = - 5"}, {"rel": "同乘除", "source": "x - 2 y = - 5", "target": "2 y - x = 5"}, {"rel": "代入", "source": "x - 2 y = - 5", "target": "35"}, {"rel": "代入", "source": "2 y - x = 5", "target": "35"}, {"rel": "被代入", "source": "4 ( x - 2 y ) ^ { 2 } - 3 ( 2 y - x ) - 50", "target": "35"}]}}
|
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{"content": "$x = 12$ is a solution to the equation $| \\frac { 1 } { 4 } x - 2 | = b$. What is the value of $b$?", "answer": "- 1", "steps": "From the given problem, we have $| \\frac { 1 } { 4 } \\times 12 - 2 | = b$, solving which gives $b = 1$.", "expr_cands": ["x + 2 y - 3 = 0", "y", "x", "2 x + 4 y - 7", "x + 2 y = 3", "2 ( x + 2 y ) - 7", "- 1"], "exprs": ["x + 2 y = 3", "2 ( x + 2 y ) - 7", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 2 y - 3 = 0"}, {"id": "x + 2 y = 3"}, {"id": "2 x + 4 y - 7"}, {"id": "2 ( x + 2 y ) - 7"}, {"id": "- 1"}], "links": [{"rel": "移项", "source": "x + 2 y - 3 = 0", "target": "x + 2 y = 3"}, {"rel": "提取因式参考", "source": "x + 2 y = 3", "target": "2 ( x + 2 y ) - 7"}, {"rel": "代入", "source": "x + 2 y = 3", "target": "- 1"}, {"rel": "提取因式", "source": "2 x + 4 y - 7", "target": "2 ( x + 2 y ) - 7"}, {"rel": "被代入", "source": "2 ( x + 2 y ) - 7", "target": "- 1"}]}}
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{"content": "$x = 1$ is a solution to the equation $x ^ 2 + ax + 2 b = 0$ in terms of $x$, then $2 a + 4 b$ = ____ ?", "answer": "x \\le 2", "steps": "Substituting $x = 1$ into the equation yields $1 + a + 2 b = 0$, which means $a + 2 b = - 1$. Therefore, $2 a + 4 b = 2 ( a + 2 b ) = - 2$.", "expr_cands": ["4 x - 6 \\ge 7 x - 12", "x", "4 x - 7 x \\ge - 12 + 6", "x \\le 2", "- 3 x \\ge - 6", "1"], "exprs": ["x \\le 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x - 6 \\ge 7 x - 12"}, {"id": "x \\le 2"}], "links": [{"rel": "不等式方程求解", "source": "4 x - 6 \\ge 7 x - 12", "target": "x \\le 2"}]}}
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{"content": "Given $- x + 2 y = 5$, what is the value of $4 ( x - 2 y ) ^ 2 - 3 ( 2 y - x ) - 50$?", "answer": "2 : 1", "steps": "Since $- x + 2 y = 5$, therefore $x - 2 y = - 5$. Substituting $x - 2 y = - 5$ and $2 y - x = 5$ into the original expression, we get the original expression $= 4 * ( - 5 ) ^ 2 - 3 * 5 - 50 = 35$.", "expr_cands": ["x = 2 y", "y", "x", "x : y", "2"], "exprs": ["2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x : y"}, {"id": "2"}, {"id": "x = 2 y"}], "links": [{"rel": "被代入", "source": "x : y", "target": "2"}, {"rel": "代入", "source": "x = 2 y", "target": "2"}]}}
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{"content": "Given $x + 2 y - 3 = 0$, what is the value of the algebraic expression $2 x + 4 y - 7$?", "answer": "9", "steps": "Because $x + 2 y - 3 = 0$, therefore $x + 2 y = 3$. Then the original expression is equal to $2 ( x + 2 y ) - 7 = 2 \\times 3 - 7 = 6 - 7 = - 1$.", "expr_cands": ["x", "2 a - 1", "a", "- a + 2", "2 a - 1 - a + 2 = 0", "a = - 1", "- 3", "x = 9"], "exprs": ["2 a - 1 - a + 2 = 0", "a = - 1", "- 3", "x = 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a - 1"}, {"id": "2 a - 1 - a + 2 = 0"}, {"id": "- a + 2"}, {"id": "一个正数 $x$ 的平方根是 $2 a - 1$ 和 $- a + 2$"}, {"id": "平方根互为相反数"}, {"id": "a = - 1"}, {"id": "- 3"}, {"id": "x"}, {"id": "x = 9"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "2 a - 1", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "被代入", "source": "2 a - 1", "target": "- 3"}, {"rel": "等式方程求解", "source": "2 a - 1 - a + 2 = 0", "target": "a = - 1"}, {"rel": "被描述", "source": "- a + 2", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "限制性描述", "source": "一个正数 $x$ 的平方根是 $2 a - 1$ 和 $- a + 2$", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "代入", "source": "a = - 1", "target": "- 3"}, {"rel": "被描述", "source": "- 3", "target": "x = 9"}, {"rel": "被描述", "source": "x", "target": "x = 9"}, {"rel": "限制性描述", "source": "平方", "target": "x = 9"}]}}
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{"content": "The solution set of the inequality $4 x - 6 \\ge 7 x - 12$ is _____.", "answer": "3", "steps": "Moving terms, we get $4 x - 7 x \\ge - 12 + 6$. Combining like terms, we get $- 3 x \\ge - 6$. Dividing by $- 3$ (and flipping the inequality sign), we get $x \\le 2$.", "expr_cands": ["x", "\\frac { m } { x - 1 } + \\frac { 3 } { 1 - x } = 1", "m", "m - 3 = x - 1", "x = m - 2", "x - 1 = 0", "x = 1", "m = 3", "3"], "exprs": ["m - 3 = x - 1", "x - 1 = 0", "x = m - 2", "x = 1", "m = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { m } { x - 1 } + \\frac { 3 } { 1 - x } = 1"}, {"id": "m - 3 = x - 1"}, {"id": "x = m - 2"}, {"id": "x - 1 = 0"}, {"id": "关于 $x$ 的分式方程 $\\frac { m } { x - 1 } + \\frac { 3 } { 1 - x } = 1$ 无解"}, {"id": "分式方程无解，则分母为0"}, {"id": "x = 1"}, {"id": "m = 3"}], "links": [{"rel": "同乘除", "source": "\\frac { m } { x - 1 } + \\frac { 3 } { 1 - x } = 1", "target": "m - 3 = x - 1"}, {"rel": "被描述", "source": "\\frac { m } { x - 1 } + \\frac { 3 } { 1 - x } = 1", "target": "x - 1 = 0"}, {"rel": "移项", "source": "m - 3 = x - 1", "target": "x = m - 2"}, {"rel": "联立", "source": "x = m - 2", "target": "m = 3"}, {"rel": "等式方程求解", "source": "x - 1 = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的分式方程 $\\frac { m } { x - 1 } + \\frac { 3 } { 1 - x } = 1$ 无解", "target": "x - 1 = 0"}, {"rel": "属性描述", "source": "分式方程无解，则分母为0", "target": "x - 1 = 0"}, {"rel": "联立", "source": "x = 1", "target": "m = 3"}]}}
|
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{"content": "If $x = 2 y$, then the value of $x : y$ is ____?", "answer": "3", "steps": "Because $x$ is equal to $2$ times $y$, therefore the ratio of $x$ to $y$ is $2$ to $1$.", "expr_cands": ["x ^ { n - 1 } \\cdot x ^ { n + 5 } = x ^ { 10 }", "x", "n", "{ x } ^ { n - 1 } \\cdot { x } ^ { n + 5 } = { x } ^ { 10 }", "{ x } ^ { n - 1 + n + 5 } = { x } ^ { 10 }", "{ x } ^ { 2 n + 4 } = { x } ^ { 10 }", "2 n + 4 = 10", "n = 3"], "exprs": ["{ x } ^ { 2 n + 4 } = { x } ^ { 10 }", "2 n + 4 = 10", "n = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { n - 1 } \\cdot x ^ { n + 5 } = x ^ { 10 }"}, {"id": "{ x } ^ { 2 n + 4 } = { x } ^ { 10 }"}, {"id": "2 n + 4 = 10"}, {"id": "n = 3"}], "links": [{"rel": "计算", "source": "x ^ { n - 1 } \\cdot x ^ { n + 5 } = x ^ { 10 }", "target": "{ x } ^ { 2 n + 4 } = { x } ^ { 10 }"}, {"rel": "同取对数", "source": "{ x } ^ { 2 n + 4 } = { x } ^ { 10 }", "target": "2 n + 4 = 10"}, {"rel": "等式方程求解", "source": "2 n + 4 = 10", "target": "n = 3"}]}}
|
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{"content": "If the square root of a positive number $x$ is $2 a - 1$ and $- a + 2$, find the value of $x$. ____?", "answer": "- 20", "steps": "$\\because$ The square roots of a positive number $x$ are $2 a - 1$ and $- a + 2$, $\\therefore$ $2 a - 1 - a + 2 = 0$. Solving for $a$, we get $a = - 1$. $\\therefore$ $2 a - 1 = - 3$. $\\therefore$ This positive number is $x = ( - 3 ) ^ 2 = 9$.", "expr_cands": ["a - b = 4", "a", "b", "\\frac { 1 } { 4 } { ( a - b ) } ^ { 2 } - 9 ( a - b ) - \\frac { 1 } { 2 } { ( a - b ) } ^ { 2 } - 5 ( b - a )", "- \\frac { 1 } { 4 } { ( a - b ) } ^ { 2 } - 4 ( a - b )", "- \\frac { 1 } { 4 } * { 4 } ^ { 2 } - 4 * 4", "- 20"], "exprs": ["- 20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - b = 4"}, {"id": "- 20"}, {"id": "\\frac { 1 } { 4 } { ( a - b ) } ^ { 2 } - 9 ( a - b ) - \\frac { 1 } { 2 } { ( a - b ) } ^ { 2 } - 5 ( b - a )"}], "links": [{"rel": "代入", "source": "a - b = 4", "target": "- 20"}, {"rel": "被代入", "source": "\\frac { 1 } { 4 } { ( a - b ) } ^ { 2 } - 9 ( a - b ) - \\frac { 1 } { 2 } { ( a - b ) } ^ { 2 } - 5 ( b - a )", "target": "- 20"}]}}
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{"content": "If the fractional equation about $x$, $\\frac { m } { x - 1 } + \\frac { 3 } { 1 - x } = 1$, has no solution, then the value of $m$ is ____?", "answer": "3", "steps": "Going to the denominator, we get $m - 3 = x - 1$, so $x = m - 2$. Since the fractional equation in terms of $x$ has no solution, the simplest common denominator is $x - 1 = 0$, so $x = 1$. When $x = 1$, we get $m = 3$, which means the value of $m$ is 3.", "expr_cands": ["m - n = - 1", "m", "n", "( m - n ) ^ { 2 } - 2 m + 2 n", "( m - n ) ^ { 2 } - 2 ( m - n )", "3"], "exprs": ["( m - n ) ^ { 2 } - 2 ( m - n )", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - n ) ^ { 2 } - 2 m + 2 n"}, {"id": "( m - n ) ^ { 2 } - 2 ( m - n )"}, {"id": "m - n = - 1"}, {"id": "3"}], "links": [{"rel": "提取因式", "source": "( m - n ) ^ { 2 } - 2 m + 2 n", "target": "( m - n ) ^ { 2 } - 2 ( m - n )"}, {"rel": "被代入", "source": "( m - n ) ^ { 2 } - 2 ( m - n )", "target": "3"}, {"rel": "提取因式参考", "source": "m - n = - 1", "target": "( m - n ) ^ { 2 } - 2 ( m - n )"}, {"rel": "代入", "source": "m - n = - 1", "target": "3"}]}}
|
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{"content": "If $x ^ { n - 1 } \\cdot x ^ { n + 5 } = x ^ { 10 }$, then $n$ = ____ ?", "answer": "- 4", "steps": "Since ${ x } ^ { n - 1 } \\cdot { x } ^ { n + 5 } = { x } ^ { 10 }$, it follows that ${ x } ^ { n - 1 + n + 5 } = { x } ^ { 10 }$. Therefore, ${ x } ^ { 2 n + 4 } = { x } ^ { 10 }$, which implies that $2 n + 4 = 10$. Solving for $n$, we get $n = 3$.", "expr_cands": ["3 - m", "m", "2 m + 1", "3 - m = - ( 2 m + 1 )", "m = - 4", "3 - m = - 2 m - 1"], "exprs": ["3 - m = - ( 2 m + 1 )", "m = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 m + 1"}, {"id": "3 - m = - ( 2 m + 1 )"}, {"id": "3 - m"}, {"id": "$3 - m$ 与 $2 m + 1$ 互为相反数"}, {"id": "根据 $3 - m$ 与 $2 m + 1$ 互为相反数"}, {"id": "m = - 4"}], "links": [{"rel": "被描述", "source": "2 m + 1", "target": "3 - m = - ( 2 m + 1 )"}, {"rel": "等式方程求解", "source": "3 - m = - ( 2 m + 1 )", "target": "m = - 4"}, {"rel": "被描述", "source": "3 - m", "target": "3 - m = - ( 2 m + 1 )"}, {"rel": "限制性描述", "source": "$3 - m$ 与 $2 m + 1$ 互为相反数", "target": "3 - m = - ( 2 m + 1 )"}, {"rel": "限制性描述", "source": "根据 $3 - m$ 与 $2 m + 1$ 互为相反数", "target": "3 - m = - ( 2 m + 1 )"}]}}
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||||
{"content": "Given $a - b = 4$, what is the value of the polynomial $\\frac { 1 } { 4 } { ( a - b ) } ^ { 2 } - 9 ( a - b ) - \\frac { 1 } { 2 } { ( a - b ) } ^ { 2 } - 5 ( b - a )$?", "answer": "- 5", "steps": "$\\frac { 1 } { 4 } { ( a - b ) } ^ { 2 } - 9 ( a - b ) - \\frac { 1 } { 2 } { ( a - b ) } ^ { 2 } - 5 ( b - a ) = - \\frac { 1 } { 4 } { ( a - b ) } ^ { 2 } - 4 ( a - b )$, when $a - b = 4$, the original expression equals $- \\frac { 1 } { 4 } * { 4 } ^ { 2 } - 4 * 4 = - 20$.", "expr_cands": ["2 ( a + 3 )", "a", "4", "2 ( a + 3 ) + 4 = 0", "a = - 5"], "exprs": ["2 ( a + 3 ) + 4 = 0", "a = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ( a + 3 )"}, {"id": "2 ( a + 3 ) + 4 = 0"}, {"id": "$2 ( a + 3 )$ 的值与 $4$ 互为相反数"}, {"id": "a = - 5"}], "links": [{"rel": "被描述", "source": "2 ( a + 3 )", "target": "2 ( a + 3 ) + 4 = 0"}, {"rel": "等式方程求解", "source": "2 ( a + 3 ) + 4 = 0", "target": "a = - 5"}, {"rel": "限制性描述", "source": "$2 ( a + 3 )$ 的值与 $4$ 互为相反数", "target": "2 ( a + 3 ) + 4 = 0"}]}}
|
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{"content": "If $m - n = - 1$, then $( m - n ) ^ 2 - 2 m + 2 n$ = ____ ?", "answer": "1", "steps": "Because $m - n = - 1$, therefore $( m - n ) ^ 2 - 2 m + 2 n = ( m - n ) ^ 2 - 2 ( m - n ) = ( - 1 ) ^ 2 - 2 * ( - 1 ) = 1 + 2 = 3$.", "expr_cands": ["a", "b", "m", "n", "( a + b ) ^ { 2015 } + ( m \\times n ) ^ { 2016 }", "a + b = 0", "mn = 1", "0 + 1", "1"], "exprs": ["a + b = 0", "mn = 1", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 是互为相反数"}, {"id": "m"}, {"id": "mn = 1"}, {"id": "n"}, {"id": "$m$ , $n$ 是互为倒数"}, {"id": "( a + b ) ^ { 2015 } + ( m \\times n ) ^ { 2016 }"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "1"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 是互为相反数", "target": "a + b = 0"}, {"rel": "被描述", "source": "m", "target": "mn = 1"}, {"rel": "代入", "source": "mn = 1", "target": "1"}, {"rel": "被描述", "source": "n", "target": "mn = 1"}, {"rel": "限制性描述", "source": "$m$ , $n$ 是互为倒数", "target": "mn = 1"}, {"rel": "被代入", "source": "( a + b ) ^ { 2015 } + ( m \\times n ) ^ { 2016 }", "target": "1"}]}}
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||||
{"content": "If $3 - m$ is the opposite of $2 m + 1$, then $m$ = ____ ?", "answer": "x \\le 2", "steps": "According to the fact that $3 - m$ and $2 m + 1$ are opposite numbers, we can write the equation $3 - m = - ( 2 m + 1 )$. Expanding the brackets, we get $3 - m = - 2 m - 1$. Rearranging and simplifying, we get $m = - 4$.", "expr_cands": ["| x - 2 | = 2 - x", "x", "x - 2 \\le 0", "x \\le 2"], "exprs": ["x - 2 \\le 0", "x \\le 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x - 2 | = 2 - x"}, {"id": "x - 2 \\le 0"}, {"id": "绝对值恒大于等于0"}, {"id": "x \\le 2"}], "links": [{"rel": "被描述", "source": "| x - 2 | = 2 - x", "target": "x - 2 \\le 0"}, {"rel": "不等式方程求解", "source": "x - 2 \\le 0", "target": "x \\le 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x - 2 \\le 0"}]}}
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{"content": "If the value of $2 ( a + 3 )$ is the opposite of $4$, then the value of $a$ is ____?", "answer": "- 1", "steps": "From the given problem, we have $2 ( a + 3 ) + 4 = 0$, which yields $a = - 5$ as the solution.", "expr_cands": ["( x + 2 ) ( x - 1 ) = x ^ { 2 } + mx + n", "n", "m", "x", "m + n", "x ^ { 2 } + x - 2 = x ^ { 2 } + mx + n", "m = 1", "n = - 2", "- 1"], "exprs": ["x ^ { 2 } + x - 2 = x ^ { 2 } + mx + n", "m = 1", "n = - 2", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + 2 ) ( x - 1 ) = x ^ { 2 } + mx + n"}, {"id": "x ^ { 2 } + x - 2 = x ^ { 2 } + mx + n"}, {"id": "m = 1"}, {"id": "n = - 2"}, {"id": "m + n"}, {"id": "- 1"}], "links": [{"rel": "展开", "source": "( x + 2 ) ( x - 1 ) = x ^ { 2 } + mx + n", "target": "x ^ { 2 } + x - 2 = x ^ { 2 } + mx + n"}, {"rel": "移项", "source": "x ^ { 2 } + x - 2 = x ^ { 2 } + mx + n", "target": "m = 1"}, {"rel": "移项", "source": "x ^ { 2 } + x - 2 = x ^ { 2 } + mx + n", "target": "n = - 2"}, {"rel": "代入", "source": "m = 1", "target": "- 1"}, {"rel": "代入", "source": "n = - 2", "target": "- 1"}, {"rel": "被代入", "source": "m + n", "target": "- 1"}]}}
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{"content": "If $a$ and $b$ are opposite numbers, and $m$ and $n$ are reciprocal numbers, then $( a + b ) ^ { 2015 } + ( m \\times n ) ^ { 2016 }$ = ____?", "answer": "\\frac { 1 } { 2 }", "steps": "From the given condition, we know that $a + b = 0$ and $mn = 1$. Therefore, the original expression is equal to $0 + 1 = 1$.", "expr_cands": ["a", "b", "\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ { n }", "m", "n", "ab", "mn", "2 m = 1", "m = \\frac { 1 } { 2 }", "n = 1", "mn = \\frac { 1 } { 2 }", "\\frac { 1 } { 2 }"], "exprs": ["2 m = 1", "n = 1", "m = \\frac { 1 } { 2 }", "\\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ { n }"}, {"id": "2 m = 1"}, {"id": "ab"}, {"id": "关于 $a$ , $b$ 的两个单项式 $\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ { n }$ 与 $ab$ 是同类项"}, {"id": "m = \\frac { 1 } { 2 }"}, {"id": "n = 1"}, {"id": "mn"}, {"id": "\\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ { n }", "target": "2 m = 1"}, {"rel": "被描述", "source": "\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ { n }", "target": "n = 1"}, {"rel": "等式方程求解", "source": "2 m = 1", "target": "m = \\frac { 1 } { 2 }"}, {"rel": "被描述", "source": "ab", "target": "2 m = 1"}, {"rel": "被描述", "source": "ab", "target": "n = 1"}, {"rel": "限制性描述", "source": "关于 $a$ , $b$ 的两个单项式 $\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ { n }$ 与 $ab$ 是同类项", "target": "2 m = 1"}, {"rel": "限制性描述", "source": "关于 $a$ , $b$ 的两个单项式 $\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ { n }$ 与 $ab$ 是同类项", "target": "n = 1"}, {"rel": "代入", "source": "m = \\frac { 1 } { 2 }", "target": "\\frac { 1 } { 2 }"}, {"rel": "代入", "source": "n = 1", "target": "\\frac { 1 } { 2 }"}, {"rel": "被代入", "source": "mn", "target": "\\frac { 1 } { 2 }"}]}}
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{"content": "If $| x - 2 | = 2 - x$, then the range of possible values for $x$ is ____?", "answer": "a < 1", "steps": "$\\because | x - 2 | = 2 - x$ , $\\therefore x - 2 \\le 0$ , $\\therefore x \\le 2$ , so the range of $x$ is $x \\le 2$.", "expr_cands": ["x + a > ax + 1", "a", "x", "x > 1", "1", "( - ax - a )", "( 1 - a ) x > 1 - a", "3", "1 - a > 0", "a < 1"], "exprs": ["( 1 - a ) x > 1 - a", "1 - a > 0", "a < 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + a > ax + 1"}, {"id": "( 1 - a ) x > 1 - a"}, {"id": "x > 1"}, {"id": "1 - a > 0"}, {"id": "a < 1"}], "links": [{"rel": "移项", "source": "x + a > ax + 1", "target": "( 1 - a ) x > 1 - a"}, {"rel": "联立", "source": "( 1 - a ) x > 1 - a", "target": "1 - a > 0"}, {"rel": "联立", "source": "x > 1", "target": "1 - a > 0"}, {"rel": "不等式方程求解", "source": "1 - a > 0", "target": "a < 1"}]}}
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{"content": "If $( x + 2 ) ( x - 1 ) = x ^ { 2 } + mx + n$, then the value of $m + n$ is ____?", "answer": "3", "steps": "Since $( x + 2 ) ( x - 1 ) = x ^ 2 + mx + n$, it follows that $x ^ 2 + x - 2 = x ^ 2 + mx + n$. Therefore, $m = 1$ and $n = - 2$. Thus, $m + n = 1 - 2 = - 1$.", "expr_cands": ["x", "- k ( x - 1 ) + 3 = 0", "k", "x = 2", "3 - k = 0", "- k ( 2 - 1 ) + 3 = 0", "k = 3"], "exprs": ["- k ( 2 - 1 ) + 3 = 0", "k = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- k ( x - 1 ) + 3 = 0"}, {"id": "- k ( 2 - 1 ) + 3 = 0"}, {"id": "x = 2"}, {"id": "k = 3"}], "links": [{"rel": "被代入", "source": "- k ( x - 1 ) + 3 = 0", "target": "- k ( 2 - 1 ) + 3 = 0"}, {"rel": "等式方程求解", "source": "- k ( 2 - 1 ) + 3 = 0", "target": "k = 3"}, {"rel": "代入", "source": "x = 2", "target": "- k ( 2 - 1 ) + 3 = 0"}]}}
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{"content": "If two monomials in terms of $a$ and $b$, $\\frac { 2 } { 5 } { a } ^ { 2 m } { b } ^ n$ and $ab$, are like terms, what is the value of $mn$?", "answer": "k < 1", "steps": "From the given information, we know that $2 m = 1$ and $n = 1$. Therefore, $mn = \\frac { 1 } { 2 } * 1 = \\frac { 1 } { 2 }$.", "expr_cands": ["y = ( k - 1 ) x - 1", "y", "k", "x", "k - 1 < 0", "k < 1"], "exprs": ["k - 1 < 0", "k < 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( k - 1 ) x - 1"}, {"id": "k - 1 < 0"}, {"id": "一次函数 $y = ( k - 1 ) x - 1$ 的函数值 $y$ 随着 $x$ 的增大而减小"}, {"id": "k < 1"}], "links": [{"rel": "被描述", "source": "y = ( k - 1 ) x - 1", "target": "k - 1 < 0"}, {"rel": "不等式方程求解", "source": "k - 1 < 0", "target": "k < 1"}, {"rel": "限制性描述", "source": "一次函数 $y = ( k - 1 ) x - 1$ 的函数值 $y$ 随着 $x$ 的增大而减小", "target": "k - 1 < 0"}]}}
|
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{"content": "If the solution set of $x + a > ax + 1$ is $x > 1$, then the range of values for $a$ is ____?", "answer": "- 6", "steps": "By the property $1$ of inequalities, adding $( - ax - a )$ to both sides yields: $( 1 - a ) x > 1 - a$, and the solution set is $x > 1$. Then, according to property $3$ of inequalities, we can obtain $1 - a > 0$, which means $a < 1$.", "expr_cands": ["( x + t ) ( x + 6 )", "t", "x", "x ^ { 2 } + ( t + 6 ) x + 6 t", "t + 6 = 0", "t = - 6"], "exprs": ["x ^ { 2 } + ( t + 6 ) x + 6 t", "t + 6 = 0", "t = - 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + t ) ( x + 6 )"}, {"id": "x ^ { 2 } + ( t + 6 ) x + 6 t"}, {"id": "x"}, {"id": "t + 6 = 0"}, {"id": "$( x + t ) ( x + 6 )$ 的结果中不含有 $x$ 的一次项"}, {"id": "t = - 6"}], "links": [{"rel": "提取因式", "source": "( x + t ) ( x + 6 )", "target": "x ^ { 2 } + ( t + 6 ) x + 6 t"}, {"rel": "被描述", "source": "x ^ { 2 } + ( t + 6 ) x + 6 t", "target": "t + 6 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "x ^ { 2 } + ( t + 6 ) x + 6 t"}, {"rel": "等式方程求解", "source": "t + 6 = 0", "target": "t = - 6"}, {"rel": "限制性描述", "source": "$( x + t ) ( x + 6 )$ 的结果中不含有 $x$ 的一次项", "target": "t + 6 = 0"}]}}
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{"content": "If the solution to the one-variable linear equation $- k ( x - 1 ) + 3 = 0$ is $x = 2$, then $k$ = ____?", "answer": "- 2", "steps": "Substituting $x = 2$ into the equation $- k ( x - 1 ) + 3 = 0$ yields $- k ( 2 - 1 ) + 3 = 0$, which can be solved to obtain $k = 3$.", "expr_cands": ["x", "9", "- 16", "9 x = x - 16", "x = - 2", "8 x = - 16"], "exprs": ["9 x = x - 16", "x = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "9"}, {"id": "9 x = x - 16"}, {"id": "x"}, {"id": "- 16"}, {"id": "$x$ 与 $9$ 的积等于 $x$ 与 $- 16$ 的和"}, {"id": "x = - 2"}], "links": [{"rel": "被描述", "source": "9", "target": "9 x = x - 16"}, {"rel": "等式方程求解", "source": "9 x = x - 16", "target": "x = - 2"}, {"rel": "被描述", "source": "x", "target": "9 x = x - 16"}, {"rel": "被描述", "source": "- 16", "target": "9 x = x - 16"}, {"rel": "限制性描述", "source": "$x$ 与 $9$ 的积等于 $x$ 与 $- 16$ 的和", "target": "9 x = x - 16"}]}}
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{"content": "If the function value $y$ of the linear function $y = ( k - 1 ) x - 1$ decreases as $x$ increases, then ____?", "answer": "- 3", "steps": "$\\because$ The function value $y$ of the linear function $y = ( k - 1 ) x - 1$ decreases as $x$ increases, $\\therefore$ $k - 1 < 0$, and solving for $k$ gives $k < 1$.", "expr_cands": ["x", "( x + a )", "a", "( x + 3 )", "x ^ { 2 } + ( a + 3 ) x + 3 a", "a + 3 = 0", "a = - 3"], "exprs": ["x ^ { 2 } + ( a + 3 ) x + 3 a", "a + 3 = 0", "a = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x"}, {"id": "x ^ { 2 } + ( a + 3 ) x + 3 a"}, {"id": "( x + a )"}, {"id": "( x + 3 )"}, {"id": "关于 $x$ 的两个多项式 $( x + a )$ 与 $( x + 3 )$ 的乘积中不含 $x$ 的一次项"}, {"id": "a + 3 = 0"}, {"id": "a = - 3"}], "links": [{"rel": "被描述", "source": "x", "target": "x ^ { 2 } + ( a + 3 ) x + 3 a"}, {"rel": "被描述", "source": "x ^ { 2 } + ( a + 3 ) x + 3 a", "target": "a + 3 = 0"}, {"rel": "被描述", "source": "( x + a )", "target": "x ^ { 2 } + ( a + 3 ) x + 3 a"}, {"rel": "被描述", "source": "( x + 3 )", "target": "x ^ { 2 } + ( a + 3 ) x + 3 a"}, {"rel": "限制性描述", "source": "关于 $x$ 的两个多项式 $( x + a )$ 与 $( x + 3 )$ 的乘积中不含 $x$ 的一次项", "target": "x ^ { 2 } + ( a + 3 ) x + 3 a"}, {"rel": "限制性描述", "source": "关于 $x$ 的两个多项式 $( x + a )$ 与 $( x + 3 )$ 的乘积中不含 $x$ 的一次项", "target": "a + 3 = 0"}, {"rel": "等式方程求解", "source": "a + 3 = 0", "target": "a = - 3"}]}}
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{"content": "If the result of $( x + t ) ( x + 6 )$ does not contain a linear term of $x$, then the value of $t$ is ____?", "answer": "- 8", "steps": "$( x + t ) ( x + 6 ) = x ^ { 2 } + ( t + 6 ) x + 6 t$ , from the absence of the $x$ linear term in the product, we get $t + 6 = 0$, that is, $t = - 6$. ", "expr_cands": ["x = 2", "x", "ax - 2", "a", "4", "x = - 2", "2 a - 2 = 4", "a = 3", "- 6 - 2", "- 8"], "exprs": ["2 a - 2 = 4", "a = 3", "- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 2"}, {"id": "2 a - 2 = 4"}, {"id": "ax - 2"}, {"id": "4"}, {"id": "当 $x = 2$ 时"}, {"id": "代数式 $ax - 2$ 的值是 $4$"}, {"id": "a = 3"}, {"id": "- 8"}, {"id": "x = - 2"}], "links": [{"rel": "被描述", "source": "x = 2", "target": "2 a - 2 = 4"}, {"rel": "等式方程求解", "source": "2 a - 2 = 4", "target": "a = 3"}, {"rel": "被描述", "source": "ax - 2", "target": "2 a - 2 = 4"}, {"rel": "被代入", "source": "ax - 2", "target": "- 8"}, {"rel": "被描述", "source": "4", "target": "2 a - 2 = 4"}, {"rel": "限制性描述", "source": "当 $x = 2$ 时", "target": "2 a - 2 = 4"}, {"rel": "限制性描述", "source": "代数式 $ax - 2$ 的值是 $4$", "target": "2 a - 2 = 4"}, {"rel": "代入", "source": "a = 3", "target": "- 8"}, {"rel": "代入", "source": "x = - 2", "target": "- 8"}]}}
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{"content": "If the product of $x$ and $9$ is equal to the sum of $x$ and $- 16$, then $x$ = ____ ?", "answer": "4", "steps": "According to the problem, we have $9 x = x - 16$. By rearranging and combining like terms, we get $8 x = - 16$. Solving for $x$, we get $x = - 2$.", "expr_cands": ["\\frac { a } { b } = \\frac { c } { d }", "b", "c", "a", "d", "a = 3", "b = 6", "c = 2", "\\frac { 3 } { 6 } = \\frac { 2 } { d }", "d = 4"], "exprs": ["\\frac { 3 } { 6 } = \\frac { 2 } { d }", "d = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a } { b } = \\frac { c } { d }"}, {"id": "\\frac { 3 } { 6 } = \\frac { 2 } { d }"}, {"id": "a = 3"}, {"id": "b = 6"}, {"id": "c = 2"}, {"id": "d = 4"}], "links": [{"rel": "被代入", "source": "\\frac { a } { b } = \\frac { c } { d }", "target": "\\frac { 3 } { 6 } = \\frac { 2 } { d }"}, {"rel": "等式方程求解", "source": "\\frac { 3 } { 6 } = \\frac { 2 } { d }", "target": "d = 4"}, {"rel": "代入", "source": "a = 3", "target": "\\frac { 3 } { 6 } = \\frac { 2 } { d }"}, {"rel": "代入", "source": "b = 6", "target": "\\frac { 3 } { 6 } = \\frac { 2 } { d }"}, {"rel": "代入", "source": "c = 2", "target": "\\frac { 3 } { 6 } = \\frac { 2 } { d }"}]}}
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||||
{"content": "If the product of two polynomials $( x + a )$ and $( x + 3 )$ with respect to $x$ does not contain a linear term, then the value of the constant $a$ is ____?", "answer": "4", "steps": "Original expression = $x ^ { 2 } + ( a + 3 ) x + 3 a$, since the result does not contain a linear term in $x$, we obtain $a + 3 = 0$, and solve for $a$: $a = - 3$.", "expr_cands": ["3 a ^ { 2 } b ^ { 2 } - 4 ab + a", "a", "b", "3 { a } ^ { 2 } { b } ^ { 2 }", "2 + 2", "4"], "exprs": ["3 { a } ^ { 2 } { b } ^ { 2 }", "2 + 2", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a ^ { 2 } b ^ { 2 } - 4 ab + a"}, {"id": "3 { a } ^ { 2 } { b } ^ { 2 }"}, {"id": "此题的最高次项是 $3 { a } ^ { 2 } { b } ^ { 2 }$"}, {"id": "多项式 $3 a ^ { 2 } b ^ { 2 } - 4 ab + a$ 的次数"}, {"id": "2 + 2"}, {"id": "此项的次数为 $2 + 2 = 4$"}, {"id": "4"}], "links": [{"rel": "被描述", "source": "3 a ^ { 2 } b ^ { 2 } - 4 ab + a", "target": "3 { a } ^ { 2 } { b } ^ { 2 }"}, {"rel": "被描述", "source": "3 { a } ^ { 2 } { b } ^ { 2 }", "target": "2 + 2"}, {"rel": "限制性描述", "source": "此题的最高次项是 $3 { a } ^ { 2 } { b } ^ { 2 }$", "target": "3 { a } ^ { 2 } { b } ^ { 2 }"}, {"rel": "限制性描述", "source": "多项式 $3 a ^ { 2 } b ^ { 2 } - 4 ab + a$ 的次数", "target": "3 { a } ^ { 2 } { b } ^ { 2 }"}, {"rel": "计算", "source": "2 + 2", "target": "4"}, {"rel": "限制性描述", "source": "此项的次数为 $2 + 2 = 4$", "target": "2 + 2"}]}}
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{"content": "When $x = 2$, the value of the algebraic expression $ax - 2$ is $4$; when $x = - 2$, the value of this algebraic expression is ____?", "answer": "\\frac { 3 } { 5 }", "steps": "Substituting $x = 2$ gives $2 a - 2 = 4$, solving for $a$ gives $a = 3$. Therefore, when $x = - 2$, the original expression equals $- 6 - 2 = - 8$.", "expr_cands": ["\\frac { 1 } { x } - \\frac { 1 } { y } = 2", "x", "y", "\\frac { 4 x + 5 xy - 4 y } { x - 3 xy - y }", "y - x = 2 xy", "x - y", "- 2 xy", "\\frac { 4 ( x - y ) + 5 xy } { ( x - y ) - 3 xy }", "\\frac { 3 } { 5 }"], "exprs": ["y - x = 2 xy", "\\frac { 4 ( x - y ) + 5 xy } { ( x - y ) - 3 xy }", "\\frac { 3 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x } - \\frac { 1 } { y } = 2"}, {"id": "y - x = 2 xy"}, {"id": "\\frac { 4 x + 5 xy - 4 y } { x - 3 xy - y }"}, {"id": "\\frac { 4 ( x - y ) + 5 xy } { ( x - y ) - 3 xy }"}, {"id": "x - y"}, {"id": "\\frac { 3 } { 5 }"}], "links": [{"rel": "同乘除", "source": "\\frac { 1 } { x } - \\frac { 1 } { y } = 2", "target": "y - x = 2 xy"}, {"rel": "代入", "source": "y - x = 2 xy", "target": "\\frac { 3 } { 5 }"}, {"rel": "提取因式", "source": "\\frac { 4 x + 5 xy - 4 y } { x - 3 xy - y }", "target": "\\frac { 4 ( x - y ) + 5 xy } { ( x - y ) - 3 xy }"}, {"rel": "被代入", "source": "\\frac { 4 ( x - y ) + 5 xy } { ( x - y ) - 3 xy }", "target": "\\frac { 3 } { 5 }"}, {"rel": "提取因式参考", "source": "x - y", "target": "\\frac { 4 ( x - y ) + 5 xy } { ( x - y ) - 3 xy }"}]}}
|
||||
{"content": "If $\\frac { a } { b } = \\frac { c } { d }$, where $a = 3$, $b = 6$, and $c = 2$, then $d$ = ____?", "answer": "16", "steps": "\\because $\\frac { a } { b } = \\frac { c } { d }$ , $a = 3$ , $b = 6$ , $c = 2$ , \\therefore $\\frac { 3 } { 6 } = \\frac { 2 } { d }$ , which gives $d = 4$ after solving.", "expr_cands": ["m + 3 n - 4 = 0", "m", "n", "2 ^ { m } \\times 8 ^ { n }", "2 ^ { m } \\times ( 2 ^ { 3 } ) ^ { n }", "2 ^ { m + 3 n }", "m + 3 n = 4", "2 ^ { 4 }", "16"], "exprs": ["2 ^ { m + 3 n }", "m + 3 n = 4", "16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 ^ { m } \\times 8 ^ { n }"}, {"id": "2 ^ { m + 3 n }"}, {"id": "m + 3 n - 4 = 0"}, {"id": "m + 3 n = 4"}, {"id": "16"}], "links": [{"rel": "计算", "source": "2 ^ { m } \\times 8 ^ { n }", "target": "2 ^ { m + 3 n }"}, {"rel": "被代入", "source": "2 ^ { m + 3 n }", "target": "16"}, {"rel": "移项", "source": "m + 3 n - 4 = 0", "target": "m + 3 n = 4"}, {"rel": "代入", "source": "m + 3 n = 4", "target": "16"}]}}
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{"content": "The degree of the polynomial $3 a ^ { 2 } b ^ { 2 } - 4 ab + a$ is ____ ?", "answer": "y = 2 x - 4", "steps": "According to the question, we know that the highest degree term of this polynomial is $3 { a } ^ { 2 } { b } ^ { 2 }$, and the degree of this term is $2 + 2 = 4$, so the degree of the polynomial is $4$.", "expr_cands": ["y = 2 x - 3", "y", "x", "2", "3", "y = 2 ( x - 2 ) - 3 + 3", "2 x - 3 = 2 ( x - 2 ) - 3 + 3", "2 x - 4"], "exprs": ["y = 2 ( x - 2 ) - 3 + 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 x - 3"}, {"id": "y = 2 ( x - 2 ) - 3 + 3"}, {"id": "2"}, {"id": "3"}, {"id": "将直线 $y = 2 x - 3$ 向右平移 $2$ 个单位"}, {"id": "再向上平移 $3$ 个单位后"}, {"id": "所得的直线的表达式"}], "links": [{"rel": "被描述", "source": "y = 2 x - 3", "target": "y = 2 ( x - 2 ) - 3 + 3"}, {"rel": "被描述", "source": "2", "target": "y = 2 ( x - 2 ) - 3 + 3"}, {"rel": "被描述", "source": "3", "target": "y = 2 ( x - 2 ) - 3 + 3"}, {"rel": "限制性描述", "source": "将直线 $y = 2 x - 3$ 向右平移 $2$ 个单位", "target": "y = 2 ( x - 2 ) - 3 + 3"}, {"rel": "限制性描述", "source": "再向上平移 $3$ 个单位后", "target": "y = 2 ( x - 2 ) - 3 + 3"}, {"rel": "限制性描述", "source": "所得的直线的表达式", "target": "y = 2 ( x - 2 ) - 3 + 3"}]}}
|
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{"content": "If the fraction $\\frac { 1 } { x } - \\frac { 1 } { y } = 2$, then the value of the fraction $\\frac { 4 x + 5 xy - 4 y } { x - 3 xy - y }$ is ____?", "answer": "12", "steps": "From $\\frac { 1 } { x } - \\frac { 1 } { y } = 2$, we have $y - x = 2 xy$. Therefore, $x - y = - 2 xy$. Thus, the original expression is $\\frac { 4 ( x - y ) + 5 xy } {( x - y ) - 3 xy } = \\frac { - 8 xy + 5 xy } { - 2 xy - 3 xy } = \\frac { - 3 xy } { - 5 xy } = \\frac { 3 } { 5 }$.", "expr_cands": ["y = x ^ { 2 } - 6 x + 5", "x", "y", "y = a ( x - h ) ^ { 2 } - k", "k", "h", "a", "hk", "y = ( x - 3 ) ^ { 2 } - 4", "h = 3", "k = 4", "12"], "exprs": ["y = ( x - 3 ) ^ { 2 } - 4", "h = 3", "k = 4", "12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 } - 6 x + 5"}, {"id": "y = ( x - 3 ) ^ { 2 } - 4"}, {"id": "将抛物线 $y = x ^ { 2 } - 6 x + 5$ 化成 $y = a ( x - h ) ^ { 2 } - k$ 的形式"}, {"id": "y = a ( x - h ) ^ { 2 } - k"}, {"id": "h = 3"}, {"id": "k = 4"}, {"id": "hk"}, {"id": "12"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 } - 6 x + 5", "target": "y = ( x - 3 ) ^ { 2 } - 4"}, {"rel": "联立", "source": "y = ( x - 3 ) ^ { 2 } - 4", "target": "h = 3"}, {"rel": "联立", "source": "y = ( x - 3 ) ^ { 2 } - 4", "target": "k = 4"}, {"rel": "限制性描述", "source": "将抛物线 $y = x ^ { 2 } - 6 x + 5$ 化成 $y = a ( x - h ) ^ { 2 } - k$ 的形式", "target": "y = ( x - 3 ) ^ { 2 } - 4"}, {"rel": "联立", "source": "y = a ( x - h ) ^ { 2 } - k", "target": "h = 3"}, {"rel": "联立", "source": "y = a ( x - h ) ^ { 2 } - k", "target": "k = 4"}, {"rel": "代入", "source": "h = 3", "target": "12"}, {"rel": "代入", "source": "k = 4", "target": "12"}, {"rel": "被代入", "source": "hk", "target": "12"}]}}
|
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{"content": "Given $m + 3 n - 4 = 0$, what is the value of $2 ^ m \\times 8 ^ n$?", "answer": "120", "steps": "Original expression = $2 ^ { m } \\times ( 2 ^ { 3 } ) ^ { n } = 2 ^ { m } \\times 2 ^ { 3 n } = 2 ^ { m + 3 n }$ . Since $m + 3 n - 4 = 0$, we have $m + 3 n = 4$. Therefore, the original expression is equal to $2 ^ { 4 } = 16$.", "expr_cands": ["a", "b", "a + b = 6", "a - b = 10", "2 a ^ { 2 } - 2 b ^ { 2 }", "2 ( a + b ) ( a - b )", "2 * 6 * 10", "120"], "exprs": ["2 ( a + b ) ( a - b )", "120"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a ^ { 2 } - 2 b ^ { 2 }"}, {"id": "2 ( a + b ) ( a - b )"}, {"id": "a + b = 6"}, {"id": "a - b = 10"}, {"id": "120"}], "links": [{"rel": "提取因式", "source": "2 a ^ { 2 } - 2 b ^ { 2 }", "target": "2 ( a + b ) ( a - b )"}, {"rel": "被代入", "source": "2 ( a + b ) ( a - b )", "target": "120"}, {"rel": "提取因式参考", "source": "a + b = 6", "target": "2 ( a + b ) ( a - b )"}, {"rel": "代入", "source": "a + b = 6", "target": "120"}, {"rel": "提取因式参考", "source": "a - b = 10", "target": "2 ( a + b ) ( a - b )"}, {"rel": "代入", "source": "a - b = 10", "target": "120"}]}}
|
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{"content": "The expression of the line obtained by translating the line $y = 2 x - 3$ $2$ units to the right and $3$ units up is _____.", "answer": "- 2", "steps": "The line $y = 2 x - 3$ is translated $2$ units to the right and $3$ units up, resulting in the expression $y = 2 ( x - 2 ) - 3 + 3$, which simplifies to $y = 2 x - 4$.", "expr_cands": ["x = 2", "x", "\\frac { x - a } { x + b }", "a", "b", "2 + b = 0", "b = - 2"], "exprs": ["2 + b = 0", "b = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 2"}, {"id": "2 + b = 0"}, {"id": "\\frac { x - a } { x + b }"}, {"id": "当 $x = 2$ 时"}, {"id": "分式 $\\frac { x - a } { x + b }$ 没有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "b = - 2"}], "links": [{"rel": "被描述", "source": "x = 2", "target": "2 + b = 0"}, {"rel": "等式方程求解", "source": "2 + b = 0", "target": "b = - 2"}, {"rel": "被描述", "source": "\\frac { x - a } { x + b }", "target": "2 + b = 0"}, {"rel": "限制性描述", "source": "当 $x = 2$ 时", "target": "2 + b = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - a } { x + b }$ 没有意义", "target": "2 + b = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "2 + b = 0"}]}}
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{"content": "Convert the parabola $y = x ^ 2 - 6 x + 5$ into the form $y = a ( x - h ) ^ 2 - k$, then $hk$ = ____?", "answer": "- 2", "steps": "Since $y = x ^ { 2 } - 6 x + 5 = x ^ { 2 } - 6 x + 9 - 4 = ( x - 3 ) ^ { 2 } - 4$, we can conclude that $h = 3$ and $k = 4$. Therefore, $hk = 3 * 4 = 12$.", "expr_cands": ["- \\frac { 1 } { 2 }", "m + 4", "m", "- 2 + m + 4 = 0", "m = - 2"], "exprs": ["- 2 + m + 4 = 0", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m + 4"}, {"id": "- 2 + m + 4 = 0"}, {"id": "- \\frac { 1 } { 2 }"}, {"id": "$- \\frac { 1 } { 2 }$ 的倒数与 $m + 4$ 互为相反数"}, {"id": "m = - 2"}], "links": [{"rel": "被描述", "source": "m + 4", "target": "- 2 + m + 4 = 0"}, {"rel": "等式方程求解", "source": "- 2 + m + 4 = 0", "target": "m = - 2"}, {"rel": "被描述", "source": "- \\frac { 1 } { 2 }", "target": "- 2 + m + 4 = 0"}, {"rel": "限制性描述", "source": "$- \\frac { 1 } { 2 }$ 的倒数与 $m + 4$ 互为相反数", "target": "- 2 + m + 4 = 0"}]}}
|
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{"content": "If real numbers $a$ and $b$ satisfy $a + b = 6$ and $a - b = 10$, then $2 a ^ 2 - 2 b ^ 2$ = ____?", "answer": "y = x - 1", "steps": "$2 a ^ { 2 } - 2 b ^ { 2 } = 2 ( a ^ { 2 } - b ^ { 2 } ) = 2 ( a + b ) ( a - b )$ , because $a + b = 6$ , $a - b = 10$ , therefore the original expression = $2 * 6 * 10 = 120$.", "expr_cands": ["y - 2", "y", "x - 3", "x", "x = 4", "y = 3", "y - 2 = k ( x - 3 )", "k", "3 - 2 = k", "k = 1", "y - 2 = x - 3", "y = x - 1"], "exprs": ["y - 2 = k ( x - 3 )", "3 - 2 = k", "k = 1", "y - 2 = x - 3", "y = x - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $y - 2 = k ( x - 3 )$"}, {"id": "y - 2 = k ( x - 3 )"}, {"id": "x = 4"}, {"id": "3 - 2 = k"}, {"id": "y = 3"}, {"id": "k = 1"}, {"id": "y - 2 = x - 3"}, {"id": "y = x - 1"}], "links": [{"rel": "假设描述", "source": "设 $y - 2 = k ( x - 3 )$", "target": "y - 2 = k ( x - 3 )"}, {"rel": "被代入", "source": "y - 2 = k ( x - 3 )", "target": "3 - 2 = k"}, {"rel": "被代入", "source": "y - 2 = k ( x - 3 )", "target": "y - 2 = x - 3"}, {"rel": "代入", "source": "x = 4", "target": "3 - 2 = k"}, {"rel": "等式方程求解", "source": "3 - 2 = k", "target": "k = 1"}, {"rel": "代入", "source": "y = 3", "target": "3 - 2 = k"}, {"rel": "代入", "source": "k = 1", "target": "y - 2 = x - 3"}, {"rel": "移项", "source": "y - 2 = x - 3", "target": "y = x - 1"}]}}
|
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{"content": "When $x = 2$, the fraction $\\frac { x - a } { x + b }$ is undefined, then $b$ = ____ ?", "answer": "x \\neq 3", "steps": "From the given information, we have $2 + b = 0$, which can be solved to obtain $b = - 2$.", "expr_cands": ["( x - 3 ) ^ { 0 } = 1", "x", "x - 3 \\neq 0", "x \\neq 3"], "exprs": ["x - 3 \\neq 0", "x \\neq 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x - 3 ) ^ { 0 } = 1"}, {"id": "x - 3 \\neq 0"}, {"id": "多项式零次方项，若底数不为0，则恒等于1"}, {"id": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0"}, {"id": "x \\neq 3"}], "links": [{"rel": "被描述", "source": "( x - 3 ) ^ { 0 } = 1", "target": "x - 3 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 3 \\neq 0", "target": "x \\neq 3"}, {"rel": "属性描述", "source": "多项式零次方项，若底数不为0，则恒等于1", "target": "x - 3 \\neq 0"}, {"rel": "属性描述", "source": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0", "target": "x - 3 \\neq 0"}]}}
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{"content": "If the reciprocal of $- \\frac { 1 } { 2 }$ is the opposite of $m + 4$, then the value of $m$ is ____?", "answer": "1", "steps": "$\\because$ The reciprocal of $- \\frac { 1 } { 2 }$ is the opposite of $m + 4$, $\\therefore$ $- 2 + m + 4 = 0$, $\\therefore$ $m = - 2$.", "expr_cands": ["2 x - 1", "x", "4 x - 5", "2 x - 1 + 4 x - 5 = 0", "x = 1"], "exprs": ["2 x - 1 + 4 x - 5 = 0", "x = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 1"}, {"id": "2 x - 1 + 4 x - 5 = 0"}, {"id": "4 x - 5"}, {"id": "代数式 $2 x - 1$ 与 $4 x - 5$ 的值互为相反数"}, {"id": "x = 1"}], "links": [{"rel": "被描述", "source": "2 x - 1", "target": "2 x - 1 + 4 x - 5 = 0"}, {"rel": "等式方程求解", "source": "2 x - 1 + 4 x - 5 = 0", "target": "x = 1"}, {"rel": "被描述", "source": "4 x - 5", "target": "2 x - 1 + 4 x - 5 = 0"}, {"rel": "限制性描述", "source": "代数式 $2 x - 1$ 与 $4 x - 5$ 的值互为相反数", "target": "2 x - 1 + 4 x - 5 = 0"}]}}
|
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{"content": "If $y - 2$ is proportional to $x - 3$, and $y = 3$ when $x = 4$, then the functional expression of $y$ in terms of $x$ is ____?", "answer": "- 33", "steps": "\\because $y - 2$ is directly proportional to $x - 3$, \\therefore let $y - 2 = k ( x - 3 )$, \\because when $x = 4$, $y = 3$, \\therefore $3 - 2 = k$, that is, $k = 1$, \\therefore $y - 2 = x - 3$, \\therefore $y = x - 1$.", "expr_cands": ["x = 89", "x", "| y | = 122", "y", "y < 0", "x + y", "y = - 122", "y = 122", "x + y = - 33", "- 33"], "exprs": ["y = - 122", "- 33"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| y | = 122"}, {"id": "y = - 122"}, {"id": "y < 0"}, {"id": "x = 89"}, {"id": "- 33"}, {"id": "x + y"}], "links": [{"rel": "联立", "source": "| y | = 122", "target": "y = - 122"}, {"rel": "代入", "source": "y = - 122", "target": "- 33"}, {"rel": "联立", "source": "y < 0", "target": "y = - 122"}, {"rel": "代入", "source": "x = 89", "target": "- 33"}, {"rel": "被代入", "source": "x + y", "target": "- 33"}]}}
|
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{"content": "If $( x - 3 ) ^ { 0 } = 1$, then $x$ should satisfy the condition ____?", "answer": "1", "steps": "From the given condition, we have $x - 3 \\neq 0$, which implies that $x \\neq 3$.", "expr_cands": ["x = 1", "x", "2 mx ^ { 3 } - 5 nx + 4", "n", "m", "7", "x = - 1", "2 m - 5 n + 4 = 7", "2 m - 5 n = 3", "- ( 2 m - 5 n ) + 4", "1"], "exprs": ["2 m - 5 n + 4 = 7", "- ( 2 m - 5 n ) + 4", "2 m - 5 n = 3", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 1"}, {"id": "2 m - 5 n + 4 = 7"}, {"id": "2 mx ^ { 3 } - 5 nx + 4"}, {"id": "7"}, {"id": "当 $x = 1$ 时"}, {"id": "整式 $2 mx ^ { 3 } - 5 nx + 4$ 的值是 $7$"}, {"id": "2 m - 5 n = 3"}, {"id": "- ( 2 m - 5 n ) + 4"}, {"id": "x = - 1"}, {"id": "这个整式的值"}, {"id": "$x = - 1$ 时"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "x = 1", "target": "2 m - 5 n + 4 = 7"}, {"rel": "移项", "source": "2 m - 5 n + 4 = 7", "target": "2 m - 5 n = 3"}, {"rel": "被描述", "source": "2 mx ^ { 3 } - 5 nx + 4", "target": "2 m - 5 n + 4 = 7"}, {"rel": "被描述", "source": "2 mx ^ { 3 } - 5 nx + 4", "target": "- ( 2 m - 5 n ) + 4"}, {"rel": "被描述", "source": "7", "target": "2 m - 5 n + 4 = 7"}, {"rel": "限制性描述", "source": "当 $x = 1$ 时", "target": "2 m - 5 n + 4 = 7"}, {"rel": "限制性描述", "source": "整式 $2 mx ^ { 3 } - 5 nx + 4$ 的值是 $7$", "target": "2 m - 5 n + 4 = 7"}, {"rel": "代入", "source": "2 m - 5 n = 3", "target": "1"}, {"rel": "被代入", "source": "- ( 2 m - 5 n ) + 4", "target": "1"}, {"rel": "被描述", "source": "x = - 1", "target": "- ( 2 m - 5 n ) + 4"}, {"rel": "限制性描述", "source": "这个整式的值", "target": "- ( 2 m - 5 n ) + 4"}, {"rel": "限制性描述", "source": "$x = - 1$ 时", "target": "- ( 2 m - 5 n ) + 4"}]}}
|
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{"content": "If the value of the algebraic expression $2 x - 1$ is the opposite of the value of $4 x - 5$, then $x$ = ____?", "answer": "- 1", "steps": "From the given information, we can derive that $2 x - 1 + 4 x - 5 = 0$. Solving for $x$, we get $x = 1$.", "expr_cands": ["\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ { 5 }", "n", "x", "y", "- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }", "m", "{ ( - 2 m + n ) } ^ { n }", "\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } y 5", "2 n - 1 = 5", "n = 3", "3 m - 1 = 5", "m = 2", "( - 2 m + n ) ^ { n }", "- 1"], "exprs": ["2 n - 1 = 5", "3 m - 1 = 5", "n = 3", "m = 2", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ { 5 }"}, {"id": "2 n - 1 = 5"}, {"id": "- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }"}, {"id": ", $\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } y 5$ 与 $- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }$ 是同类项"}, {"id": "$\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ { 5 }$ 与 $- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }$ 相加的结果是一项"}, {"id": "3 m - 1 = 5"}, {"id": "n = 3"}, {"id": "m = 2"}, {"id": "( - 2 m + n ) ^ { n }"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ { 5 }", "target": "2 n - 1 = 5"}, {"rel": "被描述", "source": "\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ { 5 }", "target": "3 m - 1 = 5"}, {"rel": "等式方程求解", "source": "2 n - 1 = 5", "target": "n = 3"}, {"rel": "被描述", "source": "- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }", "target": "2 n - 1 = 5"}, {"rel": "被描述", "source": "- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }", "target": "3 m - 1 = 5"}, {"rel": "限制性描述", "source": ", $\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } y 5$ 与 $- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }$ 是同类项", "target": "2 n - 1 = 5"}, {"rel": "限制性描述", "source": ", $\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } y 5$ 与 $- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }$ 是同类项", "target": "3 m - 1 = 5"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ { 5 }$ 与 $- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }$ 相加的结果是一项", "target": "2 n - 1 = 5"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ { 5 }$ 与 $- \\frac { 1 } { 2 } { x } ^ { 5 } { y } ^ { 3 m - 1 }$ 相加的结果是一项", "target": "3 m - 1 = 5"}, {"rel": "等式方程求解", "source": "3 m - 1 = 5", "target": "m = 2"}, {"rel": "代入", "source": "n = 3", "target": "- 1"}, {"rel": "代入", "source": "m = 2", "target": "- 1"}, {"rel": "被代入", "source": "( - 2 m + n ) ^ { n }", "target": "- 1"}]}}
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{"content": "If $x = 89$, $| y | = 122$, and $y < 0$, then $x + y$ = ____?", "answer": "- 4", "steps": "Since $| y | = 122$ and $y < 0$, we know that $y = - 122$. Therefore, $x + y = 89 - 122 = - 33$.", "expr_cands": ["a ^ { 2 } + 3 b = 2", "b", "a", "2 a ^ { 2 } + 6 b - 8", "2 ( a ^ { 2 } + 3 b ) - 8", "- 4"], "exprs": ["2 ( a ^ { 2 } + 3 b ) - 8", "- 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a ^ { 2 } + 6 b - 8"}, {"id": "2 ( a ^ { 2 } + 3 b ) - 8"}, {"id": "a ^ { 2 } + 3 b = 2"}, {"id": "- 4"}], "links": [{"rel": "提取因式", "source": "2 a ^ { 2 } + 6 b - 8", "target": "2 ( a ^ { 2 } + 3 b ) - 8"}, {"rel": "被代入", "source": "2 ( a ^ { 2 } + 3 b ) - 8", "target": "- 4"}, {"rel": "提取因式参考", "source": "a ^ { 2 } + 3 b = 2", "target": "2 ( a ^ { 2 } + 3 b ) - 8"}, {"rel": "代入", "source": "a ^ { 2 } + 3 b = 2", "target": "- 4"}]}}
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{"content": "If $x = 1$, the value of the polynomial $2 mx ^ 3 - 5 nx + 4$ is $7$. What is the value of the polynomial when $x = - 1$?", "answer": "x \\ge \\frac { 5 } { 2 }", "steps": "Substituting $x = 1$ gives $2 m - 5 n + 4 = 7$, which simplifies to $2 m - 5 n = 3$. Therefore, when $x = - 1$, the original expression is equal to $- ( 2 m - 5 n ) + 4 = - 3 + 4 = 1$.", "expr_cands": ["\\frac { 4 x + 5 } { 3 }", "x", "2 x", "\\frac { 4 x + 5 } { 3 } \\le 2 x", "\\frac { 5 } { 2 } \\le x", "4 x + 5 \\le 6 x", "4 x - 6 x \\le - 5", "- 2 x \\le - 5", "1", "x \\ge \\frac { 5 } { 2 }"], "exprs": ["\\frac { 4 x + 5 } { 3 } \\le 2 x", "x \\ge \\frac { 5 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 4 x + 5 } { 3 }"}, {"id": "\\frac { 4 x + 5 } { 3 } \\le 2 x"}, {"id": "2 x"}, {"id": "$\\frac { 4 x + 5 } { 3 }$ 不大于 $2 x$"}, {"id": "x \\ge \\frac { 5 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\frac { 4 x + 5 } { 3 }", "target": "\\frac { 4 x + 5 } { 3 } \\le 2 x"}, {"rel": "不等式方程求解", "source": "\\frac { 4 x + 5 } { 3 } \\le 2 x", "target": "x \\ge \\frac { 5 } { 2 }"}, {"rel": "被描述", "source": "2 x", "target": "\\frac { 4 x + 5 } { 3 } \\le 2 x"}, {"rel": "限制性描述", "source": "$\\frac { 4 x + 5 } { 3 }$ 不大于 $2 x$", "target": "\\frac { 4 x + 5 } { 3 } \\le 2 x"}]}}
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{"content": "Given $\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } { y } ^ 5$ added to $- \\frac { 1 } { 2 } { x } ^ 5 { y } ^ { 3 m - 1 }$ results in a single term, then the value of ${( - 2 m + n )} ^ n$ is ____?", "answer": "64", "steps": "Because $\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } y ^ 5$ and $- \\frac { 1 } { 2 } { x } ^ 5 { y } ^ { 3 m - 1 }$ add up to one term, therefore $\\frac { 1 } { 3 } { x } ^ { 2 n - 1 } y ^ 5$ and $- \\frac { 1 } { 2 } { x } ^ 5 { y } ^ { 3 m - 1 }$ are like terms, therefore $2 n - 1 = 5$, $3 m - 1 = 5$, therefore $n = 3$, $m = 2$, therefore $( - 2 m + n ) ^ n = ( - 4 + 3 ) ^ 3 = ( - 1 ) ^ 3 = - 1$.", "expr_cands": ["8 x ^ { m } y", "x", "m", "y", "6 x ^ { 3 } y ^ { n }", "n", "( m + n ) ^ { 3 }", "m = 3", "n = 1", "64"], "exprs": ["m = 3", "n = 1", "64"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "8 x ^ { m } y"}, {"id": "m = 3"}, {"id": "6 x ^ { 3 } y ^ { n }"}, {"id": "$8 x ^ { m } y$ 与 $6 x ^ { 3 } y ^ { n }$ 的和是单项式"}, {"id": ", $8 x ^ { m } y$ 与 $6 x ^ { 3 } y ^ { n }$ 是同类项"}, {"id": "n = 1"}, {"id": "( m + n ) ^ { 3 }"}, {"id": "64"}], "links": [{"rel": "被描述", "source": "8 x ^ { m } y", "target": "m = 3"}, {"rel": "被描述", "source": "8 x ^ { m } y", "target": "n = 1"}, {"rel": "代入", "source": "m = 3", "target": "64"}, {"rel": "被描述", "source": "6 x ^ { 3 } y ^ { n }", "target": "m = 3"}, {"rel": "被描述", "source": "6 x ^ { 3 } y ^ { n }", "target": "n = 1"}, {"rel": "限制性描述", "source": "$8 x ^ { m } y$ 与 $6 x ^ { 3 } y ^ { n }$ 的和是单项式", "target": "m = 3"}, {"rel": "限制性描述", "source": "$8 x ^ { m } y$ 与 $6 x ^ { 3 } y ^ { n }$ 的和是单项式", "target": "n = 1"}, {"rel": "限制性描述", "source": ", $8 x ^ { m } y$ 与 $6 x ^ { 3 } y ^ { n }$ 是同类项", "target": "m = 3"}, {"rel": "限制性描述", "source": ", $8 x ^ { m } y$ 与 $6 x ^ { 3 } y ^ { n }$ 是同类项", "target": "n = 1"}, {"rel": "代入", "source": "n = 1", "target": "64"}, {"rel": "被代入", "source": "( m + n ) ^ { 3 }", "target": "64"}]}}
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{"content": "If $a ^ 2 + 3 b = 2$, then the algebraic expression $2 a ^ 2 + 6 b - 8$ equals ____?", "answer": "y = - x ^ { 2 } + 12 x - 16", "steps": "Because $a ^ 2 + 3 b = 2$, therefore $2 a ^ 2 + 6 b - 8 = 2 ( a ^ 2 + 3 b ) - 8 = - 4$.", "expr_cands": ["y = x ^ { 2 } - 12 x + 16", "x", "y", "- y = x ^ { 2 } - 12 x + 16", "- x ^ { 2 } + 12 x - 16 = x ^ { 2 } - 12 x + 16", "y = - x ^ { 2 } + 12 x - 16", "x ^ { 2 } - 12 x + 16 = - x ^ { 2 } + 12 x - 16", "x ^ { 2 } - 12 x + 16"], "exprs": ["- y = x ^ { 2 } - 12 x + 16", "y = - x ^ { 2 } + 12 x - 16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 } - 12 x + 16"}, {"id": "- y = x ^ { 2 } - 12 x + 16"}, {"id": "x"}, {"id": "将抛物线 $y = x ^ { 2 } - 12 x + 16$ 作关于 $x$ 轴对称"}, {"id": "y = - x ^ { 2 } + 12 x - 16"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 } - 12 x + 16", "target": "- y = x ^ { 2 } - 12 x + 16"}, {"rel": "同乘除", "source": "- y = x ^ { 2 } - 12 x + 16", "target": "y = - x ^ { 2 } + 12 x - 16"}, {"rel": "被描述", "source": "x", "target": "- y = x ^ { 2 } - 12 x + 16"}, {"rel": "限制性描述", "source": "将抛物线 $y = x ^ { 2 } - 12 x + 16$ 作关于 $x$ 轴对称", "target": "- y = x ^ { 2 } - 12 x + 16"}]}}
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{"content": "Given that $\\frac { 4 x + 5 } { 3 }$ is less than or equal to $2 x$, what is the range of possible values for $x$?", "answer": "\\frac { 13 } { 5 }", "steps": "According to the problem, we have $\\frac { 4 x + 5 } { 3 } \\leq 2 x$. Multiplying both sides by 3, we get $4 x + 5 \\leq 6 x$. Subtracting 4x from both sides, we get $5 \\leq 2 x$. Dividing both sides by 2, we get $x \\geq \\frac { 5 } { 2 }$.", "expr_cands": ["\\frac { x + 1 } { 2 }", "x", "\\frac { 5 - x } { 3 }", "1", "\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1", "x = \\frac { 13 } { 5 }"], "exprs": ["\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1", "x = \\frac { 13 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x + 1 } { 2 }"}, {"id": "\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1"}, {"id": "\\frac { 5 - x } { 3 }"}, {"id": "1"}, {"id": "$\\frac { x + 1 } { 2 }$ 比 $\\frac { 5 - x } { 3 }$ 大 $1$"}, {"id": "x = \\frac { 13 } { 5 }"}], "links": [{"rel": "被描述", "source": "\\frac { x + 1 } { 2 }", "target": "\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1"}, {"rel": "等式方程求解", "source": "\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1", "target": "x = \\frac { 13 } { 5 }"}, {"rel": "被描述", "source": "\\frac { 5 - x } { 3 }", "target": "\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1"}, {"rel": "被描述", "source": "1", "target": "\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1"}, {"rel": "限制性描述", "source": "$\\frac { x + 1 } { 2 }$ 比 $\\frac { 5 - x } { 3 }$ 大 $1$", "target": "\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1"}]}}
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{"content": "If $8 x ^ { m } y$ and $6 x ^ { 3 } y ^ { n }$ have a sum that is a monomial, then the value of $( m + n ) ^ { 3 }$ is ____ ?", "answer": "x = 8", "steps": "Since $8 x ^ { m } y$ and $6 x ^ { 3 } y ^ { n }$ form a monomial when added together, it follows that $8 x ^ { m } y$ and $6 x ^ { 3 } y ^ { n }$ are like terms. Therefore, $m = 3$ and $n = 1$. Thus, $( m + n ) ^ { 3 } = ( 3 + 1 ) ^ { 3 } = 64$.", "expr_cands": ["( 3 x + 2 ) + 2 [ ( x - 1 ) - ( 2 x + 1 ) ] = 6", "x", "3 x + 2 + 2 x - 2 - 4 x - 2 = 6", "x = 8"], "exprs": ["x = 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 3 x + 2 ) + 2 [ ( x - 1 ) - ( 2 x + 1 ) ] = 6"}, {"id": "x = 8"}], "links": [{"rel": "等式方程求解", "source": "( 3 x + 2 ) + 2 [ ( x - 1 ) - ( 2 x + 1 ) ] = 6", "target": "x = 8"}]}}
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{"content": "The equation of the parabola obtained by reflecting the parabola $y = x ^ { 2 } - 12 x + 16$ about the $x$-axis is _____.", "answer": "- 1", "steps": "$\\because$ The parabola $y = x ^ 2 - 12 x + 16$ is symmetric about the $x$-axis, and the equation of the parabola obtained by reflecting it about the $x$-axis is $- y = x ^ 2 - 12 x + 16$, $\\therefore$ the equation we seek is $y = - x ^ 2 + 12 x - 16$.", "expr_cands": ["x ^ { 2 } - 3 x - 3 = 0", "x", "\\frac { 1 } { \\alpha } + \\frac { 1 } { \\beta }", "\\alpha", "\\beta", "\\alpha + \\beta = 3", "\\alpha \\beta = - 3", "\\frac { \\alpha + \\beta } { \\alpha \\beta }", "- 1"], "exprs": ["\\alpha + \\beta = 3", "\\alpha \\beta = - 3", "\\frac { \\alpha + \\beta } { \\alpha \\beta }", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 x - 3 = 0"}, {"id": "\\alpha + \\beta = 3"}, {"id": "\\alpha"}, {"id": "\\beta"}, {"id": "一元二次方程 $x ^ { 2 } - 3 x - 3 = 0$ 的两根为 \\alpha 与 \\beta"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "\\alpha \\beta = - 3"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "\\frac { 1 } { \\alpha } + \\frac { 1 } { \\beta }"}, {"id": "\\frac { \\alpha + \\beta } { \\alpha \\beta }"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 3 x - 3 = 0", "target": "\\alpha + \\beta = 3"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 x - 3 = 0", "target": "\\alpha \\beta = - 3"}, {"rel": "代入", "source": "\\alpha + \\beta = 3", "target": "- 1"}, {"rel": "被描述", "source": "\\alpha", "target": "\\alpha + \\beta = 3"}, {"rel": "被描述", "source": "\\alpha", "target": "\\alpha \\beta = - 3"}, {"rel": "被描述", "source": "\\beta", "target": "\\alpha + \\beta = 3"}, {"rel": "被描述", "source": "\\beta", "target": "\\alpha \\beta = - 3"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - 3 x - 3 = 0$ 的两根为 \\alpha 与 \\beta", "target": "\\alpha + \\beta = 3"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - 3 x - 3 = 0$ 的两根为 \\alpha 与 \\beta", "target": "\\alpha \\beta = - 3"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "\\alpha + \\beta = 3"}, {"rel": "代入", "source": "\\alpha \\beta = - 3", "target": "- 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "\\alpha \\beta = - 3"}, {"rel": "计算", "source": "\\frac { 1 } { \\alpha } + \\frac { 1 } { \\beta }", "target": "\\frac { \\alpha + \\beta } { \\alpha \\beta }"}, {"rel": "被代入", "source": "\\frac { \\alpha + \\beta } { \\alpha \\beta }", "target": "- 1"}]}}
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{"content": "If $\\frac { x + 1 } { 2 }$ is $1$ greater than $\\frac { 5 - x } { 3 }$, then $x$ = ____ ?", "answer": "a \\neq 1", "steps": "\\because $\\frac { x + 1 } { 2 }$ is $1$ greater than $\\frac { 5 - x } { 3 }$ , \\therefore $\\frac { x + 1 } { 2 } - \\frac { 5 - x } { 3 } = 1$ and solving gives $x = \\frac { 13 } { 5 }$ .", "expr_cands": ["\\frac { { a } ^ { 2 } - a } { a - 1 }", "a", "a - 1 \\neq 0", "a \\neq 1"], "exprs": ["a - 1 \\neq 0", "a \\neq 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { { a } ^ { 2 } - a } { a - 1 }"}, {"id": "a - 1 \\neq 0"}, {"id": "分式有意义，则分母不为0"}, {"id": "使分式 $\\frac { { a } ^ { 2 } - a } { a - 1 }$ 有意义的 $a$ 取值应"}, {"id": "a \\neq 1"}], "links": [{"rel": "被描述", "source": "\\frac { { a } ^ { 2 } - a } { a - 1 }", "target": "a - 1 \\neq 0"}, {"rel": "不等式方程求解", "source": "a - 1 \\neq 0", "target": "a \\neq 1"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "a - 1 \\neq 0"}, {"rel": "限制性描述", "source": "使分式 $\\frac { { a } ^ { 2 } - a } { a - 1 }$ 有意义的 $a$ 取值应", "target": "a - 1 \\neq 0"}]}}
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{"content": "The equation $( 3 x + 2 ) + 2 [( x - 1 ) - ( 2 x + 1 )] = 6$ has a solution of ____?", "answer": "- 10", "steps": "Removing the parentheses, we get: $3 x + 2 + 2 x - 2 - 4 x - 2 = 6$. Combining like terms and moving terms to opposite sides, we get: $x = 8$.", "expr_cands": ["y = x ^ { 2 } + 6 x + 2", "x", "y", "y = ( x - h ) ^ { 2 } + k", "h", "k", "h + k", "y = ( x + 3 ) ^ { 2 } - 7", "h = - 3", "k = - 7", "- 10"], "exprs": ["y = ( x + 3 ) ^ { 2 } - 7", "h = - 3", "k = - 7", "- 10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = x ^ { 2 } + 6 x + 2"}, {"id": "y = ( x + 3 ) ^ { 2 } - 7"}, {"id": "把二次函数 $y = x ^ { 2 } + 6 x + 2$ 化为 $y = ( x - h ) ^ { 2 } + k$ 的形式"}, {"id": "y = ( x - h ) ^ { 2 } + k"}, {"id": "h = - 3"}, {"id": "k = - 7"}, {"id": "h + k"}, {"id": "- 10"}], "links": [{"rel": "被描述", "source": "y = x ^ { 2 } + 6 x + 2", "target": "y = ( x + 3 ) ^ { 2 } - 7"}, {"rel": "联立", "source": "y = ( x + 3 ) ^ { 2 } - 7", "target": "h = - 3"}, {"rel": "联立", "source": "y = ( x + 3 ) ^ { 2 } - 7", "target": "k = - 7"}, {"rel": "限制性描述", "source": "把二次函数 $y = x ^ { 2 } + 6 x + 2$ 化为 $y = ( x - h ) ^ { 2 } + k$ 的形式", "target": "y = ( x + 3 ) ^ { 2 } - 7"}, {"rel": "联立", "source": "y = ( x - h ) ^ { 2 } + k", "target": "h = - 3"}, {"rel": "联立", "source": "y = ( x - h ) ^ { 2 } + k", "target": "k = - 7"}, {"rel": "代入", "source": "h = - 3", "target": "- 10"}, {"rel": "代入", "source": "k = - 7", "target": "- 10"}, {"rel": "被代入", "source": "h + k", "target": "- 10"}]}}
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{"content": "Given a quadratic equation $x ^ 2 - 3 x - 3 = 0$ with roots $\\alpha$ and $\\beta$, the value of $\\frac { 1 } { \\alpha } + \\frac { 1 } { \\beta }$ is ____?", "answer": "\\frac { 1 } { 4 }", "steps": "According to the problem, we have $\\alpha + \\beta = 3$ and $\\alpha \\beta = - 3$. Therefore, $\\frac { 1 } { \\alpha } + \\frac { 1 } { \\beta } = \\frac { \\alpha + \\beta } { \\alpha \\beta } = \\frac { 3 } { - 3 } = - 1$.", "expr_cands": ["\\frac { 1 } { a - 3 } = 1", "a", "a ^ { - 1 }", "a = 4", "\\frac { 1 } { 4 }"], "exprs": ["a = 4", "\\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { a - 3 } = 1"}, {"id": "a = 4"}, {"id": "a ^ { - 1 }"}, {"id": "\\frac { 1 } { 4 }"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 1 } { a - 3 } = 1", "target": "a = 4"}, {"rel": "代入", "source": "a = 4", "target": "\\frac { 1 } { 4 }"}, {"rel": "被代入", "source": "a ^ { - 1 }", "target": "\\frac { 1 } { 4 }"}]}}
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{"content": "The value of $a$ that makes the fraction $\\frac { { a } ^ { 2 } - a } { a - 1 }$ meaningful is ____ ?", "answer": "- 1", "steps": "From the given condition, we have $a - 1 \\neq 0$, which implies that $a \\neq 1$ after solving.", "expr_cands": ["x ^ { n + 2 } = 7", "x", "n", "n + 2 = 1", "n = - 1"], "exprs": ["n + 2 = 1", "n = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { n + 2 } = 7"}, {"id": "n + 2 = 1"}, {"id": "方程 $x ^ { n + 2 } = 7$ 是关于 $x$ 的一元一次方程"}, {"id": "n = - 1"}], "links": [{"rel": "被描述", "source": "x ^ { n + 2 } = 7", "target": "n + 2 = 1"}, {"rel": "等式方程求解", "source": "n + 2 = 1", "target": "n = - 1"}, {"rel": "限制性描述", "source": "方程 $x ^ { n + 2 } = 7$ 是关于 $x$ 的一元一次方程", "target": "n + 2 = 1"}]}}
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{"content": "If the quadratic function $y = x ^ { 2 } + 6 x + 2$ is expressed in the form $y = ( x - h ) ^ { 2 } + k$, where $h$ and $k$ are constants, then $h + k$ = ____ ?", "answer": "- 10", "steps": "Since $y = x ^ 2 + 6 x + 2 = x ^ 2 + 6 x + 9 - 9 + 2 = ( x + 3 ) ^ 2 - 7$, therefore $h = - 3$, $k = - 7$, and thus $h + k = - 3 - 7 = - 10$.", "expr_cands": ["x", "x ^ { 2 } + mx + n = 0", "n", "m", "x _ { 1 } = - 2", "x _ { 1 }", "x _ { 2 } = 4", "x _ { 2 }", "m + n", "- 2 + 4 = - m", "m = - 2", "- 2 * 4 = n", "n = - 8", "- 10"], "exprs": ["- 2 + 4 = - m", "- 2 * 4 = n", "m = - 2", "n = - 8", "- 10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + mx + n = 0"}, {"id": "- 2 + 4 = - m"}, {"id": "x _ { 1 } = - 2"}, {"id": "x _ { 2 } = 4"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } + mx + n = 0$ 的两个实数根分别为 $x _ { 1 } = - 2$ , $x _ { 2 } = 4$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "m = - 2"}, {"id": "- 2 * 4 = n"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "n = - 8"}, {"id": "m + n"}, {"id": "- 10"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + mx + n = 0", "target": "- 2 + 4 = - m"}, {"rel": "被描述", "source": "x ^ { 2 } + mx + n = 0", "target": "- 2 * 4 = n"}, {"rel": "等式方程求解", "source": "- 2 + 4 = - m", "target": "m = - 2"}, {"rel": "被描述", "source": "x _ { 1 } = - 2", "target": "- 2 + 4 = - m"}, {"rel": "被描述", "source": "x _ { 1 } = - 2", "target": "- 2 * 4 = n"}, {"rel": "被描述", "source": "x _ { 2 } = 4", "target": "- 2 + 4 = - m"}, {"rel": "被描述", "source": "x _ { 2 } = 4", "target": "- 2 * 4 = n"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } + mx + n = 0$ 的两个实数根分别为 $x _ { 1 } = - 2$ , $x _ { 2 } = 4$", "target": "- 2 + 4 = - m"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } + mx + n = 0$ 的两个实数根分别为 $x _ { 1 } = - 2$ , $x _ { 2 } = 4$", "target": "- 2 * 4 = n"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "- 2 + 4 = - m"}, {"rel": "代入", "source": "m = - 2", "target": "- 10"}, {"rel": "等式方程求解", "source": "- 2 * 4 = n", "target": "n = - 8"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "- 2 * 4 = n"}, {"rel": "代入", "source": "n = - 8", "target": "- 10"}, {"rel": "被代入", "source": "m + n", "target": "- 10"}]}}
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{"content": "If $\\frac { 1 } { a - 3 } = 1$, then the value of $a ^ { - 1 }$ is:", "answer": "- 2", "steps": "Since $\\frac { 1 } { a - 3 } = 1$, it follows that $a - 3 = 1$. Therefore, $a = 4$. Consequently, $a ^ { - 1 } = \\frac { 1 } { 4 }$.", "expr_cands": ["a + b = 1", "a", "b", "3 a + 3 b - 5", "3 ( a + b ) - 5", "- 2"], "exprs": ["3 ( a + b ) - 5", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a + 3 b - 5"}, {"id": "3 ( a + b ) - 5"}, {"id": "a + b = 1"}, {"id": "- 2"}], "links": [{"rel": "提取因式", "source": "3 a + 3 b - 5", "target": "3 ( a + b ) - 5"}, {"rel": "被代入", "source": "3 ( a + b ) - 5", "target": "- 2"}, {"rel": "提取因式参考", "source": "a + b = 1", "target": "3 ( a + b ) - 5"}, {"rel": "代入", "source": "a + b = 1", "target": "- 2"}]}}
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{"content": "The equation $x ^ { n + 2 } = 7$ is a one-variable linear equation in $x$. What is the value of $n$?", "answer": "\\frac { 1 } { 8 }", "steps": "$\\because$ The equation $x ^ { n + 2 } = 7$ is a one-variable linear equation in $x$, $\\therefore$ $n + 2 = 1$, solving for $n$ gives $n = - 1$.", "expr_cands": ["2 x + 5 y + 3 = 0", "y", "x", "4 ^ { x } \\cdot 32 ^ { y }", "2 ^ { 2 x } \\cdot 2 ^ { 5 y }", "2 ^ { 2 x + 5 y }", "2 x + 5 y = - 3", "2 ^ { - 3 }", "\\frac { 1 } { 8 }"], "exprs": ["2 ^ { 2 x + 5 y }", "2 x + 5 y = - 3", "\\frac { 1 } { 8 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 ^ { x } \\cdot 32 ^ { y }"}, {"id": "2 ^ { 2 x + 5 y }"}, {"id": "2 x + 5 y + 3 = 0"}, {"id": "2 x + 5 y = - 3"}, {"id": "\\frac { 1 } { 8 }"}], "links": [{"rel": "提取因式", "source": "4 ^ { x } \\cdot 32 ^ { y }", "target": "2 ^ { 2 x + 5 y }"}, {"rel": "被代入", "source": "2 ^ { 2 x + 5 y }", "target": "\\frac { 1 } { 8 }"}, {"rel": "提取因式参考", "source": "2 x + 5 y + 3 = 0", "target": "2 ^ { 2 x + 5 y }"}, {"rel": "移项", "source": "2 x + 5 y + 3 = 0", "target": "2 x + 5 y = - 3"}, {"rel": "代入", "source": "2 x + 5 y = - 3", "target": "\\frac { 1 } { 8 }"}]}}
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{"content": "Given a quadratic equation in one variable $x$, $x ^ 2 + mx + n = 0$, with two real roots $x _ 1 = - 2$ and $x _ 2 = 4$, find the value of $m + n$.", "answer": "- 7", "steps": "$\\because$ The two real roots of the quadratic equation in one variable $x ^ 2 + mx + n = 0$ are $x _ 1 = - 2$ and $x _ 2 = 4$, $\\therefore$ $- 2 + 4 = - m$ and $- 2 \\times 4 = n$, solving for $m = - 2$ and $n = - 8$, $\\therefore$ $m + n = - 10$.", "expr_cands": ["a - b = - 10", "a", "b", "c + d = 3", "d", "c", "( a + d ) - ( b - c )", "a - b + c + d", "- 7"], "exprs": ["a - b + c + d", "- 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a + d ) - ( b - c )"}, {"id": "a - b + c + d"}, {"id": "- 7"}, {"id": "a - b = - 10"}, {"id": "c + d = 3"}], "links": [{"rel": "展开", "source": "( a + d ) - ( b - c )", "target": "a - b + c + d"}, {"rel": "被代入", "source": "a - b + c + d", "target": "- 7"}, {"rel": "代入", "source": "a - b = - 10", "target": "- 7"}, {"rel": "代入", "source": "c + d = 3", "target": "- 7"}]}}
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{"content": "If $a + b = 1$, then $3 a + 3 b - 5$ = ____ ?", "answer": "- \\frac { 2 } { 3 }", "steps": "When $a + b = 1$, the original expression equals $3 ( a + b ) - 5 = 3 - 5 = - 2$.", "expr_cands": ["3 x - 7", "x", "6 x + 13", "3 x - 7 = - ( 6 x + 13 )", "x = - \\frac { 2 } { 3 }", "3 x + 6 x = - 13 + 7", "9 x = - 6", "1"], "exprs": ["3 x - 7 = - ( 6 x + 13 )", "x = - \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "6 x + 13"}, {"id": "3 x - 7 = - ( 6 x + 13 )"}, {"id": "3 x - 7"}, {"id": "代数式 $3 x - 7$ 和 $6 x + 13$ 互为相反数"}, {"id": "x = - \\frac { 2 } { 3 }"}], "links": [{"rel": "被描述", "source": "6 x + 13", "target": "3 x - 7 = - ( 6 x + 13 )"}, {"rel": "等式方程求解", "source": "3 x - 7 = - ( 6 x + 13 )", "target": "x = - \\frac { 2 } { 3 }"}, {"rel": "被描述", "source": "3 x - 7", "target": "3 x - 7 = - ( 6 x + 13 )"}, {"rel": "限制性描述", "source": "代数式 $3 x - 7$ 和 $6 x + 13$ 互为相反数", "target": "3 x - 7 = - ( 6 x + 13 )"}]}}
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{"content": "$2 x + 5 y + 3 = 0$, then the value of $4 ^ { x } \\cdot 32 ^ { y }$ is ____?", "answer": "4", "steps": "Original expression = $2 ^ { 2 x } \\cdot 2 ^ { 5 y } = 2 ^ { 2 x + 5 y }$ , since $2 x + 5 y + 3 = 0$ , therefore $2 x + 5 y = - 3$ , substituting it into the original expression: $2 ^ { - 3 } = \\frac { 1 } { 8 }$ .", "expr_cands": ["\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 7 }", "a - b + c = 12", "a", "b", "c", "2 a - 3 b + c", "\\frac { a } { 2 } = k", "k", "a = 2 k", "b = 3 k", "c = 7 k", "6 k = 12", "2 k - 3 k + 7 k = 12", "k = 2", "a = 4", "b = 6", "c = 14", "4"], "exprs": ["a = 2 k", "b = 3 k", "c = 7 k", "2 k - 3 k + 7 k = 12", "k = 2", "a = 4", "b = 6", "c = 14", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 7 } = k$"}, {"id": "a = 2 k"}, {"id": "b = 3 k"}, {"id": "c = 7 k"}, {"id": "a - b + c = 12"}, {"id": "2 k - 3 k + 7 k = 12"}, {"id": "k = 2"}, {"id": "a = 4"}, {"id": "b = 6"}, {"id": "c = 14"}, {"id": "2 a - 3 b + c"}, {"id": "4"}], "links": [{"rel": "假设描述", "source": "设 $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 7 } = k$", "target": "a = 2 k"}, {"rel": "假设描述", "source": "设 $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 7 } = k$", "target": "b = 3 k"}, {"rel": "假设描述", "source": "设 $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 7 } = k$", "target": "c = 7 k"}, {"rel": "代入", "source": "a = 2 k", "target": "2 k - 3 k + 7 k = 12"}, {"rel": "被代入", "source": "a = 2 k", "target": "a = 4"}, {"rel": "代入", "source": "b = 3 k", "target": "2 k - 3 k + 7 k = 12"}, {"rel": "被代入", "source": "b = 3 k", "target": "b = 6"}, {"rel": "代入", "source": "c = 7 k", "target": "2 k - 3 k + 7 k = 12"}, {"rel": "被代入", "source": "c = 7 k", "target": "c = 14"}, {"rel": "被代入", "source": "a - b + c = 12", "target": "2 k - 3 k + 7 k = 12"}, {"rel": "等式方程求解", "source": "2 k - 3 k + 7 k = 12", "target": "k = 2"}, {"rel": "代入", "source": "k = 2", "target": "a = 4"}, {"rel": "代入", "source": "k = 2", "target": "b = 6"}, {"rel": "代入", "source": "k = 2", "target": "c = 14"}, {"rel": "代入", "source": "a = 4", "target": "4"}, {"rel": "代入", "source": "b = 6", "target": "4"}, {"rel": "代入", "source": "c = 14", "target": "4"}, {"rel": "被代入", "source": "2 a - 3 b + c", "target": "4"}]}}
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{"content": "Given $a - b = - 10$, $c + d = 3$, then $( a + d ) - ( b - c )$ = ____ ?", "answer": "4", "steps": "When $a - b = - 10$, $c + d = 3$, the original expression $a + d - b + c = a - b + c + d = - 10 + 3 = - 7$.", "expr_cands": ["3 x ^ { 2 } - 4 x + 1 = 0", "x", "a", "3 a ^ { 2 } - 4 a + 5", "3 a ^ { 2 } - 4 a + 1 = 0", "a = \\frac { 1 } { 3 }", "a = 1", "3 a ^ { 2 } - 4 a = - 1", "4"], "exprs": ["3 a ^ { 2 } - 4 a + 1 = 0", "3 a ^ { 2 } - 4 a = - 1", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { 2 } - 4 x + 1 = 0"}, {"id": "3 a ^ { 2 } - 4 a + 1 = 0"}, {"id": "a"}, {"id": "x"}, {"id": "方程 $3 x ^ { 2 } - 4 x + 1 = 0$ 的一个根为 $a$"}, {"id": "3 a ^ { 2 } - 4 a = - 1"}, {"id": "3 a ^ { 2 } - 4 a + 5"}, {"id": "4"}], "links": [{"rel": "被描述", "source": "3 x ^ { 2 } - 4 x + 1 = 0", "target": "3 a ^ { 2 } - 4 a + 1 = 0"}, {"rel": "移项", "source": "3 a ^ { 2 } - 4 a + 1 = 0", "target": "3 a ^ { 2 } - 4 a = - 1"}, {"rel": "被描述", "source": "a", "target": "3 a ^ { 2 } - 4 a + 1 = 0"}, {"rel": "被描述", "source": "x", "target": "3 a ^ { 2 } - 4 a + 1 = 0"}, {"rel": "限制性描述", "source": "方程 $3 x ^ { 2 } - 4 x + 1 = 0$ 的一个根为 $a$", "target": "3 a ^ { 2 } - 4 a + 1 = 0"}, {"rel": "代入", "source": "3 a ^ { 2 } - 4 a = - 1", "target": "4"}, {"rel": "被代入", "source": "3 a ^ { 2 } - 4 a + 5", "target": "4"}]}}
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{"content": "If the algebraic expressions $3 x - 7$ and $6 x + 13$ are opposite, then the value of $x$ is ____?", "answer": "- 2", "steps": "$\\because$ The algebraic expressions $3 x - 7$ and $6 x + 13$ are opposite to each other, $\\therefore$ $3 x - 7 = - ( 6 x + 13 )$, moving terms, we get $3 x + 6 x = - 13 + 7$, combining like terms, we get $9 x = - 6$, coefficient is reduced to $1$, we get $x = - \\frac { 2 } { 3 }$.", "expr_cands": ["x", "{ x } ^ { 2 } - kx - 15 = ( x + 5 ) ( x - 3 )", "k", "x ^ { 2 } - kx - 15 = ( x + 5 ) ( x - 3 )", "x ^ { 2 } - kx - 15", "x ^ { 2 } + 2 x - 15", "- k = 2", "k = - 2"], "exprs": ["- k = 2", "k = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { 2 } - kx - 15 = ( x + 5 ) ( x - 3 )"}, {"id": "- k = 2"}, {"id": "k = - 2"}], "links": [{"rel": "移项", "source": "{ x } ^ { 2 } - kx - 15 = ( x + 5 ) ( x - 3 )", "target": "- k = 2"}, {"rel": "等式方程求解", "source": "- k = 2", "target": "k = - 2"}]}}
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{"content": "If $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 7 }$ and $a - b + c = 12$, then $2 a - 3 b + c$ is ____?", "answer": "0", "steps": "Suppose $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 7 } = k$. Then $a = 2 k$, $b = 3 k$, and $c = 7 k$. Substituting into the equation $a - b + c = 12$, we get $2 k - 3 k + 7 k = 12$, which gives $k = 2$. Therefore, $a = 4$, $b = 6$, and $c = 14$. Thus, $2 a - 3 b + c = 2 \\times 4 - 3 \\times 6 + 14 = 4$.", "expr_cands": ["x", "\\frac { x - m } { 2 } = x + \\frac { m } { 3 }", "m", "\\frac { x + 1 } { 2 } = 3 x - 2", "1", "x = - \\frac { 5 } { 3 } m", "x = 1", "1 - \\frac { 5 } { 3 } m = 1", "m = 0"], "exprs": ["x = - \\frac { 5 } { 3 } m", "x = 1", "1 - \\frac { 5 } { 3 } m = 1", "m = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - m } { 2 } = x + \\frac { m } { 3 }"}, {"id": "x = - \\frac { 5 } { 3 } m"}, {"id": "\\frac { x + 1 } { 2 } = 3 x - 2"}, {"id": "x = 1"}, {"id": "1 - \\frac { 5 } { 3 } m = 1"}, {"id": "1"}, {"id": "关于 $x$ 的一元一次方程 : $\\frac { x - m } { 2 } = x + \\frac { m } { 3 }$ 与 $\\frac { x + 1 } { 2 } = 3 x - 2$ 的解和为 $1$"}, {"id": "m = 0"}], "links": [{"rel": "等式方程部分求解", "source": "\\frac { x - m } { 2 } = x + \\frac { m } { 3 }", "target": "x = - \\frac { 5 } { 3 } m"}, {"rel": "被描述", "source": "x = - \\frac { 5 } { 3 } m", "target": "1 - \\frac { 5 } { 3 } m = 1"}, {"rel": "等式方程求解", "source": "\\frac { x + 1 } { 2 } = 3 x - 2", "target": "x = 1"}, {"rel": "被描述", "source": "x = 1", "target": "1 - \\frac { 5 } { 3 } m = 1"}, {"rel": "等式方程求解", "source": "1 - \\frac { 5 } { 3 } m = 1", "target": "m = 0"}, {"rel": "被描述", "source": "1", "target": "1 - \\frac { 5 } { 3 } m = 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元一次方程 : $\\frac { x - m } { 2 } = x + \\frac { m } { 3 }$ 与 $\\frac { x + 1 } { 2 } = 3 x - 2$ 的解和为 $1$", "target": "1 - \\frac { 5 } { 3 } m = 1"}]}}
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{"content": "If one root of the equation $3 x ^ 2 - 4 x + 1 = 0$ is $a$, then the value of $3 a ^ 2 - 4 a + 5$ is ____?", "answer": "1", "steps": "From the given problem, we have $3 a ^ 2 - 4 a + 1 = 0$. Therefore, $3 a ^ 2 - 4 a = - 1$. Thus, $3 a ^ 2 - 4 a + 5 = - 1 + 5 = 4$.", "expr_cands": ["x", "\\frac { 3 x + a } { 2 } < 1", "a", "1 - 3 x > 0", "x < \\frac { 2 - a } { 3 }", "x < \\frac { 1 } { 3 }", "\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }", "a = 1"], "exprs": ["x < \\frac { 2 - a } { 3 }", "x < \\frac { 1 } { 3 }", "\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 x + a } { 2 } < 1"}, {"id": "x < \\frac { 2 - a } { 3 }"}, {"id": "1 - 3 x > 0"}, {"id": "x < \\frac { 1 } { 3 }"}, {"id": "\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }"}, {"id": "关于 $x$ 的两个不等式 $\\frac { 3 x + a } { 2 } < 1$ 与 $1 - 3 x > 0$ 的解集相同"}, {"id": "a = 1"}], "links": [{"rel": "不等式方程部分求解", "source": "\\frac { 3 x + a } { 2 } < 1", "target": "x < \\frac { 2 - a } { 3 }"}, {"rel": "被描述", "source": "x < \\frac { 2 - a } { 3 }", "target": "\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }"}, {"rel": "不等式方程求解", "source": "1 - 3 x > 0", "target": "x < \\frac { 1 } { 3 }"}, {"rel": "被描述", "source": "x < \\frac { 1 } { 3 }", "target": "\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }"}, {"rel": "等式方程求解", "source": "\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }", "target": "a = 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的两个不等式 $\\frac { 3 x + a } { 2 } < 1$ 与 $1 - 3 x > 0$ 的解集相同", "target": "\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }"}]}}
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{"content": "Given that for any value of $x$, ${ x } ^ { 2 } - kx - 15 = ( x + 5 ) ( x - 3 )$, the value of $k$ is ____?", "answer": "30", "steps": "Since $x ^ 2 - kx - 15 = ( x + 5 ) ( x - 3 )$, we have $x ^ 2 - kx - 15 = x ^ 2 + 2 x - 15$. Therefore, $- k = 2$, so $k = - 2$.", "expr_cands": ["25 ^ { 9 } + 5 ^ { 17 }", "n", "5 ^ { 16 } * 30", "30"], "exprs": ["5 ^ { 16 } * 30", "30"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "25 ^ { 9 } + 5 ^ { 17 }"}, {"id": "5 ^ { 16 } * 30"}, {"id": "30"}, {"id": "n"}, {"id": "$25 ^ { 9 } + 5 ^ { 17 }$ 能被 $n$ 整除"}], "links": [{"rel": "计算", "source": "25 ^ { 9 } + 5 ^ { 17 }", "target": "5 ^ { 16 } * 30"}, {"rel": "被描述", "source": "5 ^ { 16 } * 30", "target": "30"}, {"rel": "被描述", "source": "n", "target": "30"}, {"rel": "限制性描述", "source": "$25 ^ { 9 } + 5 ^ { 17 }$ 能被 $n$ 整除", "target": "30"}]}}
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{"content": "If the sum of the solutions of the one-variable linear equations in $x$: $\\frac { x - m } { 2 } = x + \\frac { m } { 3 }$ and $\\frac { x + 1 } { 2 } = 3 x - 2$ is $1$, then $m$ = ____?", "answer": "- 7", "steps": "Since $\\frac { x - m } { 2 } = x + \\frac { m } { 3 }$, we can solve for $x$ to get $x = - \\frac { 5 } { 3 } m$. Also, since $\\frac { x + 1 } { 2 } = 3 x - 2$, we can solve for $x$ to get $x = 1$. Therefore, $1 - \\frac { 5 } { 3 } m = 1$, and solving for $m$ gives $m = 0$.", "expr_cands": ["\\frac { | m | - 7 } { 7 - m }", "m", "0", "| m | - 7 = 0", "m = - 7", "m = 7", "7 - m \\neq 0", "m \\neq 7"], "exprs": ["| m | - 7 = 0", "7 - m \\neq 0", "m = - 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { | m | - 7 } { 7 - m }"}, {"id": "| m | - 7 = 0"}, {"id": "0"}, {"id": "分式 $\\frac { | m | - 7 } { 7 - m }$ 的值为 $0$"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "7 - m \\neq 0"}, {"id": "m = - 7"}], "links": [{"rel": "被描述", "source": "\\frac { | m | - 7 } { 7 - m }", "target": "| m | - 7 = 0"}, {"rel": "被描述", "source": "\\frac { | m | - 7 } { 7 - m }", "target": "7 - m \\neq 0"}, {"rel": "联立", "source": "| m | - 7 = 0", "target": "m = - 7"}, {"rel": "被描述", "source": "0", "target": "| m | - 7 = 0"}, {"rel": "被描述", "source": "0", "target": "7 - m \\neq 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { | m | - 7 } { 7 - m }$ 的值为 $0$", "target": "| m | - 7 = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { | m | - 7 } { 7 - m }$ 的值为 $0$", "target": "7 - m \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "| m | - 7 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "7 - m \\neq 0"}, {"rel": "联立", "source": "7 - m \\neq 0", "target": "m = - 7"}]}}
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{"content": "The solution set of the two inequalities $\\frac { 3 x + a } { 2 } < 1$ and $1 - 3 x > 0$ are the same. Find the value of $a$.", "answer": "- 1", "steps": "From $\\frac { 3 x + a } { 2 } < 1$, we get $x < \\frac { 2 - a } { 3 }$. From $1 - 3 x > 0$, we get $x < \\frac { 1 } { 3 }$. Since the solution sets of the two inequalities are the same, we have $\\frac { 2 - a } { 3 } = \\frac { 1 } { 3 }$, which gives $a = 1$.", "expr_cands": ["x = 2", "x", "mx + 2 = 0", "m", "2 m + 2 = 0", "m = - 1"], "exprs": ["2 m + 2 = 0", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 2"}, {"id": "2 m + 2 = 0"}, {"id": "mx + 2 = 0"}, {"id": "m = - 1"}], "links": [{"rel": "代入", "source": "x = 2", "target": "2 m + 2 = 0"}, {"rel": "等式方程求解", "source": "2 m + 2 = 0", "target": "m = - 1"}, {"rel": "被代入", "source": "mx + 2 = 0", "target": "2 m + 2 = 0"}]}}
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{"content": "If $25 ^ { 9 } + 5 ^ { 17 }$ is divisible by $n$, then the possible values of $n$ are ____?", "answer": "- 7", "steps": "$25 ^ { 9 } + 5 ^ { 17 } = 5 ^ { 18 } + 5 ^ { 17 } = 5 ^ { 17 } * ( 5 + 1 ) = 5 ^ { 17 } * 6 = 5 ^ { 16 } * 30$ . $\\therefore$ If $25 ^ { 9 } + 5 ^ { 17 }$ is divisible by $n$, then $n$ could be $30$.", "expr_cands": ["8 { x } ^ { 2 } - ( x - 1 ) x - k - 7 = 0", "k", "x", "0", "x = 0", "- k - 7 = 0", "k = - 7"], "exprs": ["x = 0", "- k - 7 = 0", "k = - 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "0"}, {"id": "x = 0"}, {"id": "8 { x } ^ { 2 } - ( x - 1 ) x - k - 7 = 0"}, {"id": "x"}, {"id": "方程 $8 { x } ^ { 2 } - ( x - 1 ) x - k - 7 = 0$ 的一个根为 $0$"}, {"id": "- k - 7 = 0"}, {"id": "k = - 7"}], "links": [{"rel": "被描述", "source": "0", "target": "x = 0"}, {"rel": "代入", "source": "x = 0", "target": "- k - 7 = 0"}, {"rel": "被描述", "source": "8 { x } ^ { 2 } - ( x - 1 ) x - k - 7 = 0", "target": "x = 0"}, {"rel": "被代入", "source": "8 { x } ^ { 2 } - ( x - 1 ) x - k - 7 = 0", "target": "- k - 7 = 0"}, {"rel": "被描述", "source": "x", "target": "x = 0"}, {"rel": "限制性描述", "source": "方程 $8 { x } ^ { 2 } - ( x - 1 ) x - k - 7 = 0$ 的一个根为 $0$", "target": "x = 0"}, {"rel": "等式方程求解", "source": "- k - 7 = 0", "target": "k = - 7"}]}}
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{"content": "If the value of the fraction $\\frac { | m | - 7 } { 7 - m }$ is $0$, then the value of $m$ is ____?", "answer": "7", "steps": "From the given information, we have $| m | - 7 = 0$, $7 - m \\neq 0$, and $m = - 7$.", "expr_cands": ["x", "3", "2", "5", "3 x - 2 - 2 x = 5", "x = 7"], "exprs": ["3 x - 2 - 2 x = 5", "x = 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "3 x - 2 - 2 x = 5"}, {"id": "x"}, {"id": "2"}, {"id": "5"}, {"id": "$x$ 的 $3$ 倍与 $2$ 的差比 $x$ 的 $2$ 倍大 $5$"}, {"id": "x = 7"}], "links": [{"rel": "被描述", "source": "3", "target": "3 x - 2 - 2 x = 5"}, {"rel": "等式方程求解", "source": "3 x - 2 - 2 x = 5", "target": "x = 7"}, {"rel": "被描述", "source": "x", "target": "3 x - 2 - 2 x = 5"}, {"rel": "被描述", "source": "2", "target": "3 x - 2 - 2 x = 5"}, {"rel": "被描述", "source": "5", "target": "3 x - 2 - 2 x = 5"}, {"rel": "限制性描述", "source": "$x$ 的 $3$ 倍与 $2$ 的差比 $x$ 的 $2$ 倍大 $5$", "target": "3 x - 2 - 2 x = 5"}]}}
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{"content": "Given that $x = 2$ is a solution to the one-variable linear equation $mx + 2 = 0$, what is the value of $m$?", "answer": "84", "steps": "Substituting $x = 2$ into the equation gives $2 m + 2 = 0$, which yields $m = - 1$.", "expr_cands": ["a", "9", "b", "a + b", "a = 3", "b = 81", "84"], "exprs": ["a = 3", "b = 81", "84"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a = 3"}, {"id": "9"}, {"id": "$a$ 是 $9$ 的算术平方根"}, {"id": "b = 81"}, {"id": "b"}, {"id": "而 $b$ 的算术平方根是 $9$"}, {"id": "a + b"}, {"id": "84"}], "links": [{"rel": "被描述", "source": "a", "target": "a = 3"}, {"rel": "代入", "source": "a = 3", "target": "84"}, {"rel": "被描述", "source": "9", "target": "a = 3"}, {"rel": "被描述", "source": "9", "target": "b = 81"}, {"rel": "限制性描述", "source": "$a$ 是 $9$ 的算术平方根", "target": "a = 3"}, {"rel": "代入", "source": "b = 81", "target": "84"}, {"rel": "被描述", "source": "b", "target": "b = 81"}, {"rel": "限制性描述", "source": "而 $b$ 的算术平方根是 $9$", "target": "b = 81"}, {"rel": "被代入", "source": "a + b", "target": "84"}]}}
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{"content": "The equation $8 { x } ^ { 2 } - ( x - 1 ) x - k - 7 = 0$ has a root of $0$, then the value of $k$ is ____?", "answer": "52", "steps": "Because one root is $0$, substituting $x = 0$ gives $- k - 7 = 0$, so $k = - 7$.", "expr_cands": ["m + n = 10", "m", "n", "mn = 24", "m ^ { 2 } + n ^ { 2 }", "( m + n ) ^ { 2 } - 2 mn", "52"], "exprs": ["( m + n ) ^ { 2 } - 2 mn", "52"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m + n = 10"}, {"id": "( m + n ) ^ { 2 } - 2 mn"}, {"id": "m ^ { 2 } + n ^ { 2 }"}, {"id": "52"}, {"id": "mn = 24"}], "links": [{"rel": "提取因式参考", "source": "m + n = 10", "target": "( m + n ) ^ { 2 } - 2 mn"}, {"rel": "代入", "source": "m + n = 10", "target": "52"}, {"rel": "被代入", "source": "( m + n ) ^ { 2 } - 2 mn", "target": "52"}, {"rel": "提取因式", "source": "m ^ { 2 } + n ^ { 2 }", "target": "( m + n ) ^ { 2 } - 2 mn"}, {"rel": "代入", "source": "mn = 24", "target": "52"}]}}
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{"content": "Given that three times $x$ minus two is five more than two times $x$, find the value of $x$.", "answer": "a > - \\frac { 3 } { 5 }", "steps": "$\\because$ The triple of $x$ subtracted by $2$ is $5$ more than double of $x$, $\\therefore$ $3 x - 2 - 2 x = 5$, $\\therefore$ $x = 7$.", "expr_cands": ["x", "5 a + x = - 3", "a", "5 a + x = - 3 - 5 a", "- 5 a - 3 = - 3", "- 3 - 5 a < 0", "- \\frac { 3 } { 5 } < a", "a > - \\frac { 3 } { 5 }"], "exprs": ["- 3 - 5 a < 0", "a > - \\frac { 3 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 a + x = - 3"}, {"id": "- 3 - 5 a < 0"}, {"id": "关于 $x$ 的方程 $5 a + x = - 3$ 的解是负数"}, {"id": "a > - \\frac { 3 } { 5 }"}], "links": [{"rel": "被描述", "source": "5 a + x = - 3", "target": "- 3 - 5 a < 0"}, {"rel": "不等式方程求解", "source": "- 3 - 5 a < 0", "target": "a > - \\frac { 3 } { 5 }"}, {"rel": "属性描述", "source": "关于 $x$ 的方程 $5 a + x = - 3$ 的解是负数", "target": "- 3 - 5 a < 0"}]}}
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{"content": "$a$ is the arithmetic square root of $9$, and $b$ is the arithmetic square root of $9$. What is $a + b$?", "answer": "- 1", "steps": "$\\because a$ is the arithmetic square root of $9$, $\\therefore a = 3$. Also, $\\because$ the arithmetic square root of $b$ is $9$, $\\therefore b = 81$. Therefore, $a + b = 3 + 81 = 84$.", "expr_cands": ["x ^ { 2 } + mx + n", "x", "n", "m", "( x + 2 ) ( x - 1 )", "m + n", "x ^ { 2 } + mx + n = x ^ { 2 } + x - 2", "m = 1", "n = - 2", "- 1"], "exprs": ["x ^ { 2 } + mx + n = x ^ { 2 } + x - 2", "m = 1", "n = - 2", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + mx + n"}, {"id": "x ^ { 2 } + mx + n = x ^ { 2 } + x - 2"}, {"id": "( x + 2 ) ( x - 1 )"}, {"id": "$x ^ { 2 } + mx + n$ 分解因式的结果是 $( x + 2 ) ( x - 1 )$"}, {"id": "m = 1"}, {"id": "n = - 2"}, {"id": "m + n"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + mx + n", "target": "x ^ { 2 } + mx + n = x ^ { 2 } + x - 2"}, {"rel": "移项", "source": "x ^ { 2 } + mx + n = x ^ { 2 } + x - 2", "target": "m = 1"}, {"rel": "移项", "source": "x ^ { 2 } + mx + n = x ^ { 2 } + x - 2", "target": "n = - 2"}, {"rel": "被描述", "source": "( x + 2 ) ( x - 1 )", "target": "x ^ { 2 } + mx + n = x ^ { 2 } + x - 2"}, {"rel": "限制性描述", "source": "$x ^ { 2 } + mx + n$ 分解因式的结果是 $( x + 2 ) ( x - 1 )$", "target": "x ^ { 2 } + mx + n = x ^ { 2 } + x - 2"}, {"rel": "代入", "source": "m = 1", "target": "- 1"}, {"rel": "代入", "source": "n = - 2", "target": "- 1"}, {"rel": "被代入", "source": "m + n", "target": "- 1"}]}}
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{"content": "If $m + n = 10$ and $mn = 24$, then $m ^ 2 + n ^ 2$ = ____ ?", "answer": "x \\ge - 3", "steps": "Because $m + n = 10$ and $mn = 24$, therefore $m ^ 2 + n ^ 2 = ( m + n ) ^ 2 - 2 mn = 100 - 48 = 52$.", "expr_cands": ["x = - 3", "x", "\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } = 1", "a", "\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } \\ge 1", "- \\frac { a } { 3 } - \\frac { 9 } { 4 } = 1", "a = - \\frac { 39 } { 4 }", "x \\ge - 3"], "exprs": ["- \\frac { a } { 3 } - \\frac { 9 } { 4 } = 1", "a = - \\frac { 39 } { 4 }", "x \\ge - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } = 1"}, {"id": "- \\frac { a } { 3 } - \\frac { 9 } { 4 } = 1"}, {"id": "x = - 3"}, {"id": "a = - \\frac { 39 } { 4 }"}, {"id": "x \\ge - 3"}, {"id": "\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } \\ge 1"}], "links": [{"rel": "被代入", "source": "\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } = 1", "target": "- \\frac { a } { 3 } - \\frac { 9 } { 4 } = 1"}, {"rel": "等式方程求解", "source": "- \\frac { a } { 3 } - \\frac { 9 } { 4 } = 1", "target": "a = - \\frac { 39 } { 4 }"}, {"rel": "代入", "source": "x = - 3", "target": "- \\frac { a } { 3 } - \\frac { 9 } { 4 } = 1"}, {"rel": "联立", "source": "a = - \\frac { 39 } { 4 }", "target": "x \\ge - 3"}, {"rel": "联立", "source": "\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } \\ge 1", "target": "x \\ge - 3"}]}}
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{"content": "Given that the equation $5 a + x = - 3$ has a negative solution for $x$, the range of possible values for $a$ is _____.", "answer": "y = 2 ( x + 1 ) ^ { 2 }", "steps": "$5 a + x = - 3 x = - 3 - 5 a$ , because the solution to the inequality $5 a + x = - 3$ in terms of $x$ is negative, therefore $- 3 - 5 a < 0$ and we get $a > - \\frac { 3 } { 5 }$.", "expr_cands": ["y = 2 x ^ { 2 }", "x", "y", "1", "y = 2 ( x + 1 ) ^ { 2 }", "2 x ^ { 2 } = 2 ( x + 1 ) ^ { 2 }"], "exprs": ["y = 2 ( x + 1 ) ^ { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 2 x ^ { 2 }"}, {"id": "y = 2 ( x + 1 ) ^ { 2 }"}, {"id": "1"}, {"id": "将抛物线 $y = 2 x ^ { 2 }$ 向左平移 $1$ 个单位长度"}], "links": [{"rel": "被描述", "source": "y = 2 x ^ { 2 }", "target": "y = 2 ( x + 1 ) ^ { 2 }"}, {"rel": "被描述", "source": "1", "target": "y = 2 ( x + 1 ) ^ { 2 }"}, {"rel": "限制性描述", "source": "将抛物线 $y = 2 x ^ { 2 }$ 向左平移 $1$ 个单位长度", "target": "y = 2 ( x + 1 ) ^ { 2 }"}]}}
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{"content": "If $x ^ { 2 } + mx + n$ can be factored as $( x + 2 ) ( x - 1 )$, then the value of $m + n$ is ____ ?", "answer": "2", "steps": "Since $x ^ 2 + mx + n$ can be factored as $( x + 2 ) ( x - 1 )$, we have $x ^ 2 + mx + n = x ^ 2 + x - 2$. Therefore, $m = 1$ and $n = - 2$. Thus, $m + n = 1 - 2 = - 1$.", "expr_cands": ["x", "\\frac { x } { x - 2 }", "x - 2 = 0", "x = 2"], "exprs": ["x - 2 = 0", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x - 2 }"}, {"id": "x - 2 = 0"}, {"id": "分式 $\\frac { x } { x - 2 }$ 无意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "\\frac { x } { x - 2 }", "target": "x - 2 = 0"}, {"rel": "等式方程求解", "source": "x - 2 = 0", "target": "x = 2"}, {"rel": "限制性描述", "source": "分式 $\\frac { x } { x - 2 }$ 无意义", "target": "x - 2 = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 2 = 0"}]}}
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{"content": "If $x = - 3$ is a solution of the equation $\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } = 1$ with respect to $x$, then the solution set of $\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } \\geq 1$ is ____?", "answer": "5", "steps": "Substituting $x = - 3$ into the equation $\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } = 1$, we get $a = - \\frac { 39 } { 4 }$. Substituting $a = - \\frac { 39 } { 4 }$ into $\\frac { x - a } { 3 } - \\frac { 2 - x } { 4 } \\geq 1$, we solve to get $x \\geq - 3$.", "expr_cands": ["3 \\times { 3 } ^ { 2 } \\times { 3 } ^ { m } = { 3 } ^ { 8 }", "m", "3 ^ { 1 + 2 + m } = 3 ^ { 8 }", "m = 5", "1 + 2 + m = 8"], "exprs": ["m = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 \\times { 3 } ^ { 2 } \\times { 3 } ^ { m } = { 3 } ^ { 8 }"}, {"id": "m = 5"}], "links": [{"rel": "等式方程求解", "source": "3 \\times { 3 } ^ { 2 } \\times { 3 } ^ { m } = { 3 } ^ { 8 }", "target": "m = 5"}]}}
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{"content": "Translate the above math content in English, you should keep the content wrapped in $unchanged.What is the expression of the parabola obtained by shifting the parabola$y = 2 x ^ { 2 }$ one unit length to the left?", "answer": "3", "steps": "The parabola $y = 2 x ^ { 2 }$ is shifted one unit to the left, and the resulting expression for the parabola is $y = 2 ( x + 1 ) ^ { 2 }$.", "expr_cands": ["2 x - 1", "x", "4 - 3 x", "2 x - 1 + 4 - 3 x = 0", "x = 3", "- x + 3 = 0"], "exprs": ["2 x - 1 + 4 - 3 x = 0", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 1"}, {"id": "2 x - 1 + 4 - 3 x = 0"}, {"id": "4 - 3 x"}, {"id": "代数式 $2 x - 1$ 与 $4 - 3 x$ 的值互为相反数"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "2 x - 1", "target": "2 x - 1 + 4 - 3 x = 0"}, {"rel": "等式方程求解", "source": "2 x - 1 + 4 - 3 x = 0", "target": "x = 3"}, {"rel": "被描述", "source": "4 - 3 x", "target": "2 x - 1 + 4 - 3 x = 0"}, {"rel": "限制性描述", "source": "代数式 $2 x - 1$ 与 $4 - 3 x$ 的值互为相反数", "target": "2 x - 1 + 4 - 3 x = 0"}]}}
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{"content": "When $x$ = ____ ?, the fraction $\\frac { x } { x - 2 }$ is undefined.", "answer": "- 4038", "steps": "When the denominator $x - 2 = 0$, the fraction is undefined, that is, the fraction is undefined when $x = 2$.", "expr_cands": ["a - 2 b = 2019", "b", "a", "4 b - 2 a", "- 2 ( a - 2 b )", "- 4038"], "exprs": ["- 2 ( a - 2 b )", "- 4038"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 b - 2 a"}, {"id": "- 2 ( a - 2 b )"}, {"id": "a - 2 b = 2019"}, {"id": "- 4038"}], "links": [{"rel": "提取因式", "source": "4 b - 2 a", "target": "- 2 ( a - 2 b )"}, {"rel": "被代入", "source": "- 2 ( a - 2 b )", "target": "- 4038"}, {"rel": "提取因式参考", "source": "a - 2 b = 2019", "target": "- 2 ( a - 2 b )"}, {"rel": "代入", "source": "a - 2 b = 2019", "target": "- 4038"}]}}
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{"content": "If $3 \\times { 3 } ^ { 2 } \\times { 3 } ^ { m } = { 3 } ^ { 8 }$, then the value of $m$ is ____?", "answer": "13", "steps": "$3 * 3 ^ { 2 } * 3 ^ { m } = 3 ^ { 1 + 2 + m } = 3 ^ { 8 }$, therefore $1 + 2 + m = 8$, which implies that $m = 5$.", "expr_cands": ["x", "y", "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3", "{ x } ^ { 2 } + { y } ^ { 2 }", "x - 2 \\ge 0", "2 \\le x", "2 - x \\ge 0", "x \\le 2", "x \\ge 2", "x = 2", "y = 3", "x ^ { 2 } + y ^ { 2 }", "13"], "exprs": ["x - 2 \\ge 0", "2 - x \\ge 0", "x \\ge 2", "x \\le 2", "x = 2", "y = 3", "13"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3"}, {"id": "x - 2 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "2 - x \\ge 0"}, {"id": "x \\ge 2"}, {"id": "x \\le 2"}, {"id": "x = 2"}, {"id": "y = 3"}, {"id": "x ^ { 2 } + y ^ { 2 }"}, {"id": "13"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3", "target": "x - 2 \\ge 0"}, {"rel": "被描述", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3", "target": "2 - x \\ge 0"}, {"rel": "被代入", "source": "y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3", "target": "y = 3"}, {"rel": "不等式方程求解", "source": "x - 2 \\ge 0", "target": "x \\ge 2"}, {"rel": "不等式方程求解", "source": "x - 2 \\ge 0", "target": "x \\le 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 2 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2 - x \\ge 0"}, {"rel": "联立", "source": "x \\ge 2", "target": "x = 2"}, {"rel": "联立", "source": "x \\le 2", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "y = 3"}, {"rel": "代入", "source": "x = 2", "target": "13"}, {"rel": "代入", "source": "y = 3", "target": "13"}, {"rel": "被代入", "source": "x ^ { 2 } + y ^ { 2 }", "target": "13"}]}}
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{"content": "The algebraic expression $2 x - 1$ and $4 - 3 x$ are opposite in value, then $x$ equals ____?", "answer": "- 3", "steps": "$\\because$ The values of the algebraic expressions $2 x - 1$ and $4 - 3 x$ are opposite to each other. $\\therefore$ $2 x - 1 + 4 - 3 x = 0$. Combining like terms, we get $- x + 3 = 0$. Solving for $x$, we get $x = 3$.", "expr_cands": ["\\frac { 2 x } { x + 3 }", "x", "x + 3 = 0", "x = - 3"], "exprs": ["x + 3 = 0", "x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 x } { x + 3 }"}, {"id": "x + 3 = 0"}, {"id": "分式 $\\frac { 2 x } { x + 3 }$ 无意义的条件是 $x$ ="}, {"id": "分式有意义，则分母不为0"}, {"id": "x = - 3"}], "links": [{"rel": "被描述", "source": "\\frac { 2 x } { x + 3 }", "target": "x + 3 = 0"}, {"rel": "等式方程求解", "source": "x + 3 = 0", "target": "x = - 3"}, {"rel": "限制性描述", "source": "分式 $\\frac { 2 x } { x + 3 }$ 无意义的条件是 $x$ =", "target": "x + 3 = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x + 3 = 0"}]}}
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{"content": "If $a - 2 b = 2019$, then $4 b - 2 a$ = ____ ?", "answer": "- 5", "steps": "Because $a - 2 b = 2019$, therefore $4 b - 2 a = - 2 ( a - 2 b ) = - 2 * 2019 = - 4038$.", "expr_cands": ["a", "b", "3 a + 3 b - 5", "a + b = 0", "3 ( a + b ) - 5", "- 5"], "exprs": ["a + b = 0", "3 ( a + b ) - 5", "- 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ , $b$ 互为相反数"}, {"id": "3 a + 3 b - 5"}, {"id": "3 ( a + b ) - 5"}, {"id": "- 5"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "提取因式参考", "source": "a + b = 0", "target": "3 ( a + b ) - 5"}, {"rel": "代入", "source": "a + b = 0", "target": "- 5"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ , $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "提取因式", "source": "3 a + 3 b - 5", "target": "3 ( a + b ) - 5"}, {"rel": "被代入", "source": "3 ( a + b ) - 5", "target": "- 5"}]}}
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{"content": "Given that $x$ and $y$ are real numbers and $y = \\sqrt { x - 2 } + \\sqrt { 2 - x } + 3$, what is ${ x } ^ 2 + { y } ^ 2$?", "answer": "\\frac { 2 } { 3 }", "steps": "From the given information, we have $x - 2 \\geq 0$ and $2 - x \\geq 0$, which implies $x \\geq 2$ and $x \\leq 2$. Therefore, $x = 2$. Thus, $y = 3$. Hence, $x ^ 2 + y ^ 2 = 2 ^ 2 + 3 ^ 2 = 13$.", "expr_cands": ["a", "b", "| a - 1 | + | b + \\frac { 1 } { 3 } | = 0", "a + b", "a - 1 = 0", "a = 1", "b + \\frac { 1 } { 3 } = 0", "b = - \\frac { 1 } { 3 }", "\\frac { 2 } { 3 }"], "exprs": ["a - 1 = 0", "b + \\frac { 1 } { 3 } = 0", "a = 1", "b = - \\frac { 1 } { 3 }", "\\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 1 | + | b + \\frac { 1 } { 3 } | = 0"}, {"id": "a - 1 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "b + \\frac { 1 } { 3 } = 0"}, {"id": "a = 1"}, {"id": "b = - \\frac { 1 } { 3 }"}, {"id": "a + b"}, {"id": "\\frac { 2 } { 3 }"}], "links": [{"rel": "被描述", "source": "| a - 1 | + | b + \\frac { 1 } { 3 } | = 0", "target": "a - 1 = 0"}, {"rel": "被描述", "source": "| a - 1 | + | b + \\frac { 1 } { 3 } | = 0", "target": "b + \\frac { 1 } { 3 } = 0"}, {"rel": "等式方程求解", "source": "a - 1 = 0", "target": "a = 1"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a - 1 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b + \\frac { 1 } { 3 } = 0"}, {"rel": "等式方程求解", "source": "b + \\frac { 1 } { 3 } = 0", "target": "b = - \\frac { 1 } { 3 }"}, {"rel": "代入", "source": "a = 1", "target": "\\frac { 2 } { 3 }"}, {"rel": "代入", "source": "b = - \\frac { 1 } { 3 }", "target": "\\frac { 2 } { 3 }"}, {"rel": "被代入", "source": "a + b", "target": "\\frac { 2 } { 3 }"}]}}
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{"content": "The condition for the fraction $\\frac { 2 x } { x + 3 }$ to be undefined is $x$ = ____ ?", "answer": "192", "steps": "$\\because$ The fraction $\\frac { 2 x } { x + 3 }$ is undefined, $\\therefore$ $x + 3 = 0$, $\\therefore$ $x = - 3$.", "expr_cands": ["x : y : z = 2 : 3 : 4", "z", "y", "x", "x + y + z = 18", "xyz", "x = 2 k", "k", "y = 3 k", "z = 4 k", "9 k = 18", "2 k + 3 k + 4 k = 18", "k = 2", "x = 4", "y = 6", "z = 8", "192"], "exprs": ["x = 2 k", "y = 3 k", "z = 4 k", "2 k + 3 k + 4 k = 18", "k = 2", "x = 4", "y = 6", "z = 8", "192"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $x = 2 k$ , $y = 3 k$ , $z = 4 k$"}, {"id": "x = 2 k"}, {"id": "y = 3 k"}, {"id": "z = 4 k"}, {"id": "x + y + z = 18"}, {"id": "2 k + 3 k + 4 k = 18"}, {"id": "k = 2"}, {"id": "x = 4"}, {"id": "y = 6"}, {"id": "z = 8"}, {"id": "xyz"}, {"id": "192"}], "links": [{"rel": "假设描述", "source": "设 $x = 2 k$ , $y = 3 k$ , $z = 4 k$", "target": "x = 2 k"}, {"rel": "假设描述", "source": "设 $x = 2 k$ , $y = 3 k$ , $z = 4 k$", "target": "y = 3 k"}, {"rel": "假设描述", "source": "设 $x = 2 k$ , $y = 3 k$ , $z = 4 k$", "target": "z = 4 k"}, {"rel": "代入", "source": "x = 2 k", "target": "2 k + 3 k + 4 k = 18"}, {"rel": "被代入", "source": "x = 2 k", "target": "x = 4"}, {"rel": "代入", "source": "y = 3 k", "target": "2 k + 3 k + 4 k = 18"}, {"rel": "被代入", "source": "y = 3 k", "target": "y = 6"}, {"rel": "代入", "source": "z = 4 k", "target": "2 k + 3 k + 4 k = 18"}, {"rel": "被代入", "source": "z = 4 k", "target": "z = 8"}, {"rel": "被代入", "source": "x + y + z = 18", "target": "2 k + 3 k + 4 k = 18"}, {"rel": "等式方程求解", "source": "2 k + 3 k + 4 k = 18", "target": "k = 2"}, {"rel": "代入", "source": "k = 2", "target": "x = 4"}, {"rel": "代入", "source": "k = 2", "target": "y = 6"}, {"rel": "代入", "source": "k = 2", "target": "z = 8"}, {"rel": "代入", "source": "x = 4", "target": "192"}, {"rel": "代入", "source": "y = 6", "target": "192"}, {"rel": "代入", "source": "z = 8", "target": "192"}, {"rel": "被代入", "source": "xyz", "target": "192"}]}}
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{"content": "If $a$ and $b$ are opposite numbers, then the expression $3 a + 3 b - 5$ equals ____?", "answer": "x \\le \\frac { 1 } { 2 }", "steps": "Since $a$ and $b$ are opposite numbers, we have $a + b = 0$. Therefore, $3 a + 3 b - 5 = 3 ( a + b ) - 5 = 3 \\times 0 - 5 = - 5$.", "expr_cands": ["\\sqrt { ( 1 - 2 x ) ^ { 2 } } = 1 - 2 x", "x", "1 - 2 x \\ge 0", "x \\le \\frac { 1 } { 2 }"], "exprs": ["1 - 2 x \\ge 0", "x \\le \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { ( 1 - 2 x ) ^ { 2 } } = 1 - 2 x"}, {"id": "1 - 2 x \\ge 0"}, {"id": "$\\sqrt { ( 1 - 2 x ) ^ { 2 } } = 1 - 2 x$ 成立的 $x$ 的取值范围"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\le \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "\\sqrt { ( 1 - 2 x ) ^ { 2 } } = 1 - 2 x", "target": "1 - 2 x \\ge 0"}, {"rel": "不等式方程求解", "source": "1 - 2 x \\ge 0", "target": "x \\le \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "$\\sqrt { ( 1 - 2 x ) ^ { 2 } } = 1 - 2 x$ 成立的 $x$ 的取值范围", "target": "1 - 2 x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "1 - 2 x \\ge 0"}]}}
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{"content": "Given $a$ and $b$ satisfy $| a - 1 | + | b + \\frac { 1 } { 3 } | = 0$, what is the value of $a + b$?", "answer": "- 1", "steps": "From the given information, we have $a - 1 = 0$ and $b + \\frac { 1 } { 3 } = 0$. Solving for $a$ and $b$, we get $a = 1$ and $b = - \\frac { 1 } { 3 }$. Therefore, $a + b = \\frac { 2 } { 3 }$.", "expr_cands": ["m", "m - 1", "2 m + 4", "m - 1 + 2 m + 4 = 0", "m = - 1", "3 m = - 3"], "exprs": ["m - 1 + 2 m + 4 = 0", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m - 1"}, {"id": "m - 1 + 2 m + 4 = 0"}, {"id": "2 m + 4"}, {"id": "代数式 $m - 1$ 与 $2 m + 4$ 互为相反数"}, {"id": "m = - 1"}], "links": [{"rel": "被描述", "source": "m - 1", "target": "m - 1 + 2 m + 4 = 0"}, {"rel": "等式方程求解", "source": "m - 1 + 2 m + 4 = 0", "target": "m = - 1"}, {"rel": "被描述", "source": "2 m + 4", "target": "m - 1 + 2 m + 4 = 0"}, {"rel": "限制性描述", "source": "代数式 $m - 1$ 与 $2 m + 4$ 互为相反数", "target": "m - 1 + 2 m + 4 = 0"}]}}
|
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{"content": "If $x : y : z = 2 : 3 : 4$ and $x + y + z = 18$, then $xyz$ = ____ ?", "answer": "4 \\sqrt { 2 }", "steps": "Since $x : y : z = 2 : 3 : 4$, we assume that $x = 2 k$, $y = 3 k$, and $z = 4 k$. Since $x + y + z = 18$, we have $2 k + 3 k + 4 k = 18$. Solving for $k$, we get $k = 2$. Therefore, $x = 4$, $y = 6$, and $z = 8$. Thus, $xyz = 4 \\times 6 \\times 8 = 192$.", "expr_cands": ["x = 3", "x", "\\sqrt { x ^ { 2 } + 6 x + 5 }", "\\sqrt { ( x + 1 ) ( x + 5 ) }", "4 \\sqrt { 2 }"], "exprs": ["4 \\sqrt { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 3"}, {"id": "4 \\sqrt { 2 }"}, {"id": "\\sqrt { x ^ { 2 } + 6 x + 5 }"}], "links": [{"rel": "代入", "source": "x = 3", "target": "4 \\sqrt { 2 }"}, {"rel": "被代入", "source": "\\sqrt { x ^ { 2 } + 6 x + 5 }", "target": "4 \\sqrt { 2 }"}]}}
|
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{"content": "What is the range of values of $x$ for which $\\sqrt {( 1 - 2 x ) ^ 2 } = 1 - 2 x$ holds true?", "answer": "x = - \\frac { 1 } { 3 }", "steps": "Because $\\sqrt {( 1 - 2 x ) ^ 2 } = 1 - 2 x$, therefore $1 - 2 x \\ge 0$. Solving for $x$, we get $x \\le \\frac { 1 } { 2 }$.", "expr_cands": ["( 3 x + 1 ) ^ { 0 }", "x", "3 x + 1 = 0", "x = - \\frac { 1 } { 3 }"], "exprs": ["3 x + 1 = 0", "x = - \\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 3 x + 1 ) ^ { 0 }"}, {"id": "3 x + 1 = 0"}, {"id": "$( 3 x + 1 ) ^ { 0 }$ 没有意义"}, {"id": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0"}, {"id": "x = - \\frac { 1 } { 3 }"}], "links": [{"rel": "被描述", "source": "( 3 x + 1 ) ^ { 0 }", "target": "3 x + 1 = 0"}, {"rel": "等式方程求解", "source": "3 x + 1 = 0", "target": "x = - \\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "$( 3 x + 1 ) ^ { 0 }$ 没有意义", "target": "3 x + 1 = 0"}, {"rel": "属性描述", "source": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0", "target": "3 x + 1 = 0"}]}}
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{"content": "When $m$ = ____ ?, the algebraic expression $m - 1$ is the opposite of $2 m + 4$.", "answer": "2", "steps": "According to the problem, we have $m - 1 + 2 m + 4 = 0$. By rearranging and combining terms, we get $3 m = - 3$. Solving for $m$, we get $m = - 1$.", "expr_cands": ["x = 1", "x", "mx + 2 = 3 x + 1", "m", "m + 2 = 4", "m + 2 = 3 + 1", "m = 2"], "exprs": ["m + 2 = 3 + 1", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "mx + 2 = 3 x + 1"}, {"id": "m + 2 = 3 + 1"}, {"id": "x = 1"}, {"id": "m = 2"}], "links": [{"rel": "被代入", "source": "mx + 2 = 3 x + 1", "target": "m + 2 = 3 + 1"}, {"rel": "等式方程求解", "source": "m + 2 = 3 + 1", "target": "m = 2"}, {"rel": "代入", "source": "x = 1", "target": "m + 2 = 3 + 1"}]}}
|
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{"content": "If $x = 3$, what is the value of $\\sqrt { x ^ 2 + 6 x + 5 }$?", "answer": "a = 1", "steps": "\\because $x = 3$, \\therefore the original expression = $\\sqrt {( x + 1 ) ( x + 5 )} = \\sqrt { 4 * 8 } = 4 \\sqrt { 2 }$.", "expr_cands": ["x", "2 x ^ { 2 } - ( a - 1 ) x + a = 0", "a", "\\frac { a - 1 } { 2 } = 0", "a = 1"], "exprs": ["\\frac { a - 1 } { 2 } = 0", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 2 } - ( a - 1 ) x + a = 0"}, {"id": "\\frac { a - 1 } { 2 } = 0"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "关于 $x$ 的一元二次方程 $2 x ^ { 2 } - ( a - 1 ) x + a = 0$ 的两个实数根互为相反数"}, {"id": "a = 1"}], "links": [{"rel": "被描述", "source": "2 x ^ { 2 } - ( a - 1 ) x + a = 0", "target": "\\frac { a - 1 } { 2 } = 0"}, {"rel": "等式方程求解", "source": "\\frac { a - 1 } { 2 } = 0", "target": "a = 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "\\frac { a - 1 } { 2 } = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $2 x ^ { 2 } - ( a - 1 ) x + a = 0$ 的两个实数根互为相反数", "target": "\\frac { a - 1 } { 2 } = 0"}]}}
|
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{"content": "If $( 3 x + 1 ) ^ { 0 }$ is undefined, then the value of $x$ is _____.", "answer": "- 1", "steps": "When $3 x + 1 = 0$, $( 3 x + 1 ) ^ 0$ is undefined. Solving for $x$ from $3 x + 1 = 0$, we get $x = - \\frac { 1 } { 3 }$.", "expr_cands": ["x", "y", "mx ^ { 2 } + 2 xy - x", "m", "mx ^ { 2 } - 2 nxy + 3 y", "n", "( mx ^ { 2 } + 2 xy - x ) - ( mx ^ { 2 } - 2 nxy + 3 y )", "( 2 + 2 n ) xy - x - 3 y", "2 + 2 n = 0", "n = - 1"], "exprs": ["( mx ^ { 2 } + 2 xy - x ) - ( mx ^ { 2 } - 2 nxy + 3 y )", "2 + 2 n = 0", "n = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "mx ^ { 2 } + 2 xy - x"}, {"id": "( mx ^ { 2 } + 2 xy - x ) - ( mx ^ { 2 } - 2 nxy + 3 y )"}, {"id": "mx ^ { 2 } - 2 nxy + 3 y"}, {"id": "关于 $x$ , $y$ 的多项式 $mx ^ { 2 } + 2 xy - x$ 与 $mx ^ { 2 } - 2 nxy + 3 y$ 的差不含有二次项"}, {"id": "2 + 2 n = 0"}, {"id": "n = - 1"}], "links": [{"rel": "被描述", "source": "mx ^ { 2 } + 2 xy - x", "target": "( mx ^ { 2 } + 2 xy - x ) - ( mx ^ { 2 } - 2 nxy + 3 y )"}, {"rel": "被描述", "source": "( mx ^ { 2 } + 2 xy - x ) - ( mx ^ { 2 } - 2 nxy + 3 y )", "target": "2 + 2 n = 0"}, {"rel": "被描述", "source": "mx ^ { 2 } - 2 nxy + 3 y", "target": "( mx ^ { 2 } + 2 xy - x ) - ( mx ^ { 2 } - 2 nxy + 3 y )"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的多项式 $mx ^ { 2 } + 2 xy - x$ 与 $mx ^ { 2 } - 2 nxy + 3 y$ 的差不含有二次项", "target": "( mx ^ { 2 } + 2 xy - x ) - ( mx ^ { 2 } - 2 nxy + 3 y )"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的多项式 $mx ^ { 2 } + 2 xy - x$ 与 $mx ^ { 2 } - 2 nxy + 3 y$ 的差不含有二次项", "target": "2 + 2 n = 0"}, {"rel": "等式方程求解", "source": "2 + 2 n = 0", "target": "n = - 1"}]}}
|
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{"content": "Given that $x = 1$ is a solution to the equation $mx + 2 = 3 x + 1$ in terms of $x$, what is the value of $m$?", "answer": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3", "steps": "Substituting $x = 1$ into the equation $mx + 2 = 3 x + 1$ yields $m + 2 = 3 + 1$, which can be solved to obtain $m = 2$.", "expr_cands": ["y = \\frac { 1 } { 2 } x ^ { 2 } - 1", "y", "x", "1", "2", "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3", "\\frac { x ^ { 2 }} { 2 } - 1 = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3", "\\frac { x ^ { 2 }} { 2 } - 1"], "exprs": ["y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1"}, {"id": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"}, {"id": "y = \\frac { 1 } { 2 } x ^ { 2 } - 1"}, {"id": "2"}, {"id": "把抛物线 $y = \\frac { 1 } { 2 } x ^ { 2 } - 1$ 先向右平移 $1$ 个单位"}, {"id": "再向下平移 $2$ 个单位"}, {"id": "到的抛物线的解析式"}], "links": [{"rel": "被描述", "source": "1", "target": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"}, {"rel": "被描述", "source": "y = \\frac { 1 } { 2 } x ^ { 2 } - 1", "target": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"}, {"rel": "被描述", "source": "2", "target": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"}, {"rel": "限制性描述", "source": "把抛物线 $y = \\frac { 1 } { 2 } x ^ { 2 } - 1$ 先向右平移 $1$ 个单位", "target": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"}, {"rel": "限制性描述", "source": "再向下平移 $2$ 个单位", "target": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"}, {"rel": "限制性描述", "source": "到的抛物线的解析式", "target": "y = \\frac { 1 } { 2 } ( x - 1 ) ^ { 2 } - 3"}]}}
|
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{"content": "Regarding the one-variable quadratic equation in $x$: $2 x ^ 2 - ( a - 1 ) x + a = 0$, if the two real roots are opposite in sign, then the value of $a$ is ____?", "answer": "2", "steps": "According to the relationship between roots and coefficients and the property of opposite numbers, we can get: $\\frac { a - 1 } { 2 } = 0$ because $a = 1$.", "expr_cands": ["{ x } ^ { a } y", "x", "a", "y", "3 { x } ^ { 2 } { y } ^ { b }", "b", "ab", "a = 2", "b = 1", "2"], "exprs": ["a = 2", "b = 1", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { a } y"}, {"id": "a = 2"}, {"id": "3 { x } ^ { 2 } { y } ^ { b }"}, {"id": "${ x } ^ { a } y$ 与 $3 { x } ^ { 2 } { y } ^ { b }$ 是同类项"}, {"id": "b = 1"}, {"id": "ab"}, {"id": "2"}], "links": [{"rel": "被描述", "source": "{ x } ^ { a } y", "target": "a = 2"}, {"rel": "被描述", "source": "{ x } ^ { a } y", "target": "b = 1"}, {"rel": "代入", "source": "a = 2", "target": "2"}, {"rel": "被描述", "source": "3 { x } ^ { 2 } { y } ^ { b }", "target": "a = 2"}, {"rel": "被描述", "source": "3 { x } ^ { 2 } { y } ^ { b }", "target": "b = 1"}, {"rel": "限制性描述", "source": "${ x } ^ { a } y$ 与 $3 { x } ^ { 2 } { y } ^ { b }$ 是同类项", "target": "a = 2"}, {"rel": "限制性描述", "source": "${ x } ^ { a } y$ 与 $3 { x } ^ { 2 } { y } ^ { b }$ 是同类项", "target": "b = 1"}, {"rel": "代入", "source": "b = 1", "target": "2"}, {"rel": "被代入", "source": "ab", "target": "2"}]}}
|
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{"content": "Regarding the polynomial in $x$ and $y$ given by $mx ^ 2 + 2 xy - x$ and the polynomial in $x$ and $y$ given by $mx ^ 2 - 2 nxy + 3 y$, if their difference does not contain any quadratic terms, then $n$ = ____?", "answer": "a \\ge 1", "steps": "According to the problem, we have $( mx ^ 2 + 2 xy - x ) - ( mx ^ 2 - 2 nxy + 3 y ) = mx ^ 2 + 2 xy - x - mx ^ 2 + 2 nxy - 3 y = ( 2 + 2 n ) xy - x - 3 y$. Since the result does not contain a quadratic term, we get $2 + 2 n = 0$, which implies $n = - 1$.", "expr_cands": ["\\sqrt { a - 1 }", "a", "a - 1 \\ge 0", "1 \\le a", "a \\ge 1"], "exprs": ["a - 1 \\ge 0", "a \\ge 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { a - 1 }"}, {"id": "a - 1 \\ge 0"}, {"id": "二次根式 $\\sqrt { a - 1 }$ 中"}, {"id": "$a$ 的取值范围"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a \\ge 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { a - 1 }", "target": "a - 1 \\ge 0"}, {"rel": "不等式方程求解", "source": "a - 1 \\ge 0", "target": "a \\ge 1"}, {"rel": "限制性描述", "source": "二次根式 $\\sqrt { a - 1 }$ 中", "target": "a - 1 \\ge 0"}, {"rel": "限制性描述", "source": "$a$ 的取值范围", "target": "a - 1 \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "a - 1 \\ge 0"}]}}
|
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{"content": "Translate the above math content in English, you should keep the content wrapped in $ unchanged. Translate the given mathematical expression: y = 1/2 x^2 - 1 to a new parabola by first translating it 1 unit to the right and then 2 units down. What is the new equation of the parabola?", "answer": "6", "steps": "Translate the above math content in English, you should keep the content wrapped in $unchanged.The parabola$y = \\frac{1}{2}x^2 - 1$is first translated one unit to the right and then two units down , resulting in the parabola with equation$y = \\frac{1}{2}(x-1)^2 - 3$.", "expr_cands": ["\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 4 }", "2 a + 5 b - 3 c = 21", "b", "a", "c", "\\frac { a } { 2 } = \\frac { c } { 4 }", "a = 2 x", "x", "b = 3 x", "c = 4 x", "7 x = 21", "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21", "x = 3", "a = 6"], "exprs": ["a = 2 x", "b = 3 x", "c = 4 x", "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21", "x = 3", "a = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $a = 2 x$ , $b = 3 x$ , $c = 4 x$"}, {"id": "a = 2 x"}, {"id": "b = 3 x"}, {"id": "c = 4 x"}, {"id": "2 a + 5 b - 3 c = 21"}, {"id": "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21"}, {"id": "x = 3"}, {"id": "a = 6"}], "links": [{"rel": "假设描述", "source": "设 $a = 2 x$ , $b = 3 x$ , $c = 4 x$", "target": "a = 2 x"}, {"rel": "假设描述", "source": "设 $a = 2 x$ , $b = 3 x$ , $c = 4 x$", "target": "b = 3 x"}, {"rel": "假设描述", "source": "设 $a = 2 x$ , $b = 3 x$ , $c = 4 x$", "target": "c = 4 x"}, {"rel": "代入", "source": "a = 2 x", "target": "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21"}, {"rel": "被代入", "source": "a = 2 x", "target": "a = 6"}, {"rel": "代入", "source": "b = 3 x", "target": "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21"}, {"rel": "代入", "source": "c = 4 x", "target": "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21"}, {"rel": "被代入", "source": "2 a + 5 b - 3 c = 21", "target": "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21"}, {"rel": "等式方程求解", "source": "2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21", "target": "x = 3"}, {"rel": "代入", "source": "x = 3", "target": "a = 6"}]}}
|
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{"content": "If ${ x } ^ { a } y$ and $3 { x } ^ { 2 } { y } ^ { b }$ are similar terms, then the value of $ab$ is ____?", "answer": "- 2", "steps": "Because ${ x } ^ { a } y$ and $3 { x } ^ { 2 } { y } ^ { b }$ are like terms, so $a = 2$, $b = 1$, therefore $ab = 2 * 1 = 2$.", "expr_cands": ["2 x + 4 = 0", "x", "3 x - 4 = 2 x + a", "a", "x = - 2", "- 10 = a - 4", "a - 4 = a - 4", "x = 2", "6 - 4 = 4 + a", "a = - 2"], "exprs": ["x = - 2", "x = 2", "6 - 4 = 4 + a", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 4 = 0"}, {"id": "x = - 2"}, {"id": "x = 2"}, {"id": "因为方程 $2 x + 4 = 0$ 与方程 $3 x - 4 = 2 x + a$ 的解互为相反数"}, {"id": "方程 $3 x - 4 = 2 x + a$ 的解为 $x = 2$"}, {"id": "3 x - 4 = 2 x + a"}, {"id": "6 - 4 = 4 + a"}, {"id": "a = - 2"}], "links": [{"rel": "等式方程求解", "source": "2 x + 4 = 0", "target": "x = - 2"}, {"rel": "被描述", "source": "x = - 2", "target": "x = 2"}, {"rel": "代入", "source": "x = 2", "target": "6 - 4 = 4 + a"}, {"rel": "限制性描述", "source": "因为方程 $2 x + 4 = 0$ 与方程 $3 x - 4 = 2 x + a$ 的解互为相反数", "target": "x = 2"}, {"rel": "限制性描述", "source": "方程 $3 x - 4 = 2 x + a$ 的解为 $x = 2$", "target": "x = 2"}, {"rel": "被代入", "source": "3 x - 4 = 2 x + a", "target": "6 - 4 = 4 + a"}, {"rel": "等式方程求解", "source": "6 - 4 = 4 + a", "target": "a = - 2"}]}}
|
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{"content": "In the second root $\\sqrt { a - 1 }$, the range of values for $a$ is ____?", "answer": "1", "steps": "From the given condition, we have $a - 1 \\ge 0$, which implies that $a \\ge 1$.", "expr_cands": ["a - 2 b = 3", "b", "a", "2 a - 4 b - 5", "2 ( a - 2 b ) - 5", "1"], "exprs": ["2 ( a - 2 b ) - 5", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a - 4 b - 5"}, {"id": "2 ( a - 2 b ) - 5"}, {"id": "a - 2 b = 3"}, {"id": "1"}], "links": [{"rel": "提取因式", "source": "2 a - 4 b - 5", "target": "2 ( a - 2 b ) - 5"}, {"rel": "被代入", "source": "2 ( a - 2 b ) - 5", "target": "1"}, {"rel": "提取因式参考", "source": "a - 2 b = 3", "target": "2 ( a - 2 b ) - 5"}, {"rel": "代入", "source": "a - 2 b = 3", "target": "1"}]}}
|
||||
{"content": "Given $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 4 }$, $2 a + 5 b - 3 c = 21$, find the value of $a$.", "answer": "x \\neq - 3", "steps": "Since $\\frac { a } { 2 } = \\frac { b } { 3 } = \\frac { c } { 4 }$, we can assume that $a = 2 x$, $b = 3 x$, and $c = 4 x$. Since $2 a + 5 b - 3 c = 21$, we have $2 \\times 2 x + 5 \\times 3 x - 3 \\times 4 x = 21$. Solving for $x$, we get $x = 3$, so $a = 2 x = 6$.", "expr_cands": ["y = \\frac { x + 1 } { x + 3 }", "y", "x", "x + 3 \\neq 0", "x \\neq - 3"], "exprs": ["x + 3 \\neq 0", "x \\neq - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { x + 1 } { x + 3 }"}, {"id": "x + 3 \\neq 0"}, {"id": "在函数 $y = \\frac { x + 1 } { x + 3 }$ 中"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq - 3"}], "links": [{"rel": "被描述", "source": "y = \\frac { x + 1 } { x + 3 }", "target": "x + 3 \\neq 0"}, {"rel": "不等式方程求解", "source": "x + 3 \\neq 0", "target": "x \\neq - 3"}, {"rel": "限制性描述", "source": "在函数 $y = \\frac { x + 1 } { x + 3 }$ 中", "target": "x + 3 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x + 3 \\neq 0"}]}}
|
||||
{"content": "If the equation $2 x + 4 = 0$ and the equation $3 x - 4 = 2 x + a$ have solutions that are opposite in sign, then $a$ = ____?", "answer": "3", "steps": "Solve the equation $2 x + 4 = 0$, we get $x = - 2$. Since the solutions of the equation $2 x + 4 = 0$ and the equation $3 x - 4 = 2 x + a$ are opposite, the solution of the equation $3 x - 4 = 2 x + a$ is $x = 2$. Substituting $x = 2$ into the equation, we get $6 - 4 = 4 + a$, so $a = - 2$.", "expr_cands": ["x", "\\frac { 7 } { x - 3 }", "x - 3 = 0", "x = 3"], "exprs": ["x - 3 = 0", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 7 } { x - 3 }"}, {"id": "x - 3 = 0"}, {"id": "分式 $\\frac { 7 } { x - 3 }$ 的值不存在"}, {"id": "分式有意义，则分母不为0"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "\\frac { 7 } { x - 3 }", "target": "x - 3 = 0"}, {"rel": "等式方程求解", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "限制性描述", "source": "分式 $\\frac { 7 } { x - 3 }$ 的值不存在", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 3 = 0"}]}}
|
||||
{"content": "If $a - 2 b = 3$, then $2 a - 4 b - 5$ = ____ ?", "answer": "- 2", "steps": "Because $a - 2 b = 3$, therefore the original expression equals $2 ( a - 2 b ) - 5 = 2 * 3 - 5 = 1$.", "expr_cands": ["{ 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } = 2", "n", "n = - 2", "4 * 2 ^ { n + 1 } = 2", "2 ^ { n + 3 } = 2", "n + 3 = 1"], "exprs": ["n = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } = 2"}, {"id": "n = - 2"}], "links": [{"rel": "等式方程求解", "source": "{ 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } + { 2 } ^ { n + 1 } = 2", "target": "n = - 2"}]}}
|
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{"content": "In the function $y = \\frac { x + 1 } { x + 3 }$, what is the range of the independent variable $x$?", "answer": "3", "steps": "According to the problem, we have $x + 3 \\neq 0$, which implies that $x \\neq - 3$.", "expr_cands": ["\\frac { 3 } { 5 } x ^ { 2 } y", "x", "y", "\\frac { 5 } { 6 } x ^ { a } y ^ { b }", "b", "a", "a + b", "a = 2", "b = 1", "3"], "exprs": ["a = 2", "b = 1", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 } { 5 } x ^ { 2 } y"}, {"id": "a = 2"}, {"id": "\\frac { 5 } { 6 } x ^ { a } y ^ { b }"}, {"id": "单项式 $\\frac { 3 } { 5 } x ^ { 2 } y$ 与 $\\frac { 5 } { 6 } x ^ { a } y ^ { b }$ 可以合并"}, {"id": "单项式 $\\frac { 3 } { 5 } x ^ { 2 } y$ 与 $\\frac { 5 } { 6 } x ^ { a } y ^ { b }$ 是同类项"}, {"id": "b = 1"}, {"id": "a + b"}, {"id": "3"}], "links": [{"rel": "被描述", "source": "\\frac { 3 } { 5 } x ^ { 2 } y", "target": "a = 2"}, {"rel": "被描述", "source": "\\frac { 3 } { 5 } x ^ { 2 } y", "target": "b = 1"}, {"rel": "代入", "source": "a = 2", "target": "3"}, {"rel": "被描述", "source": "\\frac { 5 } { 6 } x ^ { a } y ^ { b }", "target": "a = 2"}, {"rel": "被描述", "source": "\\frac { 5 } { 6 } x ^ { a } y ^ { b }", "target": "b = 1"}, {"rel": "限制性描述", "source": "单项式 $\\frac { 3 } { 5 } x ^ { 2 } y$ 与 $\\frac { 5 } { 6 } x ^ { a } y ^ { b }$ 可以合并", "target": "a = 2"}, {"rel": "限制性描述", "source": "单项式 $\\frac { 3 } { 5 } x ^ { 2 } y$ 与 $\\frac { 5 } { 6 } x ^ { a } y ^ { b }$ 可以合并", "target": "b = 1"}, {"rel": "限制性描述", "source": "单项式 $\\frac { 3 } { 5 } x ^ { 2 } y$ 与 $\\frac { 5 } { 6 } x ^ { a } y ^ { b }$ 是同类项", "target": "a = 2"}, {"rel": "限制性描述", "source": "单项式 $\\frac { 3 } { 5 } x ^ { 2 } y$ 与 $\\frac { 5 } { 6 } x ^ { a } y ^ { b }$ 是同类项", "target": "b = 1"}, {"rel": "代入", "source": "b = 1", "target": "3"}, {"rel": "被代入", "source": "a + b", "target": "3"}]}}
|
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{"content": "When $x$ = ____ ?, the value of the fraction $\\frac { 7 } { x - 3 }$ does not exist.", "answer": "3", "steps": "$\\because$ The value of the fraction $\\frac { 7 } { x - 3 }$ does not exist, $\\therefore$ $x - 3 = 0$, which solves to $x = 3$.", "expr_cands": ["k", "{ x } ^ { 2 } - ( k - 3 ) xy - 8", "x", "y", "xy", "0", "k - 3 = 0", "k = 3"], "exprs": ["k - 3 = 0", "k = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ x } ^ { 2 } - ( k - 3 ) xy - 8"}, {"id": "k - 3 = 0"}, {"id": "代数式 ${ x } ^ { 2 } - ( k - 3 ) xy - 8$ 不含 $xy$ 项"}, {"id": "k = 3"}], "links": [{"rel": "被描述", "source": "{ x } ^ { 2 } - ( k - 3 ) xy - 8", "target": "k - 3 = 0"}, {"rel": "等式方程求解", "source": "k - 3 = 0", "target": "k = 3"}, {"rel": "限制性描述", "source": "代数式 ${ x } ^ { 2 } - ( k - 3 ) xy - 8$ 不含 $xy$ 项", "target": "k - 3 = 0"}]}}
|
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{"content": "If $2 ^ { n + 1 } + 2 ^ { n + 1 } + 2 ^ { n + 1 } + 2 ^ { n + 1 } = 2$, what is the value of $n$?", "answer": "0", "steps": "Because $2 ^ { n + 1 } + 2 ^ { n + 1 } + 2 ^ { n + 1 } + 2 ^ { n + 1 } = 2$, therefore $4 * 2 ^ { n + 1 } = 2$, therefore $2 ^ { n + 3 } = 2$, therefore $n + 3 = 1$, therefore $n = - 2$.", "expr_cands": ["- 2 a ^ { 2 } + 3 b + 8", "a", "b", "2", "- 4 a ^ { 2 } + 6 b + 12", "- 2 a ^ { 2 } + 3 b + 8 = 2", "- 2 a ^ { 2 } + 3 b = - 6", "2 ( - 2 a ^ { 2 } + 3 b ) + 12", "0"], "exprs": ["- 2 a ^ { 2 } + 3 b + 8 = 2", "- 2 a ^ { 2 } + 3 b = - 6", "2 ( - 2 a ^ { 2 } + 3 b ) + 12", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 a ^ { 2 } + 3 b + 8"}, {"id": "- 2 a ^ { 2 } + 3 b + 8 = 2"}, {"id": "2"}, {"id": "代数式 $- 2 a ^ { 2 } + 3 b + 8$ 的值为 $2$"}, {"id": "- 2 a ^ { 2 } + 3 b = - 6"}, {"id": "- 4 a ^ { 2 } + 6 b + 12"}, {"id": "2 ( - 2 a ^ { 2 } + 3 b ) + 12"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "- 2 a ^ { 2 } + 3 b + 8", "target": "- 2 a ^ { 2 } + 3 b + 8 = 2"}, {"rel": "移项", "source": "- 2 a ^ { 2 } + 3 b + 8 = 2", "target": "- 2 a ^ { 2 } + 3 b = - 6"}, {"rel": "被描述", "source": "2", "target": "- 2 a ^ { 2 } + 3 b + 8 = 2"}, {"rel": "限制性描述", "source": "代数式 $- 2 a ^ { 2 } + 3 b + 8$ 的值为 $2$", "target": "- 2 a ^ { 2 } + 3 b + 8 = 2"}, {"rel": "提取因式参考", "source": "- 2 a ^ { 2 } + 3 b = - 6", "target": "2 ( - 2 a ^ { 2 } + 3 b ) + 12"}, {"rel": "代入", "source": "- 2 a ^ { 2 } + 3 b = - 6", "target": "0"}, {"rel": "提取因式", "source": "- 4 a ^ { 2 } + 6 b + 12", "target": "2 ( - 2 a ^ { 2 } + 3 b ) + 12"}, {"rel": "被代入", "source": "2 ( - 2 a ^ { 2 } + 3 b ) + 12", "target": "0"}]}}
|
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{"content": "The monomial $\\frac { 3 } { 5 } x ^ 2 y$ can be combined with $\\frac { 5 } { 6 } x ^ ay ^ b$. What is the value of $a + b$?", "answer": "x = 6", "steps": "$\\because$ Monomials $\\frac { 3 } { 5 } x ^ 2 y$ and $\\frac { 5 } { 6 } x ^ ay ^ b$ can be combined, $\\therefore$ monomials $\\frac { 3 } { 5 } x ^ 2 y$ and $\\frac { 5 } { 6 } x ^ ay ^ b$ are like terms, $\\therefore$ $a = 2$, $b = 1$. $\\therefore$ $a + b = 2 + 1 = 3$.", "expr_cands": ["\\frac { 1 } { x - 1 } - \\frac { 3 } { 2 x + 3 } = 0", "x", "2 x + 3 - 3 ( x - 1 ) = 0", "x = 6", "2 x + 3 - 3 x + 3 = 0"], "exprs": ["x = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x - 1 } - \\frac { 3 } { 2 x + 3 } = 0"}, {"id": "x = 6"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 1 } { x - 1 } - \\frac { 3 } { 2 x + 3 } = 0", "target": "x = 6"}]}}
|
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{"content": "When $k$ = ____ ?, the algebraic expression ${ x } ^ { 2 } - ( k - 3 ) xy - 8$ does not contain the $xy$ term.", "answer": "4", "steps": "$\\because$ There is no $xy$ term in the algebraic expression, $\\therefore$ the coefficient of the $xy$ term is $0$, that is, $k - 3 = 0$. Solving for $k$, we get $k = 3$.", "expr_cands": ["- 2 x ^ { n } y + xy - 3", "y", "n", "x", "n + 1 = 5", "n = 4", "x = 4"], "exprs": ["n + 1 = 5", "n = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 x ^ { n } y + xy - 3"}, {"id": "n + 1 = 5"}, {"id": "$- 2 x ^ { n } y + xy - 3$ 是五次多项式"}, {"id": "n = 4"}], "links": [{"rel": "被描述", "source": "- 2 x ^ { n } y + xy - 3", "target": "n + 1 = 5"}, {"rel": "等式方程求解", "source": "n + 1 = 5", "target": "n = 4"}, {"rel": "限制性描述", "source": "$- 2 x ^ { n } y + xy - 3$ 是五次多项式", "target": "n + 1 = 5"}]}}
|
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{"content": "If the value of the algebraic expression $- 2 a ^ 2 + 3 b + 8$ is $2$, then the value of the algebraic expression $- 4 a ^ 2 + 6 b + 12$ is ____?", "answer": "4", "steps": "\\because $- 2 a ^ { 2 } + 3 b + 8 = 2$, which means $- 2 a ^ { 2 } + 3 b = - 6$, \\therefore the original expression is equal to $2 ( - 2 a ^ { 2 } + 3 b ) + 12 = - 12 + 12 = 0$.", "expr_cands": ["\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0", "m", "x", "| m | - 3 = 1", "m = - 4", "m = 4", "m + 4 \\neq 0", "m \\neq - 4"], "exprs": ["| m | - 3 = 1", "m + 4 \\neq 0", "m = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0"}, {"id": "| m | - 3 = 1"}, {"id": "$\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0$ 是关于 $x$ 的一元一次不等式"}, {"id": "m + 4 \\neq 0"}, {"id": "m = 4"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0", "target": "| m | - 3 = 1"}, {"rel": "被描述", "source": "\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0", "target": "m + 4 \\neq 0"}, {"rel": "联立", "source": "| m | - 3 = 1", "target": "m = 4"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0$ 是关于 $x$ 的一元一次不等式", "target": "| m | - 3 = 1"}, {"rel": "限制性描述", "source": "$\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0$ 是关于 $x$ 的一元一次不等式", "target": "m + 4 \\neq 0"}, {"rel": "联立", "source": "m + 4 \\neq 0", "target": "m = 4"}]}}
|
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{"content": "The solution to the equation $\\frac { 1 } { x - 1 } - \\frac { 3 } { 2 x + 3 } = 0$ is _____.", "answer": "6", "steps": "Dividing both sides by the denominator, we get: $2 x + 3 - 3 ( x - 1 ) = 0$. Expanding the brackets, we get: $2 x + 3 - 3 x + 3 = 0$. Solving for $x$, we get: $x = 6$. Checking the solution, we find that $x = 6$ satisfies the original equation.", "expr_cands": ["x = 2", "x", "\\frac { 5 a - x } { 2 } = 14", "a", "\\frac { 5 a } { 2 } - 1 = 14", "\\frac { 5 a - 2 } { 2 } = 14", "a = 6"], "exprs": ["\\frac { 5 a - 2 } { 2 } = 14", "a = 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 5 a - x } { 2 } = 14"}, {"id": "\\frac { 5 a - 2 } { 2 } = 14"}, {"id": "x = 2"}, {"id": "a = 6"}], "links": [{"rel": "被代入", "source": "\\frac { 5 a - x } { 2 } = 14", "target": "\\frac { 5 a - 2 } { 2 } = 14"}, {"rel": "等式方程求解", "source": "\\frac { 5 a - 2 } { 2 } = 14", "target": "a = 6"}, {"rel": "代入", "source": "x = 2", "target": "\\frac { 5 a - 2 } { 2 } = 14"}]}}
|
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{"content": "If $- 2 x ^ n y + xy - 3$ is a quintic polynomial, what is the value of $n$?", "answer": "7", "steps": "Since $- 2 x ^ { n } y + xy - 3$ is a fifth degree polynomial, we have $n + 1 = 5$, which implies $x = 4$.", "expr_cands": ["x ^ { 2 } - 3 x + 1 = 0", "x", "m", "n", "m ^ { 2 } + n ^ { 2 }", "m + n = 3", "mn = 1", "( m + n ) ^ { 2 } - 2 mn", "7"], "exprs": ["m + n = 3", "mn = 1", "( m + n ) ^ { 2 } - 2 mn", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 x + 1 = 0"}, {"id": "m + n = 3"}, {"id": "m"}, {"id": "n"}, {"id": "一元二次方程 $x ^ { 2 } - 3 x + 1 = 0$ 的两实数根是 $m$ , $n$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "mn = 1"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "m ^ { 2 } + n ^ { 2 }"}, {"id": "( m + n ) ^ { 2 } - 2 mn"}, {"id": "配方"}, {"id": "7"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 3 x + 1 = 0", "target": "m + n = 3"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 x + 1 = 0", "target": "mn = 1"}, {"rel": "代入", "source": "m + n = 3", "target": "7"}, {"rel": "被描述", "source": "m", "target": "m + n = 3"}, {"rel": "被描述", "source": "m", "target": "mn = 1"}, {"rel": "被描述", "source": "n", "target": "m + n = 3"}, {"rel": "被描述", "source": "n", "target": "mn = 1"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - 3 x + 1 = 0$ 的两实数根是 $m$ , $n$", "target": "m + n = 3"}, {"rel": "限制性描述", "source": "一元二次方程 $x ^ { 2 } - 3 x + 1 = 0$ 的两实数根是 $m$ , $n$", "target": "mn = 1"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "m + n = 3"}, {"rel": "代入", "source": "mn = 1", "target": "7"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "mn = 1"}, {"rel": "被描述", "source": "m ^ { 2 } + n ^ { 2 }", "target": "( m + n ) ^ { 2 } - 2 mn"}, {"rel": "被代入", "source": "( m + n ) ^ { 2 } - 2 mn", "target": "7"}, {"rel": "限制性描述", "source": "配方", "target": "( m + n ) ^ { 2 } - 2 mn"}]}}
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{"content": "Given that $\\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0$ is a linear inequality in $x$, the value of $m$ is ____?", "answer": "x > - 2", "steps": "$\\because \\frac { 1 } { 2 } ( m + 4 ) x ^ { | m | - 3 } + 6 > 0$ is a one-variable linear inequality in terms of $x$, $\\therefore | m | - 3 = 1$, $m + 4 \\neq 0$, solving for $m$ yields: $m = 4$.", "expr_cands": ["\\frac { 5 } { \\sqrt { 3 x + 6 } }", "x", "3 x + 6 > 0", "- 2 < x", "x > - 2"], "exprs": ["3 x + 6 > 0", "x > - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 5 } { \\sqrt { 3 x + 6 } }"}, {"id": "3 x + 6 > 0"}, {"id": "代数式 $\\frac { 5 } { \\sqrt { 3 x + 6 } }$ 在实数范围内有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x > - 2"}], "links": [{"rel": "被描述", "source": "\\frac { 5 } { \\sqrt { 3 x + 6 } }", "target": "3 x + 6 > 0"}, {"rel": "不等式方程求解", "source": "3 x + 6 > 0", "target": "x > - 2"}, {"rel": "限制性描述", "source": "代数式 $\\frac { 5 } { \\sqrt { 3 x + 6 } }$ 在实数范围内有意义", "target": "3 x + 6 > 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "3 x + 6 > 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "3 x + 6 > 0"}]}}
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{"content": "Given that $x = 2$ is a solution to the equation $\\frac { 5 a - x } { 2 } = 14$, what is the value of $a$?", "answer": "5", "steps": "Substituting $x = 2$ into $\\frac { 5 a - x } { 2 } = 14$, we get $\\frac { 5 a - 2 } { 2 } = 14$, which implies $a = 6$.", "expr_cands": ["m = 3", "m", "m + \\sqrt { 1 - 2 m + m ^ { 2 } }", "m + \\sqrt { ( m - 1 ) ^ { 2 } } = m + | m - 1 |", "3 + | 3 - 1 |", "5"], "exprs": ["5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m = 3"}, {"id": "5"}, {"id": "m + \\sqrt { 1 - 2 m + m ^ { 2 } }"}], "links": [{"rel": "代入", "source": "m = 3", "target": "5"}, {"rel": "被代入", "source": "m + \\sqrt { 1 - 2 m + m ^ { 2 } }", "target": "5"}]}}
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{"content": "Given a quadratic equation $x ^ 2 - 3 x + 1 = 0$ with two real roots $m$ and $n$, what is the value of $m ^ 2 + n ^ 2$?", "answer": "m \\ge - \\frac { 1 } { 4 }", "steps": "According to the problem, we have $m + n = 3$ and $mn = 1$. Therefore, $m ^ 2 + n ^ 2 = ( m + n ) ^ 2 - 2 mn = 3 ^ 2 - 2 \\times 1 = 7$.", "expr_cands": ["x", "x ^ { 2 } + x - m = 0", "m", "1 ^ { 2 } - 4 * 1 * ( - m ) \\ge 0", "- \\frac { 1 } { 4 } \\le m", "m \\ge - \\frac { 1 } { 4 }"], "exprs": ["1 ^ { 2 } - 4 * 1 * ( - m ) \\ge 0", "m \\ge - \\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + x - m = 0"}, {"id": "1 ^ { 2 } - 4 * 1 * ( - m ) \\ge 0"}, {"id": "关于 $x$ 的方程 $x ^ { 2 } + x - m = 0$ 有实数解"}, {"id": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0"}, {"id": "m \\ge - \\frac { 1 } { 4 }"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } + x - m = 0", "target": "1 ^ { 2 } - 4 * 1 * ( - m ) \\ge 0"}, {"rel": "不等式方程求解", "source": "1 ^ { 2 } - 4 * 1 * ( - m ) \\ge 0", "target": "m \\ge - \\frac { 1 } { 4 }"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x ^ { 2 } + x - m = 0$ 有实数解", "target": "1 ^ { 2 } - 4 * 1 * ( - m ) \\ge 0"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0", "target": "1 ^ { 2 } - 4 * 1 * ( - m ) \\ge 0"}]}}
|
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{"content": "If the algebraic expression $\\frac { 5 } { \\sqrt { 3 x + 6 }}$ is meaningful in the range of real numbers, then the range of values of $x$ is ____?", "answer": "\\sqrt { 2 }", "steps": "From the given condition, we have $3 x + 6 > 0$, which implies $x > - 2$.", "expr_cands": ["y = \\frac { 2 \\sqrt { 2 } } { \\sqrt { - 3 x - 1 } }", "x", "y", "\\sqrt { - 3 x - 1 }", "- 3 x - 1 > 0", "x < - \\frac { 1 } { 3 }", "- 1", "2", "\\frac { 2 \\sqrt { 2 } } { 2 }", "\\sqrt { 2 }"], "exprs": ["- 3 x - 1 > 0", "x < - \\frac { 1 } { 3 }", "- 1", "\\sqrt { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 2 \\sqrt { 2 } } { \\sqrt { - 3 x - 1 } }"}, {"id": "- 3 x - 1 > 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x < - \\frac { 1 } { 3 }"}, {"id": "- 1"}, {"id": "$y$ 要取得最大值"}, {"id": "$\\sqrt { - 3 x - 1 }$ 最小"}, {"id": ", $x$ 是整数"}, {"id": ", $x$ 最大是 $- 1$"}, {"id": "\\sqrt { 2 }"}, {"id": "$y$ 的最大值"}], "links": [{"rel": "被描述", "source": "y = \\frac { 2 \\sqrt { 2 } } { \\sqrt { - 3 x - 1 } }", "target": "- 3 x - 1 > 0"}, {"rel": "被描述", "source": "y = \\frac { 2 \\sqrt { 2 } } { \\sqrt { - 3 x - 1 } }", "target": "\\sqrt { 2 }"}, {"rel": "不等式方程求解", "source": "- 3 x - 1 > 0", "target": "x < - \\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "- 3 x - 1 > 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "- 3 x - 1 > 0"}, {"rel": "被描述", "source": "x < - \\frac { 1 } { 3 }", "target": "- 1"}, {"rel": "被描述", "source": "- 1", "target": "\\sqrt { 2 }"}, {"rel": "限制性描述", "source": "$y$ 要取得最大值", "target": "- 1"}, {"rel": "限制性描述", "source": "$\\sqrt { - 3 x - 1 }$ 最小", "target": "- 1"}, {"rel": "限制性描述", "source": ", $x$ 是整数", "target": "- 1"}, {"rel": "限制性描述", "source": ", $x$ 最大是 $- 1$", "target": "- 1"}, {"rel": "限制性描述", "source": "$y$ 的最大值", "target": "\\sqrt { 2 }"}]}}
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{"content": "When $m = 3$, what is the value of $m + \\sqrt { 1 - 2 m + m ^ 2 }$?", "answer": "\\frac { 2 } { 3 }", "steps": "Original expression = $m + \\sqrt { ( m - 1 ) ^ { 2 } } = m + | m - 1 |$ , when $m = 3$ , the original expression = $3 + | 3 - 1 | = 3 + 2 = 5$.", "expr_cands": ["x", "\\frac { 2 } { a ( x - 1 ) } = 3", "a", "x = 2", "\\frac { 2 } { a } = 3", "a = \\frac { 2 } { 3 }"], "exprs": ["\\frac { 2 } { a } = 3", "a = \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { a ( x - 1 ) } = 3"}, {"id": "\\frac { 2 } { a } = 3"}, {"id": "x = 2"}, {"id": "a = \\frac { 2 } { 3 }"}], "links": [{"rel": "被代入", "source": "\\frac { 2 } { a ( x - 1 ) } = 3", "target": "\\frac { 2 } { a } = 3"}, {"rel": "等式方程求解", "source": "\\frac { 2 } { a } = 3", "target": "a = \\frac { 2 } { 3 }"}, {"rel": "代入", "source": "x = 2", "target": "\\frac { 2 } { a } = 3"}]}}
|
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{"content": "If the equation $x ^ 2 + x - m = 0$ has real solutions, then the possible values of $m$ are _____.", "answer": "- 5", "steps": "According to the problem, we have $\\Delta = 1 ^ 2 - 4 \\times 1 \\times ( - m ) \\ge 0$. Solving for $m$, we get $m \\ge - \\frac { 1 } { 4 }$.", "expr_cands": ["| x + 2 |", "x", "| y + 7 |", "y", "- x + y", "| x + 2 | + | y + 7 | = 0", "x + 2 = 0", "x = - 2", "y + 7 = 0", "y = - 7", "- 5"], "exprs": ["| x + 2 | + | y + 7 | = 0", "x + 2 = 0", "y + 7 = 0", "x = - 2", "y = - 7", "- 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x + 2 |"}, {"id": "| x + 2 | + | y + 7 | = 0"}, {"id": "| y + 7 |"}, {"id": "$| x + 2 |$ 与 $| y + 7 |$ 互为相反数"}, {"id": "x + 2 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "y + 7 = 0"}, {"id": "x = - 2"}, {"id": "y = - 7"}, {"id": "- x + y"}, {"id": "- 5"}], "links": [{"rel": "被描述", "source": "| x + 2 |", "target": "| x + 2 | + | y + 7 | = 0"}, {"rel": "被描述", "source": "| x + 2 | + | y + 7 | = 0", "target": "x + 2 = 0"}, {"rel": "被描述", "source": "| x + 2 | + | y + 7 | = 0", "target": "y + 7 = 0"}, {"rel": "被描述", "source": "| y + 7 |", "target": "| x + 2 | + | y + 7 | = 0"}, {"rel": "限制性描述", "source": "$| x + 2 |$ 与 $| y + 7 |$ 互为相反数", "target": "| x + 2 | + | y + 7 | = 0"}, {"rel": "等式方程求解", "source": "x + 2 = 0", "target": "x = - 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x + 2 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "y + 7 = 0"}, {"rel": "等式方程求解", "source": "y + 7 = 0", "target": "y = - 7"}, {"rel": "代入", "source": "x = - 2", "target": "- 5"}, {"rel": "代入", "source": "y = - 7", "target": "- 5"}, {"rel": "被代入", "source": "- x + y", "target": "- 5"}]}}
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{"content": "Given $y = \\frac { 2 \\sqrt { 2 } } { \\sqrt { - 3 x - 1 } }$, if $x$ is an integer, then the maximum value of $y$ is _____.", "answer": "3", "steps": "To maximize $y$, we need to minimize $\\sqrt { - 3 x - 1 }$. Since $- 3 x - 1 > 0$, we have $x < - \\frac { 1 } { 3 }$. Since $x$ is an integer, the largest possible value for $x$ is $- 1$. At this value, $\\sqrt { - 3 x - 1 }$ is minimized at $2$. Therefore, the maximum value of $y$ is $\\frac { 2 \\sqrt { 2 }} { 2 } = \\sqrt { 2 }$.", "expr_cands": ["2 y - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } y -", "y", "y = - \\frac { 5 } { 3 }", "a", "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a", "a = 3"], "exprs": ["a", "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设这个数是 $a$"}, {"id": "a"}, {"id": "2 y - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } y -"}, {"id": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a"}, {"id": "y = - \\frac { 5 } { 3 }"}, {"id": "被污染的方程是 : $2 y - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } y -$"}, {"id": "此方程的解是 $y = - \\frac { 5 } { 3 }$"}, {"id": "a = 3"}], "links": [{"rel": "假设描述", "source": "设这个数是 $a$", "target": "a"}, {"rel": "假设描述", "source": "设这个数是 $a$", "target": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a"}, {"rel": "被描述", "source": "a", "target": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a"}, {"rel": "被描述", "source": "2 y - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } y -", "target": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a"}, {"rel": "等式方程求解", "source": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a", "target": "a = 3"}, {"rel": "被描述", "source": "y = - \\frac { 5 } { 3 }", "target": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a"}, {"rel": "限制性描述", "source": "被污染的方程是 : $2 y - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } y -$", "target": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a"}, {"rel": "限制性描述", "source": "此方程的解是 $y = - \\frac { 5 } { 3 }$", "target": "2 * ( - \\frac { 5 } { 3 } ) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } * ( - \\frac { 5 } { 3 } ) - a"}]}}
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{"content": "If the solution to the equation $\\frac { 2 } { a ( x - 1 )} = 3$ for $x$ is $x = 2$, then $a$ = ____?", "answer": "- 11", "steps": "Substituting $x = 2$ into the equation, we get $\\frac { 2 } { a } = 3$. Solving for $a$, we have $a = \\frac { 2 } { 3 }$. After checking that it satisfies the given condition, we conclude that $a = \\frac { 2 } { 3 }$.", "expr_cands": ["2 a + 3 b + 1", "a", "b", "- 5", "6 a + 9 b + 7", "2 a + 3 b + 1 = - 5", "2 a + 3 b = - 6", "3 ( 2 a + 3 b )", "- 18", "6 a + 9 b = - 18", "- 11"], "exprs": ["2 a + 3 b + 1 = - 5", "2 a + 3 b = - 6", "- 11"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a + 3 b + 1"}, {"id": "2 a + 3 b + 1 = - 5"}, {"id": "- 5"}, {"id": "代数式 $2 a + 3 b + 1$ 的值等于 $- 5$"}, {"id": "2 a + 3 b = - 6"}, {"id": "6 a + 9 b + 7"}, {"id": "- 11"}, {"id": "代数式 $6 a + 9 b + 7$ 的值"}], "links": [{"rel": "被描述", "source": "2 a + 3 b + 1", "target": "2 a + 3 b + 1 = - 5"}, {"rel": "移项", "source": "2 a + 3 b + 1 = - 5", "target": "2 a + 3 b = - 6"}, {"rel": "被描述", "source": "- 5", "target": "2 a + 3 b + 1 = - 5"}, {"rel": "限制性描述", "source": "代数式 $2 a + 3 b + 1$ 的值等于 $- 5$", "target": "2 a + 3 b + 1 = - 5"}, {"rel": "被描述", "source": "2 a + 3 b = - 6", "target": "- 11"}, {"rel": "被描述", "source": "6 a + 9 b + 7", "target": "- 11"}, {"rel": "限制性描述", "source": "代数式 $6 a + 9 b + 7$ 的值", "target": "- 11"}]}}
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{"content": "Given that $| x + 2 |$ and $| y + 7 |$ are opposite in sign, what is the value of $- x + y$?", "answer": "\\frac { 1 } { 4 }", "steps": "According to the problem, we have: since $| x + 2 |$ and $| y + 7 |$ are opposite in sign, then $| x + 2 | + | y + 7 | = 0$. Therefore, $x + 2 = 0$ and $y + 7 = 0$, which gives us $x = - 2$ and $y = - 7$. So, $- x + y = 2 - 7 = - 5$.", "expr_cands": ["y = 4 x ^ { 2 } + 2 x + m", "m", "y", "x", "\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0", "m = \\frac { 1 } { 4 }"], "exprs": ["\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0", "m = \\frac { 1 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = 4 x ^ { 2 } + 2 x + m"}, {"id": "\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0"}, {"id": "x"}, {"id": "抛物线 $y = 4 x ^ { 2 } + 2 x + m$ 的顶点在 $x$ 轴上"}, {"id": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0"}, {"id": "m = \\frac { 1 } { 4 }"}], "links": [{"rel": "被描述", "source": "y = 4 x ^ { 2 } + 2 x + m", "target": "\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0"}, {"rel": "等式方程求解", "source": "\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0", "target": "m = \\frac { 1 } { 4 }"}, {"rel": "被描述", "source": "x", "target": "\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0"}, {"rel": "限制性描述", "source": "抛物线 $y = 4 x ^ { 2 } + 2 x + m$ 的顶点在 $x$ 轴上", "target": "\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0", "target": "\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0"}]}}
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{"content": "Xiao Ming was doing his equation homework when he accidentally smudged one of the constants in the equation and couldn't read it clearly. The polluted equation is: $2 y - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } y -$, what should he do? After thinking for a while, Xiao Ming looked up the answer in the book. The solution to this equation is $y = - \\frac { 5 } { 3 }$. He quickly fixed the constant and completed the homework. Can you figure out what the constant should be? It should be ____?", "answer": "x > - 2", "steps": "Let the number be $a$. Substituting $y = - \\frac { 5 } { 3 }$ into the equation gives $2 \\cdot ( - \\frac { 5 } { 3 }) - \\frac { 1 } { 2 } = \\frac { 1 } { 2 } \\cdot ( - \\frac { 5 } { 3 }) - a$. Solving for $a$ gives $a = 3$.", "expr_cands": ["2 x + 4 > 0", "x", "2 x > - 4", "- 2 < x", "x > - 2"], "exprs": ["x > - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 4 > 0"}, {"id": "x > - 2"}], "links": [{"rel": "不等式方程求解", "source": "2 x + 4 > 0", "target": "x > - 2"}]}}
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{"content": "If the value of the algebraic expression $2 a + 3 b + 1$ is $- 5$, then the value of the algebraic expression $6 a + 9 b + 7$ is ____?", "answer": "2", "steps": "According to the problem, we know that $2 a + 3 b + 1 = - 5$, which means $2 a + 3 b = - 6$. Therefore, $3 ( 2 a + 3 b ) = 3 * ( - 6 )$, which is $6 a + 9 b = - 18$. So, $6 a + 9 b + 7 = - 18 + 7 = - 11$.", "expr_cands": ["- \\frac { 3 } { 2 } x ^ { 2 } y ^ { 2 x - 1 }", "y", "x", "2 + 2 x - 1 = 5", "x = 2"], "exprs": ["2 + 2 x - 1 = 5", "x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- \\frac { 3 } { 2 } x ^ { 2 } y ^ { 2 x - 1 }"}, {"id": "2 + 2 x - 1 = 5"}, {"id": "$- \\frac { 3 } { 2 } x ^ { 2 } y ^ { 2 x - 1 }$ 是五次单项式"}, {"id": "x = 2"}], "links": [{"rel": "被描述", "source": "- \\frac { 3 } { 2 } x ^ { 2 } y ^ { 2 x - 1 }", "target": "2 + 2 x - 1 = 5"}, {"rel": "等式方程求解", "source": "2 + 2 x - 1 = 5", "target": "x = 2"}, {"rel": "限制性描述", "source": "$- \\frac { 3 } { 2 } x ^ { 2 } y ^ { 2 x - 1 }$ 是五次单项式", "target": "2 + 2 x - 1 = 5"}]}}
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{"content": "The parabola $y = 4 x ^ { 2 } + 2 x + m$ has its vertex on the $x$-axis. Find the value of $m$.", "answer": "a = - 1", "steps": "$\\because$ The vertex of the parabola $y = 4 x ^ { 2 } + 2 x + m$ is on the $x$-axis, $\\therefore$ $\\frac { 4 * 4 * m - 2 ^ { 2 } } { 4 \\times 4 } = 0$, solving for $m$, we get $m = \\frac { 1 } { 4 }$.", "expr_cands": ["\\frac { 4 } { a - 3 } = \\frac { 1 } { a }", "a", "4 a = a - 3", "a = - 1"], "exprs": ["a = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 4 } { a - 3 } = \\frac { 1 } { a }"}, {"id": "a = - 1"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 4 } { a - 3 } = \\frac { 1 } { a }", "target": "a = - 1"}]}}
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{"content": "The solution of the linear inequality $2 x + 4 > 0$ is ____ ?", "answer": "2", "steps": "Because 2x is greater than negative 4, therefore x is greater than negative 2.", "expr_cands": ["a ^ { 5 } \\times ( a ^ { m } ) ^ { 3 } = a ^ { 11 }", "a", "m", "a ^ { 5 } \\times ( a ^ { m } ) ^ { 3 }", "a ^ { 3 m + 5 }", "3 m + 5 = 11", "m = 2"], "exprs": ["3 m + 5 = 11", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 5 } \\times ( a ^ { m } ) ^ { 3 } = a ^ { 11 }"}, {"id": "3 m + 5 = 11"}, {"id": "m = 2"}], "links": [{"rel": "同取对数", "source": "a ^ { 5 } \\times ( a ^ { m } ) ^ { 3 } = a ^ { 11 }", "target": "3 m + 5 = 11"}, {"rel": "等式方程求解", "source": "3 m + 5 = 11", "target": "m = 2"}]}}
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{"content": "If $- \\frac { 3 } { 2 } x ^ { 2 } y ^ { 2 x - 1 }$ is a monomial of degree five, then the value of $x$ is ____?", "answer": "- 3", "steps": "Since $- \\frac { 3 } { 2 } x ^ { 2 } y ^ { 2 x - 1 }$ is a quintic monomial, we have $2 + 2 x - 1 = 5$. Solving for $x$, we get $x = 2$.", "expr_cands": ["x = - 1", "x", "2 x - 1", "- 2 - 1", "- 3"], "exprs": ["- 2 - 1", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = - 1"}, {"id": "- 2 - 1"}, {"id": "2 x - 1"}, {"id": "- 3"}], "links": [{"rel": "代入", "source": "x = - 1", "target": "- 2 - 1"}, {"rel": "计算", "source": "- 2 - 1", "target": "- 3"}, {"rel": "被代入", "source": "2 x - 1", "target": "- 2 - 1"}]}}
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{"content": "The root of the fractional equation $\\frac { 4 } { a - 3 } = \\frac { 1 } { a }$ is ____?", "answer": "a \\neq 1", "steps": "Going to the denominator, we get $4 a = a - 3$. Solving for $a$, we get $a = - 1$. Upon checking, we find that $a = - 1$ is a solution to the fractional equation.", "expr_cands": ["x", "( 1 - a ) x + 2 = 6", "a", "1 - a \\neq 0", "a \\neq 1"], "exprs": ["1 - a \\neq 0", "a \\neq 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 1 - a ) x + 2 = 6"}, {"id": "1 - a \\neq 0"}, {"id": "关于 $x$ 的方程 $( 1 - a ) x + 2 = 6$ 是一元一次方程"}, {"id": "a \\neq 1"}], "links": [{"rel": "被描述", "source": "( 1 - a ) x + 2 = 6", "target": "1 - a \\neq 0"}, {"rel": "不等式方程求解", "source": "1 - a \\neq 0", "target": "a \\neq 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $( 1 - a ) x + 2 = 6$ 是一元一次方程", "target": "1 - a \\neq 0"}]}}
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{"content": "Given: $a ^ { 5 } \\times ( a ^ { m } ) ^ { 3 } = a ^ { 11 }$, the value of $m$ is ____?", "answer": "8", "steps": "Because $a ^ { 5 } \\times ( a ^ { m } ) ^ { 3 } = a ^ { 5 } \\times a ^ { 3 m } = a ^ { 3 m + 5 }$ , therefore $3 m + 5 = 11$ , solving for $m$ yields $m = 2$.", "expr_cands": ["x ^ { 2 } + 3 x - 4 = 0", "x", "- 2 x ^ { 2 } - 6 x + 16", "x = - 4", "x = 1", "x ^ { 2 } + 3 x = 4", "- 2 ( { x } ^ { 2 } + 3 x ) + 16", "8"], "exprs": ["x ^ { 2 } + 3 x = 4", "- 2 ( { x } ^ { 2 } + 3 x ) + 16", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + 3 x - 4 = 0"}, {"id": "x ^ { 2 } + 3 x = 4"}, {"id": "- 2 x ^ { 2 } - 6 x + 16"}, {"id": "- 2 ( { x } ^ { 2 } + 3 x ) + 16"}, {"id": "8"}], "links": [{"rel": "移项", "source": "x ^ { 2 } + 3 x - 4 = 0", "target": "x ^ { 2 } + 3 x = 4"}, {"rel": "提取因式参考", "source": "x ^ { 2 } + 3 x = 4", "target": "- 2 ( { x } ^ { 2 } + 3 x ) + 16"}, {"rel": "代入", "source": "x ^ { 2 } + 3 x = 4", "target": "8"}, {"rel": "提取因式", "source": "- 2 x ^ { 2 } - 6 x + 16", "target": "- 2 ( { x } ^ { 2 } + 3 x ) + 16"}, {"rel": "被代入", "source": "- 2 ( { x } ^ { 2 } + 3 x ) + 16", "target": "8"}]}}
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{"content": "When $x = - 1$, what is the value of the algebraic expression $2 x - 1$?", "answer": "- 4", "steps": "When $x = - 1$, the original expression equals $- 2 - 1 = - 3$.", "expr_cands": ["4 x + 4 = 3 x + 1", "x", "3 x + 2 m = 6 x + 1", "m", "x = - 3", "2 m - 9 = - 17", "- 9 + 2 m = - 18 + 1", "m = - 4"], "exprs": ["x = - 3", "- 9 + 2 m = - 18 + 1", "m = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x + 4 = 3 x + 1"}, {"id": "x = - 3"}, {"id": "3 x + 2 m = 6 x + 1"}, {"id": "- 9 + 2 m = - 18 + 1"}, {"id": "方程 $4 x + 4 = 3 x + 1$ 和方程 $3 x + 2 m = 6 x + 1$ 的解相同"}, {"id": "m = - 4"}], "links": [{"rel": "等式方程求解", "source": "4 x + 4 = 3 x + 1", "target": "x = - 3"}, {"rel": "被描述", "source": "x = - 3", "target": "- 9 + 2 m = - 18 + 1"}, {"rel": "被描述", "source": "3 x + 2 m = 6 x + 1", "target": "- 9 + 2 m = - 18 + 1"}, {"rel": "等式方程求解", "source": "- 9 + 2 m = - 18 + 1", "target": "m = - 4"}, {"rel": "限制性描述", "source": "方程 $4 x + 4 = 3 x + 1$ 和方程 $3 x + 2 m = 6 x + 1$ 的解相同", "target": "- 9 + 2 m = - 18 + 1"}]}}
|
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{"content": "If the equation $( 1 - a ) x + 2 = 6$ is a linear equation in one variable $x$, then the condition that $a$ should satisfy is ____?", "answer": "- 8", "steps": "From the given problem, we have $1 - a \\neq 0$. Solving for $a$, we get $a \\neq 1$.", "expr_cands": ["x ^ { 2 } + y ^ { 2 } + 4 x - 6 y + 13 = 0", "y", "x", "x ^ { y }", "( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 0", "x + 2 = 0", "x = - 2", "y - 3 = 0", "y = 3", "- 8"], "exprs": ["( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 0", "x + 2 = 0", "y - 3 = 0", "x = - 2", "y = 3", "- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } + y ^ { 2 } + 4 x - 6 y + 13 = 0"}, {"id": "( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 0"}, {"id": "x + 2 = 0"}, {"id": "y"}, {"id": "x"}, {"id": "且 $x$ , $y$ 是实数"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "y - 3 = 0"}, {"id": "x = - 2"}, {"id": "y = 3"}, {"id": "x ^ { y }"}, {"id": "- 8"}], "links": [{"rel": "提取因式", "source": "x ^ { 2 } + y ^ { 2 } + 4 x - 6 y + 13 = 0", "target": "( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 0"}, {"rel": "被描述", "source": "( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 0", "target": "x + 2 = 0"}, {"rel": "被描述", "source": "( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 0", "target": "y - 3 = 0"}, {"rel": "等式方程求解", "source": "x + 2 = 0", "target": "x = - 2"}, {"rel": "被描述", "source": "y", "target": "x + 2 = 0"}, {"rel": "被描述", "source": "y", "target": "y - 3 = 0"}, {"rel": "被描述", "source": "x", "target": "x + 2 = 0"}, {"rel": "被描述", "source": "x", "target": "y - 3 = 0"}, {"rel": "限制性描述", "source": "且 $x$ , $y$ 是实数", "target": "x + 2 = 0"}, {"rel": "限制性描述", "source": "且 $x$ , $y$ 是实数", "target": "y - 3 = 0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "x + 2 = 0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "y - 3 = 0"}, {"rel": "等式方程求解", "source": "y - 3 = 0", "target": "y = 3"}, {"rel": "代入", "source": "x = - 2", "target": "- 8"}, {"rel": "代入", "source": "y = 3", "target": "- 8"}, {"rel": "被代入", "source": "x ^ { y }", "target": "- 8"}]}}
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{"content": "If $x ^ 2 + 3 x - 4 = 0$, then the value of $- 2 x ^ 2 - 6 x + 16$ is ____?", "answer": "- 9", "steps": "Since $x ^ 2 + 3 x - 4 = 0$, it follows that $x ^ 2 + 3 x = 4$. Therefore, $- 2 x ^ 2 - 6 x + 16 = - 2 ( x ^ 2 + 3 x ) + 16 = - 2 * 4 + 16 = 8$.", "expr_cands": ["| x | = 4", "x", "| y | = 5", "y", "x + y", "x = - 4", "x = 4", "y = - 5", "y = 5", "- 9"], "exprs": ["x = - 4", "y = - 5", "- 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x | = 4"}, {"id": "x = - 4"}, {"id": "且 $x$ , $y$ 均为负数"}, {"id": "绝对值恒大于等于0"}, {"id": "| y | = 5"}, {"id": "y = - 5"}, {"id": "x + y"}, {"id": "- 9"}], "links": [{"rel": "被描述", "source": "| x | = 4", "target": "x = - 4"}, {"rel": "代入", "source": "x = - 4", "target": "- 9"}, {"rel": "限制性描述", "source": "且 $x$ , $y$ 均为负数", "target": "x = - 4"}, {"rel": "限制性描述", "source": "且 $x$ , $y$ 均为负数", "target": "y = - 5"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x = - 4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "y = - 5"}, {"rel": "被描述", "source": "| y | = 5", "target": "y = - 5"}, {"rel": "代入", "source": "y = - 5", "target": "- 9"}, {"rel": "被代入", "source": "x + y", "target": "- 9"}]}}
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{"content": "Given the equation $4 x + 4 = 3 x + 1$ and the equation $3 x + 2 m = 6 x + 1$ have the same solution, what is the value of $m$?", "answer": "2", "steps": "Solve the equation $4 x + 4 = 3 x + 1$ to get $x = - 3$. Substitute $x = - 3$ into $3 x + 2 m = 6 x + 1$ to get $- 9 + 2 m = - 18 + 1$. Solve for $m$ to get $m = - 4$.", "expr_cands": ["| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0", "y", "x", "z", "x + y + z", "x = - 1", "y = 5", "x + y - z = 0", "4 - z = 0", "z = 4", "8", "2"], "exprs": ["x = - 1", "y = 5", "x + y - z = 0", "4 - z = 0", "z = 4", "8", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0"}, {"id": "x = - 1"}, {"id": "绝对值恒大于等于0"}, {"id": "y = 5"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x + y - z = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "4 - z = 0"}, {"id": "z = 4"}, {"id": "x + y + z"}, {"id": "8"}, {"id": "2"}, {"id": "$| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0$ , $x + y + z$ 的立方根"}, {"id": ", $8$ 的立方根是 $2$"}], "links": [{"rel": "被描述", "source": "| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0", "target": "x = - 1"}, {"rel": "被描述", "source": "| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0", "target": "y = 5"}, {"rel": "被描述", "source": "| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0", "target": "x + y - z = 0"}, {"rel": "代入", "source": "x = - 1", "target": "4 - z = 0"}, {"rel": "代入", "source": "x = - 1", "target": "8"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x = - 1"}, {"rel": "代入", "source": "y = 5", "target": "4 - z = 0"}, {"rel": "代入", "source": "y = 5", "target": "8"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "y = 5"}, {"rel": "被代入", "source": "x + y - z = 0", "target": "4 - z = 0"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "x + y - z = 0"}, {"rel": "等式方程求解", "source": "4 - z = 0", "target": "z = 4"}, {"rel": "代入", "source": "z = 4", "target": "8"}, {"rel": "被代入", "source": "x + y + z", "target": "8"}, {"rel": "被描述", "source": "x + y + z", "target": "2"}, {"rel": "被描述", "source": "8", "target": "2"}, {"rel": "限制性描述", "source": "$| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0$ , $x + y + z$ 的立方根", "target": "2"}, {"rel": "限制性描述", "source": ", $8$ 的立方根是 $2$", "target": "2"}]}}
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{"content": "Given $x ^ { 2 } + y ^ { 2 } + 4 x - 6 y + 13 = 0$ and $x$ , $y$ are real numbers, then $x ^ { y }$ = ____ ?", "answer": "a > - 7", "steps": "$x ^ { 2 } + y ^ { 2 } + 4 x - 6 y + 13 = ( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } = 0$ , therefore $x + 2 = 0$ , $y - 3 = 0$ , which gives $x = - 2$ , $y = 3$ , so $x ^ { y } = - 8$.", "expr_cands": ["x", "2 x + a = x - 7", "a", "x = - 7 - a", "- 7 - a < 0", "- 7 < a", "a > - 7"], "exprs": ["x = - 7 - a", "- 7 - a < 0", "a > - 7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + a = x - 7"}, {"id": "x = - 7 - a"}, {"id": "- 7 - a < 0"}, {"id": "关于 $x$ 的方程 $2 x + a = x - 7$ 的解为负数"}, {"id": "a > - 7"}], "links": [{"rel": "等式方程部分求解", "source": "2 x + a = x - 7", "target": "x = - 7 - a"}, {"rel": "被描述", "source": "x = - 7 - a", "target": "- 7 - a < 0"}, {"rel": "不等式方程求解", "source": "- 7 - a < 0", "target": "a > - 7"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $2 x + a = x - 7$ 的解为负数", "target": "- 7 - a < 0"}]}}
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{"content": "Given $| x | = 4$, $| y | = 5$, and both $x$ and $y$ are negative, what is the value of $x + y$?", "answer": "3", "steps": "Since $| x | = 4$ and $| y | = 5$, and both $x$ and $y$ are negative, we have $x = - 4$ and $y = - 5$. Therefore, $x + y = - 9$.", "expr_cands": ["\\frac { x - 3 } { x + 4 }", "x", "0", "x - 3 = 0", "x = 3", "x + 4 \\neq 0", "x \\neq - 4"], "exprs": ["x - 3 = 0", "x + 4 \\neq 0", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 3 } { x + 4 }"}, {"id": "x - 3 = 0"}, {"id": "分式 $\\frac { x - 3 } { x + 4 }$ 的值为 $0$"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "x + 4 \\neq 0"}, {"id": "x = 3"}], "links": [{"rel": "被描述", "source": "\\frac { x - 3 } { x + 4 }", "target": "x - 3 = 0"}, {"rel": "被描述", "source": "\\frac { x - 3 } { x + 4 }", "target": "x + 4 \\neq 0"}, {"rel": "联立", "source": "x - 3 = 0", "target": "x = 3"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - 3 } { x + 4 }$ 的值为 $0$", "target": "x - 3 = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - 3 } { x + 4 }$ 的值为 $0$", "target": "x + 4 \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x - 3 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x + 4 \\neq 0"}, {"rel": "联立", "source": "x + 4 \\neq 0", "target": "x = 3"}]}}
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{"content": "Given $| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0$ , the cube root of $x + y + z$ is ____ ?", "answer": "- 2", "steps": "From $| x + 1 | + \\sqrt { 5 - y } + ( x + y - z ) ^ { 2 } = 0$, we know that $x = - 1$, $y = 5$, $x + y - z = 0$, therefore $z = 4$, therefore $x + y + z = - 1 + 5 + 4 = 8$, therefore the cube root of 8 is 2.", "expr_cands": ["2 x", "x", "2 - x", "2 x + ( 2 - x ) = 0", "x = - 2", "2 x + 2 - x = 0", "2 x - x = - 2"], "exprs": ["2 x + ( 2 - x ) = 0", "x = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x"}, {"id": "2 x + ( 2 - x ) = 0"}, {"id": "2 - x"}, {"id": "$2 x$ 和 $2 - x$ 互为相反数"}, {"id": "x = - 2"}], "links": [{"rel": "被描述", "source": "2 x", "target": "2 x + ( 2 - x ) = 0"}, {"rel": "等式方程求解", "source": "2 x + ( 2 - x ) = 0", "target": "x = - 2"}, {"rel": "被描述", "source": "2 - x", "target": "2 x + ( 2 - x ) = 0"}, {"rel": "限制性描述", "source": "$2 x$ 和 $2 - x$ 互为相反数", "target": "2 x + ( 2 - x ) = 0"}]}}
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{"content": "The equation $2 x + a = x - 7$ has a negative solution for $x$. The possible values of real number $a$ are _____.", "answer": "25", "steps": "From $2 x + a = x - 7$, we get $x = - 7 - a$. Since the solution of the equation is negative, we have $- 7 - a < 0$, which implies $a > - 7$.)", "expr_cands": ["x", "x ^ { 2 } - m = 0", "m", "5", "x = 5", "25 - m = 0", "m = 25"], "exprs": ["x = 5", "25 - m = 0", "m = 25"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x"}, {"id": "x = 5"}, {"id": "x ^ { 2 } - m = 0"}, {"id": "5"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } - m = 0$ 的一个解为 $5$"}, {"id": "25 - m = 0"}, {"id": "m = 25"}], "links": [{"rel": "被描述", "source": "x", "target": "x = 5"}, {"rel": "代入", "source": "x = 5", "target": "25 - m = 0"}, {"rel": "被描述", "source": "x ^ { 2 } - m = 0", "target": "x = 5"}, {"rel": "被代入", "source": "x ^ { 2 } - m = 0", "target": "25 - m = 0"}, {"rel": "被描述", "source": "5", "target": "x = 5"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - m = 0$ 的一个解为 $5$", "target": "x = 5"}, {"rel": "等式方程求解", "source": "25 - m = 0", "target": "m = 25"}]}}
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{"content": "If the value of the fraction $\\frac { x - 3 } { x + 4 }$ is $0$, then the value of $x$ is ____?", "answer": "6", "steps": "From the condition that the value of the fraction is zero, we have $x - 3 = 0$, $x + 4 \\neq 0$. From $x - 3 = 0$, we get $x = 3$. From $x + 4 \\neq 0$, we get $x \\neq - 4$. Therefore, we have $x = 3$, and the value of the fraction $\\frac { x - 3 } { x + 4 }$ is zero.", "expr_cands": ["x + 1 = - 1", "x", "2 x - k = - x", "k", "- k", "x = - 2", "- k - 4 = 2", "- 4 - k = 2", "k = - 6", "6"], "exprs": ["x = - 2", "- 4 - k = 2", "k = - 6", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 1 = - 1"}, {"id": "x = - 2"}, {"id": "2 x - k = - x"}, {"id": "- 4 - k = 2"}, {"id": "k = - 6"}, {"id": "- k"}, {"id": "6"}], "links": [{"rel": "等式方程求解", "source": "x + 1 = - 1", "target": "x = - 2"}, {"rel": "代入", "source": "x = - 2", "target": "- 4 - k = 2"}, {"rel": "被代入", "source": "2 x - k = - x", "target": "- 4 - k = 2"}, {"rel": "等式方程求解", "source": "- 4 - k = 2", "target": "k = - 6"}, {"rel": "代入", "source": "k = - 6", "target": "6"}, {"rel": "被代入", "source": "- k", "target": "6"}]}}
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{"content": "If $2 x$ and $2 - x$ are opposite numbers, then $x$ equals ____?", "answer": "9", "steps": "According to the problem, $2 x + ( 2 - x ) = 0$. Removing the parentheses, we get $2 x + 2 - x = 0$. Rearranging the terms, we get $2 x - x = - 2$. Combining like terms, we get $x = - 2$.", "expr_cands": ["5 x + 3 a = - 3", "a", "x", "x = - 6", "- 30 + 3 a = - 3", "a = 9", "3 a = 27"], "exprs": ["- 30 + 3 a = - 3", "a = 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x + 3 a = - 3"}, {"id": "- 30 + 3 a = - 3"}, {"id": "x = - 6"}, {"id": "a = 9"}], "links": [{"rel": "被代入", "source": "5 x + 3 a = - 3", "target": "- 30 + 3 a = - 3"}, {"rel": "等式方程求解", "source": "- 30 + 3 a = - 3", "target": "a = 9"}, {"rel": "代入", "source": "x = - 6", "target": "- 30 + 3 a = - 3"}]}}
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{"content": "If one of the solutions of the quadratic equation $x ^ 2 - m = 0$ with respect to $x$ is $5$, then the value of $m$ is ____?", "answer": "3", "steps": "Substituting $x = 5$ into $x ^ 2 - m = 0$ yields $25 - m = 0$, solving for $m$ gives $m = 25$.", "expr_cands": ["m = - 1", "m", "- 2 m ^ { 2 } - [ - 4 m ^ { 2 } + ( - m ^ { 2 } ) ]", "- 2 m ^ { 2 } + 4 m ^ { 2 } + m ^ { 2 }", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m = - 1"}, {"id": "3"}, {"id": "- 2 m ^ { 2 } - [ - 4 m ^ { 2 } + ( - m ^ { 2 } ) ]"}], "links": [{"rel": "代入", "source": "m = - 1", "target": "3"}, {"rel": "被代入", "source": "- 2 m ^ { 2 } - [ - 4 m ^ { 2 } + ( - m ^ { 2 } ) ]", "target": "3"}]}}
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{"content": "Given the equation $x + 1 = - 1$ and the equation $2 x - k = - x$ have the same solution, what is $- k$?", "answer": "6", "steps": "$\\because x + 1 = - 1$, $\\therefore x = - 2$. Substituting $x = - 2$ into $2 x - k = - x$, we get: $- 4 - k = 2$, $\\therefore k = - 6$, $\\therefore - k = 6$.", "expr_cands": ["x = - 2", "x", "2 x ^ { 2 } + mx + 4", "m", "18", "x = 2", "2 x ^ { 2 } + mx + 4 = 18", "12 - 2 m = 18", "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18", "m = - 3", "2 x ^ { 2 } - 3 x + 4", "2 * 2 ^ { 2 } - 3 * 2 + 4", "6"], "exprs": ["2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18", "m = - 3", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = - 2"}, {"id": "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18"}, {"id": "2 x ^ { 2 } + mx + 4"}, {"id": "18"}, {"id": "当 $x = - 2$ 时"}, {"id": "式子 $2 x ^ { 2 } + mx + 4$ 的值为 $18$"}, {"id": "m = - 3"}, {"id": "x = 2"}, {"id": "6"}], "links": [{"rel": "被描述", "source": "x = - 2", "target": "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18"}, {"rel": "等式方程求解", "source": "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18", "target": "m = - 3"}, {"rel": "被描述", "source": "2 x ^ { 2 } + mx + 4", "target": "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18"}, {"rel": "被代入", "source": "2 x ^ { 2 } + mx + 4", "target": "6"}, {"rel": "被描述", "source": "18", "target": "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18"}, {"rel": "限制性描述", "source": "当 $x = - 2$ 时", "target": "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18"}, {"rel": "限制性描述", "source": "式子 $2 x ^ { 2 } + mx + 4$ 的值为 $18$", "target": "2 * ( - 2 ) ^ { 2 } - 2 m + 4 = 18"}, {"rel": "代入", "source": "m = - 3", "target": "6"}, {"rel": "代入", "source": "x = 2", "target": "6"}]}}
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{"content": "If the solution to the equation $5 x + 3 a = - 3$ is $x = - 6$, then $a$ = ____?", "answer": "8", "steps": "Substituting $x = - 6$ into the equation, we get $- 30 + 3 a = - 3$. Rearranging and combining like terms, we get $3 a = 27$. Solving for $a$, we get $a = 9$.", "expr_cands": ["x ^ { 2 } - 3 x - 2 = 0", "x", "x _ { 1 }", "x _ { 2 }", "( x _ { 1 } + 2 ) ( x _ { 2 } + 2 )", "x _ { 1 } + x _ { 2 } = 3", "x _ { 1 } x _ { 2 } = - 2", "x _ { 1 } x _ { 2 } + 2 ( x _ { 1 } + x _ { 2 } ) + 4", "8"], "exprs": ["x _ { 1 } + x _ { 2 } = 3", "x _ { 1 } x _ { 2 } = - 2", "x _ { 1 } x _ { 2 } + 2 ( x _ { 1 } + x _ { 2 } ) + 4", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 3 x - 2 = 0"}, {"id": "x _ { 1 } + x _ { 2 } = 3"}, {"id": "x _ { 1 }"}, {"id": "x _ { 2 }"}, {"id": "方程 $x ^ { 2 } - 3 x - 2 = 0$ 的两实根为 $x _ { 1 }$ , $x _ { 2 }$"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "x _ { 1 } x _ { 2 } = - 2"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "( x _ { 1 } + 2 ) ( x _ { 2 } + 2 )"}, {"id": "x _ { 1 } x _ { 2 } + 2 ( x _ { 1 } + x _ { 2 } ) + 4"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 3 x - 2 = 0", "target": "x _ { 1 } + x _ { 2 } = 3"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 x - 2 = 0", "target": "x _ { 1 } x _ { 2 } = - 2"}, {"rel": "代入", "source": "x _ { 1 } + x _ { 2 } = 3", "target": "8"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } + x _ { 2 } = 3"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } x _ { 2 } = - 2"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } + x _ { 2 } = 3"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } x _ { 2 } = - 2"}, {"rel": "限制性描述", "source": "方程 $x ^ { 2 } - 3 x - 2 = 0$ 的两实根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "x _ { 1 } + x _ { 2 } = 3"}, {"rel": "限制性描述", "source": "方程 $x ^ { 2 } - 3 x - 2 = 0$ 的两实根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "x _ { 1 } x _ { 2 } = - 2"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + x _ { 2 } = 3"}, {"rel": "代入", "source": "x _ { 1 } x _ { 2 } = - 2", "target": "8"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "x _ { 1 } x _ { 2 } = - 2"}, {"rel": "展开", "source": "( x _ { 1 } + 2 ) ( x _ { 2 } + 2 )", "target": "x _ { 1 } x _ { 2 } + 2 ( x _ { 1 } + x _ { 2 } ) + 4"}, {"rel": "被代入", "source": "x _ { 1 } x _ { 2 } + 2 ( x _ { 1 } + x _ { 2 } ) + 4", "target": "8"}]}}
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{"content": "When $m = - 1$, what is $- 2 m ^ 2 - [ - 4 m ^ 2 + ( - m ^ 2 )]$ equal to?", "answer": "\\frac { 1 } { 2 }", "steps": "When $m = - 1$, the original expression becomes $- 2 m ^ 2 + 4 m ^ 2 + m ^ 2 = 3 m ^ 2 = 3 * ( - 1 ) ^ 2 = 3 * 1 = 3$.", "expr_cands": ["a + b = \\frac { 3 } { 2 }", "a", "b", "ab = 1", "( a - 1 ) ( b - 1 )", "ab - ( a + b ) + 1", "1 - \\frac { 3 } { 2 } + 1", "\\frac { 1 } { 2 }"], "exprs": ["ab - ( a + b ) + 1", "\\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 1 ) ( b - 1 )"}, {"id": "ab - ( a + b ) + 1"}, {"id": "a + b = \\frac { 3 } { 2 }"}, {"id": "ab = 1"}, {"id": "\\frac { 1 } { 2 }"}], "links": [{"rel": "提取因式", "source": "( a - 1 ) ( b - 1 )", "target": "ab - ( a + b ) + 1"}, {"rel": "被代入", "source": "ab - ( a + b ) + 1", "target": "\\frac { 1 } { 2 }"}, {"rel": "提取因式参考", "source": "a + b = \\frac { 3 } { 2 }", "target": "ab - ( a + b ) + 1"}, {"rel": "代入", "source": "a + b = \\frac { 3 } { 2 }", "target": "\\frac { 1 } { 2 }"}, {"rel": "提取因式参考", "source": "ab = 1", "target": "ab - ( a + b ) + 1"}, {"rel": "代入", "source": "ab = 1", "target": "\\frac { 1 } { 2 }"}]}}
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{"content": "When $x = - 2$, the value of the expression $2 x ^ 2 + mx + 4$ is $18$. What is the value of the expression when $x = 2$?", "answer": "a \\le 3", "steps": "Substituting $x = - 2$ into $2 x ^ 2 + mx + 4 = 18$, we get $2 * ( - 2 ) ^ 2 - 2 m + 4 = 18$, which gives $m = - 3$. Substituting $m = - 3$ into $2 x ^ 2 + mx + 4$, we get $2 x ^ 2 - 3 x + 4$. Substituting $x = 2$ into the equation, we get $2 * 2 ^ 2 - 3 * 2 + 4 = 6$.", "expr_cands": ["\\sqrt { 3 - a }", "a", "3 - a \\ge 0", "a \\le 3"], "exprs": ["3 - a \\ge 0", "a \\le 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 3 - a }"}, {"id": "3 - a \\ge 0"}, {"id": "$\\sqrt { 3 - a }$ 在实数范围内有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a \\le 3"}], "links": [{"rel": "被描述", "source": "\\sqrt { 3 - a }", "target": "3 - a \\ge 0"}, {"rel": "不等式方程求解", "source": "3 - a \\ge 0", "target": "a \\le 3"}, {"rel": "限制性描述", "source": "$\\sqrt { 3 - a }$ 在实数范围内有意义", "target": "3 - a \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "3 - a \\ge 0"}]}}
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{"content": "If the two real roots of the equation $x ^ 2 - 3 x - 2 = 0$ are $x _ 1$ and $x _ 2$, then the value of $( x _ 1 + 2 ) ( x _ 2 + 2 )$ is ____?", "answer": "2009", "steps": "From the given information, we have $x _ 1 + x _ 2 = 3$ and $x _ 1 x _ 2 = - 2$. The original expression is equal to $x _ 1 x _ 2 + 2 ( x _ 1 + x _ 2 ) + 4 = - 2 + 6 + 4 = 8$.", "expr_cands": ["{ a } ^ { 2 } - 3 b = 4", "b", "a", "6 b - 2 { a } ^ { 2 } + 2017", "a ^ { 2 } - 3 b = 4", "- 2 ( a ^ { 2 } - 3 b ) + 2017", "2009"], "exprs": ["- 2 ( a ^ { 2 } - 3 b ) + 2017", "2009"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "6 b - 2 { a } ^ { 2 } + 2017"}, {"id": "- 2 ( a ^ { 2 } - 3 b ) + 2017"}, {"id": "a ^ { 2 } - 3 b = 4"}, {"id": "2009"}], "links": [{"rel": "提取因式", "source": "6 b - 2 { a } ^ { 2 } + 2017", "target": "- 2 ( a ^ { 2 } - 3 b ) + 2017"}, {"rel": "被代入", "source": "- 2 ( a ^ { 2 } - 3 b ) + 2017", "target": "2009"}, {"rel": "提取因式参考", "source": "a ^ { 2 } - 3 b = 4", "target": "- 2 ( a ^ { 2 } - 3 b ) + 2017"}, {"rel": "代入", "source": "a ^ { 2 } - 3 b = 4", "target": "2009"}]}}
|
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{"content": "Given: $a + b = \\frac { 3 } { 2 }$, $ab = 1$, the result of the expression $( a - 1 ) ( b - 1 )$ is ____?", "answer": "10", "steps": "$( a - 1 ) ( b - 1 ) = ab - ( a + b ) + 1$, when $a + b = \\frac { 3 } { 2 }$, $ab = 1$,the original expression $= 1 - \\frac { 3 } { 2 } + 1 = \\frac { 1 } { 2 }$. ", "expr_cands": ["a ^ { m } b ^ { 4 }", "b", "a", "m", "2 a ^ { 6 } b ^ { n }", "n", "m + n", "m = 6", "n = 4", "10"], "exprs": ["m = 6", "n = 4", "10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { m } b ^ { 4 }"}, {"id": "m = 6"}, {"id": "2 a ^ { 6 } b ^ { n }"}, {"id": "$a ^ { m } b ^ { 4 }$ 与 $2 a ^ { 6 } b ^ { n }$ 是同类项"}, {"id": "n = 4"}, {"id": "m + n"}, {"id": "10"}], "links": [{"rel": "被描述", "source": "a ^ { m } b ^ { 4 }", "target": "m = 6"}, {"rel": "被描述", "source": "a ^ { m } b ^ { 4 }", "target": "n = 4"}, {"rel": "代入", "source": "m = 6", "target": "10"}, {"rel": "被描述", "source": "2 a ^ { 6 } b ^ { n }", "target": "m = 6"}, {"rel": "被描述", "source": "2 a ^ { 6 } b ^ { n }", "target": "n = 4"}, {"rel": "限制性描述", "source": "$a ^ { m } b ^ { 4 }$ 与 $2 a ^ { 6 } b ^ { n }$ 是同类项", "target": "m = 6"}, {"rel": "限制性描述", "source": "$a ^ { m } b ^ { 4 }$ 与 $2 a ^ { 6 } b ^ { n }$ 是同类项", "target": "n = 4"}, {"rel": "代入", "source": "n = 4", "target": "10"}, {"rel": "被代入", "source": "m + n", "target": "10"}]}}
|
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{"content": "If $\\sqrt { 3 - a }$ is meaningful in the real number range, then the range of values that $a$ can take is ____?", "answer": "- 3", "steps": "Since $\\sqrt { 3 - a }$ is meaningful in the real number range, therefore $3 - a \\ge 0$, which leads to $a \\le 3$.", "expr_cands": ["\\frac { x } { x - 3 }", "x", "\\frac { x + 1 } { x - 1 }", "\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }", "x = - 3", "x ^ { 2 } - x = x ^ { 2 } - 2 x - 3"], "exprs": ["\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }", "x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x - 3 }"}, {"id": "\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }"}, {"id": "\\frac { x + 1 } { x - 1 }"}, {"id": "使分式 $\\frac { x } { x - 3 }$ 和分式 $\\frac { x + 1 } { x - 1 }$ 相等的 $x$ 值"}, {"id": "x = - 3"}], "links": [{"rel": "被描述", "source": "\\frac { x } { x - 3 }", "target": "\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }"}, {"rel": "等式方程求解", "source": "\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }", "target": "x = - 3"}, {"rel": "被描述", "source": "\\frac { x + 1 } { x - 1 }", "target": "\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }"}, {"rel": "限制性描述", "source": "使分式 $\\frac { x } { x - 3 }$ 和分式 $\\frac { x + 1 } { x - 1 }$ 相等的 $x$ 值", "target": "\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }"}]}}
|
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{"content": "If ${ a } ^ { 2 } - 3 b = 4$, then $6 b - 2 { a } ^ { 2 } + 2017$ = ____ ?", "answer": "115", "steps": "When $a ^ 2 - 3 b = 4$, the original expression is equal to $- 2 ( a ^ 2 - 3 b ) + 2017 = - 8 + 2017 = 2009$.", "expr_cands": ["x - 2 y = 5", "x", "y", "10 ( x - 2 y ) + x - 2 y + 60", "10 ( x - 2 y ) + ( x - 2 y ) + 60", "115"], "exprs": ["10 ( x - 2 y ) + ( x - 2 y ) + 60", "115"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "10 ( x - 2 y ) + x - 2 y + 60"}, {"id": "10 ( x - 2 y ) + ( x - 2 y ) + 60"}, {"id": "x - 2 y = 5"}, {"id": "115"}], "links": [{"rel": "提取因式", "source": "10 ( x - 2 y ) + x - 2 y + 60", "target": "10 ( x - 2 y ) + ( x - 2 y ) + 60"}, {"rel": "被代入", "source": "10 ( x - 2 y ) + ( x - 2 y ) + 60", "target": "115"}, {"rel": "提取因式参考", "source": "x - 2 y = 5", "target": "10 ( x - 2 y ) + ( x - 2 y ) + 60"}, {"rel": "代入", "source": "x - 2 y = 5", "target": "115"}]}}
|
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{"content": "If $a ^ { m } b ^ { 4 }$ and $2 a ^ { 6 } b ^ { n }$ are similar terms, then the value of $m + n$ is ____?", "answer": "- 4", "steps": "From the given information, we know that $m = 6$ and $n = 4$. Therefore, $m + n = 6 + 4 = 10$.", "expr_cands": ["( x + a ) ^ { 2 } = x ^ { 2 } - 8 x + b", "a", "b", "x", "2 a = - 8", "a = - 4"], "exprs": ["2 a = - 8", "a = - 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + a ) ^ { 2 } = x ^ { 2 } - 8 x + b"}, {"id": "2 a = - 8"}, {"id": "a = - 4"}], "links": [{"rel": "移项", "source": "( x + a ) ^ { 2 } = x ^ { 2 } - 8 x + b", "target": "2 a = - 8"}, {"rel": "等式方程求解", "source": "2 a = - 8", "target": "a = - 4"}]}}
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{"content": "The value of $x$ that makes the fraction $\\frac { x } { x - 3 }$ equal to the fraction $\\frac { x + 1 } { x - 1 }$ is ____ ?", "answer": "m > 1", "steps": "According to the problem, we have $\\frac { x } { x - 3 } = \\frac { x + 1 } { x - 1 }$. Clearing denominators, we get $x ^ 2 - x = x ^ 2 - 2 x - 3$. Solving for $x$, we get $x = - 3$. Checking, we see that $x = - 3$ is indeed a solution to the fractional equation.", "expr_cands": ["y = ( m - 1 ) x ^ { 2 }", "x", "y", "m", "m - 1 > 0", "1 < m", "m > 1"], "exprs": ["m - 1 > 0", "m > 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( m - 1 ) x ^ { 2 }"}, {"id": "m - 1 > 0"}, {"id": "抛物线 $y = ( m - 1 ) x ^ { 2 }$ 的开口向上"}, {"id": "m > 1"}], "links": [{"rel": "被描述", "source": "y = ( m - 1 ) x ^ { 2 }", "target": "m - 1 > 0"}, {"rel": "不等式方程求解", "source": "m - 1 > 0", "target": "m > 1"}, {"rel": "限制性描述", "source": "抛物线 $y = ( m - 1 ) x ^ { 2 }$ 的开口向上", "target": "m - 1 > 0"}]}}
|
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{"content": "Given $x - 2 y = 5$, what is the value of $10 ( x - 2 y ) + x - 2 y + 60$?", "answer": "1", "steps": "When $x - 2 y = 5$, the original expression equals $10 ( x - 2 y ) + ( x - 2 y ) + 60 = 50 + 5 + 60 = 115$.", "expr_cands": ["y = - x ^ { 2 } + 4 x + k", "y", "x", "k", "3", "y = - ( x - 2 ) ^ { 2 } + 4 + k", "4 + k = 3", "k = - 1", "k = 1"], "exprs": ["y = - ( x - 2 ) ^ { 2 } + 4 + k", "4 + k = 3", "k = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - x ^ { 2 } + 4 x + k"}, {"id": "y = - ( x - 2 ) ^ { 2 } + 4 + k"}, {"id": "3"}, {"id": "4 + k = 3"}, {"id": "二次函数 $y = - x ^ { 2 } + 4 x + k$ 的最大值等于 $3$"}, {"id": "k = - 1"}], "links": [{"rel": "提取因式", "source": "y = - x ^ { 2 } + 4 x + k", "target": "y = - ( x - 2 ) ^ { 2 } + 4 + k"}, {"rel": "被描述", "source": "y = - ( x - 2 ) ^ { 2 } + 4 + k", "target": "4 + k = 3"}, {"rel": "被描述", "source": "3", "target": "4 + k = 3"}, {"rel": "等式方程求解", "source": "4 + k = 3", "target": "k = - 1"}, {"rel": "限制性描述", "source": "二次函数 $y = - x ^ { 2 } + 4 x + k$ 的最大值等于 $3$", "target": "4 + k = 3"}]}}
|
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{"content": "If $( x + a ) ^ { 2 } = x ^ { 2 } - 8 x + b$, then the value of $a$ is ____?", "answer": "1", "steps": "Because $( x + a ) ^ 2 = x ^ 2 + 2 ax + a ^ 2 = x ^ 2 - 8 x + b$, therefore $2 a = - 8$, therefore $a = - 4$.", "expr_cands": ["\\frac { x } { 0.3 } - \\frac { 5 + x } { 0.07 } = 1", "x", "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 }", "x = - 6.61304347826087", "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 } = 1", "x = - \\frac { 1521 } { 230 }"], "exprs": ["\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 } = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { 0.3 } - \\frac { 5 + x } { 0.07 } = 1"}, {"id": "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 } = 1"}, {"id": "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 }"}, {"id": "把方程 $\\frac { x } { 0.3 } - \\frac { 5 + x } { 0.07 } = 1$ 变形为 $\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 }$ ="}], "links": [{"rel": "被描述", "source": "\\frac { x } { 0.3 } - \\frac { 5 + x } { 0.07 } = 1", "target": "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 } = 1"}, {"rel": "被描述", "source": "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 }", "target": "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 } = 1"}, {"rel": "限制性描述", "source": "把方程 $\\frac { x } { 0.3 } - \\frac { 5 + x } { 0.07 } = 1$ 变形为 $\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 }$ =", "target": "\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 } = 1"}]}}
|
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{"content": "If the parabola $y = ( m - 1 ) x ^ 2$ opens upwards, then the range of possible values for $m$ is ____?", "answer": "2", "steps": "Because the parabola $y = ( m - 1 ) x ^ { 2 }$ opens upwards, $m - 1 > 0$, i.e. $m > 1$. Therefore, the range of values for $m$ is $m > 1$.", "expr_cands": ["\\sqrt { 7 - 2 a }", "a", "2 \\sqrt { 3 }", "7 - 2 a = 3", "a = 2"], "exprs": ["7 - 2 a = 3", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 7 - 2 a }"}, {"id": "7 - 2 a = 3"}, {"id": "2 \\sqrt { 3 }"}, {"id": "最简二次根式 $\\sqrt { 7 - 2 a }$ 与 $2 \\sqrt { 3 }$ 可以合并"}, {"id": "a = 2"}], "links": [{"rel": "被描述", "source": "\\sqrt { 7 - 2 a }", "target": "7 - 2 a = 3"}, {"rel": "等式方程求解", "source": "7 - 2 a = 3", "target": "a = 2"}, {"rel": "被描述", "source": "2 \\sqrt { 3 }", "target": "7 - 2 a = 3"}, {"rel": "限制性描述", "source": "最简二次根式 $\\sqrt { 7 - 2 a }$ 与 $2 \\sqrt { 3 }$ 可以合并", "target": "7 - 2 a = 3"}]}}
|
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{"content": "If the maximum value of the quadratic function $y = - x ^ 2 + 4 x + k$ is equal to $3$, then the value of $k$ is ____?", "answer": "3", "steps": "$y = - x ^ { 2 } + 4 x + k = - ( x - 2 ) ^ { 2 } + 4 + k$ , because the maximum value is $3$ , therefore $4 + k = 3$ , which gives $k = 1$ .", "expr_cands": ["\\frac { x - 1 } { 2 } < 1", "x", "x < 3", "1", "2", "1 + 2", "3"], "exprs": ["x < 3", "1", "2", "1 + 2", "3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 1 } { 2 } < 1"}, {"id": "x < 3"}, {"id": "1"}, {"id": "不等式的正整数解"}, {"id": "2"}, {"id": "1 + 2"}, {"id": "不等式 $\\frac { x - 1 } { 2 } < 1$ 的正整数解的和"}, {"id": "3"}], "links": [{"rel": "不等式方程求解", "source": "\\frac { x - 1 } { 2 } < 1", "target": "x < 3"}, {"rel": "被描述", "source": "x < 3", "target": "1"}, {"rel": "被描述", "source": "x < 3", "target": "2"}, {"rel": "被描述", "source": "1", "target": "1 + 2"}, {"rel": "限制性描述", "source": "不等式的正整数解", "target": "1"}, {"rel": "限制性描述", "source": "不等式的正整数解", "target": "2"}, {"rel": "被描述", "source": "2", "target": "1 + 2"}, {"rel": "计算", "source": "1 + 2", "target": "3"}, {"rel": "限制性描述", "source": "不等式 $\\frac { x - 1 } { 2 } < 1$ 的正整数解的和", "target": "1 + 2"}]}}
|
||||
{"content": "Transform the equation $\\frac { x } { 0.3 } - \\frac { 5 + x } { 0.07 } = 1$ into $\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 }$ = ____ ?", "answer": "x = 20", "steps": "The equation $\\frac { x } { 0.3 } - \\frac { 5 + x } { 0.07 } = 1$ is transformed to $\\frac { 10 x } { 3 } - \\frac { 500 + 100 x } { 7 } = 1$.", "expr_cands": ["\\frac { 7 x } { 20 } = \\frac { x } { 5 } + 3", "x", "7 x = 4 x + 60", "x = 20", "3 x = 60"], "exprs": ["x = 20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 7 x } { 20 } = \\frac { x } { 5 } + 3"}, {"id": "x = 20"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 7 x } { 20 } = \\frac { x } { 5 } + 3", "target": "x = 20"}]}}
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{"content": "Given the simplest quadratic radical $\\sqrt { 7 - 2 a }$ can be combined with $2 \\sqrt { 3 }$, what is the value of $a$?", "answer": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2", "steps": "The simplest quadratic radical $\\sqrt { 7 - 2 a }$ can be combined with $2 \\sqrt { 3 }$, resulting in $7 - 2 a = 3$. Solving for $a$, we get $a = 2$.", "expr_cands": ["y = - 0.5 x ^ { 2 }", "y", "x", "1", "2", "y = - 0.5 ( x + 1 ) ^ { 2 } - 2", "- 0.5 x ^ { 2 } = - 0.5 ( x + 1 ) ^ { 2 } - 2", "- 0.5 x ^ { 2 }"], "exprs": ["y = - 0.5 ( x + 1 ) ^ { 2 } - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - 0.5 x ^ { 2 }"}, {"id": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2"}, {"id": "1"}, {"id": "2"}, {"id": "把抛物线 $y = - 0.5 x ^ { 2 }$ 先向左平移 $1$ 个单位"}, {"id": "再向下平移 $2$ 个单位长度后"}, {"id": "所得的函数表达式"}], "links": [{"rel": "被描述", "source": "y = - 0.5 x ^ { 2 }", "target": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2"}, {"rel": "被描述", "source": "1", "target": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2"}, {"rel": "被描述", "source": "2", "target": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2"}, {"rel": "限制性描述", "source": "把抛物线 $y = - 0.5 x ^ { 2 }$ 先向左平移 $1$ 个单位", "target": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2"}, {"rel": "限制性描述", "source": "再向下平移 $2$ 个单位长度后", "target": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2"}, {"rel": "限制性描述", "source": "所得的函数表达式", "target": "y = - 0.5 ( x + 1 ) ^ { 2 } - 2"}]}}
|
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{"content": "The sum of positive integer solutions to the inequality $\\frac { x - 1 } { 2 } < 1$ is ____?", "answer": "x \\neq \\frac { 5 } { 7 }", "steps": "$\\frac { x - 1 } { 2 } < 1$ , solving gives: $x < 3$ , so the positive integer solutions are: $1$ , $2$ , $1 + 2 = 3$", "expr_cands": ["y = \\frac { 2 x } { 7 x - 5 }", "x", "y", "7 x - 5 \\neq 0", "x \\neq \\frac { 5 } { 7 }"], "exprs": ["7 x - 5 \\neq 0", "x \\neq \\frac { 5 } { 7 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\frac { 2 x } { 7 x - 5 }"}, {"id": "7 x - 5 \\neq 0"}, {"id": "在函数 $y = \\frac { 2 x } { 7 x - 5 }$ 中"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq \\frac { 5 } { 7 }"}], "links": [{"rel": "被描述", "source": "y = \\frac { 2 x } { 7 x - 5 }", "target": "7 x - 5 \\neq 0"}, {"rel": "不等式方程求解", "source": "7 x - 5 \\neq 0", "target": "x \\neq \\frac { 5 } { 7 }"}, {"rel": "限制性描述", "source": "在函数 $y = \\frac { 2 x } { 7 x - 5 }$ 中", "target": "7 x - 5 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "7 x - 5 \\neq 0"}]}}
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{"content": "The solution to the equation $\\frac { 7 x } { 20 } = \\frac { x } { 5 } + 3$ is ____?", "answer": "1", "steps": "To eliminate the denominator, we get $7 x = 4 x + 60$. By rearranging and combining terms, we get $3 x = 60$, and solving for $x$, we get $x = 20$.", "expr_cands": ["| a - 6 | + | b + 5 | = 0", "a", "b", "a + b", "a = 6", "b = - 5", "1"], "exprs": ["a = 6", "b = - 5", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 6 | + | b + 5 | = 0"}, {"id": "a = 6"}, {"id": "绝对值恒大于等于0"}, {"id": "b = - 5"}, {"id": "a + b"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "| a - 6 | + | b + 5 | = 0", "target": "a = 6"}, {"rel": "被描述", "source": "| a - 6 | + | b + 5 | = 0", "target": "b = - 5"}, {"rel": "代入", "source": "a = 6", "target": "1"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a = 6"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b = - 5"}, {"rel": "代入", "source": "b = - 5", "target": "1"}, {"rel": "被代入", "source": "a + b", "target": "1"}]}}
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{"content": "Translate the above math content in English, you should keep the content wrapped in $ unchanged.", "answer": "72", "steps": "According to the principle of adding to the left and subtracting from the right, adding to the top and subtracting from the bottom, we know that the function obtained by first shifting the parabola $y = - 0.5 x ^ 2$ one unit to the left and then two units down has the analytical expression: $y = - 0.5 ( x + 1 ) ^ 2 - 2$.", "expr_cands": ["a ^ { m } = 6", "a", "m", "a ^ { n } = 12", "n", "a ^ { m + n }", "72"], "exprs": ["72"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { m + n }"}, {"id": "72"}, {"id": "a ^ { m } = 6"}, {"id": "a ^ { n } = 12"}], "links": [{"rel": "被代入", "source": "a ^ { m + n }", "target": "72"}, {"rel": "代入", "source": "a ^ { m } = 6", "target": "72"}, {"rel": "代入", "source": "a ^ { n } = 12", "target": "72"}]}}
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{"content": "In the function $y = \\frac { 2 x } { 7 x - 5 }$, the range of the independent variable $x$ is ____?", "answer": "2", "steps": "$7 x - 5 \\neq 0$, $x \\neq \\frac { 5 } { 7 }$ means The expression $7 x - 5$ is not equal to zero, and $x$ cannot be equal to $\\frac { 5 } { 7 }$.", "expr_cands": ["- 3 ab ^ { 2 m - 1 }", "a", "m", "b", "9 ab ^ { m + 1 }", "2 m - 1 = m + 1", "m = 2"], "exprs": ["2 m - 1 = m + 1", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 3 ab ^ { 2 m - 1 }"}, {"id": "2 m - 1 = m + 1"}, {"id": "9 ab ^ { m + 1 }"}, {"id": "$- 3 ab ^ { 2 m - 1 }$ 与 $9 ab ^ { m + 1 }$ 是同类项"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "- 3 ab ^ { 2 m - 1 }", "target": "2 m - 1 = m + 1"}, {"rel": "等式方程求解", "source": "2 m - 1 = m + 1", "target": "m = 2"}, {"rel": "被描述", "source": "9 ab ^ { m + 1 }", "target": "2 m - 1 = m + 1"}, {"rel": "限制性描述", "source": "$- 3 ab ^ { 2 m - 1 }$ 与 $9 ab ^ { m + 1 }$ 是同类项", "target": "2 m - 1 = m + 1"}]}}
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{"content": "If $| a - 6 | + | b + 5 | = 0$, then the value of $a + b$ is ____?", "answer": "- 1", "steps": "Since $| a - 6 | + | b + 5 | = 0$, it follows that $a = 6$ and $b = - 5$. Therefore, $a + b = 6 + ( - 5 ) = 1$.", "expr_cands": ["x", "m + 3 x = 1 + x", "m", "2 x + m = 3 m", "2", "x = \\frac { 1 - m } { 2 }", "1 = 3 m", "x = m", "\\frac { 1 } { 2 } - \\frac { m } { 2 } = m", "\\frac { 1 - m } { 2 } - m = 2", "m = - 1"], "exprs": ["x = \\frac { 1 - m } { 2 }", "x = m", "\\frac { 1 - m } { 2 } - m = 2", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m + 3 x = 1 + x"}, {"id": "x = \\frac { 1 - m } { 2 }"}, {"id": "2 x + m = 3 m"}, {"id": "x = m"}, {"id": "\\frac { 1 - m } { 2 } - m = 2"}, {"id": "关于 $x$ 的方程 $m + 3 x = 1 + x$ 的解比关于 $x$ 的方程 $2 x + m = 3 m$ 的解大 $2$"}, {"id": "m = - 1"}], "links": [{"rel": "等式方程部分求解", "source": "m + 3 x = 1 + x", "target": "x = \\frac { 1 - m } { 2 }"}, {"rel": "被描述", "source": "x = \\frac { 1 - m } { 2 }", "target": "\\frac { 1 - m } { 2 } - m = 2"}, {"rel": "等式方程部分求解", "source": "2 x + m = 3 m", "target": "x = m"}, {"rel": "被描述", "source": "x = m", "target": "\\frac { 1 - m } { 2 } - m = 2"}, {"rel": "等式方程求解", "source": "\\frac { 1 - m } { 2 } - m = 2", "target": "m = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $m + 3 x = 1 + x$ 的解比关于 $x$ 的方程 $2 x + m = 3 m$ 的解大 $2$", "target": "\\frac { 1 - m } { 2 } - m = 2"}]}}
|
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{"content": "Given $a ^ { m } = 6$ and $a ^ { n } = 12$, what is $a ^ { m + n }$?", "answer": "9", "steps": "Because $a ^ { m } = 6$ and $a ^ { n } = 12$, therefore $a ^ { m + n } = a ^ { m } \\times a ^ { n } = 6 * 12 = 72$.", "expr_cands": ["x + y = \\frac { 2 } { 3 }", "y", "x", "xy = 2", "( x - 3 ) ( y - 3 )", "xy - 3 ( x + y ) + 9", "9"], "exprs": ["xy - 3 ( x + y ) + 9", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x - 3 ) ( y - 3 )"}, {"id": "xy - 3 ( x + y ) + 9"}, {"id": "x + y = \\frac { 2 } { 3 }"}, {"id": "xy = 2"}, {"id": "9"}], "links": [{"rel": "提取因式", "source": "( x - 3 ) ( y - 3 )", "target": "xy - 3 ( x + y ) + 9"}, {"rel": "被代入", "source": "xy - 3 ( x + y ) + 9", "target": "9"}, {"rel": "提取因式参考", "source": "x + y = \\frac { 2 } { 3 }", "target": "xy - 3 ( x + y ) + 9"}, {"rel": "代入", "source": "x + y = \\frac { 2 } { 3 }", "target": "9"}, {"rel": "提取因式参考", "source": "xy = 2", "target": "xy - 3 ( x + y ) + 9"}, {"rel": "代入", "source": "xy = 2", "target": "9"}]}}
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{"content": "If $- 3 ab ^ { 2 m - 1 }$ and $9 ab ^ { m + 1 }$ are like terms, then $m$ is equal to ____?", "answer": "17", "steps": "Because $- 3 ab ^ { 2 m - 1 }$ and $9 ab ^ { m + 1 }$ are like terms, therefore $2 m - 1 = m + 1$, solving for $m$, we get $m = 2$.", "expr_cands": ["A = 2 x ^ { 2 } + ax - y + 6", "y", "x", "a", "A", "B = bx ^ { 2 } - 3 x + 5 y - 1", "b", "B", "A - B", "x ^ { 2 }", "a ^ { 2 } + b ^ { 3 }", "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7", "2 - b = 0", "b = 2", "a + 3 = 0", "a = - 3", "17"], "exprs": ["A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7", "2 - b = 0", "a + 3 = 0", "b = 2", "a = - 3", "17"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "A = 2 x ^ { 2 } + ax - y + 6"}, {"id": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7"}, {"id": "B = bx ^ { 2 } - 3 x + 5 y - 1"}, {"id": "A - B"}, {"id": "x"}, {"id": "且 $A - B$ 中不含有 $x ^ { 2 }$ 项和 $x$ 项"}, {"id": "2 - b = 0"}, {"id": "a + 3 = 0"}, {"id": "b = 2"}, {"id": "a = - 3"}, {"id": "a ^ { 2 } + b ^ { 3 }"}, {"id": "17"}], "links": [{"rel": "被描述", "source": "A = 2 x ^ { 2 } + ax - y + 6", "target": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7"}, {"rel": "被描述", "source": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7", "target": "2 - b = 0"}, {"rel": "被描述", "source": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7", "target": "a + 3 = 0"}, {"rel": "被描述", "source": "B = bx ^ { 2 } - 3 x + 5 y - 1", "target": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7"}, {"rel": "被描述", "source": "A - B", "target": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7"}, {"rel": "被描述", "source": "x", "target": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7"}, {"rel": "限制性描述", "source": "且 $A - B$ 中不含有 $x ^ { 2 }$ 项和 $x$ 项", "target": "A - B = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7"}, {"rel": "限制性描述", "source": "且 $A - B$ 中不含有 $x ^ { 2 }$ 项和 $x$ 项", "target": "2 - b = 0"}, {"rel": "限制性描述", "source": "且 $A - B$ 中不含有 $x ^ { 2 }$ 项和 $x$ 项", "target": "a + 3 = 0"}, {"rel": "等式方程求解", "source": "2 - b = 0", "target": "b = 2"}, {"rel": "等式方程求解", "source": "a + 3 = 0", "target": "a = - 3"}, {"rel": "代入", "source": "b = 2", "target": "17"}, {"rel": "代入", "source": "a = - 3", "target": "17"}, {"rel": "被代入", "source": "a ^ { 2 } + b ^ { 3 }", "target": "17"}]}}
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{"content": "If the solution to the equation $m + 3 x = 1 + x$ is $2$ greater than the solution to the equation $2 x + m = 3 m$ in terms of $x$, then the value of $m$ is ____?", "answer": "2", "steps": "Solve the equation $m + 3 x = 1 + x$ to get $x = \\frac { 1 - m } { 2 }$. Solve the equation $2 x + m = 3 m$ to get $x = m$. According to the problem, we have $\\frac { 1 - m } { 2 } - m = 2$. Solving for $m$, we get $m = - 1$.", "expr_cands": ["a", "b", "| a + b + 2 |", "a + b = 0", "| 0 + 2 |", "2"], "exprs": ["a + b = 0", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a + b = 0"}, {"id": "b"}, {"id": "$a$ 与 $b$ 互为相反数"}, {"id": "| a + b + 2 |"}, {"id": "2"}], "links": [{"rel": "被描述", "source": "a", "target": "a + b = 0"}, {"rel": "代入", "source": "a + b = 0", "target": "2"}, {"rel": "被描述", "source": "b", "target": "a + b = 0"}, {"rel": "限制性描述", "source": "$a$ 与 $b$ 互为相反数", "target": "a + b = 0"}, {"rel": "被代入", "source": "| a + b + 2 |", "target": "2"}]}}
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{"content": "Given: $x + y = \\frac { 2 } { 3 }$, $xy = 2$, calculate the result of $( x - 3 ) ( y - 3 )$ is ____?", "answer": "2", "steps": "\\because $x + y = \\frac { 2 } { 3 }$ , $xy = 2$ , \\therefore the original expression = $xy - 3 ( x + y ) + 9 = 2 - 3 * \\frac { 2 } { 3 } + 9 = 9$ .", "expr_cands": ["| x - 1 | - | x + 6 | - 5", "x", "| x - 1 | - | x + 6 |", "1 - ( - 6 )", "7", "7 - 5", "2"], "exprs": ["2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x - 1 | - | x + 6 | - 5"}, {"id": "2"}, {"id": "代数式 $| x - 1 | - | x + 6 | - 5$ 的最大值"}], "links": [{"rel": "被描述", "source": "| x - 1 | - | x + 6 | - 5", "target": "2"}, {"rel": "限制性描述", "source": "代数式 $| x - 1 | - | x + 6 | - 5$ 的最大值", "target": "2"}]}}
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{"content": "Given $A = 2 x ^ { 2 } + ax - y + 6$ , $B = bx ^ { 2 } - 3 x + 5 y - 1$ , and $A - B$ does not contain $x ^ { 2 }$ and $x$ terms, then $a ^ { 2 } + b ^ { 3 }$ equals ____ ?", "answer": "9", "steps": "$\\because A = 2 x ^ { 2 } + ax - y + 6$ , $B = bx ^ { 2 } - 3 x + 5 y - 1$ , and $A - B$ does not contain $x ^ { 2 }$ and $x$ terms, $\\therefore A - B = 2 x ^ { 2 } + ax - y + 6 - ( bx ^ { 2 } - 3 x + 5 y - 1 ) = ( 2 - b ) x ^ { 2 } + ( a + 3 ) x + 4 y + 7$ . Thus, $2 - b = 0$ and $a + 3 = 0$ . Solving for $b$ and $a$ gives $b = 2$ and $a = - 3$ , so $a ^ { 2 } + b ^ { 3 } = 9 + 8 = 17$ .", "expr_cands": ["x - 2 y = - 2", "y", "x", "5 - 2 x + 4 y", "5 - 2 ( x - 2 y )", "9"], "exprs": ["5 - 2 ( x - 2 y )", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 - 2 x + 4 y"}, {"id": "5 - 2 ( x - 2 y )"}, {"id": "x - 2 y = - 2"}, {"id": "9"}], "links": [{"rel": "提取因式", "source": "5 - 2 x + 4 y", "target": "5 - 2 ( x - 2 y )"}, {"rel": "被代入", "source": "5 - 2 ( x - 2 y )", "target": "9"}, {"rel": "提取因式参考", "source": "x - 2 y = - 2", "target": "5 - 2 ( x - 2 y )"}, {"rel": "代入", "source": "x - 2 y = - 2", "target": "9"}]}}
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{"content": "If $a$ and $b$ are opposite numbers, then $| a + b + 2 |$ is equal to ____?", "answer": "\\frac { 2 } { 5 }", "steps": "$\\because a$ and $b$ are opposite numbers, $\\therefore a + b = 0$, then the original expression $= | 0 + 2 | = 2$.", "expr_cands": ["m - n = \\frac { 1 } { 5 }", "m", "n", "- 2 ( n - m )", "\\frac { 2 } { 5 }"], "exprs": ["\\frac { 2 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 ( n - m )"}, {"id": "\\frac { 2 } { 5 }"}, {"id": "m - n = \\frac { 1 } { 5 }"}], "links": [{"rel": "被代入", "source": "- 2 ( n - m )", "target": "\\frac { 2 } { 5 }"}, {"rel": "代入", "source": "m - n = \\frac { 1 } { 5 }", "target": "\\frac { 2 } { 5 }"}]}}
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{"content": "What is the maximum value of the algebraic expression $| x - 1 | - | x + 6 | - 5$?", "answer": "6", "steps": "The maximum value of $| x - 1 | - | x + 6 |$ is $1 - ( - 6 ) = 1 + 6 = 7$, so the maximum value of the algebraic expression is $7 - 5 = 2$.", "expr_cands": ["x = 2", "x", "( m + 2 ) x ^ { 2 } + 2 x - m ^ { 2 } = 0", "m", "- m ^ { 2 } + 4 m + 12 = 0", "4 ( m + 2 ) + 4 - m ^ { 2 } = 0", "m = - 2", "m = 6", "m ^ { 2 } - 4 m - 12 = 0", "m _ { 1 } = - 2", "m _ { 1 }", "m _ { 2 } = 6", "m _ { 2 }", "m + 2 \\neq 0", "m \\neq - 2", "6"], "exprs": ["4 ( m + 2 ) + 4 - m ^ { 2 } = 0", "m + 2 \\neq 0", "m ^ { 2 } - 4 m - 12 = 0", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m + 2 ) x ^ { 2 } + 2 x - m ^ { 2 } = 0"}, {"id": "4 ( m + 2 ) + 4 - m ^ { 2 } = 0"}, {"id": "x = 2"}, {"id": "x"}, {"id": "$x = 2$ 是关于 $x$ 的一元二次方程 $( m + 2 ) x ^ { 2 } + 2 x - m ^ { 2 } = 0$ 的一个根"}, {"id": "m ^ { 2 } - 4 m - 12 = 0"}, {"id": "m + 2 \\neq 0"}, {"id": "6"}, {"id": "$m$ 的值"}], "links": [{"rel": "被描述", "source": "( m + 2 ) x ^ { 2 } + 2 x - m ^ { 2 } = 0", "target": "4 ( m + 2 ) + 4 - m ^ { 2 } = 0"}, {"rel": "被描述", "source": "( m + 2 ) x ^ { 2 } + 2 x - m ^ { 2 } = 0", "target": "m + 2 \\neq 0"}, {"rel": "计算", "source": "4 ( m + 2 ) + 4 - m ^ { 2 } = 0", "target": "m ^ { 2 } - 4 m - 12 = 0"}, {"rel": "被描述", "source": "x = 2", "target": "4 ( m + 2 ) + 4 - m ^ { 2 } = 0"}, {"rel": "被描述", "source": "x", "target": "4 ( m + 2 ) + 4 - m ^ { 2 } = 0"}, {"rel": "限制性描述", "source": "$x = 2$ 是关于 $x$ 的一元二次方程 $( m + 2 ) x ^ { 2 } + 2 x - m ^ { 2 } = 0$ 的一个根", "target": "4 ( m + 2 ) + 4 - m ^ { 2 } = 0"}, {"rel": "限制性描述", "source": "$x = 2$ 是关于 $x$ 的一元二次方程 $( m + 2 ) x ^ { 2 } + 2 x - m ^ { 2 } = 0$ 的一个根", "target": "m + 2 \\neq 0"}, {"rel": "被描述", "source": "m ^ { 2 } - 4 m - 12 = 0", "target": "6"}, {"rel": "被描述", "source": "m + 2 \\neq 0", "target": "6"}, {"rel": "限制性描述", "source": "$m$ 的值", "target": "6"}]}}
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{"content": "Given $x - 2 y = - 2$, what is the value of the algebraic expression $5 - 2 x + 4 y$?", "answer": "2021", "steps": "Because $x - 2 y = - 2$, therefore $5 - 2 x + 4 y = 5 - 2 ( x - 2 y ) = 5 - 2 * ( - 2 ) = 5 + 4 = 9$.", "expr_cands": ["m - n", "m", "n", "1", "3 m - 3 n + 2018", "m - n = 1", "3 ( m - n ) + 2018", "2021"], "exprs": ["m - n = 1", "3 ( m - n ) + 2018", "2021"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m - n"}, {"id": "m - n = 1"}, {"id": "1"}, {"id": "代数式 $m - n$ 的值是 $1$"}, {"id": "3 m - 3 n + 2018"}, {"id": "3 ( m - n ) + 2018"}, {"id": "2021"}], "links": [{"rel": "被描述", "source": "m - n", "target": "m - n = 1"}, {"rel": "提取因式参考", "source": "m - n = 1", "target": "3 ( m - n ) + 2018"}, {"rel": "代入", "source": "m - n = 1", "target": "2021"}, {"rel": "被描述", "source": "1", "target": "m - n = 1"}, {"rel": "限制性描述", "source": "代数式 $m - n$ 的值是 $1$", "target": "m - n = 1"}, {"rel": "提取因式", "source": "3 m - 3 n + 2018", "target": "3 ( m - n ) + 2018"}, {"rel": "被代入", "source": "3 ( m - n ) + 2018", "target": "2021"}]}}
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{"content": "If $m - n = \\frac { 1 } { 5 }$, then the value of $- 2 ( n - m )$ is ____?", "answer": "k \\ge 1", "steps": "Because $m - n = \\frac { 1 } { 5 }$, therefore $- 2 ( n - m ) = 2 ( m - n ) = 2 * \\frac { 1 } { 5 } = \\frac { 2 } { 5 }$.", "expr_cands": ["x", "x - 2 + 3 k = 0", "k", "- 1", "x = - 3 k + 2", "- 3 k + 2 \\le - 1", "1 \\le k", "k \\ge 1"], "exprs": ["x = - 3 k + 2", "- 3 k + 2 \\le - 1", "k \\ge 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x - 2 + 3 k = 0"}, {"id": "x = - 3 k + 2"}, {"id": "- 3 k + 2 \\le - 1"}, {"id": "- 1"}, {"id": "关于 $x$ 的方程 $x - 2 + 3 k = 0$ 的解不大于 $- 1$"}, {"id": "k \\ge 1"}], "links": [{"rel": "等式方程部分求解", "source": "x - 2 + 3 k = 0", "target": "x = - 3 k + 2"}, {"rel": "被描述", "source": "x = - 3 k + 2", "target": "- 3 k + 2 \\le - 1"}, {"rel": "不等式方程求解", "source": "- 3 k + 2 \\le - 1", "target": "k \\ge 1"}, {"rel": "被描述", "source": "- 1", "target": "- 3 k + 2 \\le - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x - 2 + 3 k = 0$ 的解不大于 $- 1$", "target": "- 3 k + 2 \\le - 1"}]}}
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{"content": "Given that $x = 2$ is a root of the quadratic equation $( m + 2 ) x ^ 2 + 2 x - m ^ 2 = 0$ in terms of $x$, what is the value of $m$?", "answer": "- 3", "steps": "Substituting $x = 2$ into the equation $( m + 2 ) x ^ 2 + 2 x - m ^ 2 = 0$ yields $4 ( m + 2 ) + 4 - m ^ 2 = 0$. Simplifying, we get $m ^ 2 - 4 m - 12 = 0$. Solving for $m$, we get $m _ 1 = - 2$ and $m _ 2 = 6$. Since $m + 2 \\neq 0$, we conclude that $m = 6$.", "expr_cands": ["x", "\\frac { x - 2 } { x - 5 } = \\frac { m } { 5 - x }", "m", "x - 2 = - m", "x = 2 - m", "x - 5 = 0", "x = 5", "m = - 3"], "exprs": ["x - 2 = - m", "x - 5 = 0", "x = 2 - m", "x = 5", "m = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 2 } { x - 5 } = \\frac { m } { 5 - x }"}, {"id": "x - 2 = - m"}, {"id": "x = 2 - m"}, {"id": "x - 5 = 0"}, {"id": "关于 $x$ 的分式方程 $\\frac { x - 2 } { x - 5 } = \\frac { m } { 5 - x }$ 无解"}, {"id": "分式方程无解，则分母为0"}, {"id": "x = 5"}, {"id": "m = - 3"}], "links": [{"rel": "同乘除", "source": "\\frac { x - 2 } { x - 5 } = \\frac { m } { 5 - x }", "target": "x - 2 = - m"}, {"rel": "被描述", "source": "\\frac { x - 2 } { x - 5 } = \\frac { m } { 5 - x }", "target": "x - 5 = 0"}, {"rel": "等式方程部分求解", "source": "x - 2 = - m", "target": "x = 2 - m"}, {"rel": "联立", "source": "x = 2 - m", "target": "m = - 3"}, {"rel": "等式方程求解", "source": "x - 5 = 0", "target": "x = 5"}, {"rel": "限制性描述", "source": "关于 $x$ 的分式方程 $\\frac { x - 2 } { x - 5 } = \\frac { m } { 5 - x }$ 无解", "target": "x - 5 = 0"}, {"rel": "属性描述", "source": "分式方程无解，则分母为0", "target": "x - 5 = 0"}, {"rel": "联立", "source": "x = 5", "target": "m = - 3"}]}}
|
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{"content": "If the algebraic expression $m - n$ has a value of $1$, then the value of the algebraic expression $3 m - 3 n + 2018$ is ____?", "answer": "0", "steps": "When $m - n = 1$, $3 m - 3 n + 2018 = 3 ( m - n ) + 2018 = 3 * 1 + 2018 = 3 + 2018 = 2021$.", "expr_cands": ["y", "x + 1", "x", "x = 1", "y = 2", "x = - 1", "y = k ( x + 1 )", "k", "2 = k * 2", "k = 1", "y = x + 1", "y = 0"], "exprs": ["y = k ( x + 1 )", "2 = k * 2", "k = 1", "y = x + 1", "y = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 1"}, {"id": "y = k ( x + 1 )"}, {"id": "y"}, {"id": "$y$ 与 $x + 1$ 成正比例"}, {"id": "设 $y = k ( x + 1 )$"}, {"id": "x = 1"}, {"id": "2 = k * 2"}, {"id": "y = 2"}, {"id": "k = 1"}, {"id": "y = x + 1"}, {"id": "x = - 1"}, {"id": "y = 0"}], "links": [{"rel": "被描述", "source": "x + 1", "target": "y = k ( x + 1 )"}, {"rel": "被代入", "source": "y = k ( x + 1 )", "target": "2 = k * 2"}, {"rel": "被代入", "source": "y = k ( x + 1 )", "target": "y = x + 1"}, {"rel": "被描述", "source": "y", "target": "y = k ( x + 1 )"}, {"rel": "限制性描述", "source": "$y$ 与 $x + 1$ 成正比例", "target": "y = k ( x + 1 )"}, {"rel": "假设描述", "source": "设 $y = k ( x + 1 )$", "target": "y = k ( x + 1 )"}, {"rel": "代入", "source": "x = 1", "target": "2 = k * 2"}, {"rel": "等式方程求解", "source": "2 = k * 2", "target": "k = 1"}, {"rel": "代入", "source": "y = 2", "target": "2 = k * 2"}, {"rel": "代入", "source": "k = 1", "target": "y = x + 1"}, {"rel": "被代入", "source": "y = x + 1", "target": "y = 0"}, {"rel": "代入", "source": "x = - 1", "target": "y = 0"}]}}
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{"content": "If the solution to the equation $x - 2 + 3 k = 0$ with respect to $x$ is not greater than $- 1$, then the range of values for $k$ is ____?", "answer": "16", "steps": "Solve the equation $x - 2 + 3 k = 0$ to get $x = - 3 k + 2$. Since the solution to the equation in terms of $x$ is not greater than $- 1$, we have $- 3 k + 2 \\leq - 1$. Solving for $k$, we get $k \\geq 1$.", "expr_cands": ["x", "5 x + 4 = 4 x - 3", "2 ( x + 1 ) - m = - 2 ( m - 2 )", "m", "x = - 7", "- m - 12 = - 2 ( m - 2 )", "2 * ( - 7 + 1 ) - m = - 2 ( m - 2 )", "m = 16"], "exprs": ["x = - 7", "2 * ( - 7 + 1 ) - m = - 2 ( m - 2 )", "m = 16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x + 4 = 4 x - 3"}, {"id": "x = - 7"}, {"id": "2 ( x + 1 ) - m = - 2 ( m - 2 )"}, {"id": "2 * ( - 7 + 1 ) - m = - 2 ( m - 2 )"}, {"id": "m = 16"}], "links": [{"rel": "等式方程求解", "source": "5 x + 4 = 4 x - 3", "target": "x = - 7"}, {"rel": "代入", "source": "x = - 7", "target": "2 * ( - 7 + 1 ) - m = - 2 ( m - 2 )"}, {"rel": "被代入", "source": "2 ( x + 1 ) - m = - 2 ( m - 2 )", "target": "2 * ( - 7 + 1 ) - m = - 2 ( m - 2 )"}, {"rel": "等式方程求解", "source": "2 * ( - 7 + 1 ) - m = - 2 ( m - 2 )", "target": "m = 16"}]}}
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{"content": "If the fractional equation about $x$, $\\frac { x - 2 } { x - 5 } = \\frac { m } { 5 - x }$, has no solution, then the value of $m$ is ____?", "answer": "x \\neq 2", "steps": "Going to the denominator, we get $x - 2 = - m$, solving for $x$ gives $x = 2 - m$. Since the original equation has no solution, the simplest common denominator is $x - 5 = 0$, which gives $x = 5$. Therefore, we have $m = - 3$.", "expr_cands": ["\\frac { 2 } { x - 2 }", "x", "x - 2 \\neq 0", "x \\neq 2"], "exprs": ["x - 2 \\neq 0", "x \\neq 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { x - 2 }"}, {"id": "x - 2 \\neq 0"}, {"id": "对于公式 $\\frac { 2 } { x - 2 }$ , $x$ 的取值范围"}, {"id": "分式有意义，则分母不为0"}, {"id": "x \\neq 2"}], "links": [{"rel": "被描述", "source": "\\frac { 2 } { x - 2 }", "target": "x - 2 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 2 \\neq 0", "target": "x \\neq 2"}, {"rel": "限制性描述", "source": "对于公式 $\\frac { 2 } { x - 2 }$ , $x$ 的取值范围", "target": "x - 2 \\neq 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 2 \\neq 0"}]}}
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||||
{"content": "Given that $y$ is directly proportional to $x + 1$, and when $x = 1$, $y = 2$. What is the value of $y$ when $x = - 1$?", "answer": "- 2", "steps": "\\because $y$ is directly proportional to $x + 1$, \\therefore let $y = k ( x + 1 )$, \\because when $x = 1$, $y = 2$, \\therefore $2 = k * 2$, which means $k = 1$, so $y = x + 1$. Therefore, when $x = - 1$, $y = - 1 + 1 = 0$.", "expr_cands": ["x + 1 < 0", "x", "x < - 1", "- 2"], "exprs": ["x < - 1", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 1 < 0"}, {"id": "x < - 1"}, {"id": "- 2"}, {"id": "即 $- 2$ 是不等式 $x + 1 < 0$ 的一个解"}, {"id": "不等式 $x + 1 < 0$ 的一个解"}], "links": [{"rel": "不等式方程求解", "source": "x + 1 < 0", "target": "x < - 1"}, {"rel": "被描述", "source": "x < - 1", "target": "- 2"}, {"rel": "限制性描述", "source": "即 $- 2$ 是不等式 $x + 1 < 0$ 的一个解", "target": "- 2"}, {"rel": "限制性描述", "source": "不等式 $x + 1 < 0$ 的一个解", "target": "- 2"}]}}
|
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{"content": "The equation $5 x + 4 = 4 x - 3$ and $2 ( x + 1 ) - m = - 2 ( m - 2 )$ have the same solution for $x$. What is the value of $m$?", "answer": "7", "steps": "Solve the equation $5 x + 4 = 4 x - 3$ to get $x = - 7$. Substitute $x = - 7$ into $2 ( x + 1 ) - m = - 2 ( m - 2 )$ to get $2 * ( - 7 + 1 ) - m = - 2 ( m - 2 )$. Solve for $m$ to get $m = 16$.", "expr_cands": ["3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }", "a", "m", "b", "( n - 2 ) a ^ { 4 } b ^ { 3 }", "n", "0", "m + n", "\\frac { 1 } { 2 } m = 4", "m = 8", "n - 2 = - 3", "n = - 1", "7"], "exprs": ["\\frac { 1 } { 2 } m = 4", "n - 2 = - 3", "m = 8", "n = - 1", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }"}, {"id": "\\frac { 1 } { 2 } m = 4"}, {"id": "( n - 2 ) a ^ { 4 } b ^ { 3 }"}, {"id": "$3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }$ 和 $( n - 2 ) a ^ { 4 } b ^ { 3 }$ 是同类项"}, {"id": "n - 2 = - 3"}, {"id": "且它们的和为 $0$"}, {"id": "m = 8"}, {"id": "n = - 1"}, {"id": "m + n"}, {"id": "7"}], "links": [{"rel": "被描述", "source": "3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }", "target": "\\frac { 1 } { 2 } m = 4"}, {"rel": "被描述", "source": "3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }", "target": "n - 2 = - 3"}, {"rel": "等式方程求解", "source": "\\frac { 1 } { 2 } m = 4", "target": "m = 8"}, {"rel": "被描述", "source": "( n - 2 ) a ^ { 4 } b ^ { 3 }", "target": "\\frac { 1 } { 2 } m = 4"}, {"rel": "被描述", "source": "( n - 2 ) a ^ { 4 } b ^ { 3 }", "target": "n - 2 = - 3"}, {"rel": "限制性描述", "source": "$3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }$ 和 $( n - 2 ) a ^ { 4 } b ^ { 3 }$ 是同类项", "target": "\\frac { 1 } { 2 } m = 4"}, {"rel": "限制性描述", "source": "$3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }$ 和 $( n - 2 ) a ^ { 4 } b ^ { 3 }$ 是同类项", "target": "n - 2 = - 3"}, {"rel": "等式方程求解", "source": "n - 2 = - 3", "target": "n = - 1"}, {"rel": "限制性描述", "source": "且它们的和为 $0$", "target": "n - 2 = - 3"}, {"rel": "代入", "source": "m = 8", "target": "7"}, {"rel": "代入", "source": "n = - 1", "target": "7"}, {"rel": "被代入", "source": "m + n", "target": "7"}]}}
|
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{"content": "For the formula $\\frac { 2 } { x - 2 }$, the range of values for $x$ is ____?", "answer": "- \\frac { 1 } { 2 }", "steps": "From the given condition, we can deduce that $x - 2 \\neq 0$, which implies that $x \\neq 2$.", "expr_cands": ["| a + 2 | + \\sqrt { b - 4 } = 0", "a", "b", "\\frac { a } { b }", "| a + 2 | \\ge 0", "\\sqrt { b - 4 } \\ge 0", "4 \\le b", "a + 2 = 0", "a = - 2", "b - 4 = 0", "b = 4", "- \\frac { 1 } { 2 }"], "exprs": ["a + 2 = 0", "b - 4 = 0", "a = - 2", "b = 4", "- \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a + 2 | + \\sqrt { b - 4 } = 0"}, {"id": "a + 2 = 0"}, {"id": "实数 $| a + 2 | + \\sqrt { b - 4 } = 0$"}, {"id": "绝对值恒大于等于0"}, {"id": "b - 4 = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a = - 2"}, {"id": "b = 4"}, {"id": "\\frac { a } { b }"}, {"id": "- \\frac { 1 } { 2 }"}], "links": [{"rel": "被描述", "source": "| a + 2 | + \\sqrt { b - 4 } = 0", "target": "a + 2 = 0"}, {"rel": "被描述", "source": "| a + 2 | + \\sqrt { b - 4 } = 0", "target": "b - 4 = 0"}, {"rel": "等式方程求解", "source": "a + 2 = 0", "target": "a = - 2"}, {"rel": "限制性描述", "source": "实数 $| a + 2 | + \\sqrt { b - 4 } = 0$", "target": "a + 2 = 0"}, {"rel": "限制性描述", "source": "实数 $| a + 2 | + \\sqrt { b - 4 } = 0$", "target": "b - 4 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a + 2 = 0"}, {"rel": "等式方程求解", "source": "b - 4 = 0", "target": "b = 4"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "b - 4 = 0"}, {"rel": "代入", "source": "a = - 2", "target": "- \\frac { 1 } { 2 }"}, {"rel": "代入", "source": "b = 4", "target": "- \\frac { 1 } { 2 }"}, {"rel": "被代入", "source": "\\frac { a } { b }", "target": "- \\frac { 1 } { 2 }"}]}}
|
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{"content": "The inequality $x + 1 < 0$ has a solution of _____.", "answer": "400", "steps": "$x + 1 < 0$, move the term $1$ to the right side, we get: $x < - 1$. Therefore, $- 2$ is a solution to the inequality $x + 1 < 0$.", "expr_cands": ["a", "b", "c", "0", "\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }", "m", "n", "( m - n ) ^ { 2 }", "m = 10", "n = - 10", "400"], "exprs": ["m = 10", "n = - 10", "400"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }"}, {"id": "m = 10"}, {"id": "m"}, {"id": "且 $\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }$ 的最大值为 $m$"}, {"id": "$a$ , $b$ , $c$ 都不等于 $0$"}, {"id": "n = - 10"}, {"id": "n"}, {"id": "最小值为 $n$"}, {"id": "( m - n ) ^ { 2 }"}, {"id": "400"}], "links": [{"rel": "被描述", "source": "\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }", "target": "m = 10"}, {"rel": "被描述", "source": "\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }", "target": "n = - 10"}, {"rel": "代入", "source": "m = 10", "target": "400"}, {"rel": "被描述", "source": "m", "target": "m = 10"}, {"rel": "被描述", "source": "m", "target": "n = - 10"}, {"rel": "限制性描述", "source": "且 $\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }$ 的最大值为 $m$", "target": "m = 10"}, {"rel": "限制性描述", "source": "且 $\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }$ 的最大值为 $m$", "target": "n = - 10"}, {"rel": "限制性描述", "source": "$a$ , $b$ , $c$ 都不等于 $0$", "target": "m = 10"}, {"rel": "限制性描述", "source": "$a$ , $b$ , $c$ 都不等于 $0$", "target": "n = - 10"}, {"rel": "代入", "source": "n = - 10", "target": "400"}, {"rel": "被描述", "source": "n", "target": "n = - 10"}, {"rel": "限制性描述", "source": "最小值为 $n$", "target": "n = - 10"}, {"rel": "被代入", "source": "( m - n ) ^ { 2 }", "target": "400"}]}}
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{"content": "If $3 a ^ { \\frac { 1 } { 2 } m } b ^ { 3 }$ and $( n - 2 ) a ^ { 4 } b ^ { 3 }$ are similar terms, and their sum is $0$, then the value of $m + n$ is ____?", "answer": "x = 2", "steps": "From the given information, we have $\\frac { 1 } { 2 } m = 4$ and $n - 2 = - 3$. Solving for $m$ and $n$, we get $m = 8$ and $n = - 1$. Therefore, $m + n = 8 + ( - 1 ) = 7$.", "expr_cands": ["x ( x - 1 ) - ( x + 1 ) ( x - 5 ) = 11", "x", "{ x } ^ { 2 } - x - { x } ^ { 2 } + 4 x + 5 = 11", "x = 2", "{ x } ^ { 2 } - x - { x } ^ { 2 } + 4 x = 11 - 5", "3 x = 6", "1"], "exprs": ["x = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ( x - 1 ) - ( x + 1 ) ( x - 5 ) = 11"}, {"id": "x = 2"}], "links": [{"rel": "等式方程求解", "source": "x ( x - 1 ) - ( x + 1 ) ( x - 5 ) = 11", "target": "x = 2"}]}}
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{"content": "If the real number $| a + 2 | + \\sqrt { b - 4 } = 0$, then $\\frac { a } { b }$ = ____ ?", "answer": "5", "steps": "Since $| a + 2 | \\ge 0$, $\\sqrt { b - 4 } \\ge 0$, it follows that $a + 2 = 0$, $b - 4 = 0$. Therefore, $a = - 2$, $b = 4$, and $\\frac { a } { b } = - \\frac { 1 } { 2 }$.", "expr_cands": ["- 7 x ^ { m } y ^ { 4 }", "y", "x", "m", "2 x ^ { 9 } y ^ { n }", "n", "| m - n |", "m = 9", "n = 4", "5"], "exprs": ["m = 9", "n = 4", "5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 7 x ^ { m } y ^ { 4 }"}, {"id": "m = 9"}, {"id": "2 x ^ { 9 } y ^ { n }"}, {"id": "$- 7 x ^ { m } y ^ { 4 }$ 与 $2 x ^ { 9 } y ^ { n }$ 是同类项"}, {"id": "n = 4"}, {"id": "| m - n |"}, {"id": "5"}], "links": [{"rel": "被描述", "source": "- 7 x ^ { m } y ^ { 4 }", "target": "m = 9"}, {"rel": "被描述", "source": "- 7 x ^ { m } y ^ { 4 }", "target": "n = 4"}, {"rel": "代入", "source": "m = 9", "target": "5"}, {"rel": "被描述", "source": "2 x ^ { 9 } y ^ { n }", "target": "m = 9"}, {"rel": "被描述", "source": "2 x ^ { 9 } y ^ { n }", "target": "n = 4"}, {"rel": "限制性描述", "source": "$- 7 x ^ { m } y ^ { 4 }$ 与 $2 x ^ { 9 } y ^ { n }$ 是同类项", "target": "m = 9"}, {"rel": "限制性描述", "source": "$- 7 x ^ { m } y ^ { 4 }$ 与 $2 x ^ { 9 } y ^ { n }$ 是同类项", "target": "n = 4"}, {"rel": "代入", "source": "n = 4", "target": "5"}, {"rel": "被代入", "source": "| m - n |", "target": "5"}]}}
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{"content": "Given that $a$, $b$, and $c$ are all non-zero, and $\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }$ has a maximum value of $m$ and a minimum value of $n$, then $( m - n ) ^ 2$ = ____?", "answer": "a = 5", "steps": "Because $a$, $b$, $c$ are all non-zero, and the maximum value of $\\frac { a } { | a | } + \\frac { 2 b } { | b | } + \\frac { 3 c } { | c | } + \\frac { 4 abc } { | abc | }$ is $m$, and the minimum value is $n$, therefore $m = 10$ and $n = - 10$, so $( m - n ) ^ 2 = ( 10 + 10 ) ^ 2 = 400$.", "expr_cands": ["2 a = 15 - a", "a", "a = 5", "2 a + a", "15", "3 a = 15", "1"], "exprs": ["a = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a = 15 - a"}, {"id": "a = 5"}], "links": [{"rel": "等式方程求解", "source": "2 a = 15 - a", "target": "a = 5"}]}}
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{"content": "The solution to the equation $x ( x - 1 ) - ( x + 1 ) ( x - 5 ) = 11$ is ____ ?", "answer": "- 2", "steps": "Expanding the brackets, we get: ${ x } ^ { 2 } - x - { x } ^ { 2 } + 4 x + 5 = 11$. Moving the constant term to the right-hand side, we get: ${ x } ^ { 2 } - x - { x } ^ { 2 } + 4 x = 11 - 5$). Combining like terms, we get: $3 x = 6$. Dividing both sides by the coefficient of $x$, we get: $x = 2$", "expr_cands": ["1", "2 x ^ { 2 } + bx - 4 = 0", "x", "b", "t", "1 * t = - \\frac { 4 } { 2 }", "t = - 2", "- 2"], "exprs": ["t", "1 * t = - \\frac { 4 } { 2 }", "t = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程的另一个根为 $t$"}, {"id": "t"}, {"id": "2 x ^ { 2 } + bx - 4 = 0"}, {"id": "1 * t = - \\frac { 4 } { 2 }"}, {"id": "x"}, {"id": "1"}, {"id": "$1$ 是方程 $2 x ^ { 2 } + bx - 4 = 0$ 的一个根"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "t = - 2"}], "links": [{"rel": "假设描述", "source": "设方程的另一个根为 $t$", "target": "t"}, {"rel": "限制性描述", "source": "设方程的另一个根为 $t$", "target": "1 * t = - \\frac { 4 } { 2 }"}, {"rel": "被描述", "source": "t", "target": "1 * t = - \\frac { 4 } { 2 }"}, {"rel": "被描述", "source": "2 x ^ { 2 } + bx - 4 = 0", "target": "1 * t = - \\frac { 4 } { 2 }"}, {"rel": "等式方程求解", "source": "1 * t = - \\frac { 4 } { 2 }", "target": "t = - 2"}, {"rel": "被描述", "source": "x", "target": "1 * t = - \\frac { 4 } { 2 }"}, {"rel": "被描述", "source": "1", "target": "1 * t = - \\frac { 4 } { 2 }"}, {"rel": "限制性描述", "source": "$1$ 是方程 $2 x ^ { 2 } + bx - 4 = 0$ 的一个根", "target": "1 * t = - \\frac { 4 } { 2 }"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "1 * t = - \\frac { 4 } { 2 }"}]}}
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{"content": "If $- 7 x ^ { m } y ^ { 4 }$ and $2 x ^ { 9 } y ^ { n }$ are like terms, then $| m - n |$ = ____ ?", "answer": "x = - 2", "steps": "From the given information, we have $m = 9$ and $n = 4$. Therefore, $| m - n | = | 9 - 4 | = 5$.", "expr_cands": ["4 x - 7 = 6 x - 3", "x", "4 x - 6 x = - 3 + 7", "x = - 2", "- 2 x = 4", "1"], "exprs": ["x = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x - 7 = 6 x - 3"}, {"id": "x = - 2"}], "links": [{"rel": "等式方程求解", "source": "4 x - 7 = 6 x - 3", "target": "x = - 2"}]}}
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{"content": "The equation $2 a = 15 - a$ has a solution of _____.", "answer": "\\frac { 9 } { 2 }", "steps": "$\\because$ $2 a = 15 - a$, moving terms, we get $2 a + a = 15$, combining like terms, we get $3 a = 15$, coefficient simplification gives $a = 5$.", "expr_cands": ["a", "b", "c", "a = 3", "b = 2", "a ^ { 2 } = bc", "3 ^ { 2 } = 2 * c", "c = \\frac { 9 } { 2 }"], "exprs": ["a ^ { 2 } = bc", "3 ^ { 2 } = 2 * c", "c = \\frac { 9 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a ^ { 2 } = bc"}, {"id": "b"}, {"id": "c"}, {"id": "线段 $a$ 是线段 $b$ , $c$ 的比例中项"}, {"id": "a = 3"}, {"id": "3 ^ { 2 } = 2 * c"}, {"id": "b = 2"}, {"id": "c = \\frac { 9 } { 2 }"}], "links": [{"rel": "被描述", "source": "a", "target": "a ^ { 2 } = bc"}, {"rel": "被代入", "source": "a ^ { 2 } = bc", "target": "3 ^ { 2 } = 2 * c"}, {"rel": "被描述", "source": "b", "target": "a ^ { 2 } = bc"}, {"rel": "被描述", "source": "c", "target": "a ^ { 2 } = bc"}, {"rel": "限制性描述", "source": "线段 $a$ 是线段 $b$ , $c$ 的比例中项", "target": "a ^ { 2 } = bc"}, {"rel": "代入", "source": "a = 3", "target": "3 ^ { 2 } = 2 * c"}, {"rel": "等式方程求解", "source": "3 ^ { 2 } = 2 * c", "target": "c = \\frac { 9 } { 2 }"}, {"rel": "代入", "source": "b = 2", "target": "3 ^ { 2 } = 2 * c"}]}}
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{"content": "If $1$ is a root of the equation $2 x ^ 2 + bx - 4 = 0$, then the other root of the equation is ____?", "answer": "6", "steps": "Assuming the other root of the equation is $t$, we have $1 \\cdot t = - \\frac { 4 } { 2 }$. Solving for $t$, we get $t = - 2$. Therefore, the other root of the equation is $- 2$.", "expr_cands": ["x = - 3", "x", "y = - 2", "y", "z = 0", "z", "w = 5", "w", "- x + y - z + w", "6"], "exprs": ["6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- x + y - z + w"}, {"id": "6"}, {"id": "x = - 3"}, {"id": "y = - 2"}, {"id": "z = 0"}, {"id": "w = 5"}], "links": [{"rel": "被代入", "source": "- x + y - z + w", "target": "6"}, {"rel": "代入", "source": "x = - 3", "target": "6"}, {"rel": "代入", "source": "y = - 2", "target": "6"}, {"rel": "代入", "source": "z = 0", "target": "6"}, {"rel": "代入", "source": "w = 5", "target": "6"}]}}
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{"content": "The solution to the linear equation $4 x - 7 = 6 x - 3$ is ____ ?", "answer": "- \\frac { 1 } { 6 }", "steps": "Moving terms yields: $4 x - 6 x = - 3 + 7$, combining terms gives: $- 2 x = 4$, dividing by the coefficient gives: $x = - 2$.", "expr_cands": ["\\frac { 1 } { m } - \\frac { 1 } { n } = 6", "m", "n", "\\frac { mn } { m - n }", "\\frac { n - m } { mn } = 6", "n - m = 6 mn", "m - n", "- 6 mn", "- \\frac { 1 } { 6 }"], "exprs": ["\\frac { n - m } { mn } = 6", "n - m = 6 mn", "- \\frac { 1 } { 6 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { m } - \\frac { 1 } { n } = 6"}, {"id": "\\frac { n - m } { mn } = 6"}, {"id": "n - m = 6 mn"}, {"id": "\\frac { mn } { m - n }"}, {"id": "- \\frac { 1 } { 6 }"}], "links": [{"rel": "计算", "source": "\\frac { 1 } { m } - \\frac { 1 } { n } = 6", "target": "\\frac { n - m } { mn } = 6"}, {"rel": "同乘除", "source": "\\frac { n - m } { mn } = 6", "target": "n - m = 6 mn"}, {"rel": "代入", "source": "n - m = 6 mn", "target": "- \\frac { 1 } { 6 }"}, {"rel": "被代入", "source": "\\frac { mn } { m - n }", "target": "- \\frac { 1 } { 6 }"}]}}
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{"content": "Given that segment $a$ is the middle term of the ratio of segments $b$ and $c$, if $a = 3$ and $b = 2$, then $c$ = ____ ?", "answer": "a \\le 1", "steps": "$\\because$ Line segment $a$ is the mean proportional between line segments $b$ and $c$. $\\therefore$ $a ^ 2 = bc$, which means $3 ^ 2 = 2 c$. $\\therefore$ $c = \\frac { 9 } { 2 }$.", "expr_cands": ["\\sqrt { 1 - a } = b", "b", "a", "1 - a \\ge 0", "a \\le 1"], "exprs": ["1 - a \\ge 0", "a \\le 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 1 - a } = b"}, {"id": "1 - a \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "a \\le 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { 1 - a } = b", "target": "1 - a \\ge 0"}, {"rel": "不等式方程求解", "source": "1 - a \\ge 0", "target": "a \\le 1"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "1 - a \\ge 0"}]}}
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{"content": "When $x = - 3$, $y = - 2$, $z = 0$, and $w = 5$, what is the value of $- x + y - z + w$?", "answer": "- 3", "steps": "Since $x = - 3$, $y = - 2$, $z = 0$, and $w = 5$, therefore $- x + y - z + w = - ( - 3 ) + ( - 2 ) - 0 + 5 = 3 - 2 - 0 + 5 = 6$.", "expr_cands": ["2 x", "x", "3 - x", "2 x + 3 - x = 0", "x = - 3"], "exprs": ["2 x + 3 - x = 0", "x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x"}, {"id": "2 x + 3 - x = 0"}, {"id": "3 - x"}, {"id": "$2 x$ 与 $3 - x$ 互为相反数"}, {"id": "x = - 3"}], "links": [{"rel": "被描述", "source": "2 x", "target": "2 x + 3 - x = 0"}, {"rel": "等式方程求解", "source": "2 x + 3 - x = 0", "target": "x = - 3"}, {"rel": "被描述", "source": "3 - x", "target": "2 x + 3 - x = 0"}, {"rel": "限制性描述", "source": "$2 x$ 与 $3 - x$ 互为相反数", "target": "2 x + 3 - x = 0"}]}}
|
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{"content": "Given $\\frac { 1 } { m } - \\frac { 1 } { n } = 6$, what is the value of $\\frac { mn } { m - n }$?", "answer": "x = - 5", "steps": "$\\because \\frac { 1 } { m } - \\frac { 1 } { n } = 6$, $\\therefore \\frac { n - m } { mn } = 6$. $\\therefore n - m = 6 mn$. That is, $m - n = - 6 mn$. $\\therefore \\frac { mn } { m - n } = \\frac { mn } { - 6 mn } = - \\frac { 1 } { 6 }$.", "expr_cands": ["\\frac { 7 } { x - 2 } = \\frac { 5 } { x }", "x", "x ( x - 2 )", "7 x = 5 ( x - 2 )", "x = - 5", "7 x = 5 x - 10", "2 x = - 10"], "exprs": ["x = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 7 } { x - 2 } = \\frac { 5 } { x }"}, {"id": "x = - 5"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 7 } { x - 2 } = \\frac { 5 } { x }", "target": "x = - 5"}]}}
|
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{"content": "If $\\sqrt { 1 - a } = b$, then the possible values of $a$ are ____?", "answer": "- 3", "steps": "From the given condition, we have $1 - a \\geq 0$, which implies that $a \\leq 1$.", "expr_cands": ["\\frac { x - 1 } { x + 2 } = \\frac { k } { x + 2 }", "k", "x", "x - 1 = k", "x + 2 = 0", "x = - 2", "k = - 3"], "exprs": ["x - 1 = k", "x + 2 = 0", "x = - 2", "k = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 1 } { x + 2 } = \\frac { k } { x + 2 }"}, {"id": "x - 1 = k"}, {"id": "x + 2 = 0"}, {"id": "在去分母解分式方程 $\\frac { x - 1 } { x + 2 } = \\frac { k } { x + 2 }$ 时产生增根"}, {"id": "分式有增根，则分母为0"}, {"id": "x = - 2"}, {"id": "k = - 3"}], "links": [{"rel": "同乘除", "source": "\\frac { x - 1 } { x + 2 } = \\frac { k } { x + 2 }", "target": "x - 1 = k"}, {"rel": "被描述", "source": "\\frac { x - 1 } { x + 2 } = \\frac { k } { x + 2 }", "target": "x + 2 = 0"}, {"rel": "被代入", "source": "x - 1 = k", "target": "k = - 3"}, {"rel": "等式方程求解", "source": "x + 2 = 0", "target": "x = - 2"}, {"rel": "限制性描述", "source": "在去分母解分式方程 $\\frac { x - 1 } { x + 2 } = \\frac { k } { x + 2 }$ 时产生增根", "target": "x + 2 = 0"}, {"rel": "属性描述", "source": "分式有增根，则分母为0", "target": "x + 2 = 0"}, {"rel": "代入", "source": "x = - 2", "target": "k = - 3"}]}}
|
||||
{"content": "If $2 x$ and $3 - x$ are opposite in sign, then $x$ is equal to ____?", "answer": "- 8", "steps": "According to the problem, we have $2 x + 3 - x = 0$, which gives us $x = - 3$.", "expr_cands": ["| a - 2 |", "a", "| b + 3 |", "b", "2 b - a", "| a - 2 | + | b + 3 | = 0", "a - 2 = 0", "a = 2", "b + 3 = 0", "b = - 3", "- 8"], "exprs": ["| a - 2 | + | b + 3 | = 0", "a - 2 = 0", "b + 3 = 0", "a = 2", "b = - 3", "- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 2 |"}, {"id": "| a - 2 | + | b + 3 | = 0"}, {"id": "| b + 3 |"}, {"id": "$| a - 2 |$ 的值与 $| b + 3 |$ 的值互为相反数"}, {"id": "a - 2 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "根据题意得 : $| a - 2 | + | b + 3 | = 0$"}, {"id": "b + 3 = 0"}, {"id": "a = 2"}, {"id": "b = - 3"}, {"id": "2 b - a"}, {"id": "- 8"}], "links": [{"rel": "被描述", "source": "| a - 2 |", "target": "| a - 2 | + | b + 3 | = 0"}, {"rel": "被描述", "source": "| a - 2 | + | b + 3 | = 0", "target": "a - 2 = 0"}, {"rel": "被描述", "source": "| a - 2 | + | b + 3 | = 0", "target": "b + 3 = 0"}, {"rel": "被描述", "source": "| b + 3 |", "target": "| a - 2 | + | b + 3 | = 0"}, {"rel": "限制性描述", "source": "$| a - 2 |$ 的值与 $| b + 3 |$ 的值互为相反数", "target": "| a - 2 | + | b + 3 | = 0"}, {"rel": "等式方程求解", "source": "a - 2 = 0", "target": "a = 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a - 2 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b + 3 = 0"}, {"rel": "限制性描述", "source": "根据题意得 : $| a - 2 | + | b + 3 | = 0$", "target": "a - 2 = 0"}, {"rel": "限制性描述", "source": "根据题意得 : $| a - 2 | + | b + 3 | = 0$", "target": "b + 3 = 0"}, {"rel": "等式方程求解", "source": "b + 3 = 0", "target": "b = - 3"}, {"rel": "代入", "source": "a = 2", "target": "- 8"}, {"rel": "代入", "source": "b = - 3", "target": "- 8"}, {"rel": "被代入", "source": "2 b - a", "target": "- 8"}]}}
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{"content": "The solution to the equation $\\frac { 7 } { x - 2 } = \\frac { 5 } { x }$ is ____ ?", "answer": "- 3", "steps": "Multiplying both sides of the equation by $x ( x - 2 )$, we get $7 x = 5 ( x - 2 )$. Simplifying, we have $7 x = 5 x - 10$. Moving terms and combining like terms, we get $2 x = - 10$. Solving for $x$, we get $x = - 5$. Checking, we see that $x = - 5$ is indeed a solution to the equation.", "expr_cands": ["- 2 ^ { 3 } a ^ { 2 } b ^ { 3 }", "a", "b", "- 2 ^ { 3 }", "- 8", "2 + 3", "5", "- 8 + 5", "- 3"], "exprs": ["- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 ^ { 3 } a ^ { 2 } b ^ { 3 }"}, {"id": "- 3"}, {"id": "$- 2 ^ { 3 } a ^ { 2 } b ^ { 3 }$ 的系数和次数的和"}, {"id": "系数和次数的和等于 $- 8 + 5 = - 3$"}], "links": [{"rel": "被描述", "source": "- 2 ^ { 3 } a ^ { 2 } b ^ { 3 }", "target": "- 3"}, {"rel": "限制性描述", "source": "$- 2 ^ { 3 } a ^ { 2 } b ^ { 3 }$ 的系数和次数的和", "target": "- 3"}, {"rel": "限制性描述", "source": "系数和次数的和等于 $- 8 + 5 = - 3$", "target": "- 3"}]}}
|
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{"content": "If increasing the root occurs when solving the fractional equation $\\frac { x - 1 } { x + 2 } = \\frac { k } { x + 2 }$ by eliminating the denominator, then $k$ = ____ ?", "answer": "- 2", "steps": "To eliminate the denominator in the fractional equation, we get $x - 1 = k$. Since the fractional equation has a proper root, we have $x + 2 = 0$, which means $x = - 2$. Substituting $x = - 2$ into the polynomial equation, we get $k = - 2 - 1 = - 3$.", "expr_cands": ["x ^ { 5 } + 32 = 0", "x", "x = - 2", "x ^ { 5 } = - 32"], "exprs": ["x = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 5 } + 32 = 0"}, {"id": "x = - 2"}], "links": [{"rel": "等式方程求解", "source": "x ^ { 5 } + 32 = 0", "target": "x = - 2"}]}}
|
||||
{"content": "If the value of $| a - 2 |$ is the opposite of the value of $| b + 3 |$, then $2 b - a$ = ____ ?", "answer": "x = 5", "steps": "According to the problem, we have $| a - 2 | + | b + 3 | = 0$. Therefore, $a - 2 = 0$ and $b + 3 = 0$. Solving for $a$ and $b$, we get $a = 2$ and $b = - 3$. Thus, $2 b - a = 2 * ( - 3 ) - 2 = - 8$.", "expr_cands": ["1 + \\frac { 3 } { 3 - x } = \\frac { 4 - x } { x - 3 }", "x", "x - 3 - 3 = 4 - x", "x = 5"], "exprs": ["x = 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "1 + \\frac { 3 } { 3 - x } = \\frac { 4 - x } { x - 3 }"}, {"id": "x = 5"}], "links": [{"rel": "等式方程求解", "source": "1 + \\frac { 3 } { 3 - x } = \\frac { 4 - x } { x - 3 }", "target": "x = 5"}]}}
|
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{"content": "The coefficient and degree of $- 2 ^ { 3 } a ^ { 2 } b ^ { 3 }$ are ____ ?", "answer": "- 5", "steps": "The coefficient of $- 2 ^ { 3 } a ^ { 2 } b ^ { 3 }$ is $- 2 ^ { 3 } = - 8$, and the degree is $2 + 3 = 5$. Therefore, the sum of the coefficient and the degree is $- 8 + 5 = - 3$.", "expr_cands": ["10 { x } ^ { 2 } - 4 x + 6", "x", "2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7", "m", "10 { x } ^ { 2 } - 4 x + 6 + 2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7", "2 { x } ^ { 3 } + ( 2 m + 10 ) { x } ^ { 2 } - 10 x + 13", "2 m + 10 = 0", "m = - 5"], "exprs": ["10 { x } ^ { 2 } - 4 x + 6 + 2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7", "2 m + 10 = 0", "m = - 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "10 { x } ^ { 2 } - 4 x + 6"}, {"id": "10 { x } ^ { 2 } - 4 x + 6 + 2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7"}, {"id": "2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7"}, {"id": "多项式 $10 { x } ^ { 2 } - 4 x + 6$ 与 $2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7$ 相加后不含二次项"}, {"id": "2 m + 10 = 0"}, {"id": "m = - 5"}], "links": [{"rel": "被描述", "source": "10 { x } ^ { 2 } - 4 x + 6", "target": "10 { x } ^ { 2 } - 4 x + 6 + 2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7"}, {"rel": "被描述", "source": "10 { x } ^ { 2 } - 4 x + 6 + 2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7", "target": "2 m + 10 = 0"}, {"rel": "被描述", "source": "2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7", "target": "10 { x } ^ { 2 } - 4 x + 6 + 2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7"}, {"rel": "限制性描述", "source": "多项式 $10 { x } ^ { 2 } - 4 x + 6$ 与 $2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7$ 相加后不含二次项", "target": "10 { x } ^ { 2 } - 4 x + 6 + 2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7"}, {"rel": "限制性描述", "source": "多项式 $10 { x } ^ { 2 } - 4 x + 6$ 与 $2 { x } ^ { 3 } + 2 m { x } ^ { 2 } - 6 x + 7$ 相加后不含二次项", "target": "2 m + 10 = 0"}, {"rel": "等式方程求解", "source": "2 m + 10 = 0", "target": "m = - 5"}]}}
|
||||
{"content": "The real solutions of the equation $x ^ 5 + 32 = 0$ are _____.", "answer": "- 11", "steps": "Because $x ^ { 5 } + 32 = 0$, therefore $x ^ { 5 } = - 32$, and solving for $x$ gives $x = - 2$.", "expr_cands": ["x = - 2018", "x", "ax ^ { 3 } - bx - 3", "b", "a", "5", "x = 2018", "ax ^ { 3 } - bx - 3 = 5", "- 8217949832 a + 2018 b - 3 = 5", "- 2018 ^ { 3 } a + 2018 b = 8", "ax ^ { 3 } - bx - 3 = - 11", "- 11"], "exprs": ["ax ^ { 3 } - bx - 3 = 5", "- 2018 ^ { 3 } a + 2018 b = 8", "- 11"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = - 2018"}, {"id": "ax ^ { 3 } - bx - 3 = 5"}, {"id": "ax ^ { 3 } - bx - 3"}, {"id": "5"}, {"id": "当 $x = - 2018$ 时"}, {"id": "式子 $ax ^ { 3 } - bx - 3$ 的值是 $5$"}, {"id": "- 2018 ^ { 3 } a + 2018 b = 8"}, {"id": "- 11"}, {"id": "x = 2018"}, {"id": "当 $x = 2018$ 时"}, {"id": "式子 $ax ^ { 3 } - bx - 3$ 的值"}], "links": [{"rel": "被描述", "source": "x = - 2018", "target": "ax ^ { 3 } - bx - 3 = 5"}, {"rel": "移项", "source": "ax ^ { 3 } - bx - 3 = 5", "target": "- 2018 ^ { 3 } a + 2018 b = 8"}, {"rel": "被描述", "source": "ax ^ { 3 } - bx - 3", "target": "ax ^ { 3 } - bx - 3 = 5"}, {"rel": "被描述", "source": "ax ^ { 3 } - bx - 3", "target": "- 11"}, {"rel": "被描述", "source": "5", "target": "ax ^ { 3 } - bx - 3 = 5"}, {"rel": "限制性描述", "source": "当 $x = - 2018$ 时", "target": "ax ^ { 3 } - bx - 3 = 5"}, {"rel": "限制性描述", "source": "式子 $ax ^ { 3 } - bx - 3$ 的值是 $5$", "target": "ax ^ { 3 } - bx - 3 = 5"}, {"rel": "被描述", "source": "- 2018 ^ { 3 } a + 2018 b = 8", "target": "- 11"}, {"rel": "被描述", "source": "x = 2018", "target": "- 11"}, {"rel": "限制性描述", "source": "当 $x = 2018$ 时", "target": "- 11"}, {"rel": "限制性描述", "source": "式子 $ax ^ { 3 } - bx - 3$ 的值", "target": "- 11"}]}}
|
||||
{"content": "Fractional equation: The solution to $1 + \\frac { 3 } { 3 - x } = \\frac { 4 - x } { x - 3 }$ is ____?", "answer": "10", "steps": "Going to the denominator, we get: $x - 3 - 3 = 4 - x$. Solving for $x$, we get $x = 5$. Upon checking, we find that $x = 5$ is a solution to the fractional equation.", "expr_cands": ["x", "3 x + 4", "- 4 x + 6", "3 x + 4 - 4 x + 6 = 0", "x = 10"], "exprs": ["3 x + 4 - 4 x + 6 = 0", "x = 10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + 4"}, {"id": "3 x + 4 - 4 x + 6 = 0"}, {"id": "- 4 x + 6"}, {"id": "$3 x + 4$ 与 $- 4 x + 6$ 互为相反数"}, {"id": "x = 10"}], "links": [{"rel": "被描述", "source": "3 x + 4", "target": "3 x + 4 - 4 x + 6 = 0"}, {"rel": "等式方程求解", "source": "3 x + 4 - 4 x + 6 = 0", "target": "x = 10"}, {"rel": "被描述", "source": "- 4 x + 6", "target": "3 x + 4 - 4 x + 6 = 0"}, {"rel": "限制性描述", "source": "$3 x + 4$ 与 $- 4 x + 6$ 互为相反数", "target": "3 x + 4 - 4 x + 6 = 0"}]}}
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||||
{"content": "If the polynomial $10 { x } ^ 2 - 4 x + 6$ added to $2 { x } ^ 3 + 2 mx ^ 2 - 6 x + 7$ does not contain a quadratic term, then the value of the constant $m$ is ____?", "answer": "- 1", "steps": "$10 { x } ^ 2 - 4 x + 6 + 2 { x } ^ 3 + 2 mx ^ 2 - 6 x + 7 = 2 { x } ^ 3 + ( 2 m + 10 ) { x } ^ 2 - 10 x + 13$, because after adding the terms, there is no quadratic term, therefore $2 m + 10 = 0$, which solves for $m = - 5$.", "expr_cands": ["\\sqrt { 1 - a }", "a", "\\sqrt { 4 + 2 a }", "1 - a = 4 + 2 a", "a = - 1"], "exprs": ["1 - a = 4 + 2 a", "a = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 1 - a }"}, {"id": "1 - a = 4 + 2 a"}, {"id": "\\sqrt { 4 + 2 a }"}, {"id": "最简二次根式 $\\sqrt { 1 - a }$ 与 $\\sqrt { 4 + 2 a }$ 可以合并"}, {"id": "即为同类二次根式"}, {"id": "a = - 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { 1 - a }", "target": "1 - a = 4 + 2 a"}, {"rel": "等式方程求解", "source": "1 - a = 4 + 2 a", "target": "a = - 1"}, {"rel": "被描述", "source": "\\sqrt { 4 + 2 a }", "target": "1 - a = 4 + 2 a"}, {"rel": "限制性描述", "source": "最简二次根式 $\\sqrt { 1 - a }$ 与 $\\sqrt { 4 + 2 a }$ 可以合并", "target": "1 - a = 4 + 2 a"}, {"rel": "限制性描述", "source": "即为同类二次根式", "target": "1 - a = 4 + 2 a"}]}}
|
||||
{"content": "If the value of the expression $ax ^ 3 - bx - 3$ is $5$ when $x = - 2018$, then the value of the expression when $x = 2018$ is ____?", "answer": "4", "steps": "From $x = - 2018$, we know that $- 2018 ^ 3 a + 2018 b = 8$ because $ax ^ 3 - bx - 3 = 5$. When $x = 2018$, we have $ax ^ 3 - bx - 3 = 2018 ^ 3 a - 2018 b - 3 = - ( - 2018 ^ 3 a + 2018 b ) - 3 = - 8 - 3 = - 11$.", "expr_cands": ["x", "x ^ { 2 } - 3 x - a = 0", "a", "- 1", "x = - 1", "4 - a = 0", "1 + 3 - a = 0", "a = 4"], "exprs": ["x = - 1", "1 + 3 - a = 0", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 1"}, {"id": "x = - 1"}, {"id": "x"}, {"id": "x ^ { 2 } - 3 x - a = 0"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 3 x - a = 0$ 有一个实数根为 $- 1$"}, {"id": "1 + 3 - a = 0"}, {"id": "a = 4"}], "links": [{"rel": "被描述", "source": "- 1", "target": "x = - 1"}, {"rel": "代入", "source": "x = - 1", "target": "1 + 3 - a = 0"}, {"rel": "被描述", "source": "x", "target": "x = - 1"}, {"rel": "被描述", "source": "x ^ { 2 } - 3 x - a = 0", "target": "x = - 1"}, {"rel": "被代入", "source": "x ^ { 2 } - 3 x - a = 0", "target": "1 + 3 - a = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } - 3 x - a = 0$ 有一个实数根为 $- 1$", "target": "x = - 1"}, {"rel": "等式方程求解", "source": "1 + 3 - a = 0", "target": "a = 4"}]}}
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{"content": "When $x$ = ____ ?, $3 x + 4$ and $- 4 x + 6$ are opposite numbers.", "answer": "x \\ge 2", "steps": "According to the problem, we have $3 x + 4 - 4 x + 6 = 0$, which gives us $x = 10$ as the solution.", "expr_cands": ["y = \\sqrt { x - 2 }", "x", "y", "x - 2 \\ge 0", "2 \\le x", "x \\ge 2"], "exprs": ["x - 2 \\ge 0", "x \\ge 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = \\sqrt { x - 2 }"}, {"id": "x - 2 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x \\ge 2"}], "links": [{"rel": "被描述", "source": "y = \\sqrt { x - 2 }", "target": "x - 2 \\ge 0"}, {"rel": "不等式方程求解", "source": "x - 2 \\ge 0", "target": "x \\ge 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 2 \\ge 0"}]}}
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{"content": "If the simplest quadratic radical $\\sqrt { 1 - a }$ and $\\sqrt { 4 + 2 a }$ can be combined, then $a$ is ____?", "answer": "\\frac { 1 } { 2 }", "steps": "$\\because$ The simplest quadratic radicals $\\sqrt { 1 - a }$ and $\\sqrt { 4 + 2 a }$ can be combined, that is, they are of the same type of quadratic radicals. $\\therefore$ $1 - a = 4 + 2 a$, solving for $a = - 1$.", "expr_cands": ["y = kx - 3", "y", "x", "k", "y = \\frac { 1 } { 2 } x + 4", "k x - 3 = \\frac { 1 } { 2 } x + 4", "k = \\frac { 1 } { 2 }"], "exprs": ["k = \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = kx - 3"}, {"id": "k = \\frac { 1 } { 2 }"}, {"id": "y = \\frac { 1 } { 2 } x + 4"}, {"id": "直线 $y = kx - 3$ 与直线 $y = \\frac { 1 } { 2 } x + 4$ 平行"}], "links": [{"rel": "被描述", "source": "y = kx - 3", "target": "k = \\frac { 1 } { 2 }"}, {"rel": "被描述", "source": "y = \\frac { 1 } { 2 } x + 4", "target": "k = \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "直线 $y = kx - 3$ 与直线 $y = \\frac { 1 } { 2 } x + 4$ 平行", "target": "k = \\frac { 1 } { 2 }"}]}}
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{"content": "Regarding the quadratic equation in one variable $x$, $x ^ 2 - 3 x - a = 0$, there is a real root of $- 1$. What is the value of $a$?", "answer": "- 2", "steps": "$\\because x = - 1$ is a root of the quadratic equation $x ^ 2 - 3 x - a = 0$, $\\therefore$ $1 + 3 - a = 0$, $\\therefore$ $a = 4$.", "expr_cands": ["x", "\\frac { x - 3 } { x - 1 } - \\frac { m } { x - 1 } = 8", "m", "x - 3 - m = 8 x - 8", "x = \\frac { 5 - m } { 7 }", "x - 1 = 0", "x = 1", "\\frac { 5 - m } { 7 } = 1", "m = - 2"], "exprs": ["x = \\frac { 5 - m } { 7 }", "x - 1 = 0", "x = 1", "\\frac { 5 - m } { 7 } = 1", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 3 } { x - 1 } - \\frac { m } { x - 1 } = 8"}, {"id": "x = \\frac { 5 - m } { 7 }"}, {"id": "x - 1 = 0"}, {"id": "解关于 $x$ 的方程 $\\frac { x - 3 } { x - 1 } - \\frac { m } { x - 1 } = 8$ 无解"}, {"id": "分式方程无解，则分母为0"}, {"id": "x = 1"}, {"id": "\\frac { 5 - m } { 7 } = 1"}, {"id": "m = - 2"}], "links": [{"rel": "等式方程部分求解", "source": "\\frac { x - 3 } { x - 1 } - \\frac { m } { x - 1 } = 8", "target": "x = \\frac { 5 - m } { 7 }"}, {"rel": "被描述", "source": "\\frac { x - 3 } { x - 1 } - \\frac { m } { x - 1 } = 8", "target": "x - 1 = 0"}, {"rel": "被代入", "source": "x = \\frac { 5 - m } { 7 }", "target": "\\frac { 5 - m } { 7 } = 1"}, {"rel": "等式方程求解", "source": "x - 1 = 0", "target": "x = 1"}, {"rel": "限制性描述", "source": "解关于 $x$ 的方程 $\\frac { x - 3 } { x - 1 } - \\frac { m } { x - 1 } = 8$ 无解", "target": "x - 1 = 0"}, {"rel": "属性描述", "source": "分式方程无解，则分母为0", "target": "x - 1 = 0"}, {"rel": "代入", "source": "x = 1", "target": "\\frac { 5 - m } { 7 } = 1"}, {"rel": "等式方程求解", "source": "\\frac { 5 - m } { 7 } = 1", "target": "m = - 2"}]}}
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{"content": "Given the function $y = \\sqrt { x - 2 }$, what is the range of possible values for $x$?", "answer": "9", "steps": "From the given condition, we have $x - 2 \\ge 0$. Solving for $x$, we get $x \\ge 2$.", "expr_cands": ["2 a - 1", "a", "- a + 2", "2 a - 1 - a + 2 = 0", "a = - 1", "- 3", "( - 3 ) ^ { 2 }", "9"], "exprs": ["2 a - 1 - a + 2 = 0", "a = - 1", "- 3", "9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a - 1"}, {"id": "2 a - 1 - a + 2 = 0"}, {"id": "- a + 2"}, {"id": "一个正数的两个平方根分别是 $2 a - 1$ 与 $- a + 2$"}, {"id": "平方根互为相反数"}, {"id": "a = - 1"}, {"id": "- 3"}, {"id": "9"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "2 a - 1", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "被代入", "source": "2 a - 1", "target": "- 3"}, {"rel": "等式方程求解", "source": "2 a - 1 - a + 2 = 0", "target": "a = - 1"}, {"rel": "被描述", "source": "- a + 2", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "限制性描述", "source": "一个正数的两个平方根分别是 $2 a - 1$ 与 $- a + 2$", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "2 a - 1 - a + 2 = 0"}, {"rel": "代入", "source": "a = - 1", "target": "- 3"}, {"rel": "被描述", "source": "- 3", "target": "9"}, {"rel": "限制性描述", "source": "平方", "target": "9"}]}}
|
||||
{"content": "Given that the line $y = kx - 3$ is parallel to the line $y = \\frac { 1 } { 2 } x + 4$, what is the value of $k$?", "answer": "y = - x - 10", "steps": "$\\because$ The line $y = kx - 3$ is parallel to the line $y = \\frac { 1 } { 2 } x + 4$, $\\therefore$ $k = \\frac { 1 } { 2 }$.", "expr_cands": ["y = - x + b", "b", "y", "x", "x = - 8", "y = - 2", "- ( - 8 ) + b = - 2", "b = - 10", "y = - x - 10"], "exprs": ["- ( - 8 ) + b = - 2", "b = - 10", "y = - x - 10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - x + b"}, {"id": "- ( - 8 ) + b = - 2"}, {"id": "x = - 8"}, {"id": "y = - 2"}, {"id": "b = - 10"}, {"id": "y = - x - 10"}], "links": [{"rel": "被代入", "source": "y = - x + b", "target": "- ( - 8 ) + b = - 2"}, {"rel": "被代入", "source": "y = - x + b", "target": "y = - x - 10"}, {"rel": "等式方程求解", "source": "- ( - 8 ) + b = - 2", "target": "b = - 10"}, {"rel": "代入", "source": "x = - 8", "target": "- ( - 8 ) + b = - 2"}, {"rel": "代入", "source": "y = - 2", "target": "- ( - 8 ) + b = - 2"}, {"rel": "代入", "source": "b = - 10", "target": "y = - x - 10"}]}}
|
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{"content": "Solve the equation $\\frac { x - 3 } { x - 1 } - \\frac { m } { x - 1 } = 8$ has no solution, then the value of $m$ is ____?", "answer": "- 2", "steps": "Equation after eliminating the denominator: $x - 3 - m = 8 x - 8$. Solving for $x$, we get $x = \\frac { 5 - m } { 7 }$. When the denominator $x - 1 = 0$, i.e. $x = 1$, the equation has no solution. Therefore, when $\\frac { 5 - m } { 7 } = 1$, the equation has no solution, which means $m = - 2$.", "expr_cands": ["a : b = 3 : 4", "a", "b", "a + b = 14", "a - b", "3 b = 4 a", "b = \\frac { 4 } { 3 } a", "\\frac { 7 a } { 3 } = 14", "a + \\frac { 4 } { 3 } a = 14", "a = 6", "b = 8", "- 2"], "exprs": ["b = \\frac { 4 } { 3 } a", "a + \\frac { 4 } { 3 } a = 14", "a = 6", "b = 8", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a : b = 3 : 4"}, {"id": "b = \\frac { 4 } { 3 } a"}, {"id": "a + b = 14"}, {"id": "a + \\frac { 4 } { 3 } a = 14"}, {"id": "a = 6"}, {"id": "b = 8"}, {"id": "a - b"}, {"id": "- 2"}], "links": [{"rel": "同乘除", "source": "a : b = 3 : 4", "target": "b = \\frac { 4 } { 3 } a"}, {"rel": "代入", "source": "b = \\frac { 4 } { 3 } a", "target": "a + \\frac { 4 } { 3 } a = 14"}, {"rel": "被代入", "source": "b = \\frac { 4 } { 3 } a", "target": "b = 8"}, {"rel": "被代入", "source": "a + b = 14", "target": "a + \\frac { 4 } { 3 } a = 14"}, {"rel": "等式方程求解", "source": "a + \\frac { 4 } { 3 } a = 14", "target": "a = 6"}, {"rel": "代入", "source": "a = 6", "target": "b = 8"}, {"rel": "代入", "source": "a = 6", "target": "- 2"}, {"rel": "代入", "source": "b = 8", "target": "- 2"}, {"rel": "被代入", "source": "a - b", "target": "- 2"}]}}
|
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{"content": "One positive number has two square roots, which are $2 a - 1$ and $- a + 2$. The number is [ ].", "answer": "2", "steps": "$\\because$ The square root of a positive number is $2 a - 1$ and $- a + 2$, $\\therefore$ $2 a - 1 - a + 2 = 0$. Solving for $a$, we get $a = - 1$. $\\therefore$ $2 a - 1 = - 3$. $\\because$ $( - 3 ) ^ 2 = 9$, $\\therefore$ this positive number is $9$.", "expr_cands": ["mx ^ { 2 m - 3 } + 4 > - 11", "m", "x", "2 m - 3 = 1", "m = 2", "m \\neq 0"], "exprs": ["2 m - 3 = 1", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "mx ^ { 2 m - 3 } + 4 > - 11"}, {"id": "2 m - 3 = 1"}, {"id": "$mx ^ { 2 m - 3 } + 4 > - 11$ 是关于 $x$ 的一元一次不等式"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "mx ^ { 2 m - 3 } + 4 > - 11", "target": "2 m - 3 = 1"}, {"rel": "等式方程求解", "source": "2 m - 3 = 1", "target": "m = 2"}, {"rel": "限制性描述", "source": "$mx ^ { 2 m - 3 } + 4 > - 11$ 是关于 $x$ 的一元一次不等式", "target": "2 m - 3 = 1"}]}}
|
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{"content": "Given a linear function $y = - x + b$, when $x = - 8$, $y = - 2$, the expression of the linear function is ____?", "answer": "x > - 2", "steps": "$\\because$ In a linear function $y = - x + b$, when $x = - 8$, $y = - 2$. $\\therefore$ $- ( - 8 ) + b = - 2$. $\\therefore$ $b = - 10$. $\\therefore$ the analytical expression of the linear function is $y = - x - 10$.", "expr_cands": ["y = - { x } ^ { 2 } - 4 x + 5", "y", "x", "y = - x ^ { 2 } - 4 x - 5", "- ( x + 2 ) ^ { 2 } - 1", "x = - 2", "x > - 2"], "exprs": ["x = - 2", "x > - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = - { x } ^ { 2 } - 4 x + 5"}, {"id": "x = - 2"}, {"id": "二次函数y=a^x+bx+c 的对称轴为 - \\frac {b} {2a}"}, {"id": "x > - 2"}, {"id": "二次函数 $y = - { x } ^ { 2 } - 4 x + 5$"}, {"id": "当 $x$ 满足时"}, {"id": "$y$ 随 $x$ 的增大而减小"}], "links": [{"rel": "被描述", "source": "y = - { x } ^ { 2 } - 4 x + 5", "target": "x = - 2"}, {"rel": "被描述", "source": "y = - { x } ^ { 2 } - 4 x + 5", "target": "x > - 2"}, {"rel": "被描述", "source": "x = - 2", "target": "x > - 2"}, {"rel": "属性描述", "source": "二次函数y=a^x+bx+c 的对称轴为 - \\frac {b} {2a}", "target": "x = - 2"}, {"rel": "限制性描述", "source": "二次函数 $y = - { x } ^ { 2 } - 4 x + 5$", "target": "x > - 2"}, {"rel": "限制性描述", "source": "当 $x$ 满足时", "target": "x > - 2"}, {"rel": "限制性描述", "source": "$y$ 随 $x$ 的增大而减小", "target": "x > - 2"}]}}
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{"content": "If $a : b = 3 : 4$ and $a + b = 14$, then the value of $a - b$ is ____?", "answer": "20", "steps": "From $a : b = 3 : 4$, we know that $3 b = 4 a$, so $b = \\frac { 4 } { 3 } a$. Therefore, from $a + b = 14$, we have $a + \\frac { 4 } { 3 } a = 14$. Solving for $a$, we get $a = 6$, so $b = 8$. Therefore, $a - b = 6 - 8 = - 2$.", "expr_cands": ["a ^ { n } = 2", "a", "n", "a ^ { m } = 5", "m", "a ^ { m + 2 n }", "20"], "exprs": ["20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { m + 2 n }"}, {"id": "20"}, {"id": "a ^ { n } = 2"}, {"id": "a ^ { m } = 5"}], "links": [{"rel": "被代入", "source": "a ^ { m + 2 n }", "target": "20"}, {"rel": "代入", "source": "a ^ { n } = 2", "target": "20"}, {"rel": "代入", "source": "a ^ { m } = 5", "target": "20"}]}}
|
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{"content": "If $mx ^ { 2 m - 3 } + 4 > - 11$ is a one-variable linear inequality, then $m$ = ____ ?", "answer": "4", "steps": "According to the problem, $2 m - 3 = 1$, and $m$ cannot be zero. Solving for $m$, we get $m = 2$.", "expr_cands": ["3 ab ^ { 3 }", "a", "b", "1", "3", "1 + 3", "4"], "exprs": ["4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 ab ^ { 3 }"}, {"id": "4"}, {"id": "单项式 $3 ab ^ { 3 }$ 的次数"}], "links": [{"rel": "被描述", "source": "3 ab ^ { 3 }", "target": "4"}, {"rel": "限制性描述", "source": "单项式 $3 ab ^ { 3 }$ 的次数", "target": "4"}]}}
|
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{"content": "Given a quadratic function $y = - { x } ^ { 2 } - 4 x + 5$, when $x$ satisfies ____ ?, $y$ decreases as $x$ increases.", "answer": "- \\frac { 1 } { 3 }", "steps": "The quadratic function $y = y = - x ^ { 2 } - 4 x - 5$ is transformed into $y = - ( x + 2 ) ^ { 2 } - 1$, with the axis of symmetry being $x = - 2$. Therefore, when $x > - 2$, $y$ decreases as $x$ increases.", "expr_cands": ["( x + m ) ( x + \\frac { 1 } { 3 } )", "m", "x", "{ x } ^ { 2 } + ( \\frac { 1 } { 3 } + m ) x + \\frac { 1 } { 3 } m", "\\frac { 1 } { 3 } + m = 0", "m = - \\frac { 1 } { 3 }"], "exprs": ["{ x } ^ { 2 } + ( \\frac { 1 } { 3 } + m ) x + \\frac { 1 } { 3 } m", "\\frac { 1 } { 3 } + m = 0", "m = - \\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x + m ) ( x + \\frac { 1 } { 3 } )"}, {"id": "{ x } ^ { 2 } + ( \\frac { 1 } { 3 } + m ) x + \\frac { 1 } { 3 } m"}, {"id": "x"}, {"id": "\\frac { 1 } { 3 } + m = 0"}, {"id": "$( x + m ) ( x + \\frac { 1 } { 3 } )$ 不含 $x$ 的一次项"}, {"id": "m = - \\frac { 1 } { 3 }"}], "links": [{"rel": "提取因式", "source": "( x + m ) ( x + \\frac { 1 } { 3 } )", "target": "{ x } ^ { 2 } + ( \\frac { 1 } { 3 } + m ) x + \\frac { 1 } { 3 } m"}, {"rel": "被描述", "source": "{ x } ^ { 2 } + ( \\frac { 1 } { 3 } + m ) x + \\frac { 1 } { 3 } m", "target": "\\frac { 1 } { 3 } + m = 0"}, {"rel": "提取因式参考", "source": "x", "target": "{ x } ^ { 2 } + ( \\frac { 1 } { 3 } + m ) x + \\frac { 1 } { 3 } m"}, {"rel": "等式方程求解", "source": "\\frac { 1 } { 3 } + m = 0", "target": "m = - \\frac { 1 } { 3 }"}, {"rel": "限制性描述", "source": "$( x + m ) ( x + \\frac { 1 } { 3 } )$ 不含 $x$ 的一次项", "target": "\\frac { 1 } { 3 } + m = 0"}]}}
|
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{"content": "Calculate: If $a ^ n = 2$ and $a ^ m = 5$, then $a ^ { m + 2 n }$ = ____?", "answer": "9", "steps": "Since $a ^ n = 2$ and $a ^ m = 5$, therefore $a ^ { m + 2 n } = a ^ m \\times a ^ { 2 n } = a ^ m \\times ( a ^ n ) ^ 2 = 5 \\times 2 ^ 2 = 5 \\times 4 = 20$.", "expr_cands": ["a + 2", "a", "2 a - 5", "m", "( a + 2 ) + ( 2 a - 5 ) = 0", "a = 1", "3", "m = 9"], "exprs": ["( a + 2 ) + ( 2 a - 5 ) = 0", "a = 1", "3", "m = 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a + 2"}, {"id": "( a + 2 ) + ( 2 a - 5 ) = 0"}, {"id": "2 a - 5"}, {"id": "$a + 2$ 与 $2 a - 5$ 都是 $m$ 的平方根"}, {"id": "平方根互为相反数"}, {"id": "a = 1"}, {"id": "3"}, {"id": "m = 9"}, {"id": "m"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "a + 2", "target": "( a + 2 ) + ( 2 a - 5 ) = 0"}, {"rel": "被代入", "source": "a + 2", "target": "3"}, {"rel": "等式方程求解", "source": "( a + 2 ) + ( 2 a - 5 ) = 0", "target": "a = 1"}, {"rel": "被描述", "source": "2 a - 5", "target": "( a + 2 ) + ( 2 a - 5 ) = 0"}, {"rel": "限制性描述", "source": "$a + 2$ 与 $2 a - 5$ 都是 $m$ 的平方根", "target": "( a + 2 ) + ( 2 a - 5 ) = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "( a + 2 ) + ( 2 a - 5 ) = 0"}, {"rel": "代入", "source": "a = 1", "target": "3"}, {"rel": "被描述", "source": "3", "target": "m = 9"}, {"rel": "被描述", "source": "m", "target": "m = 9"}, {"rel": "限制性描述", "source": "平方", "target": "m = 9"}]}}
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{"content": "The degree of the monomial $3 ab ^ 3$ is ____ ?", "answer": "- \\frac { 1 } { 2 }", "steps": "$\\because$ In the monomial $3 ab ^ 3$, the exponent of $a$ is $1$ and the exponent of $b$ is $3$, $\\therefore$ the degree of this monomial is $1 + 3 = 4$.", "expr_cands": ["( x - 2 ) ( 1 - kx ) - ( 2 x - 3 ) ( 2 x + 3 )", "k", "x", "- ( k + 4 ) x ^ { 2 } + ( 2 k + 1 ) x + 7", "2 k + 1 = 0", "k = - \\frac { 1 } { 2 }"], "exprs": ["- ( k + 4 ) x ^ { 2 } + ( 2 k + 1 ) x + 7", "2 k + 1 = 0", "k = - \\frac { 1 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( x - 2 ) ( 1 - kx ) - ( 2 x - 3 ) ( 2 x + 3 )"}, {"id": "- ( k + 4 ) x ^ { 2 } + ( 2 k + 1 ) x + 7"}, {"id": "x"}, {"id": "2 k + 1 = 0"}, {"id": "$( x - 2 ) ( 1 - kx ) - ( 2 x - 3 ) ( 2 x + 3 )$ 的结果中不含有 $x$ 的一次式"}, {"id": "k = - \\frac { 1 } { 2 }"}], "links": [{"rel": "提取因式", "source": "( x - 2 ) ( 1 - kx ) - ( 2 x - 3 ) ( 2 x + 3 )", "target": "- ( k + 4 ) x ^ { 2 } + ( 2 k + 1 ) x + 7"}, {"rel": "被描述", "source": "- ( k + 4 ) x ^ { 2 } + ( 2 k + 1 ) x + 7", "target": "2 k + 1 = 0"}, {"rel": "提取因式参考", "source": "x", "target": "- ( k + 4 ) x ^ { 2 } + ( 2 k + 1 ) x + 7"}, {"rel": "等式方程求解", "source": "2 k + 1 = 0", "target": "k = - \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "$( x - 2 ) ( 1 - kx ) - ( 2 x - 3 ) ( 2 x + 3 )$ 的结果中不含有 $x$ 的一次式", "target": "2 k + 1 = 0"}]}}
|
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{"content": "If $( x + m ) ( x + \\frac { 1 } { 3 } )$ does not contain the linear term of $x$, then $m$ = ____ ?", "answer": "\\frac { 1 } { 32 }", "steps": "$$(x+m)(x+\\frac{1}{3})={x}^2+(\\frac{1}{3}+m)x+\\frac{1}{3}m$$Because the result does not contain a linear term in $x$, we have $\\frac { 1 } { 3 } + m = 0$. Solving for $m$, we get $m = - \\frac { 1 } { 3 }$.", "expr_cands": ["a = 4", "a", "b = \\frac { 1 } { 8 }", "b", "a ^ { 2 } b ^ { 3 }", "\\frac { 1 } { 32 }"], "exprs": ["\\frac { 1 } { 32 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 2 } b ^ { 3 }"}, {"id": "\\frac { 1 } { 32 }"}, {"id": "a = 4"}, {"id": "b = \\frac { 1 } { 8 }"}], "links": [{"rel": "被代入", "source": "a ^ { 2 } b ^ { 3 }", "target": "\\frac { 1 } { 32 }"}, {"rel": "代入", "source": "a = 4", "target": "\\frac { 1 } { 32 }"}, {"rel": "代入", "source": "b = \\frac { 1 } { 8 }", "target": "\\frac { 1 } { 32 }"}]}}
|
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{"content": "Given that $a + 2$ and $2 a - 5$ are both square roots of $m$, what is the value of $m$?", "answer": "1", "steps": "Because $a + 2$ and $2 a - 5$ are both square roots of $m$, $a + 2$ and $2 a - 5$ are opposite in sign. That is, $( a + 2 ) + ( 2 a - 5 ) = 0$. Solving for $a$, we get $a = 1$. Therefore, $a + 2 = 3$, and $m = 9$.", "expr_cands": ["{ y } = { 2 x } - { a } + { 1 }", "a", "y", "x", "{ a }", "- a + 1 = 0", "a = 1"], "exprs": ["- a + 1 = 0", "a = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "{ y } = { 2 x } - { a } + { 1 }"}, {"id": "- a + 1 = 0"}, {"id": "函数 ${ y } = { 2 x } - { a } + { 1 }$ 是正比例函数"}, {"id": "a = 1"}], "links": [{"rel": "被描述", "source": "{ y } = { 2 x } - { a } + { 1 }", "target": "- a + 1 = 0"}, {"rel": "等式方程求解", "source": "- a + 1 = 0", "target": "a = 1"}, {"rel": "限制性描述", "source": "函数 ${ y } = { 2 x } - { a } + { 1 }$ 是正比例函数", "target": "- a + 1 = 0"}]}}
|
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{"content": "Given $( x - 2 ) ( 1 - kx ) - ( 2 x - 3 ) ( 2 x + 3 )$ does not contain a linear term in $x$, then $k$ = ____?", "answer": "- 2", "steps": "$\\because$ $( x - 2 ) ( 1 - kx ) - ( 2 x - 3 ) ( 2 x + 3 ) = x - kx ^ 2 - 2 + 2 kx - ( 4 x ^ 2 - 9 ) = x - kx ^ 2 - 2 + 2 kx - 4 x ^ 2 + 9 = - ( k + 4 ) x ^ 2 + ( 2 k + 1 ) x + 7$, $\\because$ there is no linear term in the result, $\\therefore$ $2 k + 1 = 0$, $\\therefore$ $k = - \\frac { 1 } { 2 }$.", "expr_cands": ["x", "x ^ { 2 } + ( k + 3 ) x + 2 = 0", "k", "- 1", "t", "- t = 2", "t = - 2"], "exprs": ["t", "- t = 2", "t = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设方程的另一根为 $t$"}, {"id": "t"}, {"id": "- 1"}, {"id": "- t = 2"}, {"id": "x ^ { 2 } + ( k + 3 ) x + 2 = 0"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } + ( k + 3 ) x + 2 = 0$ 的一个根是 $- 1$"}, {"id": "一元二次方程根与系数关系，两根之积"}, {"id": "t = - 2"}], "links": [{"rel": "假设描述", "source": "设方程的另一根为 $t$", "target": "t"}, {"rel": "限制性描述", "source": "设方程的另一根为 $t$", "target": "- t = 2"}, {"rel": "被描述", "source": "t", "target": "- t = 2"}, {"rel": "被描述", "source": "- 1", "target": "- t = 2"}, {"rel": "等式方程求解", "source": "- t = 2", "target": "t = - 2"}, {"rel": "被描述", "source": "x ^ { 2 } + ( k + 3 ) x + 2 = 0", "target": "- t = 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } + ( k + 3 ) x + 2 = 0$ 的一个根是 $- 1$", "target": "- t = 2"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之积", "target": "- t = 2"}]}}
|
||||
{"content": "If $a = 4$ and $b = \\frac { 1 } { 8 }$, then $a ^ 2 b ^ 3$ is equal to ____?", "answer": "4", "steps": "Since $a = 4$ and $b = \\frac { 1 } { 8 }$, therefore $a ^ 2 b ^ 3 = 4 ^ 2 * ( \\frac { 1 } { 8 }) ^ 3 = 16 * \\frac { 1 } { 512 } = \\frac { 1 } { 32 }$.", "expr_cands": ["y - 2", "y", "x", "x = 2", "y = 4", "y = 3", "y - 2 = \\frac { k } { x }", "k", "4 - 2 = \\frac { k } { 2 }", "k = 4", "y = \\frac { 4 } { x } + 2", "\\frac { 4 } { x } + 2 = 3", "x = 4"], "exprs": ["y - 2 = \\frac { k } { x }", "4 - 2 = \\frac { k } { 2 }", "k = 4", "y = \\frac { 4 } { x } + 2", "\\frac { 4 } { x } + 2 = 3", "x = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $y - 2 = \\frac { k } { x }$"}, {"id": "y - 2 = \\frac { k } { x }"}, {"id": "x = 2"}, {"id": "4 - 2 = \\frac { k } { 2 }"}, {"id": "y = 4"}, {"id": "k = 4"}, {"id": "y = \\frac { 4 } { x } + 2"}, {"id": "y = 3"}, {"id": "\\frac { 4 } { x } + 2 = 3"}, {"id": "x = 4"}], "links": [{"rel": "假设描述", "source": "设 $y - 2 = \\frac { k } { x }$", "target": "y - 2 = \\frac { k } { x }"}, {"rel": "被代入", "source": "y - 2 = \\frac { k } { x }", "target": "4 - 2 = \\frac { k } { 2 }"}, {"rel": "联立", "source": "y - 2 = \\frac { k } { x }", "target": "y = \\frac { 4 } { x } + 2"}, {"rel": "代入", "source": "x = 2", "target": "4 - 2 = \\frac { k } { 2 }"}, {"rel": "等式方程求解", "source": "4 - 2 = \\frac { k } { 2 }", "target": "k = 4"}, {"rel": "代入", "source": "y = 4", "target": "4 - 2 = \\frac { k } { 2 }"}, {"rel": "联立", "source": "k = 4", "target": "y = \\frac { 4 } { x } + 2"}, {"rel": "被代入", "source": "y = \\frac { 4 } { x } + 2", "target": "\\frac { 4 } { x } + 2 = 3"}, {"rel": "代入", "source": "y = 3", "target": "\\frac { 4 } { x } + 2 = 3"}, {"rel": "等式方程求解", "source": "\\frac { 4 } { x } + 2 = 3", "target": "x = 4"}]}}
|
||||
{"content": "If the function $y = 2 x - a + 1$ is a direct proportion function, then $a$ = ____?", "answer": "0", "steps": "From the given information, we have $- a + 1 = 0$, which implies $a = 1$.", "expr_cands": ["2 x - 3", "x", "- \\frac { 1 } { 3 }", "( 2 x - 3 ) * ( - \\frac { 1 } { 3 } ) = 1", "x = 0", "2 x - 3 = - 3"], "exprs": ["( 2 x - 3 ) * ( - \\frac { 1 } { 3 } ) = 1", "x = 0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x - 3"}, {"id": "( 2 x - 3 ) * ( - \\frac { 1 } { 3 } ) = 1"}, {"id": "- \\frac { 1 } { 3 }"}, {"id": "$2 x - 3$ 与 $- \\frac { 1 } { 3 }$ 互为倒数"}, {"id": "x = 0"}], "links": [{"rel": "被描述", "source": "2 x - 3", "target": "( 2 x - 3 ) * ( - \\frac { 1 } { 3 } ) = 1"}, {"rel": "等式方程求解", "source": "( 2 x - 3 ) * ( - \\frac { 1 } { 3 } ) = 1", "target": "x = 0"}, {"rel": "被描述", "source": "- \\frac { 1 } { 3 }", "target": "( 2 x - 3 ) * ( - \\frac { 1 } { 3 } ) = 1"}, {"rel": "限制性描述", "source": "$2 x - 3$ 与 $- \\frac { 1 } { 3 }$ 互为倒数", "target": "( 2 x - 3 ) * ( - \\frac { 1 } { 3 } ) = 1"}]}}
|
||||
{"content": "If one root of the quadratic equation $x ^ 2 + ( k + 3 ) x + 2 = 0$ with respect to $x$ is $- 1$, then the other root is ____?", "answer": "2", "steps": "If the other root of the equation is $t$, then $- t = 2$, which gives $t = - 2$.", "expr_cands": ["m + 1 = 3 ( m - 1 )", "m", "m + 1 = 3 m - 3", "m = 2", "m - 3 m = - 3 - 1", "- 2 m = - 4", "1"], "exprs": ["m + 1 = 3 m - 3", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m + 1 = 3 ( m - 1 )"}, {"id": "m + 1 = 3 m - 3"}, {"id": "m = 2"}], "links": [{"rel": "展开", "source": "m + 1 = 3 ( m - 1 )", "target": "m + 1 = 3 m - 3"}, {"rel": "等式方程求解", "source": "m + 1 = 3 m - 3", "target": "m = 2"}]}}
|
||||
{"content": "Given that $y - 2$ is inversely proportional to $x$, and when $x = 2$, $y = 4$, what is the value of $x$ when $y = 3$?", "answer": "2", "steps": "Given the problem, let $y - 2 = \\frac { k } { x }$. Substituting $x = 2$ and $y = 4$, we get $4 - 2 = \\frac { k } { 2 }$, which gives us $k = 4$. Therefore, the relationship between $y$ and $x$ is $y = \\frac { 4 } { x } + 2$. When $y = 3$, we have $\\frac { 4 } { x } + 2 = 3$, which gives us $x = 4$.", "expr_cands": ["x + 4 y = 13", "y", "x", "x = 5", "4 y + 5 = 13", "5 + 4 y = 13", "y = 2"], "exprs": ["5 + 4 y = 13", "y = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 4 y = 13"}, {"id": "5 + 4 y = 13"}, {"id": "x = 5"}, {"id": "y = 2"}], "links": [{"rel": "被代入", "source": "x + 4 y = 13", "target": "5 + 4 y = 13"}, {"rel": "等式方程求解", "source": "5 + 4 y = 13", "target": "y = 2"}, {"rel": "代入", "source": "x = 5", "target": "5 + 4 y = 13"}]}}
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{"content": "If $2 x - 3$ is the reciprocal of $- \\frac { 1 } { 3 }$, then $x$ = ____?", "answer": "2018", "steps": "According to the problem, we have $( 2 x - 3 ) * ( - \\frac { 1 } { 3 }) = 1$. Simplifying this expression, we get $2 x - 3 = - 3$. Solving for $x$, we get $x = 0$.", "expr_cands": ["m - 3 n", "n", "m", "1", "2020 - 2 m + 6 n", "m - 3 n = 1", "2020 - 2 ( m - 3 n )", "2018"], "exprs": ["m - 3 n = 1", "2020 - 2 ( m - 3 n )", "2018"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m - 3 n"}, {"id": "m - 3 n = 1"}, {"id": "1"}, {"id": "代数式 $m - 3 n$ 的值是 $1$"}, {"id": "2020 - 2 m + 6 n"}, {"id": "2020 - 2 ( m - 3 n )"}, {"id": "2018"}], "links": [{"rel": "被描述", "source": "m - 3 n", "target": "m - 3 n = 1"}, {"rel": "提取因式参考", "source": "m - 3 n = 1", "target": "2020 - 2 ( m - 3 n )"}, {"rel": "代入", "source": "m - 3 n = 1", "target": "2018"}, {"rel": "被描述", "source": "1", "target": "m - 3 n = 1"}, {"rel": "限制性描述", "source": "代数式 $m - 3 n$ 的值是 $1$", "target": "m - 3 n = 1"}, {"rel": "提取因式", "source": "2020 - 2 m + 6 n", "target": "2020 - 2 ( m - 3 n )"}, {"rel": "被代入", "source": "2020 - 2 ( m - 3 n )", "target": "2018"}]}}
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{"content": "Given $m + 1 = 3 ( m - 1 )$, what is the value of $m$?", "answer": "4", "steps": "Removing the parentheses, we get $m + 1 = 3 m - 3$. Moving terms, we get $m - 3 m = - 3 - 1$. Combining like terms, we get $- 2 m = - 4$. Dividing by the coefficient, we get $m = 2$.", "expr_cands": ["3 \\cdot 9 ^ { m } \\cdot 27 ^ { m } = 3 ^ { 21 }", "m", "3 ^ { 1 + 2 m + 3 m } = 3 ^ { 21 }", "m = 4", "1 + 2 m + 3 m = 21"], "exprs": ["m = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 \\cdot 9 ^ { m } \\cdot 27 ^ { m } = 3 ^ { 21 }"}, {"id": "m = 4"}], "links": [{"rel": "等式方程求解", "source": "3 \\cdot 9 ^ { m } \\cdot 27 ^ { m } = 3 ^ { 21 }", "target": "m = 4"}]}}
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{"content": "In the quadratic equation $x + 4 y = 13$, when $x = 5$, $y$ = ____ ?", "answer": "4", "steps": "Substituting $x = 5$ into the equation $x + 4 y = 13$, we get $5 + 4 y = 13$. Solving for $y$, we get $y = 2$.", "expr_cands": ["x = 5", "x", "3 x - 2 a = 7", "a", "15 - 2 a = 7", "a = 4"], "exprs": ["15 - 2 a = 7", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 5"}, {"id": "15 - 2 a = 7"}, {"id": "3 x - 2 a = 7"}, {"id": "a = 4"}], "links": [{"rel": "代入", "source": "x = 5", "target": "15 - 2 a = 7"}, {"rel": "等式方程求解", "source": "15 - 2 a = 7", "target": "a = 4"}, {"rel": "被代入", "source": "3 x - 2 a = 7", "target": "15 - 2 a = 7"}]}}
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{"content": "If the value of the algebraic expression $m - 3 n$ is $1$, then the value of the algebraic expression $2020 - 2 m + 6 n$ is ____?", "answer": "- 1", "steps": "Since $m - 3 n = 1$, therefore $2020 - 2 m + 6 n = 2020 - 2 ( m - 3 n ) = 2020 - 2 = 2018$.", "expr_cands": ["x", "y", "x ^ { 3 } + ax ^ { 2 } + 3 xy + x ^ { 2 } - bxy + y ^ { 2 }", "b", "a", "x ^ { 2 }", "xy", "a ^ { b }", "x ^ { 3 } + ( a + 1 ) x ^ { 2 } + ( 3 - b ) xy + y ^ { 2 }", "a + 1 = 0", "a = - 1", "3 - b = 0", "b = 3", "- 1"], "exprs": ["x ^ { 3 } + ( a + 1 ) x ^ { 2 } + ( 3 - b ) xy + y ^ { 2 }", "a + 1 = 0", "3 - b = 0", "a = - 1", "b = 3", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 3 } + ax ^ { 2 } + 3 xy + x ^ { 2 } - bxy + y ^ { 2 }"}, {"id": "x ^ { 3 } + ( a + 1 ) x ^ { 2 } + ( 3 - b ) xy + y ^ { 2 }"}, {"id": "a + 1 = 0"}, {"id": "关于 $x$ , $y$ 的多项式 $x ^ { 3 } + ax ^ { 2 } + 3 xy + x ^ { 2 } - bxy + y ^ { 2 }$ 中不含 $x ^ { 2 }$ 项和 $xy$ 项"}, {"id": "a = - 1"}, {"id": "3 - b = 0"}, {"id": "b = 3"}, {"id": "a ^ { b }"}, {"id": "- 1"}], "links": [{"rel": "提取因式", "source": "x ^ { 3 } + ax ^ { 2 } + 3 xy + x ^ { 2 } - bxy + y ^ { 2 }", "target": "x ^ { 3 } + ( a + 1 ) x ^ { 2 } + ( 3 - b ) xy + y ^ { 2 }"}, {"rel": "被描述", "source": "x ^ { 3 } + ( a + 1 ) x ^ { 2 } + ( 3 - b ) xy + y ^ { 2 }", "target": "a + 1 = 0"}, {"rel": "被描述", "source": "x ^ { 3 } + ( a + 1 ) x ^ { 2 } + ( 3 - b ) xy + y ^ { 2 }", "target": "3 - b = 0"}, {"rel": "等式方程求解", "source": "a + 1 = 0", "target": "a = - 1"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的多项式 $x ^ { 3 } + ax ^ { 2 } + 3 xy + x ^ { 2 } - bxy + y ^ { 2 }$ 中不含 $x ^ { 2 }$ 项和 $xy$ 项", "target": "a + 1 = 0"}, {"rel": "限制性描述", "source": "关于 $x$ , $y$ 的多项式 $x ^ { 3 } + ax ^ { 2 } + 3 xy + x ^ { 2 } - bxy + y ^ { 2 }$ 中不含 $x ^ { 2 }$ 项和 $xy$ 项", "target": "3 - b = 0"}, {"rel": "代入", "source": "a = - 1", "target": "- 1"}, {"rel": "等式方程求解", "source": "3 - b = 0", "target": "b = 3"}, {"rel": "代入", "source": "b = 3", "target": "- 1"}, {"rel": "被代入", "source": "a ^ { b }", "target": "- 1"}]}}
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{"content": "If $3 \\cdot 9 ^ { m } \\cdot 27 ^ { m } = 3 ^ { 21 }$, then the value of $m$ is ____?", "answer": "\\frac { 19 } { 10 }", "steps": "$3 \\times 9 ^ { m } \\times 27 ^ { m } = 3 \\times 3 ^ { 2 m } \\times 3 ^ { 3 m } = 3 ^ { 1 + 2 m + 3 m } = 3 ^ { 21 }$, therefore $1 + 2 m + 3 m = 21$, and solving for $m$ gives $m = 4$.", "expr_cands": ["x \\neq - \\frac { 5 } { b }", "b", "x", "\\frac { a + x } { - bx - 5 }", "a", "2", "\\frac { 1 } { a } - \\frac { 1 } { b }", "\\frac { a + x } { - bx - 5 } = 2", "a + x = - 2 bx - 10", "- 2 b = 1", "b = - \\frac { 1 } { 2 }", "a = - 10", "\\frac { 19 } { 10 }"], "exprs": ["\\frac { a + x } { - bx - 5 } = 2", "a + x = - 2 bx - 10", "- 2 b = 1", "a = - 10", "b = - \\frac { 1 } { 2 }", "\\frac { 19 } { 10 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x \\neq - \\frac { 5 } { b }"}, {"id": "\\frac { a + x } { - bx - 5 } = 2"}, {"id": "\\frac { a + x } { - bx - 5 }"}, {"id": "2"}, {"id": "当 $x \\neq - \\frac { 5 } { b }$ 时"}, {"id": "无论 $x$ 为何值"}, {"id": "$\\frac { a + x } { - bx - 5 }$ 的值恒为 $2$"}, {"id": "a + x = - 2 bx - 10"}, {"id": "- 2 b = 1"}, {"id": "考虑最特殊情形等式两边 $x$ 的系数相"}, {"id": "常数项相等得到 $- 2 b = 1$ , $a = - 10$"}, {"id": "b = - \\frac { 1 } { 2 }"}, {"id": "a = - 10"}, {"id": "\\frac { 1 } { a } - \\frac { 1 } { b }"}, {"id": "\\frac { 19 } { 10 }"}], "links": [{"rel": "被描述", "source": "x \\neq - \\frac { 5 } { b }", "target": "\\frac { a + x } { - bx - 5 } = 2"}, {"rel": "同乘除", "source": "\\frac { a + x } { - bx - 5 } = 2", "target": "a + x = - 2 bx - 10"}, {"rel": "被描述", "source": "\\frac { a + x } { - bx - 5 }", "target": "\\frac { a + x } { - bx - 5 } = 2"}, {"rel": "被描述", "source": "2", "target": "\\frac { a + x } { - bx - 5 } = 2"}, {"rel": "限制性描述", "source": "当 $x \\neq - \\frac { 5 } { b }$ 时", "target": "\\frac { a + x } { - bx - 5 } = 2"}, {"rel": "限制性描述", "source": "无论 $x$ 为何值", "target": "\\frac { a + x } { - bx - 5 } = 2"}, {"rel": "限制性描述", "source": "无论 $x$ 为何值", "target": "- 2 b = 1"}, {"rel": "限制性描述", "source": "无论 $x$ 为何值", "target": "a = - 10"}, {"rel": "限制性描述", "source": "$\\frac { a + x } { - bx - 5 }$ 的值恒为 $2$", "target": "\\frac { a + x } { - bx - 5 } = 2"}, {"rel": "限制性描述", "source": "$\\frac { a + x } { - bx - 5 }$ 的值恒为 $2$", "target": "- 2 b = 1"}, {"rel": "限制性描述", "source": "$\\frac { a + x } { - bx - 5 }$ 的值恒为 $2$", "target": "a = - 10"}, {"rel": "被描述", "source": "a + x = - 2 bx - 10", "target": "- 2 b = 1"}, {"rel": "被描述", "source": "a + x = - 2 bx - 10", "target": "a = - 10"}, {"rel": "等式方程求解", "source": "- 2 b = 1", "target": "b = - \\frac { 1 } { 2 }"}, {"rel": "限制性描述", "source": "考虑最特殊情形等式两边 $x$ 的系数相", "target": "- 2 b = 1"}, {"rel": "限制性描述", "source": "考虑最特殊情形等式两边 $x$ 的系数相", "target": "a = - 10"}, {"rel": "限制性描述", "source": "常数项相等得到 $- 2 b = 1$ , $a = - 10$", "target": "- 2 b = 1"}, {"rel": "限制性描述", "source": "常数项相等得到 $- 2 b = 1$ , $a = - 10$", "target": "a = - 10"}, {"rel": "代入", "source": "b = - \\frac { 1 } { 2 }", "target": "\\frac { 19 } { 10 }"}, {"rel": "代入", "source": "a = - 10", "target": "\\frac { 19 } { 10 }"}, {"rel": "被代入", "source": "\\frac { 1 } { a } - \\frac { 1 } { b }", "target": "\\frac { 19 } { 10 }"}]}}
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{"content": "Given that $x = 5$ is a solution to the linear equation $3 x - 2 a = 7$, then $a$ = ____ ?", "answer": "3 a ( a + 2 ) ( a - 2 )", "steps": "By substituting $x = 5$ into the equation, we get $15 - 2 a = 7$, and solving for $a$ gives $a = 4$.", "expr_cands": ["3 { a } ^ { 3 } - 12 a", "a", "3 a ^ { 3 } - 12 a", "3 a ( a ^ { 2 } - 4 )", "3 a ( a + 2 ) ( a - 2 )"], "exprs": ["3 a ( a + 2 ) ( a - 2 )"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 { a } ^ { 3 } - 12 a"}, {"id": "3 a ( a + 2 ) ( a - 2 )"}], "links": [{"rel": "提取因式", "source": "3 { a } ^ { 3 } - 12 a", "target": "3 a ( a + 2 ) ( a - 2 )"}]}}
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{"content": "Regarding the polynomial in $x$ and $y$, $x ^ 3 + ax ^ 2 + 3 xy + x ^ 2 - bxy + y ^ 2$, which does not contain the $x ^ 2$ term and the $xy$ term, then $a ^ b$ = ____?", "answer": "x = 3", "steps": "$x ^ { 3 } + ax ^ { 2 } + 3 xy + x ^ { 2 } - bxy + y ^ { 2 } = x ^ { 3 } + ( a + 1 ) x ^ { 2 } + ( 3 - b ) xy + y ^ { 2 }$ because it does not contain $x ^ { 2 }$ term and $xy$ term, therefore $a + 1 = 0$, $3 - b = 0$, which gives $a = - 1$, $b = 3$, so $a ^ { b } = - 1$.", "expr_cands": ["\\frac { 3 - x } { x - 4 } = 1 - \\frac { 1 } { 4 - x }", "x", "3 - x = x - 4 + 1", "x = 3"], "exprs": ["x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 3 - x } { x - 4 } = 1 - \\frac { 1 } { 4 - x }"}, {"id": "x = 3"}], "links": [{"rel": "等式方程求解", "source": "\\frac { 3 - x } { x - 4 } = 1 - \\frac { 1 } { 4 - x }", "target": "x = 3"}]}}
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{"content": "When $x \\neq - \\frac { 5 } { b }$ , no matter what value $x$ takes, the value of $\\frac { a + x } { - bx - 5 }$ is always $2$. Then $\\frac { 1 } { a } - \\frac { 1 } { b }$ = ____ ?", "answer": "1", "steps": "From $\\frac { a + x } { - bx - 5 } = 2$, we get $a + x = - 2 bx - 10$. Since the equation holds for any value of $x$, we consider the most special case where the coefficients and constants on both sides are equal. Thus, we have $- 2 b = 1$ and $a = - 10$. Therefore, $b = - \\frac { 1 } { 2 }$ and $a = - 10$. Hence, $\\frac { 1 } { a } - \\frac { 1 } { b } = \\frac { 19 } { 10 }$.", "expr_cands": ["x = 9", "x", "\\frac { 2 } { x + 5 } = \\frac { k } { x - 2 }", "k", "\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }", "k = 1"], "exprs": ["\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }", "k = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 9"}, {"id": "\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }"}, {"id": "\\frac { 2 } { x + 5 } = \\frac { k } { x - 2 }"}, {"id": "$x = 9$ 是分式方程 $\\frac { 2 } { x + 5 } = \\frac { k } { x - 2 }$ 的解"}, {"id": "k = 1"}], "links": [{"rel": "被描述", "source": "x = 9", "target": "\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }"}, {"rel": "等式方程求解", "source": "\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }", "target": "k = 1"}, {"rel": "被描述", "source": "\\frac { 2 } { x + 5 } = \\frac { k } { x - 2 }", "target": "\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }"}, {"rel": "限制性描述", "source": "$x = 9$ 是分式方程 $\\frac { 2 } { x + 5 } = \\frac { k } { x - 2 }$ 的解", "target": "\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }"}]}}
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{"content": "Factorization: $3 { a } ^ { 3 } - 12 a$ = ____ ?", "answer": "2", "steps": "$3 a ^ { 3 } - 12 a = 3 a ( a ^ { 2 } - 4 )$ (Extract common factor) = $3 a ( a + 2 ) ( a - 2 )$.", "expr_cands": ["\\frac { 2 } { 5 } x ^ { 2 m } y ^ { 3 } z", "y", "z", "x", "m", "8", "2 m + 3 + 1 = 8", "m = 2"], "exprs": ["2 m + 3 + 1 = 8", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 2 } { 5 } x ^ { 2 m } y ^ { 3 } z"}, {"id": "2 m + 3 + 1 = 8"}, {"id": "8"}, {"id": "$\\frac { 2 } { 5 } x ^ { 2 m } y ^ { 3 } z$ 是 $8$ 次单项式"}, {"id": "m = 2"}], "links": [{"rel": "被描述", "source": "\\frac { 2 } { 5 } x ^ { 2 m } y ^ { 3 } z", "target": "2 m + 3 + 1 = 8"}, {"rel": "等式方程求解", "source": "2 m + 3 + 1 = 8", "target": "m = 2"}, {"rel": "被描述", "source": "8", "target": "2 m + 3 + 1 = 8"}, {"rel": "限制性描述", "source": "$\\frac { 2 } { 5 } x ^ { 2 m } y ^ { 3 } z$ 是 $8$ 次单项式", "target": "2 m + 3 + 1 = 8"}]}}
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{"content": "Equation: $\\frac { 3 - x } { x - 4 } = 1 - \\frac { 1 } { 4 - x }$, the root is ____?", "answer": "a \\neq 1", "steps": "Going to the denominator, we get: $3 - x = x - 4 + 1$. Solving for $x$, we get $x = 3$. After checking, we find that $x = 3$ is a solution to the fractional equation.", "expr_cands": ["( a - 1 ) ^ { 0 } = 1", "a", "a - 1 \\neq 0", "a \\neq 1"], "exprs": ["a - 1 \\neq 0", "a \\neq 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - 1 ) ^ { 0 } = 1"}, {"id": "a - 1 \\neq 0"}, {"id": "$( a - 1 ) ^ { 0 } = 1$ 成立"}, {"id": "多项式零次方项，若底数不为0，则恒等于1"}, {"id": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0"}, {"id": "a \\neq 1"}], "links": [{"rel": "被描述", "source": "( a - 1 ) ^ { 0 } = 1", "target": "a - 1 \\neq 0"}, {"rel": "不等式方程求解", "source": "a - 1 \\neq 0", "target": "a \\neq 1"}, {"rel": "限制性描述", "source": "$( a - 1 ) ^ { 0 } = 1$ 成立", "target": "a - 1 \\neq 0"}, {"rel": "属性描述", "source": "多项式零次方项，若底数不为0，则恒等于1", "target": "a - 1 \\neq 0"}, {"rel": "属性描述", "source": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0", "target": "a - 1 \\neq 0"}]}}
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{"content": "Given that $x = 9$ is a solution of the fractional equation $\\frac { 2 } { x + 5 } = \\frac { k } { x - 2 }$, what is the value of $k$?", "answer": "- \\frac { 9 } { 2 }", "steps": "Substituting $x = 9$ into the original equation, we get $\\frac { 2 } { 9 + 5 } = \\frac { k } { 9 - 2 }$. Solving for $k$, we get $k = 1$.", "expr_cands": ["\\frac { 1 } { x } - \\frac { 1 } { y } = 2", "y", "x", "\\frac { 4 ( y - x ) + xy } { x - y }", "\\frac { x - y } { xy } = - 2", "x - y = - 2 xy", "\\frac { 4 * 2 xy + xy } { - 2 xy }", "- \\frac { 9 } { 2 }"], "exprs": ["x - y = - 2 xy", "- \\frac { 9 } { 2 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { x } - \\frac { 1 } { y } = 2"}, {"id": "x - y = - 2 xy"}, {"id": "\\frac { 4 * 2 xy + xy } { - 2 xy }"}, {"id": "- \\frac { 9 } { 2 }"}], "links": [{"rel": "同乘除", "source": "\\frac { 1 } { x } - \\frac { 1 } { y } = 2", "target": "x - y = - 2 xy"}, {"rel": "代入", "source": "x - y = - 2 xy", "target": "- \\frac { 9 } { 2 }"}, {"rel": "被代入", "source": "\\frac { 4 * 2 xy + xy } { - 2 xy }", "target": "- \\frac { 9 } { 2 }"}]}}
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{"content": "Given that $\\frac { 2 } { 5 } x ^ { 2 m } y ^ 3 z$ is an 8th degree monomial, what is the value of $m$?", "answer": "- 1", "steps": "Because $\\frac { 2 } { 5 } x ^ { 2 m } y ^ { 3 } z$ is an 8th degree monomial, therefore $2 m + 3 + 1 = 8$, solving for $m$ gives $m = 2$.", "expr_cands": ["| a - 3 |", "a", "( b + 4 ) ^ { 2 }", "b", "a + b", "| a - 3 | + ( b + 4 ) ^ { 2 } = 0", "a - 3 = 0", "a = 3", "b + 4 = 0", "b = - 4", "- 1"], "exprs": ["| a - 3 | + ( b + 4 ) ^ { 2 } = 0", "a - 3 = 0", "b + 4 = 0", "a = 3", "b = - 4", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a - 3 |"}, {"id": "| a - 3 | + ( b + 4 ) ^ { 2 } = 0"}, {"id": "( b + 4 ) ^ { 2 }"}, {"id": "$| a - 3 |$ 与 $( b + 4 ) ^ { 2 }$ 互为相反数"}, {"id": "a - 3 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "b + 4 = 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "a = 3"}, {"id": "b = - 4"}, {"id": "a + b"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "| a - 3 |", "target": "| a - 3 | + ( b + 4 ) ^ { 2 } = 0"}, {"rel": "被描述", "source": "| a - 3 | + ( b + 4 ) ^ { 2 } = 0", "target": "a - 3 = 0"}, {"rel": "被描述", "source": "| a - 3 | + ( b + 4 ) ^ { 2 } = 0", "target": "b + 4 = 0"}, {"rel": "被描述", "source": "( b + 4 ) ^ { 2 }", "target": "| a - 3 | + ( b + 4 ) ^ { 2 } = 0"}, {"rel": "限制性描述", "source": "$| a - 3 |$ 与 $( b + 4 ) ^ { 2 }$ 互为相反数", "target": "| a - 3 | + ( b + 4 ) ^ { 2 } = 0"}, {"rel": "等式方程求解", "source": "a - 3 = 0", "target": "a = 3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a - 3 = 0"}, {"rel": "等式方程求解", "source": "b + 4 = 0", "target": "b = - 4"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "b + 4 = 0"}, {"rel": "代入", "source": "a = 3", "target": "- 1"}, {"rel": "代入", "source": "b = - 4", "target": "- 1"}, {"rel": "被代入", "source": "a + b", "target": "- 1"}]}}
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||||
{"content": "If $( a - 1 ) ^ 0 = 1$ holds, then the possible values of $a$ are ____?", "answer": "x > \\frac { 2 } { 3 }", "steps": "From the given condition, we have $a - 1 \\neq 0$, which implies that $a \\neq 1$ after solving.", "expr_cands": ["x", "\\frac { 1 } { 2 - 3 x }", "\\frac { 1 } { 2 - 3 x } < 0", "\\frac { 2 } { 3 } < x", "2 - 3 x < 0", "x > \\frac { 2 } { 3 }"], "exprs": ["2 - 3 x < 0", "x > \\frac { 2 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 2 - 3 x }"}, {"id": "2 - 3 x < 0"}, {"id": "分式 $\\frac { 1 } { 2 - 3 x }$ 的值为负数"}, {"id": "分式为负数，则分子分母异号"}, {"id": "x > \\frac { 2 } { 3 }"}], "links": [{"rel": "被描述", "source": "\\frac { 1 } { 2 - 3 x }", "target": "2 - 3 x < 0"}, {"rel": "不等式方程求解", "source": "2 - 3 x < 0", "target": "x > \\frac { 2 } { 3 }"}, {"rel": "限制性描述", "source": "分式 $\\frac { 1 } { 2 - 3 x }$ 的值为负数", "target": "2 - 3 x < 0"}, {"rel": "属性描述", "source": "分式为负数，则分子分母异号", "target": "2 - 3 x < 0"}]}}
|
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{"content": "When $\\frac { 1 } { x } - \\frac { 1 } { y } = 2$, the algebraic expression $\\frac { 4 ( y - x ) + xy } { x - y }$ = ____?", "answer": "4", "steps": "Given the equation is rearranged as: $\\frac { x - y } { xy } = - 2$, which means $x - y = - 2 xy$. The original expression is $\\frac { 4 \\times 2 xy + xy } { - 2 xy } = - \\frac { 9 } { 2 }$.", "expr_cands": ["2 x + 1", "x", "- 9", "2 x + 1 = 9", "x = 4"], "exprs": ["2 x + 1 = 9", "x = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 1"}, {"id": "2 x + 1 = 9"}, {"id": "- 9"}, {"id": "$2 x + 1$ 是 $- 9$ 的相反数"}, {"id": "x = 4"}], "links": [{"rel": "被描述", "source": "2 x + 1", "target": "2 x + 1 = 9"}, {"rel": "等式方程求解", "source": "2 x + 1 = 9", "target": "x = 4"}, {"rel": "被描述", "source": "- 9", "target": "2 x + 1 = 9"}, {"rel": "限制性描述", "source": "$2 x + 1$ 是 $- 9$ 的相反数", "target": "2 x + 1 = 9"}]}}
|
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{"content": "If $| a - 3 |$ and $( b + 4 ) ^ 2$ are opposite in sign, then the value of $a + b$ is ____?", "answer": "72", "steps": "From the given information, we have $| a - 3 | + ( b + 4 ) ^ { 2 } = 0$ , $a - 3 = 0$ , $b + 4 = 0$ , which yields $a = 3$ and $b = - 4$ . Therefore, $a + b = - 1$.", "expr_cands": ["\\frac { x - 2 } { 6 } - \\frac { 1 } { 8 } = \\frac { x } { 9 }", "x", "x = \\frac { 33 } { 4 }", "72"], "exprs": ["72"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - 2 } { 6 } - \\frac { 1 } { 8 } = \\frac { x } { 9 }"}, {"id": "72"}, {"id": "解方程 $\\frac { x - 2 } { 6 } - \\frac { 1 } { 8 } = \\frac { x } { 9 }$ 时"}, {"id": "需在方程两边乘公分母"}], "links": [{"rel": "被描述", "source": "\\frac { x - 2 } { 6 } - \\frac { 1 } { 8 } = \\frac { x } { 9 }", "target": "72"}, {"rel": "限制性描述", "source": "解方程 $\\frac { x - 2 } { 6 } - \\frac { 1 } { 8 } = \\frac { x } { 9 }$ 时", "target": "72"}, {"rel": "限制性描述", "source": "需在方程两边乘公分母", "target": "72"}]}}
|
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{"content": "When $x$ ____ ?, the value of the fraction $\\frac { 1 } { 2 - 3 x }$ is negative.", "answer": "a > 2", "steps": "Since $\\frac { 1 } { 2 - 3 x } < 0$, therefore $2 - 3 x < 0$, solving for $x$ gives $x > \\frac { 2 } { 3 }$.", "expr_cands": ["\\frac { a + 2 } { \\sqrt { a - 2 } }", "a", "a - 2 \\ge 0", "2 \\le a", "a \\ge 2", "0", "a - 2 \\neq 0", "a \\neq 2", "a > 2"], "exprs": ["a - 2 \\ge 0", "a - 2 \\neq 0", "a > 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a + 2 } { \\sqrt { a - 2 } }"}, {"id": "a - 2 \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "当 $\\frac { a + 2 } { \\sqrt { a - 2 } }$ 有意义时"}, {"id": "a - 2 \\neq 0"}, {"id": "分式有意义，则分母不为0"}, {"id": "a > 2"}], "links": [{"rel": "被描述", "source": "\\frac { a + 2 } { \\sqrt { a - 2 } }", "target": "a - 2 \\ge 0"}, {"rel": "被描述", "source": "\\frac { a + 2 } { \\sqrt { a - 2 } }", "target": "a - 2 \\neq 0"}, {"rel": "联立", "source": "a - 2 \\ge 0", "target": "a > 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "a - 2 \\ge 0"}, {"rel": "限制性描述", "source": "当 $\\frac { a + 2 } { \\sqrt { a - 2 } }$ 有意义时", "target": "a - 2 \\ge 0"}, {"rel": "联立", "source": "a - 2 \\neq 0", "target": "a > 2"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "a - 2 \\neq 0"}]}}
|
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{"content": "If $2 x + 1$ is the opposite of $- 9$, then $x$ = ____?", "answer": "6", "steps": "$\\because$ $2 x + 1$ is the opposite of $- 9$, $\\therefore$ $2 x + 1 = 9$. Solving for $x$, we get $x = 4$.", "expr_cands": ["\\sqrt { 49 } = x", "x", "3 \\sqrt { y } = 3", "y", "x - y", "\\sqrt { 49 }", "7", "x = 7", "y = 1", "\\sqrt { y } = 1", "6"], "exprs": ["x = 7", "y = 1", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 49 } = x"}, {"id": "x = 7"}, {"id": "3 \\sqrt { y } = 3"}, {"id": "y = 1"}, {"id": "x - y"}, {"id": "6"}], "links": [{"rel": "等式方程求解", "source": "\\sqrt { 49 } = x", "target": "x = 7"}, {"rel": "代入", "source": "x = 7", "target": "6"}, {"rel": "等式方程求解", "source": "3 \\sqrt { y } = 3", "target": "y = 1"}, {"rel": "代入", "source": "y = 1", "target": "6"}, {"rel": "被代入", "source": "x - y", "target": "6"}]}}
|
||||
{"content": "When solving the equation $\\frac { x - 2 } { 6 } - \\frac { 1 } { 8 } = \\frac { x } { 9 }$, we need to multiply both sides of the equation by the common denominator _____.", "answer": "- 1", "steps": "When solving the equation $\\frac { x - 2 } { 6 } - \\frac { 1 } { 8 } = \\frac { x } { 9 }$, it is necessary to multiply both sides of the equation by the common denominator $72$.", "expr_cands": ["x", "x ^ { 2 } + px - 6 = 0", "p", "3", "x = 3", "9 + 3 p - 6 = 0", "p = - 1"], "exprs": ["x = 3", "9 + 3 p - 6 = 0", "p = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "x = 3"}, {"id": "x"}, {"id": "x ^ { 2 } + px - 6 = 0"}, {"id": "关于 $x$ 的一元二次方程 $x ^ { 2 } + px - 6 = 0$ 的一个根为 $3$"}, {"id": "9 + 3 p - 6 = 0"}, {"id": "p = - 1"}], "links": [{"rel": "被描述", "source": "3", "target": "x = 3"}, {"rel": "代入", "source": "x = 3", "target": "9 + 3 p - 6 = 0"}, {"rel": "被描述", "source": "x", "target": "x = 3"}, {"rel": "被描述", "source": "x ^ { 2 } + px - 6 = 0", "target": "x = 3"}, {"rel": "被代入", "source": "x ^ { 2 } + px - 6 = 0", "target": "9 + 3 p - 6 = 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的一元二次方程 $x ^ { 2 } + px - 6 = 0$ 的一个根为 $3$", "target": "x = 3"}, {"rel": "等式方程求解", "source": "9 + 3 p - 6 = 0", "target": "p = - 1"}]}}
|
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{"content": "When $\\frac { a + 2 } { \\sqrt { a - 2 } }$ is meaningful, the range of values for $a$ is ____?", "answer": "a = 3", "steps": "According to the meaning of quadratic radicals, the radicand $a - 2 \\ge 0$, so we get $a \\ge 2$. According to the denominator not being zero, we have $a - 2 \\neq 0$, so we get $a \\neq 2$. Therefore, $a > 2$.", "expr_cands": ["\\frac { { a } ^ { 2 } - 9 } { a + 3 }", "a", "a ^ { 2 } - 9 = 0", "a = - 3", "a = 3", "a + 3 \\neq 0", "a \\neq - 3"], "exprs": ["a ^ { 2 } - 9 = 0", "a + 3 \\neq 0", "a \\neq - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { { a } ^ { 2 } - 9 } { a + 3 }"}, {"id": "a ^ { 2 } - 9 = 0"}, {"id": "要使分式 $\\frac { { a } ^ { 2 } - 9 } { a + 3 }$ 的值为零"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "a + 3 \\neq 0"}, {"id": "a \\neq - 3"}], "links": [{"rel": "被描述", "source": "\\frac { { a } ^ { 2 } - 9 } { a + 3 }", "target": "a ^ { 2 } - 9 = 0"}, {"rel": "被描述", "source": "\\frac { { a } ^ { 2 } - 9 } { a + 3 }", "target": "a + 3 \\neq 0"}, {"rel": "联立", "source": "a ^ { 2 } - 9 = 0", "target": "a \\neq - 3"}, {"rel": "限制性描述", "source": "要使分式 $\\frac { { a } ^ { 2 } - 9 } { a + 3 }$ 的值为零", "target": "a ^ { 2 } - 9 = 0"}, {"rel": "限制性描述", "source": "要使分式 $\\frac { { a } ^ { 2 } - 9 } { a + 3 }$ 的值为零", "target": "a + 3 \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "a ^ { 2 } - 9 = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "a + 3 \\neq 0"}, {"rel": "联立", "source": "a + 3 \\neq 0", "target": "a \\neq - 3"}]}}
|
||||
{"content": "Given $\\sqrt { 49 } = x$, $3 \\sqrt { y } = 3$, then $x - y$ = ____ ?", "answer": "- 1", "steps": "$\\sqrt { 49 } = 7$ , therefore $x = 7$ , $3 \\sqrt { y } = 3$ , $\\sqrt { y } = 1$ , $y = 1$ , so $x - y = 7 - 1 = 6$.", "expr_cands": ["4 x + 2", "x", "3 x - 9", "4 x - 5", "( 4 x + 2 ) + ( 3 x - 9 ) = 0", "x = 1", "4 x + 2 + 3 x - 9 = 0", "4 * 1 - 5", "- 1"], "exprs": ["( 4 x + 2 ) + ( 3 x - 9 ) = 0", "x = 1", "4 * 1 - 5", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4 x + 2"}, {"id": "( 4 x + 2 ) + ( 3 x - 9 ) = 0"}, {"id": "3 x - 9"}, {"id": "当代数式 $4 x + 2$ 与 $3 x - 9$ 的值互为相反数时"}, {"id": "x = 1"}, {"id": "4 x - 5"}, {"id": "4 * 1 - 5"}, {"id": "- 1"}], "links": [{"rel": "被描述", "source": "4 x + 2", "target": "( 4 x + 2 ) + ( 3 x - 9 ) = 0"}, {"rel": "等式方程求解", "source": "( 4 x + 2 ) + ( 3 x - 9 ) = 0", "target": "x = 1"}, {"rel": "被描述", "source": "3 x - 9", "target": "( 4 x + 2 ) + ( 3 x - 9 ) = 0"}, {"rel": "限制性描述", "source": "当代数式 $4 x + 2$ 与 $3 x - 9$ 的值互为相反数时", "target": "( 4 x + 2 ) + ( 3 x - 9 ) = 0"}, {"rel": "代入", "source": "x = 1", "target": "4 * 1 - 5"}, {"rel": "被代入", "source": "4 x - 5", "target": "4 * 1 - 5"}, {"rel": "计算", "source": "4 * 1 - 5", "target": "- 1"}]}}
|
||||
{"content": "If one root of the quadratic equation $x ^ 2 + px - 6 = 0$ is $3$, then the value of $p$ is ____?", "answer": "2 ( x - 3 ) ( x + 4 )", "steps": "Because 3 is a root of the equation $x ^ 2 + px - 6 = 0$, therefore substituting $x = 3$ into the equation gives $9 + 3 p - 6 = 0$, and $p = - 1$.", "expr_cands": ["2 x ^ { 2 } + px + q = 0", "q", "x", "p", "3", "- 4", "2 x ^ { 2 } + px + q", "2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )", "0 = 2 ( x - 3 ) ( x + 4 )", "0", "2 ( x - 3 ) ( x + 4 )"], "exprs": ["2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x ^ { 2 } + px + q = 0"}, {"id": "2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )"}, {"id": "2 x ^ { 2 } + px + q"}, {"id": "一元二次方程 $2 x ^ { 2 } + px + q = 0$ 的两个根是 $3$ , $- 4$"}, {"id": "二次三项式 $2 x ^ { 2 } + px + q$ 可分解"}], "links": [{"rel": "被描述", "source": "2 x ^ { 2 } + px + q = 0", "target": "2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )"}, {"rel": "被描述", "source": "2 x ^ { 2 } + px + q", "target": "2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )"}, {"rel": "限制性描述", "source": "一元二次方程 $2 x ^ { 2 } + px + q = 0$ 的两个根是 $3$ , $- 4$", "target": "2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )"}, {"rel": "限制性描述", "source": "二次三项式 $2 x ^ { 2 } + px + q$ 可分解", "target": "2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )"}]}}
|
||||
{"content": "To make the value of the fraction $\\frac {{ a } ^ { 2 } - 9 } { a + 3 }$ equal to zero, the value of $a$ is ____?", "answer": "x > 3", "steps": "From the given information, we have $a ^ 2 - 9 = 0$ and $a + 3 \\neq 0$. Solving for $a$, we get $a = 3$.", "expr_cands": ["x", "\\frac { \\sqrt { x - 3 } } { x - 3 }", "x - 3 \\ge 0", "3 \\le x", "x - 3 \\neq 0", "x \\neq 3", "x \\ge 3", "x > 3"], "exprs": ["x - 3 \\ge 0", "x - 3 \\neq 0", "x > 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { \\sqrt { x - 3 } } { x - 3 }"}, {"id": "x - 3 \\ge 0"}, {"id": "式子 $\\frac { \\sqrt { x - 3 } } { x - 3 }$ 在实数范围内有意义"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x - 3 \\neq 0"}, {"id": "分式有意义，则分母不为0"}, {"id": "x > 3"}], "links": [{"rel": "被描述", "source": "\\frac { \\sqrt { x - 3 } } { x - 3 }", "target": "x - 3 \\ge 0"}, {"rel": "被描述", "source": "\\frac { \\sqrt { x - 3 } } { x - 3 }", "target": "x - 3 \\neq 0"}, {"rel": "联立", "source": "x - 3 \\ge 0", "target": "x > 3"}, {"rel": "限制性描述", "source": "式子 $\\frac { \\sqrt { x - 3 } } { x - 3 }$ 在实数范围内有意义", "target": "x - 3 \\ge 0"}, {"rel": "限制性描述", "source": "式子 $\\frac { \\sqrt { x - 3 } } { x - 3 }$ 在实数范围内有意义", "target": "x - 3 \\neq 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 3 \\ge 0"}, {"rel": "联立", "source": "x - 3 \\neq 0", "target": "x > 3"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "x - 3 \\neq 0"}]}}
|
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{"content": "When the value of the algebraic expression $4 x + 2$ is the opposite of the value of $3 x - 9$, the value of $4 x - 5$ is ____?", "answer": "m \\neq - 2", "steps": "According to the problem, we have $( 4 x + 2 ) + ( 3 x - 9 ) = 0$. Simplifying this expression, we get $4 x + 2 + 3 x - 9 = 0$. Solving for $x$, we get $x = 1$. Substituting $x = 1$ into $4 x - 5$, we get $4 * 1 - 5 = - 1$.", "expr_cands": ["( m + 2 ) x ^ { 2 } + 5 x - 7 = 0", "x", "m", "m + 2 \\neq 0", "m \\neq - 2"], "exprs": ["m + 2 \\neq 0", "m \\neq - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m + 2 ) x ^ { 2 } + 5 x - 7 = 0"}, {"id": "m + 2 \\neq 0"}, {"id": "方程 $( m + 2 ) x ^ { 2 } + 5 x - 7 = 0$ 是关于 $x$ 的一元二次方程"}, {"id": "m \\neq - 2"}], "links": [{"rel": "被描述", "source": "( m + 2 ) x ^ { 2 } + 5 x - 7 = 0", "target": "m + 2 \\neq 0"}, {"rel": "不等式方程求解", "source": "m + 2 \\neq 0", "target": "m \\neq - 2"}, {"rel": "限制性描述", "source": "方程 $( m + 2 ) x ^ { 2 } + 5 x - 7 = 0$ 是关于 $x$ 的一元二次方程", "target": "m + 2 \\neq 0"}]}}
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{"content": "Given a quadratic equation $2 x ^ { 2 } + px + q = 0$ with roots $3$ and $- 4$, the quadratic trinomial $2 x ^ { 2 } + px + q$ can be factored as ____?", "answer": "x = - 3", "steps": "$\\because$ The two roots of the quadratic equation $2 x ^ { 2 } + px + q = 0$ are $3$ and $- 4$, $\\therefore$ the quadratic trinomial $2 x ^ { 2 } + px + q = 2 ( x - 3 ) ( x + 4 )$.", "expr_cands": ["\\sqrt { x + 3 } = 0", "x", "x + 3 = 0", "x = - 3"], "exprs": ["x = - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + 3 } = 0"}, {"id": "x = - 3"}], "links": [{"rel": "等式方程求解", "source": "\\sqrt { x + 3 } = 0", "target": "x = - 3"}]}}
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{"content": "When $x$ satisfies certain conditions, the expression $\\frac { \\sqrt { x - 3 }} { x - 3 }$ is meaningful in the real number range. What is this condition?", "answer": "x < 2", "steps": "From the given equation, we can deduce that $x - 3 \\ge 0$ and $x - 3 \\neq 0$. Solving for $x$, we get $x \\ge 3$ and $x \\neq 3$. Therefore, $x > 3$. This means that the expression $\\frac { \\sqrt { x - 3 }} { x - 3 }$ is defined for all real numbers greater than 3.", "expr_cands": ["x", "mx - n > 0", "n", "m", "x < 15", "( - 5 m + n ) x > n + 5 m", "m < 0", "\\frac { n } { m } = 15", "n = 15 m", "( - 5 m + 15 m ) x > 15 m + 5 m", "10 mx > 20 m", "10 m < 0", "x < 2"], "exprs": ["m < 0", "\\frac { n } { m } = 15", "n = 15 m", "( - 5 m + 15 m ) x > 15 m + 5 m", "x < 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "mx - n > 0"}, {"id": "m < 0"}, {"id": "x < 15"}, {"id": "关于 $x$ 的不等式 $mx - n > 0$ 的解集是 $x < 15$"}, {"id": "不等式两边都乘或除同一个负数，不等号的方向改变"}, {"id": "\\frac { n } { m } = 15"}, {"id": "n = 15 m"}, {"id": "( - 5 m + n ) x > n + 5 m"}, {"id": "( - 5 m + 15 m ) x > 15 m + 5 m"}, {"id": "x < 2"}], "links": [{"rel": "被描述", "source": "mx - n > 0", "target": "m < 0"}, {"rel": "被描述", "source": "mx - n > 0", "target": "\\frac { n } { m } = 15"}, {"rel": "被描述", "source": "x < 15", "target": "m < 0"}, {"rel": "被描述", "source": "x < 15", "target": "\\frac { n } { m } = 15"}, {"rel": "限制性描述", "source": "关于 $x$ 的不等式 $mx - n > 0$ 的解集是 $x < 15$", "target": "m < 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的不等式 $mx - n > 0$ 的解集是 $x < 15$", "target": "\\frac { n } { m } = 15"}, {"rel": "属性描述", "source": "不等式两边都乘或除同一个负数，不等号的方向改变", "target": "m < 0"}, {"rel": "同乘除", "source": "\\frac { n } { m } = 15", "target": "n = 15 m"}, {"rel": "代入", "source": "n = 15 m", "target": "( - 5 m + 15 m ) x > 15 m + 5 m"}, {"rel": "被代入", "source": "( - 5 m + n ) x > n + 5 m", "target": "( - 5 m + 15 m ) x > 15 m + 5 m"}, {"rel": "同乘除", "source": "( - 5 m + 15 m ) x > 15 m + 5 m", "target": "x < 2"}]}}
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{"content": "If the equation $( m + 2 ) x ^ 2 + 5 x - 7 = 0$ is a quadratic equation in $x$, then the range of values for $m$ is ____?", "answer": "k < 1", "steps": "Since the equation $( m + 2 ) x ^ 2 + 5 x - 7 = 0$ is a quadratic equation in terms of $x$, it follows that $m + 2 \\neq 0$. Therefore, $m \\neq - 2$.", "expr_cands": ["y = ( k - 1 ) x", "k", "y", "x", "k - 1 < 0", "k < 1"], "exprs": ["k - 1 < 0", "k < 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y = ( k - 1 ) x"}, {"id": "k - 1 < 0"}, {"id": "正比例函数 $y = ( k - 1 ) x$ 中"}, {"id": "$y$ 的值随自变量 $x$ 的值增大而减小"}, {"id": "k < 1"}], "links": [{"rel": "被描述", "source": "y = ( k - 1 ) x", "target": "k - 1 < 0"}, {"rel": "不等式方程求解", "source": "k - 1 < 0", "target": "k < 1"}, {"rel": "限制性描述", "source": "正比例函数 $y = ( k - 1 ) x$ 中", "target": "k - 1 < 0"}, {"rel": "限制性描述", "source": "$y$ 的值随自变量 $x$ 的值增大而减小", "target": "k - 1 < 0"}]}}
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{"content": "If $\\sqrt { x + 3 } = 0$, then the value of $x$ is ____ ?", "answer": "\\frac { 27 } { 25 }", "steps": "From the given information, we have $x + 3 = 0$, which implies that $x = - 3$.", "expr_cands": ["{ a } ^ { m } = 3", "a", "m", "{ a } ^ { n } = 5", "n", "{ a } ^ { 3 m - 2 n }", "a ^ { m } = 3", "a ^ { n } = 5", "a ^ { 3 m - 2 n }", "\\frac { 27 } { 25 }"], "exprs": ["\\frac { 27 } { 25 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ^ { 3 m - 2 n }"}, {"id": "\\frac { 27 } { 25 }"}, {"id": "a ^ { m } = 3"}, {"id": "a ^ { n } = 5"}], "links": [{"rel": "被代入", "source": "a ^ { 3 m - 2 n }", "target": "\\frac { 27 } { 25 }"}, {"rel": "代入", "source": "a ^ { m } = 3", "target": "\\frac { 27 } { 25 }"}, {"rel": "代入", "source": "a ^ { n } = 5", "target": "\\frac { 27 } { 25 }"}]}}
|
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{"content": "If the solution set of the inequality $mx - n > 0$ with respect to $x$ is $x < 15$, then the solution set of the inequality $( - 5 m + n ) x > n + 5 m$ with respect to $x$ is ____?", "answer": "- 2", "steps": "Because the solution set of the inequality $mx - n > 0$ with respect to $x$ is $x < 15$, it follows that $m < 0$ and $\\frac { n } { m } = 15$. Solving for $n$, we get $n = 15 m$. Therefore, the inequality $( - 5 m + n ) x > n + 5 m$ with respect to $x$ can be simplified to $( - 5 m + 15 m ) x > 15 m + 5 m$. Thus, $10 mx > 20 m$. Since $m < 0$, we have $10 m < 0$, which implies $x < 2$.", "expr_cands": ["x", "3 x ^ { 2 n + 5 } - 2 = \\frac { 1 } { 3 }", "n", "2 n + 5 = 1", "n = - 2"], "exprs": ["2 n + 5 = 1", "n = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x ^ { 2 n + 5 } - 2 = \\frac { 1 } { 3 }"}, {"id": "2 n + 5 = 1"}, {"id": "关于 $x$ 的方程 $3 x ^ { 2 n + 5 } - 2 = \\frac { 1 } { 3 }$ 是一元一次方程"}, {"id": "n = - 2"}], "links": [{"rel": "被描述", "source": "3 x ^ { 2 n + 5 } - 2 = \\frac { 1 } { 3 }", "target": "2 n + 5 = 1"}, {"rel": "等式方程求解", "source": "2 n + 5 = 1", "target": "n = - 2"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $3 x ^ { 2 n + 5 } - 2 = \\frac { 1 } { 3 }$ 是一元一次方程", "target": "2 n + 5 = 1"}]}}
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{"content": "Given the proportional function $y = ( k - 1 ) x$, if the value of $y$ decreases as the value of the independent variable $x$ increases, then the range of possible values for $k$ is ____?", "answer": "10", "steps": "$\\because$ In the proportional function $y = ( k - 1 ) x$, the value of $y$ decreases as the value of the independent variable $x$ increases. $\\therefore$ $k - 1 < 0$, and solving for $k$, we get $k < 1$.", "expr_cands": ["m", "a", "b", "c", "a = 4", "b = 5", "c = 8", "\\frac { a } { b } = \\frac { c } { m }", "\\frac { 4 } { 5 } = \\frac { 8 } { m }", "m = 10"], "exprs": ["\\frac { a } { b } = \\frac { c } { m }", "\\frac { 4 } { 5 } = \\frac { 8 } { m }", "m = 10"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m"}, {"id": "\\frac { a } { b } = \\frac { c } { m }"}, {"id": "a"}, {"id": "b"}, {"id": "c"}, {"id": "线段 $m$ 是线段 $a$ , $b$ , $c$ 的第四比例项"}, {"id": "a = 4"}, {"id": "\\frac { 4 } { 5 } = \\frac { 8 } { m }"}, {"id": "b = 5"}, {"id": "c = 8"}, {"id": "m = 10"}], "links": [{"rel": "被描述", "source": "m", "target": "\\frac { a } { b } = \\frac { c } { m }"}, {"rel": "被代入", "source": "\\frac { a } { b } = \\frac { c } { m }", "target": "\\frac { 4 } { 5 } = \\frac { 8 } { m }"}, {"rel": "被描述", "source": "a", "target": "\\frac { a } { b } = \\frac { c } { m }"}, {"rel": "被描述", "source": "b", "target": "\\frac { a } { b } = \\frac { c } { m }"}, {"rel": "被描述", "source": "c", "target": "\\frac { a } { b } = \\frac { c } { m }"}, {"rel": "限制性描述", "source": "线段 $m$ 是线段 $a$ , $b$ , $c$ 的第四比例项", "target": "\\frac { a } { b } = \\frac { c } { m }"}, {"rel": "代入", "source": "a = 4", "target": "\\frac { 4 } { 5 } = \\frac { 8 } { m }"}, {"rel": "等式方程求解", "source": "\\frac { 4 } { 5 } = \\frac { 8 } { m }", "target": "m = 10"}, {"rel": "代入", "source": "b = 5", "target": "\\frac { 4 } { 5 } = \\frac { 8 } { m }"}, {"rel": "代入", "source": "c = 8", "target": "\\frac { 4 } { 5 } = \\frac { 8 } { m }"}]}}
|
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{"content": "Given: $a ^ m = 3$, $a ^ n = 5$, then the value of $a ^ { 3 m - 2 n }$ is ____?", "answer": "- 2", "steps": "Because $a ^ { m } = 3$ and $a ^ { n } = 5$, therefore $a ^ { 3 m - 2 n } = ( a ^ { m } ) ^ { 3 } \\div ( a ^ { n } ) ^ { 2 } = 3 ^ { 3 } \\div 5 ^ { 2 } = \\frac { 27 } { 25 }$.", "expr_cands": ["a + b = - 1", "a", "b", "3 a ^ { 2 } + 6 ab + 3 b ^ { 2 } - 5", "3 ( a + b ) ^ { 2 } - 5", "- 2"], "exprs": ["3 ( a + b ) ^ { 2 } - 5", "- 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a ^ { 2 } + 6 ab + 3 b ^ { 2 } - 5"}, {"id": "3 ( a + b ) ^ { 2 } - 5"}, {"id": "a + b = - 1"}, {"id": "- 2"}], "links": [{"rel": "提取因式", "source": "3 a ^ { 2 } + 6 ab + 3 b ^ { 2 } - 5", "target": "3 ( a + b ) ^ { 2 } - 5"}, {"rel": "被代入", "source": "3 ( a + b ) ^ { 2 } - 5", "target": "- 2"}, {"rel": "提取因式参考", "source": "a + b = - 1", "target": "3 ( a + b ) ^ { 2 } - 5"}, {"rel": "代入", "source": "a + b = - 1", "target": "- 2"}]}}
|
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{"content": "If the equation $3 x ^ { 2 n + 5 } - 2 = \\frac { 1 } { 3 }$ is a linear equation in one variable $x$, then $n$ = ____?", "answer": "4", "steps": "According to the problem, we have $2 n + 5 = 1$, and solving for $n$ gives $n = - 2$.", "expr_cands": ["( 2 m - 4 )", "m", "( 3 m - 1 )", "2 m - 4 + 3 m - 1 = 0", "m = 1", "2 m - 4", "- 2", "{ ( - 2 ) } ^ { 2 }", "4"], "exprs": ["2 m - 4 + 3 m - 1 = 0", "m = 1", "- 2", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 2 m - 4 )"}, {"id": "2 m - 4 + 3 m - 1 = 0"}, {"id": "( 3 m - 1 )"}, {"id": "某个数的平方根是 $( 2 m - 4 )$ 与 $( 3 m - 1 )$"}, {"id": "平方根互为相反数"}, {"id": "m = 1"}, {"id": "2 m - 4"}, {"id": "- 2"}, {"id": "4"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "( 2 m - 4 )", "target": "2 m - 4 + 3 m - 1 = 0"}, {"rel": "等式方程求解", "source": "2 m - 4 + 3 m - 1 = 0", "target": "m = 1"}, {"rel": "被描述", "source": "( 3 m - 1 )", "target": "2 m - 4 + 3 m - 1 = 0"}, {"rel": "限制性描述", "source": "某个数的平方根是 $( 2 m - 4 )$ 与 $( 3 m - 1 )$", "target": "2 m - 4 + 3 m - 1 = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "2 m - 4 + 3 m - 1 = 0"}, {"rel": "代入", "source": "m = 1", "target": "- 2"}, {"rel": "被代入", "source": "2 m - 4", "target": "- 2"}, {"rel": "被描述", "source": "- 2", "target": "4"}, {"rel": "限制性描述", "source": "平方", "target": "4"}]}}
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{"content": "If segment $m$ is the fourth proportional to segments $a$, $b$, and $c$, and $a = 4$, $b = 5$, and $c = 8$, then the length of segment $m$ is _____.", "answer": "7", "steps": "Since line segment $m$ is the fourth proportional to line segments $a$, $b$, and $c$, we have $\\frac { a } { b } = \\frac { c } { m }$. Given $a = 4$, $b = 5$, and $c = 8$, we can solve for $m$ to get $m = 10$.", "expr_cands": ["m ^ { 2 } + 3 n - 1", "m", "n", "4", "2 m ^ { 2 } + 6 n - 3", "m ^ { 2 } + 3 n - 1 = 4", "m ^ { 2 } + 3 n = 5", "2 ( m ^ { 2 } + 3 n ) - 3", "7"], "exprs": ["m ^ { 2 } + 3 n - 1 = 4", "m ^ { 2 } + 3 n = 5", "2 ( m ^ { 2 } + 3 n ) - 3", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m ^ { 2 } + 3 n - 1"}, {"id": "m ^ { 2 } + 3 n - 1 = 4"}, {"id": "4"}, {"id": "$m ^ { 2 } + 3 n - 1$ 的值为 $4$"}, {"id": "m ^ { 2 } + 3 n = 5"}, {"id": "2 m ^ { 2 } + 6 n - 3"}, {"id": "2 ( m ^ { 2 } + 3 n ) - 3"}, {"id": "7"}], "links": [{"rel": "被描述", "source": "m ^ { 2 } + 3 n - 1", "target": "m ^ { 2 } + 3 n - 1 = 4"}, {"rel": "移项", "source": "m ^ { 2 } + 3 n - 1 = 4", "target": "m ^ { 2 } + 3 n = 5"}, {"rel": "被描述", "source": "4", "target": "m ^ { 2 } + 3 n - 1 = 4"}, {"rel": "限制性描述", "source": "$m ^ { 2 } + 3 n - 1$ 的值为 $4$", "target": "m ^ { 2 } + 3 n - 1 = 4"}, {"rel": "提取因式参考", "source": "m ^ { 2 } + 3 n = 5", "target": "2 ( m ^ { 2 } + 3 n ) - 3"}, {"rel": "代入", "source": "m ^ { 2 } + 3 n = 5", "target": "7"}, {"rel": "提取因式", "source": "2 m ^ { 2 } + 6 n - 3", "target": "2 ( m ^ { 2 } + 3 n ) - 3"}, {"rel": "被代入", "source": "2 ( m ^ { 2 } + 3 n ) - 3", "target": "7"}]}}
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{"content": "If $a + b = - 1$, then the value of $3 a ^ 2 + 6 ab + 3 b ^ 2 - 5$ is ____?", "answer": "- 2", "steps": "Since $a + b = - 1$, therefore $3 a ^ 2 + 6 ab + 3 b ^ 2 - 5 = 3 ( a + b ) ^ 2 - 5 = 3 * ( - 1 ) ^ 2 - 5 = 3 - 5 = - 2$.", "expr_cands": ["| a + 2 | = 0", "a", "a = - 2", "a + 2 = 0"], "exprs": ["a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| a + 2 | = 0"}, {"id": "a = - 2"}, {"id": "绝对值恒大于等于0"}], "links": [{"rel": "被描述", "source": "| a + 2 | = 0", "target": "a = - 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a = - 2"}]}}
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{"content": "If the square root of a certain number is $( 2 m - 4 )$ and $( 3 m - 1 )$, then the number is ____?", "answer": "2", "steps": "According to the problem, we have $2 m - 4 + 3 m - 1 = 0$. Solving for $m$, we get $m = 1$. Therefore, $2 m - 4 = - 2$, which means the number is equal to ${( - 2 )} ^ 2 = 4$.", "expr_cands": ["\\sqrt { 75 }", "\\sqrt { m + 1 }", "m", "5 \\sqrt { 3 }", "m + 1 = 3", "m = 2"], "exprs": ["5 \\sqrt { 3 }", "m + 1 = 3", "m = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 75 }"}, {"id": "5 \\sqrt { 3 }"}, {"id": "\\sqrt { m + 1 }"}, {"id": "m + 1 = 3"}, {"id": "$\\sqrt { 75 }$ 与最简二次根式 $\\sqrt { m + 1 }$ 是同类二次根式"}, {"id": "m = 2"}], "links": [{"rel": "计算", "source": "\\sqrt { 75 }", "target": "5 \\sqrt { 3 }"}, {"rel": "被描述", "source": "5 \\sqrt { 3 }", "target": "m + 1 = 3"}, {"rel": "被描述", "source": "\\sqrt { m + 1 }", "target": "m + 1 = 3"}, {"rel": "等式方程求解", "source": "m + 1 = 3", "target": "m = 2"}, {"rel": "限制性描述", "source": "$\\sqrt { 75 }$ 与最简二次根式 $\\sqrt { m + 1 }$ 是同类二次根式", "target": "m + 1 = 3"}]}}
|
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{"content": "If the value of $m ^ 2 + 3 n - 1$ is $4$, then the value of the algebraic expression $2 m ^ 2 + 6 n - 3$ is ____?", "answer": "27", "steps": "From the given condition, we have $m ^ 2 + 3 n - 1 = 4$, which implies $m ^ 2 + 3 n = 5$. Therefore, the original expression is equal to $2 ( m ^ 2 + 3 n ) - 3 = 10 - 3 = 7$.", "expr_cands": ["x ^ { m - 1 } + 2 x + 8", "x", "m", "m ^ { 3 }", "m - 1 = 2", "m = 3", "27"], "exprs": ["m - 1 = 2", "m = 3", "27"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { m - 1 } + 2 x + 8"}, {"id": "m - 1 = 2"}, {"id": "多项式 $x ^ { m - 1 } + 2 x + 8$ 是关于 $x$ 的二次三项式"}, {"id": "m = 3"}, {"id": "m ^ { 3 }"}, {"id": "27"}], "links": [{"rel": "被描述", "source": "x ^ { m - 1 } + 2 x + 8", "target": "m - 1 = 2"}, {"rel": "等式方程求解", "source": "m - 1 = 2", "target": "m = 3"}, {"rel": "限制性描述", "source": "多项式 $x ^ { m - 1 } + 2 x + 8$ 是关于 $x$ 的二次三项式", "target": "m - 1 = 2"}, {"rel": "代入", "source": "m = 3", "target": "27"}, {"rel": "被代入", "source": "m ^ { 3 }", "target": "27"}]}}
|
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{"content": "If $| a + 2 | = 0$, then the value of $a$ is ____?", "answer": "m < - 1", "steps": "Since $| a + 2 | = 0$, therefore $a + 2 = 0$, which implies that $a = - 2$.", "expr_cands": ["x", "\\sqrt { x + 1 } = 1 + m", "m", "1 + m < 0", "m < - 1"], "exprs": ["1 + m < 0", "m < - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + 1 } = 1 + m"}, {"id": "1 + m < 0"}, {"id": "关于 $x$ 的无理方程 $\\sqrt { x + 1 } = 1 + m$ 没有实数解"}, {"id": "即方程 $\\sqrt { x + 1 } = 1 + m$ 无解"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "m < - 1"}], "links": [{"rel": "被描述", "source": "\\sqrt { x + 1 } = 1 + m", "target": "1 + m < 0"}, {"rel": "不等式方程求解", "source": "1 + m < 0", "target": "m < - 1"}, {"rel": "限制性描述", "source": "关于 $x$ 的无理方程 $\\sqrt { x + 1 } = 1 + m$ 没有实数解", "target": "1 + m < 0"}, {"rel": "限制性描述", "source": "即方程 $\\sqrt { x + 1 } = 1 + m$ 无解", "target": "1 + m < 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "1 + m < 0"}]}}
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{"content": "If $\\sqrt { 75 }$ and the simplest quadratic radical $\\sqrt { m + 1 }$ are of the same type, then the value of $m$ is ____?", "answer": "200", "steps": "$\\because \\sqrt { 75 } = 5 \\sqrt { 3 }$ and the simplest quadratic radical $\\sqrt { m + 1 }$ are of the same type of quadratic radicals, $\\therefore m + 1 = 3$, solving for $m$, we get: $m = 2$.", "expr_cands": ["95 ^ { 2 } + 190 * 5 + 5 ^ { 2 } = m + 99 ^ { 2 } - 1", "m", "m = 200", "( 95 + 5 ) ^ { 2 } = m + ( 100 - 1 ) ^ { 2 } - 1", "100 ^ { 2 } = m + 100 ^ { 2 } - 200 + 1 - 1", "100 ^ { 2 } = m + 100 ^ { 2 } - 200"], "exprs": ["m = 200"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "95 ^ { 2 } + 190 * 5 + 5 ^ { 2 } = m + 99 ^ { 2 } - 1"}, {"id": "m = 200"}], "links": [{"rel": "等式方程求解", "source": "95 ^ { 2 } + 190 * 5 + 5 ^ { 2 } = m + 99 ^ { 2 } - 1", "target": "m = 200"}]}}
|
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{"content": "Given that the polynomial $x ^ { m - 1 } + 2 x + 8$ is a quadratic trinomial in terms of $x$, then $m ^ 3$ = ____?", "answer": "7", "steps": "Since $x ^ { m - 1 } + 2 x + 8$ is a quadratic trinomial in terms of $x$, we have $m - 1 = 2$. Therefore, $m = 3$. Thus, $m ^ 3 = 3 ^ 3 = 27$.", "expr_cands": ["m - 2 n = - 2", "n", "m", "3 - 2 m + 4 n", "3 - 2 ( m - 2 n )", "7"], "exprs": ["3 - 2 ( m - 2 n )", "7"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 - 2 m + 4 n"}, {"id": "3 - 2 ( m - 2 n )"}, {"id": "m - 2 n = - 2"}, {"id": "7"}], "links": [{"rel": "提取因式", "source": "3 - 2 m + 4 n", "target": "3 - 2 ( m - 2 n )"}, {"rel": "被代入", "source": "3 - 2 ( m - 2 n )", "target": "7"}, {"rel": "提取因式参考", "source": "m - 2 n = - 2", "target": "3 - 2 ( m - 2 n )"}, {"rel": "代入", "source": "m - 2 n = - 2", "target": "7"}]}}
|
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{"content": "If the irrational equation about $x$, $\\sqrt { x + 1 } = 1 + m$, has no real solutions, then the range of values for $m$ is ____?", "answer": "- 2", "steps": "As the equation has no real roots, i.e. the equation $\\sqrt { x + 1 } = 1 + m$ has no solution, $1 + m < 0$. Solving for $m$, we get $m < - 1$. Therefore, the range of $m$ is $m < - 1$.", "expr_cands": ["( a + 2 ) ^ { 0 }", "a", "a + 2 = 0", "a = - 2"], "exprs": ["a + 2 = 0", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a + 2 ) ^ { 0 }"}, {"id": "a + 2 = 0"}, {"id": "$( a + 2 ) ^ { 0 }$ 无意义"}, {"id": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0"}, {"id": "a = - 2"}], "links": [{"rel": "被描述", "source": "( a + 2 ) ^ { 0 }", "target": "a + 2 = 0"}, {"rel": "等式方程求解", "source": "a + 2 = 0", "target": "a = - 2"}, {"rel": "限制性描述", "source": "$( a + 2 ) ^ { 0 }$ 无意义", "target": "a + 2 = 0"}, {"rel": "属性描述", "source": "多项式零次方项，若在实数范围内有意义，则底数不为0，无意义则底数为0", "target": "a + 2 = 0"}]}}
|
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{"content": "If $95 ^ { 2 } + 190 * 5 + 5 ^ { 2 } = m + 99 ^ { 2 } - 1$, then the value of $m$ is ____?", "answer": "- 8", "steps": "Because $95 ^ 2 + 190 * 5 + 5 ^ 2 = m + 99 ^ 2 - 1$, therefore $( 95 + 5 ) ^ 2 = m + ( 100 - 1 ) ^ 2 - 1$, therefore $100 ^ 2 = m + 100 ^ 2 - 200 + 1 - 1$, therefore $100 ^ 2 = m + 100 ^ 2 - 200$, therefore $m = 200$.", "expr_cands": ["\\sqrt { x + 2 } + | y - 3 | = 0", "y", "x", "x ^ { y }", "x + 2 = 0", "x = - 2", "y - 3 = 0", "y = 3", "- 8"], "exprs": ["x + 2 = 0", "y - 3 = 0", "x = - 2", "y = 3", "- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { x + 2 } + | y - 3 | = 0"}, {"id": "x + 2 = 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x = - 2"}, {"id": "y - 3 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "y = 3"}, {"id": "x ^ { y }"}, {"id": "- 8"}], "links": [{"rel": "被描述", "source": "\\sqrt { x + 2 } + | y - 3 | = 0", "target": "x + 2 = 0"}, {"rel": "被描述", "source": "\\sqrt { x + 2 } + | y - 3 | = 0", "target": "y - 3 = 0"}, {"rel": "等式方程求解", "source": "x + 2 = 0", "target": "x = - 2"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x + 2 = 0"}, {"rel": "代入", "source": "x = - 2", "target": "- 8"}, {"rel": "等式方程求解", "source": "y - 3 = 0", "target": "y = 3"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "y - 3 = 0"}, {"rel": "代入", "source": "y = 3", "target": "- 8"}, {"rel": "被代入", "source": "x ^ { y }", "target": "- 8"}]}}
|
||||
{"content": "Given $m - 2 n = - 2$, what is the value of $3 - 2 m + 4 n$?", "answer": "2", "steps": "Since $m - 2 n = - 2$, therefore $3 - 2 m + 4 n = 3 - 2 ( m - 2 n ) = 3 - 2 * ( - 2 ) = 7$.", "expr_cands": ["x = 3", "x", "a ( x - 1 ) = 3 x - 5", "a", "2 a = 4", "2 a = 9 - 5", "a = 2"], "exprs": ["2 a = 9 - 5", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a ( x - 1 ) = 3 x - 5"}, {"id": "2 a = 9 - 5"}, {"id": "x = 3"}, {"id": "a = 2"}], "links": [{"rel": "被代入", "source": "a ( x - 1 ) = 3 x - 5", "target": "2 a = 9 - 5"}, {"rel": "等式方程求解", "source": "2 a = 9 - 5", "target": "a = 2"}, {"rel": "代入", "source": "x = 3", "target": "2 a = 9 - 5"}]}}
|
||||
{"content": "If $( a + 2 ) ^ 0$ is undefined, then $a$ = ____?", "answer": "20", "steps": "Since $( a + 2 ) ^ 0$ is undefined, it follows that $a + 2 = 0$, and therefore $a = - 2$.", "expr_cands": ["a", "b", "c", "d", "a = 2", "b = 8", "c = 5", "a : b = c : d", "ad = bc", "2 d = 8 * 5", "d = 20"], "exprs": ["a : b = c : d", "ad = bc", "2 d = 8 * 5", "d = 20"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a"}, {"id": "a : b = c : d"}, {"id": "b"}, {"id": "c"}, {"id": "d"}, {"id": "$a$ , $b$ , $c$ , $d$ 是成比例线段"}, {"id": "ad = bc"}, {"id": "a = 2"}, {"id": "2 d = 8 * 5"}, {"id": "b = 8"}, {"id": "c = 5"}, {"id": "d = 20"}], "links": [{"rel": "被描述", "source": "a", "target": "a : b = c : d"}, {"rel": "同乘除", "source": "a : b = c : d", "target": "ad = bc"}, {"rel": "被描述", "source": "b", "target": "a : b = c : d"}, {"rel": "被描述", "source": "c", "target": "a : b = c : d"}, {"rel": "被描述", "source": "d", "target": "a : b = c : d"}, {"rel": "限制性描述", "source": "$a$ , $b$ , $c$ , $d$ 是成比例线段", "target": "a : b = c : d"}, {"rel": "被代入", "source": "ad = bc", "target": "2 d = 8 * 5"}, {"rel": "代入", "source": "a = 2", "target": "2 d = 8 * 5"}, {"rel": "等式方程求解", "source": "2 d = 8 * 5", "target": "d = 20"}, {"rel": "代入", "source": "b = 8", "target": "2 d = 8 * 5"}, {"rel": "代入", "source": "c = 5", "target": "2 d = 8 * 5"}]}}
|
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{"content": "Given $\\sqrt { x + 2 } + | y - 3 | = 0$, what is $x ^ { y }$?", "answer": "3", "steps": "Because $\\sqrt { x + 2 } + | y - 3 | = 0$, therefore $x + 2 = 0$, which solves to $x = - 2$; $y - 3 = 0$, which solves to $y = 3$. Therefore, $x ^ y = ( - 2 ) ^ 3 = - 8$.", "expr_cands": ["\\sqrt { 12 n }", "n", "2 \\sqrt { 3 n }", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 12 n }"}, {"id": "3"}, {"id": "$\\sqrt { 12 n }$ 的结果为正整数"}, {"id": "正整数 $n$ 的最小值"}, {"id": "正整数 $n$ 的最小值是 $3$ 时"}, {"id": "$\\sqrt { 12 n }$ 是正整数"}], "links": [{"rel": "被描述", "source": "\\sqrt { 12 n }", "target": "3"}, {"rel": "限制性描述", "source": "$\\sqrt { 12 n }$ 的结果为正整数", "target": "3"}, {"rel": "限制性描述", "source": "正整数 $n$ 的最小值", "target": "3"}, {"rel": "限制性描述", "source": "正整数 $n$ 的最小值是 $3$ 时", "target": "3"}, {"rel": "限制性描述", "source": "$\\sqrt { 12 n }$ 是正整数", "target": "3"}]}}
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{"content": "Given that $x = 3$ is a solution to the equation $a ( x - 1 ) = 3 x - 5$ in terms of $x$, what is the value of $a$?", "answer": "- 2", "steps": "Substituting $x = 3$ into $a ( x - 1 ) = 3 x - 5$ yields $2 a = 9 - 5$, which solves for $a = 2$.", "expr_cands": ["x = - 3", "x", "a - 5 x = 13", "a", "a + 15 = 13", "a = - 2"], "exprs": ["a + 15 = 13", "a = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a - 5 x = 13"}, {"id": "a + 15 = 13"}, {"id": "x = - 3"}, {"id": "$x = - 3$ 是方程 $a - 5 x = 13$ 的解"}, {"id": "a = - 2"}], "links": [{"rel": "被描述", "source": "a - 5 x = 13", "target": "a + 15 = 13"}, {"rel": "等式方程求解", "source": "a + 15 = 13", "target": "a = - 2"}, {"rel": "被描述", "source": "x = - 3", "target": "a + 15 = 13"}, {"rel": "限制性描述", "source": "$x = - 3$ 是方程 $a - 5 x = 13$ 的解", "target": "a + 15 = 13"}]}}
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{"content": "Given $a$, $b$, $c$, $d$ are proportional line segments and $a = 2$, $b = 8$, $c = 5$, then $d$ = ____?", "answer": "- 3", "steps": "Since $a : b = c : d$, it follows that $ad = bc$. Given that $a = 2$, $b = 8$, and $c = 5$, we can conclude that $2 d = 8 \\cdot 5$, and therefore $d = 20$.", "expr_cands": ["y ^ { 2 } + 2 y + 7", "y", "6", "2 y ^ { 2 } + 4 y - 1", "y ^ { 2 } + 2 y + 7 = 6", "y = - 1", "y ^ { 2 } + 2 y = - 1", "2 ( y ^ { 2 } + 2 y ) - 1", "- 3"], "exprs": ["y ^ { 2 } + 2 y + 7 = 6", "y ^ { 2 } + 2 y = - 1", "2 ( y ^ { 2 } + 2 y ) - 1", "- 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "y ^ { 2 } + 2 y + 7"}, {"id": "y ^ { 2 } + 2 y + 7 = 6"}, {"id": "6"}, {"id": "代数式 $y ^ { 2 } + 2 y + 7$ 的值是 $6$"}, {"id": "y ^ { 2 } + 2 y = - 1"}, {"id": "2 y ^ { 2 } + 4 y - 1"}, {"id": "2 ( y ^ { 2 } + 2 y ) - 1"}, {"id": "- 3"}], "links": [{"rel": "被描述", "source": "y ^ { 2 } + 2 y + 7", "target": "y ^ { 2 } + 2 y + 7 = 6"}, {"rel": "移项", "source": "y ^ { 2 } + 2 y + 7 = 6", "target": "y ^ { 2 } + 2 y = - 1"}, {"rel": "被描述", "source": "6", "target": "y ^ { 2 } + 2 y + 7 = 6"}, {"rel": "限制性描述", "source": "代数式 $y ^ { 2 } + 2 y + 7$ 的值是 $6$", "target": "y ^ { 2 } + 2 y + 7 = 6"}, {"rel": "提取因式参考", "source": "y ^ { 2 } + 2 y = - 1", "target": "2 ( y ^ { 2 } + 2 y ) - 1"}, {"rel": "代入", "source": "y ^ { 2 } + 2 y = - 1", "target": "- 3"}, {"rel": "提取因式", "source": "2 y ^ { 2 } + 4 y - 1", "target": "2 ( y ^ { 2 } + 2 y ) - 1"}, {"rel": "被代入", "source": "2 ( y ^ { 2 } + 2 y ) - 1", "target": "- 3"}]}}
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{"content": "If the result of $\\sqrt { 12 n }$ is a positive integer, what is the smallest positive integer value of $n$?", "answer": "4", "steps": "$\\sqrt { 12 n } = 2 \\sqrt { 3 n }$ means that the smallest positive integer value of $n$ is $3$ and $\\sqrt { 12 n }$ is a positive integer.", "expr_cands": ["( 4 x - a ) ( x + 1 )", "a", "x", "4 x ^ { 2 } + ( 4 - a ) x - a", "4 - a = 0", "a = 4"], "exprs": ["4 x ^ { 2 } + ( 4 - a ) x - a", "4 - a = 0", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 4 x - a ) ( x + 1 )"}, {"id": "4 x ^ { 2 } + ( 4 - a ) x - a"}, {"id": "x"}, {"id": "4 - a = 0"}, {"id": "要使 $( 4 x - a ) ( x + 1 )$ 的积中不含有 $x$ 的一次项"}, {"id": "a = 4"}], "links": [{"rel": "提取因式", "source": "( 4 x - a ) ( x + 1 )", "target": "4 x ^ { 2 } + ( 4 - a ) x - a"}, {"rel": "被描述", "source": "4 x ^ { 2 } + ( 4 - a ) x - a", "target": "4 - a = 0"}, {"rel": "提取因式参考", "source": "x", "target": "4 x ^ { 2 } + ( 4 - a ) x - a"}, {"rel": "等式方程求解", "source": "4 - a = 0", "target": "a = 4"}, {"rel": "限制性描述", "source": "要使 $( 4 x - a ) ( x + 1 )$ 的积中不含有 $x$ 的一次项", "target": "4 - a = 0"}]}}
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{"content": "If $x = - 3$ is a solution of the equation $a - 5 x = 13$, then the value of $a$ is ____?", "answer": "1", "steps": "Substituting $x = - 3$ into the equation $a - 5 x = 13$ yields $a + 15 = 13$. Solving for $a$ gives $a = - 2$.", "expr_cands": ["( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ { 3 } - 1", "a", "b", "x", "\\frac { | a | } { | b | }", "| a | - | b | + 5 = 5", "a - b \\neq 0", "| a | = | b |", "a = - b", "1"], "exprs": ["| a | - | b | + 5 = 5", "a - b \\neq 0", "| a | = | b |", "a = - b", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ { 3 } - 1"}, {"id": "| a | - | b | + 5 = 5"}, {"id": "$( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ { 3 } - 1$ 是关于 $x$ 的五次三项式"}, {"id": "a - b \\neq 0"}, {"id": "| a | = | b |"}, {"id": "a = - b"}, {"id": "\\frac { | a | } { | b | }"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ { 3 } - 1", "target": "| a | - | b | + 5 = 5"}, {"rel": "被描述", "source": "( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ { 3 } - 1", "target": "a - b \\neq 0"}, {"rel": "移项", "source": "| a | - | b | + 5 = 5", "target": "| a | = | b |"}, {"rel": "限制性描述", "source": "$( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ { 3 } - 1$ 是关于 $x$ 的五次三项式", "target": "| a | - | b | + 5 = 5"}, {"rel": "限制性描述", "source": "$( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ { 3 } - 1$ 是关于 $x$ 的五次三项式", "target": "a - b \\neq 0"}, {"rel": "联立", "source": "a - b \\neq 0", "target": "a = - b"}, {"rel": "联立", "source": "| a | = | b |", "target": "a = - b"}, {"rel": "代入", "source": "a = - b", "target": "1"}, {"rel": "被代入", "source": "\\frac { | a | } { | b | }", "target": "1"}]}}
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{"content": "The value of the algebraic expression $y ^ { 2 } + 2 y + 7$ is $6$, then the value of $2 y ^ { 2 } + 4 y - 1$ is ____?", "answer": "3", "steps": "From $y ^ { 2 } + 2 y + 7 = 6$, we get $y ^ { 2 } + 2 y = - 1$. Therefore, the original expression is equal to $2 ( y ^ { 2 } + 2 y ) - 1 = - 2 - 1 = - 3$.", "expr_cands": ["3 x + 2 = 11", "x", "x = 3", "3 x = 11 - 2", "1"], "exprs": ["x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + 2 = 11"}, {"id": "x = 3"}], "links": [{"rel": "等式方程求解", "source": "3 x + 2 = 11", "target": "x = 3"}]}}
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{"content": "To make the product of $( 4 x - a ) ( x + 1 )$ not contain the linear term $x$, $a$ should be equal to ____?", "answer": "- 1", "steps": "$( 4 x - a ) ( x + 1 ) = 4 x ^ 2 + 4 x - ax - a = 4 x ^ 2 + ( 4 - a ) x - a$, because there is no linear term in $x$ in the product, therefore $4 - a = 0$, which solves for $a = 4$.", "expr_cands": ["m", "4", "a - 3", "a", "4 + a - 3 = 0", "a = - 1"], "exprs": ["4 + a - 3 = 0", "a = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "4"}, {"id": "4 + a - 3 = 0"}, {"id": "a - 3"}, {"id": "一个正数 $m$ 的两个平方根分别是 $4$ 和 $a - 3$"}, {"id": "a = - 1"}], "links": [{"rel": "被描述", "source": "4", "target": "4 + a - 3 = 0"}, {"rel": "等式方程求解", "source": "4 + a - 3 = 0", "target": "a = - 1"}, {"rel": "被描述", "source": "a - 3", "target": "4 + a - 3 = 0"}, {"rel": "限制性描述", "source": "一个正数 $m$ 的两个平方根分别是 $4$ 和 $a - 3$", "target": "4 + a - 3 = 0"}]}}
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{"content": "Given $( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ 3 - 1$ is a fifth degree trinomial in $x$, then $\\frac { | a | } { | b | }$ = ____?", "answer": "3", "steps": "$\\because$ $( a - b ) x ^ { | a | - | b | + 5 } + 3 x ^ 3 - 1$ is a fifth degree polynomial in $x$, $\\therefore$ $| a | - | b | + 5 = 5$, $a - b \\neq 0$, so $| a | = | b |$, $a = - b$, thus $\\frac { | a | } { | b | } = 1$.", "expr_cands": ["m = - 2", "m", "\\sqrt { { m } ^ { 2 } + 5 }", "3"], "exprs": ["3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { { m } ^ { 2 } + 5 }"}, {"id": "3"}, {"id": "m = - 2"}], "links": [{"rel": "被代入", "source": "\\sqrt { { m } ^ { 2 } + 5 }", "target": "3"}, {"rel": "代入", "source": "m = - 2", "target": "3"}]}}
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{"content": "If $3 x + 2 = 11$, then the value of $x$ is ____?", "answer": "\\frac { 1 } { 5 }", "steps": "$3 x + 2 = 11$, move the constant term to the right side, we get: $3 x = 11 - 2$. Divide both sides by the coefficient of $x$, which is $3$, we get: $x = 3$.", "expr_cands": ["5 m + 2", "m", "3", "5 m + 2 = 3", "m = \\frac { 1 } { 5 }"], "exprs": ["5 m + 2 = 3", "m = \\frac { 1 } { 5 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 m + 2"}, {"id": "5 m + 2 = 3"}, {"id": "3"}, {"id": "代数式 $5 m + 2$ 的值为 $3$"}, {"id": "m = \\frac { 1 } { 5 }"}], "links": [{"rel": "被描述", "source": "5 m + 2", "target": "5 m + 2 = 3"}, {"rel": "等式方程求解", "source": "5 m + 2 = 3", "target": "m = \\frac { 1 } { 5 }"}, {"rel": "被描述", "source": "3", "target": "5 m + 2 = 3"}, {"rel": "限制性描述", "source": "代数式 $5 m + 2$ 的值为 $3$", "target": "5 m + 2 = 3"}]}}
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{"content": "If the two square roots of a positive number $m$ are $4$ and $a - 3$, then $a$ = ____ ?", "answer": "a \\le 6", "steps": "$\\because$ A positive number $m$ has two square roots, which are $4$ and $a - 3$. $\\therefore$ $4 + a - 3 = 0$, solving for $a$ gives $a = - 1$.", "expr_cands": ["\\sqrt { ( { a - 6 } ) ^ { { 2 } } } { = 6 - a }", "a", "{ a }", "\\sqrt { ( a - 6 ) ^ { 2 } } = 6 - a", "6 - a \\ge 0", "a \\le 6"], "exprs": ["6 - a \\ge 0", "a \\le 6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { ( { a - 6 } ) ^ { { 2 } } } { = 6 - a }"}, {"id": "6 - a \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "${ a }$ 的取值范围"}, {"id": "a \\le 6"}], "links": [{"rel": "被描述", "source": "\\sqrt { ( { a - 6 } ) ^ { { 2 } } } { = 6 - a }", "target": "6 - a \\ge 0"}, {"rel": "不等式方程求解", "source": "6 - a \\ge 0", "target": "a \\le 6"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "6 - a \\ge 0"}, {"rel": "限制性描述", "source": "${ a }$ 的取值范围", "target": "6 - a \\ge 0"}]}}
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{"content": "When $m = - 2$, what is the value of the quadratic radical $\\sqrt { m ^ 2 + 5 }$?", "answer": "4", "steps": "When $m = - 2$, $\\sqrt { m ^ 2 + 5 } = \\sqrt {( - 2 ) ^ 2 + 5 } = \\sqrt { 4 + 5 } = \\sqrt { 9 } = 3$.", "expr_cands": ["a + 3", "a", "2 a - 15", "a + 3 + ( 2 a - 15 ) = 0", "a = 4"], "exprs": ["a + 3 + ( 2 a - 15 ) = 0", "a = 4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a + 3"}, {"id": "a + 3 + ( 2 a - 15 ) = 0"}, {"id": "2 a - 15"}, {"id": "一个正数的平方根是 $a + 3$ 和 $2 a - 15$"}, {"id": "平方根互为相反数"}, {"id": "a = 4"}], "links": [{"rel": "被描述", "source": "a + 3", "target": "a + 3 + ( 2 a - 15 ) = 0"}, {"rel": "等式方程求解", "source": "a + 3 + ( 2 a - 15 ) = 0", "target": "a = 4"}, {"rel": "被描述", "source": "2 a - 15", "target": "a + 3 + ( 2 a - 15 ) = 0"}, {"rel": "限制性描述", "source": "一个正数的平方根是 $a + 3$ 和 $2 a - 15$", "target": "a + 3 + ( 2 a - 15 ) = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "a + 3 + ( 2 a - 15 ) = 0"}]}}
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{"content": "If the value of the algebraic expression $5 m + 2$ is $3$, then $m$ = ____ ?", "answer": "4 x ^ { n }", "steps": "According to the problem, we have $5 m + 2 = 3$. Solving for $m$, we get $m = \\frac { 1 } { 5 }$.", "expr_cands": ["8 x ^ { 2 n } - 4 x ^ { n }", "x", "n", "4 x ^ { n } ( 2 x ^ { n } - 1 )", "4 x ^ { n }"], "exprs": ["4 x ^ { n }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "8 x ^ { 2 n } - 4 x ^ { n }"}, {"id": "4 x ^ { n }"}, {"id": "多项式 $8 x ^ { 2 n } - 4 x ^ { n }$ 的公因式"}], "links": [{"rel": "被描述", "source": "8 x ^ { 2 n } - 4 x ^ { n }", "target": "4 x ^ { n }"}, {"rel": "限制性描述", "source": "多项式 $8 x ^ { 2 n } - 4 x ^ { n }$ 的公因式", "target": "4 x ^ { n }"}]}}
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{"content": "If $\\sqrt {( a - 6 ) ^ 2 } = 6 - a$, then the possible values of $a$ are ____?", "answer": "2", "steps": "Because $\\sqrt {( a - 6 ) ^ 2 } = | a - 6 | = 6 - a$, therefore $6 - a \\ge 0$, which implies that $a \\le 6$.", "expr_cands": ["\\frac { a } { a - 2 }", "a", "a - 2 = 0", "a = 2"], "exprs": ["a - 2 = 0", "a = 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { a } { a - 2 }"}, {"id": "a - 2 = 0"}, {"id": "分式 $\\frac { a } { a - 2 }$ 无意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "a = 2"}], "links": [{"rel": "被描述", "source": "\\frac { a } { a - 2 }", "target": "a - 2 = 0"}, {"rel": "等式方程求解", "source": "a - 2 = 0", "target": "a = 2"}, {"rel": "限制性描述", "source": "分式 $\\frac { a } { a - 2 }$ 无意义", "target": "a - 2 = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "a - 2 = 0"}]}}
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{"content": "If the square root of a positive number is $a + 3$ and $2 a - 15$, then $a$ is ____?", "answer": "\\frac { 1 } { 2014 }", "steps": "According to the problem, the square root of a positive number is $a + 3$ and $2 a - 15$, so we have $a + 3 + ( 2 a - 15 ) = 0$. Solving for $a$, we get $a = 4$.", "expr_cands": ["x", "\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x }", "2014 - x \\ge 0", "x \\le 2014", "x - 2014 \\ge 0", "2014 \\le x", "x = 2014", "\\frac { 1 } { 2014 }"], "exprs": ["2014 - x \\ge 0", "x - 2014 \\ge 0", "x = 2014", "\\frac { 1 } { 2014 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x }"}, {"id": "2014 - x \\ge 0"}, {"id": "二次根式有意义，则根式恒大于等于0"}, {"id": "x - 2014 \\ge 0"}, {"id": "x = 2014"}, {"id": "\\frac { 1 } { 2014 }"}], "links": [{"rel": "被描述", "source": "\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x }", "target": "2014 - x \\ge 0"}, {"rel": "被描述", "source": "\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x }", "target": "x - 2014 \\ge 0"}, {"rel": "被代入", "source": "\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x }", "target": "\\frac { 1 } { 2014 }"}, {"rel": "联立", "source": "2014 - x \\ge 0", "target": "x = 2014"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "2014 - x \\ge 0"}, {"rel": "属性描述", "source": "二次根式有意义，则根式恒大于等于0", "target": "x - 2014 \\ge 0"}, {"rel": "联立", "source": "x - 2014 \\ge 0", "target": "x = 2014"}, {"rel": "代入", "source": "x = 2014", "target": "\\frac { 1 } { 2014 }"}]}}
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{"content": "The common factor of the polynomial $8 x ^ { 2 n } - 4 x ^ { n }$ is ____ ?", "answer": "- 1", "steps": "$8 x ^ { 2 n } - 4 x ^ { n } = 4 x ^ { n } ( 2 x ^ { n } - 1 )$ , \\therefore $4 x ^ { n }$ is a common factor.", "expr_cands": ["( 2 x + 3 y - 4 ) ( 2 x + ay + b )", "b", "x", "a", "y", "a - b", "4 x ^ { 2 } + ( 2 a + 6 ) xy + 3 ay ^ { 2 } + ( 3 b - 4 a ) y + ( 2 b - 8 ) x - 4 b", "3 b - 4 a = 0", "2 b - 8 = 0", "b = 4", "a = 3", "- 1"], "exprs": ["4 x ^ { 2 } + ( 2 a + 6 ) xy + 3 ay ^ { 2 } + ( 3 b - 4 a ) y + ( 2 b - 8 ) x - 4 b", "3 b - 4 a = 0", "2 b - 8 = 0", "b = 4", "a = 3", "- 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 2 x + 3 y - 4 ) ( 2 x + ay + b )"}, {"id": "4 x ^ { 2 } + ( 2 a + 6 ) xy + 3 ay ^ { 2 } + ( 3 b - 4 a ) y + ( 2 b - 8 ) x - 4 b"}, {"id": "3 b - 4 a = 0"}, {"id": "计算 $( 2 x + 3 y - 4 ) ( 2 x + ay + b )$ 得到的多项式不含一次项"}, {"id": "2 b - 8 = 0"}, {"id": "b = 4"}, {"id": "a = 3"}, {"id": "a - b"}, {"id": "- 1"}], "links": [{"rel": "展开", "source": "( 2 x + 3 y - 4 ) ( 2 x + ay + b )", "target": "4 x ^ { 2 } + ( 2 a + 6 ) xy + 3 ay ^ { 2 } + ( 3 b - 4 a ) y + ( 2 b - 8 ) x - 4 b"}, {"rel": "被描述", "source": "4 x ^ { 2 } + ( 2 a + 6 ) xy + 3 ay ^ { 2 } + ( 3 b - 4 a ) y + ( 2 b - 8 ) x - 4 b", "target": "3 b - 4 a = 0"}, {"rel": "被描述", "source": "4 x ^ { 2 } + ( 2 a + 6 ) xy + 3 ay ^ { 2 } + ( 3 b - 4 a ) y + ( 2 b - 8 ) x - 4 b", "target": "2 b - 8 = 0"}, {"rel": "联立", "source": "3 b - 4 a = 0", "target": "a = 3"}, {"rel": "限制性描述", "source": "计算 $( 2 x + 3 y - 4 ) ( 2 x + ay + b )$ 得到的多项式不含一次项", "target": "3 b - 4 a = 0"}, {"rel": "限制性描述", "source": "计算 $( 2 x + 3 y - 4 ) ( 2 x + ay + b )$ 得到的多项式不含一次项", "target": "2 b - 8 = 0"}, {"rel": "等式方程求解", "source": "2 b - 8 = 0", "target": "b = 4"}, {"rel": "联立", "source": "b = 4", "target": "a = 3"}, {"rel": "代入", "source": "b = 4", "target": "- 1"}, {"rel": "代入", "source": "a = 3", "target": "- 1"}, {"rel": "被代入", "source": "a - b", "target": "- 1"}]}}
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{"content": "If the fraction $\\frac { a } { a - 2 }$ is undefined, then $a$ = ____ ?", "answer": "- 8", "steps": "$\\because$ The fraction $\\frac { a } { a - 2 }$ is undefined, $\\therefore$ $a - 2 = 0$, solving for $a$ gives $a = 2$.", "expr_cands": ["| x + 4 | + | 2 - y | = 0", "y", "x", "xy", "x + 4 = 0", "x = - 4", "2 - y = 0", "y = 2", "- 8"], "exprs": ["x + 4 = 0", "2 - y = 0", "x = - 4", "y = 2", "- 8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "| x + 4 | + | 2 - y | = 0"}, {"id": "x + 4 = 0"}, {"id": "绝对值恒大于等于0"}, {"id": "2 - y = 0"}, {"id": "x = - 4"}, {"id": "y = 2"}, {"id": "xy"}, {"id": "- 8"}], "links": [{"rel": "被描述", "source": "| x + 4 | + | 2 - y | = 0", "target": "x + 4 = 0"}, {"rel": "被描述", "source": "| x + 4 | + | 2 - y | = 0", "target": "2 - y = 0"}, {"rel": "等式方程求解", "source": "x + 4 = 0", "target": "x = - 4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "x + 4 = 0"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "2 - y = 0"}, {"rel": "等式方程求解", "source": "2 - y = 0", "target": "y = 2"}, {"rel": "代入", "source": "x = - 4", "target": "- 8"}, {"rel": "代入", "source": "y = 2", "target": "- 8"}, {"rel": "被代入", "source": "xy", "target": "- 8"}]}}
|
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{"content": "For a rational number $x$, what is the value of $\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x }$?", "answer": "- \\frac { 5 } { 4 }", "steps": "Because $\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x }$ , therefore $2014 - x \\ge 0$ , $x - 2014 \\ge 0$ , therefore $x = 2014$ , therefore $\\sqrt { 2014 - x } + \\sqrt { x - 2014 } + \\frac { 1 } { x } = 0 + 0 + \\frac { 1 } { 2014 } = \\frac { 1 } { 2014 }$.", "expr_cands": ["\\frac { 1 } { 3 m } - \\frac { 1 } { 2 n } = 1", "m", "n", "\\frac { 4 n + 3 mn - 6 m } { 9 m + 6 mn - 6 n }", "2 n - 3 m = 6 mn", "\\frac { 2 ( 2 n - 3 m ) + 3 mn } { - 3 ( 2 n - 3 m ) + 6 mn }", "- \\frac { 5 } { 4 }"], "exprs": ["2 n - 3 m = 6 mn", "\\frac { 2 ( 2 n - 3 m ) + 3 mn } { - 3 ( 2 n - 3 m ) + 6 mn }", "- \\frac { 5 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { 1 } { 3 m } - \\frac { 1 } { 2 n } = 1"}, {"id": "2 n - 3 m = 6 mn"}, {"id": "\\frac { 4 n + 3 mn - 6 m } { 9 m + 6 mn - 6 n }"}, {"id": "\\frac { 2 ( 2 n - 3 m ) + 3 mn } { - 3 ( 2 n - 3 m ) + 6 mn }"}, {"id": "- \\frac { 5 } { 4 }"}], "links": [{"rel": "同乘除", "source": "\\frac { 1 } { 3 m } - \\frac { 1 } { 2 n } = 1", "target": "2 n - 3 m = 6 mn"}, {"rel": "提取因式参考", "source": "2 n - 3 m = 6 mn", "target": "\\frac { 2 ( 2 n - 3 m ) + 3 mn } { - 3 ( 2 n - 3 m ) + 6 mn }"}, {"rel": "代入", "source": "2 n - 3 m = 6 mn", "target": "- \\frac { 5 } { 4 }"}, {"rel": "提取因式", "source": "\\frac { 4 n + 3 mn - 6 m } { 9 m + 6 mn - 6 n }", "target": "\\frac { 2 ( 2 n - 3 m ) + 3 mn } { - 3 ( 2 n - 3 m ) + 6 mn }"}, {"rel": "被代入", "source": "\\frac { 2 ( 2 n - 3 m ) + 3 mn } { - 3 ( 2 n - 3 m ) + 6 mn }", "target": "- \\frac { 5 } { 4 }"}]}}
|
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{"content": "Calculate the polynomial obtained by $( 2 x + 3 y - 4 ) ( 2 x + ay + b )$, which does not contain linear terms. Here, $a$ and $b$ are constants. The value of $a - b$ is ____?", "answer": "x = 0", "steps": "$( 2 x + 3 y - 4 ) ( 2 x + ay + b ) = 4 x ^ 2 + 2 axy + 2 bx + 6 xy + 3 ay ^ 2 + 3 by - 8 x - 4 ay - 4 b = 4 x ^ 2 + ( 2 a + 6 ) xy + 3 ay ^ 2 + ( 3 b - 4 a ) y + ( 2 b - 8 ) x - 4 b$, because this polynomial does not contain a linear term, so $3 b - 4 a = 0$, $2 b - 8 = 0$, which gives $a = 3$, $b = 4$. Therefore, $a - b = 3 - 4 = - 1$.", "expr_cands": ["\\frac { x } { x - 3 }", "x", "x = 0", "x - 3 \\neq 0", "x \\neq 3"], "exprs": ["x = 0", "x - 3 \\neq 0", "x \\neq 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x } { x - 3 }"}, {"id": "x = 0"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "代数式 $\\frac { x } { x - 3 }$ 的值为零"}, {"id": "x - 3 \\neq 0"}, {"id": "x \\neq 3"}], "links": [{"rel": "被描述", "source": "\\frac { x } { x - 3 }", "target": "x = 0"}, {"rel": "被描述", "source": "\\frac { x } { x - 3 }", "target": "x - 3 \\neq 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "x - 3 \\neq 0"}, {"rel": "限制性描述", "source": "代数式 $\\frac { x } { x - 3 }$ 的值为零", "target": "x = 0"}, {"rel": "限制性描述", "source": "代数式 $\\frac { x } { x - 3 }$ 的值为零", "target": "x - 3 \\neq 0"}, {"rel": "不等式方程求解", "source": "x - 3 \\neq 0", "target": "x \\neq 3"}]}}
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{"content": "If $| x + 4 | + | 2 - y | = 0$, then $xy$ = ____ ?", "answer": "8", "steps": "From the given information, we have $x + 4 = 0$ and $2 - y = 0$. Solving for $x$ and $y$, we get $x = - 4$ and $y = 2$. Therefore, $xy = ( - 4 ) \\cdot ( 2 ) = - 8$.", "expr_cands": ["- 2 x ^ { a + 2 } y ^ { 3 }", "y", "a", "x", "5 x ^ { 4 } y ^ { b }", "b", "a ^ { b }", "a + 2 = 4", "a = 2", "b = 3", "8"], "exprs": ["a + 2 = 4", "b = 3", "a = 2", "8"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 2 x ^ { a + 2 } y ^ { 3 }"}, {"id": "a + 2 = 4"}, {"id": "5 x ^ { 4 } y ^ { b }"}, {"id": "单项式 $- 2 x ^ { a + 2 } y ^ { 3 }$ 与 $5 x ^ { 4 } y ^ { b }$ 是同类项"}, {"id": "a = 2"}, {"id": "b = 3"}, {"id": "a ^ { b }"}, {"id": "8"}], "links": [{"rel": "被描述", "source": "- 2 x ^ { a + 2 } y ^ { 3 }", "target": "a + 2 = 4"}, {"rel": "被描述", "source": "- 2 x ^ { a + 2 } y ^ { 3 }", "target": "b = 3"}, {"rel": "等式方程求解", "source": "a + 2 = 4", "target": "a = 2"}, {"rel": "被描述", "source": "5 x ^ { 4 } y ^ { b }", "target": "a + 2 = 4"}, {"rel": "被描述", "source": "5 x ^ { 4 } y ^ { b }", "target": "b = 3"}, {"rel": "限制性描述", "source": "单项式 $- 2 x ^ { a + 2 } y ^ { 3 }$ 与 $5 x ^ { 4 } y ^ { b }$ 是同类项", "target": "a + 2 = 4"}, {"rel": "限制性描述", "source": "单项式 $- 2 x ^ { a + 2 } y ^ { 3 }$ 与 $5 x ^ { 4 } y ^ { b }$ 是同类项", "target": "b = 3"}, {"rel": "代入", "source": "a = 2", "target": "8"}, {"rel": "代入", "source": "b = 3", "target": "8"}, {"rel": "被代入", "source": "a ^ { b }", "target": "8"}]}}
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{"content": "Given $\\frac { 1 } { 3 m } - \\frac { 1 } { 2 n } = 1$, find the value of $\\frac { 4 n + 3 mn - 6 m } { 9 m + 6 mn - 6 n }$.", "answer": "0", "steps": "When $\\frac { 1 } { 3 m } - \\frac { 1 } { 2 n } = 1$, therefore $2 n - 3 m = 6 mn$. Therefore, the original expression is $\\frac { 2 ( 2 n - 3 m ) + 3 mn } { - 3 ( 2 n - 3 m ) + 6 mn } = \\frac { 12 mn + 3 mn } { - 18 mn + 6 mn } = - \\frac { 5 } { 4 }$.", "expr_cands": ["2 x + y", "y", "x", "- 2", "( 2 x + y ) ^ { 2 } - 4", "2 x + y = - 2", "4 - 4", "0"], "exprs": ["2 x + y = - 2", "4 - 4", "0"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + y"}, {"id": "2 x + y = - 2"}, {"id": "- 2"}, {"id": "代数式 $2 x + y$ 的值是 $- 2$"}, {"id": "( 2 x + y ) ^ { 2 } - 4"}, {"id": "4 - 4"}, {"id": "0"}], "links": [{"rel": "被描述", "source": "2 x + y", "target": "2 x + y = - 2"}, {"rel": "代入", "source": "2 x + y = - 2", "target": "4 - 4"}, {"rel": "被描述", "source": "- 2", "target": "2 x + y = - 2"}, {"rel": "限制性描述", "source": "代数式 $2 x + y$ 的值是 $- 2$", "target": "2 x + y = - 2"}, {"rel": "被代入", "source": "( 2 x + y ) ^ { 2 } - 4", "target": "4 - 4"}, {"rel": "计算", "source": "4 - 4", "target": "0"}]}}
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{"content": "If the value of the algebraic expression $\\frac { x } { x - 3 }$ is zero, then the value of the real number $x$ is _____.", "answer": "- 2", "steps": "According to the problem, we have $x = 0$ and $x - 3 \\neq 0$, which implies that $x = 0$.", "expr_cands": ["( m - 2 ) x ^ { 3 } y ^ { | m | }", "y", "m", "x", "3 + | m | = 5", "m = - 2", "m = 2", "m - 2 \\neq 0", "m \\neq 2"], "exprs": ["3 + | m | = 5", "m - 2 \\neq 0", "m = - 2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( m - 2 ) x ^ { 3 } y ^ { | m | }"}, {"id": "3 + | m | = 5"}, {"id": "$( m - 2 ) x ^ { 3 } y ^ { | m | }$ 是关于 $x$ , $y$ 的五次单项式"}, {"id": "m - 2 \\neq 0"}, {"id": "m = - 2"}], "links": [{"rel": "被描述", "source": "( m - 2 ) x ^ { 3 } y ^ { | m | }", "target": "3 + | m | = 5"}, {"rel": "被描述", "source": "( m - 2 ) x ^ { 3 } y ^ { | m | }", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "3 + | m | = 5", "target": "m = - 2"}, {"rel": "限制性描述", "source": "$( m - 2 ) x ^ { 3 } y ^ { | m | }$ 是关于 $x$ , $y$ 的五次单项式", "target": "3 + | m | = 5"}, {"rel": "限制性描述", "source": "$( m - 2 ) x ^ { 3 } y ^ { | m | }$ 是关于 $x$ , $y$ 的五次单项式", "target": "m - 2 \\neq 0"}, {"rel": "联立", "source": "m - 2 \\neq 0", "target": "m = - 2"}]}}
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{"content": "If the monomial $- 2 x ^ { a + 2 } y ^ 3$ is a like term with $5 x ^ 4 y ^ b$, then the value of $a ^ b$ is ____?", "answer": "4", "steps": "$\\because$ The monomials $- 2 x ^ { a + 2 } y ^ 3$ and $5 x ^ 4 y ^ b$ are like terms, $\\therefore$ $a + 2 = 4$, $b = 3$, solving for $a$ and $b$ gives $a = 2$, $b = 3$, $\\therefore$ $a ^ b = 2 ^ 3 = 8$.", "expr_cands": ["a", "3", "| a - 1 |", "a + 3 = 0", "a = - 3", "4"], "exprs": ["a + 3 = 0", "a = - 3", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3"}, {"id": "a + 3 = 0"}, {"id": "a"}, {"id": "$a$ 是 $3$ 的相反数"}, {"id": "a = - 3"}, {"id": "| a - 1 |"}, {"id": "4"}], "links": [{"rel": "被描述", "source": "3", "target": "a + 3 = 0"}, {"rel": "等式方程求解", "source": "a + 3 = 0", "target": "a = - 3"}, {"rel": "被描述", "source": "a", "target": "a + 3 = 0"}, {"rel": "限制性描述", "source": "$a$ 是 $3$ 的相反数", "target": "a + 3 = 0"}, {"rel": "代入", "source": "a = - 3", "target": "4"}, {"rel": "被代入", "source": "| a - 1 |", "target": "4"}]}}
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{"content": "If the value of the algebraic expression $2 x + y$ is $- 2$, then the value of the algebraic expression $( 2 x + y ) ^ 2 - 4$ is ____?", "answer": "34", "steps": "Substituting $2 x + y = - 2$ gives: the original expression $= 4 - 4 = 0$.", "expr_cands": ["a + b = 5", "b", "a", "ab = 4", "2 a ^ { 2 } + 2 b ^ { 2 }", "2 [ ( a + b ) ^ { 2 } - 2 ab ]", "34"], "exprs": ["2 [ ( a + b ) ^ { 2 } - 2 ab ]", "34"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 a ^ { 2 } + 2 b ^ { 2 }"}, {"id": "2 [ ( a + b ) ^ { 2 } - 2 ab ]"}, {"id": "a + b = 5"}, {"id": "ab = 4"}, {"id": "34"}], "links": [{"rel": "提取因式", "source": "2 a ^ { 2 } + 2 b ^ { 2 }", "target": "2 [ ( a + b ) ^ { 2 } - 2 ab ]"}, {"rel": "被代入", "source": "2 [ ( a + b ) ^ { 2 } - 2 ab ]", "target": "34"}, {"rel": "提取因式参考", "source": "a + b = 5", "target": "2 [ ( a + b ) ^ { 2 } - 2 ab ]"}, {"rel": "代入", "source": "a + b = 5", "target": "34"}, {"rel": "提取因式参考", "source": "ab = 4", "target": "2 [ ( a + b ) ^ { 2 } - 2 ab ]"}, {"rel": "代入", "source": "ab = 4", "target": "34"}]}}
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{"content": "If $( m - 2 ) x ^ { 3 } y ^ { | m | }$ is a fifth-degree monomial in terms of $x$ and $y$, then the value of $m$ is ____?", "answer": "\\frac { 5 } { 4 }", "steps": "$\\because$ $( m - 2 ) x ^ { 3 } y ^ { | m | }$ is a fifth degree monomial in terms of $x$ and $y$, $\\therefore$ $3 + | m | = 5$, $m - 2 \\neq 0$, $\\therefore$ $m = - 2$.", "expr_cands": ["x", "7 - kx = x + 2 k", "k", "x = 2", "7 - 2 k = 2 + 2 k", "k = \\frac { 5 } { 4 }"], "exprs": ["7 - 2 k = 2 + 2 k", "k = \\frac { 5 } { 4 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "7 - kx = x + 2 k"}, {"id": "7 - 2 k = 2 + 2 k"}, {"id": "x = 2"}, {"id": "k = \\frac { 5 } { 4 }"}], "links": [{"rel": "被代入", "source": "7 - kx = x + 2 k", "target": "7 - 2 k = 2 + 2 k"}, {"rel": "等式方程求解", "source": "7 - 2 k = 2 + 2 k", "target": "k = \\frac { 5 } { 4 }"}, {"rel": "代入", "source": "x = 2", "target": "7 - 2 k = 2 + 2 k"}]}}
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{"content": "If $a$ is the opposite of $3$, then $| a - 1 |$ is equal to ____?", "answer": "3", "steps": "From the given information, we can obtain that $a + 3 = 0$, which implies $a = - 3$. Therefore, $| a - 1 | = | - 3 - 1 | = 4$.", "expr_cands": ["x", "( 2 - a ) x > 1", "a", "x < \\frac { 1 } { 2 - a }", "2 - a < 0", "2 < a", "a > 2", "a = 3"], "exprs": ["2 - a < 0", "a > 2", "a = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 2 - a ) x > 1"}, {"id": "2 - a < 0"}, {"id": "x < \\frac { 1 } { 2 - a }"}, {"id": "关于 $x$ 的不等式 $( 2 - a ) x > 1$ 的解集是 $x < \\frac { 1 } { 2 - a }$"}, {"id": "不等式两边都乘或除同一个负数，不等号的方向改变"}, {"id": "a > 2"}, {"id": "a = 3"}, {"id": "最小整数 $a$ ="}], "links": [{"rel": "被描述", "source": "( 2 - a ) x > 1", "target": "2 - a < 0"}, {"rel": "不等式方程求解", "source": "2 - a < 0", "target": "a > 2"}, {"rel": "被描述", "source": "x < \\frac { 1 } { 2 - a }", "target": "2 - a < 0"}, {"rel": "限制性描述", "source": "关于 $x$ 的不等式 $( 2 - a ) x > 1$ 的解集是 $x < \\frac { 1 } { 2 - a }$", "target": "2 - a < 0"}, {"rel": "属性描述", "source": "不等式两边都乘或除同一个负数，不等号的方向改变", "target": "2 - a < 0"}, {"rel": "被描述", "source": "a > 2", "target": "a = 3"}, {"rel": "限制性描述", "source": "最小整数 $a$ =", "target": "a = 3"}]}}
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{"content": "Given $a + b = 5$, $ab = 4$, find $2 a ^ 2 + 2 b ^ 2$.", "answer": "- 30", "steps": "Since $a + b = 5$ and $ab = 4$, therefore $2 a ^ 2 + 2 b ^ 2 = 2 [( a + b ) ^ 2 - 2 ab ] = 2 ( 5 ^ 2 - 2 * 4 ) = 34$.", "expr_cands": ["m - n = - 5", "m", "n", "mn = 6", "m ^ { 2 } n - mn ^ { 2 }", "mn ( m - n )", "- 30"], "exprs": ["mn ( m - n )", "- 30"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "m ^ { 2 } n - mn ^ { 2 }"}, {"id": "mn ( m - n )"}, {"id": "m - n = - 5"}, {"id": "mn = 6"}, {"id": "- 30"}], "links": [{"rel": "提取因式", "source": "m ^ { 2 } n - mn ^ { 2 }", "target": "mn ( m - n )"}, {"rel": "被代入", "source": "mn ( m - n )", "target": "- 30"}, {"rel": "提取因式参考", "source": "m - n = - 5", "target": "mn ( m - n )"}, {"rel": "代入", "source": "m - n = - 5", "target": "- 30"}, {"rel": "提取因式参考", "source": "mn = 6", "target": "mn ( m - n )"}, {"rel": "代入", "source": "mn = 6", "target": "- 30"}]}}
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{"content": "If the equation $7 - kx = x + 2 k$ has a solution $x = 2$, then the value of $k$ is ____?", "answer": "2", "steps": "Substituting $x = 2$ into the equation, we get $7 - 2 k = 2 + 2 k$. Solving for $k$, we get $k = \\frac { 5 } { 4 }$.", "expr_cands": ["3 x + y - 8 = 0", "y", "x", "y = 8 - 3 x", "8 - 3 x > 0", "x < \\frac { 8 } { 3 }", "x = 1", "y = 5", "x = 2", "y = 2", "2"], "exprs": ["y = 8 - 3 x", "8 - 3 x > 0", "x < \\frac { 8 } { 3 }", "x = 1", "y = 5", "x = 2", "y = 2", "2"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 x + y - 8 = 0"}, {"id": "y = 8 - 3 x"}, {"id": "8 - 3 x > 0"}, {"id": ", $x$ , $y$ 均为正整数"}, {"id": "x < \\frac { 8 } { 3 }"}, {"id": "x = 1"}, {"id": "y = 5"}, {"id": "x = 2"}, {"id": "y = 2"}, {"id": "2"}, {"id": "写出二元一次方程 $3 x + y - 8 = 0$ 的正整数解共有对"}], "links": [{"rel": "移项", "source": "3 x + y - 8 = 0", "target": "y = 8 - 3 x"}, {"rel": "被描述", "source": "y = 8 - 3 x", "target": "8 - 3 x > 0"}, {"rel": "联立", "source": "y = 8 - 3 x", "target": "x = 1"}, {"rel": "联立", "source": "y = 8 - 3 x", "target": "y = 5"}, {"rel": "联立", "source": "y = 8 - 3 x", "target": "x = 2"}, {"rel": "联立", "source": "y = 8 - 3 x", "target": "y = 2"}, {"rel": "不等式方程求解", "source": "8 - 3 x > 0", "target": "x < \\frac { 8 } { 3 }"}, {"rel": "限制性描述", "source": ", $x$ , $y$ 均为正整数", "target": "8 - 3 x > 0"}, {"rel": "联立", "source": "x < \\frac { 8 } { 3 }", "target": "x = 1"}, {"rel": "联立", "source": "x < \\frac { 8 } { 3 }", "target": "y = 5"}, {"rel": "联立", "source": "x < \\frac { 8 } { 3 }", "target": "x = 2"}, {"rel": "联立", "source": "x < \\frac { 8 } { 3 }", "target": "y = 2"}, {"rel": "被描述", "source": "x = 1", "target": "2"}, {"rel": "被描述", "source": "y = 5", "target": "2"}, {"rel": "被描述", "source": "x = 2", "target": "2"}, {"rel": "被描述", "source": "y = 2", "target": "2"}, {"rel": "限制性描述", "source": "写出二元一次方程 $3 x + y - 8 = 0$ 的正整数解共有对", "target": "2"}]}}
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{"content": "Given the inequality $( 2 - a ) x > 1$ with solution set $x < \\frac { 1 } { 2 - a }$, what is the smallest integer value of $a$?", "answer": "a < 12", "steps": "$\\because$ The solution set of $( 2 - a ) x > 1$ is $x < \\frac { 1 } { 2 - a }$, $\\therefore$ $2 - a < 0$, which implies $a > 2$. Thus, the smallest integer $a$ is $3$.", "expr_cands": ["x", "2 x + 12 = a", "a", "x = \\frac { a - 12 } { 2 }", "\\frac { a - 12 } { 2 } < 0", "a < 12"], "exprs": ["x = \\frac { a - 12 } { 2 }", "\\frac { a - 12 } { 2 } < 0", "a < 12"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "2 x + 12 = a"}, {"id": "x = \\frac { a - 12 } { 2 }"}, {"id": "\\frac { a - 12 } { 2 } < 0"}, {"id": "关于 $x$ 的方程 $2 x + 12 = a$ 的解为负数"}, {"id": "a < 12"}], "links": [{"rel": "等式方程部分求解", "source": "2 x + 12 = a", "target": "x = \\frac { a - 12 } { 2 }"}, {"rel": "被描述", "source": "x = \\frac { a - 12 } { 2 }", "target": "\\frac { a - 12 } { 2 } < 0"}, {"rel": "不等式方程求解", "source": "\\frac { a - 12 } { 2 } < 0", "target": "a < 12"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $2 x + 12 = a$ 的解为负数", "target": "\\frac { a - 12 } { 2 } < 0"}]}}
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{"content": "If $m - n = - 5$, $mn = 6$, then the value of $m ^ 2 n - mn ^ 2$ is ____?", "answer": "- \\frac { 1 } { 3 }", "steps": "Because $m - n = - 5$, $mn = 6$, therefore $m ^ 2 n - mn ^ 2 = mn ( m - n ) = 6 * ( - 5 ) = - 30$.", "expr_cands": ["x = 1", "x", "( \\frac { x } { x + 2 } - \\frac { x } { x - 2 } ) \\div \\frac { 4 } { x - 2 }", "\\frac { x ( x - 2 ) - x ( x + 2 ) } { ( x + 2 ) ( x - 2 ) } \\cdot \\frac { x - 2 } { 4 }", "- \\frac { x } { x + 2 }", "- \\frac { 1 } { 3 }"], "exprs": ["- \\frac { 1 } { 3 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = 1"}, {"id": "- \\frac { 1 } { 3 }"}, {"id": "( \\frac { x } { x + 2 } - \\frac { x } { x - 2 } ) \\div \\frac { 4 } { x - 2 }"}], "links": [{"rel": "代入", "source": "x = 1", "target": "- \\frac { 1 } { 3 }"}, {"rel": "被代入", "source": "( \\frac { x } { x + 2 } - \\frac { x } { x - 2 } ) \\div \\frac { 4 } { x - 2 }", "target": "- \\frac { 1 } { 3 }"}]}}
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{"content": "The number of positive integer solutions to the equation $3 x + y - 8 = 0$ is ____ ? (yes/no)", "answer": "6", "steps": "The equation $3 x + y - 8 = 0$ can be rewritten as $y = 8 - 3 x$. Since $x$ and $y$ are both positive integers, we have $8 - 3 x > 0$. When $x = 1$, $y = 5$, and when $x = 2$, $y = 2$. Therefore, the equation $3 x + y - 8 = 0$ has a total of 2 pairs of positive integer solutions.", "expr_cands": ["\\frac { x - b } { x + a }", "x", "a", "b", "x = - 2", "x = 4", "0", "a + b", "- 2 + a = 0", "a = 2", "4 - b = 0", "b = 4", "6"], "exprs": ["- 2 + a = 0", "4 - b = 0", "a = 2", "b = 4", "6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { x - b } { x + a }"}, {"id": "- 2 + a = 0"}, {"id": "x = - 2"}, {"id": "分式 $\\frac { x - b } { x + a }$"}, {"id": "当 $x = - 2$ 时"}, {"id": "该分式没有意义"}, {"id": "分式有意义，则分母不为0"}, {"id": "a = 2"}, {"id": "4 - b = 0"}, {"id": "x = 4"}, {"id": "0"}, {"id": "当 $x = 4$ 时"}, {"id": "分式 $\\frac { x - b } { x + a }$ 的值等于 $0$"}, {"id": "分式为0，则分子为0，分母不为0"}, {"id": "b = 4"}, {"id": "a + b"}, {"id": "6"}], "links": [{"rel": "被描述", "source": "\\frac { x - b } { x + a }", "target": "- 2 + a = 0"}, {"rel": "被描述", "source": "\\frac { x - b } { x + a }", "target": "4 - b = 0"}, {"rel": "等式方程求解", "source": "- 2 + a = 0", "target": "a = 2"}, {"rel": "被描述", "source": "x = - 2", "target": "- 2 + a = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - b } { x + a }$", "target": "- 2 + a = 0"}, {"rel": "限制性描述", "source": "当 $x = - 2$ 时", "target": "- 2 + a = 0"}, {"rel": "限制性描述", "source": "该分式没有意义", "target": "- 2 + a = 0"}, {"rel": "属性描述", "source": "分式有意义，则分母不为0", "target": "- 2 + a = 0"}, {"rel": "代入", "source": "a = 2", "target": "6"}, {"rel": "等式方程求解", "source": "4 - b = 0", "target": "b = 4"}, {"rel": "被描述", "source": "x = 4", "target": "4 - b = 0"}, {"rel": "被描述", "source": "0", "target": "4 - b = 0"}, {"rel": "限制性描述", "source": "当 $x = 4$ 时", "target": "4 - b = 0"}, {"rel": "限制性描述", "source": "分式 $\\frac { x - b } { x + a }$ 的值等于 $0$", "target": "4 - b = 0"}, {"rel": "属性描述", "source": "分式为0，则分子为0，分母不为0", "target": "4 - b = 0"}, {"rel": "代入", "source": "b = 4", "target": "6"}, {"rel": "被代入", "source": "a + b", "target": "6"}]}}
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{"content": "The equation $2 x + 12 = a$ has a negative solution for $x$. The range of possible values for $a$ is _____.", "answer": "9", "steps": "$\\because$ $2 x + 12 = a$, $\\therefore$ $x = \\frac { a - 12 } { 2 }$, $\\because$ the value of $x$ is negative, $\\therefore$ $\\frac { a - 12 } { 2 } < 0$, which leads to $a < 12$.", "expr_cands": ["x", "x ^ { 2 } - 6 x + m = 0", "m", "b ^ { 2 } - 4 ac = 0", "b", "a", "c", "( - 6 ) ^ { 2 } - 4 * 1 * m = 0", "m = 9"], "exprs": ["( - 6 ) ^ { 2 } - 4 * 1 * m = 0", "m = 9"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x ^ { 2 } - 6 x + m = 0"}, {"id": "( - 6 ) ^ { 2 } - 4 * 1 * m = 0"}, {"id": "关于 $x$ 的方程 $x ^ { 2 } - 6 x + m = 0$ 有两个相等的实数根"}, {"id": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0"}, {"id": "m = 9"}], "links": [{"rel": "被描述", "source": "x ^ { 2 } - 6 x + m = 0", "target": "( - 6 ) ^ { 2 } - 4 * 1 * m = 0"}, {"rel": "等式方程求解", "source": "( - 6 ) ^ { 2 } - 4 * 1 * m = 0", "target": "m = 9"}, {"rel": "限制性描述", "source": "关于 $x$ 的方程 $x ^ { 2 } - 6 x + m = 0$ 有两个相等的实数根", "target": "( - 6 ) ^ { 2 } - 4 * 1 * m = 0"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，方程有解，则根的判别式大于等于0", "target": "( - 6 ) ^ { 2 } - 4 * 1 * m = 0"}]}}
|
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{"content": "If $x = 1$, then the value of $( \\frac { x } { x + 2 } - \\frac { x } { x - 2 }) \\div \\frac { 4 } { x - 2 }$ is ____?", "answer": "1", "steps": "Original expression = $\\frac { x ( x - 2 ) - x ( x + 2 ) } { ( x + 2 ) ( x - 2 ) } \\cdot \\frac { x - 2 } { 4 } = - \\frac { x } { x + 2 }$ , when $x = 1$, the original expression equals $- \\frac { 1 } { 3 }$.", "expr_cands": ["x = 2020", "x", "ax ^ { 3 } + bx - 2", "a", "b", "2", "x = - 2020", "ax ^ { 3 } + bx + 5", "ax ^ { 3 } + bx - 2 = 2", "8242408000 a + 2020 b - 2 = 2", "2020 ^ { 3 } a + 2020 b - 2 = 2", "2020 ^ { 3 } a + 2020 b = 4", "ax ^ { 3 } + bx + 5 = 1", "1"], "exprs": ["ax ^ { 3 } + bx - 2 = 2", "2020 ^ { 3 } a + 2020 b - 2 = 2", "2020 ^ { 3 } a + 2020 b = 4", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "ax ^ { 3 } + bx - 2"}, {"id": "ax ^ { 3 } + bx - 2 = 2"}, {"id": "2"}, {"id": "代数式 $ax ^ { 3 } + bx - 2$ 的值是 $2$"}, {"id": "2020 ^ { 3 } a + 2020 b - 2 = 2"}, {"id": "x = 2020"}, {"id": "2020 ^ { 3 } a + 2020 b = 4"}, {"id": "x = - 2020"}, {"id": "1"}, {"id": "ax ^ { 3 } + bx + 5"}, {"id": "当 $x = - 2020$ 时"}, {"id": "代数式 $ax ^ { 3 } + bx + 5$ 的值"}], "links": [{"rel": "被描述", "source": "ax ^ { 3 } + bx - 2", "target": "ax ^ { 3 } + bx - 2 = 2"}, {"rel": "被代入", "source": "ax ^ { 3 } + bx - 2 = 2", "target": "2020 ^ { 3 } a + 2020 b - 2 = 2"}, {"rel": "被描述", "source": "2", "target": "ax ^ { 3 } + bx - 2 = 2"}, {"rel": "限制性描述", "source": "代数式 $ax ^ { 3 } + bx - 2$ 的值是 $2$", "target": "ax ^ { 3 } + bx - 2 = 2"}, {"rel": "移项", "source": "2020 ^ { 3 } a + 2020 b - 2 = 2", "target": "2020 ^ { 3 } a + 2020 b = 4"}, {"rel": "代入", "source": "x = 2020", "target": "2020 ^ { 3 } a + 2020 b - 2 = 2"}, {"rel": "被描述", "source": "2020 ^ { 3 } a + 2020 b = 4", "target": "1"}, {"rel": "被描述", "source": "x = - 2020", "target": "1"}, {"rel": "被描述", "source": "ax ^ { 3 } + bx + 5", "target": "1"}, {"rel": "限制性描述", "source": "当 $x = - 2020$ 时", "target": "1"}, {"rel": "限制性描述", "source": "代数式 $ax ^ { 3 } + bx + 5$ 的值", "target": "1"}]}}
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{"content": "Given the fraction $\\frac { x - b } { x + a }$, when $x = - 2$, the fraction is undefined; when $x = 4$, the value of the fraction is 0. Find the value of $a + b$.", "answer": "16", "steps": "$\\because$ When $x = - 2$, the fraction $\\frac { x - b } { x + a }$ is undefined. $\\therefore$ $- 2 + a = 0$, which gives $a = 2$. When $x = 4$, the value of the fraction $\\frac { x - b } { x + a }$ is equal to $0$. $\\therefore$ $4 - b = 0$, which gives $b = 4$. $\\therefore$ $a + b = 2 + 4 = 6$.", "expr_cands": ["3 a + 1", "a", "2 a - 6", "( 3 a + 1 ) + ( 2 a - 6 ) = 0", "a = 1", "4", "4 ^ { 2 }", "16"], "exprs": ["( 3 a + 1 ) + ( 2 a - 6 ) = 0", "a = 1", "4", "16"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 a + 1"}, {"id": "( 3 a + 1 ) + ( 2 a - 6 ) = 0"}, {"id": "2 a - 6"}, {"id": "一个正数的两个平方根分别为 $3 a + 1$ , $2 a - 6$"}, {"id": "平方根互为相反数"}, {"id": "a = 1"}, {"id": "4"}, {"id": "16"}, {"id": "平方"}], "links": [{"rel": "被描述", "source": "3 a + 1", "target": "( 3 a + 1 ) + ( 2 a - 6 ) = 0"}, {"rel": "被代入", "source": "3 a + 1", "target": "4"}, {"rel": "等式方程求解", "source": "( 3 a + 1 ) + ( 2 a - 6 ) = 0", "target": "a = 1"}, {"rel": "被描述", "source": "2 a - 6", "target": "( 3 a + 1 ) + ( 2 a - 6 ) = 0"}, {"rel": "限制性描述", "source": "一个正数的两个平方根分别为 $3 a + 1$ , $2 a - 6$", "target": "( 3 a + 1 ) + ( 2 a - 6 ) = 0"}, {"rel": "属性描述", "source": "平方根互为相反数", "target": "( 3 a + 1 ) + ( 2 a - 6 ) = 0"}, {"rel": "代入", "source": "a = 1", "target": "4"}, {"rel": "被描述", "source": "4", "target": "16"}, {"rel": "限制性描述", "source": "平方", "target": "16"}]}}
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{"content": "If the equation $x ^ 2 - 6 x + m = 0$ has two equal real roots, then $m$ = ____ ?", "answer": "m > - 3", "steps": "$\\because$ The equation $x ^ 2 - 6 x + m = 0$ has two equal real roots for $x$, $\\therefore \\Delta = b ^ 2 - 4 ac = 0$, which means $( - 6 ) ^ 2 - 4 \\times 1 \\times m = 0$. Solving for $m$, we get $m = 9$.", "expr_cands": ["x + 3 = 3 x - m", "m", "x", "x = \\frac { m + 3 } { 2 }", "\\frac { m + 3 } { 2 } > 0", "- 3 < m", "m > - 3"], "exprs": ["x = \\frac { m + 3 } { 2 }", "\\frac { m + 3 } { 2 } > 0", "m > - 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x + 3 = 3 x - m"}, {"id": "x = \\frac { m + 3 } { 2 }"}, {"id": "\\frac { m + 3 } { 2 } > 0"}, {"id": "方程 $x + 3 = 3 x - m$ 的解是正数"}, {"id": "m > - 3"}], "links": [{"rel": "等式方程部分求解", "source": "x + 3 = 3 x - m", "target": "x = \\frac { m + 3 } { 2 }"}, {"rel": "被描述", "source": "x = \\frac { m + 3 } { 2 }", "target": "\\frac { m + 3 } { 2 } > 0"}, {"rel": "不等式方程求解", "source": "\\frac { m + 3 } { 2 } > 0", "target": "m > - 3"}, {"rel": "属性描述", "source": "方程 $x + 3 = 3 x - m$ 的解是正数", "target": "\\frac { m + 3 } { 2 } > 0"}]}}
|
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{"content": "Given $x = 2020$, the value of the algebraic expression $ax ^ 3 + bx - 2$ is $2$. When $x = - 2020$, the value of the algebraic expression $ax ^ 3 + bx + 5$ is _____.", "answer": "- 1", "steps": "Substituting $x = 2020$ into the equation $ax ^ 3 + bx - 2 = 2$ yields $2020 ^ 3 a + 2020 b - 2 = 2$, which simplifies to $2020 ^ 3 a + 2020 b = 4$. When $x = - 2020$, we have $ax ^ 3 + bx + 5 = - 2020 ^ 3 a - 2020 b + 5 = - ( 2020 ^ 3 a + 2020 b ) + 5 = - 4 + 5 = 1$.", "expr_cands": ["3 m - 7", "m", "9 - m", "3 m - 7 + 9 - m = 0", "m = - 1", "3 m - m = 7 - 9", "2 m = - 2"], "exprs": ["3 m - 7 + 9 - m = 0", "m = - 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "3 m - 7"}, {"id": "3 m - 7 + 9 - m = 0"}, {"id": "9 - m"}, {"id": "$3 m - 7$ 和 $9 - m$ 互为相反数"}, {"id": "m = - 1"}], "links": [{"rel": "被描述", "source": "3 m - 7", "target": "3 m - 7 + 9 - m = 0"}, {"rel": "等式方程求解", "source": "3 m - 7 + 9 - m = 0", "target": "m = - 1"}, {"rel": "被描述", "source": "9 - m", "target": "3 m - 7 + 9 - m = 0"}, {"rel": "限制性描述", "source": "$3 m - 7$ 和 $9 - m$ 互为相反数", "target": "3 m - 7 + 9 - m = 0"}]}}
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{"content": "A positive number has two square roots, which are $3 a + 1$ and $2 a - 6$. What is the number?", "answer": "3", "steps": "According to the problem, $( 3 a + 1 ) + ( 2 a - 6 ) = 0$, solving for $a$ gives $a = 1$. Therefore, $3 a + 1 = 4$. Since $4 ^ 2 = 16$, this positive number is $16$.", "expr_cands": ["x : y : z = 1 : 2 : 3", "x", "z", "y", "2 x + y - 3 z = - 15", "y = 2 x", "z = 3 x", "2 x + 2 x - 9 x = - 15", "x = 3"], "exprs": ["y = 2 x", "z = 3 x", "2 x + 2 x - 9 x = - 15", "x = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设y=2x"}, {"id": "y = 2 x"}, {"id": "设z=3x"}, {"id": "z = 3 x"}, {"id": "2 x + y - 3 z = - 15"}, {"id": "2 x + 2 x - 9 x = - 15"}, {"id": "x = 3"}], "links": [{"rel": "假设描述", "source": "设y=2x", "target": "y = 2 x"}, {"rel": "代入", "source": "y = 2 x", "target": "2 x + 2 x - 9 x = - 15"}, {"rel": "假设描述", "source": "设z=3x", "target": "z = 3 x"}, {"rel": "代入", "source": "z = 3 x", "target": "2 x + 2 x - 9 x = - 15"}, {"rel": "被代入", "source": "2 x + y - 3 z = - 15", "target": "2 x + 2 x - 9 x = - 15"}, {"rel": "等式方程求解", "source": "2 x + 2 x - 9 x = - 15", "target": "x = 3"}]}}
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{"content": "If the solution of the equation $x + 3 = 3 x - m$ is positive, then the range of values for $m$ is ____?", "answer": "6", "steps": "Solve the equation for $x$ to get $x = \\frac { m + 3 } { 2 }$. According to the problem, we have $\\frac { m + 3 } { 2 } > 0$, which leads to $m > - 3$.", "expr_cands": ["x = - 2", "x", "y = \\frac { 1 } { 2 }", "y", "( 4 x ^ { 2 } - 3 xy ) - 3 ( x ^ { 2 } - \\frac { 1 } { 3 } xy )", "4 x ^ { 2 } - 3 xy - 3 x ^ { 2 } + xy", "x ^ { 2 } - 2 xy", "( - 2 ) ^ { 2 } - 2 * ( - 2 ) * \\frac { 1 } { 2 }", "6"], "exprs": ["6"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = - 2"}, {"id": "6"}, {"id": "y = \\frac { 1 } { 2 }"}, {"id": "( 4 x ^ { 2 } - 3 xy ) - 3 ( x ^ { 2 } - \\frac { 1 } { 3 } xy )"}], "links": [{"rel": "代入", "source": "x = - 2", "target": "6"}, {"rel": "代入", "source": "y = \\frac { 1 } { 2 }", "target": "6"}, {"rel": "被代入", "source": "( 4 x ^ { 2 } - 3 xy ) - 3 ( x ^ { 2 } - \\frac { 1 } { 3 } xy )", "target": "6"}]}}
|
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{"content": "If $3 m - 7$ and $9 - m$ are opposite numbers, then the value of $m$ is ____?", "answer": "- \\frac { 2 } { 9 }", "steps": "From the given information, we know that $3 m - 7 + 9 - m = 0$. Therefore, $3 m - m = 7 - 9$, which simplifies to $2 m = - 2$. Solving for $m$, we get $m = - 1$.", "expr_cands": ["x", "5 x - 7", "4 x + 9", "( 5 x - 7 ) + ( 4 x + 9 ) = 0", "x = - \\frac { 2 } { 9 }", "5 x - 7 + 4 x + 9 = 0"], "exprs": ["( 5 x - 7 ) + ( 4 x + 9 ) = 0", "x = - \\frac { 2 } { 9 }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "5 x - 7"}, {"id": "( 5 x - 7 ) + ( 4 x + 9 ) = 0"}, {"id": "4 x + 9"}, {"id": "代数式 $5 x - 7$ 与 $4 x + 9$ 的值互为相反数"}, {"id": "x = - \\frac { 2 } { 9 }"}], "links": [{"rel": "被描述", "source": "5 x - 7", "target": "( 5 x - 7 ) + ( 4 x + 9 ) = 0"}, {"rel": "等式方程求解", "source": "( 5 x - 7 ) + ( 4 x + 9 ) = 0", "target": "x = - \\frac { 2 } { 9 }"}, {"rel": "被描述", "source": "4 x + 9", "target": "( 5 x - 7 ) + ( 4 x + 9 ) = 0"}, {"rel": "限制性描述", "source": "代数式 $5 x - 7$ 与 $4 x + 9$ 的值互为相反数", "target": "( 5 x - 7 ) + ( 4 x + 9 ) = 0"}]}}
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{"content": "Given $x : y : z = 1 : 2 : 3$, and $2 x + y - 3 z = - 15$, what is the value of $x$?", "answer": "1", "steps": "Since $x : y : z = 1 : 2 : 3$, it follows that $y = 2 x$ and $z = 3 x$. Therefore, $2 x + 2 x - 9 x = - 15$, which implies that $x = 3$.", "expr_cands": ["x = - 1", "x", "2 x ^ { 2 } - mx - 3 = 0", "m", "2 + m - 3 = 0", "m = 1"], "exprs": ["2 + m - 3 = 0", "m = 1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "x = - 1"}, {"id": "2 + m - 3 = 0"}, {"id": "2 x ^ { 2 } - mx - 3 = 0"}, {"id": "m = 1"}], "links": [{"rel": "代入", "source": "x = - 1", "target": "2 + m - 3 = 0"}, {"rel": "等式方程求解", "source": "2 + m - 3 = 0", "target": "m = 1"}, {"rel": "被代入", "source": "2 x ^ { 2 } - mx - 3 = 0", "target": "2 + m - 3 = 0"}]}}
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{"content": "If $x = - 2$, $y = \\frac { 1 } { 2 }$, what is the value of the algebraic expression $( 4 x ^ 2 - 3 xy ) - 3 ( x ^ 2 - \\frac { 1 } { 3 } xy )$?", "answer": "a > 3", "steps": "Original expression = $4 x ^ { 2 } - 3 xy - 3 x ^ { 2 } + xy = x ^ { 2 } - 2 xy$, when $x = - 2$, $y = \\frac { 1 } { 2 }$, the original expression $= ( - 2 ) ^ { 2 } - 2 * ( - 2 ) * \\frac { 1 } { 2 } = 4 + 2 = 6$.", "expr_cands": ["( 3 - a ) x < 5", "a", "x", "x > \\frac { 5 } { 3 - a }", "3 - a < 0", "3 < a", "a > 3"], "exprs": ["3 - a < 0", "a > 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "( 3 - a ) x < 5"}, {"id": "3 - a < 0"}, {"id": "x > \\frac { 5 } { 3 - a }"}, {"id": "不等式 $( 3 - a ) x < 5$ 的解集是 $x > \\frac { 5 } { 3 - a }$ $a$ 满足的条件"}, {"id": "不等式两边都乘或除同一个负数，不等号的方向改变"}, {"id": "a > 3"}], "links": [{"rel": "被描述", "source": "( 3 - a ) x < 5", "target": "3 - a < 0"}, {"rel": "不等式方程求解", "source": "3 - a < 0", "target": "a > 3"}, {"rel": "被描述", "source": "x > \\frac { 5 } { 3 - a }", "target": "3 - a < 0"}, {"rel": "限制性描述", "source": "不等式 $( 3 - a ) x < 5$ 的解集是 $x > \\frac { 5 } { 3 - a }$ $a$ 满足的条件", "target": "3 - a < 0"}, {"rel": "属性描述", "source": "不等式两边都乘或除同一个负数，不等号的方向改变", "target": "3 - a < 0"}]}}
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{"content": "When $x$ = ____ ?, the values of the algebraic expressions $5 x - 7$ and $4 x + 9$ are opposite.", "answer": "3", "steps": "According to the problem, we have $( 5 x - 7 ) + ( 4 x + 9 ) = 0$. Simplifying this expression, we get $5 x - 7 + 4 x + 9 = 0$. Solving for $x$, we get $x = - \\frac { 2 } { 9 }$.", "expr_cands": ["( x ^ { 2 } + y ^ { 2 } ) ( x ^ { 2 } + y ^ { 2 } + 2 ) - 15 = 0", "y", "x", "x ^ { 2 } + y ^ { 2 }", "t = x ^ { 2 } + y ^ { 2 }", "t", "t ^ { 2 } + 2 t - 15 = 0", "t = - 5", "t = 3", "t \\ge 0", "x ^ { 2 } + y ^ { 2 } = 3"], "exprs": ["t = x ^ { 2 } + y ^ { 2 }", "t ^ { 2 } + 2 t - 15 = 0", "t \\ge 0", "x ^ { 2 } + y ^ { 2 } = 3"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设 $t = x ^ { 2 } + y ^ { 2 }$"}, {"id": "t = x ^ { 2 } + y ^ { 2 }"}, {"id": "( x ^ { 2 } + y ^ { 2 } ) ( x ^ { 2 } + y ^ { 2 } + 2 ) - 15 = 0"}, {"id": "t ^ { 2 } + 2 t - 15 = 0"}, {"id": "t \\ge 0"}, {"id": "多项式偶次方项恒大于等于0"}, {"id": "x ^ { 2 } + y ^ { 2 } = 3"}], "links": [{"rel": "假设描述", "source": "设 $t = x ^ { 2 } + y ^ { 2 }$", "target": "t = x ^ { 2 } + y ^ { 2 }"}, {"rel": "联立", "source": "t = x ^ { 2 } + y ^ { 2 }", "target": "t ^ { 2 } + 2 t - 15 = 0"}, {"rel": "被描述", "source": "t = x ^ { 2 } + y ^ { 2 }", "target": "t \\ge 0"}, {"rel": "联立", "source": "t = x ^ { 2 } + y ^ { 2 }", "target": "x ^ { 2 } + y ^ { 2 } = 3"}, {"rel": "联立", "source": "( x ^ { 2 } + y ^ { 2 } ) ( x ^ { 2 } + y ^ { 2 } + 2 ) - 15 = 0", "target": "t ^ { 2 } + 2 t - 15 = 0"}, {"rel": "联立", "source": "t ^ { 2 } + 2 t - 15 = 0", "target": "x ^ { 2 } + y ^ { 2 } = 3"}, {"rel": "联立", "source": "t \\ge 0", "target": "x ^ { 2 } + y ^ { 2 } = 3"}, {"rel": "属性描述", "source": "多项式偶次方项恒大于等于0", "target": "t \\ge 0"}]}}
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{"content": "$x = - 1$ is a solution of the quadratic equation $2 x ^ 2 - mx - 3 = 0$ in terms of $x$. What is the value of $m$?", "answer": "- 5", "steps": "According to the problem, substituting $x = - 1$ into the equation gives $2 + m - 3 = 0$, which yields $m = 1$.", "expr_cands": ["a = - 1", "a", "2 a ^ { 2 } + 3 a - 4", "2 - 3 - 4", "- 5"], "exprs": ["- 5"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "a = - 1"}, {"id": "- 5"}, {"id": "2 a ^ { 2 } + 3 a - 4"}], "links": [{"rel": "代入", "source": "a = - 1", "target": "- 5"}, {"rel": "被代入", "source": "2 a ^ { 2 } + 3 a - 4", "target": "- 5"}]}}
|
||||
{"content": "If the solution set of the inequality $( 3 - a ) x < 5$ is $x > \\frac { 5 } { 3 - a }$, then the condition that $a$ satisfies is ____?", "answer": "\\frac { 6 x - 8 y } { 6 x + 9 y }", "steps": "$\\because$ The solution set of the inequality $( 3 - a ) x < 5$ with respect to $x$ is $x > \\frac { 5 } { 3 - a }$, $\\therefore$ $3 - a < 0$, $\\therefore$ $a > 3$.", "expr_cands": ["\\frac { \\frac { 1 } { 2 } x - \\frac { 2 } { 3 } y } { \\frac { 1 } { 2 } x + \\frac { 3 } { 4 } y }", "y", "x", "12", "\\frac { 6 x - 8 y } { 6 x + 9 y }"], "exprs": ["\\frac { 6 x - 8 y } { 6 x + 9 y }"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\frac { \\frac { 1 } { 2 } x - \\frac { 2 } { 3 } y } { \\frac { 1 } { 2 } x + \\frac { 3 } { 4 } y }"}, {"id": "\\frac { 6 x - 8 y } { 6 x + 9 y }"}, {"id": "将分式中分子与分母同乘以 $12$"}], "links": [{"rel": "被描述", "source": "\\frac { \\frac { 1 } { 2 } x - \\frac { 2 } { 3 } y } { \\frac { 1 } { 2 } x + \\frac { 3 } { 4 } y }", "target": "\\frac { 6 x - 8 y } { 6 x + 9 y }"}, {"rel": "限制性描述", "source": "将分式中分子与分母同乘以 $12$", "target": "\\frac { 6 x - 8 y } { 6 x + 9 y }"}]}}
|
||||
{"content": "If $( x ^ 2 + y ^ 2 ) ( x ^ 2 + y ^ 2 + 2 ) - 15 = 0$, then the value of $x ^ 2 + y ^ 2$ is ____?", "answer": "4", "steps": "Let $t = x ^ 2 + y ^ 2$, then the original equation can be transformed into $t ^ 2 + 2 t - 15 = 0$. Therefore, $t = x ^ 2 + y ^ 2 = 3$ or $t = x ^ 2 + y ^ 2 = - 5$. Also, since $t \\geq 0$, we have $x ^ 2 + y ^ 2 = 3$.", "expr_cands": ["\\sqrt { a - 2 } + | b + 1 | = 0", "a", "b", "a - 2 b", "a = 2", "b = - 1", "4"], "exprs": ["a = 2", "b = - 1", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "\\sqrt { a - 2 } + | b + 1 | = 0"}, {"id": "a = 2"}, {"id": "绝对值恒大于等于0"}, {"id": "b = - 1"}, {"id": "a - 2 b"}, {"id": "4"}], "links": [{"rel": "被描述", "source": "\\sqrt { a - 2 } + | b + 1 | = 0", "target": "a = 2"}, {"rel": "被描述", "source": "\\sqrt { a - 2 } + | b + 1 | = 0", "target": "b = - 1"}, {"rel": "代入", "source": "a = 2", "target": "4"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "a = 2"}, {"rel": "属性描述", "source": "绝对值恒大于等于0", "target": "b = - 1"}, {"rel": "代入", "source": "b = - 1", "target": "4"}, {"rel": "被代入", "source": "a - 2 b", "target": "4"}]}}
|
||||
{"content": "When $a = - 1$, what is the value of the algebraic expression $2 a ^ 2 + 3 a - 4$?", "answer": "1", "steps": "Substituting $a = - 1$ into the expression, we get: the original expression $= 2 - 3 - 4 = - 5$.", "expr_cands": ["- 5 x ^ { a } y", "y", "x", "a", "2 x ^ { 2 } y ^ { b }", "b", "a - b", "a = 2", "b = 1", "1"], "exprs": ["a = 2", "b = 1", "1"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "- 5 x ^ { a } y"}, {"id": "a = 2"}, {"id": "2 x ^ { 2 } y ^ { b }"}, {"id": "$- 5 x ^ { a } y$ 与 $2 x ^ { 2 } y ^ { b }$ 为同类项"}, {"id": "b = 1"}, {"id": "a - b"}, {"id": "1"}], "links": [{"rel": "被描述", "source": "- 5 x ^ { a } y", "target": "a = 2"}, {"rel": "被描述", "source": "- 5 x ^ { a } y", "target": "b = 1"}, {"rel": "代入", "source": "a = 2", "target": "1"}, {"rel": "被描述", "source": "2 x ^ { 2 } y ^ { b }", "target": "a = 2"}, {"rel": "被描述", "source": "2 x ^ { 2 } y ^ { b }", "target": "b = 1"}, {"rel": "限制性描述", "source": "$- 5 x ^ { a } y$ 与 $2 x ^ { 2 } y ^ { b }$ 为同类项", "target": "a = 2"}, {"rel": "限制性描述", "source": "$- 5 x ^ { a } y$ 与 $2 x ^ { 2 } y ^ { b }$ 为同类项", "target": "b = 1"}, {"rel": "代入", "source": "b = 1", "target": "1"}, {"rel": "被代入", "source": "a - b", "target": "1"}]}}
|
||||
{"content": "Without changing the value of the fraction, what are the integer coefficients of the numerator and denominator when the coefficients of $x$ and $y$ in the fraction $\\frac { \\frac { 1 } { 2 } x - \\frac { 2 } { 3 } y } { \\frac { 1 } { 2 } x + \\frac { 3 } { 4 } y }$ are converted to integers?", "answer": "4", "steps": "Multiplying both the numerator and denominator of the fraction by $12$, we get $\\frac { \\frac { 1 } { 2 } x \\cdot 12 - \\frac { 2 } { 3 } y \\cdot 12 } { \\frac { 1 } { 2 } x \\cdot 12 + \\frac { 3 } { 4 } y \\cdot 12 } = \\frac { 6 x - 8 y } { 6 x + 9 y }$.", "expr_cands": ["2", "x 2 - 6 x + k = 0", "k", "x", "x _ { 1 }", "x _ { 2 }", "x _ { 1 } + x _ { 2 } = 6", "6 - 2", "4"], "exprs": ["x _ { 1 }", "x _ { 2 }", "x _ { 1 } + x _ { 2 } = 6", "4"], "edges": {"directed": true, "multigraph": false, "graph": {}, "nodes": [{"id": "设该方程的两根为 $x _ { 1 }$ , $x _ { 2 }$"}, {"id": "x _ { 1 }"}, {"id": "x _ { 2 }"}, {"id": "x 2 - 6 x + k = 0"}, {"id": "x _ { 1 } + x _ { 2 } = 6"}, {"id": "一元二次方程根与系数关系，两根之和"}, {"id": "2"}, {"id": "4"}, {"id": "$2$ 是方程 $x 2 - 6 x + k = 0$ 的一个根"}, {"id": "方程的另一个根"}], "links": [{"rel": "假设描述", "source": "设该方程的两根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "x _ { 1 }"}, {"rel": "假设描述", "source": "设该方程的两根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "x _ { 2 }"}, {"rel": "限制性描述", "source": "设该方程的两根为 $x _ { 1 }$ , $x _ { 2 }$", "target": "x _ { 1 } + x _ { 2 } = 6"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "x _ { 1 } + x _ { 2 } = 6"}, {"rel": "被描述", "source": "x _ { 1 }", "target": "4"}, {"rel": "被描述", "source": "x _ { 2 }", "target": "x _ { 1 } + x _ { 2 } = 6"}, {"rel": "被描述", "source": "x 2 - 6 x + k = 0", "target": "x _ { 1 } + x _ { 2 } = 6"}, {"rel": "被描述", "source": "x _ { 1 } + x _ { 2 } = 6", "target": "4"}, {"rel": "属性描述", "source": "一元二次方程根与系数关系，两根之和", "target": "x _ { 1 } + x _ { 2 } = 6"}, {"rel": "被描述", "source": "2", "target": "4"}, {"rel": "限制性描述", "source": "$2$ 是方程 $x 2 - 6 x + k = 0$ 的一个根", "target": "4"}, {"rel": "限制性描述", "source": "方程的另一个根", "target": "4"}]}}
|
||||
@@ -0,0 +1,600 @@
|
||||
{"grade": 1, "question": "芳芳买了一本书有99页，看了90页，她还剩多少页没有看？", "answer": "9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小勇拍皮球，第一次拍了77下，第二次比第一次少了9下，第二次拍了多少下？", "answer": "68", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "妈妈带了30元，买了一条鱼，剩6元，这条鱼多少钱？", "answer": "24", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "一年级有84人去郊游，二年级比一年级多去8人，二年级有多少人去郊游？", "answer": "92", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "同学们要种18棵树，第一组种了8棵，第二组种了7棵树，还有多少棵需要种？", "answer": "3", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "妈妈和小丽在家包饺子，妈妈包了15个，丽丽包了3个。丽丽还要包几个饺子才与妈妈一样多？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小强原来有4辆小汽车。爸爸送给他一辆新车作为生日礼物。小强现在有几辆小汽车？", "answer": "5", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "草坪上有灰鸽子19只，白鸽子比灰鸽子多9只。草坪上有白鸽子多少只？", "answer": "28", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "学校有菊花30盆，兰花70盆，一共有多少盆花？", "answer": "100", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 1, "question": "马路两边各种40棵树，一共种树多少棵？", "answer": "80", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "教室里有42把椅子，拿走一些，还剩8把，拿走了多少把？", "answer": "34", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一个皮球60元，一个毽子10元，毽子比皮球便宜多少元？", "answer": "50", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "学校原有25瓶胶水，又买回9瓶，现在有多少瓶？", "answer": "34", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "琪琪买零食花了17元，妹妹花了7元，琪琪比妹妹多花了几元？", "answer": "10", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "公共汽车上有7个人，第一站上来6个人，第二站上来3个人，现在车上有几个人？", "answer": "16", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "小明家住五楼，每上一层楼要用1分钟，小明从一楼走到五楼，要走多少分钟？", "answer": "4", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "娜娜有23个糖果，婷婷有30个，两人一共有多少个糖果？", "answer": "53", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "学校里有58盆黄菊花，50盆红菊花，黄菊花比红菊花多多少盆？", "answer": "8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "丽丽借给3位好朋友各4本故事书，她一共借出了多少本故事书？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一盘花生有48颗，梅梅拿出几颗后，还剩下40颗？", "answer": "8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "喜羊羊和灰太狼比赛骑木马，喜羊羊骑了50下，灰太狼骑了40下。喜羊羊比灰太狼多骑了多少下？", "answer": "10", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "妈妈买了一些梨，小刚吃了3个，妈妈吃了2个，小刚和妈妈一共吃了多少个？", "answer": "5", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "小苹种7盆红花，又种了同样多的黄花，两种花共多少盆？", "answer": "14", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "光明小区棋牌院有69副象棋，60副跳棋，象棋比跳棋多多少副？", "answer": "9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "有67颗五角星，第一大组发了14颗，第二大组发了6颗，还剩多少颗五角星？", "answer": "47", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "妈妈买来18个鸡蛋，吃了5个后，还剩多少个鸡蛋？", "answer": "13", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小明有9只纸飞机，比小军多3只。小军有多少只纸飞机？", "answer": "6", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "汽车站有40辆汽车，第一次开走20辆，第二次开走10辆，还剩多少辆？", "answer": "10", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "学雷锋小组上午修了20把椅子，下午修了10把，一天总共修了多少把椅子？", "answer": "30", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "老师买回14盒白粉笔和2盒彩粉笔，上课用了1盒粉笔，还剩几盒粉笔？", "answer": "15", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "同学们要做29个灯笼，已经做好了9个，还要做多少个？", "answer": "20", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "森林运动会上，小狐狸比小狗多跑9米，小兔比小狗多跑4米，小狐狸比小兔多跑多少米？", "answer": "5", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 1, "question": "小灰兔拔了9个萝卜，小白兔拔的是小灰兔的2倍。它们一共拔了多少个萝卜？", "answer": "27", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "树上有21只小松鼠，树下有8只，一共有多少只小松鼠？", "answer": "29", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小红已经写了8个字，还差5个字没写，小红要多少个字？", "answer": "13", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "杉杉捡了20个贝壳，还要捡39个，一共要捡多少个贝壳？", "answer": "59", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "学校体育室里有23个足球，被人借走了10个，还有多少个？", "answer": "13", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小兔拔了26个萝卜，比小猪少6个，小猪拔了多少个？", "answer": "32", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "商店里有16个蛋糕，卖出去5个，还剩多少个？", "answer": "11", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一艘船可以坐50人，船上已经有30人，还可以坐几人？", "answer": "20", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "马场上有20匹马，又来了21匹，现在马场上有多少匹？", "answer": "41", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "节日里，欢欢买了7个气球，回家路上爆了3个，他带回家几个气球？", "answer": "4", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "乐乐有19枚邮票，文文有8枚邮票，一共有多少枚邮票？", "answer": "27", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "农民伯伯上午摘了16筐番茄，运送到食堂3筐后，下午又摘了6筐。现在农民伯伯需要把多少筐番茄运到市场上卖？", "answer": "19", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "亮亮有15张画片，送给弟弟5张，还剩下几张画片？", "answer": "10", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "电饭锅卖了8个，还剩下7个，原来一共有多少个？", "answer": "15", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "从前面数，贝贝排9，从后面数他排12，这一排一共有几人？", "answer": "20", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "阳阳有102张1元的纸币，他最多能换几张50元的纸币？", "answer": "2", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 1, "question": "活动课上打乒乓球的有8人，做操的有24人。打乒乓球球和做操的同学共有多少人？", "answer": "32", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "苗苗家里有50个苹果，有30个梨，有20个桃。桃比苹果少多少个？", "answer": "30", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小明有8本书，小丽有9本书，小强有6本书。小明和小丽的书总数比小强多几本？", "answer": "11", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "爸爸买一条毛巾要付4元，买一支笔用去同样多的钱，一共要付多少元钱?", "answer": "8", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "婷婷买了12只气球，娜娜买了10只，她们一共买了多少只气球？", "answer": "22", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "商店新进了79台电脑，卖出9台后，还剩多少台电脑？", "answer": "70", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "弟弟今年4岁，哥哥比弟弟大8岁，2年前哥哥多少岁？", "answer": "10", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "修一条99米的路，第一天修了22米，第二天修了8米，还要修多少米才能修完？", "answer": "69", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "学校运动会中，参加游泳的有32名同学，参加跑步的有30名，参加这两项比赛的一共有多少名同学？", "answer": "62", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "伟伟现在身高98厘米，比去年长高了7厘米，伟伟去年身高多少厘米?", "answer": "91", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "停车场里有4辆小轿车，后来又开来3辆，停车场里一共有多少辆小轿车？", "answer": "7", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "有2个小猴子各摘了8个桃子，一个小猴子吃了2个后，他们把剩下的都放进盒子里。盒子里有几个桃子？", "answer": "14", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "晚上停电，小文在家点了8支蜡烛，先被风吹灭了1支蜡烛，后来又被风吹灭了2支，最后还剩多少支蜡烛？", "answer": "5", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 1, "question": "李伯伯家养了33只鸡，30只鸭，9只鹅。鸭和鹅一共多少只？", "answer": "39", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一列火车的第10号车厢原有85人，到下一站后有23人下车，又到一站后有7人下车。再开车时，这节车厢有多少人？", "answer": "55", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "飞机场上有75架飞机，飞走了50架，现在机场上有飞机多少架？", "answer": "25", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "妈咪带了98元，买菜花了23元，买水果花了7元，还剩多少元？", "answer": "68", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "果园有96棵梨树，90棵桃树。果园里梨树比桃树多多少棵？", "answer": "6", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "水果店运进19个西瓜，卖出7个后，又运进了5个，水果店现在有多少个西瓜？", "answer": "17", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "小路要买一本书，要付13元，他只有4元，还差多少元钱？", "answer": "9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小红有7本新书，又买了4本。一共有多少本？", "answer": "11", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "狗妈妈生了9只小狗，这9只小狗分别住宅3个小房子里，最小的房子里住了2只小狗，稍大些的房子里住了3只小狗，那么最大的房子里住几只小狗？", "answer": "4", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 1, "question": "一根彩带长67米，第一次用去14米，第二次用去6米，现在还剩多少米？", "answer": "47", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "幼儿园小朋友叠纸鹤，小刚叠了7个，小丽叠了5个，小红叠了6个，三个小朋友共叠多少个？", "answer": "18", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "一个饭盒7元，一条毛巾4元，妈妈只带了10元，买这两样东西还差多少元钱？", "answer": "1", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "一次上体育课排队，从左边开始报数，明明报了“7”，林林报了“10”;从右边开始报数，明明报了“7”，林林应该报几？", "answer": "3", "reasoning_step": 4, "num_digits": 1}
|
||||
{"grade": 1, "question": "妈妈买来30个苹果，吃了一些，还剩下20个，那么妈妈吃了多少个苹果？", "answer": "10", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "买一个书包需要30元，买一本书需要20元，买一个玩具小汽车需要40元。买一个书包和一本书一共用多少元？", "answer": "50", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "树上一共有32个桃子，猴哥哥摘了13个，猴弟弟摘了7个，树上还剩多少个桃子？", "answer": "12", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "豆豆买一个羊肉串用了4元，买一瓶汽水用了2元，一共用了多少元钱？", "answer": "6", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 1, "question": "我和小朋友玩捉迷藏的游戏。一共有16个小朋友藏起来了，我已经找到了5个，还有几个小朋友没找到？", "answer": "11", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "王叔叔家有白兔70只，黑兔20只，灰兔40只。白兔和黑兔共有多少只？", "answer": "90", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一个贝壳37元，一个海螺20元，各买一个需要多少钱？", "answer": "57", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小狮子今年8岁，狮妈妈是小狮子年龄的3倍，狮妈妈今年多少岁？", "answer": "24", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小红爸爸给她买了4支铅笔，妈妈又给她买了一盒，小红现在有16支铅笔，妈妈买的那一盒铅笔有多少根？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "老师奖励小红花，小红有4朵，王利有6朵，张鹏有5朵，他们三人一共有多少朵？", "answer": "15", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "妈咪有39元，买梨子用了8元，还剩多少元？", "answer": "31", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一只羽毛球要2元，波波付了50元，应找回多少元？", "answer": "48", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "学校舞蹈小组有33人，其中女生有20人，男生有多少人？", "answer": "13", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "同学们参加劳动，摘黄瓜46筐，摘的白瓜比黄瓜少10筐，摘白瓜多少筐？", "answer": "36", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一（1）班上男生有26人，女生有20人。女生比男生少多少人？", "answer": "6", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "小苹种7盆红花，又种了同样多的黄花和紫花，三种花共多少盆？", "answer": "21", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "玩具飞机价格是30元，玩具汽车价格是15元，布娃娃价格是20元。一架飞机比一辆汽车贵多少钱？", "answer": "15", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "公交车上原来有10名乘客，到站后下来4个人，又上去8个人，现在车上有多少人？", "answer": "14", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "菲菲书包有15本书，我书包有10本，菲菲书包里的书比我的多多少本?", "answer": "5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "鸟巢里有10只小鸟，先飞走了7只，又飞来了2只，鸟巢现在有几只小鸟？", "answer": "5", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "一双球鞋29元，一双布鞋比一双球鞋便宜9元。买一双球鞋和一双布鞋一共要用多少元?", "answer": "49", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "欣欣上午吃了26块饼干，下午吃了6块，一天一共吃了多少块饼干？", "answer": "32", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "一共有14个足球，小熊抱走了4，小狗抱来了3个，现在有多少个？", "answer": "13", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "奇奇和美美一共拍了80下篮球，其中奇奇拍了50下，那么美美拍了多少下？", "answer": "30", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 1, "question": "姐姐今年8岁，姐姐比弟弟大2岁，哥哥比弟弟大5岁，哥哥今年几岁？", "answer": "11", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 1, "question": "一（1）班有男生23人，女生20人，一共有多少人？", "answer": "43", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "张师傅上午修了18把椅子，下午修了29把椅子，一天共修了多少把椅子？", "answer": "47", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "小小图书室有52本故事书，科技书比故事书多6本，科技书有多少本？", "answer": "58", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "李师傅用一根长52米的绳子做跳绳，每根跳绳长6米，最多可以做多少根跳绳？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一个皮球65元，一个球拍38元，皮球比球拍多几元？", "answer": "27", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "2个茶杯装一盒，19个茶杯可以装满多少盒？", "answer": "9", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小明和小芳一起做跳绳，小明跳了67下，小芳跳了42下。小芳再跳多少下就和小明跳的一样多？", "answer": "25", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "有60只灯泡，每盏吊灯需装7只灯泡。这些灯泡可以装多少盏吊灯？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "有32名同学排队，每4名同学站一排，可以站几排？", "answer": "8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "小丽的妈妈买回来两根绳，一根长15米，另一根长21米，截去一些做跳绳，还剩8米，做跳绳用去多少米？", "answer": "28", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "花店运来19朵玫瑰花，每2朵扎成一束，可以扎成多少束？", "answer": "6", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一个三角形纸片有3个角，一个四边形纸片有4个角，那么6个三角形纸片和1个四边形共有多少个角？", "answer": "22", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "有25本笔记本，每5本笔记本放在一个盒子里，需要多少个盒子？", "answer": "5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "元旦到了，二（4）班买了42个气球装点教室。如果每7个气球扎成一束，能扎成几束？", "answer": "6", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "妈妈买了3千克苹果和5千克梨，每千克梨6元，每千克苹果3元，苹果和梨共多少钱？", "answer": "39", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "李立用彩纸做花，已经用了45张，还差8张，李立要用多少张彩纸？", "answer": "53", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "书店有3捆书，每捆2本，每本8元，书店把这两捆书全部卖完，可以卖多少元？", "answer": "48", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小方家有45只鸡，养的鸭比鸡少5只，小方家有鸭多少只？", "answer": "40", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "关村要修一条长92米的马路，已经修了54米，还有多少米没修？", "answer": "38", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "老师买了48根跳绳，平均分给6个班，每个班分几根？", "answer": "8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "哥哥把21本书平均分给小里、小洛和小骏，小骏分到多少本？", "answer": "7", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "一本笔记本卖4元。欢欢有27元，可以买多少本笔记本？", "answer": "6", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "我们小组共有12人，我的左边有7人，我的右边有多少人？", "answer": "4", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小华有95枚邮票，小明有83枚邮票，小明的邮票比小华多多少枚？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "小兰的妈妈做一套衣服用去3米布，25米布可以做多少套这样的衣服？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "二年级买来科技书18本，故事书24本。把这些书分给二年级6个班，平均每个班分多少本？", "answer": "7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "同学们今天种了25棵树，前天种了12棵树，昨天种了22棵树，昨天和今天一共种了多少棵？", "answer": "47", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "一个篮球的售价是48元，一个足球的售价是39元，一个足球的售价比一个篮球少多少元？", "answer": "9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "杯子里有72颗糖，拿走38颗，杯子里还有多少颗糖？", "answer": "34", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "红、黄、蓝三种气球一样多，一共15个，红气球有多少个？", "answer": "5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "小平种了26盒红花，又种了同样多的黄花，两种花共种了多少盒？", "answer": "52", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "7辆小汽车收费多少元？(停车收费标准:小汽车：6元;面包车：8元)", "answer": "42", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "爸爸用50元买了7本笔记本，商店的阿姨找了爸爸1元，那么一本笔记本多少元？", "answer": "7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一个足球要36元，小丽有24元，还差多少元才能购买一个足球？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "学校买来74本课外书，平均分给9个班，每个班分多少本？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小松鼠摘松果，每个筐里放了6个松果，一共放了8筐，小松鼠一共摘了多少个松果？", "answer": "48", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "2.二（2）班筹集班费87元，买一个篮球需要52元，买清洁工具需要46元，还差多少元？", "answer": "11", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小丽家离学校53米远，一天她上学走了23米，想起忘记带手工纸，立即返回家中，拿了手工纸再上学，这次上学她一共走了多少米？", "answer": "99", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "同学们做了5盒大红花，每盒装9朵，送三好学生32朵，还剩下多少朵？", "answer": "13", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "刘老师有5盒乒乓球，每盒装6个，同学们借走了23个，还剩多少个？", "answer": "7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "5头牛1天吃了30筐草，平均每头牛吃几筐草?", "answer": "6", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "二年级有男生20人，女生25人，每9人一组参加跳绳比赛，一共可以分成几组？", "answer": "5", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一只兔子有4只脚，一只鸡有2只脚，4只兔子和一只鸡一共有多少只脚？", "answer": "18", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "合唱队有男同学38人，比女同学少15人，合唱队共有多少人？", "answer": "91", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "露露：我踢了45下。佳佳：我比露露多踢18下。乐乐：我比佳佳多踢21下。乐乐踢了多少下？", "answer": "84", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "9只兔子有多少条腿？", "answer": "36", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "一批产品，4人做9天可以完成。照这样计算，这批产品要在6天完成，需要多少人？", "answer": "6", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 2, "question": "李明红剪五角星，第一次剪了13个，第二次和第一次剪的同样多，两次一共剪了多少个？", "answer": "26", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "27个小朋友去划船，每条船可以坐3人，需要几条船？", "answer": "9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "一辆公交车原来有乘客46人，到西门站下车17人，又上车11人，公交车上现在有多少人？", "answer": "40", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "某商店新进一批洗衣粉，共有33袋，每7袋装一箱，可以装多少箱？", "answer": "4", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "李奶奶养了3只公鸡，15只母鸡。6只鸡住一个鸡笼，一共需要多少个鸡笼？", "answer": "3", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "二年级（1）班借来15本书。平均分给5个小组，每组分多少本？", "answer": "3", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "妈妈今年32岁，小明比妈妈小19岁，小明今年多少岁？", "answer": "13", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "妈妈买来9个桃，爸爸买来15个桃，把这些桃平均放在4个盘里，每盘放多少个桃?", "answer": "6", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "饲养员养了20只公鸡，12只母鸡，每4只放入一个笼子，需要多少个笼子?", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "果园里有苹果树26棵，桃树35棵，苹果树和桃树一共有多少棵？", "answer": "61", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "绘画组有4组，每组7人，书法组比绘画组少20人，书法组有多少人？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一个花店有菊花35朵，玫瑰花27朵，百合花29朵，这三种花一共有多少朵？", "answer": "91", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "同学们分4组做风车，每组做9只。送给幼儿园18只，还有多少只？", "answer": "18", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "有36盆盆景。把这些盆景摆成4排，平均每排摆几盆？", "answer": "9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "一筐桃的总重量是38千克，其中筐的重量是2千克，那么桃的重量是多少千克？", "answer": "36", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "公交车上原有32人，到站后上来15人，又下车4人，现在公交车上有多少人？", "answer": "43", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "共有86个西瓜，李大爷运走了39个，王叔叔运走了27个，还剩多少个？", "answer": "20", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "李老师去商店买粉笔，每盒粉笔7元，老师买了8盒，老师应支付多少元？", "answer": "56", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "养兔场养兔8000只，昨天卖出3000只。还剩下多少只？", "answer": "5000", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 2, "question": "食品加工厂某工作人员一上午生产了59个月饼，将这些月饼每8个装一袋，可以装满多少袋？", "answer": "7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "曾老师买回33个桔子，每个住校生分得4个，可以分给多少个住校生？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "操场上打篮球的有21人，打排球的有25人，走了18人，现在操场上打篮球和排球的一共有多少人？", "answer": "28", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "食堂买回30袋大米，吃了4天后，还剩2袋，平均每天吃多少袋？", "answer": "7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小明有23张卡片，小丽的卡片比小明少18张，小丽和小明一共有多少张卡片？", "answer": "28", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "有46个人住旅馆，每间房住6个人，一共可以住满多少间房？", "answer": "7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一筐松果有39个，每个篮子里装5个松果，至少需要多少个篮子？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小巧把59张彩色照片贴在相册上。每页贴了7张，可以贴多少页？", "answer": "8", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小明今年12岁，在小明出生时妈妈23岁，妈妈今年多少岁？", "answer": "35", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "妈妈买了3块蛋糕，每块6元，买蛋糕用去多少元？", "answer": "18", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "小萌在超市买了3瓶饮料花了9元，如果买5瓶饮料需要多少元？", "answer": "15", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小明看一本故事书，每天看5页，已经看了4天，还剩16页没看完，这本故事书一共有多少页？", "answer": "36", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "从教室搬11张桌子布置会场，每个老师搬一张，两个同学抬一张。现在有2位老师还要多少个同学帮忙才能一次搬完？", "answer": "18", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小图书室有90本故事书，借出去一些后，还剩43本，借出多少本？", "answer": "47", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "妈妈今年42岁，小明今年16岁，30年后妈妈比小明大多少岁？", "answer": "26", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "一辆汽车上午运货23箱，下午运货12箱，下午比上午少运多少箱？", "answer": "11", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "二（1）班有25名学生。平均分成5个小组，每个小组有多少名学生？", "answer": "5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "小林妈妈有98元，给小林买书包花了26元，还剩下多少元？", "answer": "72", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "二年级一班的4名老师带领23名学生租两辆车去旅游，其中一辆车上坐了13人，另一辆车上坐了多少人？", "answer": "14", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "爷爷有11个西瓜，一个篮子正好能装3个西瓜。爷爷的西瓜能装满多少个篮子？", "answer": "3", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一大袋洗衣粉重10千克，一大袋洗衣粉里面有2小袋，一小袋洗衣粉重多少千克？", "answer": "5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "松鼠每天吃5个松果，吃了3天，还剩20个。原来一共有多少个松果？", "answer": "35", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "王老师做了一些花奖励给明明和东东，明明得到15朵，东东得到23朵，王老师一共做了多少朵花？", "answer": "38", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "小明有63张卡片，小丽有45张，小丽给小明多少张后两人的卡片一样多？", "answer": "18", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "同学们排成一队，小丽前面有23人，后面有19人，这一队共有多少人？", "answer": "43", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "一年级有52名同学，准备乘两辆车去公园，一辆车上已经坐了21名同学，另一辆车上坐了多少人？", "answer": "31", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "一只猴子重32千克，一只山羊比它重7千克，这只山羊重多少千克？", "answer": "39", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "有12个小朋友参加跳绳比赛，4个人一组，可以分成几组？", "answer": "3", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "实验小学三，四年级的同学去植树，三年级种了37棵，四年级种了45棵，这两个年级一共种了多少棵树？", "answer": "82", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "花店运来19朵玫瑰花，每2朵扎成一束，还剩多少朵？", "answer": "1", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "猴妈妈摘来26个桃，平均分给4只小猴，每只小猴分得多少个桃？", "answer": "6", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "小红和小明去钓鱼，小红钓了27条鱼，小明钓的比小红多钓9条鱼，小明钓了多少条？", "answer": "36", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "学校体育室有92个篮球，每班借5个，共有9个班。还剩多少个篮球？", "answer": "47", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 2, "question": "把20粒珠子，平均放入红、黄、蓝、白四个袋子里。每个袋子放多少粒珠子？", "answer": "5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 2, "question": "同学们擦玻璃，每间教室有4扇窗户，共有8间教室，擦了28扇，还剩多少扇窗子没擦？", "answer": "4", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "小猴摘了84个桃子，平均分给6只猴子，每只猴子能吃到几个桃子？", "answer": "14", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "李师傅4天加工40个零件，照这样计算，今年4月份全月一共可加工多少个零件？", "answer": "300", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "某班学生48人，其中21人参加数学竞赛，13人参加作文竞赛，而且有7人既参加数学竞赛又参加作文竞赛。那么只参加数学竞赛的有多少人？", "answer": "14", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "做一套校服用布3米。66米布最多能做多少套？", "answer": "22", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "同学们收集图片。张明、李红、蔡正明、王丹、熊伟、高伟、梅芳7人收集了名山图片，吴凤、李红、王丹、戴月红、高伟这5人收集了河流图片，吴心怡、张冬、李可这3人收集了奥运图片。收集名山图片和奥运图片的共有多少人？", "answer": "10", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "食堂运来一批蔬菜，原计划每天吃45千克，10天可以消费完这批蔬菜，后来根据大家意见，每天比原计划多吃5千克，这批蔬菜可以吃多少天？", "answer": "9", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "菜店运来6吨大白菜，上午卖出3000千克，下午全部卖完．下午卖出大白菜多少千克？", "answer": "3000", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 3, "question": "李师傅加工490个零件，前5天共加工了350个零件，照这样计算，这批零件共需要多少天能加工完？", "answer": "7", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "“金秋”体育节的检阅方队中，每班36人，某小学有71个班，一共有多少人参加检阅？", "answer": "2556", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 3, "question": "小豪家有个书架共5层，每层放36本书，现在要空出两层放碟片，把这些书放入3层中，每层比原来多放多少本？", "answer": "24", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "妈妈把一块长1米的布分成六块。第一块长3/6米，第二块长1/6米，剩下的那块长多少米？", "answer": "2/6", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "停车场上停放着5辆面包车，停放的轿车数量是面包车的4倍。停车场上一共停放着多少辆轿车和面包车？", "answer": "25", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "某荔枝园有一块长方形的果园，长16米，宽7米。这块荔枝园的面积是多少平方米？", "answer": "112", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "王叔叔和他的爸爸、妈妈去风景区游玩。如果风景区的每张门票售价是108元，王叔叔买所有人的门票一共应付多少钱？", "answer": "324", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "一项工程，8个人工作15小时可以完成，如果12个人工作，那么多少小时可以完成？", "answer": "10", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "阿亮7：35从家步行上学，7：44到达学校。他家到学校的距离是306米。阿亮每分钟走多少米？", "answer": "34", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "希望小学三年级8个班为地震灾区捐书656本，平均每班捐书多少本？", "answer": "82", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "李春从学校放学回家走了20分钟，他平均每分钟走110米，李春家离学校有多远？", "answer": "2200", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 3, "question": "某小区的绿化带3/9的地种了柏树，剩下的种松树，种松树的地占整个绿化带的几分之几？", "answer": "6/9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "一个没关紧的水龙头，一星期能流失约670千克的水。平均每天大约流失多少千克的水？", "answer": "96", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "王老师在教师节收到138朵花，如果每3朵扎成一束，可以扎几束？", "answer": "46", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "刘阳今年6岁，妈妈的年龄是刘阳年龄的5倍，奶奶的年龄又是妈妈年龄的2倍，奶奶今年多少岁？", "answer": "60", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "小华带了20元钱去文具店买学习用品，买一支钢笔用去总钱数的1/2，买一本笔记本用去总钱数的1/5，一支钢笔比一只笔记本贵多少钱？", "answer": "6", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "一套明信片14张，每张售价5元，今天一共卖出52套，一共收入多少元？", "answer": "3640", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 3, "question": "为了美化校园环境，学校买了160盆菊花，分别摆在校门口两侧，平均每侧放多少盆？", "answer": "80", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "兴兴玩具厂生产了408个布娃娃，如果每3个布娃娃装一箱，一共可以装多少箱？", "answer": "136", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "王老师买了一根长绳，现在要把这根长绳剪成几根同样长的短绳。王老师把长绳连续对折三次，然后沿折痕剪开，每根短绳占这根长绳的几分之几", "answer": "1/8", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "货车平均每小时行87千米，从甲城到乙城行了16小时，甲、乙两城相距多少千米？", "answer": "1392", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 3, "question": "电影院共有1400个座位。东风小学共有31个班，平均每班有43人，请问这个学校的所有学生都坐下后还剩多少座位？", "answer": "67", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 3, "question": "某工地的一项工程，原计划由30人工作，每天工作8小时，50天完工，为了提前完工，实际由40人工作，每天工作10小时，可以提前几天完工？", "answer": "20", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 3, "question": "民兵军训，4小时走16千米，为了达到目的地，每小时多走1千米，剩下的20千米要几小时？", "answer": "4", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "商店新进560千克苹果，如果5天卖完，平均每天卖多少千克？", "answer": "112", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "李明家去年养鸭600只，今年比去年多248只，今年养了多少只鸭？", "answer": "848", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "某剧场楼上有16排座位，每排可以坐42人，平均每个座位票价5元，若座位全部坐满，这个剧院可以收入多少元？", "answer": "3360", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 3, "question": "一块边长为2米的正方形花布，它的面积是多少平方米？", "answer": "4", "reasoning_step": 1, "num_digits": 1}
|
||||
{"grade": 3, "question": "新华书店购进故事书860本，文艺书410本，购进的故事书比文艺书多多少本?", "answer": "450", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "在一块边长8米的正方形水池外围上一条宽1米的小路，求小路的面积。", "answer": "36", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 3, "question": "学校食堂运进1240千克食用油，用了8天，还剩40千克，平均每天用多少千克食用油？", "answer": "150", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 3, "question": "丽丽从一楼走到三楼用了18秒，照这样计算，他从一楼到七楼需要多长时间？", "answer": "54", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "小强写一篇作文一共写了608个字，正好4面，每面有8行。平均每面写多少个字？", "answer": "152", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "用篱笆围成一个长方形的养鸡场，一边利用18米长的墙壁，篱笆共长40米，求养鸡场的面积。", "answer": "198", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 3, "question": "有6个小组，每组4人，共摘了216千克苹果，平均每人摘了多少个苹果？", "answer": "9", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "一根绳子长25米，剪10米做一根长跳绳，剩下的每2米做一根短跳绳，可以做多少根短跳绳？", "answer": "7", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "冷饮公司每天生产960支奶油冰淇淋。每盒装6支，可以装几盒？", "answer": "160", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "5个工人10天修了500米，若速度不变，20人要修4000米，需要多少天？", "answer": "20", "reasoning_step": 4, "num_digits": 4}
|
||||
{"grade": 3, "question": "学校新买来780本图书，分给6个年级的学生，平均每个年级可以分到多少本？", "answer": "130", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "超市里一个书包32元，买21个这样的书包需要多少元？", "answer": "672", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "在一块长30米，宽15米得长方形地理栽白薯，平均15平方分米栽一棵，平均每棵收白薯2.5千克，这块地共收白薯多少千克？", "answer": "7500", "reasoning_step": 4, "num_digits": 4}
|
||||
{"grade": 3, "question": "为了绿化校园，学校买来了60盒鲜花，这些鲜花如果摆6排，每排可以摆多少盒？", "answer": "10", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "明明跑了450米，爷爷跑的米数相当于明明跑的2倍，爸爸跑的比明明和爷爷跑的总数还多850米。爸爸跑了多少米？", "answer": "2200", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 3, "question": "羊圈里有黑羊8只，白羊448只。白羊是黑羊的多少倍？", "answer": "56", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "一本书有66页，芳芳准备3天看完，平均每天看多少页？", "answer": "22", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "小明的爸爸在河边挖了一块长方形菜地，长8米，宽5米．他用篱笆把不靠河边的三面围了起来，至少得用多少米篱笆？", "answer": "18", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "自行车厂4名工人5小时能安装自行车80辆，现在要10小时内安装800辆自行车，需增加多少名工人？", "answer": "16", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 3, "question": "一个长方形运动场，长250米，宽200米，早操时三年级同学绕操场跑了两圈，同学们跑了多少米？", "answer": "1800", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 3, "question": "一块长方形菜地，长6米，宽5米，四周围上篱笆，其中有一面靠墙，篱笆至少长多少米？", "answer": "16", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "一根铁丝可以围一个边长10分米的正方形，现在用它来围一个长方形，如果这个长方形的宽是7分米，那么长是多少分米？", "answer": "13", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "第一根电线长10.4米，第二根电线比第一根短1.8米。两根电线共长多少米？", "answer": "19", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "有160名学生，每两人用一张桌子，如果把这些桌子平均放在4间教室里，每间教室放多少张桌子？", "answer": "20", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "安装一条水管，头4天安装了120米，还要15天可装完，这条水管总长多少米？", "answer": "450", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 3, "question": "某商店里手套12.4元一副，帽子35.7元一顶。买一副手套和一顶帽子一共要多少钱?", "answer": "48.1", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "两只熊猫4天吃了56千克竹子，平均每天一只熊猫吃多少千克竹子？", "answer": "14", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "三只大象和1只小河马共重16吨，2只大象和1只小河马共重11吨。请你算出1只小河马的体重。", "answer": "1", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "停车场里卡车的辆数是客车的6倍，客车有8辆，卡车有多少辆？", "answer": "48", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "学校举行大型团体表演。有18个班参加，每班站6行，每行9人。参加表演的一共有多少人？", "answer": "972", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "学校跑道内圈200米，若沿着内圈走5圈，一共走了多少米？", "answer": "1000", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 3, "question": "三（1）班有50人，其中25人喜欢吃苹果，22人喜欢吃橘子，13人喜欢吃苹果又喜欢吃橘子。两种水果都不喜欢的有多少人？", "answer": "16", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "从学校到木兰湖有416千米，乘客车4小时可以到达，这两客车平均每小时行驶多少千米？", "answer": "104", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "果园里有桃树120棵，梨树的棵数是桃树的3倍，梨树有多少棵？", "answer": "360", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "小林看一本故事书，3天看了24页．照这样计算，7天可以看多少页？", "answer": "56", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "一捆树苗有12棵。三(1)班要栽13捆树苗，一共要栽多少棵？", "answer": "156", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "为响应“退耕还林”的号召，学校组织植树活动，同学们共领来435棵树苗。如果每行栽8棵，可以栽多少行？", "answer": "54", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "一张写字台的长是13分米，宽是6分米。它的面积是多少平方分米？", "answer": "78", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "甲、乙两个牧童在草地上放羊。乙有5只羊，甲说：“把你的羊给我2只，我的羊就是你的5倍了。”你能猜出甲原有多少只羊吗？", "answer": "13", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "张叔叔单位每加班一次给加班费65元，这个月张叔叔一共加班14次。张叔叔这个月可得加班费多少元？", "answer": "910", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "马拉松比赛，从起点开始设服务站，以后每隔1000米设一个。当小伟跑到第5个服务站时，他跑了多少米？", "answer": "4000", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 3, "question": "商店上午卖出上衣48件，是下午卖出的4倍，下午卖出多少件？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "3千克鲜鱼可以制成1千克鱼干。750千克鲜鱼可以制成多少千克鱼干？", "answer": "250", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "小林看一本故事书，3天看了24页．照这样计算，全书160页，多少天可以看完？", "answer": "20", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "王洋的卧室地面长是8米，宽是5米，如果用面积为4平方分米的方砖铺地，需要方砖多少块", "answer": "1000", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 3, "question": "学校共有科技图书760本，平均放在8个书架上，每个书架有5层，每层放多少本图书？", "answer": "19", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "学校图书室新买了24套儿童读物，每套12本，这批儿童读物共有多少本？", "answer": "288", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "小东看一本故事书，第一天看了2/8，第二天看了3/8，小东已经看了这本书的几分之几？", "answer": "5/8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "三年级(1)班有学生56人，老师把全班同学的一半平均分成4组去擦玻璃，平均每组有多少人？", "answer": "7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "饲养小组养了425只兔子，每6只关一个笼子，最少需要多少个笼子？", "answer": "71", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "一桶油连油带桶共重200千克，倒出一半油以后，连桶重110千克，原来油重多少千克？", "answer": "180", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 3, "question": "爷爷今年63岁，小明今年7岁，爷爷的年龄是小明的多少倍？", "answer": "9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "某制衣车间的8名工人4小时做了160件衣服，平均每人每小时做多少套衣服？", "answer": "5", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "一张长方形纸的2/7涂蓝色,3/7涂红色,没有涂色的部分占这张纸的几分之几?", "answer": "2/7", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "有大米84袋，每袋质量相同。用卡车一次最多运7袋，至少要运多少次？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "一辆汽车从甲地开往乙地，每小时行60千米，4小时到达。若要3小时到达，则每小时需要多行多少千米？", "answer": "20", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 3, "question": "陈敏在期中考试中语文的93分，数学得98分．如果她语文、数学、英语三科的平均成绩是97分，你知道她英语得了多少分吗？", "answer": "100", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 3, "question": "一辆洒水车，每分钟前进200米，洒水的宽度是7米．洒水车行驶8分钟，能给多大的地面洒水？", "answer": "11200", "reasoning_step": 2, "num_digits": 5}
|
||||
{"grade": 3, "question": "池塘里有360条鲤鱼，有240条鲫鱼，池塘里一共有多少条鱼？", "answer": "600", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "爷爷今年77岁，欢欢今年7岁。今年爷爷的年龄是欢欢的几倍？", "answer": "11", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "一个长方形草坪的面积是42平方米，宽是3米，现在长不变，将宽增加到5米后，面积是多少平方米？", "answer": "70", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 3, "question": "妙可的祖奶奶今年100周岁了，她是2月29日出生的．请你算一下，妙可的祖奶奶一共过了多少个生日？", "answer": "25", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 3, "question": "王大爷养了210只兔子，平均装入7个笼子里，每个笼子里装多少只兔子？", "answer": "30", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 3, "question": "三年级60个同学参加植树活动，如果每5人一组，可以分成几组？", "answer": "12", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 3, "question": "有128个苹果，平均分给8个小朋友，每个小朋友分几个？", "answer": "16", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "用面包机烤面包时，第一面烤2分钟，第二面只要烤1分钟，即烤一片面包需要3分钟，小勤的面包机一次只能放2片，他每天早上吃3片面包，至少需要烤多少分钟？", "answer": "5", "reasoning_step": 4, "num_digits": 1}
|
||||
{"grade": 4, "question": "一个数的千位上是2，个位上是3，十位和十分位上都是1，百位和百分位上都是8，这个数是多少？", "answer": "2813.18", "reasoning_step": 5, "num_digits": 6}
|
||||
{"grade": 4, "question": "一件上衣156元，一条裤子144元，一件上衣和一条裤子一共多少元？", "answer": "300", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "水果店购回香梨和苹果各36箱，香梨每箱45元，苹果每箱55元。香梨和苹果一共花了多少元？", "answer": "3600", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "钱红和顾颖用计算机同时合打一份1820个字的材料。钱红每分钟打字120个，顾颖每分钟打字140个。她俩打完这份材料，需用多少分钟？", "answer": "7", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "张村挖一条长1200米的水渠，计划15天挖完。实际每天比计划多挖20米，实际用了多少天？", "answer": "12", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 4, "question": "煤场上午运来煤1.5吨，下午又运来了一些，一天共运来煤4.36吨，下午运来多少吨？", "answer": "2.86", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "某花店里菊花有300朵，是百合花的12倍，百合花有多少朵？", "answer": "25", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "运输队运水泥，一辆大车每次运90包，一辆小车每次运35包，大车和小车各运16次，一共运水泥多少包？", "answer": "2000", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "水果店运来26箱橙子，每箱36千克，运来的苹果比橙子多278千克，水果店运来橙子和苹果共多少千克？", "answer": "2150", "reasoning_step": 4, "num_digits": 4}
|
||||
{"grade": 4, "question": "一辆汽车从甲地开往乙地，每小时行驶76千米，行驶了7个小时，甲、乙两地的距离是多少千米？", "answer": "532", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "把2.01扩大到原数的100倍是多少？", "answer": "201", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "莉莉有129张邮票，明明有71张邮票，每张邮票售价2元，莉莉和明明可以卖多少元？", "answer": "400", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "汽车上山的速度为每小时36千米，行了5小时到达山顶，下山时按原路返回只用了4小时。汽车下山时平均每小时行多少千米？", "answer": "45", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 4, "question": "某工厂去年生产农具3657件，今年比去年多生产679件，今年生产了多少件农具？", "answer": "4336", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "可可在家里烙饼，锅里每次可烙两张饼，两面都要烙，每面要烙2分钟，烙7张饼要用几分钟？", "answer": "14", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 4, "question": "1吨海水含盐20千克，100千克这样的海水含盐多少吨？", "answer": "0.002", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "妈妈到超市买苹果用去21.76元，买梨用去19.39元，一共用去多少元？", "answer": "41.15", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "春节前，工厂后勤工作人员带1000元买桔树，买了13盆桔树，还剩155元，平均每盆桔树的价格是多少元？", "answer": "65", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "春节快到了，小明准备送给几个好朋友祝福贺卡，他了解到每张贺卡2元，于是他带着买15张贺卡的钱来到商店，发现他想买的这种贺卡每张涨了1元．请帮小明算一算，他所带的钱现在可以买多少张贺卡？", "answer": "10", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 4, "question": "一瓶橙汁是4元钱，一箱24瓶，王叔叔这个月卖了25箱，一共卖了多少元钱？", "answer": "2400", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "修路工人要修一条850米长的公路，已经修了19天，还有185米没修，平均每天修了多少米？", "answer": "35", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "学校买足球用去96.8，买排球用去56.6，买篮球用去103.2元。学校买足球、排球和篮球共用了多少元？", "answer": "256.6", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "由4个一、4个十分之一、4个千分之一组成的数是多少？", "answer": "4.404", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "王老师带了386元钱，买足球用去了130元，剩下的钱又买了8个篮球，平均每个篮球多少元？", "answer": "32", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "张兰在读一个小数时，把小数点丢了，结果读成了一万四千零二。如果原来的小数要读出两个零。那么原来的小数是多少？", "answer": "14.002", "reasoning_step": 1, "num_digits": 5}
|
||||
{"grade": 4, "question": "学校举行团体操表演，每行25位学生，一共有36行，参加团体操表演的学生一共有多少位？", "answer": "900", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "水果店运来24箱水果，每箱25千克，这些水果一共多少千克？", "answer": "600", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "小华的四次跳远成绩分别为141厘米、160厘米、161厘米、162厘米。请你帮小华算出她跳远的平均成绩是多少厘米。", "answer": "156", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 4, "question": "一块长方形地的面积是0.8公顷，它的宽是100米．那么它的长是多少米？", "answer": "80", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "学校买来练习本3640本，分给三年级256本，分给四年级378本，分给五年级444本，分给六年级322本。这四个年级一共分得多少本练习本？", "answer": "1400", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 4, "question": "甲在乙的后面28千米，两人同时相向而行，甲每小时行16千米，乙每小时行9千米，问甲几小时追上乙？", "answer": "4", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 4, "question": "王老师买香蕉用了10.75元，买橘子用了9.7元，他付给售货员25元，应找回多少元？", "answer": "4.55", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "新华书店运到2车图书，每辆车装125包，每包有80本。新华书店运到图书多少本？", "answer": "20000", "reasoning_step": 2, "num_digits": 5}
|
||||
{"grade": 4, "question": "102，103，105，101，109，102，106，104这8个数的平均数是？", "answer": "104", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 4, "question": "每箱可口可乐有18瓶，每瓶3元，爸爸拿80元钱买了一箱，还剩下多少元？", "answer": "26", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 4, "question": "把21.06扩大到原数的1000倍是多少？", "answer": "21060", "reasoning_step": 1, "num_digits": 5}
|
||||
{"grade": 4, "question": "10千克花生可榨油4.3千克，1千克花生可榨油多少千克？", "answer": "0.43", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "果园里种梨树18行，每行35棵。苹果树25行，每行32棵。果园里共有果树多少棵？", "answer": "1430", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 4, "question": "小明和小芳同院，小芳上学每分走50米，12分到学校，小明上学每分比小芳多走10米，小明几分到学校？", "answer": "10", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 4, "question": "张明看一本572页的故事书，如果他每天看22页，需要多少天才能看完？", "answer": "26", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "修一段长960米的公路，修了6天完成了全长的一半，余下的平均每天修80米，修完这段公路一共需要多少天？", "answer": "12", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 4, "question": "海沧野生动物园的狮子一天要吃37千克的食物，十月份一个月要吃多少千克食物？", "answer": "1147", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "小明在计算除法时，错将除数36看成63，结果得到商24。请你帮他算一算，正确的商应该是多少？", "answer": "42", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 4, "question": "一台磨面机每小时磨面800千克，照这样计算，7台磨面机每小时能磨面粉多少千克？", "answer": "5600", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "同学们进行跳绳比赛，规定每人跳3分钟。小红平均比小米多跳8个，小明共跳了321个。求小红一共跳了多少个？", "answer": "345", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 4, "question": "一束鲜花30元，买5束送一束。王阿姨一次买5束，每束便宜多少元？", "answer": "5", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 4, "question": "学校要进行作业展览，一共要展示270件作品，每块展板最多放25件，需要多少块展板？", "answer": "11", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "四年级同学们采集松树种子，（1）班采集了31.56千克，（2）班采集了33.45千克，（3）班采集了30.2千克。三个班一共采集了多少千克？", "answer": "95.21", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "用一个平底锅煎饼，每次最多能同时放2张饼，如果煎一张饼需要2分钟（假定正、反面各需1分钟）．煎3张饼至少需要几分钟？", "answer": "3", "reasoning_step": 3, "num_digits": 1}
|
||||
{"grade": 4, "question": "甲数是100，比乙数的6倍少20，乙数是多少？", "answer": "20", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "一辆大巴车从张村出发，如果每小时行驶60千米，4小时就可以到达李庄，结果只用了3个小时就到达了，这辆车实际平均每小时行驶多少千米？", "answer": "80", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 4, "question": "光明小学花了270元买新华字典，每本新华字典6元，能买多少本新华字典？", "answer": "45", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "南京到北京的公路长840千米，一辆汽车从南京开往北京，每小时行70千米，行11小时后，还剩多少千米？", "answer": "70", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "10千克油菜籽可榨油4.1千克，100千克油菜籽可榨油多少千克？", "answer": "41", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "莉莉每天装订图书31本，照这样计算，莉莉一个月可以装订多少本图书？（一个月按30天算）", "answer": "930", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "一条新建的高速公路，长100千米，宽50米．那么这条公路占地多少公顷？", "answer": "500", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "自行车3小时行45千米，汽车4小时行216千米。汽车平均每小时比自行车平均每小时多行多少千米？", "answer": "39", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 4, "question": "一辆汽车从沈阳开往大连，4小时后行了480千米．照这样的速度，再行3小时才能到大连．沈阳到大连有多少千米？", "answer": "840", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 4, "question": "一瓶油连瓶重3.5千克，用去油的一半后连瓶重1.8千克，原来这瓶里的油重多少千克？", "answer": "3.4", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 4, "question": "1000张纸叠起来厚10.2厘米，平均每张纸厚多少毫米？", "answer": "0.102", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "钢铁厂计划全年生产钢铁265.8万吨，结果上半年生产了155.7万吨，下半年生产了129.9万吨。全年超过计划多少万吨？", "answer": "19.8", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "一本漫画书有316页，林林第一天看了129页，第二天看了71页，还剩下多少页没看完？", "answer": "116", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "蓝天电影城每排有125个座位，一共有76排，每张票8元，若所有的票都已经售完，电影院可以收入多少元钱？", "answer": "76000", "reasoning_step": 2, "num_digits": 5}
|
||||
{"grade": 4, "question": "将一个小数先扩大到原来的100倍，再缩小到原来的1/1000，又缩小到原来的1/10后是0.038，这个小数原来是多少？", "answer": "3.8", "reasoning_step": 3, "num_digits": 5}
|
||||
{"grade": 4, "question": "小欣每天坚持练习毛笔字。他4个星期共写了448个毛笔字，平均每天写多少个毛笔字？", "answer": "16", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "一个养鸡场四月份卖出3500只鸡，五月份卖出5200只鸡，这两个月平均每个月卖出了多少只鸡？", "answer": "4350", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "李雷今年重42千克，比去年增加了3.6千克，他去年的体重是多少千克？", "answer": "38.4", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "一个工程队修路，第一天修了147米，第二天修了203米，第三天修了153米，三天一共修路多少米？", "answer": "503", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "10千克小麦可磨面粉8.3千克，照这样计算，1吨小麦可磨面粉多少千克？", "answer": "830", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "一块长方形菜地，它的长是50米，宽是40米，面积是多少平方米？", "answer": "2000", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "农场有一块长200米，宽150米的长方形试验田，平均每公顷产稻谷14吨，这块试验田一共产稻谷多少吨？", "answer": "42", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "公园里有牡丹花117棵，菊花59棵，兰花83棵，公园中这三种花一共有多少棵？", "answer": "259", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "一种播种机的播种宽度是3米，播种机每小时行5千米，照这样计算，2小时可以播种多少公顷？", "answer": "3", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 4, "question": "一只平底锅上只能煎两条鱼，用它煎一条鱼需要4分钟。（正反面各2分钟），那么，煎三条鱼至少需要几分钟？", "answer": "6", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 4, "question": "一个电脑安装小组26天安装了226台电脑，比原计划多安装了18台，原计划每天安装多少台？", "answer": "8", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "张老师带领四年级（1）班共45人去栽树，张老师一人栽5棵，男生一人栽3棵，女生一人栽2棵，总共栽树115棵，问有多少名男生？", "answer": "22", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 4, "question": "李伟昨天先从家出发走了400米买了张报纸，然后又走了600米去超市买了一瓶洗涤剂，共用了20分钟．他的平均速度是每分钟行多少米？", "answer": "50", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "公园举办玫瑰花展览，红玫瑰和黄玫瑰都摆了22行，红玫瑰每行56盆，黄玫瑰每行44盆，一共摆了红玫瑰和黄玫瑰多少盆？", "answer": "2200", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 4, "question": "我国发射的“嫦娥一号”探月卫星，以每秒11千米的速度飞向月球，5分钟后离开地球有多少千米？", "answer": "3300", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "一条新建的高速公路，长200千米，宽40米。这条公路占地多少公顷？", "answer": "800", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "四年级同学去参观博物馆，平均分成17队，每队都是27人，参观博物馆的学生共有多少人？", "answer": "459", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "某超市一次进了164箱纯牛奶和136箱酸牛奶，每箱纯牛奶和酸牛奶都是16盒，一共进了多少盒牛奶？", "answer": "4800", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "小红早晨7时从家步行到学校去，出发2分钟后，因事又立即返回家，然后又立即从家里赶往学校，到校门口时一看钟表正好是7时20分．如果小红每分钟走60米，她家到学校有多少米？", "answer": "960", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 4, "question": "果园里有1586棵果树，其中梨树517棵，枣树383棵，剩下的是苹果树，苹果树有多少棵？", "answer": "686", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "一支钢笔的价钱是10.50元，是一支铅笔价格的10倍，一块手表的价钱是这支铅笔价钱的1000倍，一块手表价钱是多少元？", "answer": "1050", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 4, "question": "王叔叔从县城出发去木兰乡送化肥。去的速度是50千米∕小时，用了4小时。返回时用了2小时。原路返回时平均每小时行多少千米？", "answer": "100", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "一筒羽毛球有8个，每个羽毛球3元，体育老师买羽毛球用了720元．他买了多少筒羽毛球？", "answer": "30", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "操场原是一个长300米，宽100米的长方形，经过拓宽，长和宽都增加了100米，现在的操场比过去增加了多少平方米？", "answer": "50000", "reasoning_step": 5, "num_digits": 5}
|
||||
{"grade": 4, "question": "南京到上海的水路长392千米，一艘轮船从南京开出每小时行49千米，经过几时船到上海？", "answer": "8", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "某市高中一年级学生进行野外军训。晴天每天行20千米，雨天行10千米，在12天内行程为200千米。这期间有多少天是晴天？", "answer": "8", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 4, "question": "一只船顺水行320千米需用8小时，水流速度为每小时15千米，这只船逆水行这段路程需用几小时？", "answer": "32", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 4, "question": "停车场停放着12辆自行车和三轮车，自行车和三轮车共有车轮26个，自行车多少辆？", "answer": "10", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 4, "question": "甲、乙两地相距425千米，王师傅开车从甲地到乙地出差，汽车每小时行85千米，已经走了170千米，还要几小时可以到达乙地？", "answer": "3", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "100千克稻谷可碾米75千克，1吨稻谷可碾米多少千克？", "answer": "750", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 4, "question": "一个三位小数四舍五入到百分位结果是6.60，这个小数最大是多少？", "answer": "6.604", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 4, "question": "有一块占地1公顷的正方形菜地，如果它的边各延长100米。那么菜地的面积增加多少公顷？", "answer": "3", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 4, "question": "有五根木条，它们的长度分别是2厘米、3厘米、4厘米、5厘米和6厘米，从它们当中选出3根木条拼成一个三角形，共有多少种不同的选法？", "answer": "7", "reasoning_step": 5, "num_digits": 1}
|
||||
{"grade": 4, "question": "一块长方形的玉米地，长600米，宽300米。如果每公顷平均收玉米10吨，这块玉米地能收玉米多少吨？", "answer": "180", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 4, "question": "一个旅游区，上午有游客367人，下午有游客298人，这一天一共有游客多少人？", "answer": "665", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 5, "question": "一组学生植树，每人栽6棵还剩4棵；如果其中3人各栽5棵，其余每人各栽7棵，正好栽完。这一组学生有多少人？", "answer": "10", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个底面是正方形的长方体纸盒，它的侧面展开图是一个边长是21厘米的正方形。这个纸盒的表面积是多少平方厘米", "answer": "2646", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 5, "question": "故宫的面积是72万平方米，比天安门广场面积的2倍少16万平方米．天安门广场的面积多少万平方米？", "answer": "44", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "一根竹竿插入水中，入水部分长5/14米，入泥部分1/14米，露出水面3/14米。这根竹竿长多少米？", "answer": "9/14", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "循环使用一本教科书，一年可以节约纸张2/5千克，如果一年循环使用3本教科书，可以节约纸张多少千克？", "answer": "6/5", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "在20米花坛的一侧，每隔4米栽一棵树苗（只栽一端）。一共需要多少棵树苗？", "answer": "5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "某校长方形操场面积2800平方米，因扩建，要把宽从40米增加到50米，长不变．扩建后的操场面积比原来增加多少平方米？", "answer": "700", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 5, "question": "一个长方体的底面是周长16厘米的正方形，高3厘米，这个长方体的体积是多少立方厘米？", "answer": "48", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "红领巾公园内一条林荫大道全长800米,在它的一侧从头到尾等距离地放着41个垃圾桶,每两个垃圾桶之间相距多少米？", "answer": "20", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "五年级同学到森林公园去春游，准备乘16人的面包车或乘24人的中巴客车，不论是专乘16人的面包车，还是专乘24人的中巴车，都正好坐满。五年级至少有多少同学去春游？", "answer": "48", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "把5克盐放入355克水中，盐的质量占盐水的几分之几？", "answer": "1/72", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 5, "question": "一根长方体木料，体积是0.078立方分米，已知木料的长是2厘米，宽是3厘米，这根木料高是多少厘米？", "answer": "13", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "有11个外形完全相同的羽毛球，其中10个是正品，一个是次品，且次品稍轻一些。如果用天平称，至少称几次就一定能把这个次品找出来？", "answer": "3", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "把一根绳子对折，又对折，再对折，其中的3小段占这根绳子全长的几分之几？", "answer": "3/8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个底面是正方形的长方体纸盒，将它的侧面展开正好是一个边长是6厘米的正方形，做这个纸盒至少需要纸板多少平方米？", "answer": "216", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "在学校组织的“有奖读书活动”中，笑笑获得了一等奖，奖品是一本科普书，共80页，她每天看10页，那么她每天看这本书的几分之几？", "answer": "1/8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "航模小组女生有4人，男生有5人，男生占小组人数的几分之几？", "answer": "5/9", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个圆形水池周围每隔2米栽一棵杨树,共栽了40棵,水池的周长是多少米?", "answer": "80", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "一座楼房每上1层要走16级台阶，到小英家要走64级台阶，小英家住在几楼？", "answer": "5", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个箱子里有20本书，其中19本质量相同，另有1本质量不足，轻一些。用天平至少称几次能保证找出这本书来？", "answer": "3", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "鸡兔同笼，鸡和兔的只数相同，两种动物的腿加起来共有42条，鸡和兔的数量是多少？", "answer": "7", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "爷爷的果园种着桃树和梨树，明明数了数，发觉一共有36棵，他还发现桃树的棵数是梨树的2倍．爷爷的果园里种着梨树多少棵？", "answer": "12", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "小熊、小羊和小鹿共同修建了一个小水池。小羊每2天到池边喝一次水，小鹿每3天到池边喝一次水，小熊每4天到池边喝一次水。7月1日的这一天，它们三个又同时来到池边喝水，并约定月底的某一天早上8:00准时到池边喝水。你知道这一天是7月几日吗？", "answer": "24", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "在一条公路上两侧每隔16米架设一根电线杆,共用电线杆52根,这条公路全长多少米？", "answer": "400", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "工程队修一条公路，原计划每天修路1.65千米，20天可以完成。实际只用了15天，实际平均每天修路多少千米？", "answer": "2.2", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "城东小学的同学们做早操，21个同学排成一排，每相邻的两个同学之间的距离相等，第一个人到最后一个人的距离是40米，相邻两个人间距多少米？", "answer": "2", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "两座楼房之间相距56米,每隔4米栽雪松一棵,一行能栽多少棵?", "answer": "13", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "某年级同学春游时租船游湖，若每只船乘10人，还多2个座位；若每只船多坐2人，可少租一条船，这时每人可节省5角钱。租一只船需要多少钱？", "answer": "24", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "有一根木料，打算把每根锯成3段，每锯开一处，需要5分钟，全部锯完需要多少分钟?", "answer": "10", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "3年前晶晶和爸爸的年龄和是45，爸爸比晶晶大35岁，今年爸爸多大？", "answer": "43", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "20以内最大的质数与最小的质数的2倍的和是多少？", "answer": "23", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个表面积为36平方分米的正方体，沿一个面切成4个长方体后，表面积增加了多少平方分米？", "answer": "36", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 5, "question": "学校举行书画竞赛，四、五年级共有75人获奖，其中五年级获奖人数是四年级的1.5倍，五年级各有多少同学获奖？", "answer": "45", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "五（3）班有男生25人，女生20人，女生人数是男生人数的几分之几？(用最简分数表示)", "answer": "4/5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个长方体，底面是个正方形，它的高是7厘米，所有的棱长之和是100厘米，这个长方体的体积是多少立方分米？", "answer": "0.567", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "长方形的砖长48厘米，宽32厘米，用这样的砖铺成一块正方形的地面，至少需要砖多少块？", "answer": "6", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "战斗机的飞行速度是4000千米/时，比超音速飞机速度的3倍还多700千米，超音速飞机每小时飞行多少千米？", "answer": "1100", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 5, "question": "跳远比赛使用的沙坑长7米，宽3米，坑内沙子的厚度为0.8米，沙坑里有沙子多少立方米？", "answer": "16.8", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "一条公路21天修完，平均每天完成这条公路的几分之几？", "answer": "1/21", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 5, "question": "食堂运来一些煤，第一天烧了5/8吨，第二天烧了1/4吨，两天一共烧了多少吨？", "answer": "7/8", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "光每秒能传播30万千米，这个距离大约比地球赤道长度的7倍还多2万千米。地球赤道大约长多少万千米？", "answer": "4", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个长方体玻璃容器内存水5.5升，放入一个苹果后，水位上升至6升，这个苹果的体积是多少立方分米？", "answer": "0.5", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "同学们去买作文书，如果每人出8元，就多出了8元；每人出7元，就多出了4元．那么有多少个同学去买书？", "answer": "4", "reasoning_step": 5, "num_digits": 1}
|
||||
{"grade": 5, "question": "王师傅加工一批零件，每天加工20个，可以提前1天完成。工作4天后，由于改进了技术，每天可多加工5个，结果提前3天完成。问：这批零件有多少个？", "answer": "280", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 5, "question": "张老师从网上下载了一些图片，一共占硬盘空间12MB，现在他准备用软盘把这些图片拷贝到学校的电脑里，每张软盘的空间是1.44MB，那么这些图片至少需要多少张这样的软盘？", "answer": "9", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 5, "question": "一个长方形的周长是4.8米，长是宽的3倍．这个长方形的宽是多少米？", "answer": "0.6", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "用100千克油菜籽可以炸出41千克菜籽油。1千克油菜籽可以炸出多少千克菜籽油？", "answer": "41/100", "reasoning_step": 1, "num_digits": 5}
|
||||
{"grade": 5, "question": "三个工程队合修一条公路，甲队修了全长的2/9，乙队修了全长的4/27，剩下的由丙队修。丙队修了全长的几分之几？", "answer": "17/27", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "一根3米长的方木，截成3断后，表面积增加了3.6平方米，这根方木的体积是多少立方米？", "answer": "0.027", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 5, "question": "在一条长50米的跑道两旁,从头到尾每隔5米插一面彩旗,一共插多少面彩旗?", "answer": "22", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个长方体水缸，长、宽、高分别是50厘米、24厘米、40厘米，若里面放进38.4升的水，水面距上口多少厘米？", "answer": "8", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 5, "question": "美心蛋糕房特制一种生日蛋糕，每个需要0.32千克面粉，李师傅领了4千克面粉做蛋糕。他最多可以做几个生日蛋糕？", "answer": "12", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "某厂九月份计划生产6000部智能手机，实际上半月完成3600部，上半月完成全月计划的几分之几？", "answer": "3/5", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 5, "question": "有一块长方形的菜地，长10.5米，宽8.2米，它的周长是多少米？", "answer": "37.4", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "张华离家到县城去上学，他以每分50米的速度走了2分后，发现按这个速度走下去就要迟到8分。于是他加快了速度，每分多走10米，结果到校时，离上课还有5分。张华家到学校的路程是多少？", "answer": "4000", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 5, "question": "放假时，老师让每位学生写毛笔字25页，现在丽丽已经写完7页，没有写完的占总数的几分之几？", "answer": "18/25", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "每千克花生仁批发价7.62元，零售价8.9元，刘大伯批发价买进这种花生仁240千克，零售价卖出后，一共可得毛利多少元？", "answer": "307.2", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "一个长方形的长是9厘米，等于宽的1.5倍，这个长方形的面积是多少平方厘米？", "answer": "54", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "仓库有18.6吨水泥，现在用卡车运到工地，每辆卡车运2.5吨，需要多少辆卡车？", "answer": "8", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "小明的体重是35千克，他的体重比爸爸轻8/15，小明爸爸的体重是多少千克？", "answer": "75", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "一个长方体的底面是正方形，如果高增加3厘米，就成为一个正方体，这时表面积比原来增加了48平方厘米，原来长方体的体积是多少立方厘米？", "answer": "16", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个长方体的体积是56立方厘米，高是4厘米，它的底面积是多少平方厘米？", "answer": "14", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "学校音乐小组有64人，现需通知外出演出，如果用打电话方式，每人通话1分钟，通知到每人需要多少分钟？", "answer": "7", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "在一个长方体上截下一个体积是72立方厘米的小长方体后，剩下的部分是一个棱长为6厘米的正方体，原来这个长方体的表面积是多少平方厘米？", "answer": "264", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 5, "question": "李大爷以相同的速度在乡间布满电话线杆的小路上散步。他从第1根电话线杆走到第12根电话线杆用了22分钟。他如果走36分钟，应走到第几根电话线杆？", "answer": "19", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个分数约分后得到的最简分数是5/7，已知原来的分数分子和分母的和是72，原来的分数是多少？", "answer": "30/42", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "红星小学陈列馆正进行二期工程改造，现工地需要52吨沙子，用一辆载重量4.5吨的汽车运8次，余下的改用一辆载重3.5吨的汽车运，还要运多少次？", "answer": "5", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": ".李明家正在给新房子装修，已知卫生间的面积是6m2，如果用边长是0.3m的正方形瓷砖铺地，每块瓷砖的价格是9.83元，那么大约需要多少钱？（得数保留整数）", "answer": "659", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 5, "question": "有一个仓库，从里面量得长是25米，宽是10米，存货高度为3米。这个仓库最多能存放货物多少立方米？", "answer": "750", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "等腰三角形一边长为4，另一边长为2，则其周长为多少？", "answer": "10", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "一根长10米的彩带，每1.5米可以包扎一个礼盒，这根彩带可以包扎几个礼盒？", "answer": "6", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "动物园里有9只骆驼，6只老虎，老虎的数量是骆驼的几分之几？", "answer": "6/9", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 5, "question": "挂钟6点钟敲6下，10秒敲完，那么9点钟敲9下，几秒敲完?", "answer": "16", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "一块长是48厘米，宽和高都是6厘米的长方体铁块，把它铸造成棱长是3厘米的正方体铁块，可以铸造多少块？（损耗不计）", "answer": "64", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "小明每天早晨以每分钟0.15千米的速度练习跑步，每天坚持练12分钟。照这样计算，2009年共跑了多少千米？（2009年365天）", "answer": "657", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "爷爷今年75岁，比小明岁数的5倍还多5岁．小明今年几岁？", "answer": "14", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个旅游团有10人出去旅行，由于临时改变计划，要紧急通知，如果用打电话的方式，每1分钟通知1人，最少要在几分钟内通知到所有人？", "answer": "4", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 5, "question": "一箱水果糖有31袋，其中30袋质量相同，另外有一袋质量轻些，用天平至少称几次才能保证找出较轻的一袋？", "answer": "4", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 5, "question": "每只小船租金30元，100元钱最多租几小时？", "answer": "3", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "甲数是3/20，乙数是3/10，甲数与乙数的和是多少？", "answer": "9/20", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 5, "question": "合唱队有若干人，若12人站一排，余5人；若15人站一排还是余5人，这个合唱队至少有多少人？", "answer": "65", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 5, "question": "有一块平行四边形钢板，底是8.4分米，高是3.5分米．如果每平方分米钢板重0.75千克，这块钢板重多少千克？", "answer": "22.05", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "一个分数是18/30，把它的分子扩大2倍，要使分数的大小不变，分母应该加上多少？", "answer": "30", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "凡凡和3位小朋友一共吃了1千克香蕉，平均每位小朋友吃了多少千克？", "answer": "250", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 5, "question": "园林工人在一段公路的一边每隔4米栽一棵树，一共栽了17棵．现在要改成每隔6米栽一棵树．那么，不用移栽的树有多少棵？", "answer": "6", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "有一条长800米的公路,在公路的一侧从头到尾每隔20米栽一棵杨树,需多少棵杨树苗?", "answer": "41", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "从一楼跑到五楼有96个台阶，小芳从一楼跑到20楼供需迈多少个台阶？", "answer": "456", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 5, "question": "有一个梯形广告牌，它的上底长4米，下底长6.5米，高为2米，在它的正面涂上广告漆，每平方米需要10元钱，涂满整个广告牌需要花费多少钱？", "answer": "105", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 5, "question": "一本书有83000个字，如果每页排25行，每行排24个字，从第1页起，排完这些字最少需要多少张纸？", "answer": "70", "reasoning_step": 3, "num_digits": 5}
|
||||
{"grade": 5, "question": "赵叔叔要完成100个零件的加工任务，已经完成了74个，剩下的零件占这批零件的几分之几？", "answer": "13/50", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "一个两位数，它既有因数2，又有因数3，还有因数5，这个两位数最小是多少？", "answer": "30", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 5, "question": "公园大门前的公路长80米，要在公路两边栽上白杨树，每两棵树相距8米（两端也要种）。园林工人共需要准备多少棵树？", "answer": "22", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "码头上有两堆货物，第一堆有29吨，第二堆有32吨，第一堆货物的重量是第二堆的几分之几？", "answer": "29/32", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 5, "question": "有一堆同样规格的小螺丝钉，不容易数出它们的个数，称得它们的质量是1.53千克，数出100个，这100个小螺丝钉的质量是0.03千克．这堆小螺丝钉一共有多少个？", "answer": "5100", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 5, "question": "有一批正方形的砖，排成一个大正方形，余下32块；如果将它们改排成每边比原来多一块砖的正方形，就要差49块。这批砖原有多少块？", "answer": "1632", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 5, "question": "买一块平行四边形的玻璃，底为80厘米，高为50厘米，每平方米的玻璃售价为23元，买这块玻璃共需要多少元？", "answer": "9.2", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "一个正方体木箱的棱长是1.5米，做这个木箱需要木板多少平方米？", "answer": "13.5", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 5, "question": "同学们去植树，3人一组或5人一组都正好分完。已知参加植树的同学在40～50人之间，则这些同学最多有多少人？", "answer": "45", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 5, "question": "合唱队有7人，暑假老师要通知合唱队的队员去演出，若每分钟通知一个人，至少需要多少分钟才能通知到每个队员？", "answer": "3", "reasoning_step": 3, "num_digits": 1}
|
||||
{"grade": 5, "question": "有7个钢珠，其中有1个较轻的是次品，用天平称至少称几次，才能保证一定能找出来？", "answer": "2", "reasoning_step": 3, "num_digits": 1}
|
||||
{"grade": 6, "question": "某小学在“献爱心--为汶川地震区捐款”活动中，六年级五个班共捐款8000元，其中一班捐款1500元，二班比一班多捐款200元，三班捐款1600元，四班与五班捐款数之比是3：5．四班捐款多少元？", "answer": "1200", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 6, "question": "小俊在东西大道上跑步，若规定向东为正。他先向东跑了800米，然后又跑了一段之后，他位于出发点西边100米处，小俊第二段跑了多少米？", "answer": "900", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 6, "question": "A车和B车同时从甲、乙两地相向开出，经过5小时相遇．然后，它们又各自按原速原方向继续行驶3小时，这时A车离乙地还有135千米，B车离甲地还有165千米．甲、乙两地相距多少千米？", "answer": "775", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "东西两个仓库共存粮480吨，东库存粮数是西库存粮数的1.4倍，求东库存粮多少吨？", "answer": "280", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "水果店运来苹果和梨共840千克，苹果的质量是梨的3倍，苹果重多少千克？", "answer": "630", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个足球场长120m，宽90m，把它画在一张纸上，长画了16cm，宽应该画多少厘米？", "answer": "12", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "在一个底面半径为10厘米的圆柱形容器内，倒入10厘米深的水，然后将一个底面直径4厘米，高6厘米的圆锥形铅锤放入水中，容器中水面上升多少厘米？", "answer": "0.08", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "商店运来桔子、苹果、梨共360千克，桔子与苹果的重量比是5：6，梨的重量是苹果的1/6，桔子有多少千克？", "answer": "150", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 6, "question": "A、B两个数的和是616，A的个位上是0，若把0去掉，就与B相同，那么A是多少？", "answer": "560", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "在学校的数学竞赛活动中，一共有126人获奖．其中获得一、二、三等奖的人数比是1：2：3．获得一等奖的有多少人？", "answer": "21", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "一份稿件，甲单独打字需6小时完成，乙单独打字需10小时完成，现在甲单独打若干小时后，因有事由乙接着打完，共用了7小时，那么甲打字用了多少小时？", "answer": "9/2", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 6, "question": "在一个比例中，两个外项互为倒数，其中一个内项是5/13，另一个内项是多少?", "answer": "13/5", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 6, "question": "某汽车制造厂上半年生产小汽车36400辆，比原计划多生产3900辆，超产百分之几？", "answer": "12%", "reasoning_step": 2, "num_digits": 5}
|
||||
{"grade": 6, "question": "蕉坝中心完小六年级三个班共植树120棵，已知六（1）、（2）、（3）班植树的棵树比为1：3：2，那（1）班植树多少棵？", "answer": "20", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "有一只羊栓在草地的木桩上，绳子的长度是6米，这只羊最多可以吃到多少平方米的草？", "answer": "113.04", "reasoning_step": 2, "num_digits": 5}
|
||||
{"grade": 6, "question": "甲乙两人分别从A、B两地同时相向而行，甲每分钟行100米，乙每分钟行120米，12.5分钟后两人已相遇，彼此相距150米．A、B两地相距多少米？", "answer": "2600", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 6, "question": "某皮鞋店调进女皮鞋225双，其中男装皮鞋的双数相当于女装皮鞋的40%，这个皮鞋店调进男装皮鞋多少双？", "answer": "90", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 6, "question": "爸爸买了5000元国债，定期两年，到期时可取回本金和利息共5873元，这种国债的年利率是多少？", "answer": "0.0873", "reasoning_step": 3, "num_digits": 5}
|
||||
{"grade": 6, "question": "人体中的血液约占体重的1/13,血液里约2/3是水.小东的体重是39千克,他的血液里大约含水多少千克?", "answer": "2", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "儿童公园有一个直径是15米的圆形金鱼池，在金鱼池周围要做4圈圆形栏杆，至少要用多少钢条？", "answer": "188.4", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 6, "question": "某工厂有一堆煤，如果每天烧0.8吨，可以烧30天。如果每天节约0.2吨，可以多烧多少天？", "answer": "10", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 6, "question": "我校在“创建绿色循环经济示范单位”活动中，打算在生物园新挖一个直径是6米，深12分米的圆形水池。这个水池的占地面积是多少平方米？", "answer": "28.26", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 6, "question": "某会议室需要粉刷的面积是500平方米，每平方米需要涂料0.6千克，但实际粉刷时会有损耗，因此要多准备10%。实际应准备多少千克涂料？", "answer": "330", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "有黑色、白色、黄色的小棒各8根，混放在一起，从这些小棒之中至少要取出才能保证有4根颜色相同的小棒子？", "answer": "10", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "贝贝、晶晶、欢欢、迎迎、妮妮五种福娃共10个，至少买多少个福娃才可以保证一定有两个一样的福娃？", "answer": "6", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "“五一”期间，小熊商场所有商品“九五折”出售。空调原价2800元，“五一”期间，空调价格比原来便宜多少元？", "answer": "140", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 6, "question": "商店以68元一件的价格购进一批衣服，售价为每件96元。一月份共售出115件，除去各种开支874元。那么这家商店还可以赚多少元？", "answer": "2346", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 6, "question": "甲乙丙三数之和是170，乙比甲的2倍少4，丙比甲的3倍多6，求乙是多少？", "answer": "52", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "育才小学有教师108人，其中女教师人数是男教师的3倍．男教师有多少人？", "answer": "27", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "一种糖水，糖和水按照1：150配制的；现有糖100克，可以配制这样的糖水多少克？", "answer": "15100", "reasoning_step": 2, "num_digits": 5}
|
||||
{"grade": 6, "question": "一种自行车前齿轮的齿数是48，后齿轮的齿数是24。如果车轮的直径是66厘米，蹬一圈大约能走多少米？", "answer": "414.48", "reasoning_step": 5, "num_digits": 5}
|
||||
{"grade": 6, "question": "某款书包打八折后售价是120元，如果打九折出售，买这款书包需要多少元？", "answer": "135", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "小丽和小明进行踢毽子比赛，小明5/6分钟踢了20个，小丽2/3分钟踢了18个，小丽每分钟比小明多踢多少个？", "answer": "3", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 6, "question": "一种衣服现在打九折出售，每件卖45元，那么原价是多少元？", "answer": "50", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 6, "question": "某件皮衣原价是1800元，现降价270元出售，该商品是打了几折出售的？", "answer": "8.5", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 6, "question": "甲乙两车汽车同时从A地开往B地，当甲车行了全程的1/3时，乙车正好行了60千米；当甲车到达B地时，乙车行了全程的3/5，AB两地相距多少千米？", "answer": "300", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个卷烟厂上半年的销售额是3000万元，如果按照销售额45%缴纳消费税，上半年应缴纳消费税款多少万元？", "answer": "1350", "reasoning_step": 1, "num_digits": 4}
|
||||
{"grade": 6, "question": "一个酒瓶的高度是30厘米，底面直径是8厘米，瓶里酒的高度是20厘米，把酒瓶塞拧紧后瓶口向下。这时酒的高度是24厘米。问酒瓶容积是多少毫升？", "answer": "1306.24", "reasoning_step": 5, "num_digits": 6}
|
||||
{"grade": 6, "question": "一件上衣300元，若上衣比裤子少2/3，一条裤子多少钱？", "answer": "900", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "一种糖水，糖和水按照1：150配制的；要配制这样的糖水15100克，需要水多少克？", "answer": "15000", "reasoning_step": 1, "num_digits": 5}
|
||||
{"grade": 6, "question": "一种自行车前齿轮的齿数是26，后齿轮的齿数是24。车轮的半径是35厘米，蹬一圈大约能走多少米？", "answer": "2.198", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 6, "question": "六一班共50名同学，其中考试数学的优秀率为56%，六一班数学得优的有多少人？", "answer": "28", "reasoning_step": 1, "num_digits": 2}
|
||||
{"grade": 6, "question": "在某高速公路上A、B两车正好相距96千米，现两车正好同时从两个不同的服务区上同向而行，A车每小时行95千米，B车每小时行107千米．经过几小时B车可以追上A车？", "answer": "8", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个圆锥形麦堆，底面直径2米，高0.6米，每立方米小麦约重500千克，这堆小麦重多少千克？", "answer": "314", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "张老师到银行存款4500元，年利率是2.25％，扣除20％利息税，一年后取回本息多少元?", "answer": "4581", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 6, "question": "六年级一共有146人，其中男生人数是女生人数的2倍多56人，男生有多少人？", "answer": "116", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 6, "question": "一根3米长的圆柱形木料，锯成3段，表面积增加16平方分米，这根木料的底面积是多少平方分米？", "answer": "4", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "红卫小学有4000本图书，育新小学有5000本图书，育新小学的图书比红卫小学多百分之几？", "answer": "25%", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 6, "question": "张阿姨在端午节一共包了蛋黄粽与肉粽75个，蛋黄粽与肉粽的比是2：3．张阿姨包了多少个肉粽？", "answer": "45", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "一只蝴蝶2时飞行15.6千米，一只蜜蜂的飞行速度是这只蝴蝶的2.4倍。这只蜜蜂每时飞行多少千米？", "answer": "18.72", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 6, "question": "爸爸的工资是6500元，扣除3500元个税免征额后的部分需要按3%的税率缴纳个人所得税，爸爸缴税后实际得到的工资是多少元？", "answer": "6410", "reasoning_step": 3, "num_digits": 4}
|
||||
{"grade": 6, "question": "用一根长8米的绳子围着一棵树绕4圈，还余1.72米。这棵树的直径约是多少米？", "answer": "0.5", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "一段圆柱形钢材，它的底面半径是4厘米，高是35厘米。已知1立方厘米的钢材重7.8克，这段钢材有多重？", "answer": "13715.52", "reasoning_step": 4, "num_digits": 7}
|
||||
{"grade": 6, "question": "天天妈妈的服装店实行薄利多销的原则，一般在进价的基础上提高二成后作为销售价。照这样计算，一件进价为220元的衣服应标价多少元？", "answer": "264", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个棱长3分米的立方体玻璃缸中装满了水。李明将其中的水全部倒入了一个长5分米，宽2分米，高4.5分米的长方体玻璃缸中。这时水与玻璃缸的接触面面积是多少？", "answer": "47.8", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "一项工程，甲独做要18天完成，乙独做要15天，二人合作6天，其余的由乙单独做，还要几天做完？", "answer": "4", "reasoning_step": 4, "num_digits": 2}
|
||||
{"grade": 6, "question": "纸箱中有同样的红、黄色圆锥体各5个，至少拿出几个，才能保证一定有2个圆锥体都是红色？", "answer": "3", "reasoning_step": 2, "num_digits": 1}
|
||||
{"grade": 6, "question": "一项工程，由甲单独做30天完成，这项工程先由甲乙两队合做8天，余下的甲队10天完成，那么乙单独做这项工程需要多少天完成？", "answer": "20", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 6, "question": "一个圆锥形稻谷堆，底面周长12.56米，高0.8米．如果每立方米重0.75吨，这堆稻谷重多少吨？（结果保留一位小数）", "answer": "2.5", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 6, "question": "粮库有94吨小麦和138吨玉米，如果每天运出小麦和玉米各是9吨，问几天后剩下的玉米是小麦的3倍？", "answer": "8", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "张华骑车从家到车站用了30分钟，前18分钟，按每分钟205m的速度行驶，最后提高了速度，按每分钟256m的速度行驶，张华家到车站有多少米？", "answer": "6762", "reasoning_step": 4, "num_digits": 4}
|
||||
{"grade": 6, "question": "一项工作，甲独做要8天才能完成，乙独做要6天才能完成。甲乙合作，每天完成这项工作的几分之几？", "answer": "7/24", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 6, "question": "明光工业园区进行基础设施建设，去年实际投资380万元，比计划投资节省20万元，节省了百分之几？", "answer": "5%", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "在1：6000000的地图上，量得甲、乙两地间的距离是6.5cm。一辆货车从甲地到乙地行驶了6.5小时。求这辆车的速度是多少千米每小时。", "answer": "60", "reasoning_step": 2, "num_digits": 7}
|
||||
{"grade": 6, "question": "盐城市区出租车的计费标准是：起步价（3千米以内，包括3千米）7元，以后每超过1千米（不足1千米的按1千米计算）另加价1.6元。请你算一算，乘车7千米要付多少钱？", "answer": "13.4", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个袋子里有红、黄、蓝色的袜子各10只，则最少要拿多少只才能保证其中至少有两双颜色不相同的袜子？", "answer": "13", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "某乡去年水稻总产量是1500吨，今年预计比去年增产一成五。今年水稻总产量预计是多少吨？", "answer": "1725", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 6, "question": "原价360，如果打八折销售现价是？", "answer": "288", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 6, "question": "春运期间，深圳到武汉的飞机票涨价10%后，票价为880元，春运前的飞机票价是多少元？", "answer": "800", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个课外小组把6米高的竹竿直立在地上，量得影子的长度是9.6米。同时测得一个烟囱的影长是32米，那么烟囱有多高？", "answer": "20", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "客车和货车分别从相距840千米的两站同时相对开出，6小时在途中相遇。已知客车每小时行80千米，货车每小时行多少千米？", "answer": "60", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "生产240个零件，师徒二人合做6天完成，师傅单独做4天能完成全部任务的2/5，徒弟每天生产零件多少个？", "answer": "16", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个底面半径是10cm的圆柱形水桶中装水，水中放一个底面半径是5cm的圆锥形铅锤，铅锤全部淹没。取出铅锤后桶面水面下降2cm，求铅锤的高。", "answer": "24", "reasoning_step": 5, "num_digits": 2}
|
||||
{"grade": 6, "question": "一个长方体玻璃鱼缸，长50厘米，宽40厘米，高30厘米。做这个鱼缸至少需要玻璃多少平方厘米？", "answer": "7400", "reasoning_step": 5, "num_digits": 4}
|
||||
{"grade": 6, "question": "一辆自行车的轮胎外直径为70厘米，如果每分钟转120圈，1小时约行多少千米？（得数保留整数）", "answer": "16", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 6, "question": "某班一天有3人请假，出勤率是94%，下午请假的3人中又有1人到校，求下午的出勤率。", "answer": "0.96", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "甲、乙两人沿一个圆形的操场的边散步，从同一地点出发，背向而行，甲每分钟走40米．乙每分钟走38.5米，4分钟相遇，这个操场的直径是多少米？", "answer": "100", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "一个手机信号发射接收塔埋在地下与露出地面部分的比是3：18，埋在地下的部分是4米，那么这个塔的全长是多少米？", "answer": "28", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "一台拖拉机每小时耕地2/7公顷，3台拖拉机14小时耕地多少公顷？", "answer": "12", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "挖一条水渠，在比例尺是1:300的地图上，量得这条水渠长40厘米。这条水渠实际长多少米？", "answer": "120", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "某野生动物园，一共有东北虎和白虎16只，东北虎的只数是白虎的7倍，白虎有多少只？", "answer": "2", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "一套桌椅的价钱共400元，其中椅子的价钱是桌子的60%．椅子的单价是多少？", "answer": "150", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "据信息产业部统计，到目前为止，我国电话用户达3.6亿户，其中移动电话用户是固定电话用户的2倍．求我国移动电话用户是多少亿户？", "answer": "2.4", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 6, "question": "一项工程，甲单独做需要20天完成，乙单独做需要15天完成，甲先做了5天后，剩下的甲乙合做几天可以完成？", "answer": "45/7", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 6, "question": "妈妈和李阿姨一起到超市购物，妈妈买了5千克大米和4千克面粉共付29.2元，李阿姨也买了同样的4千克大米和5千克面粉则共付28.4元，每千克大米多少钱？", "answer": "3.6", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "甲乙两班共有学生88人，甲班比乙班多8人，求甲班有多少人？", "answer": "48", "reasoning_step": 3, "num_digits": 2}
|
||||
{"grade": 6, "question": "某自来水公司为鼓励节约用水，采取按月分段计算的方法收取水费，12吨以内的每吨2.5元，超过12吨的部分，每吨3.8元，李奶奶家上个月的用水量是19吨，应缴水费多少元？", "answer": "56.6", "reasoning_step": 4, "num_digits": 3}
|
||||
{"grade": 6, "question": "在一个半径为10米的圆柱体储水池里，把一个长5米，宽4米的长方体铁块全部放入水中，桶里的水面上升0.5米。求这段铁块的高是多少米。", "answer": "7.85", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "广场中央有一个圆柱形喷水池，底面内直径是20米，深0.8米。若在池内的侧面和池底抹一层水泥，水泥面的面积是多少平方米？（得数保留整平方米）", "answer": "365", "reasoning_step": 5, "num_digits": 3}
|
||||
{"grade": 6, "question": "六年级学生参加植树劳动，男生植了160棵，女生植的树比男生3/4的多5棵。 女生植树多少棵？", "answer": "125", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "某商场一件上衣打八折后卖400元，如果打7折出售，可以卖多少元？", "answer": "350", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "爸爸买了一辆售价12万元的家用轿车，按照规定缴纳10%的车辆购置税。爸爸买这辆车一共花了多少万元？", "answer": "13.2", "reasoning_step": 2, "num_digits": 3}
|
||||
{"grade": 6, "question": "华的妈妈把1000元钱存入银行，定期三年。如果年利率按5.22%计算，到期时可以取回多少元？", "answer": "1156.6", "reasoning_step": 4, "num_digits": 5}
|
||||
{"grade": 6, "question": "给直径为0.75米的水缸做一个木盖，木盖的直径比缸口直径大5厘米，这个木盖的面积是多少平方米？", "answer": "0.5024", "reasoning_step": 5, "num_digits": 5}
|
||||
{"grade": 6, "question": "美元和人民币的汇率为7.19：1（1美元可兑换人民币7.19元）现在有140美元，可以兑换人民币多少元？", "answer": "1006.6", "reasoning_step": 1, "num_digits": 5}
|
||||
{"grade": 6, "question": "猴山上6只猴分桃，总有1只猴至少分到了5个桃。这堆桃至少有多少个呢？", "answer": "25", "reasoning_step": 2, "num_digits": 2}
|
||||
{"grade": 6, "question": "一杯约250毫升的鲜牛奶大约含有3/10克的钙质，小华每天喝2杯这样的牛奶，他在整个九月份通过喝牛奶可以摄取钙质多少克？", "answer": "18", "reasoning_step": 3, "num_digits": 3}
|
||||
{"grade": 6, "question": "每辆汽车平均每千米排放160克二氧化碳。赵老师为了响应市政府“绿色出行”的号召，上下班由自驾车方式改为骑自行车方式。已知赵老师家距学校20千米。赵老师每天可以减少排放多少克的二氧化碳？", "answer": "6400", "reasoning_step": 2, "num_digits": 4}
|
||||
{"grade": 6, "question": "一箱香蕉重1/20吨，15箱这样的香蕉重多少吨?", "answer": "3/4", "reasoning_step": 1, "num_digits": 3}
|
||||
{"grade": 6, "question": "学校运来480本图书，分别分给8个班级，每个班级分多少本？", "answer": "60", "reasoning_step": 1, "num_digits": 3}
|
||||
@@ -0,0 +1,101 @@
|
||||
{"idx": 0, "question": "下列四个数中，最小的数是()\nA.-2 B.0 C. 3 D. -1/2", "answer": "-2", "choice_answer": "A"}
|
||||
{"idx": 1, "question": "已知点(-3,2)在反比例函数y=k/x (k≠0)的图象上，则k的值为()\nA.-3 B.3 C.-6 D.6", "answer": "-6", "choice_answer": "C"}
|
||||
{"idx": 2, "question": "若两个相似三角形的相似比是1:3，则这两个相似三角形的面积比是()\nA.1:3 B.1:4 C.1:6 D.1:9", "answer": "1:9", "choice_answer": "D"}
|
||||
{"idx": 3, "question": "已知m=√27-√3，则实数m的范围是（）\nA.2<m<3 B.3<m<4 C.4<m<5 D.5<m<6", "answer": "3<m<4", "choice_answer": "B"}
|
||||
{"idx": 4, "question": "已知整式M:a_n x^n+a_(n-1) x^(n-1)+⋯+a_1 x+a_0，其中n,a_(n-1),⋯,a_0为自然数，a_n为正整数，且n+a_n+a_(n-1)+⋯+a_1+a_0=5．下列说法：①满足条件的整式M中有5个单项式；②不存在任何一个n，使得满足条件的整式M有且只有3个；③满足条件的整式M共有16个．其中正确的个数是（）\nA.0 B.1 C.2 D.3", "answer": "3", "choice_answer": "D"}
|
||||
{"idx": 5, "question": "重庆是一座魔幻都市，有着丰富的旅游资源．甲、乙两人相约来到重庆旅游，两人分别从A、B、C三个景点中随机选择一个景点游览，甲、乙两人同时选择景点B的概率为", "answer": "1/9", "choice_answer": ""}
|
||||
{"idx": 6, "question": "计算：(π-3)^0 + (1/2)^(-1)＝", "answer": "3", "choice_answer": ""}
|
||||
{"idx": 7, "question": "如果一个多边形的每一个外角都是40°，那么这个多边形的边数为", "answer": "9", "choice_answer": ""}
|
||||
{"idx": 8, "question": "随着经济复苏，某公司近两年的总收入逐年递增．该公司2021年缴税40万元，2023年缴税48.4万元，该公司这两年缴税的年平均增长率是", "answer": "10%", "choice_answer": ""}
|
||||
{"idx": 9, "question": "若关于x的不等式组\\begin{equation}\n{\\begin{cases}\n{\n (4x-1)/3<x+1 \n 2(x+1)≥-x+a)\n}\\end{cases}\n}\n\\end{equation}至少有2个整数解，且关于y的分式方程(a-1)/(y-1)=2-3/(1-y)的解为非负整数，则所有满足条件的整数a的值之和为", "answer": "16", "choice_answer": ""}
|
||||
{"idx": 10, "question": "我们规定：若一个正整数A能写成m^2-n，其中m与n都是两位数，且m与n的十位数字相同，个位数字之和为8，则称A为“方减数”，并把A分解成m^2-n的过程，称为“方减分解”．例如：因为602=25^2-23，25与23的十位数字相同，个位数字5与3的和为8，所以602是“方减数”，602分解成602=25^2-23的过程就是“方减分解”．按照这个规定，最小的“方减数”是____．把一个“方减数”A进行“方减分解”，即A=m^2-n，将m放在n的左边组成一个新的四位数B，若B除以19余数为1，且2m+n=k^2（k为整数），则满足条件的正整数A为____", "answer": "82,4564", "choice_answer": ""}
|
||||
{"idx": 11, "question": "计算：x(x-2y)+(x+y)^2", "answer": "2x^2+y^2", "choice_answer": ""}
|
||||
{"idx": 12, "question": "计算：(1+1/a)÷(a^2-1)/(a^2+a)", "answer": "\\frac{(a+1)}{(a-1)}", "choice_answer": ""}
|
||||
{"idx": 13, "question": "为促进新质生产力的发展，某企业决定投入一笔资金对现有甲、乙两类共30条生产线的设备进行更新换代．(1)为鼓励企业进行生产线的设备更新，某市出台了相应的补贴政策．根据相关政策，更新1条甲类生产线的设备可获得3万元的补贴，更新1条乙类生产线的设备可获得2万元的补贴．这样更新完这30条生产线的设备，该企业可获得70万元的补贴．该企业甲、乙两类生产线各有多少条？(2)经测算，购买更新1条甲类生产线的设备比购买更新1条乙类生产线的设备需多投入5万元，用200万元购买更新甲类生产线的设备数量和用180万元购买更新乙类生产线的设备数量相同，那么该企业在获得70万元的补贴后，还需投入多少资金更新生产线的设备？", "answer": "10, 20, 13300000", "choice_answer": ""}
|
||||
{"idx": 14, "question": "反比例函数y=-10/x的图象一定经过的点是（）\nA.(1,10) B.(-2,5) C.(2,5) D.(2,8)", "answer": "(-2,5)", "choice_answer": "B"}
|
||||
{"idx": 15, "question": "估计\\sqrt{12}(\\sqrt{2}+\\sqrt{3})的值应在（）\nA.8和9之间 B.9和10之间 C.10和11之间 D.11和12之间", "answer": "10和11之间", "choice_answer": "C"}
|
||||
{"idx": 16, "question": "计算：|-2|+3^0=", "answer": "3", "choice_answer": ""}
|
||||
{"idx": 17, "question": "若正多边形的一个外角是45°，则该正多边形的边数是", "answer": "8", "choice_answer": ""}
|
||||
{"idx": 18, "question": "重庆在低空经济领域实现了新的突破．今年第一季度低空飞行航线安全运行了200架次，预计第三季度低空飞行航线安全运行将达到401架次．设第二、第三两个季度安全运行架次的平均增长率为x，根据题意，可列方程为", "answer": "200(1+x)^2=401", "choice_answer": ""}
|
||||
{"idx": 19, "question": "若∠A=55°，则∠A的补角为（）\nA.35° B.45° C.115° D.125°", "answer": "125°", "choice_answer": "D"}
|
||||
{"idx": 20, "question": "因式分解：2x^2-8=", "answer": "2(x+2)(x-2)", "choice_answer": ""}
|
||||
{"idx": 21, "question": "已知一次函数y=-2x+4，当自变量x>2时，函数y的值可以是____（写出一个合理的值即可）", "answer": "2", "choice_answer": ""}
|
||||
{"idx": 22, "question": "定义一种新运算*，规定运算法则为：m*n=m^n-mn（m，n均为整数，且m≠0）．例：2*3=2^3-2×3=2，则(-2)*2=____", "answer": "8", "choice_answer": ""}
|
||||
{"idx": 23, "question": "解不等式组：\\begin{equation}\n{\\begin{cases}\n{2(x-2)<x+3 \\ \n(x+1)/2<2x} \n \\end{cases}\n}\\end{equation}", "answer": "1/3<x<7", "choice_answer": ""}
|
||||
{"idx": 24, "question": "先化简，再求值：[(2a+b)^2-(2a+b)(2a-b)]÷2b=，其中a=2，b=-1", "answer": "3", "choice_answer": ""}
|
||||
{"idx": 25, "question": "在一只不透明的布袋中，装有质地、大小均相同的四个小球，小球上分别标有数字1，2，3，4．甲乙两人玩摸球游戏，规则为：两人同时从袋中随机各摸出1个小球，若两球上的数字之和为奇数，则甲胜；若两球上的数字之和为偶数，则乙胜．请用画树状图或列表的方法，求甲获胜的概率", "answer": "7/12", "choice_answer": ""}
|
||||
{"idx": 26, "question": "下列各数中，是无理数的是（）\nA.π/2 B.1/3 C.\\sqrt[3]{27} D. 0.13133", "answer": "π/2", "choice_answer": "A"}
|
||||
{"idx": 27, "question": "据央视财经《经济信息联播》消息：甘肃天水凭借一碗香喷喷的麻辣烫成为最“热辣滚烫”的顶流．2024年3月份，天水市累计接待游客464万人次，旅游综合收入27亿元．将数据“27亿”用科学记数法表示为（）\nA.2.7×10^8 B. 0.27×10^10 C.2.7×10^9 D.27×10^8", "answer": "2.7×10^9", "choice_answer": "C"}
|
||||
{"idx": 28, "question": "下列各式运算结果为a^5的是（ ）\nA.a^2+a^3 B.a^2a^3 C.(a^2)^3 D.a^10÷a^2", "answer": "a^2a^3", "choice_answer": "B"}
|
||||
{"idx": 29, "question": "一次函数 y=kx-1(k≠0)，若y随x的增大而减小，则它的图象不经过（ ）\nA. 第一象限 B. 第二象限 C. 第三象限 D. 第四象限", "answer": "第一象限", "choice_answer": "A"}
|
||||
{"idx": 30, "question": "端午节期间，某商家推出“优惠酬宾”活动，决定每袋粽子降价2元销售．细心的小夏发现，降价后用240元可以比降价前多购买10袋，求：每袋粽子的原价是多少元？设每袋粽子的原价是x元，所得方程正确的是（ ）\nA.240/x-240/(x+2)=10 B.240/x-240/(x-2)=10 C.240/(x-2)-240/x=10 D.240/(x+2)-240/x=10", "answer": "240/(x-2)-240/x=10", "choice_answer": "C"}
|
||||
{"idx": 31, "question": "因式分解：x^2-1/4=____", "answer": "(x-1/2)(x+1/2)", "choice_answer": ""}
|
||||
{"idx": 32, "question": "若关于x的一元二次方程x2+2x﹣m=0有两个相等的实数根，则m的值为", "answer": "-1", "choice_answer": ""}
|
||||
{"idx": 33, "question": "化简：(a+1+1/(a-1))÷(a^2+a)/(a-1)", "answer": "a/(a+1)", "choice_answer": ""}
|
||||
{"idx": 34, "question": "学习小组在延时课上制作了A，B，C，D四张卡片，四张卡片除图片内容不同外，其他没有区别，放置于暗箱中摇匀。小临从四张卡片中随机抽取一张，抽中C卡片的概率是______", "answer": "1/4", "choice_answer": ""}
|
||||
{"idx": 35, "question": "学习小组在延时课上制作了A，B，C，D四张卡片，四张卡片除图片内容不同外，其他没有区别，放置于暗箱中摇匀。小夏从四张卡片中随机抽取两张，用列表法或画树状图法求小夏抽取两张卡片内容均为化学变化的概率", "answer": "1/6", "choice_answer": ""}
|
||||
{"idx": 36, "question": "下列运算正确的是（ ）．\nA.(-m^3 )^2=-m^5 B.m^2n\\dot m=m^3\\cdot n C. 3mn-m=3n D.(m-1)^2=m^2-1", "answer": "m^2n\\cdot m=m^3\\cdot", "choice_answer": "B"}
|
||||
{"idx": 37, "question": "二十四节气，它基本概括了一年中四季交替的准确时间以及大自然中一些物候等自然现象发生的规律，二十四个节气分别为：春季（立春、雨水、惊蛰、春分、清明、谷雨），夏季（立夏、小满、芒种、夏至、小暑、大暑），秋季（立秋、处暑、白露、秋分、寒露、霜降），冬季（立冬、小雪、大雪、冬至、小寒、大寒），若从二十四个节气中选一个节气，则抽到的节气在夏季的概率为（ ）．\nA. 1/2 B. 1/12 C. 1/6 D. 1/4", "answer": "1/4", "choice_answer": "D"}
|
||||
{"idx": 38, "question": "已知一元二次方程x^2-3x+m=0一个根为1，则m=______ ", "answer": "2", "choice_answer": ""}
|
||||
{"idx": 39, "question": "计算: -2 \\cdot \\cos 45^{\\circ}+(\\pi-3.14)^{0}+|1-\\sqrt{2}|+\\left(\\frac{1}{4}\\right)^{-1}.", "answer": "4", "choice_answer": ""}
|
||||
{"idx": 40, "question": "先化简, 再求值: \\left(1-\\frac{2}{a+1}\\right) \\div \\frac{a^{2}-2 a+1}{a+1}, 其中 a=\\sqrt{2}+1", "answer": "\\frac{1}{a-1}, \\frac{\\sqrt{2}}{2}", "choice_answer": ""}
|
||||
{"idx": 41, "question": "今年 618各大电商平台促销火热, 线下购物中心也亮出大招, 年中大促进入 “白热化” . 深圳各大购物中心早在5月就开始推出618活动, 进入6月更是持续加码, 如图, 某商场为迎接即将到来的 618优惠节, 采购了若干辆购物车. 如图为某商场叠放的购物车, 右图为购物车叠放在一起的示意图, 若一辆购物车车身长 1 m , 每增加一辆购物车, 车身增加 0.2 m . (1) 若某商场采购了 n 辆购物车, 求车身总长 L 与购物车辆数 n 的表达式. (2) 若该商场用直立电梯从一楼运输该批购物车到二楼, 已知该商场的直立电梯长为2.6 m , 且一次可以运输两列购物车, 求直立电梯一次性最多可以运输多少辆购物车? (3) 若该商场扶手电梯一次性可以运输 24 辆购物车, 若要运输 100 辆购物车, 且最多只能使用电梯 5次, 求: 共有多少种运输方案?", "answer": "(0.8+0.2 n) m, 18, 3", "choice_answer": ""}
|
||||
{"idx": 42, "question": "计算 -5+3 的结果是 ( ) . A. 2 B. -2 C. 8 D. -8", "answer": "-2", "choice_answer": "B"}
|
||||
{"idx": 43, "question": "2024年6月6日，嫦娥六号在距离地球约 384000千米外上演 “太空牵手”, 完成月球轨道的交会对接. 数据384000用科学记数法表示为(). A. 3.84 \\times 10^{4} B. 3.84 \\times 10^{5} C. 3.84 \\times 10^{6} D. 38.4 \\times 10^{5}", "answer": "3.84 \\times 10^{5}", "choice_answer": "B"}
|
||||
{"idx": 44, "question": "下列计算正确的是（）. A. a^{2} \\cdot a^{5}=a^{10} B. a^{8} \\div a^{2}=a^{4} C. -2 a+5 a=7 a D. \\left(a^{2}\\right)^{5}=a^{10}", "answer": "\\left(a^{2}\\right)^{5}=a^{10}3", "choice_answer": "D"}
|
||||
{"idx": 45, "question": "长江是中华民族的母亲河, 长江流域孕育出藏羌文化、巴蜀文化、荆楚文化、吴越文化等区域文化. 若从上述四种区域文化中随机选一种文化开展专题学习, 则选中 “巴蜀文化” 的概率是 ). A. \\frac{1}{4} B. \\frac{1}{3} C. \\frac{1}{2} D. \\frac{3}{4}", "answer": "\\frac{1}{4}", "choice_answer": "A"}
|
||||
{"idx": 46, "question": "完全相同的 4 个正方形面积之和是 100 , 则正方形的边长是（）. A. 2 B. 5 C. 10 D. 20", "answer": "5", "choice_answer": "B"}
|
||||
{"idx": 47, "question": "若点 \\left(0, y_{1}\\right),\\left(1, y_{2}\\right),\\left(2, y_{3}\\right) 都在二次函数 y=x^{2} 的图象上，则（ ）. A. y_{3}>y_{2}>y_{1} B. y_{2}>y_{1}>y_{3} C. y_{1}>y_{3}>y_{2} D. y_{3}>y_{1}>y_{2}", "answer": "y_{3}>y_{2}>y_{1}", "choice_answer": "A"}
|
||||
{"idx": 48, "question": "方程 \\frac{2}{x-3}=\\frac{3}{x} 的解为 ( ). A. x=3 B. x=-9 C. x=9 D. x=-3", "answer": "x=9", "choice_answer": "C"}
|
||||
{"idx": 49, "question": "数据 2,3,5,5,4 众数是____", "answer": "5", "choice_answer": ""}
|
||||
{"idx": 50, "question": "若关于 x 的一元二次方程 x^{2}+2 x+c=0 有两个相等的实数根, 则 c=____", "answer": "1", "choice_answer": ""}
|
||||
{"idx": 51, "question": "计算: \\frac{a}{a-3}-\\frac{3}{a-3}=", "answer": "1", "choice_answer": ""}
|
||||
{"idx": 52, "question": "计算: 2^{0} \\times\\left|-\\frac{1}{3}\\right|+\\sqrt{4}-3^{-1}", "answer": "2", "choice_answer": ""}
|
||||
{"idx": 53, "question": "广东省全力实施 “百县千镇万村高质量发展工程” , 2023年农产品进出口总额居全国首位，其中荔枝鲜果远销欧美．某果商以每吨2 万元的价格收购早熟荔枝, 销往国外. 若按每吨5万元出售, 平均每天可售出 100吨．市场调查反映：如果每吨降价 1 万元，每天销售量相应增加 50吨. 该果商如何定价才能使每天的 “利润” 或 “销售收入”最大？并求出其最大值", "answer": "4.5, 312.5", "choice_answer": ""}
|
||||
{"idx": 54, "question": "下列各数中, 比-2小的数是（） A. -1 B. -4 C. 4 D. 1", "answer": "-4", "choice_answer": "B"}
|
||||
{"idx": 55, "question": "定义一种新运算 *, 规定运算法则为: m * n=m^{n}-m n （ \\mathrm{~m}, \\mathrm{n} 均为整数, 且 m \neq 0 ）. 例: 2 * 3=2^{3}-2 \\times 3=2, 则 (-2) * 2=____", "answer": "8", "choice_answer": ""}
|
||||
{"idx": 56, "question": "甘肃临夏砖雕是一种历史悠久的古建筑装饰艺术，是第一批国家级非物质文化遗产. 如图 1 是一块扇面形的临夏砖雕作品, 它的部分设计图如图2, 其中扇形 O B C 和扇形 O A D 有相同的圆心 O, 且圆心角 \\angle O=100^{\\circ}, 若 O A=120 \\mathrm{~cm}, O B= 60 cm , 则阴影部分的面积是 ____ \\mathrm{cm}^{2}. （结果用 \\pi 表示）", "answer": "3000 \\pi", "choice_answer": ""}
|
||||
{"idx": 57, "question": "计算: \\sqrt{18}-\\sqrt{12} \\times \\sqrt{\\frac{3}{2}}", "answer": "0", "choice_answer": ""}
|
||||
{"idx": 58, "question": "解不等式组: \\left\\{\\begin{array}{c}2(x-2)<x+3 \\ \\frac{x+1}{2}<2 x\\end{array}\\right.", "answer": "\\frac{1}{3}<x<7", "choice_answer": ""}
|
||||
{"idx": 59, "question": "先化简, 再求值: \\left[(2 a+b)^{2}-(2 a+b)(2 a-b)\\right] \\div 2 b, 其中 a=2, b=-1", "answer": "2a+b, 3", "choice_answer": ""}
|
||||
{"idx": 60, "question": "在一只不透明的布袋中，装有质地、大小均相同的四个小球，小球上分别标有数字 1 ， 2 ， 3 ， 4.甲乙两人玩摸球游戏, 规则为：两人同时从袋中随机各摸出 1 个小球, 若两球上的数字之和为奇数 , 则甲胜; 若两球上的数字之和为偶数, 则乙胜. 求甲获胜的概率.", "answer": "\\frac{7}{12}", "choice_answer": ""}
|
||||
{"idx": 61, "question": "下列各数中, 是无理数的是（） A. \\frac{\\pi}{2} B. \\frac{1}{3} C. \\sqrt[3]{27} D. 0.13133", "answer": "\\frac{\\pi}{2}", "choice_answer": "A"}
|
||||
{"idx": 62, "question": "据央视财经《经济信息联播》消息：甘肃天水凭借一碗香喷喷的麻辣䖿成为最 “热辣滚煼” 的顶流. 2024 年3月份, 天水市累计接待游客464万人次，旅游综合收入27亿元．将数据 “ 27 亿” 用科学记数法表示为 A. 2.7 \\times 10^{8} B. 0.27 \\times 10^{10} C. 2.7 \\times 10^{9} D. 27 \\times 10^{8}", "answer": "2.7 \\times 10^{9}", "choice_answer": "C"}
|
||||
{"idx": 63, "question": "下列各式运算结果为 a^{5} 的是 ( ) A. a^{2}+a^{3} B. a^{2} a^{3} C. \\left(a^{2}\\right)^{3} D. a^{10} \\div a^{2}", "answer": "a^{2} a^{3}", "choice_answer": "B"}
|
||||
{"idx": 64, "question": "一次函数 y=k x-1(k \neq 0), 若 y 随 x 的增大而减小, 则它的图象不经过 ( ) A. 第一象限 B. 第二象限 C. 第三象限 D. 第四象限", "answer": "第一象限", "choice_answer": "A"}
|
||||
{"idx": 65, "question": "端午节期间，某商家推出“优惠酬宾” 活动，决定每袋粽子降价 2 元销售. 细心的小夏发现，降价后用240元可以比降价前多购买10袋，求：每袋粽子的原价是多少元？设每袋粽子的原价是 x 元, 所得方程正确的是 A. \\frac{240}{x}-\\frac{240}{x+2}=10 B. \\frac{240}{x}-\\frac{240}{x-2}=10 C. \\frac{240}{x-2}-\\frac{240}{x}=10 D. \\frac{240}{x+2}-\\frac{240}{x}=10", "answer": "\\frac{240}{x-2}-\\frac{240}{x}=10", "choice_answer": "C"}
|
||||
{"idx": 66, "question": "若关于 x 的一元二次方程 x^{2}+2 x-m=0 有两个相等的实数根, 则 m 的值为 ___", "answer": "-1", "choice_answer": ""}
|
||||
{"idx": 67, "question": "计算: |-\\sqrt{4}|-\\left(\\frac{1}{3}\\right)^{-1}+2025^{0}.", "answer": "0", "choice_answer": ""}
|
||||
{"idx": 68, "question": "化简: \\left(a+1+\\frac{1}{a-1}\\right) \\div \\frac{a^{2}+a}{a-1}.", "answer": "\\frac{a}{a+1}", "choice_answer": ""}
|
||||
{"idx": 69, "question": "解不等式组: \\left\\{\\begin{array}{c}2 x+1 \\geq x+2, \\ 2 x-1<\\frac{1}{2}(x+4)\\end{array}\\right..", "answer": "x\\in[1,2)", "choice_answer": ""}
|
||||
{"idx": 70, "question": "在平面直角坐标系中, 抛物线 y=-x^{2}+b x+c 与 x 轴交于 A(-1,0), B(3,0) 两点,求拋物线的解析式", "answer": "y=-x^{2}+2x+3", "choice_answer": ""}
|
||||
{"idx": 71, "question": "下列运算正确的是（） A. a^{7}-a^{3}=a^{4} B. 3 a^{2} \\cdot 2 a^{2}=6 a^{2} C. (-2 a)^{3}=-8 a^{3} D. a^{4} \\div a^{4}=a", "answer": "(-2a)^{3}=-8a^{3}", "choice_answer": "C"}
|
||||
{"idx": 72, "question": "下列数中, 能使不等式 5 x-1<6 成立的 x的值为 ( ) A. 1 B. 2 C. 3 D. 4", "answer": "1", "choice_answer": "A"}
|
||||
{"idx": 73, "question": "节能环保已成为人们的共识. 淇淇家计划购买 500 度电，若平均每天用电 x 度，则能使用 y 天．下列说法错误的是（） A. 若 x=5, 则 y=100 B. 若 y=125, 则 x=4 C. 若 x减小,则 y也减小 D. 若 x减小一半, 则 y 增大一倍", "answer": "若x减小,则y也减小", "choice_answer": "C"}
|
||||
{"idx": 74, "question": "淇淇在计算正数 a 的平方时, 误算成 a 与 2 的积, 求得的答案比正确答案小 1 , 则 a=____ A. 1 B. \\sqrt{2}-1 C. \\sqrt{2}+1 D. 1 或 \\sqrt{2}+1", "answer": "\\sqrt{2}+1", "choice_answer": "C"}
|
||||
{"idx": 75, "question": "已知 A 为整式, 若计算 \\frac{A}{x y+y^{2}}-\\frac{y}{x^{2}+x y} 的结果为 \\frac{x-y}{x y}, 则 A=(____) A. x B. y C. x+y D. x-y", "answer": "3", "choice_answer": ""}
|
||||
{"idx": 76, "question": "某校生物小组的 9 名同学各用 100 粒种子做发芽实验, 几天后观察并记录种子的发芽数分别为: 89 ，73，90，86，75，86，89，95，89, 以上数据的众数为 ___ .", "answer": "89", "choice_answer": ""}
|
||||
{"idx": 77, "question": "已知 \\mathrm{a}, \\mathrm{b}, \\mathrm{n} 均为正整数. (1) 若 n<\\sqrt{10}<n+1, 则 n= ___ ; (2) 若 n-1<\\sqrt{a}< n, n<\\sqrt{b}<n+1, 则满足条件的 a 的个数总比 b 的个数少 ___ 个.", "answer": "3, 2", "choice_answer": ""}
|
||||
{"idx": 78, "question": "甲、乙、丙三张卡片正面分别写有 a+b, 2 a+b, a-b, 除正面的代数式不同外, 其余均相同. \\begin{table}[] \\begin{tabular}{|c|c|c|c|l|l|l|l|l|l|} \\hline 第一次 & $a+b$ & $a+b$ & $a+b$ & $2a+b$ & $2a+b$ & $2a+b$ & $a-b$ & $a-b$ & $a-b$ \\ \\hline 第二次 & $a+b$ & $2a+b$ & $a-b$ & $a+b$ & $2a+b$ & $a-b$ & $a+b$ & $2a+b$ & $a-b$ \\ \\hline 和 & $2a+2b$ & & $2a$ & & & & $2a$ & & \\ \\hline \\end{tabular} \\end{table} (1)将三张卡片背面向上并洗匀, 从中随机抽取一张, 当 a=1,b=-2 时, 求取出的卡片上代数式的值为负数的概率; (2)将三张卡片背面向上并洗匀, 从中随机抽取一张, 放回后重新洗匀, 再随机抽取一张. 请在表格中补全两次取出的卡片上代数式之和的所有可能结果（化为最简）, 并求出和为单项式的概率.", "answer": "\\frac13, \\frac{4}{9}", "choice_answer": ""}
|
||||
{"idx": 79, "question": "拋物线 C_{1}: y=a x^{2}-2 x 过点 (4,0), 顶点为 Q . 抛物线 C_{2}: y=-\\frac{1}{2}(x-t)^{2}+\\frac{1}{2} t^{2}- 2 （其中 t 为常数, 且 t>2 ), 顶点为 P.直接写出 a 的值和点 Q 的坐标.", "answer": "a=\\frac{1}{2}, Q(2,-2)", "choice_answer": ""}
|
||||
{"idx": 80, "question": "据统计，2023年我国人工智能核心产业规模达5784亿元，数据 “5784亿”用科学记数法表示为 A. 5784 \\times 10^{8} B. 5.784 \\times 10^{10} C. 5.784 \\times 10^{11} D. 0.5784 \\times 10^{12}", "answer": "5.784 \\times 10^{11}", "choice_answer": "C"}
|
||||
{"idx": 81, "question": "下列不等式中，与 -x>1 组成的不等式组无解的是（） A. x>2 B. x<0 C. x<-2 D. x>-3", "answer": "x>2", "choice_answer": "A"}
|
||||
{"idx": 82, "question": "计算 (a \\cdot a \\cdots \\cdots \\cdot a)^{a} \\uparrow= 的结果是 ( ) A. a^{5} B. a^{6} C. a^{a+3} D. a^{3 a}", "answer": "a^{3 a}", "choice_answer": "D"}
|
||||
{"idx": 83, "question": "豫剧是国家级非物质文化遗产, 因其雅俗共赏, 深受大众喜爱．正面印有豫剧经典剧目人物的三张卡片如图所示, 它们除正面外完全相同. 把这三张卡片背面朝上洗匀, 从中随机抽取一张, 放回洗匀后，再从中随机抽取一张，两次抽取的卡片正面相同的概率为___ A. \\frac{1}{9} B. \\frac{1}{6} C. \\frac{1}{5} D. \\frac{1}{3}", "answer": "\\frac{1}{3}", "choice_answer": "D"}
|
||||
{"idx": 84, "question": "若关于 x 的方程 \\frac{1}{2} x^{2}-x+c=0 有两个相等的实数根, 则 c 的值为___ .", "answer": "0.5", "choice_answer": ""}
|
||||
{"idx": 85, "question": "计算: \\sqrt{2} \\times \\sqrt{50}-(1-\\sqrt{3})^{0}", "answer": "9", "choice_answer": ""}
|
||||
{"idx": 86, "question": "从地面坚直向上发射的物体离地面的高度 h(\\mathrm{~m}) 满足关系式 h=-5 t^{2}+v_{0} t, 其中 t(\\mathrm{~s}) 是物体运动的时间, v_{0}(\\mathrm{~m} / s )是物体被发射时的速度. 社团活动时, 科学小组在实验楼前从地面坚直向上发射小球. (1)小球被发射后 ___ s 时离地面的高度最大 (用含 v_{0} 的式子表示).", "answer": "\\frac{v_{0}}{10}", "choice_answer": ""}
|
||||
{"idx": 87, "question": "从地面坚直向上发射的物体离地面的高度 h(\\mathrm{~m}) 满足关系式 h=-5 t^{2}+v_{0} t, 其中 t(\\mathrm{~s}) 是物体运动的时间, v_{0}(\\mathrm{~m} / s )是物体被发射时的速度. 社团活动时, 科学小组在实验楼前从地面坚直向上发射小球.若小球离地面的最大高度为 20 m , 求小球被发射时的速度.", "answer": "20", "choice_answer": ""}
|
||||
{"idx": 88, "question": "某校八年级3班承担下周学校升旗任务, 老师从备选的甲、乙、丙、丁四名同学中, 选择两名担任升旗手, 则甲、乙两名同学同时被选中的概率是 A. \\frac{1}{6} B. \\frac{1}{8} C. \\frac{1}{4} D. \\frac{2}{3}", "answer": "\\frac{1}{6}", "choice_answer": "A"}
|
||||
{"idx": 89, "question": "一种药品原价每盒 48 元, 经过两次降价后每盒 27 元, 两次降价的百分率相同, 则每次降价的百分率为 A. 20 \\% B. 22 \\% C. 25 \\% D. 28 \\%", "answer": "25 \\%", "choice_answer": "C"}
|
||||
{"idx": 90, "question": "函数 y=\\frac{\\sqrt{x+3}}{x} 中, 自变量 x 的取值范围是 ___", "answer": "x \\geq -3, x \neq 0", "choice_answer": ""}
|
||||
{"idx": 91, "question": "将抛物线 y=a x^{2}+b x+3 向下平移 5 个单位长度后, 经过点 (-2,4), 则 6 a-3 b- 7= ____ .", "answer": "2", "choice_answer": ""}
|
||||
{"idx": 92, "question": "已知一组正整数 \\mathrm{a}, 1, \\mathrm{~b}, \\mathrm{~b}, 3 有唯一众数 8 , 中位数是 5 , 则这一组数据的平均数为 ___", "answer": "5", "choice_answer": ""}
|
||||
{"idx": 93, "question": "若分式方程 \\frac{x}{x-1}=3-\\frac{m x}{1-x} 的解为正整数, 则整数 m 的值为 ___ ", "answer": "-1", "choice_answer": ""}
|
||||
{"idx": 94, "question": "二次函数 y=\\frac{1}{2} x^{2}+b x+c 的图象与 x 轴交于 \\mathrm{A} 、 \\mathrm{~B} 两点, 与 y 轴交于点 C , 点 A 的坐标为 (-1,0), 点 C 的坐标为 (0,-3).求该二次函数的解析式", "answer": "y=\\frac{1}{2} x^{2}-\\frac{5}{2} x-3", "choice_answer": ""}
|
||||
{"idx": 95, "question": "牡丹江某县市作为猴头菇生产的 “黄金地带” ，年总产量占全国总产量的 50 \\% 以上，黑龙江省发布的 “九珍十八品” 名录将猴头菇列为首位. 某商店准备在该地购进特级鲜品、特级干品两种猴头菇, 购进鲜品猴头菇3箱、干品猴头菇2箱需 420 元, 购进鲜品猴头菇 4 箱、干品猴头菇 5 箱需 910 元. 请解答下列问题: 特级鲜品猴头菇和特级干品猴头菇每箱的进价各是多少元？", "answer": "40, 150", "choice_answer": ""}
|
||||
{"idx": 96, "question": "在平面直角坐标系中, 直线 y=x+ b 与 x 轴的正半轴交于点 A , 与 y 轴的负半轴交于点 D , 点 B 在 x 轴的正半轴上, 四边形 A B C D 是平行四边形, 线段 O A 的长是一元二次方程 x^{2}-4 x-12=0 的一个根. 请解答下列问题: 求点 D 的坐标", "answer": "(0,-6)", "choice_answer": ""}
|
||||
{"idx": 97, "question": "下列实数中，最小的是（） A. \\pi B. -(-2) C. \\sqrt{8} D. |-\\sqrt{3}|", "answer": "|-\\sqrt{3}|", "choice_answer": "D"}
|
||||
{"idx": 98, "question": "下列计算正确的是 ( ) A. x^{5}-x^{2}=x^{3} B. (x-2)^{2}=x^{2}-4 C. \\left(-3 x^{2}\\right)^{3}=-9 x^{6} D. 3 x^{2} y \\div 3 x y=x", "answer": "3 x^{2} y \\div 3 x y=x", "choice_answer": "D"}
|
||||
{"idx": 99, "question": "下列说法中正确的是（） A. 1000件产品中只有一件是次品, 从中随机抽取一件, “是次品” 是不可能事件 B. “在一张纸上任意画两个直角三角形, 这两个直角三角形相似” 是随机事件 C. 天气预报明天武汉有雨, “武汉明天下雨” 是必然事件 D. 了解汉江襄阳段的水质情况，适合用全面调查", "answer": "“在一张纸上任意画两个直角三角形, 这两个直角三角形相似” 是随机事件", "choice_answer": "B"}
|
||||
{"idx": 100, "question": "我国古代数学名著《算法统宗》里有这样一首诗: “我问开店李三公, 众客都来到店中, 一房七客多七客, 一房九客一房空. ”诗中后面两句的意思是: 如果一间客房住7人, 那么有7人无房可住; 如果一间客房住 9 人, 那么就空出一间客房. 若设该店有客房x间, 则可列方程为 ___ , 求出客房数量为 ___ 间.", "answer": "7x+7=9(x-1),8", "choice_answer": ""}
|
||||
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{"source": "2023BJ_2023年新高考北京卷-数学/01Q.en", "question": "Given sets $M=\\{x|x+2\\geq 0\\},N=\\{x|x-1<0\\}$, find $M \\cap N$.", "lang": "en", "answer": "$\\{x|-2\\leq x < 1\\}$"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/02Q.en", "question": "In the complex plane, complex number $z$ has coordinates $(-1,\\sqrt{3})$, what is its conjugate $\\bar{z}$?", "lang": "en", "answer": "$-1-\\sqrt{3}$"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/03Q.en", "question": "Vectors $\\vec{a},\\vec{b}$ satisfy $\\vec{a}+\\vec{b} = (2,3), \\vec{a}-\\vec{b}=(-2,1)$. Find $\\lvert\\vec{a}\\rvert^{2} - \\lvert\\vec{b}\\rvert^{2}$.", "lang": "en", "answer": "-1"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/05Q.en", "question": "What is the coefficient of $x$ in the full expansion of $(2x-\\frac{1}{x})^{5}$?", "lang": "en", "answer": "80"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/06Q.en", "question": "$C: y^{2}=8x$ has focus at point $F$. Point $M$ is on parabola $C$．If the distance between $M$ and the line $x=-3$ is5, then $\\lvert MF \\rvert$ = ?", "lang": "en", "answer": "4"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/07Q.en", "question": "In $\\triangle ABC$, $(a+c)(\\sin A - \\sin C) = b(\\sin A - \\sin B)$, what is the value of $\\angle C$ in radian?", "lang": "en", "answer": "$\\frac{\\pi}{3}$"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/11Q.en", "question": "If $f(x)=4^{x}+\\log_{2} x$, $f(\\frac{1}{2})$=?", "lang": "en", "answer": "1"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/12Q.en", "question": "The foci of the hyperbola are $(-2,0)$ and $(2,0)$. Its eccentricity is $\\sqrt{2}$. Find C.", "lang": "en", "answer": "$\\frac{x^{2}}{2} - \\frac{y^{2}}{2} = 1$"}
|
||||
{"source": "2023BJ_2023年新高考北京卷-数学/14Q.en", "question": "There are 9 positive numbers in sequence $\\{a_{n}\\}$, and they are in increasing order. The first 3 numbers form an arithmetic progression, and the last 7 numbers form a geometric progression. Also, $a_{1}=1,a_{5}=12,a_{9}=192$. What is the value of $a_{7}$? What is the sum of all the numbers in $\\{a_{n}\\}$?", "lang": "en", "answer": "$48$, $384$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/01Q.en", "question": "If the universal set is $U=\\{1,2,3,4,5\\}$, and $M=\\{1,4\\},N=\\{2,5\\}$, find $N \\cup \\overline{M}$.", "lang": "en", "answer": "$\\{2,3,5\\}$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/02Q.en", "question": "$\\frac{5(1+i^{3})}{(2+i)(2-i)}=?$", "lang": "en", "answer": "$1-i$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/03Q.en", "question": "Vectors $\\vec{a},\\vec{b}$ are $\\vec{a}=(3,1),\\vec{b}=(2,2)$. Find $\\cos \\langle \\vec{a}+\\vec{b},\\vec{a}-\\vec{b} \\rangle$", "lang": "en", "answer": "$\\frac{\\sqrt{17}}{17}$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/04Q.en", "question": "A team of 2 students is to be chosen from 4 candidates. Among the 4 candidates, there are 2 juniors and 2 seniors. What is the probability that the 2 selected students are in different grades?", "lang": "en", "answer": "$\\frac{2}{3}$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/05Q.en", "question": "Let $S_{n}$ be the sum of the first $n$ members of an arithmetic sequence $\\{a_{n}\\}$. If $a_{2}+a_{8}=10$ and $a_{4}a_{8}=45$, find $S_{5}$.", "lang": "en", "answer": "$20$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/07Q.en", "question": "Let $F_{1},F_{2}$ be the two focus points on ellipse $C: \\frac{x^{2}}{5}+y^{2}=1$. Point P is on C and $\\overrightarrow{PF_{1}}\\cdot \\overrightarrow{PF_{2}}=0$. Find $|\\overrightarrow{PF_{1}}| \\cdot |\\overrightarrow{PF_{2}}|$.", "lang": "en", "answer": "$2$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/08Q.en", "question": "Find the tangent line to the function $y=\\frac{e^{x}}{x+1}$ at point $\\left(1,\\frac{e}{2}\\right)$.", "lang": "en", "answer": "$y=\\frac{e}{4}x+\\frac{e}{4}$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/09Q.en", "question": "The eccentricity of hyperbola $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1,\\;(a>0,b>0)$ is $\\sqrt{5}$. One of its asymptotes intersects the circle $(x-2)^{2}+(y-3)^{2}=1$ at points A and B. Find $|AB|$", "lang": "en", "answer": "$\\frac{4\\sqrt{5}}{5}$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/10Q.en", "question": "In the triangular pyramid $P-ABC$, the base $\\triangle ABC$is an equilateral triangle with a side length of 2. $PA=PB=2$ and $PC= \\sqrt{6}$. Find the volume of the pyramid.", "lang": "en", "answer": "$1$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/12Q.en", "question": "The graph of $y=f(x)$ is obtained by shifting the graph of $y=\\cos \\left(2x+\\frac{\\pi}{6}\\right)$ to the left $\\frac{\\pi}{6}$ units. How many intersection points does the graph of $y=f(x)$ have with line $y=\\frac{1}{2}x-\\frac{1}{2}$?", "lang": "en", "answer": "$3$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/13Q.en", "question": "Let $S_{n}$ be the sum of the first $n$ members of a geometric sequence $\\{a_{n}\\}$. If $8S_{6}=7S_{3}$, find the common ratio of $\\{a_{n}\\}$.", "lang": "en", "answer": "$-\\frac{1}{2}$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/14Q.en", "question": "If $f(x)=(x-1)^{2}+ax+\\sin \\left(x+\\frac{\\pi}{2}\\right)$ is an even function, what is the value of $a$?", "lang": "en", "answer": "$2$"}
|
||||
{"source": "2023NationalAARTS_2023年全国甲卷-数学文理/15Q.en", "question": "Variables x and y satisfy the system $\\begin{cases} 3x-2y \\leq 3 \\\\ -2x+3y \\leq 3 \\\\ x+y \\geq 1 \\end{cases}$. What is the maximum value of $z=3x+2y$?", "lang": "en", "answer": "$15$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/01Q.en", "question": "Let sets $A$ and $B$ be defined as $A=\\{x|x=3k+1,k\\in \\mathbb{Z}\\},B=\\{x|x=3k+2,k \\in \\mathbb{Z}\\}$. U, the universal set, is the set of integers. Find $\\overline{A \\cup B}$.", "lang": "en", "answer": "$\\{x|x=3k,k\\in \\mathbb{Z}\\}$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/02Q.en", "question": "Given complex number $(a+i)(1-ai)=2,\\;a \\in \\mathbb{R}$, find $a$.", "lang": "en", "answer": "$1$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/04Q.en", "question": "Vectors $\\vec{a}$, $\\vec{b}$, and $\\vec{c}$ have the following properties: $|\\vec{a}|=|\\vec{b}|=1$, $|\\vec{c}|=\\sqrt{2}$, and $\\vec{a}+\\vec{b}+\\vec{c}=\\vec{0}$. Find $\\cos \\langle\\vec{a}-\\vec{c},\\vec{b}-\\vec{c}\\rangle$", "lang": "en", "answer": "$\\frac{4}{5}$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/05Q.en", "question": "In the geometric sequence $\\{a_{n}\\}$, $a_{1}=1$ and the common ratio is positive. Let $S_{n}$ be the sum of the first n members. If $S_{5}=5S_{3}-4$, find $S_{4}$.", "lang": "en", "answer": "$15$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/06Q.en", "question": "50 people sign up for a football club, 60 people sign up for a table tennis club, and 70 people sign up for either the football or the table tennis club. There are no other clubs and everyone signs up for at least one of the clubs. If someone signs up for the football club, what is the probability that the same person signs up for the table tennis club?", "lang": "en", "answer": "$0.8$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/07Q.en", "question": "The statement \"$\\sin^{2} \\alpha + \\sin^{2} \\beta = 1$\" is what kind of condition for the statement \"$\\sin \\alpha + \\cos\\beta=0$\"? (sufficient, necessary, both, or neither)", "lang": "en", "answer": "$\\text{a necessary but not a sufficient condition}$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/09Q.en", "question": "Five volunteers participate in a community service on Saturday and Sunday. Each day, two people among the five are randomly chosen for that day. In how many ways can one person be chosen for both days and the other person varies on the two days?", "lang": "en", "answer": "$60$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/11Q.en", "question": "Pyramid $P-ABCD$ has a square base ABCD and P is the top vertex. $AB=4$, $PC=PD=3$, and $\\angle PCA=45^{\\degree}$. Find the area of $\\triangle PBC$.", "lang": "en", "answer": "$4\\sqrt{2}$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/12Q.en", "question": "The two focus points of ellipse $\\frac{x^{2}}{9}+\\frac{y^{2}}{6}=1$ are $F_{1}$ and $F_{2}$. O is the origin. Point P is a point on the ellipse such that $\\cos\\angle F_{1}PF_{2}=\\frac{3}{5}$. Find $|PO|$", "lang": "en", "answer": "$\\frac{\\sqrt{30}}{2}$"}
|
||||
{"source": "2023NationalASTEM_2023年高考全国甲卷-数学（理）/16Q.en", "question": "In $\\triangle ABC$, $AB=2$, $\\angle BAC=60^{\\degree}$, and $BC=\\sqrt{6}$. Point D is on BC and AD bisects $\\angle BAC$. Find AD.", "lang": "en", "answer": "$2$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/01Q.en", "question": "$\\lvert 2+i^{2}+2i^{3} \\rvert = ?$", "lang": "en", "answer": "$\\sqrt{5}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/02Q.en", "question": "Suppose the universe set is $U=\\{0,1,2,4,6,8\\}$. Two of its subsets are $M=\\{0,4,6\\},\\;N=\\{0,1,6\\}$. Find $M \\cup \\bar{N}$.", "lang": "en", "answer": "$\\{0,2,4,6,8\\}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/04Q.en", "question": "In $\\triangle ABC$，angles $A,B,C$ correspond to sides $a,b,c$ respectively. If $a\\cos B - b\\cos A=c$, and $C=\\frac{\\pi}{5}$, find $\\angle B$ in radian.", "lang": "en", "answer": "$\\frac{3\\pi}{10}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/05Q.en", "question": "$f(x)=\\frac{xe^{x}}{e^{ax}-1}$ is an even function, what is the value of $a$?", "lang": "en", "answer": "2"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/06Q.en", "question": "Square ABCD has side length 2, E is the midpoint on AB. Find $\\overrightarrow{EC}\\cdot\\overrightarrow{ED}$.", "lang": "en", "answer": "3"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/07Q.en", "question": "In the standard $xy$ coordinate plane, a point is chose randomly in the area defined by $\\{(x,y)|1 \\leq x^{2}+y^{2}\\leq 4\\}$. What is the probability that the angle between the line OA and the positive $x-$axis is no bigger than $\\frac{\\pi}{4}$?", "lang": "en", "answer": "$\\frac{1}{8}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/08Q.en", "question": "Function $f(x)=x^{3}+2ax+2$ has 3 roots. What is the range of $a$?", "lang": "en", "answer": "$(-\\infty,-3)$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/09Q.en", "question": "There are 6 topics to choose from in an essay competition. Each contestant chooses 1 topic. If there are 2 contestants, what is the probability that the topics they choose are different?", "lang": "en", "answer": "$\\frac{5}{6}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/10Q.en", "question": "Function $f(x)=\\sin(\\omega x + \\varphi)$ is increasing on the interval $(\\frac{\\pi}{6}, \\frac{2\\pi}{3})$. And the points at $x=\\frac{\\pi}{6}$ and $x=\\frac{2\\pi}{3}$ are vertices of $y=f(x)$. What is the value of $f(-\\frac{5\\pi}{12})?$", "lang": "en", "answer": "$\\frac{\\sqrt{3}}{2}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/11Q.en", "question": "If real numbers $x,y$ satisfy $x^{2}+y^{2}-4x-2y-4=0$, what is the maximum value of $x-y?$", "lang": "en", "answer": "$1+3\\sqrt{2}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/13Q.en", "question": "$A(1,\\sqrt{5})$ is on the parabola $C: y^{2}=2px$. What is the distance between A and its directrix?", "lang": "en", "answer": "$\\frac{9}{4}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/14Q.en", "question": "If $\\theta \\in (0, \\frac{\\pi}{2})$, find the value of $\\sin \\theta - \\cos \\theta$", "lang": "en", "answer": "$-\\frac{\\sqrt{5}}{5}$"}
|
||||
{"source": "2023NationalBARTS_2023年全国乙卷-文科数学/15Q.en", "question": "Variables x and y satisfy the system $\\begin{cases} x-2y \\leq -1 \\\\ x+2y \\leq 9 \\\\ 3x+y \\geq 7 \\end{cases}$. What is the maximum value of $z=2x-y$?", "lang": "en", "answer": "$8$"}
|
||||
{"source": "2023NationalBSTEM_2023年全国乙卷-理科数学/01Q.en", "question": "If $z=\\frac{2+i}{1+i^{2}+i^{5}}$, what is the value of $\\bar{z}=?$", "lang": "en", "answer": "$1+2i$"}
|
||||
{"source": "2023NationalBSTEM_2023年全国乙卷-理科数学/07Q.en", "question": "Jack and Kirk each selects 2 topics from 6 reading topics. How many ways can they pick 1 and only 1 common topic?", "lang": "en", "answer": "$120$"}
|
||||
{"source": "2023NationalBSTEM_2023年全国乙卷-理科数学/10Q.en", "question": "Arithmetic sequence $\\{a_{n}\\}$ has common difference $\\frac{2\\pi}{3}$. Let set $S$ be $S=\\{\\cos a_{n} | n \\in \\mathbb{N}^{+}\\}$. If $S=\\{a,b\\}$, find the value of $ab$.", "lang": "en", "answer": "$-\\frac{1}{2}$"}
|
||||
{"source": "2023NationalBSTEM_2023年全国乙卷-理科数学/12Q.en", "question": "Circle $\\odot O$ has radius 1. P is a point outside the circle. Line PA is tangent to $\\odot O$ at point $A$. Line PB intersects $\\odot O$ at points B and C. D is the midpoint of BC. If $\\lvert OP \\rvert = \\sqrt{2}, find the maximum value of $\\overrightarrow{PA} \\cdot \\overrightarrow{PD}$.", "lang": "en", "answer": "$\\frac{1+\\sqrt{2}}{2}$"}
|
||||
{"source": "2023NationalBSTEM_2023年全国乙卷-理科数学/15Q.en", "question": "Sequence $\\{a_{n}\\}$ is a geometric sequence in which $a_{2}a_{4}a_{5} = a_{3}a_{6}$ and $a_{9}a_{10}=-8$. Find $a_{7}$", "lang": "en", "answer": "$-2$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/01Q.en", "question": "If sets $M=\\{-2,-1,0,1,2\\}$ and $N=\\{x | x^{2} - x - 6 \\geq 0\\}$, what is in the set $M \\cap N$?", "lang": "en", "answer": "$\\{-2\\}$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/02Q.en", "question": "If $z=\\frac{1-i}{2+2i}$, find $z-\\bar{z}$.", "lang": "en", "answer": "$-i$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/04Q.en", "question": "Function $f(x)=2^{x(x-a)}$ is monotonically increasing on the interval $(0,1)$, what is the range of $a$?", "lang": "en", "answer": "$[2,+\\infty)$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/05Q.en", "question": "Ellipses $C_{1}: \\frac{x^{2}}{a^{2}} + y^{2} = 1\\;(a>1)$ and $C_{2}=\\frac{x^{2}}{4} + y^{2} = 1$ have eccentricity $e_{1}$ and $e_{2}$ respectively. If $e_{2}=\\sqrt{3}e_{1}$, find $a$.", "lang": "en", "answer": "$\\frac{2\\sqrt{3}}{3}$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/06Q.en", "question": "The angle between the two lines that are tangent to the circle $x^{2}+y^{2}-4x-1=0$ and that pass through point $(0,-2)$is $\\alpha$. Find $\\sin \\alpha$.", "lang": "en", "answer": "$\\frac{\\sqrt{15}}{4}$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/08Q.en", "question": "If $\\sin(\\alpha - \\beta)=\\frac{1}{3}$ and $\\cos \\alpha \\sin \\beta=\\frac{1}{6}$, find $\\cos (2\\alpha + 2\\beta).$", "lang": "en", "answer": "$\\frac{1}{9}$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/13Q.en", "question": "A school has 4 science classes and 4 art classes. Each student is required to choose 2 or 3 classes from the total 8 classes and must have at least 1 class in each group (science and art). How many ways can a student select his classes?", "lang": "en", "answer": "$64$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/15Q.en", "question": "Function $f(x)=\\cos \\omega x -1\\;(\\omega>0)$ has 3 roots on the interval $[0, 2\\pi]$, what is the range of $\\omega$?", "lang": "en", "answer": "$[2,3)$"}
|
||||
{"source": "2023NewNational1_2023年新课标全国I卷-数学/16Q.en", "question": "Hyperbola $C$ is given by $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} = 1\\;(a>0,b>0).$ Its left and right foci points are $F_{1},F_{2}$ respectively．Point $A$ is on $C$, and point $B$ is on the $y-$axis. $\\overrightarrow{F_{1}A} \\perp \\overrightarrow{F_{1}B}$ and $\\overrightarrow{F_{2}A} = -\\frac{2}{3}\\overrightarrow{F_{2}B}$. What is the eccentricity of $C$?", "lang": "en", "answer": "$\\frac{3\\sqrt{5}}{5}$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/01Q.en", "question": "In the complex plane, which quadrant lies $(1+3i)(3-i)?$", "lang": "en", "answer": "$Quadrant I$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/02Q.en", "question": "Let sets $A$ and $B$ be $A=\\{0,-a\\}$ and $B=\\{1,a-2,2a-2\\}$. If $A \\subseteq B$, find a.", "lang": "en", "answer": "$1$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/03Q.en", "question": "Jackson school district samples its students to participate in a survey. The district has 400 middle school students and 200 high school students. The number of students sampled from each school is proportional to each school's total population. If there are total 60 students selected, in how many ways can these students be selected? (express your answer in the combination notation.)", "lang": "en", "answer": "$C_{400}^{40} \\cdot C_{200}^{20}$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/04Q.en", "question": "If $f(x)=(x+a)\\ln \\frac{2x-1}{2x+1}$ is an even function, what is the value of $a=?$", "lang": "en", "answer": "$0$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/05Q.en", "question": "The left and the right foci points of the ellipse $C:\\frac{x^{2}}{3}+y^{2}=1$ are $F_{1}$ and $F_{2}$ respectively. Line $y=x+m$ intersects $C$ at points A and B. If the area of $\\triangle F_{1}AB$ is twice the area of $\\triangle F_{2}AB$, what is the value of $m$?", "lang": "en", "answer": "$-\\frac{\\sqrt{2}}{3}$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/06Q.en", "question": "Function $f(x)=ae^{x}-\\ln x$ is increasing on the interval $(1,2)$. What is the minimum value of a?", "lang": "en", "answer": "$e^{-1}$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/07Q.en", "question": "Given that $\\alpha$ is an acute angle and $\\cos \\alpha = \\frac{1+\\sqrt{5}}{4}$. Find $\\sin \\frac{\\alpha}{2}$", "lang": "en", "answer": "$\\frac{-1+\\sqrt{5}}{4}$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/08Q.en", "question": "Let $S_{n}$ be the sum of the first $n$ items of a geometric sequence $\\{a_{n}\\}$. If $S_{4}=-5$ and $S_{6}=21S_{2}$, find $S_{8}$.", "lang": "en", "answer": "$-85$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/13Q.en", "question": "Vectors $\\vec{a}$ and $\\vec{b}$ satisfy $|\\vec{a}-\\vec{b}|=\\sqrt{3}$ and $|\\vec{a}+\\vec{b}|=|2\\vec{a}-\\vec{b}|$. Find $|\\vec{b}|$.", "lang": "en", "answer": "$\\sqrt{3}$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/14Q.en", "question": "A pyramid has a square base whose side is 4 units long. The pyramid has its upper portion cut off. The portion that is cut off is a smaller pyramid with a height of 3 units tall and a smaller square base whose side is 2 units long. The lower portion of the pyramid after its top is cut off is called a frustum. What is the volume of the frustum?", "lang": "en", "answer": "$28$"}
|
||||
{"source": "2023NewNational2_2023年新课标全国II卷-数学/15Q.en", "question": "Line $l: x-my+1=0$ intersects the circle $\\odot C: (x-1)^{2}+y^{2}=4$ at points A and B. The area of the triangle $\\triangle ABC$ is $\\frac{8}{5}. Find one possible value of $m$.", "lang": "en", "answer": "$2$ or $-2$ or $\\frac{1}{2}$ or $-\\frac{1}{2}$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/01Q.en", "question": "Solve $|x-2|<1$.", "lang": "en", "answer": "$(1,3)$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/02Q.en", "question": "If $\\vec{a}=(-2,3)$, and $\\vec{b}=(1,2)$, find $\\vec{a} \\cdot \\vec{b}$.", "lang": "en", "answer": "$4$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/03Q.en", "question": "In geometric sequence $\\{a_{n}\\}$, the common ratio $r$ is 2 and the first item $a_{1}=3$. Find $S_{6}$, the sum of the first 6 items.", "lang": "en", "answer": "$189$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/04Q.en", "question": "If $\\tan\\alpha=3$, find $\\tan 2\\alpha$.", "lang": "en", "answer": "$-\\frac{3}{4}$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/05Q.en", "question": "Find the range of function $f(x)=\\begin{cases} 2^{x},\\;&x>0 \\\\ 1,\\;&x \\leq 0 \\end{cases}$.", "lang": "en", "answer": "$[1,+\\infty)$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/06Q.en", "question": "If $z=1+i$, find $|1-i\\cdot z|$.", "lang": "en", "answer": "$\\sqrt{5}$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/07Q.en", "question": "The area of the circle given by $x^{2}+y^{2}-4y-4m=0$ is $\\pi$. Find $m$.", "lang": "en", "answer": "$-3$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/08Q.en", "question": "In $\\triangle ABC$, the three sides are $a=4,b=5,c=6$. Find $\\sin A$.", "lang": "en", "answer": "$\\frac{\\sqrt{7}}{4}$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/09Q.en", "question": "Gross domestic product (GDP) is an indicator to measure regional economic conditions. According to statistics, a city experienced GDP growth quarter to quarter in 2020. The GDP in the first quarter and the fourth quarter were 231 and 242 (in millions of dollars) respectively. The median and the average of the GDP in the four quarters are equal, what is the total GDP in 2020 (in millions of dollars)?", "lang": "en", "answer": "$946$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/10Q.en", "question": "Given that $(1+2023x)^{100}+(2023-x)^{100}=a_{0}+a_{1}x+a_{2}x^{2}+\\ldots+a_{100}x^{100}$, where $a_{0},a_{1},a_{2},\\ldots,a_{100} \\in \\mathbb{R}$ and $0 \\leq k \\leq 100$ and $k \\in \\mathbb{N}$, what is maximum value of $k$ such that $a_{k} < 0$?", "lang": "en", "answer": "$49$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/12Q.en", "question": "Three points A,B,C satisfy $AB=BC=AC=1$. How many ways are there to pick another two points (in any order) such that the 5 points (A,B,C and the two additional points) form the vertices of a regular square pyramid?", "lang": "en", "answer": "$9$"}
|
||||
{"source": "2023SH_2023年上海卷-数学/13Q.en", "question": "Sets P and Q are defined as follows: $P=\\{1,2\\},Q=\\{2,3\\}$. If $M=\\{x|x \\in P\\;and\\;x \\notin Q\\}$, find $M$.", "lang": "en", "answer": "$\\{1\\}$"}
|
||||
{"source": "2023TJ_2023天津卷-数学/01Q.en", "question": "Given sets $U=\\{1,2,3,4,5\\}, A=\\{1,3\\}, B=\\{1,2,4\\}$ ，what is $\\overline{B} \\cup A$?", "lang": "en", "answer": "$\\{1,3,5\\}$"}
|
||||
{"source": "2023TJ_2023天津卷-数学/03Q.en", "question": "Given $a=1.01^{0.5}, b=1.01^{0.6}, c=0.6^{0.5}$，put a,b,c in decreasing order.", "lang": "en", "answer": "b > a > c"}
|
||||
{"source": "2023TJ_2023天津卷-数学/06Q.en", "question": "Sequence $\\{a_{n}\\}$ is an geometric sequence. $S_n$ is the sum of the first $n$ numbers of $\\{a_{n}\\}$ and $a_{n+1} = 2S_{n}+2$. What is the value of $a_{4}$?", "lang": "en", "answer": "54"}
|
||||
{"source": "2023TJ_2023天津卷-数学/09Q.en", "question": "The left and right foci of hyperbola $\\frac{x^{2}}{a^{2}} - \\frac{y^{2}}{b^{2}} (a>0, b>0)$ are $F_{1}, F_{2}$ respectively．A perpendicular line is drawn from $F_{2}$ to one of the asymptotes and the perpendicular foot is point P. If $PF_{2} = 2$, and the slope of $PF_{1}$ is $\\frac{\\sqrt{2}}{4}$, find the equation of the hyperbola.", "lang": "en", "answer": "$\\frac{x^{2}}{2} - \\frac{y^{2}}{4}=1$"}
|
||||
{"source": "2023TJ_2023天津卷-数学/10Q.en", "question": "Simplify $\\frac{5+14i}{2+3i}$ where $i$ is the imaginary number.", "lang": "en", "answer": "$4+i$"}
|
||||
{"source": "2023TJ_2023天津卷-数学/11Q.en", "question": "When $(2x^{3}-\\frac{1}{x})^{6}$ is multiplied out，what is the coefficient of the $x^{2}$ term?", "lang": "en", "answer": "60"}
|
||||
{"source": "2023TJ_2023天津卷-数学/12Q.en", "question": "A line passes through the origin and is tangent to the circle $C: (x+2)^{2}+y^{2}=3$. The line intersects the curve $y^{2} = 2px (p>0)$ at point P. If $|OP|=8$ ，find $p$.", "lang": "en", "answer": "6"}
|
||||
{"source": "2023TJ_2023天津卷-数学/13Q.en", "question": "Three boxes A, B, C contain certain number of black and white balls. The total number of balls in each box is in the ratio of 5:4:6．The percentage of black balls in each box is respectively 40%, 25%, and 50%．A ball is picked out randomly from each of the three boxes. What is the probability that all three balls are black? If the balls in the three boxes are merged into one box and a ball is drawn randomly, what is the probability of drawing a white ball?", "lang": "en", "answer": "$0.05$ and $0.6$"}
|
||||
{"source": "2023TJ_2023天津卷-数学/14Q.en", "question": "In $\\triangle ABC$，$\\angle A = 60^{\\degree}$ ，$BC=1$ ，Point D is the midpoint of AB and point E is the midpoint of CD. Let $\\overline{AB} = \\overrightarrow{a},\\overline{AC} = \\overrightarrow{b}$. Express $\\overline{AE}$ in terms of $\\overrightarrow{a},\\;\\overrightarrow{b}$. If $\\overline{BF} = \\frac{1}{3}\\overline{BC}$, find the maximum value of $\\overline{AE} \\cdot \\overline{AF}$.", "lang": "en", "answer": "$\\frac{1}{4}\\overrightarrow{a} + \\frac{1}{2}\\overrightarrow{b}$ and $\\frac{13}{24}$"}
|
||||
{"source": "2023TJ_2023天津卷-数学/15Q.en", "question": "If the function $f(x)=ax^{2}-2x-|x^{2}-ax+1|$ has only 2 real roots, what is the range of $a$?", "lang": "en", "answer": "$(-\\infty, 0)\\cup(0,1)\\cup(1,+\\infty)$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Find the sum of the four least positive integers each of whose digits add to 12.", "id": "PurpleMeet2023_1", "lang": "en", "answer": "$210$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "There are positive real numbers a, b, c, d, and p such that a is 62.5\\% of b, b is 64\\% of c, c is 125\\% of d, and d is p\\% of a. Find p.", "id": "PurpleMeet2023_2", "lang": "en", "answer": "$200$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Mike has two similar pentagons. The first pentagon has a perimeter of 18 and an area of $8\\frac{7}{16}$. The second pentagon has a perimeter of 24. Find the area of the second pentagon.", "id": "PurpleMeet2023_3", "lang": "en", "answer": "$15$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "cdrst has digits a, b, c, d, r, s, and t, in that order, and none of the digits is 0. The two-digit numbers ab, bc, cd, and dr, and the three-digit number rst are all perfect squares. Find $a b c d r s t$.", "id": "PurpleMeet2023_4", "lang": "en", "answer": "$8164961$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "m n satisfy $(m+n)(24mn+1)=2023$. Find m+n+12mn.", "id": "PurpleMeet2023_5", "lang": "en", "answer": "$151$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Find the least positive integer such that the product of its digits is $8! = 8 \\cdot 7 \\cdot 6 \\cdot 5 \\cdot 4 \\cdot 3 \\cdot 2 \\cdot 1$.", "id": "PurpleMeet2023_6", "lang": "en", "answer": "$257889$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Elijah went on a four-mile journey. He walked the first mile at 3 miles per hour and the second mile at 4 miles per hour. Then he ran the third mile at 5 miles per hour and the fourth mile at 6 miles per hour. Elijah’s average speed for this journey in miles per hour was $\\frac{m}{n}$, where m and n are relatively prime positive integers. Find m + n.", "id": "PurpleMeet2023_7", "lang": "en", "answer": "$99$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Find the number of ways to write 24 as the sum of at least three positive integer multiples of 3. For example, count 3+18+3, 18+3+3, and 3+6+3+9+3, but not 18+6 or 24.", "id": "PurpleMeet2023_8", "lang": "en", "answer": "$120$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Find the positive integer n such that $1+2+3+\\ldots+n=(n+1)+(n+2)+\\ldots+(n+35)$.", "id": "PurpleMeet2023_9", "lang": "en", "answer": "$84$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "A smaller square in completely inside a larger square. No side of the smaller square touches any side of the large square. Both squares have integer side lengths. The region inside the larger square but outside the smaller square has area 52. Find the area of the larger square.", "id": "PurpleMeet2023_10", "lang": "en", "answer": "$196$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Three of the 16 squares from a $4 \\times 4$ grid of squares are selected at random. The probability that at least one corner square of the grid is selected is mn , where m and n are relatively prime positive integers. Find m + n.", "id": "PurpleMeet2023_11", "lang": "en", "answer": "$45$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Find the greatest prime that divides $1^{2}-2^{2}+3^{2}-4^{2}+\\ldots-98^{2}+99^{2}$.", "id": "PurpleMeet2023_12", "lang": "en", "answer": "$11$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "In convex quadrilateral ABCD, $\\angle BAD = \\angle BCD = 90^{\\circ}$, and BC = CD. Let E be the intersection of diagonals AC and BD. Given that $\\angle AED = 123^{\\circ}$, find the degree measure of $\\angle ABD$.", "id": "PurpleMeet2023_13", "lang": "en", "answer": "$78$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "A square, a regular pentagon, and a regular hexagon are all inscribed in the same circle. The 15 vertices of these polygons divide the circle into at most 15 arcs. Let M be the degree measure of the longest of these arcs. Find the minimum possible value for M.", "id": "PurpleMeet2023_14", "lang": "en", "answer": "$48$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "A rectangle with integer side lengths has the property that its area minus 5 times its perimeter equals 2023. Find the minimum possible perimeter of this rectangle.", "id": "PurpleMeet2023_15", "lang": "en", "answer": "$448$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "A sequence of 28 letters consists of 14 of each of the letters A and B arranged in random order. The expected number of times that ABBA appears as four consecutive letters in that sequence is $\\frac{m}{n}$ , where m and n are relatively prime positive integers. Find m + n.", "id": "PurpleMeet2023_16", "lang": "en", "answer": "$145$solution:"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Let x, y, and z be positive integers satisfying the following system of equations: $\\begin{cases} x^{2} + \\frac{2023}{x}=2y^{2} \\\\ y + \\frac{2028}{y^{2}} = z^{2} \\\\ 2z + \\frac{2025}{z^{2}} = xy \\end{cases}$. Find x+y+z. Calculators are not allowed to solve the equations directly. Caculators are allowed for adding, subtracting, multiplying, and dividing numbers.", "id": "PurpleMeet2023_17", "lang": "en", "answer": "$25$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "For real number x, let $\\lfloor x \\rfloor$ denote the greatest integer less than or equal to x, and let {x} denote the fractional part of x, that is $\\{x\\}=x-\\lfloor x \\rfloor$. The sum of the solutions to the equation $2\\lfloor x \\rfloor^{2}+3\\{x\\}^{2}=\\frac74 x \\lfloor x \\rfloor$ can be written as $\\frac{p}{q}$, where p and q are prime numbers. Find 10p+q.", "id": "PurpleMeet2023_18", "lang": "en", "answer": "$232$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "A trapezoid has side lengths 24, 25, 26, and 27 in some order. Find its area.", "id": "PurpleMeet2023_19", "lang": "en", "answer": "$612$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "PurpleMeet2023", "question": "Nine light bulbs are equally spaced around a circle. When the power is turned on, each of the nine light bulbs turns blue or red, where the color of each bulb is determined randomly and independently of the colors of the other bulbs. Each time the power is turned on, the probability that the color of each bulb will be the same as at least one of the two adjacent bulbs on the circle is $\\frac{m}{n}$, where m and n are relatively prime positive integers. Find m + n.", "id": "PurpleMeet2023_20", "lang": "en", "answer": "$293$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "What is the total cost of 3 jars of peanut butter priced $5 each and 5 jars of jelly priced $3 each?", "id": "Mathcounts2023ChapterSprint_1", "lang": "en", "answer": "30"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "What is the absolute difference between $2^{5}$ and $5^{2}$?", "id": "Mathcounts2023ChapterSprint_2", "lang": "en", "answer": "7"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "What is the value of the expression $\\frac15 + \\frac{11}{15}$? Express your answer as a common fraction. Calculate without using a calculator.", "id": "Mathcounts2023ChapterSprint_3", "lang": "en", "answer": "$\\frac{14}{15}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Bob has \\$50 in his pocket when he sets off for the movie theatre. After he pays \\$10 for a ticket, \\$8 for popcorn and \\$7 for a soda, how much money does Bob have left in his pocket?", "id": "Mathcounts2023ChapterSprint_4", "lang": "en", "answer": "25"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "What is the area of the right triangle shown with legs of lengths 5 cm and 8 cm?", "id": "Mathcounts2023ChapterSprint_5", "lang": "en", "answer": "20 $cm^{2}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "A large square is divided into 25 congruent squares. The middle $3\\times3$ grid is another square, smaller than the original square. What percentage of the original big square is the middle smaller square?", "id": "Mathcounts2023ChapterSprint_6", "lang": "en", "answer": "36"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "On the number line, the number n is one-third of the way from 0 and 6 and lies in between 0 and 6. What is the value of $n$?", "id": "Mathcounts2023ChapterSprint_7", "lang": "en", "answer": "2"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "What is the value of $\\frac12$ of $\\frac13$ of $\\frac14$ of 240?", "id": "Mathcounts2023ChapterSprint_8", "lang": "en", "answer": "10"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "If $4n+1=250$, what is the nearest integer to the value of n?", "id": "Mathcounts2023ChapterSprint_10", "lang": "en", "answer": "62"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "What is the value of the expression $\\sqrt{\\frac14}+\\sqrt{\\frac19}+\\sqrt{\\frac{1}{16}}$? Express your answer as a common fraction. Perform any fraction additions without caculator.", "id": "Mathcounts2023ChapterSprint_11", "lang": "en", "answer": "$\\frac{13}{12}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "The sum of two numbers is 9 and their absolute difference is 3. What is their product?", "id": "Mathcounts2023ChapterSprint_12", "lang": "en", "answer": "18"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Of the fish in Ari’s aquarium, $\\frac12$ are red, $\\frac14$ are blue and $\\frac18$ are black. The remaining 4 fish are yellow. Given that all of Ari’s fish are a single color, how many fish are in Ari’s aquarium?", "id": "Mathcounts2023ChapterSprint_13", "lang": "en", "answer": "32"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "A bag is filled with 100 marbles each colored red, white or blue. The table shows the results when Cia randomly draws 10 marbles. Based on this data, how many of the marbles in the bag are expected to be red?\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlineRed & // \\\\ \\hlineWhite & ///// \\\\ \\hlineBlue & /// \\\\ \\hline\\end{tabular}\\end{table}", "id": "Mathcounts2023ChapterSprint_14", "lang": "en", "answer": "20"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Margaret holds tea parties every Tuesday afternoon for the purpose of using her collection of 100 teacups. If she invites n people, she will use n + 1 teacups: one for each invited guest and one for herself. If she has already had 24 tea parties, each with two guests, how many tea parties with three guests should she host to ensure each teacup is used exactly once?", "id": "Mathcounts2023ChapterSprint_15", "lang": "en", "answer": "7"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Evan gets stuck in an elevator. At 12:11 a.m., the elevator repair company dispatches a technician who is 75 miles away. The technician drives at an average speed of 50 mi/h, and after arriving, takes 10 minutes to enter the building, and then an additional 7 minutes to unlock the elevator. At what time is Evan released from the elevator?", "id": "Mathcounts2023ChapterSprint_16", "lang": "en", "answer": "1:58 am"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "If the median of the data set {x + 2, x + 3, x − 4, x − 1, x + 1} is 6, what is the value of x?", "id": "Mathcounts2023ChapterSprint_17", "lang": "en", "answer": "$7$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Let x, y and z be positive integers with x < y < z. If the mean of x, y and z is 99, what is the greatest possible value of z?", "id": "Mathcounts2023ChapterSprint_18", "lang": "en", "answer": "$294$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Which digit should replace E in the units place of the number 1,234,56E so that the number is divisible by 9?", "id": "Mathcounts2023ChapterSprint_19", "lang": "en", "answer": "$6$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "During P.E. class, twelve students take turns playing an 8-player game. If class lasts 1 hour and each student plays the same amount of time, what is the maximum number of minutes each student can actively be playing the game?", "id": "Mathcounts2023ChapterSprint_20", "lang": "en", "answer": "$40$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Triangle ABC has AB = 10 and BC = 6. How many different integer lengths are possible for side AC?", "id": "Mathcounts2023ChapterSprint_21", "lang": "en", "answer": "$11$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Consider the geometric sequence $a_{1}=1, a_{2}=2, a_{3}=4, a_{4}=8, \\ldots$. What is the average of the first 10 terms of this sequence? Express your answer as a decimal to the nearest tenth.", "id": "Mathcounts2023ChapterSprint_22", "lang": "en", "answer": "$102.3$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Isosceles trapezoid MATH has bases TH and AM. AT and HM are the other two sides. MATH has height 4 cm. If AT = HM = 5 cm and TH = 4 cm, what is the perimeter of trapezoid MATH?", "id": "Mathcounts2023ChapterSprint_23", "lang": "en", "answer": "$24$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "If x + 2y = 9 and 5x + 4y = −4, what is the value of 8x + 10y?", "id": "Mathcounts2023ChapterSprint_24", "lang": "en", "answer": "$23$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Grace made over four dozen cupcakes. If she makes packages of 2 cupcakes, then there is 1 left over. Packaging in groups of 3 cupcakes leaves 2 left over, and packaging in groups of 4 cupcakes leaves 3 left over. What is the fewest number of cupcakes Grace could have made?", "id": "Mathcounts2023ChapterSprint_25", "lang": "en", "answer": "$59$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "What is the value of $\\frac{\\sqrt{45}+2\\sqrt{15}}{\\sqrt{128}+2\\sqrt{24}}$? Conversion via decimal point numbers not allowed. Express your answer as a common fraction in simplest radical form.", "id": "Mathcounts2023ChapterSprint_26", "lang": "en", "answer": "$\\frac{\\sqrt{30}}{8}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Each face of a right square pyramid has a perimeter of 24 cm. What is the volume of the pyramid? Express your answer in simplest radical form.", "id": "Mathcounts2023ChapterSprint_27", "lang": "en", "answer": "$36\\sqrt{7}cm^{3}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "If each of a, b and c is the cube of some nonzero integer, what is the product of all of the different possible values of $\\frac{abc}{|abc|}+\\frac{ab}{|ab|}+\\frac{ac}{|ac|}+\\frac{bc}{|bc|}+\\frac{a}{|a|}+\\frac{b}{|b|}+\\frac{c}{|c|}$?", "id": "Mathcounts2023ChapterSprint_28", "lang": "en", "answer": "$-7$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "The area of a rectangle is $32cm^{2}$ . What is the least possible integer perimeter of the rectangle?", "id": "Mathcounts2023ChapterSprint_29", "lang": "en", "answer": "$23$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterSprint", "question": "Consider the six distinct points on a circle, three on one semicicle and three on the other semicicle. If George draws line segments connecting pairs of points so that each point is connected to exactly two other points, what is the probability that the resulting figure is a convex hexagon? Express your answer as a common fraction.", "id": "Mathcounts2023ChapterSprint_30", "lang": "en", "answer": "$\\frac{1}{70}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "Alton paid \\$31.50 for 15 coloring books and 3 boxes of crayons. A coloring book costs half as much as a box of crayons. How much does one coloring book cost?", "id": "Mathcounts2023ChapterTarget_1", "lang": "en", "answer": "$1.50$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "A fortnight is a period of time measuring two weeks. If a microfortnight is defined to be one millionth of a fortnight, how many seconds long is a microfortnight? Express your answer as a decimal to the nearest hundredth.", "id": "Mathcounts2023ChapterTarget_2", "lang": "en", "answer": "$1.21$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "In the following sequence, each term is the sum of the two previous terms: a, b, c, d, 8, 15, 23, 38, 61. What is the value of a + c?", "id": "Mathcounts2023ChapterTarget_3", "lang": "en", "answer": "$-4$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "In trapezoid ABCD, CD and AB are the top and bottom bases respectively. AD and CB are the other two sides. Vertex A is the center of a circle, vertex D lies on the circle, and vertices B and C are right angles. If AB = 14.9 cm, BC = 5.3 cm and DC = 9.6 cm, what is the area of the sector of the circle that is the intersection of the circle and the trapezoid? Express your answer as a decimal to the nearest tenth.", "id": "Mathcounts2023ChapterTarget_4", "lang": "en", "answer": "$22.1cm^{2}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "Four friends, Wally, Eli, Yuko, and Sam, live in different neighborhoods and commute to school every day by bus. Eli’s neighborhood is half as far from school as Wally’s. Yuko travels as far as the total distance traveled by Wally and Eli. Sam travels 3 times the distance that Wally travels. How many blocks does Wally travel to school if the friends together travel 888 blocks?", "id": "Mathcounts2023ChapterTarget_5", "lang": "en", "answer": "$148$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "Suppose that 4 letters are chosen at random without replacement from the phrase “MAZAMORRA MORADA.” What is the probability that the four letters chosen can be arranged to spell AMOR? Express your answer as a common fraction.", "id": "Mathcounts2023ChapterTarget_6", "lang": "en", "answer": "$\\frac{6}{91}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "Two parallel planes a distance of 8 meters apart intersect a sphere. Each of the circles where the planes intersects the sphere has an area of $128\\pi\\;m^{2}$. What is the surface area of the sphere? Express your answer in terms of $\\pi$.", "id": "Mathcounts2023ChapterTarget_7", "lang": "en", "answer": "$576\\pi\\;m^{2}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTarget", "question": "Each workday when Kate arrives at the office, she does one of the following: either Kate says “Good morning” to her coworkers, or Kate says nothing to her coworkers. On Monday, Kate says “Good morning” to her coworkers. If on each subsequent day there is a 75% chance that she will repeat her previous day’s behavior, what is the probability that on Friday Kate says “Good morning” to her coworkers? Express your answer as a common fraction.", "id": "Mathcounts2023ChapterTarget_8", "lang": "en", "answer": "$\\frac{17}{32}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "In dynastic China, a kè was a unit of time measuring 1 of one hour. How many kè are in one (non-leap) year?", "id": "Mathcounts2023ChapterTeam_1", "lang": "en", "answer": "$35040$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "The students at Somerset School are going on a field trip. There will be 270 students and 24 teachers on the trip. Each bus can carry 42 passengers. How many buses will be needed?", "id": "Mathcounts2023ChapterTeam_2", "lang": "en", "answer": "$7$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "An abundant number is a number for which the sum of its positive proper factors is greater than the number itself. For example, because the sum of the positive proper factors of 24 is 36, it follows that 24 is an abundant number. What is the least abundant number greater than 24?", "id": "Mathcounts2023ChapterTeam_3", "lang": "en", "answer": "$30$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "The town of Heterochromia has 1200 residents. Each resident has two eyes, and each eye is green, blue or brown. If 400 residents have at least one green eye, 600 residents have at least one blue eye, and 900 residents have at least one brown eye, how many residents have two eyes of the same color?", "id": "Mathcounts2023ChapterTeam_4", "lang": "en", "answer": "$500$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "The exam scores of twenty-four students are in the stem-and-leaf plot shown, where 7 | 2 represents 72 points. Two students, Erica and Makawee, took the test late, and Makawee earned 6 points more than Erica. It turns out that when Erica’s and Makawee’s scores are included with the other twenty-four, the overall average does not change. What was Makawee’s score on the exam?\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{ll}\\hline\\multicolumn{1}{|l|}{Stem} & \\multicolumn{1}{l|}{Leaf} \\\\ \\hline\\multicolumn{1}{|l|}{5} & \\multicolumn{1}{l|}{8} \\\\ \\hline\\multicolumn{1}{|l|}{6} & \\multicolumn{1}{l|}{7} \\\\ \\hline7 & 0 2 2 4 4 5 6 6 8 8 8 \\\\8 & 0 0 5 5 8 9 \\\\9 & 2 2 3 3 5\\end{tabular}\\end{table}", "id": "Mathcounts2023ChapterTeam_5", "lang": "en", "answer": "$83$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "Veronica runs on a treadmill for 1 minute at 5 mi/h. She increases her speed by x mi/h and runs for 30 seconds, and then she decreases her speed back to 5 mi/h and runs for 1 minute. Veronica continues in this manner, alternating between the slower speed for 1 minute and the faster speed for 30 seconds. Assume that all changes in speed are instantaneous. She stops when she has reached 5 miles, which occurs after exactly 55 minutes. What is the value of x? Express your answer as a decimal to the nearest tenth.", "id": "Mathcounts2023ChapterTeam_6", "lang": "en", "answer": "$1.4$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "How many different non-congruent right triangles with integer side lengths have a perimeter less than 75?", "id": "Mathcounts2023ChapterTeam_7", "lang": "en", "answer": "11"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "ABCD is a quadrilateral with AB = 3, BC = 4, CD = 12 andAD = 13. If $AB \\perp BC$ and $AC \\perp CD$, what is the least possible area of ABCD?", "id": "Mathcounts2023ChapterTeam_8", "lang": "en", "answer": "$24$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "x-sided dice. Given that the sum of the three numbers rolled is 9, what is the probability that all the dice showed different numbers? Express your answer as a common fraction.", "id": "Mathcounts2023ChapterTeam_9", "lang": "en", "answer": "$\\frac{18}{25}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "Mathcounts2023ChapterTeam", "question": "How many two-digit positive integers have exactly three positive one-digit factors?", "id": "Mathcounts2023ChapterTeam_10", "lang": "en", "answer": "16"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "Four people are playing rock-paper-scissors. They each play one of the three options (rock, paper, or scissors) independently at random, with equal probability of each choice. Compute the probability that someone beats everyone else.", "id": "HMMT2023General_1", "lang": "en", "answer": "$\\frac{4}{27}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "A regular n-gon $P_{1}P_{2} \\ldots P_{n}$ satisfies $\\angle P_{1}P_{7}P_{8} = 178^{\\circ}$. Compute n.", "id": "HMMT2023General_2", "lang": "en", "answer": "630"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "Compute the number of positive four-digit multiples of 11 whose sum of digits (in base ten) is divisible by 11.", "id": "HMMT2023General_3", "lang": "en", "answer": "72"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "Suppose that a and b are real numbers such that the line $y=ax+b$ intersects the graph of $y=x^{2}$ at two distinct points A and B. If the coordinates of the midpoint of AB are $(5,101)$, compute $a+b$.", "id": "HMMT2023General_4", "lang": "en", "answer": "61"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "On an $8 \\times 8$ chessboard, 6 black rooks and $k$ white rooks are placed on different cells so that each rook only attacks rooks of the opposite color. Compute the maximum possible value of $k$. (Two rooks attack each other if they are in the same row or column and no rooks are between them.)", "id": "HMMT2023General_5", "lang": "en", "answer": "14"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "Let ABCD be a square of side length 5. A circle passing through A is tangent to segment CD at T and meets AB and AD again at $X \\neq A$ and $Y \\neq A$, espectively. Given that XY = 6, compute AT.", "id": "HMMT2023General_6", "lang": "en", "answer": "$\\sqrt{30}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "Compute all ordered triples (x, y, z) of real numbers satisfying the following system of equations: $\\begin{cases} xy+z=40 \\\\ xz+y=51 \\\\ x+y+z=19 \\end{cases}$", "id": "HMMT2023General_7", "lang": "en", "answer": "(12,3,4), (6,5.4,7.6)"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "Mark writes the expression $\\sqrt{d}$ for each positive divisor d of 8! on the board. Seeing that these expressions might not be worth points on HMMT, Rishabh simplifies each expression to the form $a\\sqrt{b}$, where a and b are integers such that b is not divisible by the square of a prime number. (For example, $\\sqrt{20}, \\sqrt{16} and $\\sqrt{6}$ simplify to $2\\sqrt{5}, 4\\sqrt{1}$, and $1\\sqrt{6}$, respectively.) Compute the sum of $a+b$ across all expressions that Rishabh writes.", "id": "HMMT2023General_8", "lang": "en", "answer": "3480"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "An entry in a grid is called a saddle point if it is the largest number in its row and the smallest number in its column. Suppose that each cell in a $3 \\times 3$ grid is filled with a real number, each chosen independently and uniformly at random from the interval [0,1]. Compute the probability that this grid has at least one saddle point.answers: $\\frac{3}{10}$", "id": "HMMT2023General_9", "lang": "en", "answer": ""}
|
||||
{"source": "part1.en_zh.json", "sourcename": "HMMT2023General", "question": "Let ABCD be a convex trapezoid such that $\\angle ABC = \\angle BCD = 90^{\\circ}$, AB = 3, BC = 6, and CD = 12. Among all points X inside the trapezoid satisfying $\\angle XBC = \\angle XDA$, compute the minimum possible value of CX.", "id": "HMMT2023General_10", "lang": "en", "answer": "$\\sqrt{113}-\\sqrt{65}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "The graph of the polynomial function f, where y=f(x), has x-intercepts of (-6,0) and (6,0). Which of the following must be true: (A) f(-6)=0 (B) f(6)=-6 (C) f(-6)=6 (D) f(0)=-6", "id": "052023SATS3_3", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "What is the solution (x,y) to the system of equations $\\begin{cases} y=4x+6 \\\\ -5x-y=21 \\end{cases}$?", "id": "052023SATS3_4", "lang": "en", "answer": "$(-3,-6)$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "What are all the possible solutions to $|x-10|=0$?", "id": "052023SATS3_5", "lang": "en", "answer": "$10$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "The given equation $q=s(r-1)^{2}$ relates the positive numbers q,r, and s. Express r in terms of q and s, when r > 1.", "id": "052023SATS3_6", "lang": "en", "answer": "$r=1+\\sqrt{\\frac{q}{s}}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "In the relationship between variablex x and y, each increas of 1 in the value of x decreases the value of y by 2. When x = 0, y = 5, express this relation by writing y as a function of x.", "id": "052023SATS3_7", "lang": "en", "answer": "$y=-2x+5$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "An architect is asked to construct an opening in a wall in the shape of a parabola. The parabola is a face-down parabola on the range $0 \\leq x \\leq 8$ with x-intercepts at (0,0) and (8,0). The vertex is at (4,8). The formula $y=\\frac{-x(x-8)}{k}$, where k is a constant, can be used to determine the height y, in feet, of the opening at a horizontal distance of x feet from the origin. Based on this information, what is the value of k?", "id": "052023SATS3_8", "lang": "en", "answer": "$2$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "An isosceles right triangle has a hypotenuse of length 4 inches. What is the perimeter, in inches, of this triangle? Express your answer in simplest radical form.", "id": "052023SATS3_9", "lang": "en", "answer": "$4+4\\sqrt{2}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "How many solutions does the equation $4(x-2)=-2(x+4)$ have?", "id": "052023SATS3_10", "lang": "en", "answer": "$1$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "Circle O is centered at C(-5,-4). Line k is tangent to the circle at point A(-7,-5). What is the slope of line k?", "id": "052023SATS3_11", "lang": "en", "answer": "$-2$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "The function R $R(t)=1830-790(2.71)^{-.18t}$ gives the predicted average rating, expressed as a number of points, in the German chess federation database for a player based on the number of years, t, the player has participated in professional chess tournments. What value of t gives the predicted average rating of a player who has just entered their first professional chess tournament?", "id": "052023SATS3_12", "lang": "en", "answer": "$0$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "Alice took 60 minutes to complete a task on her first trial. The time it took Alice to complete the task decreased by 10% of the previous time for each additional trial. Approximately how many minutes will it take Alice to complete the task on her fifth trial: (A) 50 (B) 42 (C) 39 (D) 35", "id": "052023SATS3_13", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "In which of the following tables are all the values of x and their corresponding values of y solutions to the system of inequalities $\\begin{cases} y <\\frac25 x + 3 \\\\ y > \\frac12 x - 6 \\end{cases}$:(A) \\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlinex & y \\\\ \\hline-2 & -8 \\\\ \\hline0 & -4 \\\\ \\hline4 & 4 \\\\ \\hline\\end{tabular}\\end{table}(B)\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlinex & y \\\\ \\hline-2 & -8 \\\\ \\hline0 & 4 \\\\ \\hline4 & 4 \\\\ \\hline\\end{tabular}\\end{table}(C) \\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlinex & y \\\\ \\hline-2 & 3 \\\\ \\hline0 & 2 \\\\ \\hline4 & -3 \\\\ \\hline\\end{tabular}\\end{table}(D) \\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlinex & y \\\\ \\hline-2 & 2 \\\\ \\hline0 & -3 \\\\ \\hline4 & 3 \\\\ \\hline\\end{tabular}\\end{table}", "id": "052023SATS3_14", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "Simplify $(\\sqrt{32})(\\sqrt[5]{64})$ to the simplest radical form. Express any number inside a radical by its prime factorization.", "id": "052023SATS3_15", "lang": "en", "answer": "$8(\\sqrt[10]{2^{7}})$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "The expression $\\frac{32x^{6}}{4x^{3}}$ is equivalent to $cx^{d}$, where c and d are constants and x > 0. What is the value of c+d?", "id": "052023SATS3_17", "lang": "en", "answer": "$11$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "What value of h is the solution to $2.1(h+3)=3h+2.1$?", "id": "052023SATS3_18", "lang": "en", "answer": "$\\frac{14}{3}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "A cylinder and a sphere both have the same radius r, where r > 0. The cylinder has a height of 16. The volume of the sphere is half the volume of the cylinder. What is the value of r?", "id": "052023SATS3_19", "lang": "en", "answer": "$6$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS3", "question": "In equation $x^{2}+bx+c=0$, b and c are constants. If $-b + \\sqrt{b^{2}-4c}=18$ and $-b - \\sqrt{b^{2}-4c}=10$, what are the possible values of x?", "id": "052023SATS3_20", "lang": "en", "answer": "$5$ or $9$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "ject has a mass of 3,300 milligrams. What is the mass of the object in grams?", "id": "052023SATS4_2", "lang": "en", "answer": "$3.3$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "On average, one square inch of human skin contains 650 sweat glands. A certain area of skin contains 1,170 sweat glands. Based on this information, what is the size of this rea, in square inches?", "id": "052023SATS4_4", "lang": "en", "answer": "$1.8$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "The table give the heights, in feet, of 5 peaks in the Rocky Mountains and 5 peaks in the Appalachian Mountains. What is the height, in meters, of Blanca Peak? (round to the nearest one)\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|l|l|l|}\\hlineRocky Mountain Peak & Height (in feet) & & Appalachian Mountain Peak & Height (in feet) \\\\ \\hlineMount Elbert & 14,439 & & Mount Mitchell & 6,684 \\\\ \\hlineMount Massive & 14,429 & & Mount Craig & 6,647 \\\\ \\hlineMount Harvard & 14,419 & & Clingman's Dome & 6,643 \\\\ \\hlineBlanca Peak & 14,350 & & Mount Guyot & 6,621 \\\\ \\hlineLa Plata Peak & 14,343 & & Balsam Cone & 6,611 \\\\ \\hline\\end{tabular}\\end{table}", "id": "052023SATS4_7", "lang": "en", "answer": "$4375$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "(Continue from the previous question.) For the given Appalachian Mountain peaks, the height of the highest peak is approximately what percent greater than the height of the lowest peak? (round to the nearest tenth)", "id": "052023SATS4_8", "lang": "en", "answer": "$1.1\\%$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "Given two data sets $\\begin{cases} A: 2,4,6,6,8,12 \\\\ B:2,4,6,6,8,12,26 \\end{cases}$, which statement best compares the medians of the two sets?(A) The median of data set A is greater than the median of data set B(B) The median of data set A is less than the median of data set B(C) The medians of data sets A and B are equal(D) There is not enough information to compare the med", "id": "052023SATS4_9", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "The mass of a solution of isopropanol and water is 100 grams. The equation $0.79x + 1.0y = 100$ represents this situation, where x is the volume of isopropanol, in cubic centimenters, and y is the volume of water, in cubic centimenters. If the volume of isopropanol is 70 cubic centimenters, what is the approximate volue of water, in cubic centimenters? (round to the nearest one's place)", "id": "052023SATS4_10", "lang": "en", "answer": "$45$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "There are 435 voting members of the US House of Representatives. If b voting members are in favor of a certain bill, which expression represents the percentage of the voting members in favor of the bill?(A) $100\\left(\\frac{b}{435}\\right)$(B) $100\\left(\\frac{435}{b}\\right)$(C) $435\\left(\\frac{b}{100}\\right)$(D) $435(100b)$Answer: A", "id": "052023SATS4_11", "lang": "en", "answer": ""}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "Which of the following equations has the same solutin as $10(x+120)=120$: (A) $x+120=12$ (B) $x+120=130$ (C) $x+12=12$ (D) $x+12=120$", "id": "052023SATS4_12", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "The given function $C(x)=4.3x+19$ models the annual soybean use in one country, in millions of metric tons, between 1995 and 2014, where x is the number of years after 1995. According to the model, what is the best interpretation of 4.3 in this context: (A) Each year between 1995 and 2014, this country used 4.3 million metric tons of soybeans (B) Each year between 1995 and 2014, this country's annual use of soybeans increased by 4.3 million metric tons (C) this country used 4.3 million metric tons of soybeans in 1995 (D) This country used a total of 4.3 million metric tons of soybeans between 1995 and 2014.", "id": "052023SATS4_15", "lang": "en", "answer": "B"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "The function $C(x)=50,000+0.75x$ models the total cost (sum of fixed cost and variable cost), in dollars, of growing and harvesting x bales of hay on a certain farm. The function $R(x)=4.75x$ models the revenue, in dollars, earned from selling x bales of hay. According to the function R, how many bales of hay would hae to be sold to earn a revenue of \\$1,425?", "id": "052023SATS4_16", "lang": "en", "answer": "$300$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "(Continue from the previous question) WHich of the following inequalities models the number of bales of hay that must be sold to earn a profit of \\$10,000 or more (profit = revenue - cost): (A) $10,000 \\leq 4x - 50,000$ (B) $10,000 \\geq 4x - 50,000$ (C) $10,000 \\leq 4x + 50,000$ (D) $10,000 \\geq 4x + 50,000$", "id": "052023SATS4_17", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "Which expression is equivalent to $(x^{2}+4)^{2}+(x-2)(x+2)$: (A) $x^{4}+x^{2}+20$ (B) $x^{4} + 5x^{2} + 16$ (C) $x^{4}+9x^{2}$ (D) $x^{4} + 9x^{2}+12$", "id": "052023SATS4_18", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "How many solutions does the system of equations $\\begin{cases} y=4x+1 \\\\ y=4x+3 \\end{cases}$ have?", "id": "052023SATS4_19", "lang": "en", "answer": "0"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "Which inequality represents all values of x for which the graph of y=h(x) in the xy-plane is above the x-axis where $h(x)=3x+3$: (A) x < 3 (B) x < -1 (C) x > -1 (D) x > 3", "id": "052023SATS4_20", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "Which quadratic equation has no real solutions: (A) $3x^{2} - 3 = 0$ (B) $3x^{2}+3x = 0$ (C) $3x^{2}+3x+3=0$ (D) $3x^{2}-6x+3=0$", "id": "052023SATS4_22", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "In a right triangle, the leg opposite angle x is 42 units long and the hypotenuse is 59 units long. Which of the following is NOT true: (A) $\\sin x = \\frac{42}{59}$ (B) $\\sin(90^{\\circ}-x)=\\frac{42}{59}$ (C) $\\cos(90^{\\circ}-x)=\\frac{42}{59}$ (D) $\\sin(90^{\\circ}-x)-cos=0$", "id": "052023SATS4_23", "lang": "en", "answer": "B"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "In 1976, there were approximately 1,000 gray wolves in northern Minnesota. The number of gray wolves in northern Minnesota in 2008 was 190% greater than in 1976. Approximately how many gray wolves were in northern Minnesota in 2008?", "id": "052023SATS4_24", "lang": "en", "answer": "$2,900$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "When the quadratic function f is graphed in the xy-plane, where y=f(x), its vertex is (-2,5). One of the x-intercepts of this graph is $\\left(-\\frac73,0\\right)$. What is the other x-intercept of the graph?", "id": "052023SATS4_25", "lang": "en", "answer": "$\\left(-\\frac53,0\\right)$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "For an exponential function g, the value of g(x) decreases by 20\\% for each 1-unit increase in the value of x. If g(2)=16, which equation could define g: (A) $g(x)=16(0.8)^{x-2}$ (B) $g(x)=16(0.8)^{x+2}$ (C) $g(x)=16(0.2)^{x-2}$ (D) $g(x)=16(0.2)^{x+2}$", "id": "052023SATS4_27", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "Micha and Rana each selected a random sample of students at their school and asked how many soft drink swervings each student had consumed the previous week. Micha enstimated that the mean number of soft drink servings was 7.1, with an associated margin of error of 1.2. Rana estimated that the mean number of soft drink servings was 8.3, with an associated margin of error of 0.8. Assuming the margins of error were calculated in the same way, which of the following best explains why Rana obtained a smaller margin of error than Micha:(A) Rana's sample contained more students than Micha's sample contained.(B) Rana's sample contained more students who drank soft drinks than Micha's sample contained.(C) Rana's sample contained more students who drank exactly seven soft drink servings than Micha's sample contained.(D) Rana's sample contained more students who drank exactly eight soft drink servings than Micha's sample contained.", "id": "052023SATS4_28", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "A circle in the xy-plane has its center at (-3,4) and point (-2,1) lies on the circle. Which equation represents this circle: (A) $(x-3)^{2} + (y+4)^{2}=\\sqrt{10}$ (B) $(x+3)^{2} + (y-4)^{2}=\\sqrt{10}$ (C) $(x-3)^{2} + (y+4)^{2}=10$ (D) $(x+3)^{2} + (y-4)^{2}=10$", "id": "052023SATS4_29", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "For the quadratic function h, the table gives three values of x and their corresponding values of h(x). At what value of x does h reach its minimum?\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlinex & h(x) \\\\ \\hline2 & 0 \\\\ \\hline4 & 0 \\\\ \\hline6 & 8 \\\\ \\hline\\end{tabular}\\end{table}", "id": "052023SATS4_30", "lang": "en", "answer": "$3$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "A company decides to sponsor an employee bowling team. The cost to form the team is \\$180 per team member plus a onetime \\$25 team-registration fee. What is the maximum number of team members who can join the team, if the company can spend \\$925 for the bowling team?", "id": "052023SATS4_32", "lang": "en", "answer": "$5$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "THe solution to the system $\\begin{cases} 3x+4y=18 \\\\ 2x-4y = 17 \\end{cases}$ is (x,y). What is the value of x?", "id": "052023SATS4_33", "lang": "en", "answer": "$7$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "If $5\\sqrt{3x}-2=13$, what is the value of 3x?", "id": "052023SATS4_35", "lang": "en", "answer": "$9$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "$\\frac{x}{y}=8$ and $\\frac{2x}{ty}=160$, what is the value of t?", "id": "052023SATS4_36", "lang": "en", "answer": "$\\frac{1}{10}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "The table summarizes the distribution of age and assigned group for participants in a study. One of these participants will be selected at random. What is the probability of selecting a participant from group A, given that the participant is at least 10 years of age? (Express your answer as a decimal or fraction, not a percent.)\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|l|l|l|}\\hline& 0-9 years & 10-19 years & 20+ years & Total \\\\ \\hlineGroup A & 20 & 18 & 22 & 60 \\\\ \\hlineGroup B & 14 & 14 & 32 & 60 \\\\ \\hlineGroup C & 26 & 28 & 6 & 60 \\\\ \\hlineTotal & 60 & 60 & 60 & 180 \\\\ \\hline\\end{tabular}\\end{table}", "id": "052023SATS4_37", "lang": "en", "answer": "$\\frac13$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "052023SATS4", "question": "A small circle lies inside a big circle and is tangent to the big circle at point A. AC is the diameter of the big circle. AB is the diameter of the small circle. AB and AC are co-linear. The ratio of AB to BC is 4 to 1. If the area of the small circle is 72. What is the area of the part of the big circle that lies outside the small circle?", "id": "052023SATS4_38", "lang": "en", "answer": "$40.5$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "What ist he positive solution to $|x+3|=6$?", "id": "102023SATS3_2", "lang": "en", "answer": "$3$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "What is the solution (x, y) to the given system of equations $\\begin{cases} y=7-4x \\\\ 15x-4y=3 \\end{cases}$?", "id": "102023SATS3_4", "lang": "en", "answer": "$(1,3)$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "In which quadrant(s) does the solution set to $\\begin{cases} y \\leq 2x+3 \\\\ y \\geq 0.5x-6$ lie?", "id": "102023SATS3_5", "lang": "en", "answer": "$I,II,III,IV$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "The given function f $f(t)=0.17t+2.54$ models the annual worldwide production of avocados, in millions of metric tons, t years after 2000. According to the function, by how many millions of metric tons did the annual worldwide production of avocados increase from 2010 to 2011?", "id": "102023SATS3_6", "lang": "en", "answer": "$0.17$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "Which equation has no solution: (A) $4(x+1)=x+4$ (B) $4(x+1)=x+1$ (C) $4(x+1)=4x+4$ (D) $4(x+1)=4x$", "id": "102023SATS3_7", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "The function f is defined by $f(x)= (x − 1)(x + 1)( + 2)$. Which of the following is NOT an x-intercept of the graph y=f(x) in the xy-plane: (A) (-2,0) (B) (-1,0) (C) (1,0) (D) (2,0)?", "id": "102023SATS3_8", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "Which expression is equivalent to $16^{\\left(\\frac12 x\\right)}$: (A) $4^{x}$ (B) $8^{x}$ (C) $8x$ (D) $16\\sqrt{x}$", "id": "102023SATS3_9", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "Trapezoid ABCD is similar to trapezoid PQRS. The length of each side of trapezoid PQRS is 3 times the length of its corresponding side of trapezoid ABCD. The area of trapezoid ABCD is 6 square centimeters. What is the area, in square centimeters, of trapezoid PQRS?", "id": "102023SATS3_10", "lang": "en", "answer": "$54$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "The function f is defined by $f(x)=(x+1)^{2}-9$. In the xy-plane, the graph of which of the following equations has no x-intercepts: (A) $y=f(x-2)$ (B) $y=f(x+2)$ (C) $y=f(x)-11$ (D) $y=f(x)+11$", "id": "102023SATS3_11", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "In the xy-plane, the graph of $x^{2}-10x+y^{2}-6y=30$ is a circle. What is the area of this circle? Express your answer in terms of $\\pi$.", "id": "102023SATS3_12", "lang": "en", "answer": "$64\\pi$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "The expression $\\frac{x^{2}+x}{x+5}$ can be rewritten as $A+\\frac{20}{x+5}$, where A is a polynomial. Find the expression of A.", "id": "102023SATS3_13", "lang": "en", "answer": "$x-4$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "An electric circuit contains three capacitors in a particular arrangement. The equation $\\frac{1}{c}=\\frac{1}{d} + \\frac{1}{e} + \\frac{1}{f}$ relates the equivalent capacitance C of the arrangement to d, e, and f, the capacitances of the individual capacitors. Express C in terms of d, e, and f.", "id": "102023SATS3_14", "lang": "en", "answer": "$C=\\frac{def}{de+df+ef}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "The cost of renting a bicycle is \\$8 for the first hour plus \\$4 for each additional hour. Write a function that gives the cost C(h), in dollars, of renting the bicycle for h hours, where h is a positive integer.", "id": "102023SATS3_15", "lang": "en", "answer": "$C(h)=4h+4$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "What value of r is the solution to $5r=3(r+1)$?", "id": "102023SATS3_16", "lang": "en", "answer": "$1.5$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "What is the positive solution to the equation $(3x)^{2}-4(3x)-12=0$?", "id": "102023SATS3_17", "lang": "en", "answer": "$2$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "In the xy-plane, line k passes through points (-4,0) and (0,-2). Line j is perpendicular to line k. What is the slope of line j?", "id": "102023SATS3_18", "lang": "en", "answer": "$2.5$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "In the xy-plane, the graph of $y=(-14)\\left(\\frac12\\right)^{x}+k$, where k is a constant, has a y-intercept of (0,2). What is the value of k?", "id": "102023SATS3_19", "lang": "en", "answer": "$16$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS3", "question": "The perimeter of a square inscribed in a circle is 30 inches. The radius of the circle is $x\\sqrt{2}$ inches. What is the value of x?", "id": "102023SATS3_20", "lang": "en", "answer": "$3.75$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "The function $k(x)=260x$ gives the total amount k(x), in dollars, Kayla earned after working x weeks at a bookstore. What is the total amount Kayla earned, in dollars, after working 10 weeks at the bookstore?", "id": "102023SATS4_1", "lang": "en", "answer": "$2,600$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "The function f is defined by $f(x)=x+4$. What is the value of $f(x)$ when x = 5?", "id": "102023SATS4_2", "lang": "en", "answer": "$9$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "What is the positive solution to the given equation $\\sqrt{m^{2}}=\\sqrt{64r^{2}}$?", "id": "102023SATS4_5", "lang": "en", "answer": "$64$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "In a group of 8 children, 2 have freckles and 6 do not have freckles. If one of the 8 children is selected at random, what is the probability of selecting a child who has freckles?", "id": "102023SATS4_8", "lang": "en", "answer": "$0.25$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "The ratio of Jupiter’s diameter to Saturn’s diameter is approximately 1.2 to 1. The diameter of Jupiter is 143,000 kilometers. What is the diameter, to the nearest ten thousand kilometers, of Saturn?", "id": "102023SATS4_9", "lang": "en", "answer": "$120,000$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "For the linear function f, the table shows three values of x and their corresponding values of f(x). Find f(x).\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlinex & f(x) \\\\ \\hline0 & 13 \\\\ \\hline10 & 9 \\\\ \\hline25 & 3 \\\\ \\hline\\end{tabular}\\end{table}", "id": "102023SATS4_11", "lang": "en", "answer": "$f(x)=-\\frac25 x + 13$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "Which of the following is equivalent to $2-x^{2}-x^{2}-2$: (A) 0 (B) 2 (C) $-x^{2}$ (D) $-2x^{2}$", "id": "102023SATS4_12", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "A data set consists of 100 values. The least value is 30, and the greatest two values are 80 and 125. If the value 125 is removed from the original data set to create a new data set with 99 values, which statement must be true:(A) The means of the two data sets will be the same.(B) The mean of the new data set will be less than the mean of the original data set.(C) The medians of the two data sets will be the same.(D) The median of the new data set will be less than the median of the original data set.", "id": "102023SATS4_13", "lang": "en", "answer": "B"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "A forestry department measures tree trunk diameter, in inches, at a constant height from the ground for each tree growing in a certain area. The data for 95 of these trees are summarized in the table.\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hline\\begin{tabular}[c]{@{}l@{}}Trunk diameter\\\\ (inch)\\end{tabular} & Frequency \\\\ \\hline(0,6) & 5 \\\\ \\hline{[}6,12) & 10 \\\\ \\hline{[}12,18) & 30 \\\\ \\hline{[}18,24) & 25 \\\\ \\hline{[}24,30) & 15 \\\\ \\hline{[}30,36) & 5 \\\\ \\hline{[}36,42) & 5 \\\\ \\hline\\end{tabular}\\end{table}Approximately what percentage of these trees have a trunk diameter less than 6 inches? Express your answer with the percent sign and round to the nearest tenth.", "id": "102023SATS4_14", "lang": "en", "answer": "$5.3\\%$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "(Continue from the previous question.) Which of the following is a possible value for the median trunk diameter, in inches, of these trees: (A) 10 (B) 15 (C) 20 (D) 25", "id": "102023SATS4_15", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "The three candidates who ran for a city council position in Memphis in 2015 received a total of 14,705 votes. The ratio of votes for the first candidate to votes for the second candidate was approximately 4 to 1. The ratio of votes for the first candidate to votes for the third candidate was approximately 100 to 13. Based on this information, which of the following is closest to the number of votes received by the first candidate: (A) 1,911 (B) 3,676 (C) 10,656 (D) 14,601", "id": "102023SATS4_16", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "Kai went on a bike ride. During the first mile of his bike ride, he rode at a constant speed, s. After the first mile of his ride, Kai increased his speed by 100\\%. Express the speed Kai rode his bike after the first mile, in terms of s.", "id": "102023SATS4_17", "lang": "en", "answer": "$2s$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "A state representative wants to increase the amount of driving practice time required before a student can earn a driver’s license. The representative surveyed a random sample of 50 from the 250 students taking the driver’s education class at Jefferson High School. The survey reported that 62\\% of students taking the class agree that driving practice time should be increased, with an associated margin of error of 5\\%. Based on the results of the survey, how many of the students surveyed agree with the proposed change?", "id": "102023SATS4_19", "lang": "en", "answer": "$31$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "Which of the following populations can the results of the survey at Jefferson High School be extended to: (A) Students at any US high school (B) Students taking a driver’s education class at any US high school (C) Students at Jefferson High School (D) Students taking a driver’s education class at Jefferson High Sc", "id": "102023SATS4_20", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "Which quadratic equation has exactly one distinct real solution: (A) $x^{2} + 4x + 16 = 0$ (B) $x^{2} -8x - 16 = 0$ (C) $x^{2} -6x - 16 = 0$ (D) $x^{2} -8x + 16 = 0$", "id": "102023SATS4_21", "lang": "en", "answer": "D"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "Rectangle X has a length of 12 centimeters (cm) and a width of 2.5 cm. Right triangle Y has a base of 10 cm. The area of rectangle X is three times the area of right triangle Y. What is the height, in cm, of right triangle Y?", "id": "102023SATS4_23", "lang": "en", "answer": "$2$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "According to a 2008 study, there were five known subspecies of tigers, including the Amur and Bengal, living in the wild. Scientists estimated that there were a total of 4,000 tigers in the wild. Of these, 450 were Amur tigers and x were Bengal tigers. Which inequality represents all possible numbers of tigers in the wild in 2008 that belonged to the Bengal tiger subspecies: (A) $5x \\geq 4000$ (B) $\\frac{x}{5} \\geq 4000$ (C) $1\\leq x \\leq 3547$ (D) $3547 < x \\leq 4000$", "id": "102023SATS4_24", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "Triangle DEF have a side DE of 3 units long and side EF of 5 units long. Which of the following additional pieces of information provides enough information to prove whether triangle DEF is a right triangle? (I) The measure of angle D (II) The length of segment DF:(A) I only(B) II only(C) Either I or II(D) Neither I nor II", "id": "102023SATS4_25", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "According to a model, if 100 people see a sequence of three letters, 87 of them will recall this sequence immediately after seeing it. The model predicts that this number will decrease by 14\\% of the number the previous second for each second that passes. Which function represents this model, where f(t) is the predicted number of people who will recall the sequence after t seconds have passed: (A) $f(t)=14(0.87)^{t}$ (B) $f(t)=87(0.14)^{t}$ (C) $f(t)=87(0.86)^{t}$ (D) $f(t)=87(1.14)^{t}$", "id": "102023SATS4_26", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "One of the two linear equations in a system is -6x+7y = -6. The system has no solution. Which equation could be the second equation in this system: (A) 6x-7y = 0 (B) $-\\frac{21x}{4}+\\frac{49y}{8}=-\\frac{21}{4}$ (C) $-\\frac{21x}{4}-14y=0$ (D) 6x - 7y = 6", "id": "102023SATS4_27", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "A cottonwood tree had a trunk diameter of 2.00 inches at the time it was planted. The tree had a trunk diameter of 22.24 inches x years after it was planted. The equation 0.64x + 2 = 22.24 represents this situation. Which statement is the best interpretation of 0.64x in this context:(A) The total increase of the tree’s trunk diameter x years after it was planted(B) The tree’s trunk diameter x years after it was planted(C) The maximum trunk diameter of the tree over its lifetime(D) The total increase of the tree’s trunk diameter each year after it was planted", "id": "102023SATS4_28", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "Given $y=-x^{2}+40x-256$. Which of the following equivalent forms of the equation shows the maximum value of y as a constant or coefficient: (A) $y=-(x-8)(x-32)$ (B) $y=-(x-20)^{2}+144$ (C) $y=-x(x-40)-256$ (D) $y=-x^{2}+8(5x-32)$", "id": "102023SATS4_29", "lang": "en", "answer": "B"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "If $\\frac{x-3}{7}=\\frac{x-3}{9}$, the value of x-3 is between which of the following pairs of values: (A) -7 and -9 (B) -1 and 1 (C) 2.5 and 3.5 (D) 6.75 and 9.25", "id": "102023SATS4_30", "lang": "en", "answer": "B"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "An object has a mass of 200 grams and a volume of 10 cubic centimeters. What is the density, in grams per cubic centimeter, of the object?", "id": "102023SATS4_31", "lang": "en", "answer": "$20$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "An artist decorates square plates using the same pattern of blue and green tiles, where the ratio of blue to green tiles is 3 to 2. For a certain plate, the artist uses 120 blue tiles and 20n green tiles. What is the value of n?", "id": "102023SATS4_32", "lang": "en", "answer": "$4$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "The solution to $\\begin{cases} y=3x+9 \\\\ y=-3x+3 \\end{cases}$ is (x,y). What is the value of y?", "id": "102023SATS4_33", "lang": "en", "answer": "$6$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "A circle has been divided into three nonoverlapping regions: I, II, and III. The area of region I is $4\\pi$ square centimeters ($cm^{2}$, the area of region II is $12\\pi cm^{2}$, and the area of region III is $16\\pi cm^{2}$. If a point in the circle is selected at random, what is the probability of selecting a point that does not lie in region II? (Express your answer as a decimal or fraction, not as a percent.)", "id": "102023SATS4_34", "lang": "en", "answer": "$\\frac58$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "In 1980, the world population was 4.44 billion people. A model created at the time predicted that this population would grow at a rate of 0.076 billion people each year after 1980. In 2015, the world population was 7.35 billion people. What is the positive difference, in billions of people, between the actual world population in 2015 and the model’s predicted world population in 2015?", "id": "102023SATS4_35", "lang": "en", "answer": "$\\frac14$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "The given function $M(t)=10\\left(\\frac12\\right)^{\\frac{t}{5730}}$ models the mass of the radioactive isotope carbon-14 in a sample, in picograms, t years after the initial measurement. How much time, in years, does it take for the mass of carbon-14 in the sample to decrease to 5 picograms?", "id": "102023SATS4_36", "lang": "en", "answer": "$5730$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "102023SATS4", "question": "A circle in the xy-plane has its center at (4, −5) and has a radius of 2. An equation of the circle is $x^{2}-8x+y^{2}+10y+c=0$, where c is a constant. What is the value of c?", "id": "102023SATS4_38", "lang": "en", "answer": "$37$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "How many yards are equivalent to 612 inches?", "id": "2023PSAT89_1.1", "lang": "en", "answer": "$17$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "The table gives the results of a survey of 200 people who were asked how often they see a movie in a theater. How many people responded either \"never\" or \"almost never\"?\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlineResponse & Frequency \\\\ \\hlineOnce a week or more & 3 \\\\ \\hlineTwo or three times a month & 16 \\\\ \\hlineAbout once a month & 26 \\\\ \\hlineA few times a year & 73 \\\\ \\hlineAlmost never & 53 \\\\ \\hlineNever & 29 \\\\ \\hlineTotal & 200 \\\\ \\hline\\end{tabular}\\end{table}", "id": "2023PSAT89_1.2", "lang": "en", "answer": "$82$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "Connor has c dollars and Maria has m dollars. Connor has 4 times as many dollars as Maria, and together they have a total of \\$25.00. Which system of equations represents this situation: (A) $\\begin{cases} c=4m \\\\ c+m=25 \\end{cases}$ (B) $\\begin{cases} m=4c \\\\ c+m=25 \\end{cases}$ (C) $\\begin{cases} c=25m \\\\ c+m=4 \\end{cases}$ (D) $\\begin{cases} m=25c \\\\ c+m=4 \\end{cases}$", "id": "2023PSAT89_1.10", "lang": "en", "answer": "A"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "The equation d=16t represents the distance d, in inches, where t represents the number of seconds since an object started moving. WHich of the following is the best interpretation of 16 in this context: (A) The object moved a total of 16 inches. (B) The object moved a total of 16t inches. (C) The object is moving at a rate of 16 inches per second. (D) The object is moving at a rate of $\\frac{1}{16}$ inches per second.", "id": "2023PSAT89_1.11", "lang": "en", "answer": "C"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "A distance 112 furlongs is equivalent to how many feet (1 furlong = 220 yards).", "id": "2023PSAT89_1.13", "lang": "en", "answer": "$73920$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "A store sells two different-sized containers of a certain Greek yogurt. The store's sales of this Greek yogurt totaled 1,277.94 dollars last month. The equation $5.48x+7.30y=1,277.94$ represents this situation, where x is the number of smaller containers sold and y is the number of larger containers sold. According tot he equation, what is the price, in dollars, of each smaller container?", "id": "2023PSAT89_1.19", "lang": "en", "answer": "$5.48$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "The number a is 70\\% less than the positive number b. The number c is 60\\% greater than a. The number c is how many times b?", "id": "2023PSAT89_1.27", "lang": "en", "answer": "$.48$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "In a box of pens, the ratio of black pens to red pens is 8 to 1. There are 40 black pens in the box. How many red pens are in the box?", "id": "2023PSAT89_2.3", "lang": "en", "answer": "$5$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "The perimeter of an isosceles triangle is 83 inches. Each of the two congruent sides of the triangle has a length of 24 inches. What is the length, in inches, of the third side?", "id": "2023PSAT89_2.6", "lang": "en", "answer": "$35$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "Davio bought some potatoes and celery. THe potatoes cost \\$0.69 per pound, and the celery cost \\$0.99 per pound. If Davio spent \\$5.34 in total and bought twice as many pounds of celery as pounds of potatoes, how many pounds of celery did Davio buy?", "id": "2023PSAT89_2.17", "lang": "en", "answer": "$4$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "For the values j and k, the ratio of j to k is 11 to 12. If j is multiplied by 17, what is k multiplied by in order to maintain the same ratio?", "id": "2023PSAT89_2.21", "lang": "en", "answer": "$17$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "2023PSAT89", "question": "A right circular cone has a height of 22 centimeters and a base with a dimeter of 6cm. The volume of this cone is $n\\pi\\;cm^{3}$. What is the value of n?", "id": "2023PSAT89_2.27", "lang": "en", "answer": "$66$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "In Amul’s yard, there was no snow on the ground at 10:15 a.m. From 10:15 a.m. until 3:30 p.m., it snowed at an average rate of $\\frac12$ inch per hour. How many inches of snow were on the ground at 3:30 p.m. ?", "id": "042023ACTF11_17", "lang": "en", "answer": "$2\\frac58$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "On each of 8 tests, Gustav scored 2 points higher than Russ. When their average test scores on these 8 tests are compared, how many points higher is Gustav’s average than Russ’s average?", "id": "042023ACTF11_18", "lang": "en", "answer": "$2$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "To determine the number of fish in a pond, a biologist catches a random sample of 200 fish, tags them, and releases them back into the pond. A week later, the biologist catches a random sample of 120 fish from this pond, and 32 of them have tags. Which of the following is the best estimate of the number of fish in the pond?", "id": "042023ACTF11_26", "lang": "en", "answer": "$750$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "A jar of solid-colored marbles contains some yellow marbles, 21 blue marbles, 40 red marbles, and 14 green marbles. The probability of randomly drawing a green marble if $\\frac{2}{15}$. How many yellow marbles are in the jar?", "id": "042023ACTF11_28", "lang": "en", "answer": "$30$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "For all the planets in the solar system, the equation $T=0.4D^{1.5}$ models the relationship between T, the orbital period in Earth days, and D, the average distance in millions of miles from the Sun, given in the table. According to this model, which of the following expressions is equal to Mercury’s orbital period in Earth days: (F) $0.4(6^{2})$ (G) $0.4(6^{3})$ (H) $(0.4\\cdot 6)^{3}$ (J) $0.4(\\sqrt[3]{36})^{2}$ (K) $(\\sqrt[3]{0.4\\cdot 36})^{2}$", "id": "042023ACTF11_32", "lang": "en", "answer": "G"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "Julia will serve grape juice from 12-ounce cans at her graduation party. She already has 3 boxes that each contain 12 such cans. How many more 12-ounce cans of grape juice does Julia need to have exactly enough juice to serve 60 guests 8 ounces each?", "id": "042023ACTF11_37", "lang": "en", "answer": "$4$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "On a typical day at Limestone College, $\\frac35$ of all students ride a bicycle to class. Among the rest of the students, $\\frac13$ ride the bus and $\\frac16$ walk. On a typical day, what fraction of the students ride the bus to class?", "id": "042023ACTF11_38", "lang": "en", "answer": "$\\frac{2}{15}$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "Suzi ran 3 laps on a circular path with a radius of 49 meters. What is the distance, in meters, that Suzi ran? Round your answer to the nearest ten.", "id": "042023ACTF11_40", "lang": "en", "answer": "$920$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "One thousand people registered to run a 5-mile road race. Each of the 1,000 people who registered paid either the early, regular, or late registration fee given in the table below. Ninety percent of all the registered people started the race, and 95% of the people who started the race also finished the race. There were 300 people who paid the early registration fee, 400people who paid the regular fee, and 300 people who paid the late fee. What is the total revenue from entry fees, in dollars, written in scientific notation?\\begin{table}[]\\caption{Table 1}\\label{tab:my-table}\\begin{tabular}{|l|l|}\\hlineType of registration & Registration fee \\\\ \\hlineEarly & 25.00 dollars \\\\ \\hlineRegular & 35.00 dollars \\\\ \\hlineLate & 45.00 dollars \\\\ \\hline\\end{tabular}\\end{table}", "id": "042023ACTF11_41", "lang": "en", "answer": "$35,000$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "(Continue from the previous question) Kirk ran the 1st x miles of the race at a rate of 7 minutes per mile and the rest of the race at a rate of 8 minutes per mile. One of the following expressions gives the total amount of minutes that it took Kirk to run the race. Which one: (F) $\\frac{x}{7}+\\frac{x-5}{8}$ (G) $\\frac{x}{7}+\\frac{5-x}{8}$ (H) $7x+8x$ (J) $7x + 8(x-5)$ (K) $7x+8(5-x)$", "id": "042023ACTF11_42", "lang": "en", "answer": "K"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "(Continue from the previous question) One registered person will be randomly selected to win a raffle prize. What is the probability that the selected person did NOT finish the race?", "id": "042023ACTF11_43", "lang": "en", "answer": "$.145$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "A solid rectangular prism has dimensions 5 units long, 4 units tall, and 5 units deep. It was built by alternating congruent black cubes and white cubes such that 2 cubes of the same color have at most 1 edge touching. Each cube's dimensions is 1 unit by 1 unit by 1 unit. What is the total number of white cubes that were used to build the prism?", "id": "042023ACTF11_45", "lang": "en", "answer": "$50$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "In the standard (x,y) coordinate plane, only 1 parabola of the form $y=(x−h)^{2} +k$ has x-intercepts of 3 and 15. What is the axis of symmetry of this parabola?", "id": "042023ACTF11_46", "lang": "en", "answer": "$x=9$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "What is the 330th digit after the decimal point in the repeating decimal $4.\\overline{6238}?$ Calculator not allowed.", "id": "042023ACTF11_47", "lang": "en", "answer": "$2$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "Which of the following is equivalent to $\\frac{a}{a+b}+\\frac{b}{a-b}+\\frac{a-b}{a+b}=$: (F) $\\frac{2(a-b)}{a+b}$ (G) $\\frac{2(a^{2}+b^{2})}{a^{2}-b^{2}}$ (H) $\\frac{2(a^{2}-ab+b^{2})}{a^{2}-b^{2}}$ (J) $\\frac{2a}{(a+b)(a-b)}$ (K) $\\frac{2a}{3a+b}$", "id": "042023ACTF11_48", "lang": "en", "answer": "H"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "Counting repetition, how many prime factors appear in any prime factorization of the integer $(63)^{5}$? Calculators are only allowed to perform addition, subtraction, multiplication, and division.", "id": "042023ACTF11_49", "lang": "en", "answer": "$15$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "For whole numbers x and y, the list $2,6,4,1,x,y$ has 4 as its mean, median, and mode. What is the value of xy ?", "id": "042023ACTF11_50", "lang": "en", "answer": "$28$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "Consider the graphs of the functions f(x) = (x + 2)(x − 8) and $g(x) = x^{2} + 5x − 12$ in the standard (x,y) coordinate plane. What is the sum of the y-intercepts of the functions?", "id": "042023ACTF11_51", "lang": "en", "answer": "$-28$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "For all real numbers x and y, the 2 operations $\\bigstar$ and $\\Diamond$ are defined as: $x \\bigstar y = xy + y$ and $x \\Diamond y = x + xy$. For all real numbers a and b, what is $(a \\Diamond b) \\bigstar b$ ?", "id": "042023ACTF11_53", "lang": "en", "answer": "$ab+ab^{2}+b$"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "If $\\left(x^{y}\\right)\\left(x^{k}\\right)=1$, what relationship must exist between y and k: (F) y+k=0 (G) y+k=1 (H) yk=0 (J) yk=1 (K) y=k", "id": "042023ACTF11_54", "lang": "en", "answer": "F"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "In the standard (x,y) coordinate plane, one of the following is an equation of the circle that can be inscribed in the ellipse $4x^{2}+9y^{2}=36$. Which equation is that? (F) $x^{2}+y^{2}=36$ (G) $x^{2}+y^{2}=9$ (H) $x^{2}+y^{2}=34$ (J) $(x-4)^{2}+(y-9)^{2}=36$ (K) $(x-9)^{2}+(y-4)^{2}=36$", "id": "042023ACTF11_56", "lang": "en", "answer": "H"}
|
||||
{"source": "part1.en_zh.json", "sourcename": "042023ACTF11", "question": "The 3 statements given below are all true about certain positive integers a, b, and c.a is an even prime numberb is an odd integer such that 6 < b < 11c is a perfect square such that 10 < c < 30How many ordered triples (a,b,c) satisfy the 3 statements?", "id": "042023ACTF11_58", "lang": "en", "answer": "$4$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "Cities 𝐴 and 𝐵 are 45 miles apart. Alice and Barbara start biking from 𝐴 and 𝐵 at speeds of 18 mph and 12 mph, respectively. How far away from city 𝐴 will they be when they meet?", "id": "2023AMC10A_1", "lang": "en", "answer": "$27$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "The weight of $\\frac13$ of a large pizza together with $3\\frac12$ cups of orange slices is the same as the weight of $\\frac34$ of a large pizza together with $\\frac12$ cup of orange slices. A cup of orange slices weighs $\\frac14$ of a pound. What is the weight, in pounds, of a large pizza?", "id": "2023AMC10A_2", "lang": "en", "answer": "$1\\frac45$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "How many positive perfect squares less than 2023 are divisible by 5?", "id": "2023AMC10A_3", "lang": "en", "answer": "$8$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "A quadrilateral has all integer sides lengths, a perimeter of 26, and one side of length 4. What is the greatest possible length of one side of this quadrilateral?", "id": "2023AMC10A_4", "lang": "en", "answer": "$12$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is 21. What is the value of the cube?", "id": "2023AMC10A_6", "lang": "en", "answer": "$126$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "Janet rolls a standard 6-sided die 4 times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal 3?", "id": "2023AMC10A_7", "lang": "en", "answer": "$\\frac{49}{216}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "Barb the baker has developed a new temperature scale for her bakery called the Breadus scale, which is a linear function of the Fahrenheit scale. Bread rises at 110 degrees Fahrenheit, which is 0 degrees on the Breadus scale. Bread is baked at 350 degrees Fahrenheit, which is 100 degrees on the Breadus scale. Bread is done when its internal temperature is 200 degrees Fahrenheit. What is this in degrees on the Breadus scale?", "id": "2023AMC10A_8", "lang": "en", "answer": "$37.5$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "A digital display shows the current date as an 8-digit integer, consisting of a 4-digit year, followed by a 2-digit month, followed by a 2-digit date within the month. For how many dates in 2023 will each digit appear an even number of times in the digital display for that date?", "id": "2023AMC10A_9", "lang": "en", "answer": "$9$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "If Mareen scores an 11 on her next test, her mean score will go up by 1. If she gets three 11's in a row, her mean score will increase by 2. What is her current mean test score?", "id": "2023AMC10A_10", "lang": "en", "answer": "$7$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "A square with area 3 has a square with area 2 inscribed in it. Each of the four corners of the small square are on each of the four sides of the big square respectively. This creates 4 smaller congruent right triangles. What is the ratio of the smaller leg to the larger leg in the shaded right triangle? Express your answer in the simplest radical form.", "id": "2023AMC10A_11", "lang": "en", "answer": "$2-\\sqrt{3}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "How many three-digit positive integers N satisfy the following properties:\\n\\begin{itemize}\\n\\item The number N is divisible by 7.\\n\\item The number formed by reversing the digits of N is divisible by 5.\\n\\end{itemize}\\nChecking all 3-digit positive integers is not allowed.", "id": "2023AMC10A_12", "lang": "en", "answer": "$14$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "Abdul and Chiang are standing 48 feet apart in a field. Bharat is standing in the same field as far from Abdul as possible so that the angle formed by his lines of sight to Abdul and Chiang measures 60°. What is the square of the distance (in feet) between Abdul and Bharat?", "id": "2023AMC10A_13", "lang": "en", "answer": "$3072$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "A number is chosen at random from among the first 100 positive integers, and a positive integer divisor of that number is then chosen at random. What is the probability that the chosen divisor is divisible by 11? Checking all positive integers less than 100 is not allowed.", "id": "2023AMC10A_14", "lang": "en", "answer": "$\\frac{9}{200}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "In a tennis tournament, each person plays every other person once. In this tournament, there are twice as many right-handed players than left-handed players, but left-handed players won 40\\% more games than right-handed players. How many total games were played?", "id": "2023AMC10A_16", "lang": "en", "answer": "$36$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "Let ABCD be a rectangle with AB = 30 and BC = 28. Points P and q lie on BC and CD respectively so that all sides of $\\triangle ABP$, $\\triangle PCQ$, and $\\triangle QDA$ have integer lengths. What is the perimeter of $\\triangle APQ$?", "id": "2023AMC10A_17", "lang": "en", "answer": "$84$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "A rhombic dodecahedron is a solid with 12 congruent rhombus faces. At every vertex, 3 or 4 edges meet, depending on the vertex. How many vertices have exactly 3 edges meet?", "id": "2023AMC10A_18", "lang": "en", "answer": "$8$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "The line segment formed by A(1, 2) and B(3, 3) is rotated to the line segment formed by A'(3, 1) and B'(4, 3) about the point P(r,s). What is |r-s|?", "id": "2023AMC10A_19", "lang": "en", "answer": "$1$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "Consider the polynomial P(x) such that:\\n\\begin{itemize}\\n\\item 1 is a root of P(x)-1,\\n\\item 2 is a root of P(x-2),\\n\\item 3 is a root of P(3x), and\\n\\item 4 is a root of 4P(x).\\n\\end{itemize}\\nAll the roots of P(x) except one are integers. If the one non-integer root can be written in the form $\\frac{m}{n}$, where m and n are relatively prime, what is m+n?", "id": "2023AMC10A_21", "lang": "en", "answer": "$47$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "Circle $C_{1}$ and $C_{2}$ have radius 1, and the distance between their centers is $\\frac12$. Circle $C_{3}$ is the largest circle internally tangent to both $C_{1}$ and $C_{2}$. Circle $C_{4}$ is internally tangent to both $C_{1}$ and $C_{2}$ and is externally tangent to $C_{3}$. What is the radius of $C_{4}$?", "id": "2023AMC10A_22", "lang": "en", "answer": "$\\frac{3}{28}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "divisors a and b of n are called complementary if ab=n. Given that N has a pair of complementary divisors that differ by 20 and a pair of complementary divisors that differ by 23, find the sum of the digits of N.", "id": "2023AMC10A_23", "lang": "en", "answer": "$15$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10A", "question": "If A and B are vertices of a polyhedron, define the distance d(A,B) to be the minimum number of edges of the polyhedron one must traverse in order to connect A and B. For example, if $\\overline{AB}$ is an edge of the polyhedron, then d(A,B) = 1, but if $\\overline{AC}$ and $\\overline{CB}$ are edges and $\\overline{AB}$ is not an edge, then d(A,B) = 2. Let Q,R, and S be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that d(Q,R) > d(R,S)?", "id": "2023AMC10A_25", "lang": "en", "answer": "$\\frac{7}{22}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely full but runs out of juice when the fourth glass is only $\\frac13$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four glasses will have the same amount of juice?", "id": "2023AMC10B_1", "lang": "en", "answer": "$\\frac16$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by 20\\% on every pair of shoes. Carlos also knew that he had to pay a 7.5\\% sales tax on the discounted price. He had 43 dollars. What is the original (before discount) price of the most expensive shoes he could afford to buy?", "id": "2023AMC10B_2", "lang": "en", "answer": "$50$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "A 3-4-5 right triangle is inscribed in circle A, and a 5-12-13 right triangle is inscribed in circle B. What is the ratio of the area of circle A to the area of circle B?", "id": "2023AMC10B_3", "lang": "en", "answer": "$\\frac{25}{169}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "Jackson’s paintbrush makes a narrow strip with a width of 6.5 millimeters. Jackson has enough paint to make a strip 25 meters long. How many square centimeters of paper could Jackson cover with paint?", "id": "2023AMC10B_4", "lang": "en", "answer": "$1,625$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "If $L_{1}=1$, $L_{2} = 3$, and $L_{n+2}=L_{n+1}$ for $n \\geq 1$. How many terms in the sequence $L_{1}, L_{2}, L_{3}, \\ldots, L_{2023}$ are even? Listing out all the terms of $L$ is not allowed.", "id": "2023AMC10B_6", "lang": "en", "answer": "$674$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "What is the units digit of $2022^{2023} + 2023^{2022}$. Calculating out the numerical value of the given expression is not allowed.", "id": "2023AMC10B_8", "lang": "en", "answer": "$7$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "The numbers 16 and 25 are a pair of consecutive positive perfect squares whose difference is 9. How many pairs of consecutive positive perfect squares have a difference of less than or equal to 2023? Listing out all pairs of consecutive positive perfect squares is not allowed.", "id": "2023AMC10B_9", "lang": "en", "answer": "$1011$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "You are playing a game. A $2 \\times 1$ rectangle covers two adjacent squares (oriented either horizontally or vertically) of a $3 \\times 3$ grid of squares, but you are not told which two squares are covered. Your goal is to find at least one square that is covered by the rectangle. A \"turn\" consists of you guessing a square, after which you are told whether that square is covered by the hidden rectangle. What is the minimum number of turns you need to ensue that at least one of your guessed squares is covered by the rectangle?", "id": "2023AMC10B_10", "lang": "en", "answer": "$4$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "Suzanne went to the bank and withdrew 800. The teller gave her this amount using 20 bills, 50 bills, and 100 bills, with at least one of each denomination. How many different collections of bills could Suzanne have received?", "id": "2023AMC10B_11", "lang": "en", "answer": "$21$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "When the roots of the polynomial $P(x)=(x-1)^{1}(x-2)^{2}(x-3)^{3}\\ldots(x-10)^{10}$ are removed from the real number line, what remains is the union of 11 disjoint open intervals. On how many of those intervals is P(x) positive?", "id": "2023AMC10B_12", "lang": "en", "answer": "$6$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "What is the area of the region in the coordinate plane defined by $||x|-1|+||y|-1| \\leq 1 $?", "id": "2023AMC10B_13", "lang": "en", "answer": "$8$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "How many ordered pairs of integers (m,n) satisfy the equation $m^{2}+mn+n^{2}=m^{2}n^{2}$?", "id": "2023AMC10B_14", "lang": "en", "answer": "$3$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "What is the least positive integer m such that $m\\cdot 2! \\cdot 3! \\cdot 4! \\cdot 5! \\ldots 16!$ is a perfect square? Listing out all possibilities of m is not allowed.", "id": "2023AMC10B_15", "lang": "en", "answer": "$70$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "Define an upno to be a positive integer of 2 or more digits where the digits are strictly increasing moving left to right. Similarly, define a downno to be a positive integer of 2 or more digits where the digits are strictly decreasing moving left to right. For instance, the number 258 is an upno and 8620 is a downno. Let U equal the total number of upnos and let d equal the total number of downnos. What is |U − d|?", "id": "2023AMC10B_16", "lang": "en", "answer": "$511$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "A rectangular box P has distinct edge lengths a, b, and c. The sum of the lengths of all 12 edges of P is 13, the sum of the areas of all 6 faces of P is $\\frac{11}{2}$, and the volume of P is $\\frac12$. What is the length of the longest interior diagonal connecting two vertices of P?", "id": "2023AMC10B_17", "lang": "en", "answer": "$\\frac94$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "Sonya the frog chooses a point uniformly at random lying within the square $[0, 6] \\times [0, 6]$ in the coordinate plane and hops to that point. She then randomly chooses a distance uniformly at random from [0, 1] and a direction uniformly at random from {north, south east, west}. All he choices are independent. She now hops the distance in the chosen direction. What is the probability that she lands outside the square?", "id": "2023AMC10B_19", "lang": "en", "answer": "$\\frac{1}{12}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "Each of 2023 balls is placed in on of 3 bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?", "id": "2023AMC10B_21", "lang": "en", "answer": "$\\frac14$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "How many distinct values of x satisfy $\\lfloor x \\rfloor^{2} - 3x + 2 = 0$, where $\\lfloor x \\rfloor$ denotes the largest integer less than or equal to x?", "id": "2023AMC10B_22", "lang": "en", "answer": "$4$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "An arithmetic sequence has $n \\geq 3$ terms, initial term a and common difference d > 1. Carl wrote down all the terms in this sequence correctly except for one term which was off by 1. The sum of the terms was 222. What was a+d+n?", "id": "2023AMC10B_23", "lang": "en", "answer": "$20$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "What is the perimeter of the boundary of the region consisting of all points which can be expressed as (2u-3w, v+4w) with $0 \\leq u \\leq 1$, $0 \\leq v \\leq 1$, and $0 \\leq w \\leq 1$?", "id": "2023AMC10B_24", "lang": "en", "answer": "$16$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC10B", "question": "A regular pentagon with area $\\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?", "id": "2023AMC10B_25", "lang": "en", "answer": "$\\sqrt{5}-1$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "A and B lie on the graph of $y=\\log_{2} x$. The midpoint of $\\overline{AB}$ is (6,2). What is the positive difference between the x-coordinates of A and B?", "id": "2023AMC12A_6", "lang": "en", "answer": "$4\\sqrt{5}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "umbers 𝑥 and 𝑦 satisfy $y^{3}=x^{2}$ and $(y-x)^{2}=4y^{2}$. What is x+y?", "id": "2023AMC12A_10", "lang": "en", "answer": "$36$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "What is the degree measure of the acute angle formed by lines with slopes 2 and $\\frac13$?", "id": "2023AMC12A_11", "lang": "en", "answer": "$45$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "What is the value of $2^{3}-1^{2}+4^{3}-3^{3}+6^{3}-5^{3}+\\ldots18^{3}-17^{3}$? Calculating out the numerical value of the given expression is not allowed.", "id": "2023AMC12A_12", "lang": "en", "answer": "$3159$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "How many complex numbers z satisfy $z^{5}=\\overline{z}$, where $\\overline{z}$ s the conjugate of the complex number z?", "id": "2023AMC12A_14", "lang": "en", "answer": "$7$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "Consider the set of complex numbers z satisfying $|1+z+z^{2}|=4$. The largest possible value for the imaginary part of z can be written as $\\frac{\\sqrt{m}}{n}$, where m and n are positive integers and m is not divisible by the square of any prime. What is m+n?", "id": "2023AMC12A_16", "lang": "en", "answer": "$21$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "Flora the frog starts at 0 on the number line and makes a sequence of jumps to the right. In any one jump, independent of previous jumps, Flora leaps a positive integer distance m with probability $\\frac{1}{2^{m}}$. What is the probability that Flora will eventually land at 10?", "id": "2023AMC12A_17", "lang": "en", "answer": "$\\frac12$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "Let f be the unique function defined on the positive integers such that $\\sum_{d|n} d\\cdot f\\left(\\frac{n}{d}\\right) = 1$ for all positive integers n, where the sum is taken over all positive divisors of n. What is f(2023)?", "id": "2023AMC12A_22", "lang": "en", "answer": "$96$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "How many ordered pairs of positive real numbers (a,b) satisfy the equation $(1+2a)(2+2b)(2a+b)=32ab$?", "id": "2023AMC12A_23", "lang": "en", "answer": "$1$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "Let K be the number of sequences $A_{1},A_{2},\\ldots,A_{n}$ such that n is a positive integer less than or equal to 10, each $A_{i}$ is a subset of $\\{1,2,3, \\ldots,10\\}$, and $A_{i-1}$ is a subset of $A_{i}$ for each i between 2 and n, inclusive. For example, {},{5,7},{2,5,7},{2,5,7},{2,5,6,7,9} is one such sequence, with n = 5. What is the remainder when K is divided by 10?", "id": "2023AMC12A_24", "lang": "en", "answer": "$5$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12A", "question": "There is a unique sequence of integers $a_{1}, a_{2}, \\ldots, a_{2023}$ such that $\\tan 2023x = \\frac{a_{1}\\tan x + a_{3}\\tan^{3} x + a_{5}\\tan^{5}x + \\ldots + a_{2023}\\tan^{2023}x}{1+a_{2}\\tan^{2}x+a_{4}\\tan^{4}x+\\ldots+a_{2022}\\tan^{2022}x}$ whenever $\\tan 2023x$ is defined. What is $a_{2023}$?", "id": "2023AMC12A_25", "lang": "en", "answer": "$-1$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "For how many integers n does the expression $\\sqrt{\\frac{\\log(n^{2})-(\\log n)^{2}}{\\log n - 3}}$ represent a real number, where log denotes the base 10 logarithm?", "id": "2023AMC12B_7", "lang": "en", "answer": "$902$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "How many nonempty subsets B of $\\{0,1,2,3,\\ldots,12\\}$ have the property that the number of elements in B is equal to the least element of B? For example, B = {4, 6, 8, 11} satisfies the condition.", "id": "2023AMC12B_8", "lang": "en", "answer": "$144$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "In the xy-plane, a circle of radius 4 with center on the positive x axis is tangent to the y-axis at the origin, and a circle of radius 10 with center on the positive y-axis is tangent to the x-axis at the origin. What is the slope of the line passing through the two points at which these circles intersect?", "id": "2023AMC12B_10", "lang": "en", "answer": "$\\frac25$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "What is the maximum area of an isosceles trapezoid that has legs of length 1 and one base twice as long as the other?", "id": "2023AMC12B_11", "lang": "en", "answer": "$\\frac32$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "For complex numbers w=a+bi and z=c+di (where $i=\\sqrt{-1}$), define the binary operator $\\circ$ by $w\\circ z = ac + bdi$. Suppose z is a complex number such at $z \\circ z = z^{2} + 40$. What is |z|?", "id": "2023AMC12B_12", "lang": "en", "answer": "$5\\sqrt{2}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "For how many ordered pairs (a,b) of integers does the polynomial $x^{3}+ax^{2}+bx+6$ have 3 distinct integer roots?", "id": "2023AMC12B_14", "lang": "en", "answer": "$5$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "In the state of Coinland, coins have values of values of 6, 10, and 15. Suppose x is the value of the most expensive item in Coinland that cannot be purchased using these coins with exact change. What is the sum of the digits of x?", "id": "2023AMC12B_16", "lang": "en", "answer": "$11$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "The three side lengths of a triangle are in arithmetic progression with shortest side of length 6. One of the interior angles measures $120^{\\circ}$. What is the area of the triangle?", "id": "2023AMC12B_17", "lang": "en", "answer": "$15\\sqrt{3}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "A real-valued function f has the property that for all real numbers a and b, f(a+b)+f(a-b)=2f(a)f(b). Which one of the following cannot be the value of f(1): (A) 0 (B) 1 (C) -1 (D) 2 (E) -2", "id": "2023AMC12B_22", "lang": "en", "answer": "E"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "When n standard six-sided dice are rolled, the product of the numbers rolled can be any of 936 possible values. What is n?", "id": "2023AMC12B_23", "lang": "en", "answer": "$11$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "2023AMC12B", "question": "Suppose that a,b,c and d are positive integers satisfying all of the following relations:\\n\\begin{align}\\nabcd &= 2^{6} \\cdot 3^{9} \\cdot 5^{7} &\\\\\\nlcm(a,b)&= 2^{3} \\cdot 3^{2} \\cdot 5^{3} &\\\\\\nlcm(a,c)&= 2^{3} \\cdot 3^{3} \\cdot 5^{3} &\\\\\\nlcm(a,d)&= 2^{3} \\cdot 3^{3} \\cdot 5^{3} &\\\\\\nlcm(b,c)&= 2^{1} \\cdot 3^{3} \\cdot 5^{2} &\\\\\\nlcm(b,d)&= 2^{2} \\cdot 3^{3} \\cdot 5^{2} &\\\\\\nlcm(c,d)&= 2^{2} \\cdot 3^{3} \\cdot 5^{2} &\\\\\\n\\end{align}\\nWhat is $\\gcd(a,b,c,d)$?", "id": "2023AMC12B_24", "lang": "en", "answer": "$3$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "A quilt maker uses 25 dollars worth of materials to produce a single quilt that sells for 100 dollars. What is the minimum number of this type of quilt the quilt maker can produce and sell to earn a profit of at least 900 dollars?", "id": "042023ACTZ18_12", "lang": "en", "answer": "$12$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "The lengths of corresponding sides of 2 similar right triangles are in the ratio 2:7. The hypotenuse of the smaller triangle is 8 inches long. How many inches long is the hypotenuse of the larger triangle?", "id": "042023ACTZ18_14", "lang": "en", "answer": "$28$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "The Student Council is preparing a budget for an upcoming fund-raising dance. They have decided to spend 225 dollars for a light and sound show, 600 dollars for refreshments, and 75 dollars for decorations. These are the only expnses. Given that the Student Council estimates 400 students will attend the dance, what should be the price, per student, for admission to the dance if the Student Council wants to raise as close as possible to 500 dollars after paying expenses?", "id": "042023ACTZ18_15", "lang": "en", "answer": "$3.50$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "The price of 4 meals for Erin and her 3 friends was 60 dollars before the addition of tax and tip. Erin and 2 friends each had meals of equal price. The price of the 3rd friend's meal was twice the price of Erin's meal. What was the price, in dollars, of Erin's meal?", "id": "042023ACTZ18_18", "lang": "en", "answer": "$12$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Camp counselors will combined $5\\frac12$ pounds of peanuts with $1\\frac14$ pounds of cashews to make trail mix for their campers. Each camper will get a small bag filled with $\\frac18$ pound of the trail mix. What is the maximum number of bags of trail mix that the camp counselors can fill?", "id": "042023ACTZ18_19", "lang": "en", "answer": "$54$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "A student took 4 chemistry tests and earned an average score of exactly 80 points. A score of how many points on the 5th test will earn the student an average score of exactly 82 points for the 5 tests?", "id": "042023ACTZ18_20", "lang": "en", "answer": "$90$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "For all positive real numbers a, $\\left(\\sqrt[3]{a^{36}}\\right)^{\\frac12}=$? Calculator is allowed only to perform addition, subtraction, multiplication, and division.", "id": "042023ACTZ18_23", "lang": "en", "answer": "$a^{6}$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "A particular brand of paper towel has 193 sheets per roll and each sheet is 11 in by 5.5 in. The paper towels come in boxes of 6 rolls with a shipping weight of 6.8 pounds. The cost for a box of 6 rolls of this paper towel is 15.99 dollars plus the shippping cost given below. For number of boxes shipped in the range [1,10], the shipping cost per box is 13.25 dollars, for [11,99] boxes, the cost is 12 dollars per box, and for 100 or more boxes, the shipping cost per box is 10 dollars. The diameter of each cylindrical roll is 6 inches. Each roll has a cylindrical hole 1.5 inches in diameter, and each roll has a height of 11 inches. The 6 rolls are shipped in a box where the interior of the box is as small as possible. What are the interior dimensions of this box, in inches?", "id": "042023ACTZ18_33", "lang": "en", "answer": "$12\\times18\\times11$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "(Continue from the previous question) The total weight of the box and packaging material is 1 pound. Which of the following numeric expressions represents the weight, in pounds, of 1 roll of these paper towels: (F) $\\frac{6.8-1}{6}$ (G) $\\frac{6.8}{6}$ (H) $\\frac{6.8+1}{6}$ (J) $6(6.8)$ (K) $6(6.8-1)$", "id": "042023ACTZ18_34", "lang": "en", "answer": "F"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "(Continue from the previous question) Assuming a use of 5 paper towel sheets per day, what is the least number of boxes of paper towels that should be ordered to ensure a 2-year supply?", "id": "042023ACTZ18_35", "lang": "en", "answer": "$4$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "What time is it exactly 230 hours after 3pm?", "id": "042023ACTZ18_38", "lang": "en", "answer": "$5am$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Vanessa averaged exactly 12 points per game for 5 basketball games. She scored at least 10 points per game in each of the 5 games. What is the maximum number of points per game she could have scored in any 1 of the 5 games?", "id": "042023ACTZ18_39", "lang": "en", "answer": "$20$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "For a certain geometric sequence of only positive terms, $a_{3}=\\frac{16}{3}$ and $a_{5}=\\frac{64}{27}$. Which of the following numeric expressions gives the value of $a_{8}$: (A) $\\frac{2^{7}}{3^{4}}$ (B) $\\frac{2^{8}}{3^{5}}$ (C) $\\frac{2^{9}}{3^{6}}$ (D) $\\frac{2^{10}}{3^{4}}$ (E) $\\frac{2^{10}}{3^{7}}$", "id": "042023ACTZ18_43", "lang": "en", "answer": "C"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Let S be the set of all numbers of the form b - a where a and b are real numbers such that a > b. Which of the following describes S:\\n(A) All negative numbers\\n(B) All positive numbers\\n(C) All real numbers\\n(D) All numbers less than or equal to 0\\n(E) All numbers greater than or equal to 0", "id": "042023ACTZ18_45", "lang": "en", "answer": "A"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "The volume of a solid object is equal to the volume of water it displaces when completely submerged in water. A solid object will be placed in a rectangular tank that has a base of 40 in by 30 in and is filled with water to a uniform depth of 10 in. When the object is completely submerged, the new depth of the water in the tank is $10\\frac12$ in. What is the volume, in cubic inches, of the object?", "id": "042023ACTZ18_46", "lang": "en", "answer": "$600$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Twenty households in a community were surveyed to determine how many bicycles they owned. the results were recorded and listed below. No household had 6 or more bicycles. Based on the data in the table, what is the expected value of the number of bicycles from a household in this community, rounded to the nearest tenth of a bicycle?\\n\\begin{table}[]\\n\\caption{Table 1}\\n\\label{tab:my-table}\\n\\begin{tabular}{|l|l|}\\n\\hline\\nNumber of bicycles & Percent of households \\\\ \\hline\\n0 & 10\\% \\\\ \\hline\\n1 & 10\\% \\\\ \\hline\\n2 & 25\\% \\\\ \\hline\\n3 & 10\\% \\\\ \\hline\\n4 & 35\\% \\\\ \\hline\\n5 & 10\\% \\\\ \\hline\\n\\end{tabular}\\n\\end{table}", "id": "042023ACTZ18_48", "lang": "en", "answer": "$2.8$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Anna will save 5 dollars in Week 1, 6 dollars in Week 2, and 7 dollars in Week 3, saving 1 dollar more each subsequent week until she saves 20 dollars in Week 6. She will then start over, saving 5 dollars in Week 17, then 6 dollars in Week 18, saving 1 dollar more each subsequent week, until she saves 20 dollars in Week 32. This pattern will continue for all 52 weeks of the year. What is the total amount of money Anna will save in 1 full year?", "id": "042023ACTZ18_50", "lang": "en", "answer": "$626$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "A student in a chemistry lab had 25 mL of a solution composed of only water and acid. The solution contained 80\\% water. The student added 50mL of 100\\% water to this solution. What is the percent of acid in the new solution?", "id": "042023ACTZ18_51", "lang": "en", "answer": "$6\\frac23$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "A ramp has a horizontal length of 12 feet and a vertical height of 1 foot. Which of the following expressions gives the measure of the acute angle the ramp makes with the horizontal: (F) $\\tan \\left(\\frac{1}{12}\\right)$ (G) $\\tan(12)$ (H) $\\tan^{-1} \\left(\\frac{1}{12}\\right)$ (J) $tan^{-1}(12)$ (K) $\\sqrt{1^{2}+12^{12}}$", "id": "042023ACTZ18_52", "lang": "en", "answer": "H"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "For a fund-raising project, your class decided to sell school supplies. They bought the following items to sell: 1,000 pencils at a cost of 5 cents each; 500 pens at 30 cents each; and 500 packages of notebook paper at 50 cents per package. Your class charged these prices: pencils 10 cents each; pens 50 cents each; and notebook paper 1 dollar per package. At the end of the school year, your class had sold 980 pencils, 462 pens, and 459 package of notebook paper. How much profit did you class make on this project? (Note: The class donated the unsold items to a youth shelter.)", "id": "042023ACTZ18_53", "lang": "en", "answer": "$338$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Aimee commutes to work on an interstate highway, and she had been averaging 60 miles per hour over her 1-hour round-trip. Lately, the trip has been taking 15 minutes longer each way because of road construction. What is her average speed, in miles per hour, for a round-trip with the construction delays?", "id": "042023ACTZ18_54", "lang": "en", "answer": "$40$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Increasing both the length and width of a rectangle by 20\\% will increase the area of the rectangle by what percent?", "id": "042023ACTZ18_55", "lang": "en", "answer": "$44$"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Li is walking from one point on a large model of the standard (x,y) coordinate plane to another point. She walks only along lilnes parallel or perpendicular to the x-asix. The expression |6-1|+|2-(-5)| gives the minimum distance, in coordinate units, Li walks between which of the following pairs of points: (F) (2,6) and (-5,1) (G) (2,6) and (-1,-5) (H) (6,-5) and (-1,2) (J) (6, -1) and (2, 5) (K) (6,1) and (2, -5)", "id": "042023ACTZ18_56", "lang": "en", "answer": "F"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "Which of the following systems when solved gives values of a and b such that $(3^{a})(3^{b})=9$ and $\\frac{2^{a}}{2^{b}}=64$:\\n(A) $\\begin{cases} a+b=2 \\\\ a-b = 6 \\end{cases}$\\n(B) $\\begin{cases} a+b=9 \\\\ a-b = 64 \\end{cases}$\\n(C) $\\begin{cases} ab=1 \\\\ \\frac{a}{b}=6 \\end{cases}$\\n(D) $\\begin{cases} ab=2 \\\\ \\frac{a}{b}=6 \\end{cases}$\\n(E) $\\begin{cases} ab=9 \\\\ \\frac{a}{b}=64 \\end{cases}$", "id": "042023ACTZ18_57", "lang": "en", "answer": "A"}
|
||||
{"source": "part2.en_zh.json", "sourcename": "042023ACTZ18", "question": "For all angle measures $A^{\\circ}$ greater than $0^{\\circ}$, which of the following expressions is equal to $\\cos A^{\\circ}$: (F) $\\cos (90+A)^{\\circ}$ (G) $\\cos (180-A)^{\\circ}$ (H) $\\cos (180+A)^{\\circ}$ (J) $\\cos (270+A)^{\\circ}$ (K) $\\cos (360-A)^{\\circ}$", "id": "042023ACTZ18_58", "lang": "en", "answer": "K"}
|
||||
@@ -0,0 +1,14 @@
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知集合 $A = {x \\mid -5 < x^3 < 5}$，$B = {-3, -1, 0, 2, 3}$，则 $A \\cap B =$ ( )\n\nA. ${-1, 0}$\nB. ${2, 3}$\nC. ${-3, -1, 0}$\nD. ${-1, 0, 2}$", "answer": "A", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n若 $\\frac{z}{z - 1} = 1 + i$，则 $z =$ ( )\n\nA. $-1 - i$\nB. $-1 + i$\nC. $1 - i$\nD. $1 + i$", "answer": "C", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知向量 $a = (0, 1)$，$b = (2, x)$，若 $b \\perp (b - 4a)$，则 $x =$ ( )\n\nA. $-2$\nB. $-1$\nC. $1$\nD. $2$", "answer": "D", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知 $\\cos(\\alpha + \\beta) = m$，$\\tan \\alpha \\tan \\beta = 2$，则 $\\cos(\\alpha - \\beta) =$ ( )\n\nA. $-3m$\nB. $-\\frac{m}{3}$\nC. $\\frac{m}{3}$\nD. $3m$", "answer": "A", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知圆柱和圆锥的底面半径相等，侧面积相等，且它们的高均为 $\\sqrt{3}$，则圆锥的体积为 ( )\n\nA. $2\\sqrt{3}\\pi$\nB. $3\\sqrt{3}\\pi$\nC. $6\\sqrt{3}\\pi$\nD. $9\\sqrt{3}\\pi$", "answer": "B", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知函数 \\( f(x) \\) 定义如下：\n$$\nf(x) = \n\\begin{cases} \ne^{-x} + \\ln(x + 1), & \\text{if } x \\geq 0 \\\\\n-x^2 - 2ax - a, & \\text{if } x < 0 \n\\end{cases}\n$$\n如果函数在实数集 \\( \\mathbb{R} \\) 上单调递增，则 \\( a \\) 的取值范围是：\nA. $(-\\infty, 0]$\nB. $[-1, 0]$\nC. $[-1, 1]$\nD. $[0, +\\infty)$", "answer": "B", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n当 $x \\in [0, 2\\pi]$ 时，曲线 $y = \\sin x$ 与 $y = 2\\sin(3x - \\frac{\\pi}{6})$ 的交点个数为 ( )\n\nA. $3$\nB. $4$\nC. $6$\nD. $8$", "answer": "C", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知函数 $f(x)$ 的定义域为 $\\mathbb{R}$，$f(x) > f(x - 1) + f(x - 2)$，且当 $x < 3$ 时，$f(x) = x$，则下列结论中一定正确的是\n\nA. $f(10) > 100$\nB. $f(20) > 1000$\nC. $f(10) < 1000$\nD. $f(20) < 10000$", "answer": "B", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，在每小题给出的选项中，有一项或多项符合题目要求，请选出所有你认为正确的选项。\n为了解推动出口后的亩收入（单位：万元）情况，从该种植区抽取样本，得到推动出口后亩收入的样本均值 $\\overline{x} = 2.1$，样本方差 $S^2 = 0.01$，已知该种植区以往的亩收入 $x$ 服从正态分布 $N(1.8, 0.1^2)$，假设推动出口后的亩收入 $Y$ 服从正态分布 $N(\\overline{x}, S^2)$，则（若随机变量 $Z$ 服从正态分布 $N(u, \\alpha^2)$，则 $P(Z < u + \\alpha) \\approx 0.8413$）：\n\nA. $P(x > 2) > 0.2$\nB. $P(x > 2) < 0.5$\nC. $P(Y > 2) > 0.5$\nD. $P(Y > 2) < 0.8$", "answer": "BC", "type": "multi"}
|
||||
{"question": "请完成下面一道选择题，在每小题给出的选项中，有一项或多项符合题目要求，请选出所有你认为正确的选项。\n设函数 $f(x) = (x-1)^2(x-4)$，则：\n\nA. $x = 3$ 是 $f(x)$ 的极小值点\nB. 当 $0 < x < 1$ 时 $f(x) < f(x^2)$\nC. 当 $1 < x < 2$ 时，$-4 < f(2x-1) < 0$\nD. 当 $-1 < x < 0$ 时，$f(2-x) > f(x)$", "answer": "ACD", "type": "multi"}
|
||||
{"question": "请完成下面一道选择题，在每小题给出的选项中，有一项或多项符合题目要求，请选出所有你认为正确的选项。\n某造型可以看作图中的曲线 $C$ 的一部分。已知 $C$ 过坐标原点 $O$，且 $C$ 上的点满足横坐标大于 $-2$，到点 $F(2,0)$ 的距离与到定直线 $x = a$ ($a < 0$) 的距离之积为 $4$，则：\nA. $a = -2$\nB. 点 $(2\\sqrt{2}, 0)$ 在 $C$ 上\nC. $C$ 在第一象限的点的纵坐标的最大值为 $1$\nD. 当点 $(x_0, y_0)$ 在 $C$ 上时，$y_0 \\leq \\frac{4}{(x_0 + 2)}$", "answer": "ABD", "type": "multi"}
|
||||
{"question": "设双曲线 $C$：$\\frac{x^2}{a^2} - \\frac{y^2}{b^2} = 1$ ($a > 0, b > 0$) 的左右焦点分别为 $F_1$ 和 $F_2$，过 $F_2$ 作平行于 $y$ 轴的直线交 $C$ 于 $A$ 和 $B$ 两点，若 $|F_1A| = 13$，$|AB| = 10$，则 $C$ 的离心率为 ___", "answer": "3/2", "type": "blank"}
|
||||
{"question": "若曲线 $y = e^x + x$ 在点 $(0, 1)$ 处的切线也是曲线 $y = \\ln(x + 1) + a$ 的切线，则 $a = ___", "answer": "ln2", "type": "blank"}
|
||||
{"question": "甲、乙两人各有四张卡片，每张卡片上标有一个数字，甲的卡片分别标有数字 $1$，$3$，$5$，$7$，乙的卡片上分别标有数字 $2$，$4$，$6$，$8$。两人进行四轮比赛，在每轮比赛中，两人各自从自己持有的卡片中随机选一张，并比较所选卡片的数字的大小，数字大的人得$1$分，数字小的人得$0$分，然后各弃置此轮所选的卡片（弃置的卡片在此后的轮次中不能使用）。则四轮比赛比赛后，甲的得分小于$2$ 的概率为____", "answer": "1/2", "type": "blank"}
|
||||
@@ -0,0 +1,14 @@
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知 $z = -1 - i$，则 $|z|=$ \nA. 0 \nB. 1 \nC. $\\sqrt{2}$ \nD. 2", "answer": "C", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知命题 $p: \\forall A \\in R，|x+1|>1$，命题 $q: \\exists x>0，x^3=x$，则 \nA. $p$ 和 $q$ 都是真命题 \nB. $\\neg p$ 和 $q$ 都是真命题 \nC. $p$ 和 $\\neg q$ 都是真命题 \nD. $\\neg p$ 和 $\\neg q$ 都是真命题", "answer": "B", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知向量 $a,b$ 满足 $|a|=1,|a+2b|=2$，且 $(b-2a) \\perp b$，则 $|b|=$ \nA. $\\frac{1}{2}$ \nB. $\\frac{\\sqrt{2}}{2}$ \nC. $\\frac{\\sqrt{3}}{2}$ \nD. 1", "answer": "B", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n某农业研究部门在面积相等的 100 块稻田上种植新型水稻，得到了各块稻田的亩产量（单位：kg）并部分整理如下表所示。\n\n| 亩产量（kg） | 频数 |\n| ------------ | ---- |\n| [900,950) | 6 |\n| [950,1000) | 12 |\n| [1000,1050) | 18 |\n| [1050,1150) | 24 |\n| [1150,1200) | 10 |\n\n根据表中数据，下列结论正确的是 \n\nA. 100 块稻田亩产量的中位数小于 1050 kg \nB. 100 块稻田亩产量低于 1100 kg 的稻田所占比例超过 40% \nC. 100 块稻田亩产量的极差介于 200 kg 到 300 kg 之间 \nD. 100 块稻田亩产量的均值介于 900 kg 到 1000 kg 之间", "answer": "C", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知曲线 $C: x^2+y^2=16 (y>0)$，从 $C$ 上任意一点 $P$ 向 $x$ 轴作垂线段 $PP'$，$P'$ 为垂足，则线段 $PP'$ 的中点 $M$ 的轨迹方程为 \nA. $\\frac{x^2}{16}+\\frac{y^2}{4}=1 (y>0)$ \nB. $\\frac{x^2}{16}+\\frac{y^2}{8}=1 (y>0)$ \nC. $\\frac{y^2}{16}+\\frac{x^2}{4}=1 (y>0)$ \nD. $\\frac{y^2}{16}+\\frac{x^2}{8}=1 (y>0)$", "answer": "A", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n设函数 $f(x) = a(x+1)^2 - 1$，$g(x) = \\cos x + 2ax$（$a$为常数），当 $x \\in (-1, 1)$ 时，曲线 $y = f(x)$ 和 $y = g(x)$ 恰有一个交点，则 $a =$ \nA. $-1$ \nB. $\\frac{1}{2}$ \nC. 1 \nD. 2", "answer": "D", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n已知正三棱台 $ABC - A'B'C'$ 的体积为 $\\frac{52}{3}$，$AB = 6$，$A_1B_1 = 2$，则 $AA'$ 与平面 $ABC$ 所成角的正切值为 \nA. $\\frac{1}{2}$ \nB. 1 \nC. 2 \nD. 3", "answer": "B", "type": "single"}
|
||||
{"question": "请完成下面一道选择题，每个小题四个选项中，只有一项是符合题目要求的。\n设函数 $f(x) = (x + a) \\ln (x + b)$，若 $f(x) \\geq 0$，则 $a^2 + b^2$ 的最小值为 \nA. $\\frac{1}{8}$ \nB. $\\frac{1}{4}$ \nC. $\\frac{1}{2}$ \nD. 1", "answer": "C", "type": "single"}
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{"question": "请完成下面一道选择题，在每小题给出的选项中，有一项或多项符合题目要求，请选出所有你认为正确的选项。\n对于函数 $f(x) = \\sin 2x$ 和 $g(x) = \\sin (2x - \\frac{\\pi}{4})$，下面正确的有： \nA. $f(x)$ 和 $g(x)$ 有相同零点 \nB. $f(x)$ 和 $g(x)$ 有相同最大值 \nC. $f(x)$ 和 $g(x)$ 有相同的最小正周期 \nD. $f(x)$ 和 $g(x)$ 有相同的对称轴", "answer": "BC", "type": "multi"}
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{"question": "请完成下面一道选择题，在每小题给出的选项中，有一项或多项符合题目要求，请选出所有你认为正确的选项。\n抛物线 $C: y^2 = 4x$ 的准线为 $l$，$P$ 为 $C$ 上的动点，过 $P$ 作圆 $A: x^2 + (y - 4)^2 = 1$ 的一条切线，$Q$ 为切点，过 $P$ 作 $C$ 的垂线，垂足为 $B$，则： \nA. $l$ 与圆 $A$ 相切 \nB. 当 $P$，$A$，$B$ 三点共线时，$|PQ| = \\sqrt{15}$ \nC. 当 $|PB| = 2$ 时，$PA \\perp AB$ \nD. 满足 $|PA| = |PB|$ 的点 $P$ 有且仅有两个", "answer": "ABD", "type": "multi"}
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{"question": "请完成下面一道选择题，在每小题给出的选项中，有一项或多项符合题目要求，请选出所有你认为正确的选项。\n函数 $f(x) = 2x^3 - 3ax^2 + 1$ ，则： \nA. 当 $a > 1$ 时，$f(x)$ 有三个零点。 \nB. 当 $a < 0$ 时 $x = 0$ 是 $f(x)$ 的极大值点。 \nC. 存在 $a, b$，使得 $x = b$ 是曲线 $y = f(x)$ 的对称轴。 \nD. 存在 $a$，使得点 $(1, f(1))$ 为曲线 $y = f(x)$ 的对称中心。", "answer": "AD", "type": "multi"}
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{"question": "记 $S_n$ 为等差数列 $\\{a_n\\}$ 的前 $n$ 项和，若$a_3 + a_4 = 7$，$3a_2 + a_5 = 5$，则$S_{10} =$ _______", "answer": "95", "type": "blank"}
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{"question": "在右图的4×4方格表中选4个方格，颜色均均匀分布，要求每行和每列均恰有一个方格被选中，则共有_____种选法，在所有符合上述要求的选择中，选中方格中的4个数之和的最大值是_____\n\n**方格表：**\n| 11 | 21 | 31 | 40 |\n|----|----|----|----|\n| 12 | 22 | 33 | 42 |\n| 13 | 22 | 33 | 43 |\n| 15 | 24 | 34 | 44 |", "answer": "\\frac{-2\\sqrt{2}}{3}", "type": "blank"}
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{"question": "甲、乙两人各有四张卡片，每张卡片上标有一个数字，甲的卡片分别标有数字 $1$，$3$，$5$，$7$，乙的卡片上分别标有数字 $2$，$4$，$6$，$8$。两人进行四轮比赛，在每轮比赛中，两人各自从自己持有的卡片中随机选一张，并比较所选卡片的数字的大小，数字大的人得$1$分，数字小的人得$0$分，然后各弃置此轮所选的卡片（弃置的卡片在此后的轮次中不能使用）。则四轮比赛比赛后，甲的得分小于$2$ 的概率为____", "answer": "24;112", "type": "blank"}
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@@ -0,0 +1,91 @@
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{"idx": 0, "question": "已知集合 M={x \\mid-3<x<1}, N={x \\mid-1 \\leq x<4}, 则 M \\cup N=(\\quad) A. {x \\mid-1 \\leq x<1} B. {x \\mid x>-3} C. {x \\mid-3<x<4} D. {x \\mid x<4}", "answer": "C. {x \\mid-3<x<4}", "choice_answer": "C"}
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{"idx": 1, "question": "已知 \\frac{z}{\\mathrm{i}}=-1-\\mathrm{i}, 则 z=(\\quad). A. -1-\\mathrm{i} B. -1+\\mathrm{i} C. 1-\\mathrm{i} D. 1+\\mathrm{i}", "answer": "C. 1-\\mathrm{i}", "choice_answer": "C"}
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{"idx": 2, "question": "圆 x^{2}+y^{2}-2 x+6 y=0 的圆心到直线 x-y+2=0 的距离为 (\\quad) A. \\sqrt{2} B. 2 C. 3 D. 3 \\sqrt{2}", "answer": "3 \\sqrt{2}", "choice_answer": "D"}
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{"idx": 3, "question": "在 (x-\\sqrt{x})^{4} 的展开式中, x^{3} 的系数为 (\\quad) A. 6 B. -6 C. 12 D. -12", "answer": "6", "choice_answer": "A"}
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{"idx": 4, "question": "设 \\vec{a}, \\vec{b} 是向量, 则“ (\\vec{a}+\\vec{b}) \\cdot(\\vec{a}-\\vec{b})=0 ”是“ \\vec{a}=-\\vec{b} 或 \\vec{a}=\\vec{b} ”的 ( ). A. 充分不必要条件 B. 必要不充分条件 C. 充要条件 D. 既不充分也不必要条件", "answer": "必要不充分条件", "choice_answer": "B"}
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{"idx": 5, "question": "设函数 f(x)=\\sin \\omega x(\\omega>0). 已知 f\\left(x_{1}\\right)=-1, f\\left(x_{2}\\right)=1, 且 \\left|x_{1}-x_{2}\\right| 的最小值为 \\frac{\\pi}{2}, 则 \\omega=(\\quad) A. 1 B. 2 C. 3 D. 4", "answer": "2", "choice_answer": "B"}
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{"idx": 6, "question": "生物丰富度指数 d=\\frac{S-1}{\\ln N} 是河流水质的一个评价指标, 其中 S, N 分别表示河流中的生物种类数与生物个体总数.生物丰富度指数 d 越大, 水质越好. 如果某河流治理前后的生物种类数 S 没有变化, 生物个体总数由 N_{1} 变为 N_{2}, 生物丰富度指数由 2.1 提高到 3.15 , 则 A. 3 N_{2}=2 N_{1} B. 2 N_{2}=3 N_{1} C. N_{2}^{2}=N_{1}^{3} D. N_{2}^{3}=N_{1}^{2}", "answer": "N_{2}^{3}=N_{1}^{2}", "choice_answer": "D"}
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{"idx": 7, "question": "如图, 在四棱雉 P-A B C D 中, 底面 A B C D 是边长为 4 的正方形, P A=P B=4, P C=P D=2 \\sqrt{2}, 该棱雉的高为 ( ). /Users/niyuge/Documents/notes/Education/Math/Chinese/2024Math/questions/2024BJ_question_8.jpg A. 1 B. 2 C. \\sqrt{2} D. \\sqrt{3}", "answer": "\\sqrt{3}", "choice_answer": "D"}
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{"idx": 8, "question": "已知 \\left(x_{1}, y_{1}\\right),\\left(x_{2}, y_{2}\\right) 是函数 y=2^{x} 的图象上两个不同的点, 则 ( ) A. \\log _{2} \\frac{y_{1}+y_{2}}{2}<\\frac{x_{1}+x_{2}}{2} B. \\log _{2} \\frac{y_{1}+y_{2}}{2}>\\frac{x_{1}+x_{2}}{2} C \\log _{2} \\frac{y_{1}+y_{2}}{2}<x_{1}+x_{2} D. \\log _{2} \\frac{y_{1}+y_{2}}{2}>x_{1}+x_{2}", "answer": "\\log _{2} \\frac{y_{1}+y_{2}}{2}>\\frac{x_{1}+x_{2}}{2}", "choice_answer": "B"}
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{"idx": 9, "question": "已知 M=\\left{(x, y) \\mid y=x+t\\left(x^{2}-x\\right), 1 \\leq x \\leq 2,0 \\leq t \\leq 1\\right} 是平面直角坐标系中的点集. 设 d 是 M 中两点间距离的最大值, S 是 M 表示的图形的面积, 则 A. d=3, S<1 B. d=3, S>1 C. d=\\sqrt{10}, S<1 D. d=\\sqrt{10}, S>1", "answer": "d=\\sqrt{10}, S<1", "choice_answer": "C"}
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{"idx": 10, "question": "抛物线 y^{2}=16 x 的焦点坐标为 \\qquad .", "answer": "(4,0)", "choice_answer": ""}
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{"idx": 11, "question": "在平面直角坐标系 x O y 中, 角 \\alpha 与角 \\beta 均以 O x 为始边, 它们的终边关于原点对称. 若 \\alpha \\in\\left[\\frac{\\pi}{6}, \\frac{\\pi}{3}\\right],则 \\cos \\beta 的最大值为 \\qquad .", "answer": "(-\\frac{1}{2}, -0.5)", "choice_answer": ""}
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{"idx": 12, "question": "若直线 y=k(x-3) 与双曲线 \\frac{x^{2}}{4}-y^{2}=1 只有一个公共点, 则 k 的一个取值为 . \\qquad", "answer": "\\frac{1}{2}, -\\frac{1}{2}", "choice_answer": ""}
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{"idx": 13, "question": "汉代刘㰴设计的“铜嘉量”是侖、合、升、斗、斛五量合一的标准量器, 其中升量器、斗量器、斛量器的形状均可视为圆柱.若升、斗、解量器的容积成公比为 10 的等比数列, 底面直径依次为 65 \\mathrm{~mm}, 325 \\mathrm{~mm}, 325 \\mathrm{~mm}, 且斛量器的高为 230 mm , 则斗量器的高为 \\qquad mm , 升量器的高为 \\qquad mm .", "answer": "23 ,57.5, \\frac{115}{2}", "choice_answer": ""}
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{"idx": 14, "question": "设 \\left{a_{n}\\right} 与 \\left{b_{n}\\right} 是两个不同的无穷数列, 且都不是常数列. 记集合 M=\\left{k \\mid a_{k}=b_{k}, k \\in \\mathrm{N}^{*}\\right}, 给出下列 4 个结论: (1)若 \\left{a_{n}\\right} 与 \\left{b_{n}\\right} 均为等差数列, 则 M 中最多有 1 个元素; (2)若 \\left{a_{n}\\right} 与 \\left{b_{n}\\right} 均为等比数列, 则 M 中最多有 2 个元素; (3)若 \\left{a_{n}\\right} 等差数列, \\left{b_{n}\\right} 为等比数列, 则 M 中最多有 3 个元素; (4)若 \\left{a_{n}\\right} 递增数列, \\left{b_{n}\\right} 为递减数列, 则 M 中最多有 1个元素; 其中正确结论的序号是 \\qquad .", "answer": "(1)(3)(4)", "choice_answer": ""}
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{"idx": 15, "question": "已知椭圆 E: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0), 以椭圆 E 的焦点和短轴端点为顶点的四边形是边长为 2 的正方形. 过点 (0, t)(t>\\sqrt{2}) 且斜率存在的直线与椭圆 E 交于不同的两点 A, B, 过点 A 和 C(0,1) 的直线 A C与椭圆 E 的另一个交点为 D. (1) 求椭圆 E 的方程及离心率; (2) 若直线 B D 斜率为 0 , 求 t 的值.", "answer": "\\frac{x^{2}}{4}+\\frac{y^{2}}{2}=1, e=\\frac{\\sqrt{2}}{2},t=2", "choice_answer": ""}
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{"idx": 16, "question": "设 z=\\sqrt{2} \\mathrm{i}, 则 z \\cdot \\bar{z}= A. -2 B. \\sqrt{2} C. -\\sqrt{2} D. 2", "answer": "2", "choice_answer": "D"}
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{"idx": 17, "question": "若集合 A={1,2,3,4,5,9}, B={x \\mid x+1 \\in A}, 则 A \\cap B= A. {1,3,4} B. {2,3,4} C. {1,2,3,4} D. {0,1,2,3,4,9}", "answer": "{1,2,3,4}", "choice_answer": "C"}
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{"idx": 18, "question": "若 x, y 满足约束条件 \\left{\\begin{array}{l}4 x-3 y-3 \\geq 0 \\ x-2 y-2 \\leq 0 \\ 2 x+6 y-9 \\leq 0\\end{array}\\right., 则 z=x-5 y 的最小值为 ( ) A. \\frac{1}{2} B. 0 C. -\\frac{5}{2} D. -\\frac{7}{2}", "answer": "-\\frac{7}{2}", "choice_answer": "D"}
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{"idx": 19, "question": "甲、乙、丙、丁四人排成一列，则丙不在排头，且甲或乙在排尾的概率是（） A. \\frac{1}{4} B. \\frac{1}{3} C. \\frac{1}{2} D. \\frac{2}{3}", "answer": "\\frac{1}{3}", "choice_answer": "B"}
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{"idx": 20, "question": "已知等差数列 \\left{a_{n}\\right} 的前 n 项和为 S_{n}, 若 S_{9}=1, 则 a_{3}+a_{7}= ( ) A. -2 B. \\frac{7}{3} C. 1 D. \\frac{2}{9}", "answer": "\\frac{2}{9}", "choice_answer": "D"}
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{"idx": 21, "question": "已知双曲线的两个焦点分别为 (0,4),(0,-4), 点 (-6,4) 在该双曲线上, 则该双曲线的离心率为（） A. 4 B. 3 C. 2 D. \\sqrt{2}", "answer": "2", "choice_answer": "C"}
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{"idx": 22, "question": "设函数 f(x)=\\frac{\\mathrm{e}^{x}+2 \\sin x}{1+x^{2}}, 则曲线 y=f(x) 在点 (0,1) 处的切线与两坐标轴所围成的三角形的面积为 A. \\frac{1}{6} B. \\frac{1}{3} C. \\frac{1}{2} D. \\frac{2}{3}", "answer": "\\frac{1}{6}", "choice_answer": "A"}
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{"idx": 23, "question": "已知 \\frac{\\cos \\alpha}{\\cos \\alpha-\\sin \\alpha}=\\sqrt{3}, 则 \\tan \\left(\\alpha+\\frac{\\pi}{4}\\right)=(\\quad) A. 2 \\sqrt{3}+1 B. 2 \\sqrt{3}-1 C. \\frac{\\sqrt{3}}{2} D. 1-\\sqrt{3}", "answer": "2\\sqrt{3}-1", "choice_answer": "B"}
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{"idx": 24, "question": "已知直线 a x+y+2-a=0 与圆 C: x^{2}+y^{2}+4 y-1=0 交于 A, B 两点, 则 |A B| 的最小值为 (\\quad) A. 2 B. 3 C. 4 D. 6", "answer": "4", "choice_answer": "C"}
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{"idx": 25, "question": "设 \\alpha 、 \\beta 为两个平面, m 、 n 为两条直线, 且 \\alpha \\cap \\beta=m. 下述四个命题: (1)若 m / / n, 则 n / / \\alpha 或 n / / \\beta (2)若 m \\perp n, 则 n \\perp \\alpha 或 n \\perp \\beta (3)若 n / / \\alpha 且 n / / \\beta, 则 m / / n (4)若 n 与 \\alpha, \\beta 所成的角相等, 则 m \\perp n 其中所有真命题的编号是 A. (1)(3) B. (2) (4) C. (1)(2)(3) D. (1)(3)(4)", "answer": "(1)(3)", "choice_answer": "A"}
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{"idx": 26, "question": "在 \\triangle A B C 中, 内角 A, B, C 所对的边分别为 a, b, c, 若 B=\\frac{\\pi}{3}, b^{2}=\\frac{9}{4} a c, 则 \\sin A+\\sin C=(\\quad) A. \\frac{2 \\sqrt{39}}{13} B. \\frac{\\sqrt{39}}{13} C. \\frac{\\sqrt{7}}{2} D. \\frac{3 \\sqrt{13}}{13}", "answer": "\\frac{\\sqrt{7}}{2}", "choice_answer": "C"}
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{"idx": 27, "question": "函数 f(x)=\\sin x-\\sqrt{3} \\cos x 在 [0, \\pi] 上的最大值是 \\qquad .", "answer": "2", "choice_answer": ""}
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{"idx": 28, "question": "已知圆台甲、乙的上底面半径均为 r_{1}, 下底面半径均为 r_{2}, 圆台的母线长分别为 2\\left(r_{2}-r_{1}\\right), 3\\left(r_{2}-r_{1}\\right), 则圆台甲与乙的体积之比为 \\qquad .", "answer": "\\frac{\\sqrt{6}}{4}", "choice_answer": ""}
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{"idx": 29, "question": "已知 a>1 且 \\frac{1}{\\log _{8} a}-\\frac{1}{\\log _{a} 4}=-\\frac{5}{2}, 则 a= \\qquad .", "answer": "64", "choice_answer": ""}
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{"idx": 30, "question": "曲线 y=x^{3}-3 x 与 y=-(x-1)^{2}+a 在 (0,+\\infty) 上有两个不同 交点, 则 a 的取值范围为 \\qquad .", "answer": "(-2, 1)", "choice_answer": ""}
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{"idx": 31, "question": "已知等比数列 \\left{a_{n}\\right} 的前 n 项和为 S_{n}, 且 2 S_{n}=3 a_{n+1}-3. (1) 求 \\left{a_{n}\\right} 的通项公式; (2) 求数列 \\left{S_{n}\\right} 的前 n 项和.", "answer": "a_{n}=\\left(\\frac{5}{3}\\right)^{n-1}, \\frac{15}{4} \\cdot\\left(\\frac{5}{3}\\right)^{n}-\\frac{3}{2} n-\\frac{15}{4}", "choice_answer": ""}
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{"idx": 32, "question": "已知椭圆 C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0) 的右焦点为 F, 点 M\\left(1, \\frac{3}{2}\\right) 在 C 上, 且 M F \\perp x 轴. (1) 求 C 的方程;", "answer": "\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1", "choice_answer": ""}
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{"idx": 33, "question": "在直角坐标系 x O y 中, 以坐标原点为极点, x 轴正半轴为极轴建立极坐标系, 曲线 C 的极坐标方程为 \\rho=\\rho \\cos \\theta+1. (1) 写出 C 的直角坐标方程; (2) 设直线 l:\\left{\\begin{array}{l}x=t \\ y=t+a\\end{array}\\right. ( t 为参数), 若 C 与 l 相交于 A 、 B 两点, 若 |A B|=2, 求 a. ", "answer": "y^{2}=2 x+1, a=\\frac{3}{4}", "choice_answer": ""}
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{"idx": 34, "question": " 集合 A={1,2,3,4}, B={2,3,4,5}, 则 A \\cap B=(\\quad) A. {1,2,3,4} B. {2,3,4} C. {2,4} D. {1}", "answer": "{2,3,4}", "choice_answer": "B"}
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{"idx": 35, "question": "设 a, b \\in \\mathbf{R}, 则“ a^{3}=b^{3 ”} 是“ 3^{a}=3^{b} ”的 ( ) A. 充分不必要条件 B. 必要不充分条件 C. 充要条件 D. 既不充分也不必要条件", "answer": "充要条件", "choice_answer": "C"}
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{"idx": 36, "question": "下列函数是偶函数的是 A. y=\\frac{\\mathrm{e}^{x}-x^{2}}{x^{2}+1} B. y=\\frac{\\cos x+x^{2}}{x^{2}+1} C. y=\\frac{\\mathrm{e}^{x}-x}{x+1} D. y=\\frac{\\sin x+4 x}{\\mathrm{e}^{|x|}}", "answer": "y=\\frac{\\cos x+x^{2}}{x^{2}+1}", "choice_answer": "B"}
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{"idx": 37, "question": "若 a=4.2^{-0.3}, b=4.2^{0.3}, c=\\log _{4.2} 0.2, 则 a, b, c 的大小关系为 (\\quad) A. a>b>c B. b>a>c C. c>a>b D. b>c>a", "answer": "b>a>c", "choice_answer": "B"}
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{"idx": 38, "question": "若 m, n 为两条不同的直线, \\alpha 为一个平面，则下列结论中正确的是（） A. 若 m / / \\alpha, n / / \\alpha, 则 m \\perp n B. 若 m / / \\alpha, n / / \\alpha, 则 m / / n C. 若 m / / \\alpha, n \\perp \\alpha, 则 m \\perp n D. 若 m / / \\alpha, n \\perp \\alpha, 则 m 与 n 相交", "answer": "若 m / / \\alpha, n \\perp \\alpha, 则 m \\perp n", "choice_answer": "C"}
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{"idx": 39, "question": "已知函数 f(x)=\\sin 3\\left(\\omega x+\\frac{\\pi}{3}\\right)(\\omega>0) 的最小正周期为 \\pi. 则 f(x) 在 \\left[-\\frac{\\pi}{12}, \\frac{\\pi}{6}\\right] 的最小值是（） A. -\\frac{\\sqrt{3}}{2} B. -\\frac{3}{2} C. 0 D. \\frac{3}{2}", "answer": "-\\frac{\\sqrt{3}}{2}", "choice_answer": "A"}
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{"idx": 40, "question": "双曲线 \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, \\quad b>0) 左、右焦点分别为 F_{1} 、 F_{2} . P 是双曲线右支上一点, 且直线 P F_{2} 的斜率为 2. \\triangle P F_{1} F_{2} 是面积为 8 的直角三角形, 则双曲线的方程为 ( ) A. \\frac{x^{2}}{8}-\\frac{y^{2}}{2}=1 B. \\frac{x^{2}}{8}-\\frac{y^{2}}{4}=1 C. \\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1 D. \\frac{x^{2}}{4}-\\frac{y^{2}}{8}=1", "answer": "\\frac{x^{2}}{2}-\\frac{y^{2}}{8}=1", "choice_answer": "C"}
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{"idx": 41, "question": "一个五面体 A B C-D E F. 已知 A D / / B E / / C F, 且两两之间距离为 1 . 并已知 A D=1, B E=2, C F=3. 则该五面体的体积为 (\\quad) /Users/niyuge/Documents/notes/Education/Math/Chinese/2024Math/questions/2024TJ_question_9.jpg A. \\frac{\\sqrt{3}}{6} B. \\frac{3 \\sqrt{3}}{4}+\\frac{1}{2} C. \\frac{\\sqrt{3}}{2} D. \\frac{3 \\sqrt{3}}{4}-\\frac{1}{2}", "answer": "\\frac{\\sqrt{3}}{2}", "choice_answer": "C"}
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{"idx": 42, "question": "已知 i 是虚数单位, 复数 (\\sqrt{5}+\\mathrm{i}) \\cdot(\\sqrt{5}-2 \\mathrm{i})= \\qquad .", "answer": "7-\\sqrt{5 \\mathrm{i}}", "choice_answer": ""}
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{"idx": 43, "question": "在 \\left(\\frac{3}{x^{3}}+\\frac{x^{3}}{3}\\right)^{6} 展开式中, 常数项为 \\qquad .", "answer": "20", "choice_answer": ""}
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{"idx": 44, "question": "圆 (x-1)^{2}+y^{2}=25 的圆心与抛物线 y^{2}=2 p x(p>0) 的焦点 F 重合, A 为两曲线的交点, 则原点到直线 A F 的距离为 \\qquad .", "answer": "\\frac{4}{5}", "choice_answer": ""}
|
||||
{"idx": 45, "question": "A, B, C, D, E 五种活动, 甲、乙都要选择三个活动参加. 甲选到 A 的概率为 \\qquad ; 已知乙选了 A 活动,他再选择 B 活动的概率为 \\qquad .", "answer": "\\frac{3}{5},\\frac{1}{2}", "choice_answer": ""}
|
||||
{"idx": 46, "question": "若函数 f(x)=2 \\sqrt{x^{2}-a x}-|a x-2|+1 恰有一个零点, 则 a 的取值范围为 \\qquad .", "answer": "(-\\sqrt{3},-1) \\cup (1, \\sqrt{3})", "choice_answer": ""}
|
||||
{"idx": 47, "question": "在 \\triangle A B C 中, 角 A, B, C 所对的边分别为 a, b, c, 已知 \\cos B=\\frac{9}{16}, b=5, \\frac{a}{c}=\\frac{2}{3}. （1）求 a; （2）求 \\sin A; (3) 求 \\cos (B-2 A) 的值.", "answer": "4, \\frac{\\sqrt{7}}{4},\\frac{57}{64}", "choice_answer": ""}
|
||||
{"idx": 48, "question": "已知椭圆 \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0) 椭圆的离心率 e=\\frac{1}{2}. 左顶点为 A , 下顶点为 B, C 是线段 O B 的中点,其中 S_{\\triangle A B C}=\\frac{3 \\sqrt{3}}{2}. 求椭圆方程.", "answer": "\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1", "choice_answer": ""}
|
||||
{"idx": 49, "question": "设函数 f(x)=x \\ln x. （1）求 f(x) 图象上点 (1, f(1)) 处的切线方程;", "answer": "\\frac{x^{2}}{12}+\\frac{y^{2}}{9}=1", "choice_answer": ""}
|
||||
{"idx": 50, "question": "已知 z=-1-\\mathrm{i}, 则 |z|=(\\quad) A. 0 B. 1 C. \\sqrt{2} D. 2", "answer": "\\sqrt{2}", "choice_answer": "C"}
|
||||
{"idx": 51, "question": "已知命题 p: \\forall x \\in \\mathbf{R},|x+1|>1; 命题 q: \\exists x>0, x^{3}=x, 则 ( ) A. p 和 q 都 真命题 B. \\neg p 和 q 都是真命题 C. p 和 \\neg q 都是真命题 D. \\neg p 和 \\neg q 都是真命题", "answer": "\\neg p 和 q 都是真命题", "choice_answer": "B"}
|
||||
{"idx": 52, "question": "已知向量 \\vec{a}, \\vec{b} 满足 |\\vec{a}|=1,|\\vec{a}+2 \\vec{b}|=2, 且 (\\vec{b}-2 \\vec{a}) \\perp \\vec{b}, 则 |\\vec{b}|=(\\quad) A. \\frac{1}{2} B. \\frac{\\sqrt{2}}{2} C. \\frac{\\sqrt{3}}{2} D. 1", "answer": "\\frac{\\sqrt{2}}{2}", "choice_answer": "B"}
|
||||
{"idx": 53, "question": "某农业研究部门在面积相等的 100 块稻田上种植一种新型水稻, 得到各块稻田的亩产量（单位： kg ）并整理如下表\n\\begin{table}[]\n\\begin{tabular}{|c|c|c|l|l|l|l|}\n\\hline\n亩产量 & [900,950) & [950,1000) & [1000,1050) & [1050,1100) & [1100,1150) & [1150,1200) \\ \\hline\n频数 & 6 & 12 & 18 & 30 & 24 & 10 \\ \\hline\n\\end{tabular}\n\\end{table}\n根据表中数据，下列结论中正确的是（） A. 100 块稻田亩产量的中位数小于 1050 kg B. 100 块稻田中亩产量低于 1100 kg 的稻田所占比例超过 80 \\% C. 100 块稻田亩产量的极差介于 200 kg 至 300kg 之间 D. 100 块稻田亩产量的平均值介于 900 kg 至 1000 kg 之间", "answer": "100 块稻田亩产量的极差介于 200 kg 至 300kg 之间", "choice_answer": "C"}
|
||||
{"idx": 54, "question": "已知曲线 C: x^{2}+y^{2}=16(y>0), 从 C 上任意一点 P 向 x 轴作垂线段 P P^{\\prime}, P^{\\prime} 为垂足, 则线段 P P^{\\prime}的中点 M 的轨迹方程为 A \\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1 \\quad(y>0) B. \\frac{x^{2}}{16}+\\frac{y^{2}}{8}=1 \\quad(y>0) C. \\frac{y^{2}}{16}+\\frac{x^{2}}{4}=1 \\quad(y>0) D. \\frac{y^{2}}{16}+\\frac{x^{2}}{8}=1 \\quad(y>0)", "answer": "\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1 (y>0)", "choice_answer": "A"}
|
||||
{"idx": 55, "question": "设函数 f(x)=a(x+1)^{2}-1, g(x)=\\cos x+2 a x, 当 x \\in(-1,1) 时, 曲线 y=f(x) 与 y=g(x) 恰有一个交点, 则 a=(\\quad) A. -1 B. \\frac{1}{2} C. 1 D. 2", "answer": "2", "choice_answer": "D"}
|
||||
{"idx": 56, "question": "已知正三棱台 A B C-A_{1} B_{1} C_{1} 的体积为 \\frac{52}{3}, A B=6, A_{1} B_{1}=2, 则 A_{1} A 与平面 A B C 所成角的正切值为 A. \\frac{1}{2} B. 1 C. 2 D. 3", "answer": "1", "choice_answer": "B"}
|
||||
{"idx": 57, "question": "设函数 f(x)=(x+a) \\ln (x+b), 若 f(x) \\geq 0, 则 a^{2}+b^{2} 的最小值为 ( ) A. \\frac{1}{8} B. \\frac{1}{4} C. \\frac{1}{2} D. 1", "answer": "\\frac{1}{2}", "choice_answer": "C"}
|
||||
{"idx": 58, "question": "记 S_{n} 为等差数列 \\left{a_{n}\\right} 的前 n 项和, 若 a_{3}+a_{4}=7,3 a_{2}+a_{5}=5, 则 S_{10}= \\qquad .", "answer": "95", "choice_answer": ""}
|
||||
{"idx": 59, "question": "已知 \\alpha 为第一象限角, \\beta 为第三象限角, \\tan \\alpha+\\tan \\beta=4, \\tan \\alpha \\tan \\beta=\\sqrt{2}+1, 则 \\sin (\\alpha+\\beta)= \\qquad .", "answer": "-\\frac{2\\sqrt{2}}{3}", "choice_answer": ""}
|
||||
{"idx": 60, "question": "在如图的 4 \\times 4 的方格表中选 4 个方格, 要求每行和每列均恰有一个方格被选中, 则共有 \\qquad种选法, 在所有符合上述要求的选法中, 选中方格中的 4 个数之和的最大值是 \\qquad . \\begin{tabular}{|l|l|l|l|} \\hline 11 & 21 & 31 & 40 \\ \\hline 12 & 22 & 33 & 42 \\ \\hline 13 & 22 & 33 & 43 \\ \\hline 15 & 24 & 34 & 44 \\ \\hline \\end{tabular}", "answer": "24,112 ", "choice_answer": ""}
|
||||
{"idx": 61, "question": "记 \\triangle A B C 的内角 A, B, C 的对边分别为 a, b, c, 已知 \\sin A+\\sqrt{3} \\cos A=2. （1）求 A. (2)若 a=2，\\sqrt{2}b\\sin C=c\\sin 2B，求\\triangle ABC的 周长.", "answer": "A=\\frac{\\pi}{6},2+\\sqrt{6}+3 \\sqrt{2}", "choice_answer": ""}
|
||||
{"idx": 62, "question": "已知函数 f(x)=\\mathrm{e}^{x}-a x-a^{3}. （1）当 a=1 时, 求曲线 y=f(x) 在点 (1, f(1)) 处的切线方程; (2) 若 f(x) 有极小值, 且极小值小于 0 , 求 a 的取值范围.", "answer": "(\\mathrm{e}-1)x-y-1=0, (1,+\\infty)", "choice_answer": ""}
|
||||
{"idx": 63, "question": "已知集合 A=\\left{x-5<x^{3}<5\\right}, B={-3,-1,0,2,3}, 则 A \\cap B= A. {-1,0} B. {2,3} C. {-3,-1,0} D. {-1,0,2}", "answer": "{-1,0}", "choice_answer": "A"}
|
||||
{"idx": 64, "question": "若 \\frac{z}{z-1}=1+\\mathrm{i}, 则 z=(\\quad) A. -1-\\mathrm{i} B. -1+\\mathrm{i} C. 1-\\mathrm{i} D. 1+\\mathrm{i}", "answer": "1-\\mathrm{i}", "choice_answer": "C"}
|
||||
{"idx": 65, "question": "已知向量 \\vec{a}=(0,1), \\vec{b}=(2, x), 若 \\vec{b} \\perp(\\vec{b}-4 \\vec{a}), 则 x=(\\quad) A. -2 B. -1 C. 1 D. 2", "answer": "2", "choice_answer": "D"}
|
||||
{"idx": 66, "question": "已知 \\cos (\\alpha+\\beta)=m, \\tan \\alpha \\tan \\beta=2, 则 \\cos (\\alpha-\\beta)= ( ) A. -3 m B. -\\frac{m}{3} C. \\frac{m}{3} D. 3 m", "answer": "-3 m", "choice_answer": "A"}
|
||||
{"idx": 67, "question": "已知圆柱和圆雉的底面半径相等, 侧面积相等, 且它们的高均为 \\sqrt{3}, 则圆雉的体积为 ( ) A. 2 \\sqrt{3} \\pi B. 3 \\sqrt{3} \\pi C. 6 \\sqrt{3} \\pi D. 9 \\sqrt{3} \\pi", "answer": "3\\sqrt{3}\\pi", "choice_answer": "B"}
|
||||
{"idx": 68, "question": "已知函数 f(x)=\\left{\\begin{array}{l}-x^{2}-2 a x-a, x<0 \\ \\mathrm{e}^{x}+\\ln (x+1), x \\geq 0\\end{array}\\right. 在 \\mathbf{R} 上单调递增, 则 a的 取值范围是 (, A. (-\\infty, 0] B. [-1,0] C. [-1,1] D. [0,+\\infty)", "answer": "[-1,0]", "choice_answer": "B"}
|
||||
{"idx": 69, "question": "当 x \\in[0,2 \\pi] 时, 曲线 y=\\sin x 与 y=2 \\sin \\left(3 x-\\frac{\\pi}{6}\\right) 的交点个数为 ( ) A. 3 B. 4 C. 6 D. 8", "answer": "6", "choice_answer": "C"}
|
||||
{"idx": 70, "question": "已知函数 f(x) 的定义域为 \\mathbf{R}, f(x)>f(x-1)+f(x-2), 且当 x<3 时 f(x)=x, 则下列结论中一定正确的是（） A. f(10)>100 B. f(20)>1000 C. f(10)<1000 D. f(20)<10000.", "answer": "f(20)>1000", "choice_answer": "B"}
|
||||
{"idx": 71, "question": "设双曲线 C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)^{1} 左右焦点分别为 F_{1} 、 F_{2}, 过 F_{2} 作平行于 y 轴的直线交 C 于 A, B两点, 若 \\left|F_{1} A\\right|=13,|A B|=10, 则 C 的离心率为 \\qquad .", "answer": "\\frac{3}{2}", "choice_answer": ""}
|
||||
{"idx": 72, "question": "若曲线 y=\\mathrm{e}^{x}+x 在点 (0,1) 处的切线也是曲线 y=\\ln (x+1)+a 的切线, 则 a= \\qquad .", "answer": "\\ln 2", "choice_answer": ""}
|
||||
{"idx": 73, "question": "甲、乙两人各有四张卡片, 每张卡片上标有一个数字, 甲的卡片上分别标有数字 1,3,5,7, 乙的卡片上分别标有数字 2,4,6,8, 两人进行四轮比赛, 在每轮比赛中, 两人各自从自己持有的卡片中随机选一张, 并比较所选卡片上数字的大小, 数字大的人得 1 分, 数字小的人得 0 分, 然后各自弃置此轮所选的卡片 (弃置的卡片在此后的轮次中不能使用). 则四轮比赛后, 甲的总得分不小于 2 的概率为 \\qquad .", "answer": "\\frac{1}{2}", "choice_answer": ""}
|
||||
{"idx": 74, "question": "记 \\triangle A B C 的内角 A 、 B 、 C 的对边分别为 a, b, c, 已知 \\sin C=\\sqrt{2} \\cos B, a^{2}+b^{2}-c^{2}=\\sqrt{2} a b （1）求 B; (2) 若 \\triangle A B C 的面积为 3+\\sqrt{3}, 求 c.", "answer": "B=\\frac{\\pi}{3},2 \\sqrt{2} ", "choice_answer": ""}
|
||||
{"idx": 75, "question": "若 z=5+\\mathrm{i}, 则 \\mathrm{i}(\\bar{z}+z)= ( ) A. 10 i B. 2 i C. 10 D. 2", "answer": "10", "choice_answer": "A"}
|
||||
{"idx": 76, "question": "已知集合 A={1,2,3,4,5,9}, B={x \\mid \\sqrt{x} \\in A}, 则 (A \\cap B)=(\\quad) A. {1,4,9} B. {3,4,9} C. {1,2,3} D. {2,3,5}", "answer": "{2,3,5}", "choice_answer": "D"}
|
||||
{"idx": 77, "question": "若 x, y 满足约束条件 \\left{\\begin{array}{l}4 x-3 y-3 \\geq 0 \\ x-2 y-2 \\leq 0 \\ 2 x+6 y-9 \\leq 0\\end{array}\\right., 则 z=x-5 y 的最小值为 （） A. \\frac{1}{2} B. 0 C. -\\frac{5}{2} D. -\\frac{7}{2}", "answer": "-\\frac{7}{2}", "choice_answer": "D"}
|
||||
{"idx": 78, "question": "记 S_{n} 为等差数列 \\left{a_{n}\\right} 的前 n 项和, 已知 S_{5}=S_{10}, a_{5}=1, 则 a_{1}=(\\quad) A. \\frac{7}{2} B. \\frac{7}{3} C. -\\frac{1}{3} D. -\\frac{7}{11}", "answer": "\\frac{7}{3}", "choice_answer": "B"}
|
||||
{"idx": 79, "question": "已知双曲线的两个焦点分别为 (0,4),(0,-4), 点 (-6,4) 在该双曲线上, 则该双曲线的离心率为 ( ) A. 4 B. 3 C. 2 D. \\sqrt{2}", "answer": "2", "choice_answer": "C"}
|
||||
{"idx": 80, "question": "设函数 f(x)=\\frac{\\mathrm{e}^{x}+2 \\sin x}{1+x^{2}}, 则曲线 y=f(x) 在点 (0,1) 处的切线与两坐标轴所围成的三角形的面积为 A. \\frac{1}{6} B. \\frac{1}{3} C. \\frac{1}{2} D. \\frac{2}{3}", "answer": "\\frac{1}{6}", "choice_answer": "A"}
|
||||
{"idx": 81, "question": "已知 \\frac{\\cos \\alpha}{\\cos \\alpha-\\sin \\alpha}=\\sqrt{3}, 则 \\tan \\left(\\alpha+\\frac{\\pi}{4}\\right)= ( ) A. 2 \\sqrt{3}+1 B. 2 \\sqrt{3}-1 C. \\frac{\\sqrt{3}}{2} D. 1-\\sqrt{3}", "answer": "2\\sqrt{3}-1", "choice_answer": "B"}
|
||||
{"idx": 82, "question": "设向量 \\vec{a}=(x+1, x), \\vec{b}=(x, 2), 则 (\\quad) A. “ x=-3 ”是“ \\vec{a} \\perp \\vec{b} ”的必要条件 B. “ x=-3 ”是“ \\vec{a} / / \\vec{b} ” 的必要条件 C. “ x=0 ”是“ \\vec{a} \\perp \\vec{b} ”的充分条件 D. “ x=-1+\\sqrt{3} ” 是“ \\vec{a} / / \\vec{b} ” 的充分条件", "answer": "x=0 是 \\vec{a} \\perp \\vec{b} 的充分条件", "choice_answer": "C"}
|
||||
{"idx": 83, "question": "在 \\triangle A B C 中, 内角 A, B, C 所对 边分别为 a, b, c, 若 B=\\frac{\\pi}{3}, b^{2}=\\frac{9}{4} a c, 则 \\sin A+\\sin C=(\\quad) A. \\frac{2 \\sqrt{39}}{13} B. \\frac{\\sqrt{39}}{13} C. \\frac{\\sqrt{7}}{2} D. \\frac{3 \\sqrt{13}}{13}", "answer": "\\frac{\\sqrt{7}}{2}", "choice_answer": "C"}
|
||||
{"idx": 84, "question": "已知 b 是 a, c 的等差中项, 直线 a x+b y+c=0 与圆 x^{2}+y^{2}+4 y-1=0 交于 A, B 两点, 则 |A B| 的最小 值为 ( ) A. 1 B. 2 C. 4 D. 2 \\sqrt{5}", "answer": "4", "choice_answer": "C"}
|
||||
{"idx": 85, "question": "\\left(\\frac{1}{3}+x\\right)^{10} 的展开式中, 各项系数中的最大值为 \\qquad .", "answer": "5", "choice_answer": ""}
|
||||
{"idx": 86, "question": "已知圆台甲、乙的上底面半径均为 r_{1}, 下底面半径均为 r_{2}, 圆台的母线长分别为 2\\left(r_{2}-r_{1}\\right), 3\\left(r_{2}-r_{1}\\right), 则圆台甲与乙的体积之比为 \\qquad .", "answer": "\\frac{\\sqrt{6}}{4}", "choice_answer": ""}
|
||||
{"idx": 87, "question": "已知 a>1 且 \\frac{1}{\\log _{8} a}-\\frac{1}{\\log _{a} 4}=-\\frac{5}{2}, 则 a= \\qquad .", "answer": "64", "choice_answer": ""}
|
||||
{"idx": 88, "question": "有 6 个相同的球, 分别标有数字 1 、 2 、 3 、 4 、 5 、 6, 从中无放回地随机取 3 次, 每次取 1 个球. 记 m 为前两次取出的球上数字的平均值, n 为取出的三个球上数字的平均值, 则 m 与 n 之差的绝对值不大于 \\frac{1}{2} 的概率为 \\qquad .", "answer": "\\frac{7}{15}", "choice_answer": ""}
|
||||
{"idx": 89, "question": "记 S_{n} 为数列 \\left{a_{n}\\right} 前 n 项和, 已知 4 S_{n}=3 a_{n}+4. (1) 求 \\left{a_{n}\\right} 的通项公式; (2) 设 b_{n}=(-1)^{n-1} n a_{n}, 求数列 \\left{b_{n}\\right} 的前 n 项和 T_{n}.", "answer": "a_{n}=4 \\cdot(-3)^{n-1},T_{n}=(2 n-1) \\cdot 3^{n}+1 ", "choice_answer": ""}
|
||||
{"idx": 90, "question": "在直角坐标系 x O y 中, 以坐标原点为极点, x 轴正半轴为极轴建立极坐标系, 曲线 C 的极坐标方程为 \\rho=\\rho \\cos \\theta+1. (1) 写出 C 的直角坐标方程; (2) 设直线 l:\\left{\\begin{array}{l}x=t \\ y=t+a\\end{array}\\right. ( t 为参数), 若 C 与 l 相交于 A 、 B 两点, 若 |A B|=2, 求 a.", "answer": "y^{2}=2 x+1,a=\\frac{3}{4}", "choice_answer": ""}
|
||||
@@ -0,0 +1,118 @@
|
||||
{"passage": null, "question": "已知 $a \\in \\mathrm{R}$, 函数 $f(x)=\\left\\{\\begin{array}{l}x^{2}-4, x>2 \\\\ |x-3|+a, x \\leq 2,\\end{array}\\right.$ 若 $f[f(\\sqrt{6})]=3$, 则 $a=(\\quad)$", "options": null, "label": null, "answer": "2", "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "已知多项式 $(x-1)^{3}+(x+1)^{4}=x^{4}+a_{1} x^{3}+a_{2} x^{2}+a_{3} x+a_{4}$, 则 $a_{1}=(\\quad)$, $a_{2}+a_{3}+a_{4}=(\\quad)$", "options": null, "label": null, "answer": "$5$;$10$", "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "袋中有 4 个红球 $m$ 个黄球, $n$ 个绿球. 现从中任取两个球, 记取出的红球数为 $\\xi$, 若取出的两个球都是 红球的概率为 $\\frac{1}{6}$, 一红一黄的概率为 $\\frac{1}{3}$, 则 $m-n=(\\quad)$, $E(\\xi)=(\\quad)$", "options": null, "label": null, "answer": "1;$\\frac{8}{9}$", "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "已知椭圆 $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$, 焦点 $F_{1}(-c, 0)$, $F_{2}(c, 0)(c>0)$, 若过 $F_{1}$ 的直线和圆 $\\left(x-\\frac{1}{2} c\\right)^{2}+y^{2}=c^{2}$ 相切, 与椭圆在第一象限交于点 $P$, 且 $P F_{2} \\perp x$ 轴, 则该直线的斜率是 $(\\quad)$, 椭圆的离心率是 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{2\\sqrt{5}}{5}$;$\\frac{\\sqrt{5}}{5}$", "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "记 $S_{n}$ 为数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和. 若 $S_{n}=2 a_{n}+1$, 则 $S_{6}=(\\quad)$.", "options": null, "label": null, "answer": "-63", "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "（5 分) 从 2 位女生, 4 位男生中选 3 人参加科技比赛, 且至少有 1 位女生 入选, 则不同的选法共有 $(\\quad)$ 种. (用数字填写答案)", "options": null, "label": null, "answer": "16", "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "（5 分）已知函数 $f(x)=2 \\sin x+\\sin 2 x$, 则 $f(x)$ 的最小值是 $(\\quad)$.", "options": null, "label": null, "answer": "$-\\frac{3\\sqrt{3}}{2}$", "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$(x+a)^{10}$ 的展开式中, $x^{7}$ 的系数为 15 , 则 $a=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{1}{2}$", "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "（5 分）已知 $\\alpha$ 是第三象限角, $\\sin \\alpha=-\\frac{1}{3}$, 则 $\\cot\\alpha=(\\quad)$.", "options": null, "label": null, "answer": "$2\\sqrt{2}$", "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "函数 $y=\\sin x-\\sqrt{3} \\cos x$ 的图象可由函数 $y=\\sin x+\\sqrt{3} \\cos x$ 的图象至少向 右平移 $(\\quad)$ 个单位长度得到.", "options": null, "label": null, "answer": "$\\frac{2\\pi}{3}$", "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "（5 分) 已知 $f(x)$ 为偶函数, 当 $x<0$ 时, $f(x)=\\ln (-x)+3 x$, 则曲线 $y=f(x)$ 在点 $(1,-3)$ 处的切线方程是 $(\\quad)$.", "options": null, "label": null, "answer": "$2x+y+1=0$", "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知直线 $\\mid: m x+y+3 m-\\sqrt{3}=0$ 与圆 $x^{2}+y^{2}=12$ 交于 $A, B$ 两点, 过 $A$,$B$ 分别作 $\\mid$ 的垂线与 $x$ 轴交于 $C, D$ 两点, 若 $|A B|=2 \\sqrt{3}$, 则 $|C D|=4$.", "options": null, "label": null, "answer": "4", "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知单位向量 $a, b$ 的夹角为 $45^{\\circ}, k a-b$ 与 $a$ 垂直, 则 $k=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{\\sqrt{2}}{2}$", "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "设复数 $z_{1}, z_{2}$ 满足 $\\left|z_{1}\\right|=\\left|z_{2}\\right|=2, \\quad z_{1}+z_{2}=\\sqrt{3}+i$, 则 $\\left|z_{1}-z_{2}\\right|=(\\quad)$", "options": null, "label": null, "answer": "$2\\sqrt{3}$", "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "已知双曲线 $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的离心率为 2 , 则该双曲线的渐近线方程为 $(\\quad)$", "options": null, "label": null, "answer": "$y=\\pm\\sqrt{3}x$", "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "已知向量 $\\vec{a}+\\vec{b}+\\vec{c}=\\overrightarrow{0},|\\vec{a}|=1,|\\vec{b}|=|\\vec{c}|=2, \\vec{a} \\cdot \\vec{b}+\\vec{b} \\cdot \\vec{c}+\\vec{c} \\cdot \\vec{a}=(\\quad)$", "options": null, "label": null, "answer": "$-\\frac{9}{2}$", "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=\\left|e^{x}-1\\right|, x_{1}<0, x_{2}>0$, 函数 $f(x)$ 的图象在点 $A\\left(x_{1}, f\\left(x_{1}\\right)\\right)$ 和点 $B\\left(x_{2}, f\\left(x_{2}\\right)\\right)$ 的两条 切线互相垂直, 且分别交 $y$ 轴于 $M, N$ 两点, 则 $\\frac{|A M|}{|B N|}$ 取值范围是 $(\\quad)$.", "options": null, "label": null, "answer": "$(0,1)$", "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "（5 分) 不等式 $\\sqrt{2 x^{2}+1}-x \\leqslant 1$ 的解集是 $(\\quad)$.", "options": null, "label": null, "answer": "$[0,2]$", "other": {"source": "2010年数学试卷（理科）（大纲版ⅰ）"}}
|
||||
{"passage": null, "question": "设向量 $\\vec{a}=(1,2), \\vec{b}=(2,3)$, 若向量 $\\lambda \\vec{a}+\\vec{b}$ 与向量 $\\vec{c}=(-4,-7)$ 共 线, 则 $\\lambda=(\\quad)$. ", "options": null, "label": null, "answer": "2", "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "设曲线 $y=e^{a x}$ 在点 $(0,1)$ 处的切线与直线 $x+2 y+1=0$ 垂直, 则 $ a=(\\quad)$.", "options": null, "label": null, "answer": "2", "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $F$ 是抛物线 $C: y^{2}=4 x$ 的焦点, 过 $F$ 且斜率为 1 的直线交 $C$ 于 $A$, B 两点. 设 $|F A|>|F B|$, 则 $|F A|$ 与 $|F B|$ 的比值等于 ($\\quad$).", "options": null, "label": null, "answer": "$3+2\\sqrt{2}$", "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "平面内的一个四边形为平行四边形的充要条件有多个, 如两组对边 分别平行, 类似地, 写出空间中的一个四棱柱为平行六面体的两个充要条件: \\\\\n充要条件$\\textcircled{1}(\\quad)$;\\\\\n充要条件$\\textcircled{2}(\\quad)$.\\\\\n(写出你认为正确的两个充要条件)", "options": null, "label": null, "answer": "三组对面分别平行的四棱柱为平行六面体;平行六面体的对角线交于一点,并且在交点处互相平分", "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "$(x-y)(x+y)^{8}$ 的展开式中 $x^{2} y^{7}$ 的系数为 $(\\quad)$. (用数字填写答 案)", "options": null, "label": null, "answer": "-20", "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "甲、乙、丙三位同学被问到是否去过 A, B, C三个城市时,\\\\\n甲说: 我去过的城市比乙多, 但没去过 $\\mathrm{B}$ 城市;\\\\\n乙说: 我没去过 C 城市;\\\\\n丙说: 我们三人去过同一城市;\\\\\n由此可判断乙去过的城市为 $(\\quad)$.", "options": null, "label": null, "answer": "$A$", "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知 $A, B, C$ 为圆 $O$ 上的三点, 若 $\\overrightarrow{\\mathrm{AO}}=\\frac{1}{2}(\\overrightarrow{\\mathrm{AB}}+\\overrightarrow{\\mathrm{AC}})$, 则 $\\overrightarrow{\\mathrm{AB}}$ 与 $\\overrightarrow{\\mathrm{AC}}$ 的夹 角为 $(\\quad)$.", "options": null, "label": null, "answer": "$90^{\\circ}$", "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知 $a, b, c$ 分别为 $\\triangle A B C$ 的三个内角 $A, B, C$ 的对边, $a=2$ 且 $(2+b)(\\sin A-\\sin B)=(c-b) \\sin C$, 则 $\\triangle A B C$ 面积的最大值为 $(\\quad)$.", "options": null, "label": null, "answer": "$\\sqrt{3}$", "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "曲线 $y=2 \\ln (x+1)$ 在点 $(0,0)$ 处的切线方程为 $(\\quad)$.", "options": null, "label": null, "answer": "$y=2x$", "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $\\sin \\alpha+\\cos \\beta=1, \\cos \\alpha+\\sin \\beta=0$, 则 $\\sin (\\alpha+\\beta)=(\\quad)$.", "options": null, "label": null, "answer": "$-\\frac{1}{2}$", "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知圆雉的顶点为 $\\mathrm{S}$, 母线 $\\mathrm{SA}, \\mathrm{SB}$ 所成角的余弦值为 $\\frac{7}{8}, \\mathrm{SA}$ 与圆 锥底面所成角为 $45^{\\circ}$, 若 $\\triangle \\mathrm{SAB}$ 的面积为 $5 \\sqrt{15}$, 则该圆雉的侧面积为 $(\\quad)$.", "options": null, "label": null, "answer": "$40\\sqrt{2}\\pi$", "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "曲线 $y=\\frac{2 x-1}{x+2}$ 在点 $(-1,-3)$ 处的切线方程为 $(\\quad)$", "options": null, "label": null, "answer": "$5x-y+2=0$", "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "已知 $F_{1}, F_{2}$ 为椭圆 $C: \\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ 的 两个焦点, $P, Q$ 为 $C$ 上关于坐标原点对称的两点, 且 $|P Q|=\\left|F_{1} F_{2}\\right|$, 则四边形 $P F_{1} Q F_{2}$ 的面积为 $(\\quad)$", "options": null, "label": null, "answer": "8", "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "已知 $\\mathrm{a}$ 是第二象限的角, $\\tan (\\pi+2 \\alpha)=-\\frac{4}{3}$, 则 $\\tan \\alpha=(\\quad)$.", "options": null, "label": null, "answer": "$-\\frac{1}{2}$", "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "若 $\\left(x-\\frac{a}{x}\\right){ }^{9}$ 的展开式中 $x^{3}$ 的系数是 -84 , 则 $a=(\\quad)$.", "options": null, "label": null, "answer": "1", "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "已知抛物线 $C: y^{2}=2 p x(p>0)$ 的准线 $\\mid$, 过 $M(1,0)$ 且斜率为 $\\sqrt{3}$ 的直线与 $\\mid$ 相交于 $A$, 与 $C$ 的一个交点为 $B$, 若 $\\overrightarrow{\\mathrm{AM}}=\\overrightarrow{\\mathrm{MB}}$, 则 $p=(\\quad)$.", "options": null, "label": null, "answer": "2", "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "已知双曲线$C:\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1\\left(a\\textgreater0,b\\textgreater0\\right)$的右顶点为 $A$, 以 $A$ 为 圆心, $b$ 为半径作圆 $A$, 圆 $A$ 与双曲线 $C$ 的一条渐近线交于 $M$、 $N$ 两点. 若 $\\angle \\mathrm{MAN}=60^{\\circ}$, 则 $\\mathrm{C}$ 的离心率为 $(\\quad)$.", "options": null, "label": null, "answer": "$\\frac{2\\sqrt{3}}{3}$", "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "若变量 $x, y$ 满足约束条件 $\\left\\{\\begin{array}{l}3 \\leqslant 2 x+y \\leqslant 9 \\\\ 6 \\leqslant x-y \\leqslant 9\\end{array}\\right.$, 则 $z=x+2 y$ 的最小值为 $(\\quad)$. ", "options": null, "label": null, "answer": "-6", "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "已知矩形 $A B C D$ 的顶点都在半径为 4 的球 $O$ 的球面上, 且 $A B=6$, $B C=2 \\sqrt{3}$, 则棱雉 $O-A B C D$ 的体积为 $(\\quad)$", "options": null, "label": null, "answer": "$8\\sqrt{3}$", "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "在 $\\triangle A B C$ 中, $B=60^{\\circ}, A C=\\sqrt{3}$, 则 $A B+2 B C$ 的最大值为 $(\\quad)$.", "options": null, "label": null, "answer": "$2\\sqrt{7}$", "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "已知随机变量 $X$ 服从正态分布 $N\\left(2, \\sigma^{2}\\right)$, 且 $P(2<X \\leq 2.5)=0.36$, 则 $P(X>2.5)=(\\quad)$", "options": null, "label": null, "answer": "$0.14$", "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "写出曲线 $y=\\ln |x|$ 过坐标原点的切线方程: $(\\quad)$, $(\\quad)$", "options": null, "label": null, "answer": "$y=\\frac{1}{\\mathrm{e}}x$;$y=-\\frac{1}{\\mathrm{e}}x$", "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "已知点 $A(-2,3), B(0, a)$, 若直线 $A B$ 关于 $y=a$ 的对称直线与圆 $(x+3)^{2}+(y+2)^{2}=1$ 存在公共点, 则实数 $a$ 的取值范围为 $(\\quad)$", "options": null, "label": null, "answer": "$\\left[\\frac{1}{3},\\frac{3}{2}\\right]$", "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "从班委会 5 名成员中选出 3 名, 分别担任班级学习委员、文娱委员与体育委员，其中甲、乙二人不能担任文娱委员，则不同的选法共有 ($\\quad$) 种. (用数字作答)", "options": null, "label": null, "answer": "$36$", "other": {"source": "2007年全国高考数学（理科）试卷（全国卷Ⅰ）"}}
|
||||
{"passage": null, "question": "等比数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和为 $S_{n}$, 已知 $S_{1}, 2 S_{2}$, $3 S_{3}$ 成等差数列, 则 $\\left\\{a_{n}\\right\\}$ 的公比为 ($\\quad$).", "options": null, "label": null, "answer": "$\\frac{1}{3}$", "other": {"source": "2007年全国高考数学（理科）试卷（全国卷Ⅰ）"}}
|
||||
{"passage": null, "question": "一个等腰直角三角形的三个顶点分别在正三棱柱 的三条侧棱上, 已知正三棱柱的底面边长为 2 , 则该三角形的斜边长为 ($\\quad$).", "options": null, "label": null, "answer": "$2\\sqrt{3}$", "other": {"source": "2007年全国高考数学（理科）试卷（全国卷Ⅰ）"}}
|
||||
{"passage": null, "question": "设向量 $\\vec{a}, \\vec{b}$ 的夹角的余弦值为 $\\frac{1}{3}$, 且 $|\\vec{a}|=1,|\\vec{b}|=3$, 则 $(2 \\vec{a}+\\vec{b}) \\cdot \\vec{b}=(\\quad)$\n<Answer>\\\\\n11\\\\\n\n\n<Next Instance>\\\\\n<Question>\\\\\n若双曲线 $y^{2}-\\frac{x^{2}}{m^{2}}=1(m>0)$ 的渐近线与圆 $x^{2}+y^{2}-4 y+3=0$ 相切, 则 $m=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{\\sqrt{3}}{3}$", "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "从正方体的 8 个顶点中任选 4 个, 则这 4 个点在同一个平面的概率为 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{6}{35}$", "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "已知 $\\triangle A B C$ 中, 点 $D$ 在边 $B C$ 上, $\\angle A D B=120^{\\circ}, A D=2, C D=2 B D$. 当 $\\frac{A C}{A B}$ 取得 最小值时, $B D=(\\quad)$", "options": null, "label": null, "answer": "$\\sqrt{3}-1$", "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "已知单位向量 $a, b$ 的夹角为 $45^{\\circ}, k a-b$ 与 $a$ 垂直, 则 $k=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{\\sqrt{2}}{2}$", "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "设复数 $z_{1}, z_{2}$ 满足 $\\left|z_{1}\\right|=\\left|z_{2}\\right|=2, \\quad z_{1}+z_{2}=\\sqrt{3}+i$, 则 $\\left|z_{1}-z_{2}\\right|=(\\quad)$", "options": null, "label": null, "answer": "$2\\sqrt{3}$", "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $\\mathrm{i}$ 是虚数单位, 化简 $\\frac{11-3 \\mathrm{i}}{1+2 \\mathrm{i}}$ 的结果为 $(\\quad)$", "options": null, "label": null, "answer": "$1-5\\mathrm{i}$", "other": {"source": "2022年新高考天津数学"}}
|
||||
{"passage": null, "question": "$\\left(\\sqrt{x}+\\frac{3}{x^{2}}\\right)^{5}$ 的展开式中的常数项为 $(\\quad)$", "options": null, "label": null, "answer": "15", "other": {"source": "2022年新高考天津数学"}}
|
||||
{"passage": null, "question": "若直线 $x-y+m=0(m>0)$ 与圆 $(x-1)^{2}+(y-1)^{2}=3$ 相交所得的弦长为 $m$, 则 $m=(\\quad)$", "options": null, "label": null, "answer": "2", "other": {"source": "2022年新高考天津数学"}}
|
||||
{"passage": null, "question": "52 张扑克牌, 没有大小王, 无放回地抽取两次, 则两次都抽到 $A$ 的概率为 $(\\quad)$；已知第一次 抽到的是 $A$, 则第二次抽取 $A$ 的概率为 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{1}{221}$;$\\frac{1}{17}$", "other": {"source": "2022年新高考天津数学"}}
|
||||
{"passage": null, "question": "一批产品的二等品率为 0.02 , 从这批产品中每次随机取一件, 有放 回地抽取 100 次. $X$ 表示抽到的二等品件数, 则 $D X=(\\quad)$", "options": null, "label": null, "answer": "1.96", "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "函数 $f(x)=\\sin ^{2} x+\\sqrt{3} \\cos x-\\frac{3}{4}\\left(x \\in\\left[0, \\frac{\\pi}{2}\\right]\\right)$ 的最大值是 $(\\quad)$.", "options": null, "label": null, "answer": "1", "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "等差数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和为 $S_{n}, a_{3}=3, S_{4}=10$, 则 $\\sum_{k=1}^{n} \\frac{1}{S_{k}}= (\\quad)$.", "options": null, "label": null, "answer": "$\\frac{2n}{n+1}$", "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "（5 分) 已知 $F$ 是抛物线 $C: y^{2}=8 x$ 的焦点, $M$ 是 $C$ 上一点, $F M$ 的延长线交 $y$ 轴于点 $N$. 若 $M$ 为 $F N$ 的中点, 则 $|F N|=(\\quad)$.", "options": null, "label": null, "answer": "6", "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "$\\left(x^{3}-\\frac{1}{x}\\right)^{4}$ 展开式中常数项为 $(\\quad)$", "options": null, "label": null, "answer": "$-4$", "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "$\\vec{a}=(2,1), \\vec{b}=(2,-1), \\vec{c}=(0,1)$, 则 $(\\vec{a}+\\vec{b}) \\cdot \\vec{c}=(\\quad)$ ;$\\vec{a} \\cdot \\vec{b}=(\\quad)$", "options": null, "label": null, "answer": "0;3", "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "从甲、乙等 5 名同学中随机选 3 名参加社区服务工作, 则甲、乙都入选的概率为 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{3}{10}$", "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "记函数 $f(x)=\\cos (\\omega x+\\varphi)(\\omega>0,0<\\varphi<\\pi)$ 的最小正周期为 $T$, 若 $f(T)=\\frac{\\sqrt{3}}{2}$, $x=\\frac{\\pi}{9}$ 为 $f(x)$ 的零点, 则 $\\omega$ 的最小值为 $(\\quad)$", "options": null, "label": null, "answer": "3", "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知 $x=x_{1}$ 和 $x=x_{2}$ 分别是函数 $f(x)=2 a^{x}-e x^{2}(a>0$ 且 $a \\neq 1)$ 的极小值点和极 大值点. 若 $x_{1}<x_{2}$, 则 $a$ 的取值范围是 $(\\quad)$", "options": null, "label": null, "answer": "$\\left(\\frac{1}{\\mathrm{e}},1\\right)$", "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知正方形 $A B C D$ 的边长为 $2, E$ 为 $C D$ 的中点, 则 $\\overrightarrow{\\mathrm{AE}} \\cdot \\overrightarrow{\\mathrm{BD}}=(\\quad)$.", "options": null, "label": null, "answer": "2", "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "从 $\\mathrm{n}$ 个正整数 $1,2, \\ldots, n$ 中任意取出两个不同的数, 若取出的两 数之和等于 5 的概率为 $\\frac{1}{14}$, 则 $n=(\\quad)$.", "options": null, "label": null, "answer": "8", "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "等差数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和为 $S_{n}$, 已知 $S_{10}=0, S_{15}=25$, 则 $n S_{n}$ 的最小值为$(\\quad)$.", "options": null, "label": null, "answer": "-49", "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "$i$ 是虚数单位, 复数 $\\frac{9+2 i}{2+i}=(\\quad)$.", "options": null, "label": null, "answer": "$4-\\mathrm{i}$", "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "在 $\\left(2 x^{3}+\\frac{1}{x}\\right)^{6}$ 的展开式中, $x^{6}$ 的系数是 $(\\quad)$.", "options": null, "label": null, "answer": "160", "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "若斜率为 $\\sqrt{3}$ 的直线与 $y$ 轴交于点 $\\mathrm{A}$, 与圆 $x^{2}+(y-1)^{2}=1$ 相切于点 $B$, 则 $|A B|=(\\quad)$.", "options": null, "label": null, "answer": "$\\sqrt{3}$", "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "若 $a>0, b>0$, 则 $\\frac{1}{a}+\\frac{a}{b^{2}}+b$ 的最小值为 $(\\quad)$.", "options": null, "label": null, "answer": "$2\\sqrt{2}$", "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "甲、乙两人在每次猜谜活动中各猜一个谜语, 若一方猜对且另一方猜错, 则猜对的 一方获胜, 否则本次平局, 已知每次活动中, 甲、乙猜对的概率分别为 $\\frac{5}{6}$ 和 $\\frac{1}{5}$, 且每 次活动中甲、乙猜对与否互不影响, 各次活动也互不影响, 则一次活动中, 甲获胜的概率为 $(\\quad)$, 3 次活动中, 甲至少获胜 2 次的概率为 $(\\quad)$.", "options": null, "label": null, "answer": "$\\frac{2}{3}$;$\\frac{20}{27}$", "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "在边长为 1 的等边三角形 $A B C$ 中, $D$ 为线段 $B C$ 上的动点, $D E \\perp A B$ 且交 $A B$ 于点E.$D F / / A B$ 且交 $A C$ 于点 $F$, 则 $|2 \\overrightarrow{B E}+\\overrightarrow{D F}|$ 的值为 $(\\quad)$; $(\\overrightarrow{D E}+\\overrightarrow{D F}) \\cdot \\overrightarrow{D A}$ 的最小值为 $(\\quad)$.", "options": null, "label": null, "answer": "1;$\\frac{11}{20}$", "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "$(x \\sqrt{y}-y \\sqrt{x})^{4}$ 的展开式中 $x^{3} y^{3}$ 的系数为($\\quad$).", "options": null, "label": null, "answer": "6", "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "若函数 $f(x)=x \\ln \\left(x+\\sqrt{a+x^{2}}\\right)$ 为偶函数, 则 $a=(\\quad)$.", "options": null, "label": null, "answer": "1", "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "一个圆经过椭圆 $\\frac{x^{2}}{16}+\\frac{y^{2}}{4}=1$ 的三个顶点. 且圆心在 $x$ 轴的正半轴 上. 则该圆标准方程为 $(\\quad)$.", "options": null, "label": null, "answer": "$\\left(x-\\frac{3}{2}\\right)^{2}+y^{2}=\\frac{25}{4}$", "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "函数 $f(x)=\\frac{1}{x}+\\sqrt{1-x}$ 的定义域是 $(\\quad)$", "options": null, "label": null, "answer": "$(-\\infty,0)\\cup(0,1]$", "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "已知双曲线 $y^{2}+\\frac{x^{2}}{m}=1$ 的渐近线方程为 $y=\\pm \\frac{\\sqrt{3}}{3} x$, 则 $m=(\\quad)$", "options": null, "label": null, "answer": "$-3$", "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "若函数 $f(x)=A \\sin x-\\sqrt{3} \\cos x$ 的一个零点为 $\\frac{\\pi}{3}$, 则 $A=(\\quad)$; $f\\left(\\frac{\\pi}{12}\\right)=(\\quad)$", "options": null, "label": null, "answer": "1;$-\\sqrt{2}$", "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "已知数列 $\\left\\{a_{n}\\right\\}$ 各项均为正数, 其前 $n$ 项和 $S_{n}$ 满足 $a_{n} \\cdot S_{n}=9(n=1,2, \\cdots)$. 给出下列四个结论:\\\\\n$\\textcircled{1}$ $\\left\\{a_{n}\\right\\}$ 的第 2 项小于$3$;\\\\\n$\\textcircled{2}$ $\\left\\{a_{n}\\right\\}$ 为等比数列;\\\\\n$\\textcircled{3}$ $\\left\\{a_{n}\\right\\}$ 为递减数列;\\\\\n$\\textcircled{4}$ $\\left\\{a_{n}\\right\\}$ 中存在小于 $\\frac{1}{100}$ 的项.\\\\\n其中所有正确结论的序号是 $(\\quad)$", "options": null, "label": null, "answer": "$\\textcircled{1}\\textcircled{3}\\textcircled{4}$", "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "$\\left(1-\\frac{y}{x}\\right)(x+y)^{8}$ 的展开式中 $x^{2} y^{6}$ 的系数为 $(\\quad)$(用数字作答).", "options": null, "label": null, "answer": "$-28$", "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "若曲线 $y=(x+a) \\mathrm{e}^{x}$ 有两条过坐标原点的切线, 则 $a$ 的取值范围是 $(\\quad)$.", "options": null, "label": null, "answer": "$(-\\infty,-4)\\cup(0,+\\infty)$", "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "我国高铁发展迅速, 技术先进. 经统计, 在经停某站的高铁列车中, 有 10 个车次的正点 率为 $0.97$, 有 20 个车次的正点率为 $0.98$, 有 10 个车次的正点率为 $0.99$, 则经停该站高铁列 车所有车次的平均正点率的估计值为 $(\\quad)$", "options": null, "label": null, "answer": "0.98", "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "$\\triangle A B C$ 的内角 $A, B, C$ 的对边分别为 $a, b, c$. 若 $b=6, a=2 c, B=\\frac{\\pi}{3}$, 则 $\\triangle A B C$ 的面积为 $(\\quad)$", "options": null, "label": null, "answer": "$6\\sqrt{3}$", "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "若 $x, y$ 满足约束条件 $\\left\\{\\begin{array}{l}x-y \\geqslant 0 \\\\ x+y-2 \\leqslant 0 \\\\ y \\geqslant 0\\end{array}\\right.$, 则 $z=3 x-4 y$ 的最小值为 $(\\quad)$.", "options": null, "label": null, "answer": "-1", "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "(5 分） 设等比数列 $\\left\\{a_{n}\\right\\}$ 满足 $a_{1}+a_{2}=-1, a_{1}-a_{3}=-3$, 则 $a_{4}=(\\quad)$.", "options": null, "label": null, "answer": "-8", "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=\\left\\{\\begin{array}{ll}x+1, & x \\leqslant 0 \\\\ 2^{x}, & x>0\\end{array}\\right.$, 则满足 $f(x)+f\\left(x-\\frac{1}{2}\\right)>1$ 的 $x$ 的 取值范围是 $(\\quad)$.", "options": null, "label": null, "answer": "$\\left(-\\frac{1}{4},+\\infty\\right)$", "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "$\\left(x^{2}+\\frac{2}{x}\\right)^{6}$ 的展开式中常数项是 $(\\quad)$ (用数字作答).", "options": null, "label": null, "answer": "240", "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "关于函数 $f(x)=\\sin x+\\frac{1}{\\sin x}$ 有如下四个命题:\\\\\n$\\textcircled{1}$ $f(x)$ 的图像关于 $y$ 轴对称.\\\\\n$\\textcircled{2}$ $f(x)$ 的图像关于原点对称.\\\\\n$\\textcircled{3}$ $f(x)$ 的图像关于直线 $x=\\frac{\\pi}{2}$ 对称.\\\\\n$\\textcircled{4}$ $f(x)$ 的最小值为2.\\\\\n其中所有真命题的序号是 $(\\quad)$", "options": null, "label": null, "answer": "$\\textcircled{2}\\textcircled{3}$", "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "曲线 $y=3\\left(x^{2}+x\\right) \\mathrm{e}^{x}$ 在点 $(0,0)$ 处的切线方程为 $(\\quad)$", "options": null, "label": null, "answer": "$3x-y=0$", "other": {"source": "2019年新课标ⅰ数学"}}
|
||||
{"passage": null, "question": "记 $S_{n}$ 为等比数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和. 若 $a_{1}=\\frac{1}{3}, a_{4}^{2}=a_{6}$, 则 $S_{5}=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{121}{3}$", "other": {"source": "2019年新课标ⅰ数学"}}
|
||||
{"passage": null, "question": "甲、乙两队进行篮球决赛, 采取七场四胜制(当一队赢得四场胜利时, 该队获胜, 决赛 结束). 根据前期比赛成绩, 甲队的主客场安排依次为“主主客客主客主”. 设甲队主场取胜 的概率为 $0.6$, 客场取胜的概率为 $0.5$, 且各场比赛结果相互独立, 则甲队以 $4: 1$ 获胜的概率是 $(\\quad)$", "options": null, "label": null, "answer": "$0.216$", "other": {"source": "2019年新课标ⅰ数学"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=x^{3}\\left(a \\cdot 2^{x}-2^{-x}\\right)$ 是偶函数, 则 $a=(\\quad)$", "options": null, "label": null, "answer": "1", "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "已知 $O$ 为坐标原点, 抛物线 $C: y^{2}=2 p x(p>0)$ 的焦点为 $F, P$ 为 $C$ 上一点, $P F$ 与 $x$ 轴垂直, $Q$ 为 $x$ 轴上一点, 且 $P Q \\perp O P$, 若 $|F Q|=6$, 则 $C$ 的准线方程为 $(\\quad)$", "options": null, "label": null, "answer": "$x=-\\frac{3}{2}$", "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "函数 $f(x)=|2 x-1|-2 \\ln x$ 的最小值为 $(\\quad)$", "options": null, "label": null, "answer": "1", "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "某校学生在研究民间剪纸艺术时, 发现剪纸时经常会沿纸的某条对称轴把纸对折, 规格为 $20 \\mathrm{dm} \\times 12 \\mathrm{dm}$ 的长方形纸, 对折 1 次共可以得到 $10 \\mathrm{dm} \\times 12 \\mathrm{dm}, 20 \\mathrm{dm} \\times 6 \\mathrm{dm}$ 两种规格的图形, 它们的面积之和 $S_{1}=240 \\mathrm{dm}^{2}$, 对折 2 次共可以得到 $5 \\mathrm{dm} \\times 12 \\mathrm{dm}, 10 \\mathrm{dm} \\times 6 \\mathrm{dm}, 20 \\mathrm{dm} \\times 3 \\mathrm{dm}$ 三种规格的图形, 它们的 面积之和 $S_{2}=180 \\mathrm{dm}^{2}$, 以此类推, 则对折 4 次共可以得到不同规格图形的种数为 $(\\quad)$; 如果对折 $n$ 次, 那么 $\\sum_{k=1}^{n} S_{k}=(\\quad)\\mathrm{dm}^{2}$.", "options": null, "label": null, "answer": "5;$720-\\frac{15(3+n)}{2^{n-4}}$", "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "$\\left(\\frac{x}{\\sqrt{y}}-\\frac{y}{\\sqrt{x}}\\right)^{8}$ 的展开式中 $x^{2} y^{2}$ 的系数为$(\\quad)$. (用数字作答)", "options": null, "label": null, "answer": "70", "other": {"source": "2014年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "直线 $l_{1}$ 和 $l_{2}$ 是圆 $x^{2}+y^{2}=2$ 的两条切线, 若 $l_{1}$ 与 $l_{2}$ 的交点为 $(1,3)$, 则 $I_{1}$ 与 $I_{2}$ 的夹角的正切值等于 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{4}{3}$", "other": {"source": "2014年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "若函数 $f(x)=\\cos 2 x+a \\sin x$ 在区间 $\\left(\\frac{\\pi}{6}, \\frac{\\pi}{2}\\right.$ ) 是减函数, 则 $a$ 的取值范围是 $(\\quad)$.", "options": null, "label": null, "answer": "$(-\\infty,2]$", "other": {"source": "2014年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "$(a+x)(1+x)^{4}$ 的展开式中 $x$ 的奇数次幂项的系数之和为 32 , 则 $a=(\\quad)$.", "options": null, "label": null, "answer": "3", "other": {"source": "2015年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设 $a, b$ 为单位向量, 且 $|\\boldsymbol{a}+\\boldsymbol{b}|=1$, 则 $|a-b|=(\\quad)$", "options": null, "label": null, "answer": "$\\sqrt{3}$", "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "已知 $F$ 为双曲线 $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的右焦点, $A$ 为 $C$ 的右顶点, $B$ 为 $C$ 上的点, 且 $B F$ 垂直于 $x$ 轴.若 $A B$ 的斜率为 3 , 则 $C$ 的离心率为 $(\\quad)$", "options": null, "label": null, "answer": "2", "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "当函数 $y=\\sin x-\\sqrt{3} \\cos x(0 \\leqslant x<2 \\pi)$ 取得最大值时, $x=(\\quad)$.", "options": null, "label": null, "answer": "$\\frac{5\\pi}{6}$", "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "我国南宋著名数学家秦九韶, 发现了从三角形三边求面积的公式, 他把这种方法称为 “三斜求积”, 它填补了我国传统数学的一个空白. 如果把这个方法写成公式, 就是 $S=\\sqrt{\\frac{1}{4}\\left[c^{2} a^{2}-\\left(\\frac{c^{2}+a^{2}-b^{2}}{2}\\right)^{2}\\right]}$, 其中 $a, b, c$ 是三角形的三边, $S$ 是三角形的面积. 设 某三角形的三边 $a=\\sqrt{2}, b=\\sqrt{3}, c=2$, 则该三角形的面积 $S=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{\\sqrt{23}}{4}$", "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "已知多项式 $(x+2)(x-1)^{4}=a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}$, 则 $a_{2}=(\\quad)$, $a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=(\\quad)$", "options": null, "label": null, "answer": "8;$-2$", "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "若 $3 \\sin \\alpha-\\sin \\beta=\\sqrt{10}, \\alpha+\\beta=\\frac{\\pi}{2}$, 则 $\\sin \\alpha=(\\quad)$, $\\cos 2 \\beta=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{3\\sqrt{10}}{10}$;$\\frac{4}{5}$", "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=\\left\\{\\begin{array}{l}-x^{2}+2, x \\leq 1, \\\\ x+\\frac{1}{x}-1, x>1,\\end{array}\\right.$ 则 $f\\left(f\\left(\\frac{1}{2}\\right)\\right)=(\\quad)$; 若当 $x \\in[a, b]$ 时, $1 \\leq f(x) \\leq 3$, 则 $b-a$ 的最大值是 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{37}{28}$;$3+\\sqrt{3}$", "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "现有 7 张卡片, 分别写上数字 $1,2,2,3,4,5,6$. 从这 7 张卡片中随机抽取 3 张, 记所抽取卡片上数字的最小值为 $\\xi$, 则 $P(\\xi=2)=(\\quad)$ , $E(\\xi)=(\\quad)$", "options": null, "label": null, "answer": "$\\frac{16}{35}$;$\\frac{12}{7}$", "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "已知双曲线 $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的左焦点为 $F$, 过 $F$ 且斜率为 $\\frac{b}{4 a}$ 的直线交双曲线于 点 $A\\left(x_{1}, y_{1}\\right)$, 交双曲线的渐近线于点 $B\\left(x_{2}, y_{2}\\right)$ 且 $x_{1}<0<x_{2}$. 若 $|F B|=3|F A|$, 则双曲 线的离心率是 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{3\\sqrt{6}}{4}$", "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "函数 $f(x)=\\frac{1}{x+1}+\\ln x$ 的定义域是 $(\\quad)$", "options": null, "label": null, "answer": "$(0,+\\infty)$", "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "已知双曲线 $C: \\frac{x^{2}}{6}-\\frac{y^{2}}{3}=1$, 则 $C$ 的右焦点的坐标为 $(\\quad)$ ; $C$ 的焦点到其渐近线的距离是 $(\\quad)$", "options": null, "label": null, "answer": "$\\left(3,0\\right)$;$\\sqrt{3}$", "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "若函数 $f(x)=\\sin (x+\\varphi)+\\cos x$ 的最大值为 2 , 则常数 $\\varphi$ 的一个取值为 $(\\quad)$", "options": null, "label": null, "answer": "$\\frac{\\pi}{2}$", "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "（5 分）已知向量 $\\vec{a}=(1,2), \\vec{b}=(2,-2), \\vec{c}=(1, \\lambda)$. 若 $\\vec{c} / /(2 \\vec{a}+\\vec{b})$, 则 $\\lambda=(\\quad)$.", "options": null, "label": null, "answer": "$\\frac{1}{2}$", "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "曲线 $y=(a x+1) e^{x}$ 在点 $(0,1)$ 处的切线的斜率为 -2 , 则 $a=(\\quad)$.", "options": null, "label": null, "answer": "-3", "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "函数 $f(x)=\\cos \\left(3 x+\\frac{\\pi}{6}\\right)$ 在 $[0, \\pi]$ 的零点个数为 $(\\quad)$.", "options": null, "label": null, "answer": "3", "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知点 $M(-1,1)$ 和抛物线 $C: y^{2}=4 x$, 过 $C$ 的焦点且斜率为 $k$ 的 直线与 $C$ 交于 $A, B$ 两点. 若 $\\angle A M B=90^{\\circ}$, 则 $k=(\\quad)$.", "options": null, "label": null, "answer": "2", "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "将数列 $\\{2 n-1\\}$ 与 $\\{3 n-2\\}$ 的公共项从小到大排列得到数列 $\\left\\{a_{n}\\right\\}$, 则 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和为 $(\\quad)$", "options": null, "label": null, "answer": "$3n^{2}-2n$", "other": {"source": "2020年高考真题数学【新高考全国Ⅰ卷】(山东卷)"}}
|
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{"passage": null, "question": "设向量 $\\vec{a}=(m, 1), \\vec{b}=(1,2)$, 且 $|\\vec{a}+\\vec{b}|^{2}=|\\vec{a}|^{2}+|\\vec{b}|^{2}$, 则 $m=(\\quad)$", "options": null, "label": null, "answer": "-2", "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$(2 x+\\sqrt{x})^{5}$ 的展开式中, $x^{3}$ 的系数是 $(\\quad)$. (用数字填写答案)", "options": null, "label": null, "answer": "10", "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设等比数列 $\\left\\{a_{n}\\right\\}$ 满足 $a_{1}+a_{3}=10, a_{2}+a_{4}=5$, 则 $a_{1} a_{2} \\ldots a_{n}$ 的最大值为 $(\\quad)$.", "options": null, "label": null, "answer": "64", "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
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@@ -0,0 +1,351 @@
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{"passage": null, "question": "设集合 $A=\\{x \\mid x \\geq 1\\}, B=\\{x \\mid-1<x<2\\}$, 则 $A \\cap B=$ ( )", "options": {"A": "$\\{x \\mid x>-1\\}$", "B": "$\\{x \\mid x \\geq 1\\}$", "C": "$\\{x \\mid-1<x<1\\}$", "D": "$\\{x \\mid 1 \\leq x<2\\}$"}, "label": "D", "answer": null, "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "已知 $a \\in R,(1+a i) i=3+i$, ( $i$ 为虚数单位), 则 $a=( )", "options": {"A": "$-1$", "B": "1", "C": "$-3$", "D": "3"}, "label": "C", "answer": null, "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "已知非零向量 $\\vec{a}, \\vec{b}, \\vec{c}$, 则“ $\\vec{a} \\cdot \\vec{c}=\\vec{b} \\cdot \\vec{c}$ ”是“ $\\vec{a}=\\vec{b}$ ”的 ( )", "options": {"A": "充分不必要条件", "B": "必要不充分条件", "C": "充分必要条件", "D": "既不充分又不必要条件"}, "label": "B", "answer": null, "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "若实数 $x, y$ 满足约束条件 $\\left\\{\\begin{array}{l}x+1 \\geq 0 \\\\ x-y \\leq 0 \\\\ 2 x+3 y-1 \\leq0\\end{array}\\right.$, 则 $z=x-\\frac{1}{2} y$ 的最小值是（( )", "options": {"A": "$-2$", "B": "$-\\frac{3}{2}$", "C": "$-\\frac{1}{2}$", "D": "$\\frac{1}{10}$"}, "label": "B", "answer": null, "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "已知 $a, b \\in \\mathrm{R}, a b>0$, 函数 $f(x)=a x^{2}+b(x \\in \\mathrm{R})$. 若 $f(s-t), f(s), f(s+t)$ 成等比数列, 则平面上点 $(s, t)$ 的轨迹是 ( )", "options": {"A": "直线和圆", "B": "直线和椭圆", "C": "直线和双曲线", "D": "直线和抛物线"}, "label": "C", "answer": null, "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "已知数列 $\\left\\{a_{n}\\right\\}$ 满足 $a_{1}=1, a_{n+1}=\\frac{a_{n}}{1+\\sqrt{a_{n}}}\\left(n \\in \\mathrm{N}^{*}\\right)$. 记数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和为 $S_{n}$, 则 ( )", "options": {"A": "$\\frac{1}{2}<S_{100}<3$", "B": "$3<S_{100}<4$", "C": "$4<S_{100}<\\frac{9}{2}$", "D": "$\\frac{9}{2}<S_{100}<5$"}, "label": "A", "answer": null, "other": {"source": "2021年浙江卷—数学"}}
|
||||
{"passage": null, "question": "设 $z=\\frac{1-i}{1+i}+2 i$, 则 $|z|=( )", "options": {"A": "0", "B": "$\\frac{1}{2}$", "C": "1", "D": "$\\sqrt{2}$"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\left\\{x \\mid x^{2}-x-2>0\\right\\}$, 则 $C_{R} A=（ ( ) ）$", "options": {"A": "$\\{x \\mid-1<x<2\\}$", "B": "$\\{x \\mid-1 \\leqslant x \\leqslant 2\\}$", "C": "$\\{x \\mid x<-1\\} \\cup\\{x \\mid x>2\\}$", "D": "$\\{x \\mid x \\leqslant -1\\} \\cup\\{x \\mid x \\geqslant 2\\}$"}, "label": "B", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "记 $S_{n}$ 为等差数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和. 若 $3 S_{3}=S_{2}+S_{4}, a_{1}=2$, 则 $a_{5}=( )", "options": {"A": "-12", "B": "-10", "C": "10", "D": "12"}, "label": "B", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "在 $\\triangle A B C$ 中, $A D$ 为 $B C$ 边上的中线, $E$ 为 $A D$ 的中点, 则 $\\overrightarrow{E B}=( )", "options": {"A": "$\\frac{3}{4} \\overrightarrow{\\mathrm{AB}}-\\frac{1}{4} \\overrightarrow{\\mathrm{AC}}$", "B": "$\\frac{1}{4} \\overrightarrow{\\mathrm{AB}}-\\frac{3}{4} \\overrightarrow{\\mathrm{AC}}$", "C": "$\\frac{3}{4} \\overrightarrow{\\mathrm{AB}}+\\frac{1}{4} \\overrightarrow{\\mathrm{AC}}$", "D": "$\\frac{1}{4} \\overrightarrow{\\mathrm{AB}}+\\frac{3}{4} \\overrightarrow{\\mathrm{AC}}$"}, "label": "A", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设抛物线 $C: y^{2}=4 x$ 的焦点为 $F$, 过点 $(-2,0)$ 且斜率为 $\\frac{2}{3}$ 的直线与 $C$ 交于 $M, N$ 两点, 则 $\\overrightarrow{F M} \\cdot \\overrightarrow{F N}=( )", "options": {"A": "5", "B": "6", "C": "7", "D": "8"}, "label": "D", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知函数 $\\left\\{\\begin{array}{l}e^{x}, x \\leqslant 0， \\\\ ln x, x>0 \\end{array}, g(x)=f(x)+x+a \\right.$.若 $g(x)$ 存在 $2$ 个零点, 则 $a$ 的取值范围是 ( )", "options": {"A": "$[-1,0)$", "B": "$[0,+\\infty)$", "C": "$[-1,+\\infty)$", "D": "$[1,+\\infty)$"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知双曲线 $C: \\frac{x^{2}}{3}-y^{2}=1, O$ 为坐标原点, $F$ 为 $C$ 的右焦点, 过 $F$ 的直线与 $C$ 的两条渐近线的交点分别为 $M, N$. 若 $\\triangle O M N$ 为直角三角形, 则 $|\\mathrm{MN}|=( )", "options": {"A": "$\\frac{3}{2}$", "B": "3", "C": "$2 \\sqrt{3}$", "D": "4"}, "label": "B", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知正方体的棱长为 1 , 每条棱所在直线与平面 $\\alpha$ 所成的角都相等, 则 $\\alpha$ 截此正方体所得截面面积的最大值为 ( )", "options": {"A": "$\\frac{3 \\sqrt{3}}{4}$", "B": "$\\frac{2 \\sqrt{3}}{3}$", "C": "$\\frac{3 \\sqrt{2}}{4}$", "D": "$\\frac{\\sqrt{3}}{2}$"}, "label": "A", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设集合 $M=\\{0,1,2\\}, N=\\left\\{x \\mid x^{2}-3 x+2 \\leqslant 0\\right\\}$, 则 $M \\cap N=( )", "options": {"A": "$\\{1\\}$", "B": "$\\{2\\}$", "C": "$\\{0,1\\}$", "D": "$\\{1,2\\}$"}, "label": "D", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设复数 $z_{1}, z_{2}$ 在复平面内的对应点关于虚轴对称, $z_{1}=2+i$, 则 $z_{1} z_{2}=$ ( )", "options": {"A": "-5", "B": "5", "C": "$-4+\\mathrm{i}$", "D": "$-4-\\mathrm{i}$"}, "label": "A", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设向量 $\\vec{a}$, $\\vec{b}$ 满足 $|\\vec{a}+\\vec{b}|=\\sqrt{10},|\\vec{a}-\\vec{b}|=\\sqrt{6}$, 则 $\\vec{a} \\vec{b}=( )", "options": {"A": "1", "B": "2", "C": "3", "D": "5"}, "label": "A", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "针角三角形 $A B C$ 的面积是 $\\frac{1}{2}, A B=1, B C=\\sqrt{2}$, 则 $A C=( )", "options": {"A": "5", "B": "$\\sqrt{5}$", "C": "2", "D": "1"}, "label": "B", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "某地区空气质量监测资料表明, 一天的空气质量为优良的概率是 0.75 , 连续两天为优良的概率是 0.6, 已知某天的空气质量为优良, 则随后 一天的空气质量为优良的概率是 ( )", "options": {"A": "0.8", "B": "0.75", "C": "0.6", "D": "0.45"}, "label": "A", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设曲线 $y=a x-\\ln (x+1)$ 在点 $(0,0)$ 处的切线方程为 $y=2 x$, 则 $a=$ ( )", "options": {"A": "0", "B": "1", "C": "2", "D": "3"}, "label": "D", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设 $x, y$ 满足约束条件 $\\left\\{\\begin{array}{l}x+y-7 \\leqslant 0 \\\\ x-3 y+1 \\leqslant 0 \\\\ 3 x-y-5 \\geqslant 0,\\end{array}\\right.$ 则 $z=2 x-y$ 的最大值为 ( )", "options": {"A": "10", "B": "8", "C": "3", "D": "2"}, "label": "B", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设 $F$ 为抛物线 $C: y^{2}=3 x$ 的焦点, 过 $F$ 且倾斜角为 $30^{\\circ}$ 的直线交 $C$ 于 $A, B$ 两点, $O$ 为坐标原点, 则 $\\triangle O A B$ 的面积为 ( )", "options": {"A": "$\\frac{3 \\sqrt{3}}{4}$", "B": "$\\frac{9 \\sqrt{3}}{8}$", "C": "$\\frac{63}{32}$", "D": "$\\frac{9}{4}$"}, "label": "D", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "直三棱柱 $A B C-A_{1} B_{1} C_{1}$ 中, $\\angle B C A=90^{\\circ}, M, N$ 分别是 $A_{1} B_{1}, A_{1} C_{1}$ 的 中点, $B C=C A=C C_{1}$, 则 $B M$ 与 $A N$ 所成角的余弦值为 ( )", "options": {"A": "$\\frac{1}{10}$", "B": "$\\frac{2}{5}$", "C": "$\\frac{\\sqrt{30}}{10}$", "D": "$\\frac{\\sqrt{2}}{2}$"}, "label": "C", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=\\sqrt{3} \\sin \\frac{\\pi x}{m}$, 若存在 $f(x)$ 的极值点 $x_{0}$ 满足 $x_{0}^{2}+[f$ $\\left.\\left(x_{0}\\right)\\right]^{2}<m^{2}$, 则 $m$ 的取值范围是 ( )", "options": {"A": "$(-\\infty,-6) \\cup(6,+\\infty)$", "B": "$(-\\infty,-4) \\cup(4,+\\infty)$", "C": "$(-\\infty,-2) \\cup(2,+\\infty)$", "D": "$(-\\infty,-1) \\cup(1,+\\infty)$"}, "label": "C", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设集合 $A=\\{1,2,3\\}, B=\\{4,5\\}, M=\\{x \\mid x=a+b, a \\in A, b \\in B\\}$, 则 $M$ 中元素的个数为 ( )", "options": {"A": "3", "B": "4", "C": "5", "D": "6"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "$(1+\\sqrt{3} i)^{3}=( )", "options": {"A": "-8", "B": "8", "C": "$-8 \\mathrm{i}$", "D": "$8 \\mathrm{i}$"}, "label": "A", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知向量 $\\vec{\\pi}=(\\lambda+1,1), \\overrightarrow{\\mathrm{n}}=(\\lambda+2,2)$, 若 $(\\vec{\\pi}+\\vec{n}) \\perp(\\vec{\\pi}-\\overrightarrow{\\mathrm{n}})$, 则 $\\lambda=( )", "options": {"A": "-4", "B": "-3", "C": "-2", "D": "-1"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)$ 的定义域为 $(-1$， $0)$, 则函数 $f(2 x+1)$ 的定义域为 ( )", "options": {"A": "$(-1,1)$", "B": "$\\left(-1,-\\frac{1}{2}\\right)$", "C": "$(-1,0)$", "D": "$\\left(\\frac{1}{2}, 1\\right)$"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "函数 $\\mathrm{f}(\\mathrm{x})=\\log _{2}\\left(1+\\frac{1}{\\mathrm{x}}\\right)(x>0)$ 的反函数 $\\mathrm{f}^{-1}(\\mathrm{x})=( )", "options": {"A": "$\\frac{1}{2^{x}-1}(x>0)$", "B": "$\\frac{1}{2^{x}-1}(x \\neq 0)$", "C": "$2^{x}-1(x \\in R)$", "D": "$2^{x}-1(x>0)$"}, "label": "A", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知数列 $\\left\\{a_{n}\\right\\}$ 满足 $3 a_{n+1}+a_{n}=0, a_{2}=-\\frac{4}{3}$, 则 $\\left\\{a_{n}\\right\\}$ 的前 10 项和等于 ( )", "options": {"A": "$-6\\left(1-3^{-10}\\right)$ ", "B": "$\\frac{1}{9}\\left(1-3^{-10}\\right)$", "C": "$3\\left(1-3^{-10}\\right)$", "D": "$3\\left(1+3^{-10}\\right)$"}, "label": "C", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "$(1+x)^{3}(1+y)^{4}$ 的展开式中 $x^{2} y^{2}$ 的系数是 ( )", "options": {"A": "5", "B": "8", "C": "12", "D": "18"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "椭圆 $C: \\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$ 的左、右顶点分别为 $A_{1}$、 $A_{2}$, 点 $P$ 在 $C$ 上且直线 $\\mathrm{PA}_{2}$ 斜率的取值范围是 $[-2,-1]$, 那么直线 $\\mathrm{PA}_{1}$ 斜率的取值范围是 ( )", "options": {"A": "$\\left[\\frac{1}{2}, \\frac{3}{4}\\right]$", "B": "$\\left[\\frac{3}{8}, \\frac{3}{4}\\right]$", "C": "$\\left[\\frac{1}{2}, 1\\right]$", "D": "$\\left[\\frac{3}{4}, 1\\right]$"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "若函数 $f(x)=x^{2}+a x+x$ 在 $\\left(\\frac{1}{2},+\\infty\\right)$ 是增函数, 则 $a$ 的取值范围是 ( )", "options": {"A": "$[-1,0]$", "B": "$[-1,+\\infty)$", "C": "$[0,3]$", "D": "$[3,+\\infty)$"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知正四棱柱 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, $A A_{1}=2 A B$, 则 $C D$ 与平面 $B D C_{1}$ 所 成角的正弦值等于 ( )", "options": {"A": "$\\frac{2}{3}$", "B": "$\\frac{\\sqrt{3}}{3}$", "C": "$\\frac{\\sqrt{2}}{3}$", "D": "$\\frac{1}{3}$"}, "label": "A", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知抛物线 $C: y^{2}=8 x$ 的焦点为 $F$, 点 $M(-2,2)$, 过点 $F$ 且斜率 为 $k$ 的直线与 $C$ 交于 $A, B$ 两点, 若 $\\overrightarrow{M A} \\cdot \\overrightarrow{M B}=0$, 则 $k=( )", "options": {"A": "$\\sqrt{2}$", "B": "$\\frac{\\sqrt{2}}{2}$", "C": "$\\frac{1}{2}$", "D": "2"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=\\cos x \\sin 2 x$, 下列结论中不正确的是 ( )", "options": {"A": "$y=f(x)$ 的图象关于 $(\\pi, 0)$ 中心对称", "B": "$y=f(x)$ 的图象关于 $x=\\frac{\\pi}{2}$ 对称", "C": "$f(x)$ 的最大值为 $\\frac{\\sqrt{3}}{2}$", "D": "$f(x)$ 既是奇函数, 又是周期函数"}, "label": "C", "answer": null, "other": {"source": "2013年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "设集合 $S=\\{x \\mid(x-2)(x-3) \\geqslant 0\\}, ~ T=\\{x \\mid x>0\\}$, 则 $S \\cap T=( )", "options": {"A": "$[2,3]$", "B": "$(-\\infty, 2] \\cup[3,+\\infty)$", "C": "$[3,+\\infty)$", "D": "$(0,2] \\cup[3,+\\infty)$"}, "label": "D", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "若 $z=1+2 i$, 则 $\\frac{4 i}{z * \\bar{z}-1}=( )", "options": {"A": "1", "B": "-1", "C": "i", "D": "- $\\mathrm{i}$"}, "label": "C", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知向量 $\\overrightarrow{B A}=\\left(\\frac{1}{2}, \\frac{\\sqrt{3}}{2}\\right), \\overrightarrow{B C}=\\left(\\frac{\\sqrt{3}}{2}, \\frac{1}{2}\\right)$, 则 $\\angle \\mathrm{ABC}=( )", "options": {"A": "$30^{\\circ}$", "B": "$45^{\\circ}$", "C": "$60^{\\circ}$", "D": "$120^{\\circ}$"}, "label": "A", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "若 $\\tan \\alpha=\\frac{3}{4}$, 则 $\\cos ^{2} \\alpha+2 \\sin 2 \\alpha=( )", "options": {"A": "$\\frac{64}{25}$", "B": "$\\frac{48}{25}$", "C": "1", "D": "$\\frac{16}{25}$"}, "label": "A", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知 $a=2^{\\frac{4}{3}}, b=3^{\\frac{2}{3}}, c=25^{\\frac{1}{3}}$, 则 ( )", "options": {"A": "$b<a<c$", "B": "$a<b<c$", "C": "$b<c<a$", "D": "$c<a<b$"}, "label": "A", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "在 $\\triangle A B C$ 中, $B=\\frac{\\pi}{4}, B C$ 边上的高等于 $\\frac{1}{3} B C$, 则 $\\cos A$ 等于 ( )", "options": {"A": "$\\frac{3 \\sqrt{10}}{10}$", "B": "$\\frac{\\sqrt{10}}{10}$", "C": "$-\\frac{\\sqrt{10}}{10}$", "D": "$-\\frac{3 \\sqrt{10}}{10}$"}, "label": "C", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "在封闭的直三棱柱 $A B C-A_{1} B_{1} C_{1}$ 内有一个体积为 $\\vee$ 的球, 若 $A B \\perp B C$, $A B=6, B C=8, ( ) A A_{1}=3$, 则 $V$ 的最大值是 ( )", "options": {"A": "$4 \\pi$", "B": "$\\frac{9 \\pi}{2}$", "C": "$6 \\pi$", "D": "$\\frac{32 \\pi}{3}$"}, "label": "B", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知 $O$ 为坐标原点, $F$ 是椭圆 $C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 \\left(a>b>0\\right)$的左焦点, $A$, $B$ 分别为 $C$ 的左, 右顶点. $P$ 为 $C$ 上一点, 且 $P F \\perp x$ 轴, 过点 $A$ 的直线 $\\mid$ 与线段 $P F$ 交于点 $M$, 与 $y$ 轴交于点 $E$. 若直线 $B M$ 经过 $O E$ 的中点, 则 $C$ 的 离心率为 ( )", "options": {"A": "$\\frac{1}{3}$", "B": "$\\frac{1}{2}$", "C": "$\\frac{2}{3}$", "D": "$\\frac{3}{4}$"}, "label": "A", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "定义 “规范 01 数列” $\\left\\{a_{n}\\right\\}$ 如下: $\\left\\{a_{n}\\right\\}$ 共有 $2 m$ 项, 其中 $m$ 项为 $0, m$ 项为 1 , 且对任意 $k \\leqslant 2 m, a_{1}, a_{2}, \\ldots, a_{k}$ 中 0 的个数不少于 1 的个数, 若 $m=4$, 则不同的“规范 01 数列”共有 ( )", "options": {"A": "18 个", "B": "16 个", "C": "14 个", "D": "12 个"}, "label": "C", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "若 $\\alpha$ 为第四象限角, 则 ( )", "options": {"A": "$\\cos 2 \\alpha>0$", "B": "$\\cos 2 \\alpha<0$", "C": "$\\sin 2 \\alpha>0$", "D": "$\\sin 2 \\alpha<0$"}, "label": "D", "answer": null, "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "在新冠肺炎疫情防控期间, 某超市开通网上销售业务, 每天能完成 1200 份订单的配货, 由 于订单量大幅增加, 导致订单积压. 为解决困难, 许多志愿者踊跃报名参加配货工作. 已知该超 市某日积压 500 份订单末配货, 预计第二天的新订单超过 1600 份的概率为 $0.05$, 志愿者每人 每天能完成 50 份订单的配货, 为使第二天完成积压订单及当日订单的配货的概率不小于 $0.95$, 则至少需要志愿者 ( )", "options": {"A": "10 名", "B": "18 名", "C": "24 名", "D": "32 名"}, "label": "B", "answer": null, "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "设 $O$ 为坐标原点, 直线 $x=a$ 与双曲线 $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的两条渐近线分别交于 $D, E$ 两点, 若 $\\square O D E$ 的面积为 8 , 则 $C$ 的焦距的最小值为 ( )", "options": {"A": "4", "B": "8", "C": "16", "D": "32"}, "label": "B", "answer": null, "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=\\ln |2 x+1|-\\ln |2 x-1|$, 则 $f(x)( )", "options": {"A": "是偶函数, 且在 $\\left(\\frac{1}{2},+\\infty\\right)$ 单调递增", "B": "是奇函数, 且在 $\\left(-\\frac{1}{2}, \\frac{1}{2}\\right)$ 单调递减", "C": "是偶函数, 且在 $\\left(-\\infty,-\\frac{1}{2}\\right)$ 单调递增", "D": "是奇函数, 且在 $\\left(-\\infty,-\\frac{1}{2}\\right)$ 单调递减"}, "label": "D", "answer": null, "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "已知 $\\triangle A B C$ 是面积为 $\\frac{9 \\sqrt{3}}{4}$ 的等边三角形, 且其顶点都在球 $O$ 的球面上. 若球 $O$ 的表面积为 $16 \\pi$, 则 $O$ 到平面 $A B C$ 的距离为 ( )", "options": {"A": "$\\sqrt{3}$", "B": "$\\frac{3}{2}$", "C": "1", "D": "$\\frac{\\sqrt{3}}{2}$"}, "label": "C", "answer": null, "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "若 $2^{x}-2^{y}<3^{-x}-3^{-y}$, 则 ( )", "options": {"A": "$\\ln (y-x+1)>0$", "B": "$\\ln (y-x+1)<0$", "C": "$\\ln |x-y|>0$", "D": "$\\ln |x-y|<0$"}, "label": "A", "answer": null, "other": {"source": "2020年新课标Ⅱ数学"}}
|
||||
{"passage": null, "question": "复数 $\\frac{2-\\mathrm{i}}{1-3 \\mathrm{i}}$ 在复平面内对应的点所在的象限为 ( )", "options": {"A": "第一象限", "B": "第二象限", "C": "第三象限", "D": "第四象限"}, "label": "A", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "设集合 $U=\\{1,2,3,4,5,6\\}, A=\\{1,3,6\\}, B=\\{2,3,4\\}$, 则 $A \\cap\\left(C_{U} B\\right)=( )", "options": {"A": "$\\{3\\}$", "B": "$\\{1,6\\}$", "C": "$\\{5,6\\}$", "D": "$\\{1,3\\}$"}, "label": "B", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "抛物线 $y^{2}=2 p x(p>0)$ 的焦点到直线 $y=x+1$ 的距离为 $\\sqrt{2}$, 则 $p=( )", "options": {"A": "1", "B": "2", "C": "$2 \\sqrt{2}$", "D": "4"}, "label": "B", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "北斗三号全球卫星导航系统是我国航天事业的重要成果. 在卫星导航系统中, 地球静止同步卫星的轨道 位于地球赤道所在平面, 轨道高度为 $36000 \\mathrm{~km}$ (轨道高度是指卫星到地球表面的距离). 将地球看作是一 个球心为 $O$, 半径 $r$ 为 $6400 \\mathrm{~km}$ 的球, 其上点 $A$ 的纬度是指 $O A$ 与赤道平面所成角的度数. 地球表面上能直 接观测到一颗地球静止同步轨道卫星点的纬度最大值为 $\\alpha$, 记卫星信号覆盖地球表面的表面积为 $S=2 \\pi r^{2}(1-\\cos \\alpha)$ （单位: $\\mathrm{km}^{2}$ ), 则 $S$ 占地球表面积的百分比约为 ( )", "options": {"A": "$26 \\%$", "B": "$34 \\%$", "C": "$42 \\%$", "D": "$50 \\%$"}, "label": "C", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "正四棱台的上、下底面的边长分别为 2,4 , 侧棱长为 2 , 则其体积为 ( )", "options": {"A": "$20+12 \\sqrt{3}$", "B": "$28 \\sqrt{2}$", "C": "$\\frac{56}{3}$", "D": "$\\frac{28 \\sqrt{2}}{3}$"}, "label": "D", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "某物理量的测量结果服从正态分布 $N\\left(10, \\sigma^{2}\\right)$, 下列结论中不正确的是 ( )", "options": {"A": "$\\sigma$ 越小, 该物理量在一次测量中在 $(9.9,10.1)$ 的概率越大", "B": "$\\sigma$ 越小, 该物理量在一次测量中大于 10 的概率为 $0.5$", "C": "$\\sigma$ 越小, 该物理量在一次测量中小于 $9.99$ 与大于 $10.01$ 的概率相等", "D": "$\\sigma$ 越小, 该物理量在一次测量中落在 $(9.9,10.2)$ 与落在 $(10,10.3)$ 的概率相等"}, "label": "D", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)$ 的定义域为 $\\mathbf{R}, f(x+2)$ 为偶函数, $f(2 x+1)$ 为奇函数, 则 ( )", "options": {"A": "$f\\left(-\\frac{1}{2}\\right)=0$", "B": "$f(-1)=0$", "C": "$f(2)=0$", "D": "$f(4)=0$"}, "label": "B", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "下列统计量中, 能度量样本 $x_{1}, x_{2}, \\cdots, x_{n}$ 的离散程度的是 ( )", "options": {"A": "样本 $x_{1}, x_{2}, \\cdots, x_{n}$ 的标准差", "B": "样本 $x_{1}, x_{2}, \\cdots, x_{n}$ 的中位数 ", "C": "样本 $x_{1}, x_{2}, \\cdots, x_{n}$ 的极差", "D": "样本 $x_{1}, x_{2}, \\cdots, x_{n}$ 的平均数"}, "label": "A", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "设正整数 $n=a_{0} \\cdot 2^{0}+a_{1} \\cdot 2+\\cdots+a_{k-1} \\cdot 2^{k-1}+a_{k} \\cdot 2^{k}$, 其中 $a_{i} \\in\\{0,1\\}$, 记 $\\omega(n)=a_{0}+a_{1}+\\cdots+a_{k}$. 则 ( )", "options": {"A": "$\\omega(2 n)=\\omega(n)$", "B": "$\\omega(2 n+3)=\\omega(n)+1$", "C": "$\\omega(8 n+5)=\\omega(4 n+3)$", "D": "$\\omega\\left(2^{n}-1\\right)=n$"}, "label": "A", "answer": null, "other": {"source": "2021全国新高考Ⅱ卷数学"}}
|
||||
{"passage": null, "question": "复数 $\\frac{3+2 i}{2-3 i}= $ ( )", "options": {"A": "i", "B": "- i", "C": "$12-13 i$", "D": "$12+13 \\mathrm{i}$"}, "label": "B", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅰ）"}}
|
||||
{"passage": null, "question": "$(1+2 \\sqrt{x}){ }^{3}(1-\\sqrt[3]{x})^{5}$ 的展开式中 $\\mathrm{x}$ 的系数是 ( )", "options": {"A": "-4", "B": "-2", "C": "2", "D": "4"}, "label": "C", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅰ）"}}
|
||||
{"passage": null, "question": "设 $a=\\log _{3} 2, b=\\ln 2, c=5^{-\\frac{1}{2}}$, 则 ( )", "options": {"A": "$a<b<c$", "B": "$b<c<a$", "C": "$c<a<b$", "D": "$c<b<a$"}, "label": "C", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅰ）"}}
|
||||
{"passage": null, "question": "已知圆 $O$ 的半径为 $1, P A$、 $P B$ 为该圆的两条切线, $A$、 $B$ 为两切点, 那么 $\\overrightarrow{\\mathrm{PA}} \\cdot \\overrightarrow{\\mathrm{PB}}$ 的最小值为 ( )", "options": {"A": "$-4+\\sqrt{2}$", "B": "$-3+\\sqrt{2}$", "C": "$-4+2 \\sqrt{2}$", "D": "$-3+2 \\sqrt{2}$"}, "label": "D", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅰ）"}}
|
||||
{"passage": null, "question": "已知在半径为 2 的球面上有 $A$、 $B$、 $C$、 $D$ 四点, 若 $A B=C D=2$, 则四 面体 $A B C D$ 的体积的最大值为 ( )", "options": {"A": "$\\frac{2 \\sqrt{3}}{3}$", "B": "$\\frac{4 \\sqrt{3}}{3}$", "C": "$2 \\sqrt{3}$", "D": "$\\frac{8 \\sqrt{3}}{3}$"}, "label": "B", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅰ）"}}
|
||||
{"passage": null, "question": "设集合 $M=\\{m \\in Z \\mid-3<m<2\\}, N=\\{n \\in Z \\mid-1 \\leqslant n \\leqslant 3\\}$, 则 $M \\cap N=$ ( )", "options": {"A": "$\\{0,1\\}$", "B": "$\\{-1,0,1\\}$", "C": "$\\{0,1,2\\}$", "D": "$\\{-1,0,1,2\\}$"}, "label": "B", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "设 $a, b \\in R$ 且 $b \\neq 0$, 若复数 $(a+b i)^{3}$ 是实数, 则( )", "options": {"A": "$b^{2}=3 a^{2}$", "B": "$a^{2}=3 b^{2}$", "C": "$b^{2}=9 a^{2}$", "D": "$a^{2}=9 b^{2}$"}, "label": "A", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "函数 $f(x)=\\frac{1}{x}-x$ 的图象关于( )", "options": {"A": "$y$ 轴对称", "B": "直线 $y=-x$ 对称", "C": "坐标原点对称", "D": "直线 $y=x$ 对称"}, "label": "C", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "若 $x \\in\\left(e^{-1}, 1\\right), a=\\ln x, b=2 \\ln x, c=\\ln ^{3} x$, 则( )", "options": {"A": "$a<b<c$", "B": "$c<a<b$", "C": "$b<a<c$", "D": "$\\mathrm{b}<\\mathrm{c}<\\mathrm{a}$"}, "label": "C", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "设变量 $x, y$ 满足约束条件: $\\left\\{\\begin{array}{l}y \\geqslant x \\\\ x+2 y \\leqslant 2 \\\\ x \\geqslant-2\\end{array}\\right.$, 则 $z=x-3 y$ 的最小值( )", "options": {"A": "-2", "B": "-4", "C": "-6", "D": "-8"}, "label": "D", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "从 20 名男同学, 10 名女同学中任选 3 名参加体能测试, 则选到的 3 名同学中既有男同学又有女同学的概率为( )", "options": {"A": "$\\frac{9}{29}$", "B": "$\\frac{10}{29}$", "C": "$\\frac{19}{29}$", "D": "$\\frac{20}{29}$"}, "label": "D", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "$(1-\\sqrt{\\mathrm{x}})^{6}(1+\\sqrt{\\mathrm{x}})^{4}$ 的展开式中 $\\mathrm{x}$ 的系数是( )", "options": {"A": "-4", "B": "-3", "C": "3", "D": "4"}, "label": "B", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "若动直线 $x=a$ 与函数 $f(x)=\\sin x$ 和 $g(x)=\\cos x$ 的图象分别交于 $M$, $N$ 两点，则 $|M N|$ 的最大值为( )", "options": {"A": "1", "B": "$\\sqrt{2}$", "C": "$\\sqrt{3}$", "D": "2"}, "label": "B", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "设 $a>1$, 则双曲线 $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{(a+1)^{2}}=1$ 的离心率 $e$ 的取值范围是 ( )", "options": {"A": "$(\\sqrt{2}, 2)$", "B": "$(\\sqrt{2}, \\sqrt{5})$", "C": "$(2,5)$", "D": "$(2, \\sqrt{5})$"}, "label": "B", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "已知正四棱雉 $S-A B C D$ 的侧棱长与底面边长都相等, $E$ 是 $S B$ 的中 点, 则 $A E$、 $S D$ 所成的角的余弦值为( )", "options": {"A": "$\\frac{1}{3}$", "B": "$\\frac{\\sqrt{2}}{3}$", "C": "$\\frac{\\sqrt{3}}{3}$", "D": "$\\frac{2}{3}$"}, "label": "C", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "等腰三角形两腰所在直线的方程分别为 $x+y-2=0$ 与 $x-7 y-4=0$, 原点在等腰三角形的底边上, 则底边所在直线的斜率为( )", "options": {"A": "3", "B": "2", "C": "$-\\frac{1}{3}$", "D": "$-\\frac{1}{2}$"}, "label": "A", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "已知球的半径为 2 , 相互垂直的两个平面分别截球面得两个圆, 若 两圆的公共弦长为 2 , 则两圆的圆心距等于( )", "options": {"A": "1", "B": "$\\sqrt{2}$", "C": "$\\sqrt{3}$", "D": "2"}, "label": "C", "answer": null, "other": {"source": "2008年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{1,2,3,4,5\\}, B=\\{(x, y) \\mid x \\in A, y \\in A, x-y \\in A\\}$, 则 $B$ 中所含元素的个数为 （( ) $）", "options": {"A": "3", "B": "6", "C": "8", "D": "10"}, "label": "D", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "将 2 名教师, 4 名学生分成 2 个小组, 分别安排到甲、乙两地参加 社会实践活动, 每个小组由 1 名教师和 2 名学生组成, 不同的安排方案共有 （( ) $）", "options": {"A": "12 种", "B": "10 种", "C": "9 种", "D": "8 种"}, "label": "A", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "下面是关于复数 $z=\\frac{2}{-1+i}$ 的四个命题: 其中的真命题为 （( ) $）,\n\n$\\mathrm{p}_{1}:|\\mathrm{z}|=2$,\n\n$p_{2}: z^{2}=2 \\mathrm{i}$,\n\n$p_{3}: z$ 的共轭复数为 $1+i$,\n\n$p_{4}: \\mathrm{z}$ 的虚部为 -1 .", "options": {"A": "$\\mathrm{p}_{2}, \\mathrm{p}_{3}$", "B": "$p_{1}, p_{2}$", "C": "$\\mathrm{p}_{2}, \\mathrm{p}_{4}$", "D": "$p_{3}, p_{4}$"}, "label": "C", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "设 $F_{1}$、 $F_{2}$ 是椭圆 $E: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左、右焦点, $P$ 为直线 $x=\\frac{3 a}{2}$ 上一点, $\\triangle F_{2} P F_{1}$ 是底角为 $30^{\\circ}$ 的等腰三角形, 则 $E$ 的离心率为 （( ) $）", "options": {"A": "$\\frac{1}{2}$", "B": "$\\frac{2}{3}$", "C": "$\\frac{3}{4}$", "D": "$\\frac{4}{5}$"}, "label": "C", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "已知 $\\left\\{a_{n}\\right\\}$ 为等比数列, $a_{4}+a_{7}=2, a_{5} a_{6}=-8$, 则 $a_{1}+a_{10}=( )", "options": {"A": "7", "B": "5", "C": "-5", "D": "-7"}, "label": "D", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "等轴双曲线 $C$ 的中心在原点, 焦点在 $x$ 轴上, $C$ 与抛物线 $y^{2}=16 x$ 的 准线交于点 $A$ 和点 $B,|A B|=4 \\sqrt{3}$, 则 $C$ 的实轴长为 （( ) $）", "options": {"A": "$\\sqrt{2}$", "B": "$2 \\sqrt{2}$", "C": "4", "D": "8"}, "label": "C", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "已知 $\\omega>0$, 函数 $f(x)=\\sin \\left(\\omega x+\\frac{\\pi}{4}\\right)$ 在区间 $\\left[\\frac{\\pi}{2}, \\pi\\right]$ 上单调递减, 则实数 $\\omega$ 的取值范围是 （( ) $）", "options": {"A": "$\\left[\\frac{1}{2}, \\frac{5}{4}\\right]$", "B": "$\\left[\\frac{1}{2}, \\frac{3}{4}\\right]$", "C": "$\\left(0, \\frac{1}{2}\\right]$", "D": "$(0,2]$"}, "label": "A", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "已知三棱雉 $S-A B C$ 的所有顶点都在球 $O$ 的表面上, $\\triangle A B C$ 是边长 为 1 的正三角形, $S C$ 为球 $O$ 的直径, 且 $S C=2$, 则此三棱雉的体积为 （( ) $）", "options": {"A": "$\\frac{1}{4}$", "B": "$\\frac{\\sqrt{2}}{4}$", "C": "$\\frac{\\sqrt{2}}{6}$", "D": "$\\frac{\\sqrt{2}}{12}$"}, "label": "C", "answer": null, "other": {"source": "2012年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\left\\{x \\mid x^{2}-2 x-3 \\geqslant 0\\right\\}, B=\\{x \\mid-2 \\leqslant x<2\\}$, 则 $A \\cap B=( )", "options": {"A": "$[1,2)$", "B": "$[-1,1]$", "C": "$[-1,2)$", "D": "$[-2,-1]$"}, "label": "D", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$(5$ 分 $) \\frac{(1+i)^{3}}{(1-i)^{2}}=( )", "options": {"A": "$1+\\mathrm{i}$", "B": "$1-\\mathrm{i}$", "C": "$-1+i$", "D": "$-1-i$"}, "label": "C", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知 $F$ 为双曲线 $C: x^{2}-m y^{2}=3 m(m>0)$ 的一个焦点, 则点 $F$ 到 $C$ 的一条渐近线的距离为 ( )", "options": {"A": "$\\sqrt{3}$", "B": "3", "C": "$\\sqrt{3} \\mathrm{~m}$", "D": "$3 m$"}, "label": "A", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$4$位同学各自在周六、周日两天中任选一天参加公益活动, 则周六、 周日都有同学参加公益活动的概率为 ( )", "options": {"A": "$\\frac{1}{8}$", "B": "$\\frac{3}{8}$", "C": "$\\frac{5}{8}$", "D": "$\\frac{7}{8}$"}, "label": "D", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设 $\\alpha \\in\\left(0, \\frac{\\pi}{2}\\right), \\beta \\in\\left(0, \\frac{\\pi}{2}\\right)$, 且 $\\tan \\alpha=\\frac{1+\\sin \\beta}{\\cos \\beta}$, 则 ( )", "options": {"A": "$3 \\alpha-\\beta=\\frac{\\pi}{2}$", "B": "$3 \\alpha+\\beta=\\frac{\\pi}{2}$", "C": "$2 \\alpha-\\beta=\\frac{\\pi}{2}$", "D": "$2 \\alpha+\\beta=\\frac{\\pi}{2}$"}, "label": "C", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知抛物线 $C: y^{2}=8 x$ 的焦点为 $F$, 准线为 $\\mid, P$ 是 $\\mid$ 上一点, $Q$ 是直 线 $P F$ 与 $C$ 的一个交点, 若 $\\overrightarrow{F P}=4 \\overrightarrow{F Q}$, 则 $|Q F|=( )", "options": {"A": "$\\frac{7}{2}$", "B": "3", "C": "$\\frac{5}{2}$", "D": "2"}, "label": "B", "answer": null, "other": {"source": "2014年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$\\frac{1+2 i}{1-2 i}=( )", "options": {"A": "$-\\frac{4}{5}-\\frac{3}{5} i$", "B": "$-\\frac{4}{5}+\\frac{3}{5} i$", "C": "$-\\frac{3}{5}-\\frac{4}{5} i$", "D": "$-\\frac{3}{5}+\\frac{4}{5} i$"}, "label": "D", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\left\\{(x, y) \\mid x^{2}+y^{2} \\leqslant 3, x \\in Z, y \\in Z\\right\\}$, 则 $A$ 中元素的个数为 ( )", "options": {"A": "9", "B": "8", "C": "5", "D": "4"}, "label": "A", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "双曲线 $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的离心率为 $\\sqrt{3}$, 则其渐近线方程为 ( )", "options": {"A": "$y= \\pm \\sqrt{2} x$", "B": "$y= \\pm \\sqrt{3} x$", "C": "$y= \\pm \\frac{\\sqrt{2}}{2} x$", "D": "$y= \\pm \\frac{\\sqrt{3}}{2}$"}, "label": "A", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "我国数学家陈景润在哥德巴赫猜想的研究中取得了世界领先的成 果. 哥德巴赫猜想是 “每个大于 2 的偶数可以表示为两个素数的和”, 如 $30=7+23$. 在不超过 30 的素数中, 随机选取两个不同的数, 其和等于 30 的 概率是 ( )", "options": {"A": "$\\frac{1}{12}$", "B": "$\\frac{1}{14}$", "C": "$\\frac{1}{15}$", "D": "$\\frac{1}{18}$"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "在长方体 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, $A B=B C=1, A A_{1}=\\sqrt{3}$, 则异面直线 $A D_{1}$ 与 $\\mathrm{DB}_{1}$ 所成角的余弦值为 ( )", "options": {"A": "$\\frac{1}{5}$", "B": "$\\frac{\\sqrt{5}}{6}$", "C": "$\\frac{\\sqrt{5}}{5}$", "D": "$\\frac{\\sqrt{2}}{2}$"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "若 $f(x)=\\cos x-\\sin x$ 在 $[-a, a]$ 是减函数, 则 $a$ 的最大值是 ( )", "options": {"A": "$\\frac{\\pi}{4}$", "B": "$\\frac{\\pi}{2}$", "C": "$\\frac{3 \\pi}{4}$", "D": "$\\pi$"}, "label": "A", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $f(x)$ 是定义域为 $(-\\infty,+\\infty)$ 的奇函数, 满足 $f(1-x)=f$ $(1+x)$, 若 $f(1)=2$, 则 $f(1)+f(2)+f(3)+\\ldots+f(50)=( )", "options": {"A": "-50", "B": "0", "C": "2", "D": "50"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $F_{1}, F_{2}$ 是椭圆 C: $\\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左、右焦点, $A$ 是 $C$ 的左顶点, 点 $P$ 在过 $A$ 且斜率为 $\\frac{\\sqrt{3}}{6}$ 的直线上, $\\triangle P F_{1} F_{2}$ 为等腰三角形, $\\angle F_{1} F_{2} P=120^{\\circ}$, 则 $C$ 的离心率为 ( )", "options": {"A": "$\\frac{2}{3}$", "B": "$\\frac{1}{2}$", "C": "$\\frac{1}{3}$", "D": "$\\frac{1}{4}$"}, "label": "D", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设集合 $M=\\{x \\mid 0<x<4\\}, N=\\left\\{x \\mid \\frac{1}{3} \\leq x \\leq 5\\right\\}$, 则 $M \\cap N=( )", "options": {"A": "$\\left\\{x \\mid 0<x \\leq \\frac{1}{3}\\right\\}$", "B": "$\\left\\{x \\mid \\frac{1}{3} \\leq x<4\\right\\}$", "C": "$\\{x \\mid 4 \\leq x<5\\}$", "D": "$\\{x \\mid 0<x \\leq 5\\}$"}, "label": "B", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "已知 $(1-i)^{2} z=3+2 i$, 则 $z=( )", "options": {"A": "$-1-\\frac{3}{2} i$", "B": "$-1+\\frac{3}{2} i$", "C": "$-\\frac{3}{2}+i$", "D": "$-\\frac{3}{2}-i$"}, "label": "B", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "青少年视力是社会普遍关注的问题, 视力情况可借助视力表测量. 通常用五分记录法和小数记录法记录 视力数据, 五分记录法的数据 $L$ 和小数记录表的数据 $V$ 的满足 $L=5+\\lg V$. 已知某同学视力的五分记录法 的数据为 $4.9$, 则其视力的小数记录法的数据为 （( ) $(\\sqrt[10]{10} \\approx 1.259)$", "options": {"A": "$1.5$", "B": "$1.2$", "C": "$0.8$", "D": "$0.6$"}, "label": "C", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "已知 $F_{1}, F_{2}$ 是双曲线 $C$ 的两个焦点, $P$ 为 $C$ 上一点, 且 $\\angle F_{1} P F_{2}=60^{\\circ},\\left|P F_{1}\\right|=3\\left|P F_{2}\\right|$, 则 $C$ 的离心率为 （( )", "options": {"A": "$\\frac{\\sqrt{7}}{2}$", "B": "$\\frac{\\sqrt{13}}{2}$", "C": "$\\sqrt{7}$", "D": "$\\sqrt{13}$"}, "label": "A", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "等比数列 $\\left\\{a_{n}\\right\\}$ 的公比为 $q$, 前 $n$ 项和为 $S_{n}$, 设甲: $q>0$, 乙: $\\left\\{S_{n}\\right\\}$ 是递增数列, 则 （( )", "options": {"A": "甲是乙的充分条件但不是必要条件", "B": "甲是乙的必要条件但不是充分条件", "C": "甲是乙的充要条件", "D": "甲既不是乙的充分条件也不是乙的必要条件"}, "label": "B", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "将 4 个 1 和 2 个 0 随机排成一行, 则 2 个 0 不相邻的概率为 （( )", "options": {"A": "$\\frac{1}{3}$", "B": "$\\frac{2}{5}$", "C": "$\\frac{2}{3}$", "D": "$\\frac{4}{5}$"}, "label": "C", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "已如 $A, B, C$ 是半径为 1 的球 $O$ 的球面上的三个点, 且 $A C \\perp B C, A C=B C=1$, 则三棱雉 $O-A B C$ 的体积为 （( )", "options": {"A": "$\\frac{\\sqrt{2}}{12}$", "B": "$\\frac{\\sqrt{3}}{12}$", "C": "$\\frac{\\sqrt{2}}{4}$", "D": "$\\frac{\\sqrt{3}}{4}$"}, "label": "A", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "设函数 $f(x)$ 的定义域为 $\\mathbf{R}, f(x+1)$ 为奇函数, $f(x+2)$ 为偶函数, 当 $x \\in[1,2]$ 时,$f(x)=a x^{2}+b$. 若 $f(0)+f(3)=6$, 则 $f\\left(\\frac{9}{2}\\right)=( )", "options": {"A": "$-\\frac{9}{4}$", "B": "$-\\frac{3}{2}$", "C": "$\\frac{7}{4}$", "D": "$\\frac{5}{2}$"}, "label": "D", "answer": null, "other": {"source": "2021全国甲卷数学"}}
|
||||
{"passage": null, "question": "复数 $\\left(\\frac{3-i}{1+i}\\right)^{2}=$ ( )", "options": {"A": "$-3-4 \\mathrm{i}$", "B": "$-3+4 i$", "C": "$3-4 i$", "D": "$3+4 i$"}, "label": "A", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "函数 $y=\\frac{1+\\ln (x-1)}{2}(x>1)$ 的反函数是 ( )", "options": {"A": "$y=e^{2 x-1}-1(x>0)$", "B": "$y=e^{2 x-1}+1 \\quad(x>0)$", "C": "$y=e^{2 x-1}-1 \\quad(x \\in R)$", "D": "$y=e^{2 x-1}+1 \\quad(x \\in R)$"}, "label": "D", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "若变量 $x, y$ 满足约束条件 $\\left\\{\\begin{array}{l}x \\geqslant-1 \\\\ y \\geqslant x \\\\ 3 x+2 y \\leqslant 5,\\end{array}\\right.$ 则 $z=2 x+y$ 的最大值为 ( )", "options": {"A": "1", "B": "2", "C": "3", "D": "4"}, "label": "C", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "如果等差数列 $\\left\\{a_{n}\\right\\}$ 中, $a_{3}+a_{4}+a_{5}=12$, 那么 $a_{1}+a_{2}+\\ldots+a_{7}=$ ( )", "options": {"A": "14", "B": "21", "C": "28", "D": "35"}, "label": "C", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "不等式 $\\frac{x^{2}-x-6}{x-1}>0$ 的解集为 ( )", "options": {"A": "$\\{x \\mid x<-2$, 或 $x>3\\}$", "B": "$\\{x \\mid x<-2$, 或 $1<x<3\\}$", "C": "$\\{x \\mid-2<x<1$, 或 $x>3\\}$", "D": "$\\{x \\mid-2<x<1$, 或 $1<x<3\\}$"}, "label": "C", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "将标号为 $1,2,3,4,5,6$ 的 6 张卡片放入 3 个不同的信封中, 若 每个信封放 2 张, 其中标号为 1,2 的卡片放入同一信封, 则不同的方法共 有 ( )", "options": {"A": "12 种", "B": "18 种", "C": "36 种", "D": "54 种"}, "label": "B", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "为了得到函数 $y=\\sin \\left(2 x-\\frac{\\pi}{3}\\right)$ 的图象, 只需把函数 $y=\\sin \\left(2 x+\\frac{\\pi}{6}\\right)$ 的图象 ( )", "options": {"A": "向左平移 $\\frac{\\pi}{4}$ 个长度单位", "B": "向右平移 $\\frac{\\pi}{4}$ 个长度单位", "C": "向左平移 $\\frac{\\pi}{2}$ 个长度单位", "D": "向右平移 $\\frac{\\pi}{2}$ 个长度单位"}, "label": "B", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "$\\triangle A B C$ 中, 点 $D$ 在边 $A B$ 上, $C D$ 平分 $\\angle A C B$, 若 $\\overrightarrow{C B}=\\vec{a}, \\overrightarrow{C A}=\\vec{b}, \\mid \\vec{a}$ $|=1,| \\vec{b} \\mid=2$, 则 $\\overrightarrow{C D}=$ ( )", "options": {"A": "$\\frac{1}{3} \\vec{a}+\\frac{2}{3} \\vec{b}$", "B": "$\\frac{2}{3} \\vec{a}+\\frac{1}{3 b}$", "C": "$\\frac{3}{5} \\vec{a}+\\frac{4}{5} \\vec{b}$", "D": "$\\frac{4}{5} \\vec{a}+\\frac{3}{5 b}$"}, "label": "B", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "已知正四棱雉 $S-A B C D$ 中, $S A=2 \\sqrt{3}$, 那么当该棱雉的体积最大时, 它的高为 ( )", "options": {"A": "1", "B": "$\\sqrt{3}$", "C": "2", "D": "3"}, "label": "C", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "若曲线 $y=x^{-\\frac{1}{2}}$ 在点 $\\left( a, a^{-\\frac{1}{2}}\\right)$ 处的切线与两个坐标围成的三角形 的面积为 18 , 则 $a=$ ( )", "options": {"A": "64", "B": "32", "C": "16", "D": "8"}, "label": "D", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "已知椭圆 $T: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1 \\left(a>b>0\\right)$ 的离心率为 $\\frac{\\sqrt{3}}{2}$, 过右焦点 $F$ 且 斜率为 $k(k>0)$ 的直线与 $T$ 相交于 $A, B$ 两点, 若 $\\overline{\\mathrm{AF}}=3 \\overline{\\mathrm{FB}}$, 则 $k=$ ( )", "options": {"A": "1", "B": "$\\sqrt{2}$", "C": "$\\sqrt{3}$", "D": "2"}, "label": "B", "answer": null, "other": {"source": "2010年数学试卷（理科）（大纲版ⅱ）"}}
|
||||
{"passage": null, "question": "设 $2(z+\\bar{z})+3(z-\\bar{z})=4+6 i$, 则 $z=( )", "options": {"A": "$1-2 i$", "B": "$1+2 i$", "C": "$1+i$", "D": "$1-i$"}, "label": "C", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知集合 $S=\\{s \\mid s=2 n+1, n \\in Z\\}, T=\\{t \\mid t=4 n+1, n \\in Z\\}$, 则 $S \\cap T=( )", "options": {"A": "$\\varnothing$", "B": "$S$", "C": "$T$", "D": "$Z$"}, "label": "C", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知命题 $p: \\exists x \\in R, \\sin x<1$; 命题 $q: \\forall x \\in R, e^{|x|} \\geq 1$, 则下列命题中为真命题的是 ( )", "options": {"A": "$p \\wedge q$", "B": "$\\neg p \\wedge q$", "C": "$p \\wedge \\neg q$", "D": "$\\neg(p \\vee q)$"}, "label": "A", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "在正方体 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, $P$ 为 $B_{1} D_{1}$ 的中点, 则直线 $P B$ 与 $A D_{1}$ 所成的角为 ( )", "options": {"A": "$\\frac{\\pi}{2}$", "B": "$\\frac{\\pi}{3}$", "C": "$\\frac{\\pi}{4}$", "D": "$\\frac{\\pi}{6}$ "}, "label": "D", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "将 5 名北京冬奥会志愿者分配到花样滑冰, 短道速滑、冰球和冰壶 4 个项目进行培训, 每名 志愿者只分配到1个项目, 每个项目至少分配1名志愿者, 则不同的分配方案共有 ( )", "options": {"A": "$60$ 种", "B": "$120$ 种", "C": "$240$ 种", "D": "$480$ 种"}, "label": "C", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "把函数 $y=f(x)$ 图像上所有点的横坐标缩短到原来的 $\\frac{1}{2}$ 倍, 纵坐标不变, 再把所得曲 线向右平移 $\\frac{\\pi}{3}$ 个单位长度, 得到函数 $y=\\sin \\left(x-\\frac{\\pi}{4}\\right)$ 的图像, 则 $f(x)=( )", "options": {"A": "$\\sin \\left(\\frac{x}{2}-\\frac{7 \\pi}{12}\\right)$", "B": "$\\sin \\left(\\frac{x}{2}+\\frac{\\pi}{12}\\right)$", "C": "$\\sin \\left(2 x-\\frac{7 \\pi}{12}\\right)$", "D": "$\\sin \\left(2 x+\\frac{\\pi}{12}\\right)$"}, "label": "A", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "设 $a \\neq 0$, 若 $x=a$ 为函数 $f(x)=a(x-a)^{2}(x-b)$ 的极大值点, 则 ( )", "options": {"A": "$a<b$", "B": "$a>b$", "C": "$a b<a^{2}$", "D": "$a b>a^{2}$"}, "label": "D", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "设 $B$ 是椭圆 $C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的上顶点, 若 $C$ 上的任意一点 $P$ 都满足, $|P B| \\leq 2 b$, 则 $C$ 的离心率的取值范围是 ( )", "options": {"A": "$\\left[\\frac{\\sqrt{2}}{2}, 1\\right)$", "B": "$\\left[\\frac{1}{2}, 1\\right)$", "C": "$\\left(0, \\frac{\\sqrt{2}}{2}\\right]$", "D": "$\\left(0, \\frac{1}{2}\\right]$"}, "label": "C", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "设 $a=2 \\ln 1.01, b=\\ln 1.02, c=\\sqrt{1.04}-1$, 则 ( )", "options": {"A": "$a<b<c$", "B": "$b<c<a$", "C": "$b<a<c$", "D": "$c<a<b$"}, "label": "B", "answer": null, "other": {"source": "2021年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{x \\mid x<1\\}, B=\\left\\{x \\mid 3^{x}<1\\right\\}$ ，则 ( )", "options": {"A": "$A \\cap B=\\{x \\mid x<0\\}$", "B": "$A \\cup B=R$", "C": "$A \\cup B=\\{x \\mid x>1\\}$", "D": "$A \\cap B=\\varnothing$"}, "label": "A", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设有下面四个命题\\\\\n$p_{1}$ : 若复数 $z$ 满足 $\\frac{1}{z} \\in R$, 则 $z \\in R$;\\\\\n$p_{2}$ : 若复数 $z$ 满足 $z^{2} \\in R$, 则 $z \\in R$;\\\\\n$p_{3}$ : 若复数 $z_{1}, z_{2}$ 满足 $z_{1} z_{2} \\in R$, 则 $z_{1}=\\bar{z_{2}}$;\\\\\n$p_{4}$ : 若复数 $z \\in R$, 则 $\\bar{z} \\in R$.\\\\\n其中的真命题为 ( )", "options": {"A": "$\\mathrm{p}_{1}, \\mathrm{p}_{3}$", "B": "$\\mathrm{p}_{1}, \\mathrm{p}_{4}$", "C": "$\\mathrm{p}_{2}, \\mathrm{p}_{3}$", "D": "$\\mathrm{p}_{2}, \\mathrm{p}_{4}$"}, "label": "B", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "记 $S_{n}$ 为等差数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和. 若 $a_{4}+a_{5}=24, S_{6}=48$, 则 $\\left\\{a_{n}\\right\\}$ 的公差为 ( )", "options": {"A": "1", "B": "2", "C": "4", "D": "8"}, "label": "C", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "函数 $f(x)$ 在 $(-\\infty,+\\infty)$ 单调递减, 且为奇函数. 若 $f(1)=-1$, 则满足 $-1 \\leqslant f(x-2) \\leqslant 1$ 的 $x$ 的取值范围是 ( )", "options": {"A": "$[-2,2]$", "B": "$[-1,1]$", "C": "$[0,4]$", "D": "$[1,3]$"}, "label": "D", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$\\left(1+\\frac{1}{x^{2}}\\right)(1+x)^{6}$ 展开式中 $x^{2}$ 的系数为 ( )", "options": {"A": "15", "B": "20", "C": "30", "D": "35"}, "label": "C", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知曲线 $C_{1}: y=\\cos x, C_{2}: y=\\sin \\left(2 x+\\frac{2 \\pi}{3}\\right)$, 则下面结论正确的是 ( )", "options": {"A": "把 $C_{1}$ 上各点的横坐标伸长到原来的 $2$ 倍, 纵坐标不变, 再把得到的曲线 向右平移 $\\frac{\\pi}{6}$ 个单位长度, 得到曲线 $C_{2}$", "B": "把 $C_{1}$ 上各点的横坐标伸长到原来的 $2$ 倍, 纵坐标不变, 再把得到的曲线 向右平移 $\\frac{\\pi}{12}$ 个单位长度, 得到曲线 $C_{2}$", "C": "把 $C_{1}$ 上各点的横坐标伸长到原来的 $\\frac{1}{2}$ 倍, 纵坐标不变, 再把得到的曲线 向右平移 $\\frac{\\pi}{6}$ 个单位长度, 得到曲线 $C_{2}$", "D": "把 $C_{1}$ 上各点的横坐标伸长到原来的 $\\frac{1}{2}$ 倍, 纵坐标不变, 再把得到的曲线 向右平移 $\\frac{\\pi}{12}$ 个单位长度, 得到曲线 $C_{2}$"}, "label": "D", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设 $x$、 $y$、 $z$ 为正数, 且 $2^{x}=3^{y}=5^{z}$, 则 ( )", "options": {"A": "$2 x<3 y<5 z$", "B": "$5 z<2 x<3 y$", "C": "$3 y<5 z<2 x$", "D": "$3 y<2 x<5 z$"}, "label": "D", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "几位大学生响应国家的创业号召, 开发了一款应用软件. 为激发大 家学习数学的兴趣, 他们推出了“解数学题获取软件激活码”的活动. 这款软 件的激活码为下面数学问题的答案: 已知数列 $1,1,2,1,2,4,1,2,4$, $8,1,2,4,8,16, \\ldots$, 其中第一项是 $2^{0}$, 接下来的两项是 $2^{0}, 2^{1}$, 再接下 来的三项是 $2^{0}, 2^{1}, 2^{2}$, 依此类推. 求满足如下条件的最小整数 $N: N>100$ 且该数列的前 $\\mathrm{N}$ 项和为 2 的整数幂. 那么该款软件的激活码是 ( )", "options": {"A": "440", "B": "330", "C": "220", "D": "110"}, "label": "A", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "复数 $\\frac{2+i}{1-2 i}$ 的共轭复数是 ( )", "options": {"A": "$-\\frac{3}{5} i$", "B": "$\\frac{3}{5} i$", "C": "- i", "D": "i"}, "label": "C", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "下列函数中, 既是偶函数又在 $(0,+\\infty)$ 上单调递增的函数是 ( )", "options": {"A": "$y=2 x^{3}$", "B": "$y=|x|+1$", "C": "$y=-x^{2}+4$", "D": "$y=2^{-|x|}$"}, "label": "B", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "有 3 个兴趣小组, 甲、乙两位同学各自参加其中一个小组, 每位同 学参加各个小组的可能性相同, 则这两位同学参加同一个兴趣小组的概率为 ( )", "options": {"A": "$\\frac{1}{3}$", "B": "$\\frac{1}{2}$", "C": "$\\frac{2}{3}$", "D": "$\\frac{3}{4}$"}, "label": "A", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "设直线 $\\mid$ 过双曲线 $C$ 的一个焦点, 且与 $C$ 的一条对称轴垂直, $\\mid$ 与 $C$ 交于 $A, B$ 两点, $|A B|$ 为 $C$ 的实轴长的 2 倍, 则 $C$ 的离心率为 ( )", "options": {"A": "$\\sqrt{2}$", "B": "$\\sqrt{3}$", "C": "2", "D": "3"}, "label": "B", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "$\\left(x+\\frac{a}{x}\\right)\\left(2 x-\\frac{1}{x}\\right)^{5}$ 的展开式中各项系数的和为 2 , 则该展开式中常数项为 ( )", "options": {"A": "-40", "B": "-20", "C": "20", "D": "40"}, "label": "D", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "由曲线 $y=\\sqrt{x}$, 直线 $y=x-2$ 及 $y$ 轴所围成的图形的面积为 ( )", "options": {"A": "$\\frac{10}{3}$", "B": "4", "C": "$\\frac{16}{3}$", "D": "6"}, "label": "A", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=\\sin (\\omega x+\\phi)+\\cos (\\omega x+\\phi)\\left(\\omega>0,|\\phi|<\\frac{\\pi}{2}\\right)$ 的最小正周期为 $\\pi$, 且 $f(-x)=f(x)$, 则 ( )", "options": {"A": "$f(x)$ 在 $\\left(0, \\frac{\\pi}{2}\\right)$ 单调递减", "B": "$f(x)$ 在 $\\left(\\frac{\\pi}{4}, \\frac{3 \\pi}{4}\\right)$ 单调递减", "C": "$f(x)$ 在 $\\left(0, \\frac{\\pi}{2}\\right)$ 单调递增", "D": "$f(x)$ 在 $\\left(\\frac{\\pi}{4}, \\frac{3 \\pi}{4}\\right)$ 单调递增"}, "label": "A", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "函数 $y=\\frac{1}{1-x}$ 的图象与函数 $y=2 \\sin \\pi x,(-2 \\leqslant x \\leqslant 4)$ 的图象所有交点 的横坐标之和等于 ( )", "options": {"A": "8", "B": "6", "C": "4", "D": "2"}, "label": "A", "answer": null, "other": {"source": "2011年数学试卷（理科）（新课标）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{-1,1,2,4\\}, B=\\{x|| x-1 \\mid \\leq 1\\}$, 则 $A \\cap B=( )", "options": {"A": "$\\{-1,2\\}$", "B": "$\\{1,2\\}$", "C": "$\\{1,4\\}$", "D": "$\\{-1,4\\}$"}, "label": "B", "answer": null, "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "$(2+2 \\mathrm{i})(1-2 \\mathrm{i})=( )", "options": {"A": "$-2+4 \\mathrm{i}$", "B": "$-2-4 \\mathrm{i}$", "C": "$6+2 i$", "D": "$6-2 i$"}, "label": "D", "answer": null, "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "有甲乙丙丁戊 5 名同学站成一排参加文艺汇演, 若甲不站在两端, 丙和丁相邻的不同排 列方式有多少种 ( )", "options": {"A": "12 种", "B": "24 种", "C": "36 种", "D": "48 种"}, "label": "B", "answer": null, "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "角 $\\alpha, \\beta$ 满足 $\\sin (\\alpha+\\beta)+\\cos (\\alpha+\\beta)=2 \\sqrt{2} \\cos \\left(\\alpha+\\frac{\\pi}{4}\\right) \\sin \\beta$, 则 ( )", "options": {"A": "$\\tan (\\alpha+\\beta)=1$", "B": "$\\tan (\\alpha+\\beta)=-1$", "C": "$\\tan (\\alpha-\\beta)=1$", "D": "$\\tan (\\alpha-\\beta)=-1$"}, "label": "D", "answer": null, "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "若函数 $f(x)$ 的定义域为 $\\mathbf{R}$, 且 $f(x+y)+f(x-y)=f(x) f(y), f(1)=1$, 则 $\\sum_{k=1}^{22} f(k)=( )", "options": {"A": "$-3$", "B": "$-2$", "C": "0", "D": "1"}, "label": "A", "answer": null, "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "函数 $f(x)=\\sin (2 x+\\varphi)(0<\\varphi<\\pi)$ 的图象以 $\\left(\\frac{2 \\pi}{3}, 0\\right)$ 中心对称, 则 ( )", "options": {"A": "$y=f(x)$ 在 $\\left(0, \\frac{5 \\pi}{12}\\right)$ 单调递减", "B": "$y=f(x)$ 在 $\\left( -\\frac{\\pi}{12}, \\frac{11 \\pi}{12}\\right)$ 有 $2$ 个极值点", "C": "直线 $x= \\frac{7 \\pi}{6} $ 是一条对称轴", "D": "直线 $y= \\frac{\\sqrt{3}}{2} - x $ 是一条切线"}, "label": "AD", "answer": null, "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "已知 $O$ 为坐标原点, 过抛物线 $C: y^{2}=2 p x(p>0)$ 的焦点 $F$ 的直线与 $C$ 交于 $A, B$ 两 点, 点 $A$ 在第一象限, 点 $M(p, 0)$, 若 $|A F|=|A M|$, 则 ( )", "options": {"A": "直线 $A B$ 的斜率为 $2 \\sqrt{6}$", "B": "$|O B|=|O F|$", "C": "$|A B|>4|O F|$", "D": "$\\angle O A M+\\angle O B M<180^{\\circ}$"}, "label": "ACD", "answer": null, "other": {"source": "2022年全国新高考II卷数学"}}
|
||||
{"passage": null, "question": "若 $z=-1+\\sqrt{3} \\mathbf{i}$, 则 $\\frac{z}{z \\bar{z}-1}=( )", "options": {"A": "$-1+\\sqrt{3} \\mathrm{i}$", "B": "$-1-\\sqrt{3} \\mathrm{i}$", "C": "$-\\frac{1}{3}+\\frac{\\sqrt{3}}{3} \\mathrm{i}$", "D": "$-\\frac{1}{3}-\\frac{\\sqrt{3}}{3} \\mathrm{i}$"}, "label": "C", "answer": null, "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "设全集 $U=\\{-2,-1,0,1,2,3\\}$, 集合 $A=\\{-1,2\\}, B=\\left\\{x \\mid x^{2}-4 x+3=0\\right\\}$, 则 $C_{U}(A \\cup B)=( )", "options": {"A": "$\\{1,3\\}$", "B": "$\\{0,3\\}$", "C": "$\\{-2,1\\}$", "D": "$\\{-2,0\\}$"}, "label": "D", "answer": null, "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "在长方体 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, 已知 $B_{1} D$ 与平面 $A B C D$ 和平面 $A A_{1} B_{1} B$ 所成的角均为 $30^{\\circ}$ ，则 ( )", "options": {"A": "$A B=2 A D$", "B": "$A B $ 与平面 $A B_{1} C_{1} D$ 所成的角为 $30^{\\circ}$", "C": "$A C=C B_{1}$", "D": "$B_{1} D$ 与平面 $B B_{1} C_{1} C$ 所成的角为 $45^{\\circ}$"}, "label": "D", "answer": null, "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "椭圆 $C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$ 的左顶点为 $A$, 点 $P, Q$ 均在 $C$ 上, 且关于 $y$ 轴对 称. 若直线 $A P, A Q$ 的斜率之积为 $\\frac{1}{4}$, 则 $C$ 的离心率为 ( )", "options": {"A": "$\\frac{\\sqrt{3}}{2}$", "B": "$\\frac{\\sqrt{2}}{2}$", "C": "$\\frac{1}{2}$", "D": "$\\frac{1}{3}$"}, "label": "A", "answer": null, "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=\\sin \\left(\\omega x+\\frac{\\pi}{3}\\right)$ 在区间 $(0, \\pi)$ 恰有三个极值点、两个零点, 则 $\\omega$ 的取值范围是 ( )", "options": {"A": "$\\left[\\frac{5}{3}, \\frac{13}{6}\\right)$", "B": "$\\left[\\frac{5}{3}, \\frac{19}{6}\\right)$", "C": "$\\left(\\frac{13}{6}, \\frac{8}{3}\\right]$", "D": "$\\left(\\frac{13}{6}, \\frac{19}{6}\\right]$"}, "label": "C", "answer": null, "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "已知 $a=\\frac{31}{32}, b=\\cos \\frac{1}{4}, c=4 \\sin \\frac{1}{4}$, 则 ( )", "options": {"A": "$c>b>a$", "B": "$b>a>c$", "C": "$a>b>c$", "D": "$a>c>b$"}, "label": "A", "answer": null, "other": {"source": "2022年全国高考甲卷数学"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\left\\{x \\mid x^{2}-2 x>0\\right\\}, B=\\{x \\mid-\\sqrt{5}<x<\\sqrt{5}\\}$, 则 ( )", "options": {"A": "$A \\cap B=\\emptyset$", "B": "$A \\cup B=R$", "C": "$B \\subseteq A$", "D": "$A \\subseteq B$"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "若复数 $z$ 满足 $\\left( 3-4 i \\right) z=|4+3 i|$, 则 $z$ 的虚部为 ( )", "options": {"A": "-4", "B": "$-\\frac{4}{5}$", "C": "4", "D": "$\\frac{4}{5}$"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "为了解某地区中小学生的视力情况, 拟从该地区的中小学生中抽取 部分学生进行调查, 事先已经了解到该地区小学、初中、高中三个学段学生 的视力情况有较大差异, 而男女生视力情况差异不大. 在下面的抽样方法中, 最合理的抽样方法是 ( )", "options": {"A": "简单的随机抽样", "B": "按性别分层抽样", "C": "按学段分层抽样", "D": "系统抽样"}, "label": "C", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知双曲线 C: $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 ( )(a>0, b>0)$ 的离心率为 $\\frac{\\sqrt{5}}{2}$, 则 $C$ 的渐近线方程为 ( )", "options": {"A": "$y= \\pm \\frac{1}{4} x$", "B": "$y= \\pm \\frac{1}{3} x$", "C": "$y= \\pm x$", "D": "$y= \\pm \\frac{1}{2} x$"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设等差数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和为 $S_{n}$, 若 $S_{m-1}=-2, S_{m}=0, S_{m+1}=3$, 则 $m=( )", "options": {"A": "3", "B": "4", "C": "5", "D": "6"}, "label": "C", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设 $m$ 为正整数, $(x+y)^{2 m}$ 展开式的二项式系数的最大值为 $a,(x+y)^{2 m+1}$ 展开式的二项式系数的最大值为 $b$, 若 $13 a=7 b$, 则 $m=( )", "options": {"A": "5", "B": "6", "C": "7", "D": "8"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知椭圆 $E: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{b^{2}}=1(a>b>0)$的右焦点为 $F(3,0)$,过点F的直线交椭圆 $E$ 于 $A$、 $B$ 两点. 若 $A B$ 的中点坐标为 $(1,-1)$, 则 $E$ 的方程为 ( )", "options": {"A": "$\\frac{x^{2}}{45}+\\frac{y^{2}}{36}=1$", "B": "$\\frac{x^{2}}{36}+\\frac{y^{2}}{27}=1$", "C": "$\\frac{x^{2}}{27}+\\frac{y^{2}}{18}=1$", "D": "$\\frac{x^{2}}{18}+\\frac{y^{2}}{9}=1$"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=\\left\\{\\begin{array}{l}-x^{2}+2 x, ( ) x \\leqslant 0 \\\\ \\ln (x+1), ( ) x>0, \\text { 若 }|f(x)| \\geqslant a x, \\text { 则 } a \\text { 的取值 }\\end{array}\\right.$ 范围是 ( )", "options": {"A": "$(-\\infty, 0]$", "B": "$(-\\infty, 1]$", "C": "$[-2,1]$", "D": "$[-2,0]$"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设 $\\triangle A_{n} B_{n} C_{n}$ 的三边长分别为 $a_{n}, b_{n}, c_{n}, \\triangle A_{n} B_{n} C_{n}$ 的面积为 $S_{n}, n=1, 2 , 3...$ 若 $b_{1}>c_{1}, ( ) b_{1}+c_{1}=2 a_{1}, ( ) a_{n+1}=a_{n}, ( ) b_{n+1}=\\frac{c_{n}+a_{n}}{2}, ( ) c_{n+1}=\\frac{b_{n}+a_{n}}{2}$, 则 ( )", "options": {"A": "$\\left\\{S_{n}\\right\\}$ 为递减数列", "B": "$\\left\\{S_{n}\\right\\}$ 为递增数列", "C": "$\\left\\{S_{2 n-1}\\right\\}$ 为递增数列, $\\left\\{S_{2 n}\\right\\}$ 为递减数列", "D": "$\\left\\{S_{2 n-1}\\right\\}$ 为递减数列, $\\left\\{S_{2 n}\\right\\}$ 为递增数列"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知集合 $U=\\{-2,-1,0,1,2,3\\}, A=\\{-1,0,1\\}, B=\\{1,2\\}$, 则 $C_{U}(A \\cup B)=$ ( )", "options": {"A": "$\\{-2,3\\}$", "B": "$\\{-2,2,3\\}$", "C": "$\\{-2,-1,0,3\\}$", "D": "$\\{-2,-1$, $0,2,3\\}$"}, "label": "A", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "若 $\\alpha$ 为第四象限角, 则 ( )", "options": {"A": "$\\cos 2 \\alpha>0$", "B": "$\\cos 2 \\alpha<0$", "C": "$\\sin 2 \\alpha>0$", "D": "$\\sin 2 \\alpha<0$"}, "label": "D", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "在新冠肺炎疫情防控期间, 某超市开通网上销售业务, 每天能完成 1200 份订单的配货, 由 于订单量大幅增加, 导致订单积压.为解决困难, 许多志愿者踊跃报名参加配货工作.已知该 超市某日积压 500 份订单末配货, 预计第二天的新订单超过 1600 份的概率为 0.05 , 志愿者每 人每天能完成 50 份订单的配货, 为使第二天完成积压订单及当日订单的配货的概率不小于 0.95 , 则至少需要志愿者 ( )", "options": {"A": "10 名", "B": "18 名", "C": "24 名", "D": "32 名"}, "label": "B", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "若过点 $(2,1)$ 的圆与两坐标轴都相切, 则圆心到直线 $2 x-y-3=0$ 的距离为 ( )", "options": {"A": "$\\frac{\\sqrt{5}}{5}$", "B": "$\\frac{2 \\sqrt{5}}{5}$", "C": "$\\frac{3 \\sqrt{5}}{5}$", "D": "$\\frac{4 \\sqrt{5}}{5}$"}, "label": "B", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "数列 $\\left\\{a_{n}\\right\\}$ 中, $a_{1}=2, a_{m+n}=a_{m} a_{n}$, 若 $a_{k+1}+a_{k+2}+\\cdots+a_{k+10}=2^{15}-2^{5}$, 则 $k=( )", "options": {"A": "2", "B": "3", "C": "4", "D": "5"}, "label": "C", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "设 $O$ 为坐标原点, 直线 $x=a$ 与双曲线 $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的两条渐近线分别交于 $D, E$ 两点, 若 $\\triangle O D E$ 的面积为 8 , 则 $C$ 的焦距的最小值为 ( )", "options": {"A": "4", "B": "8", "C": "16", "D": "32"}, "label": "B", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=\\ln |2 x+1|-\\ln |2 x-1|$, 则 $f(x)( )", "options": {"A": "是偶函数, 且在 $\\left(\\frac{1}{2},+\\infty\\right)$ 单调递增", "B": "是奇函数, 且在 $\\left(-\\frac{1}{2}, \\frac{1}{2}\\right)$ 单调递减", "C": "是偶函数, 且在 $\\left(-\\infty,-\\frac{1}{2}\\right)$ 单调递增", "D": "是奇函数, 且在 $\\left(-\\infty,-\\frac{1}{2}\\right)$ 单调递减"}, "label": "D", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $\\triangle A B C$ 是面积为 $\\frac{9 \\sqrt{3}}{4}$ 的等边三角形, 且其顶点都在球 $O$ 的球面上.若球 $O$ 的表面积为 $16 \\pi$, 则 $O$ 到平面 $A B C$ 的距离为 ( )", "options": {"A": "$\\sqrt{3}$", "B": "$\\frac{3}{2}$", "C": "1", "D": "$\\frac{\\sqrt{3}}{2}$"}, "label": "C", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "若 $2^{x}-2^{y}<3^{-x}-3^{-y}$, 则 ( )", "options": {"A": "$\\ln (y-x+1)>0$", "B": "$\\ln (y-x+1)<0$", "C": "$\\ln |x-y|>0$", "D": "$\\ln |x-y|<0$"}, "label": "A", "answer": null, "other": {"source": "2020年数学试卷（理科）（新课标Ⅱ）"}}
|
||||
{"passage": null, "question": "设全集 $U=\\{-2,-1,0,1,2\\}$, 集合 $A=\\{0,1,2\\}, B=\\{-1,2\\}$, 则 $A \\cap\\left(\\partial_{U} B\\right)=( )", "options": {"A": "$\\{0,1\\}$", "B": "$\\{0,1,2\\}$", "C": "$\\{-1,1,2\\}$", "D": "$\\{0,-1,1,2\\}$"}, "label": "A", "answer": null, "other": {"source": "2022年新高考天津数学"}}
|
||||
{"passage": null, "question": "“ $x$ 为整数”是“ $2 x+1$ 为整数”的 ( )", "options": {"A": "充分不必要", "B": "必要不充分", "C": "充分必要", "D": "既不允分也不必要"}, "label": "A", "answer": null, "other": {"source": "2022年新高考天津数学"}}
|
||||
{"passage": null, "question": "$\\frac{3+i}{1+i}=( )", "options": {"A": "$1+2 i$", "B": "$1-2 i$", "C": "$2+\\mathrm{i}$", "D": "$2-\\mathrm{i}$"}, "label": "D", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设集合 $A=\\{1,2,4\\}, B=\\left\\{x \\mid x^{2}-4 x+m=0\\right\\}$. 若 $A \\cap B=\\{1\\}$, 则 $B=( )", "options": {"A": "$\\{1,-3\\}$", "B": "$\\{1,0\\}$", "C": "$\\{1,3\\}$", "D": "$\\{1,5\\}$"}, "label": "C", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "我国古代数学名著《算法统宗》中有如下问题: “远看巍巍塔七层, 红光点点倍加增, 共灯三百八十一, 请问尖头几或灯？\"意思是: 一座 7 层 塔共挂了 381 盏灯, 且相邻两层中的下一层灯数是上一层灯数的 2 倍, 则塔 的顶层共有灯 ( )", "options": {"A": "1 盏", "B": "3 或", "C": "5 盏", "D": "9 盏"}, "label": "B", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设 $x, y$ 满足约束条件 $\\left\\{\\begin{array}{l}2 x+3 y-3 \\leqslant 0 \\\\ 2 x-3 y+3 \\geqslant 0 \\\\ y+3 \\geqslant 0\\end{array}, ( )right.$ 则 $z=2 x+y$ 的最小值是 ( )", "options": {"A": "-15", "B": "-9", "C": "1", "D": "9"}, "label": "A", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "安排 3 名志愿者完成 4 项工作, 每人至少完成 1 项, 每项工作由 1 人完成, 则不同的安排方式共有 ( )", "options": {"A": "12 种", "B": "18 种", "C": "24 种", "D": "36 种"}, "label": "D", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "甲、乙、丙、丁四位同学一起去问老师询问成语竞赛的成绩. 老师 说: 你们四人中有 2 位优秀, 2 位良好, 我现在给甲看乙、丙的成绩, 给乙 看丙的成绩, 给丁看甲的成绩. 看后甲对大家说: 我还是不知道我的成绩. 根据以上信息, 则 ( )", "options": {"A": "乙可以知道四人的成绩", "B": "丁可以知道四人的成绩", "C": "乙、丁可以知道对方的成绩", "D": "乙、丁可以知道自己的成绩"}, "label": "D", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "若双曲线 $c: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的一条渐近线被圆 $(x-2)$ ${ }^{2}+y^{2}=4$ 所截得的弦长为 2 , 则 $C$ 的离心率为 ( )", "options": {"A": "2", "B": "$\\sqrt{3}$", "C": "$\\sqrt{2}$", "D": "$\\frac{2 \\sqrt{3}}{3}$"}, "label": "A", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知直三棱柱 $A B C-A_{1} B_{1} C_{1}$ 中, $\\angle A B C=120^{\\circ}, A B=2, B C=C C_{1}=1$, 则 异面直线 $A B_{1}$ 与 $B C_{1}$ 所成角的余弦值为 ( )", "options": {"A": "$\\frac{\\sqrt{3}}{2}$", "B": "$\\frac{\\sqrt{15}}{5}$", "C": "$\\frac{\\sqrt{10}}{5}$", "D": "$\\frac{\\sqrt{3}}{3}$"}, "label": "C", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "若 $x=-2$ 是函数 $f(x)=\\left(x^{2}+a x-1\\right) e^{x-1}$ 的极值点, 则 $f(x)$ 的极 小值为( )", "options": {"A": "-1", "B": "$-2 e^{-3}$", "C": "$5 e^{-3}$", "D": "1"}, "label": "A", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $\\triangle A B C$ 是边长为 2 的等边三角形, $P$ 为平面 $A B C$ 内一点, 则 $\\overrightarrow{P A}\\cdot(\\overrightarrow{\\mathrm{PB}}+\\overrightarrow{\\mathrm{PC}})$ 的最小值是 ( )", "options": {"A": "-2", "B": "$-\\frac{3}{2}$", "C": "$-\\frac{4}{3}$", "D": "-1"}, "label": "B", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{x \\mid-1<x<1\\}, B=\\{x \\mid 0 \\leq x \\leq 2\\}$, 则 $A \\cup B=( )", "options": {"A": "$(-1,2)$", "B": "$(-1,2]$", "C": "$[0,1)$", "D": "$[0,1]$"}, "label": "B", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "在复平面内, 复数 $z$ 满足 $(1-i) z=2$, 则 $z=( )", "options": {"A": "$2+i$", "B": "$2-i$", "C": "$1-i$", "D": "$1+i$"}, "label": "D", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "已知 $f(x)$ 是定义在上 $[0,1]$ 的函数, 那么 “函数 $f(x)$ 在 $[0,1]$ 上单调递增” 是 “函数 $f(x)$ 在 $[0,1]$ 上的最大值为 $f(1) ”$ 的 ( )", "options": {"A": "充分而不必要条件", "B": "必要而不充分条件", "C": "充分必要条件", "D": "既不充分也不必要条件"}, "label": "A", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "双曲线 $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1$ 过点 $(\\sqrt{2}, \\sqrt{3})$, 且离心率为 2 , 则该双曲线的标准方程为 ( )", "options": {"A": "$x^{2}-\\frac{y^{2}}{3}=1$", "B": "$\\frac{x^{2}}{3}-y^{2}=1$", "C": "$x^{2}-\\frac{\\sqrt{3} y^{2}}{3}=1$", "D": "$\\frac{\\sqrt{3} x^{2}}{3}-y^{2}=1$"}, "label": "A", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "$\\left\\{a_{n}\\right\\}$ 和 $\\left\\{b_{n}\\right\\}$ 是两个等差数列, 其中 $\\frac{a_{k}}{b_{k}}(1 \\leq k \\leq 5)$ 为常值, $a_{1}=288, a_{5}=96, b_{1}=192$, 则 $b_{3}= ( )", "options": {"A": "64", "B": "128", "C": "256", "D": "512"}, "label": "B", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "函数 $f(x)=\\cos x-\\cos 2 x$, 试判断函数的奇偶性及最大值 ( )", "options": {"A": "奇函数, 最大值为 2", "B": "偶函数, 最大值为 2", "C": "奇函数, 最大值为 $\\frac{9}{8}$", "D": "偶函数，最大值为 $\\frac{9}{8}$"}, "label": "D", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "已知圆 $C: x^{2}+y^{2}=4$, 直线 $l: y=k x+m$, 当 $k$ 变化时, $l$ 截得圆 $C$ 弦长的最小值为 2 , 则 $m=( )", "options": {"A": "$\\pm 2$", "B": "$\\pm \\sqrt{2}$", "C": "$\\pm \\sqrt{3}$", "D": "$\\pm \\sqrt{5}$"}, "label": "C", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "数列 $\\left\\{a_{n}\\right\\}$ 是递增的整数数列, 且 $a_{1} \\geq 3, a_{1}+a_{2}+\\cdots+a_{n}=100$, 则 $n$ 的最大值为 ( )", "options": {"A": "9", "B": "10", "C": "11", "D": "12"}, "label": "C", "answer": null, "other": {"source": "2021北京高考数学"}}
|
||||
{"passage": null, "question": "设全集 $U=\\{1,2,3,4,5\\}$, 集合 $M$ 满足 $C_{U} M=\\{1,3\\}$, 则 ( )", "options": {"A": "$2 \\in M$", "B": "$3 \\in M$", "C": "$4 \\notin M$", "D": "$5 \\notin M$"}, "label": "A", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知 $z=1-2 i$, 且 $z+a \\bar{z}+b=0$, 其中 $a, b$ 为实数, 则 ( )", "options": {"A": "$a=1, b=-2$", "B": "$a=-1, b=2$", "C": "$a=1, b=2$", "D": "$a=-1, b=-2$"}, "label": "A", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知向量 $\\vec{a}, \\vec{b}$ 满足 $|\\vec{a}|=1,|\\vec{b}|=\\sqrt{3},|\\vec{a}-2 \\vec{b}|=3$, 则 $\\vec{a} \\cdot \\vec{b}=( )", "options": {"A": "$-2$", "B": "$-1$", "C": "1", "D": "2"}, "label": "C", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "嫦娥二号卫星在完成探月任务后, 继续进行深空探测, 成为我国第一颗环绕太阳飞行 人造行星, 为研究嫦娥二号绕日周期与地球绕日周期的比值, 用到数列 $\\left\\{b_{n}\\right\\}$ : $b_{1}=1+\\frac{1}{\\alpha_{1}}, ( ) b_{2}=1+\\frac{1}{\\alpha_{1}+\\frac{1}{\\alpha_{2}}}, ( ) b_{3}=1+\\frac{1}{\\alpha_{1}+\\frac{1}{\\alpha_{2}+\\frac{1}{\\alpha_{3}}}}, \\ldots$, 依此类推, 其中 $\\alpha_{k} \\in \\mathbf{N}^{*}(k=1,2, \\cdots)$. 则 ( )", "options": {"A": "$b_{1}<b_{5}$", "B": "$b_{3}<b_{8}$", "C": "$b_{6}<b_{2}$", "D": "$b_{4}<b_{7}$"}, "label": "D", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "设 $F$ 为抛物线 $C: y^{2}=4 x$ 的焦点, 点 $A$ 在 $C$ 上, 点 $B(3,0)$, 若 $|A F|=|B F|$, 则 $|A B|=( )", "options": {"A": "2", "B": "$2 \\sqrt{2}$", "C": "3", "D": "$3 \\sqrt{2}$"}, "label": "B", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "在正方体 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, $E, F$ 分别为 $A B, B C$ 的中点, 则 ( )", "options": {"A": "平面 $B_{1} E F \\perp$ 平面 $B D D_{1}$", "B": "平面 $B_{1} E F \\perp$ 平面 $A_{1} B D$", "C": "平面 $B_{1} E F / /$ 平面 $A_{1} A C$", "D": "平面 $B_{1} E F / /$ 平面 $A_{1} C_{1} D$"}, "label": "A", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知等比数列 $\\left\\{a_{n}\\right\\}$ 的前 3 项和为 $168, a_{2}-a_{5}=42$, 则 $a_{6}= ( )", "options": {"A": "14", "B": "12", "C": "6", "D": "3"}, "label": "D", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知球 $O$ 的半径为 1 , 四棱雉的顶点为 $O$, 底面的四个顶点均在球 $O$ 的球面上, 则当该四棱雉的体积最大时, 其高为 ( )", "options": {"A": "$\\frac{1}{3}$", "B": "$\\frac{1}{2}$", "C": "$\\frac{\\sqrt{3}}{3}$", "D": "$\\frac{\\sqrt{2}}{2}$"}, "label": "C", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "某棋手与甲、乙、丙三位棋手各比赛一盘, 各盘比赛结果相互独立. 已知该棋手与甲、乙、丙比赛获胜 概率分别为 $p_{1}, p_{2}, p_{3}$, 且 $p_{3}>p_{2}>p_{1}>0$. 记该棋手连胜两盘的 概率为 $p$, 则 ( )", "options": {"A": "$p$ 与该棋手和甲、乙、丙的比赛次序无关", "B": "该棋手在第二盘与甲比赛, $p$ 最大", "C": "该棋手在第二盘与乙比赛, $p$ 最大", "D": "该棋手在第二盘与丙比赛, $p$ 最大"}, "label": "D", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "双曲线 $C$ 的两个焦点为 $F_{1}, F_{2}$, 以 $C$ 的实轴为直径的圆记为 $D$, 过 $F_{1}$ 作 $D$ 的切线与 $C$ 的两支交于 $M, N$ 两点, 且 $\\cos \\angle F_{1} N F_{2}=\\frac{3}{5}$, 则 $C$ 的离心率为 ( )", "options": {"A": "$\\frac{\\sqrt{5}}{2}$", "B": "$\\frac{3}{2}$", "C": "$\\frac{\\sqrt{13}}{2}$", "D": "$\\frac{\\sqrt{17}}{2}$"}, "label": "D", "answer": null, "other": {"source": "2022年全国高考乙卷数学"}}
|
||||
{"passage": null, "question": "已知集合 $M=\\left\\{x \\mid(x-1)^{2}<4, x \\in R\\right\\}, N=\\{-1,0,1,2,3\\}$, 则 $M \\cap N=( )", "options": {"A": "$\\{0,1,2\\}$", "B": "$\\{-1,0,1,2\\}$", "C": "$\\{-1,0,2,3\\}$", "D": "$\\{0,1,2,3\\}$"}, "label": "A", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设复数 $z$ 满足 $(1-i) z=2 i$, 则 $z=( )", "options": {"A": "$-1+i$", "B": "$-1-i$", "C": "$1+i$", "D": "$1-\\mathrm{i}$"}, "label": "C", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $m, n$ 为异面直线, $m \\perp$ 平面 $\\alpha, n \\perp$ 平面 $\\beta$. 直线 $\\mid$ 满足 $\\mid \\perp m$, $\\mathrm{I} \\perp \\mathrm{n},|\\not \\subset \\alpha, ( )| \\not \\subset \\beta, ( ) 则 ( )", "options": {"A": "$\\alpha / / \\beta$ 且 $\\mathrm{l} / / \\alpha$", "B": "$\\alpha \\perp \\beta$ 且 $\\mid \\perp \\beta$", "C": "$\\alpha$ 与 $\\beta$ 相交, 且交线垂直于$\\mid$", "D": "$\\alpha$ 与 $\\beta$ 相交, 且交线平行于$\\mid$"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $(1+a x)(1+x){ }^{5}$ 的展开式中 $x^{2}$ 的系数为 5 , 则 $a=( )", "options": {"A": "-4", "B": "-3", "C": "-2", "D": "-1"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设 $a=\\log _{3} 6, b=\\log _{5} 10, c=\\log _{7} 14$, 则 ( )", "options": {"A": "$c>b>a$", "B": "$b>c>a$", "C": "$a>c>b$", "D": "$a>b>c$"}, "label": "D", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=x^{3}+a x^{2}+b x+c$, 下列结论中错误的是 ( )", "options": {"A": "$\\exists x_{0} \\in R, f\\left(x_{0}\\right)=0$", "B": "函数 $y=f(x)$ 的图象是中心对称图形", "C": "若 $x_{0}$ 是 $f(x)$ 的极小值点, 则 $f(x)$ 在区间 $\\left(-\\infty, x_{0}\\right)$ 单调递减", "D": "若 $x_{0}$ 是 $f(x)$ 的极值点, 则 $f^{\\prime}\\left(x_{0}\\right)=0$"}, "label": "C", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设抛物线 $C: y^{2}=2 p x(p>0)$ 的焦点为 $F$, 点 $M$ 在 $C$ 上, $|M F|=5$, 若以 MF 为直径的圆过点 $(0,2)$, 则 $C$ 的方程为 ( )", "options": {"A": "$y^{2}=4 x$ 或 $y^{2}=8 x$", "B": "$y^{2}=2 x$ 或 $y^{2}=8 x$", "C": "$y^{2}=4 x$ 或 $y^{2}=16 x$", "D": "$y^{2}=2 x$ 或 $y^{2}=16 x$"}, "label": "C", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知点 $A(-1,0), B(1,0), C(0,1)$, 直线 $y=a x+b(a>0)$ 将 $\\triangle A B C$ 分割为面积相等的两部分, 则 $b$ 的取值范围是 ( )", "options": {"A": "$(0,1)$", "B": "$\\left(1-\\frac{\\sqrt{2}}{2}, \\frac{1}{2}\\right)$ ", "C": "$\\left(1-\\frac{\\sqrt{2}}{2}, \\frac{1}{3}\\right]$", "D": "$\\left[\\frac{1}{3}, \\frac{1}{2}\\right)$"}, "label": "B", "answer": null, "other": {"source": "2013年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设集合 $A=\\{-1,0,1\\}, B=\\{1,3,5\\}, C=\\{0,2,4\\}$, 则 $(A \\cap B) \\cup C=$ ( )", "options": {"A": "$\\{0\\}$", "B": "$\\{0,1,3,5\\}$", "C": "$\\{0,1,2,4\\}$", "D": "$\\{0,2,3,4\\}$"}, "label": "C", "answer": null, "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "已知 $a \\in \\mathbf{R}$, 则 “ $a>6$ ”是“ $a^{2}>36$ ”的 ( )", "options": {"A": "充分不必要条件", "B": "必要不充分条件", "C": "充要条件", "D": "既不允分也不必要条件 "}, "label": "A", "answer": null, "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "设 $a=\\log _{2} 0.3, b=\\log _{\\frac{1}{2}} 0.4, c=0.4^{0.3}$, 则 $a, b, c$ 的大小关系为 ( )", "options": {"A": "$a<b<c$", "B": "$c<a<b$", "C": "$b<c<a$", "D": "$a<c<b$"}, "label": "D", "answer": null, "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "两个圆雉的底面是一个球的同一截面, 顶点均在球面上, 若球的体积为 $\\frac{32 \\pi}{3}$, 两个圆雉的高之比为 $1: 3$, 则这两个圆雉的体积之和为 ( )", "options": {"A": "$3 \\pi$", "B": "$4 \\pi$", "C": "$9 \\pi$", "D": "$12 \\pi$"}, "label": "B", "answer": null, "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "若 $2^{a}=5^{b}=10$, 则 $\\frac{1}{a}+\\frac{1}{b}=( )", "options": {"A": "$-1$", "B": "$\\lg 7$", "C": "1", "D": "$\\log _{7} 10$"}, "label": "C", "answer": null, "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "已知双曲线 $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的右焦点与抛物线 $y^{2}=2 p x(p>0)$ 的焦点重 合, 抛物线的准线交双曲线于 $A, B$ 两点, 交双曲线的渐近线于 $C 、 D$ 两点, 若 $|C D|=\\sqrt{2}|A B|$. 则双曲线的离心率为 ( )", "options": {"A": "$\\sqrt{2}$", "B": "$\\sqrt{3}$", "C": "$2$", "D": "$3$"}, "label": "A", "answer": null, "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "设 $a \\in \\mathbf{R}$, 函数 $f(x)=\\left\\{\\begin{array}{ll}\\cos (2 \\pi x-2 \\pi a) . & x<a \\\\ x^{2}-2(a+1) x+a^{2}+5, & x \\geq a\\end{array}\\right.$, 若 $f(x)$ 在区间 $(0,+\\infty)$ 内 恰有 6 个零点, 则 $a$ 的取值范围是 ( )", "options": {"A": "$\\left(2, \\frac{9}{4}\\right] \\cup\\left(\\frac{5}{2}, \\frac{11}{4}\\right]$", "B": "$\\left(\\frac{7}{4}, 2\\right) \\cup\\left(\\frac{5}{2}, \\frac{11}{4}\\right)$", "C": "$\\left(2, \\frac{9}{4}\\right] \\cup\\left[\\frac{11}{4}, 3\\right)$", "D": "$\\left(\\frac{7}{4}, 2\\right) \\cup\\left[\\frac{11}{4}, 3\\right)$"}, "label": "A", "answer": null, "other": {"source": "2021年天津市高考数学"}}
|
||||
{"passage": null, "question": "$\\frac{10 i}{2-i}=( )", "options": {"A": "$-2+4 i$", "B": "$-2-4 i$", "C": "$2+4 \\mathrm{i}$", "D": "$2-4 i$"}, "label": "A", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "设集合 $A=\\{x|| x \\mid>3\\}, B=\\left\\{x \\mid \\frac{x-1}{x-4}<0\\right\\}$, 则 $A \\cap B=$( )", "options": {"A": "$\\phi$", "B": "$(3,4)$", "C": "$(-2,1)$", "D": "$(4,+\\infty)$"}, "label": "B", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "函数 $y=\\frac{x}{2 x-1}$ 在点 $(1,1)$ 处的切线方程为 ( )", "options": {"A": "$x-y-2=0$", "B": "$x+y-2=0$", "C": "$x+4 y-5=0$", "D": "$x-4 y+3=0$"}, "label": "B", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "已知正四棱柱 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, $A A_{1}=2 A B, E$ 为 $A A_{1}$ 中点, 则异 面直线 $\\mathrm{BE}$ 与 $\\mathrm{CD}_{1}$ 所形成角的余弦值为( )", "options": {"A": "$\\frac{\\sqrt{10}}{10}$", "B": "$\\frac{1}{5}$", "C": "$\\frac{3 \\sqrt{10}}{10}$", "D": "$\\frac{3}{5}$"}, "label": "C", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "已知向量 $\\vec{a}=(2,1), \\vec{a} \\cdot \\vec{b}=10,|\\vec{a}+\\vec{b}|=5 \\sqrt{2}$, 则 $|\\vec{b}|=$( )", "options": {"A": "$\\sqrt{5}$", "B": "$\\sqrt{10}$", "C": "5", "D": "25"}, "label": "C", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "设 $a=\\log _{3} \\pi, b=\\log _{2} \\sqrt{3}, c=\\log _{3} \\sqrt{2}$, 则( )", "options": {"A": "$a>b>c$", "B": "$a>c>b$", "C": "$b>a>c$", "D": "$b>c>a$"}, "label": "D", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "已知直线 $y=k(x+2)(k>0)$ 与抛物线 $C: y^{2}=8 x$ 相交于 $A$ 、 $B$ 两点, $F$ 为 $C$ 的焦点, 若 $|F A|=2|F B|$, 则 $k=$( )", "options": {"A": "$\\frac{1}{3}$", "B": "$\\frac{\\sqrt{2}}{3}$", "C": "$\\frac{2}{3}$", "D": "$\\frac{2 \\sqrt{2}}{3}$"}, "label": "D", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅱ）"}}
|
||||
{"passage": null, "question": "设复数 $z$ 满足 $\\frac{1+z}{1-z}=i$, 则 $|z|=( )", "options": {"A": "1", "B": "$\\sqrt{2}$", "C": "$\\sqrt{3}$", "D": "2"}, "label": "A", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$\\sin 20^{\\circ} \\cos 10^{\\circ}-\\cos 160^{\\circ} \\sin 10^{\\circ}=( )", "options": {"A": "$\\frac{\\sqrt{3}}{2}$", "B": "$\\frac{\\sqrt{3}}{2}$", "C": "$-\\frac{1}{2}$", "D": "$\\frac{1}{2}$"}, "label": "D", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设命题 $p: \\exists n \\in N, n^{2}>2^{n}$, 则 $\\neg p$ 为 ( )", "options": {"A": "$\\forall n \\in N, n^{2}>2^{n}$", "B": "$\\exists n \\in N, n^{2} \\leqslant 2^{n}$", "C": "$\\forall n \\in N, n^{2} \\leqslant 2^{n}$", "D": "$\\exists n \\in N, n^{2}=2^{n}$"}, "label": "C", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "投篮测试中, 每人投 3 次, 至少投中 2 次才能通过测试. 已知某同 学每次投篮投中的概率为 0.6 , 且各次投篮是否投中相互独立, 则该同学通 过测试的概率为 ( )", "options": {"A": "0.648", "B": "0.432", "C": "0.36", "D": "0.312"}, "label": "A", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设 $\\mathrm{D}$ 为 $\\triangle \\mathrm{ABC}$ 所在平面内一点, $\\overrightarrow{\\mathrm{BC}}=3 \\overrightarrow{\\mathrm{CD}}$, 则 ( )", "options": {"A": "$\\overrightarrow{\\mathrm{AD}}=-\\frac{1}{3} \\overrightarrow{\\mathrm{AB}}+\\frac{4}{3} \\overrightarrow{\\mathrm{AC}}$", "B": "$\\overrightarrow{\\mathrm{AD}}=\\frac{1}{3} \\overrightarrow{\\mathrm{AB}}-\\frac{4}{3} \\overrightarrow{\\mathrm{AC}}$", "C": "$\\overrightarrow{\\mathrm{AD}}=\\frac{4}{3} \\overrightarrow{\\mathrm{AB}}+\\frac{1}{3} \\overrightarrow{\\mathrm{AC}}$", "D": "$\\overrightarrow{\\mathrm{AD}}=\\frac{4}{3} \\overrightarrow{\\mathrm{AB}}-\\frac{1}{3} \\overrightarrow{\\mathrm{AC}}$"}, "label": "A", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "$\\left(x^{2}+x+y\\right){ }^{5}$ 的展开式中, $x^{5} y^{2}$ 的系数为 ( )", "options": {"A": "10", "B": "20", "C": "30", "D": "60"}, "label": "C", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=e^{x}(2 x-1)-a x+a$, 其中 $a<1$, 若存在唯一的整数 $x_{0}$ 使得 $\\mathrm{f}\\left(\\mathrm{x}_{0}\\right)<0$, 则 $\\mathrm{a}$ 的取值范围是 ( )", "options": {"A": "$\\left[-\\frac{3}{2 \\mathrm{e}}, 1\\right)$", "B": "$\\left[-\\frac{3}{2 \\mathrm{e}}, \\frac{3}{4}\\right)$", "C": "$\\left[\\frac{3}{2 \\mathrm{e}}, \\frac{3}{4}\\right)$", "D": "$\\left[\\frac{3}{2 \\mathrm{e}}, 1\\right)$"}, "label": "D", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知全集 $U=\\{x \\mid-3<x<3\\}$, 集合 $A=\\{x \\mid-2<x \\leq 1\\}$, 则 $C_{U} A=( )", "options": {"A": "$(-2,1]$", "B": "$(-3,-2) \\cup[1,3)$", "C": "$[-2,1)$", "D": "$(-3,-2] \\cup(1,3)$"}, "label": "D", "answer": null, "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "若复数 $z$ 满足 $\\mathrm{i} \\cdot z=3-4 \\mathrm{i}$, 则 $|z|=( )", "options": {"A": "1", "B": "5", "C": "7", "D": "25"}, "label": "B", "answer": null, "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "若直线 $2 x+y-1=0$ 是圆 $(x-a)^{2}+y^{2}=1$ 的一条对称轴, 则 $a=( )", "options": {"A": "$\\frac{1}{2}$", "B": "$-\\frac{1}{2}$", "C": "1", "D": "$-1$"}, "label": "A", "answer": null, "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "己知函数 $f(x)=\\frac{1}{1+2^{x}}$, 则对任意实数 $x$, 有 ( )", "options": {"A": "$f(-x)+f(x)=0$", "B": "$f(-x)-f(x)=0$", "C": "$f(-x)+f(x)=1$", "D": "$f(-x)-f(x)=\\frac{1}{3}$"}, "label": "C", "answer": null, "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "若 $(2 x-1)^{4}=a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0}$, 则 $a_{0}+a_{2}+a_{4}=( )", "options": {"A": "40", "B": "41", "C": "$-40$", "D": "$-41$"}, "label": "B", "answer": null, "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "在 $\\triangle A B C$ 中, $A C=3, B C=4, \\angle C=90^{\\circ} . P$ 为 $\\triangle A B C$ 所在平面内的动点, 且 $P C=1$, 则 $\\overrightarrow{P A} \\cdot \\overrightarrow{P B}$ 的取值范围是 ( )", "options": {"A": "$[-5,3]$", "B": "$[-3,5]$", "C": "$[-6,4]$", "D": "$[-4,6]$"}, "label": "D", "answer": null, "other": {"source": "2022年北京市高考数学"}}
|
||||
{"passage": null, "question": "若集合 $M=\\{x \\mid \\sqrt{x}<4\\}, N=\\{x \\mid 3 x \\geq 1\\}$, 则 $M \\cap N=( )", "options": {"A": "$\\{x \\mid 0 \\leq x<2\\}$", "B": "$\\left\\{x \\mid \\frac{1}{3} \\leq x<2\\right\\}$", "C": "$\\{x \\mid 3 \\leq x<16\\}$", "D": "$\\left\\{x \\mid \\frac{1}{3} \\leq x<16\\right\\}$"}, "label": "D", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "在 $\\triangle A B C$ 中, 点 $D$ 在边 $A B$ 上, $B D=2 D A$. 记 $\\overrightarrow{C A}=\\vec{m}, \\overrightarrow{C D}=\\vec{n}$, 则 $\\overrightarrow{C B}=( )", "options": {"A": "$3 \\vec{m}-2 \\vec{n}$", "B": "$-2 \\vec{m}+3 \\vec{n}$", "C": "$3 \\vec{m}+2 \\vec{n}$", "D": "$2 \\vec{m}+3 \\vec{n}$"}, "label": "B", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "南水北调工程缓解了北方一些地区水资源短缺问题, 其中一部分水蓄入某水库.已知该水 库水位为海拔 $148.5 \\mathrm{~m}$ 时, 相应水面的面积为 $140.0 \\mathrm{~km}^{2}$; 水位为海拔 $157.5 \\mathrm{~m}$ 时, 相应水 面的面积为 $180.0 \\mathrm{~km}^{2}$, 将该水库在这两个水位间的形状看作一个棱台, 则该水库水位从 海拔 $148.5 \\mathrm{~m}$ 上升到 $157.5 \\mathrm{~m}$ 时, 增加的水量约为 $(\\sqrt{7} \\approx 2.65)( )", "options": {"A": "$1.0 \\times 10^{9} \\mathrm{~m}^{3}$", "B": "$1.2 \\times 10^{9} \\mathrm{~m}^{3}$", "C": "$1.4 \\times 10^{9} \\mathrm{~m}^{3}$", "D": "$1.6 \\times 10^{9} \\mathrm{~m}^{3}$"}, "label": "C", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "从 2 至 8 的 7 个整数中随机取 2 个不同的数, 则这 2 个数互质的概率为 ( )", "options": {"A": "$\\frac{1}{6}$", "B": "$\\frac{1}{3}$", "C": "$\\frac{1}{2}$", "D": "$\\frac{2}{3}$"}, "label": "A", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "设 $a=0.1 \\mathrm{e}^{0.1}, b=\\frac{1}{9}, c=-\\ln 0.9$, 则 ( )", "options": {"A": "$a<b<c$", "B": "$c<b<a$", "C": "$c<a<b$", "D": "$a<c<b$"}, "label": "C", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "已知正四棱雉的侧棱长为 $l$, 其各顶点都在同一球面上. 若该球的体积为 $36 \\pi$, 且 $3 \\leq l \\leq 3 \\sqrt{3}$, 则该正四棱雉体积的取值范围是 ( )", "options": {"A": "$\\left[18, \\frac{81}{4}\\right]$", "B": "$\\left[\\frac{27}{4}, \\frac{81}{4}\\right]$", "C": "$\\left[\\frac{27}{4}, \\frac{64}{3}\\right]$", "D": "[18, 27]"}, "label": "C", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "已知正方体 $A B C D-A_{1} B_{1} C_{1} D_{1}$, 则 ( )", "options": {"A": "直线 $B C_{1}$ 与 $D A_{1}$ 所成的角为 $90^{\\circ}$", "B": "直线 $B C_{1}$ 与 $C A_{1}$ 所成的角为 $90^{\\circ}$", "C": "直线 $B C_{1}$ 与平面 $B B_{1} D_{1} D$ 所成的角为 $45^{\\circ}$", "D": "直线 $B C_{1}$ 与平面 $A B C D $ 所成的角为 $45^{\\circ}$"}, "label": "A B D", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=x^{3}-x+1$, 则 ( )", "options": {"A": "$f(x)$ 有两个极值点", "B": "$f(x)$ 有三个零点", "C": "点 $(0,1)$ 是曲线 $y=f(x)$ 的对称中心", "D": "直线 $y=2 x$ 是曲线 $y=f(x)$ 的切"}, "label": "A C", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "已知 $O$ 为坐标原点, 点 $A(1,1)$ 在抛物线 $C: x^{2}=2 p y(p>0)$ 上, 过点 $B(0,-1)$ 的直线 交 $C$ 于 $P, Q$ 两点, 则 ( )", "options": {"A": "$C$ 的准线为 $y=-1$", "B": "直线 $A B$ 与 $C$ 相切", "C": "$|O P| \\cdot|O Q|>|O A|^{2}$", "D": "$|B P| \\cdot|B Q|>|B A|^{2}$"}, "label": "B C D", "answer": null, "other": {"source": "2022年全国新高考I卷数学"}}
|
||||
{"passage": null, "question": "设集合 $A=\\left\\{x \\mid x^{2}-5 x+6>0\\right\\}, B=\\{x \\mid x-1<0\\}$, 则 $A \\cap B=$", "options": {"A": "$(-\\infty, 1)$", "B": "$(-2,1)$", "C": "$(-3,-1)$", "D": "$(3,+\\infty)$"}, "label": "A", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "设 $z=-3+2 \\mathrm{i}$, 则在复平面内 $\\bar{z}$ 对应的点位于", "options": {"A": "第一象限", "B": "第二象限", "C": "第三象限", "D": "第四象限"}, "label": "C", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
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{"passage": null, "question": "演讲比赛共有 9 位评委分别给出某选手的原始评分, 评定该选手的成绩时, 从 9 个原 始评分中去掉 1 个最高分、 1 个最低分, 得到 7 个有效评分. 7 个有效评分与 9 个原始评分 相比, 不变的数字特征是", "options": {"A": "中位数", "B": "平均数", "C": "方差", "D": "极差"}, "label": "A", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "若 $a>b$, 则", "options": {"A": "$\\ln (a-b)>0$", "B": "$3^{a}<3^{b}$", "C": "$a^{3}-b^{3}>0$", "D": "$|a|>|b|$"}, "label": "C", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "设 $\\alpha, \\beta$ 为两个平面, 则 $\\alpha / / \\beta$ 的充要条件是", "options": {"A": "$\\alpha$ 内有无数条直线与 $\\beta$ 平行", "B": "$\\alpha$ 内有两条相交直线与 $\\beta$ 平行", "C": "$\\alpha, \\beta$ 平行于同一条直线", "D": "$\\alpha, \\beta$ 垂直于同一平面"}, "label": "B", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "下列函数中, 以 $\\frac{\\pi}{2}$ 为周期且在区间 $\\left(\\frac{\\pi}{4}, \\frac{\\pi}{2}\\right)$ 单调递增的是", "options": {"A": "$f(x)=|\\cos 2 x|$", "B": "$f(x)=|\\sin 2 x|$", "C": "$f(x)=\\cos |x|$", "D": "$f(x)=\\sin |x|$"}, "label": "A", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "设 $F$ 为双曲线 $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的右焦点, $O$ 为坐标原点, 以 $O F$ 为直径的 圆与圆 $x^{2}+y^{2}=a^{2}$ 交于 $P$、 $Q$ 两点. 若 $|P Q|=|O F|$, 则 $C$ 的离心率为", "options": {"A": "$\\sqrt{2}$", "B": "$\\sqrt{3}$", "C": "2", "D": "$\\sqrt{5}$"}, "label": "A", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "设函数 $f(x)$ 的定义域为 $\\mathbf{R}$, 满足 $f(x+1)=2 f(x)$, 且当 $x \\in(0,1]$ 时, $f(x)=x(x-1)$. 若对任意 $x \\in(-\\infty, m]$, 都有 $f(x) \\geq-\\frac{8}{9}$, 则 $m$ 的取值范围是", "options": {"A": "$\\left(-\\infty, \\frac{9}{4}\\right]$", "B": "$\\left(-\\infty, \\frac{7}{3}\\right]$", "C": "$\\left(-\\infty, \\frac{5}{2}\\right]$", "D": "$\\left(-\\infty, \\frac{8}{3}\\right]$"}, "label": "B", "answer": null, "other": {"source": "2019年新课标ⅱ数学"}}
|
||||
{"passage": null, "question": "已知集合 $\\left.A=\\left\\{(x, y) \\mid x^{2}+y^{2}=1\\right\\}, B=\\{（ x, y) \\mid y=x\\right\\}$, 则 $A \\cap B$ 中元 素的个数为 ( )", "options": {"A": "3", "B": "2", "C": "1", "D": "0"}, "label": "B", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "设复数 $z$ 满足 $(1+i) z=2 i$, 则 $|z|=( )", "options": {"A": "$\\frac{1}{2}$", "B": "$\\frac{\\sqrt{2}}{2}$", "C": "$\\sqrt{2}$", "D": "2"}, "label": "C", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "$(x+y)(2 x-y)^{5}$ 的展开式中的 $x^{3} y^{3}$ 系数为 ( )", "options": {"A": "-80", "B": "-40", "C": "40", "D": "80"}, "label": "C", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知双曲线 $c: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 ( )(a>0, b>0)$ 的一条渐近线方程为 $y=$ $\\frac{\\sqrt{5}}{2} x$, 且与椭圆 $\\frac{x^{2}}{12}+\\frac{y^{2}}{3}=1$ 有公共焦点, 则 $C$ 的方程为 ( )", "options": {"A": "$\\frac{x^{2}}{8}-\\frac{y^{2}}{10}=1$", "B": "$\\frac{x^{2}}{4}-\\frac{y^{2}}{5}=1$", "C": "$\\frac{x^{2}}{5}-\\frac{y^{2}}{4}=1$", "D": "$\\frac{x^{2}}{4}-\\frac{y^{2}}{3}=1$"}, "label": "B", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "设函数 $f(x)=\\cos \\left(x+\\frac{\\pi}{3}\\right)$ ，则下列结论错误的是 ( )", "options": {"A": "$f(x)$ 的一个周期为 $-2 \\pi$", "B": "$y=f(x)$ 的图象关于直线 $x=\\frac{8 \\pi}{3}$ 对称", "C": "$f(x+\\pi)$ 的一个零点为 $x=\\frac{\\pi}{6}$", "D": "$f(x)$ 在 $\\left(\\frac{\\pi}{2}, \\pi\\right)$ 单调递减"}, "label": "D", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知圆柱的高为 1 , 它的两个底面的圆周在直径为 2 的同一个球的球 面上，则该圆柱的体积为 ( )", "options": {"A": "$\\pi$", "B": "$\\frac{3 \\pi}{4}$", "C": "$\\frac{\\pi}{2}$", "D": "$\\frac{\\pi}{4}$"}, "label": "B", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "等差数列 $\\left\\{a_{n}\\right\\}$ 的首项为 1 , 公差不为 0 . 若 $a_{2}, a_{3}, a_{6}$ 成等比数列, 则 $\\left\\{a_{n}\\right\\}$ 前 6 项的和为 ( )", "options": {"A": "-24", "B": "-3", "C": "3", "D": "8"}, "label": "A", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=x^{2}-2 x+a\\left(e^{x-1}+e^{-x+1}\\right)$ 有唯一零点, 则 $a=( )", "options": {"A": "$-\\frac{1}{2}$", "B": "$\\frac{1}{3}$", "C": "$\\frac{1}{2}$", "D": "1"}, "label": "C", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "在矩形 $A B C D$ 中, $A B=1, A D=2$, 动点 $P$ 在以点 $C$ 为圆心且与 $B D$ 相 切的圆上. 若 $\\overrightarrow{\\mathrm{AP}}=\\lambda \\overrightarrow{\\mathrm{AB}}+\\mu \\overrightarrow{\\mathrm{AD}}$, 则 $\\lambda+\\mu$ 的最大值为 ( )", "options": {"A": "3", "B": "$2 \\sqrt{2}$", "C": "$\\sqrt{5}$", "D": "2"}, "label": "A", "answer": null, "other": {"source": "2017年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\left\\{(x, y) \\mid x, y \\in \\mathbf{N}^{*}, y \\geq x\\right\\}, B=\\{(x, y) \\mid x+y=8\\}$, 则 $A \\cap B$ 中元素的个数为 ( )", "options": {"A": "2", "B": "3", "C": "4", "D": "6"}, "label": "C", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "在一组样本数据中, $1,2,3,4$ 出现的频率分别为 $p_{1}, p_{2}, p_{3}, p_{4}$, 且 $\\sum_{i=1}^{4} p_{i}=1$, 则下面四种 情形中, 对应样本的标准差最大的一组是 ( )", "options": {"A": "$p_{1}=p_{4}=0.1, p_{2}=p_{3}=0.4$", "B": "$p_{1}=p_{4}=0.4, p_{2}=p_{3}=0.1$", "C": "$p_{1}=p_{4}=0.2, p_{2}=p_{3}=0.3$", "D": "$p_{1}=p_{4}=0.3, p_{2}=p_{3}=0.2$"}, "label": "B", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "设 $O$ 为坐标原点, 直线 $x=2$ 与抛物线 $C: y^{2}=2 p x(p>0)$ 交于 $D, E$ 两点, 若 $O D \\perp O E$, 则 $C$ 的 焦点坐标为 ( )", "options": {"A": "$\\left(\\frac{1}{4}, 0\\right)$", "B": "$\\left(\\frac{1}{2}, 0\\right)$", "C": "$(1,0)$", "D": "$(2,0)$"}, "label": "B", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "已知向量 $\\boldsymbol{a}, \\boldsymbol{b}$ 满足 $|a|=5,|b|=6, \\boldsymbol{a} \\cdot b=-6$, 则 $\\cos \\langle\\boldsymbol{a}, \\boldsymbol{a}+\\boldsymbol{b}\\rangle=( )", "options": {"A": "$-\\frac{31}{35}$", "B": "$-\\frac{19}{35}$", "C": "$\\frac{17}{35}$", "D": "$\\frac{19}{35}$"}, "label": "D", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "在 $\\triangle A B C$ 中, $\\cos C=\\frac{2}{3}, A C=4, B C=3$, 则 $\\cos B=( )", "options": {"A": "$\\frac{1}{9}$", "B": "$\\frac{1}{3}$", "C": "$\\frac{1}{2}$", "D": "$\\frac{2}{3}$"}, "label": "A", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "已知 $2 \\tan \\theta-\\tan \\left(\\theta+\\frac{\\pi}{4}\\right)=7$, 则 $\\tan \\theta=( )", "options": {"A": "$-2$", "B": "$-1$", "C": "1", "D": "2"}, "label": "D", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "若直线 $l$ 与曲线 $y=\\sqrt{x}$ 和 $x^{2}+y^{2}=\\frac{1}{5}$ 都相切, 则 $l$ 的方程为 ( )", "options": {"A": "$y=2 x+1$", "B": "$y=2 x+\\frac{1}{2}$", "C": "$y=\\frac{1}{2} x+1$", "D": "$y=\\frac{1}{2} x+\\frac{1}{2}$"}, "label": "D", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "设双曲线 $C: \\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1(a>0, b>0)$ 的左、右焦点分别为 $F_{1}, F_{2}$, 离心率为 $\\sqrt{5} . P$ 是 $C$ 上一点, 且 $F_{1} P \\perp F_{2} P$. 若 $\\triangle P F_{1} F_{2}$ 的面积为 4 , 则 $a=( )", "options": {"A": "1", "B": "2", "C": "4", "D": "8"}, "label": "A", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "已知 $5^{5}<8^{4}, 13^{4}<8^{5}$. 设 $a=\\log _{5} 3, b=\\log _{8} 5, c=\\log _{13} 8$, 则 ( )", "options": {"A": "$a<b<c$", "B": "$b<a<c$", "C": "$b<c<a$", "D": "$c<a<b$"}, "label": "A", "answer": null, "other": {"source": "2020年高考全国卷Ⅲ数学"}}
|
||||
{"passage": null, "question": "设复数 $z$ 满足 $|z-i|=1, z$ 在复平面内对应的点为 $(x, y)$, 则", "options": {"A": "$(x+1)^{2}+y^{2}=1$", "B": "$(x-1)^{2}+y^{2}=1$", "C": "$x^{2}+(y-1)^{2}=1$", "D": "$x^{2}+(y+1)^{2}=1$"}, "label": "C", "answer": null, "other": {"source": "2019年新课标ⅰ数学"}}
|
||||
{"passage": null, "question": "记 $S_{n}$ 为等差数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和. 已知 $S_{4}=0, a_{5}=5$, 则", "options": {"A": "$a_{n}=2 n-5$", "B": "$a_{n}=3 n-10$", "C": "$S_{n}=2 n^{2}-8 n$", "D": "$S_{n}=\\frac{1}{2} n^{2}-2 n$"}, "label": "A", "answer": null, "other": {"source": "2019年新课标ⅰ数学"}}
|
||||
{"passage": null, "question": "已知椭圆 $C$ 的焦点为 $F_{1}(-1,0), F_{2}(1,0)$, 过 $F_{2}$ 的直线与 $C$ 交于 $A, B$ 两点. 若 $\\left|A F_{2}\\right|=2\\left|F_{2} B\\right|,|A B|=\\left|B F_{1}\\right|$, 则 $C$ 的方程为", "options": {"A": "$\\frac{x^{2}}{2}+y^{2}=1$", "B": "$\\frac{x^{2}}{3}+\\frac{y^{2}}{2}=1$", "C": "$\\frac{x^{2}}{4}+\\frac{y^{2}}{3}=1$", "D": "$\\frac{x^{2}}{5}+\\frac{y^{2}}{4}=1$"}, "label": "B", "answer": null, "other": {"source": "2019年新课标ⅰ数学"}}
|
||||
{"passage": null, "question": "关于函数 $f(x)=\\sin |x|+|\\sin x|$ 有下述四个结论:\\\\\n$\\textcircled{1}f(x)$ 是偶函数\\\\\n$\\textcircled{2}f(x)$ 在区间 $\\left(\\frac{\\pi}{2}, \\pi\\right)$ 单调递增\\\\\n$\\textcircled{3}f(x)$ 在 $[-\\pi, \\pi]$ 有 4 个零点\\\\\n$\\textcircled{4}f(x)$ 的最大值为 2\\\\\n其中所有正确结论的编号是", "options": {"A": "\\textcircled{1}\\textcircled{2}\\textcircled{4}", "B": "\\textcircled{2}\\textcircled{4}", "C": "\\textcircled{1}\\textcircled{4}", "D": "\\textcircled{1}\\textcircled{3}"}, "label": "C", "answer": null, "other": {"source": "2019年新课标ⅰ数学"}}
|
||||
{"passage": null, "question": "设集合 $A=\\{x \\mid-2<x<4\\}, B=\\{2,3,4,5\\}$, 则 $A \\cap B=( )", "options": {"A": "$\\{2\\}$", "B": "$\\{2,3\\}$", "C": "$\\{3,4\\}$", "D": "$\\{2,3,4\\}$"}, "label": "B", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "已知 $z=2-i$, 则 $z(\\bar{z}+\\mathrm{i})=( )", "options": {"A": "$6-2 \\mathrm{i}$", "B": "$4-2 \\mathrm{i}$", "C": "$6+2 i$", "D": "$4+2 i$"}, "label": "C", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "下列区间中，函数 $f(x)=7 \\sin \\left(x-\\frac{\\pi}{6}\\right)$ 单调递增的区间是 ( )", "options": {"A": "$\\left(0, \\frac{\\pi}{2}\\right)$", "B": "$\\left(\\frac{\\pi}{2}, \\pi\\right)$", "C": "$\\left(\\pi, \\frac{3 \\pi}{2}\\right)$", "D": "$\\left(\\frac{3 \\pi}{2}, 2 \\pi\\right)$"}, "label": "A", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "已知 $F_{1}, F_{2}$ 是椭圆 $C: \\frac{x^{2}}{9}+\\frac{y^{2}}{4}=1$ 的两个焦点, 点 $M$ 在 $C$ 上, 则 $\\left|M F_{1}\\right| \\cdot\\left|M F_{2}\\right|$ 的最大值为 ( )", "options": {"A": "13", "B": "12", "C": "9", "D": "6"}, "label": "C", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "若 $\\tan \\theta=-2$, 则 $\\frac{\\sin \\theta(1+\\sin 2 \\theta)}{\\sin \\theta+\\cos \\theta}=( )", "options": {"A": "$-\\frac{6}{5}$", "B": "$-\\frac{2}{5}$", "C": "$\\frac{2}{5}$", "D": "$\\frac{6}{5}$"}, "label": "C", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "若过点 $(a, b)$ 可以作曲线 $y=\\mathrm{e}^{x}$ 的两条切线, 则 ( )", "options": {"A": "$\\mathrm{e}^{b}<a$", "B": "$\\mathrm{e}^{a}<b$", "C": "$0<a<\\mathrm{e}^{b}$", "D": "$0<b<\\mathrm{e}^{a}$"}, "label": "D", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "有 6 个相同的球, 分别标有数字 $1,2,3,4,5,6$, 从中有放回的随机取两次, 每次取 1 个球, 甲表示 事件“第一次取出的球的数字是 1”，乙表示事件“第二次取出的球的数字是 2”，丙表示事件“两次取出的球的 数字之和是 8 ”, 丁表示事件“两次取出的球的数字之和是 7”, 则 ( )", "options": {"A": "甲与丙相互独立", "B": "甲与丁相互独立", "C": "乙与丙相互独立", "D": "丙与丁相互独立"}, "label": "B", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "有一组样本数据 $x_{1}, x_{2}, \\ldots, x_{n}$, 由这组数据得到新样本数据 $y_{1}, y_{2}, \\ldots, y_{n}$, 其中 $y_{i}=x_{i}+c(i=1,2, \\cdots, n), c$ 为非零常数, 则 ( )", "options": {"A": "两组样本数据的样本平均数相同", "B": "两组样本数据的样本中位数相同", "C": "两组样本数据的样本标准差相同", "D": "两组样数据的样本极差相同"}, "label": "CD", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "已知 $O$ 为坐标原点, 点 $P_{1}(\\cos \\alpha, \\sin \\alpha), P_{2}(\\cos \\beta,-\\sin \\beta), P_{3}(\\cos (\\alpha+\\beta), \\sin (\\alpha+\\beta)), A(1,0)$, 则 ( )", "options": {"A": "$|\\overrightarrow{O P}|=\\left|\\overrightarrow{O P_{2}}\\right|$", "B": "$\\left|\\overrightarrow{A P_{1}}\\right|=\\left|\\overrightarrow{A P_{2}}\\right|$", "C": "$\\overrightarrow{\\mathrm{OA}} \\cdot \\overrightarrow{\\mathrm{OP}}_{3}=\\overrightarrow{\\mathrm{OP}}_{1} \\cdot \\overrightarrow{\\mathrm{OP}_{3}}$", "D": "$\\overrightarrow{O A} \\cdot \\overrightarrow{O P_{1}}=\\overrightarrow{O P_{2}} \\cdot \\overrightarrow{O P_{3}}$"}, "label": "AC", "answer": null, "other": {"source": "2021新高考1卷数学"}}
|
||||
{"passage": null, "question": "设集合 $A=\\{4,5,7,9\\}, B=\\{3,4,7,8,9\\}$, 全集 $U=A \\cup B$, 则集 合 $C_{U}(A \\cap B)$ 中的元素共有 ( )", "options": {"A": "3 个", "B": "4 个", "C": "5 个", "D": "6 个"}, "label": "A", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "已知 $\\frac{\\bar{Z}}{1+i}=2+i$, 则复数 $z=( )", "options": {"A": "$-1+3 i$", "B": "$1-3 i$", "C": "$3+\\mathrm{i}$", "D": "$3-\\mathrm{i}$"}, "label": "B", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "不等式 $\\left|\\frac{x+1}{x-1}\\right|<1$ 的解集为 ( )", "options": {"A": "$\\{x \\mid 0<x<1\\} \\cup\\{x \\mid x>1\\}$", "B": "$\\{x \\mid 0<x<1\\}$", "C": "$\\{x \\mid-1<x<0\\}$ ", "D": "$\\{x \\mid x<0\\}$ "}, "label": "D", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "已知双曲线 $\\frac{x^{2}}{a^{2}}-\\frac{y^{2}}{b^{2}}=1 \\left(a>0 , b>0 \\right)$ 的渐近线与抛物线 $y=x^{2}+1$ 相 切，则该双曲线的离心率为 ( )", "options": {"A": "$\\sqrt{3}$", "B": "2", "C": "$\\sqrt{5}$", "D": "$\\sqrt{6}$"}, "label": "C", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "甲组有 5 名男同学, 3 名女同学; 乙组有 6 名男同学、2 名女同 学. 若从甲、乙两组中各选出 2 名同学, 则选出的 4 人中恰有 1 名女同学的 不同选法共有 ( )", "options": {"A": "150 种", "B": "180 种", "C": "300 种", "D": "345 种"}, "label": "D", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "设 $\\vec{a}$、$\\vec{b}$、$\\vec{c}$ 是单位向量, 且 $\\vec{a} \\cdot \\vec{b}=0$, 则 $(\\vec{a}-\\vec{c}) \\cdot(\\vec{b}-\\vec{c})$ 的最小值为 ( )", "options": {"A": "-2", "B": "$\\sqrt{2}-2$", "C": "-1", "D": "$1-\\sqrt{2}$"}, "label": "D", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "已知直线 $y=x+1$ 与曲线 $y=\\ln (x+a)$ 相切, 则 $a$ 的值为 ( )", "options": {"A": "1", "B": "2", "C": "- 1", "D": "-2"}, "label": "B", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "函数 $f(x)$ 的定义域为 $R$, 若 $f(x+1)$ 与 $f(x-1)$ 都是奇函数, 则 ( )", "options": {"A": "$f(x)$ 是偶函数", "B": "$f(x)$ 是奇函数", "C": "$f(x)=f(x+2)$", "D": "$f(x+3)$ 是奇函数"}, "label": "D", "answer": null, "other": {"source": "2009年数学试卷（理科）（全国卷ⅰ）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{-2,-1,0,1,2\\}, B=\\{x \\mid （ x-1)(x+2)<0\\}$, 则 $A \\cap B=( )", "options": {"A": "$\\{-1,0\\}$", "B": "$\\{0,1\\}$", "C": "$\\{-1,0,1\\}$", "D": "$\\{0,1,2\\}$"}, "label": "A", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "若 $a$ 为实数, 且 $(2+a i)(a-2 i)=-4 i$, 则 $a=( )", "options": {"A": "-1", "B": "0", "C": "1", "D": "2"}, "label": "B", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知等比数列 $\\left\\{a_{n}\\right\\}$ 满足 $a_{1}=3, a_{1}+a_{3}+a_{5}=21$, 则 $a_{3}+a_{5}+a_{7}=( )", "options": {"A": "21", "B": "42", "C": "63", "D": "84"}, "label": "B", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "过三点 $A(1,3), B(4,2), C(1,-7)$ 的圆交 $y$ 轴于 $M, N$ 两 点, 则 $|\\mathrm{MN}|=( )", "options": {"A": "$2 \\sqrt{6}$", "B": "8", "C": "$4 \\sqrt{6}$", "D": "10"}, "label": "C", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "已知 $A, B$ 是球 $O$ 的球面上两点, $\\angle A O B=90^{\\circ}, C$ 为该球面上的动点, 若三棱雉 $O-A B C$ 体积的最大值为 36 , 则球 $O$ 的表面积为 ( )", "options": {"A": "$36 \\pi$", "B": "$64 \\pi$", "C": "$144 \\pi$", "D": "$256 \\pi$"}, "label": "C", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "设函数 $f^{\\prime}(x)$ 是奇函数 $f(x)(x \\in R)$ 的导函数, $f(-1)=0$, 当 $x$ $>0$ 时, $x f^{\\prime}(x)-f(x)<0$, 则使得 $f(x)>0$ 成立的 $x$ 的取值范围是 ( )", "options": {"A": "$(-\\infty,-1) \\cup(0,1)$", "B": "$(-1,0) \\cup(1,+\\infty)$", "C": "$(-\\infty,-1) \\cup(-1,0)$", "D": "$(0,1) \\cup(1,+\\infty)$"}, "label": "A", "answer": null, "other": {"source": "2015年数学试卷（理科）（新课标ⅱ）"}}
|
||||
{"passage": null, "question": "若 $\\mathrm{z}=1+i$, 则 $\\left|\\mathrm{z}^{2}-2 z\\right|=( )", "options": {"A": "0", "B": "1", "C": "$\\sqrt{2}$", "D": "2"}, "label": "D", "answer": null, "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "设集合 $A=\\left\\{x \\mid x^{2}-4 \\leq 0\\right\\}, B=\\{x \\mid 2 x+a \\leq 0\\}$, 且 $A \\cap B=\\{x \\mid-2 \\leq x \\leq 1\\}$, 则 $a=( )", "options": {"A": "$-4$", "B": "$-2$", "C": "2", "D": "4"}, "label": "B", "answer": null, "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "已知 $A$ 为抛物线 $C: y^{2}=2 p x(p>0)$ 上一点, 点 $A$ 到 $C$ 的焦点的距离为 12 , 到 $y$ 轴的距离为 9 , 则 $p=( )", "options": {"A": "2", "B": "3", "C": "6", "D": "9"}, "label": "C", "answer": null, "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "函数 $f(x)=x^{4}-2 x^{3}$ 的图像在点 $(1, f(1))$ 处的切线方程为 ( )", "options": {"A": "$y=-2 x-1$", "B": "$y=-2 x+1$", "C": "$y=2 x-3$", "D": "$y=2 x+1$"}, "label": "B", "answer": null, "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "$\\left(x+\\frac{y^{2}}{x}\\right)(x+y)^{5}$ 的展开式中 $x^{3} y^{3}$ 的系数为 ( )", "options": {"A": "5", "B": "10", "C": "15", "D": "20"}, "label": "C", "answer": null, "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "已知 $\\alpha \\in(0, \\pi)$, 且 $3 \\cos 2 \\alpha-8 \\cos \\alpha=5$, 则 $\\sin \\alpha=( )", "options": {"A": "$\\frac{\\sqrt{5}}{3}$", "B": "$\\frac{2}{3}$", "C": "$\\frac{1}{3}$", "D": "$\\frac{\\sqrt{5}}{9}$"}, "label": "A", "answer": null, "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "若 $2^{a}+\\log _{2} a=4^{b}+2 \\log _{4} b$, 则 ( )", "options": {"A": "$a>2 b$", "B": "$a<2 b$", "C": "$a>b^{2}$", "D": "$a<b^{2}$"}, "label": "B", "answer": null, "other": {"source": "2020年全国卷Ⅰ数学"}}
|
||||
{"passage": null, "question": "复数 $\\frac{-1+3 i}{1+i}=( )", "options": {"A": "$2+i$", "B": "$2-\\mathrm{i}$", "C": "$1+2 i$", "D": "$1-2 \\mathrm{i}$"}, "label": "C", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{1,3, \\sqrt{\\pi}\\}, B=\\{1, m\\}, A \\cup B=A$, 则 $m$ 的值为 ( )", "options": {"A": "0 或 $\\sqrt{3}$", "B": "0 或 3", "C": "1 或 $\\sqrt{3}$", "D": "1 或 3"}, "label": "B", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "椭圆的中心在原点, 焦距为 4 , 一条准线为 $x=-4$, 则该椭圆的方程为 ( )", "options": {"A": "$\\frac{x^{2}}{16}+\\frac{y^{2}}{12}=1$", "B": "$\\frac{x^{2}}{12}+\\frac{y^{2}}{8}=1$", "C": "$\\frac{x^{2}}{8}+\\frac{y^{2}}{4}=1$", "D": "$\\frac{x^{2}}{12}+\\frac{y^{2}}{4}=1$"}, "label": "C", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知正四棱柱 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 中, $A B=2, C C_{1}=2 \\sqrt{2}, E$ 为 $C C_{1}$ 的中 点, 则直线 $A C_{1}$ 与平面 $\\mathrm{BED}$ 的距离为 ( )", "options": {"A": "2", "B": "$\\sqrt{3}$", "C": "$\\sqrt{2}$", "D": "1"}, "label": "D", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知等差数列 $\\left\\{a_{n}\\right\\}$ 的前 $n$ 项和为 $S_{n}, a_{5}=5, S_{5}=15$, 则数列 $\\left\\{\\frac{1}{a_{n} a_{n+1}}\\right\\}$ 的前 100 项和为 ( )", "options": {"A": "$\\frac{100}{101}$", "B": "$\\frac{99}{101}$", "C": "$\\frac{99}{100}$", "D": "$\\frac{101}{100}$"}, "label": "A", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "$\\triangle A B C$ 中, $A B$ 边的高为 $C D$, 若 $\\overrightarrow{C B}=\\vec{a}, \\overrightarrow{C A}=\\vec{b}, \\vec{a} \\cdot \\vec{b}=0,|\\vec{a}|=1, \\mid \\vec{b}$ $\\mid=2$ ，则 $\\overrightarrow{\\mathrm{AD}}=( )", "options": {"A": "$\\frac{1}{3} \\vec{a}-\\frac{1}{3} \\vec{b}$", "B": "$\\frac{2}{3} \\vec{a}-\\frac{2}{3} \\vec{b}$", "C": "$\\frac{3}{5} \\vec{a}-\\frac{3}{5} \\vec{b}$", "D": "$\\frac{4}{5} \\vec{a}-\\frac{4}{5} \\vec{b}$"}, "label": "D", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知 $\\alpha$ 为第二象限角, $\\sin \\alpha+\\cos \\alpha=\\frac{\\sqrt{3}}{3}$, 则 $\\cos 2 \\alpha=( )", "options": {"A": "$-\\frac{\\sqrt{5}}{3}$", "B": "$-\\frac{\\sqrt{5}}{9}$", "C": "$\\frac{\\sqrt{5}}{9}$", "D": "$\\frac{\\sqrt{5}}{3}$"}, "label": "A", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知 $F_{1}$、 $F_{2}$ 为双曲线 $C: x^{2}-y^{2}=2$ 的左、右焦点, 点 $P$ 在 $C$ 上, $\\left|P F_{1}\\right|=2\\left|P F_{2}\\right|$, 则 $\\cos \\angle F_{1} P F_{2}=( )", "options": {"A": "$\\frac{1}{4}$", "B": "$\\frac{3}{5}$", "C": "$\\frac{3}{4}$", "D": "$\\frac{4}{5}$"}, "label": "C", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知 $x=\\ln \\pi, y=\\log _{5} 2, z=e^{-\\frac{1}{2}}$, 则 ( )", "options": {"A": "$x<y<z$", "B": "$z<x<y$", "C": "$z<y<x$", "D": "$\\mathrm{y}<\\mathrm{z}<\\mathrm{x}$"}, "label": "D", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "已知函数 $y=x^{3}-3 x+c$ 的图象与 $x$ 轴恰有两个公共点, 则 $c=( )", "options": {"A": "-2 或 2", "B": "-9 或 3", "C": "-1 或 1", "D": "-3 或 1"}, "label": "A", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "将字母 $a, a, b, b, c, c$ 排成三行两列, 要求每行的字母互不相 同, 每列的字母也互不相同, 则不同的排列方法共有 ( )", "options": {"A": "12 种", "B": "18 种", "C": "24 种", "D": "36 种"}, "label": "A", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "正方形 $A B C D$ 的边长为 1 , 点 $E$ 在边 $A B$ 上, 点 $F$ 在边 $B C$ 上, $\\mathrm{AE}=\\mathrm{BF}=\\frac{3}{7}$, 动点 $\\mathrm{P}$ 从 $\\mathrm{E}$ 出发沿直线向 $\\mathrm{F}$ 运动, 每当碰到正方形的边时反弹, 反弹时反射角等于入射角, 当点 $P$ 第一次碰到 $E$ 时, $P$ 与正方形的边碰撞的次数为 ( )", "options": {"A": "16", "B": "14", "C": "12", "D": "10"}, "label": "B", "answer": null, "other": {"source": "2012年数学试卷（理科）（大纲版）"}}
|
||||
{"passage": null, "question": "设集合 $A=\\{1,2\\}, B=\\{2,4,6\\}$, 则 $A \\cup B=( )", "options": {"A": "$\\{2\\}$", "B": "$\\{1,2\\}$", "C": "$\\{2,4,6\\}$", "D": "$\\{1,2,4,6\\}$"}, "label": "B", "answer": null, "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "若实数 $x, y$ 满足约束条件 $\\left\\{\\begin{array}{l}x-2 \\geq 0, \\\\ 2 x+y-7 \\leq 0, \\text { 则 } z=3 x+4 y \\text { 的最大值是 }( ) \\\\ x-y-2 \\leq 0,\\end{array}\\right.$", "options": {"A": "20", "B": "18", "C": "13", "D": "6"}, "label": "B", "answer": null, "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "设 $x \\in \\mathbf{R}$, 则“ $\\sin x=1$ ”是“ $\\cos x=0$ ”的 ( )", "options": {"A": "充分不必要条件", "B": "必要不充分条件", "C": "充分必要条件", "D": "既不充 分也不必要条件"}, "label": "A", "answer": null, "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "为了得到函数 $y=2 \\sin 3 x$ 的图象, 只要把函数 $y=2 \\sin \\left(3 x+\\frac{\\pi}{5}\\right)$ 图象上所有的点 ( )", "options": {"A": "向左平移 $\\frac{\\pi}{5}$ 个单位长度", "B": "向右平移 $\\frac{\\pi}{5}$ 个单位长度", "C": "向左平移 $\\frac{\\pi}{15}$ 个单位长度", "D": "向右平移 $\\frac{\\pi}{15}$ 个单位长度"}, "label": "D", "answer": null, "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "已知 $2^{a}=5, \\log _{8} 3=b$, 则 $4^{a-3 b}=( )", "options": {"A": "25", "B": "5", "C": "$\\frac{25}{9}$", "D": "$\\frac{5}{3}$"}, "label": "C", "answer": null, "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "已知 $a, b \\in \\mathbf{R}$, 若对任意 $x \\in \\mathbf{R}, a|x-b|+|x-4|-|2 x-5| \\geq 0$, 则 ( )", "options": {"A": "$a \\leq 1, b \\geq 3$", "B": "$a \\leq 1, b \\leq 3$", "C": "$a \\geq 1, b \\geq 3$", "D": "$a \\geq 1, b \\leq 3$"}, "label": "D", "answer": null, "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "已知数列 $\\left\\{a_{n}\\right\\}$ 满足 $a_{1}=1, a_{n+1}=a_{n}-\\frac{1}{3} a_{n}^{2}\\left(n \\in \\mathbf{N}^{*}\\right)$, 则 ( )", "options": {"A": "$2<100 a_{100}<\\frac{5}{2}$", "B": "$\\frac{5}{2}<100 a_{100}<3$", "C": "$3<100 a_{100}<\\frac{7}{2}$", "D": "$\\frac{7}{2}<100 a_{100}<4$"}, "label": "B", "answer": null, "other": {"source": "2022年浙江省高考数学"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{-1,0,1,2\\}, B=\\{x \\mid 0<x<3\\}$, 则 $A \\cap B=$ ( ).", "options": {"A": "$\\{-1,0,1\\}$", "B": "$\\{0,1\\}$", "C": "$\\{-1,1,2\\}$", "D": "$\\{1,2\\}$"}, "label": "D", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "在复平面内, 复数 $z$ 对应的点的坐标是 $(1,2)$, 则 $i \\cdot z=()$ ( ).", "options": {"A": "$1+2 i$", "B": "$-2+i$", "C": "$1-2 i$", "D": "$-2-i$"}, "label": "B", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "已知半径为 1 的圆经过点 $(3,4)$, 则其圆心到原点的距离的最小值为 ( ).", "options": {"A": "4", "B": "5", "C": "6", "D": "7"}, "label": "A", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=2^{x}-x-1$, 则不等式 $f(x)>0$ 的解集是 ( ).", "options": {"A": "$(-1,1)$", "B": "$(-\\infty,-1) \\cup(1,+\\infty)$", "C": "$(0,1)$", "D": "$(-\\infty, 0) \\cup(1,+\\infty)$"}, "label": "D", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "设抛物线的顶点为 $O$, 焦点为 $F$, 准线为 $l . P$ 是抛物线上异于 $O$ 的一点, 过 $P$ 作 $P Q \\perp l$ 于 $Q$, 则线段 $F Q$ 的垂直平分线 ( ).", "options": {"A": "经过点 $O$", "B": "经过点 $P$", "C": "平行于直线 $O P$", "D": "垂直于直线 $O P$"}, "label": "B", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "在等差数列 $\\left\\{a_{n}\\right\\}$ 中, $a_{1}=-9, a_{3}=-1$. 记 $T_{n}=a_{1} a_{2} \\ldots a_{n}(n=1,2, \\ldots)$, 则数列 $\\left\\{T_{n}\\right\\}$ ( )", "options": {"A": "有最大项, 有最小项", "B": "有最大项, 无最小项", "C": "无最大项, 有最小项", "D": "无最大项, 无最小项"}, "label": "B", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "已知 $\\alpha, \\beta \\in R$, 则“存在 $k \\in Z$ 使得 $\\alpha=k \\pi+(-1)^{k} \\beta$ ”是“ $\\sin \\alpha=\\sin \\beta$ ”的 ( ).", "options": {"A": "充分而不必要条件", "B": "必要而不充分条件", "C": "充分必要条件", "D": "既不充分也不必要条件"}, "label": "C", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "2020 年 3 月 14 日是全球首个国际圆周率日（ $\\pi$ Day ). 历史上, 求圆周率 $\\pi$ 的方法有多 种, 与中国传统数学中的“割圆术”相似. 数学家阿尔. 卡西的方法是：当正整数 $n$ 充分大时, 计算单位圆的内接正 $6 n$ 边形的周长和外切正 $6 n$ 边形 (各边均与圆相切的正 $6 n$ 边形) 的周 长, 将它们的算术平均数作为 $2 \\pi$ 的近似值. 按照阿尔. 卡西的方法, $\\pi$ 的近似值的表达式是 ( ).", "options": {"A": "$3 n\\left(\\sin \\frac{30^{\\circ}}{n}+\\tan \\frac{30^{\\circ}}{n}\\right)$", "B": "$6 n\\left(\\sin \\frac{30^{\\circ}}{n}+\\tan \\frac{30^{\\circ}}{n}\\right)$", "C": "$3 n\\left(\\sin \\frac{60^{\\circ}}{n}+\\tan \\frac{60^{\\circ}}{n}\\right)$", "D": "$6 n\\left(\\sin \\frac{60^{\\circ}}{n}+\\tan \\frac{60^{\\circ}}{n}\\right)$"}, "label": "A", "answer": null, "other": {"source": "2020年北京市高考理科数学试卷"}}
|
||||
{"passage": null, "question": "已知集合 $A=\\{x \\mid x-1 \\geqslant 0\\}, B=\\{0,1 ， 2\\}$, 则 $A \\cap B=$ ( )", "options": {"A": "$\\{0\\}$", "B": "$\\{1\\}$", "C": "$\\{1,2\\}$", "D": "$\\{0,1,2\\}$"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "$(5$ 分 $)(1+i)(2-i)=( )", "options": {"A": "$-3-\\mathrm{i}$", "B": "$-3+i$", "C": "$3-\\mathrm{i}$", "D": "$3+i$"}, "label": "D", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "若 $\\sin \\alpha=\\frac{1}{3}$, 则 $\\cos 2 \\alpha=( )", "options": {"A": "$\\frac{8}{9}$", "B": "$\\frac{7}{9}$", "C": "$-\\frac{7}{9}$", "D": "$-\\frac{8}{9}$"}, "label": "B", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "$\\left(\\mathrm{x}^{2}+\\frac{2}{\\mathrm{x}}\\right){ }^{5}$ 的展开式中 $\\mathrm{x}^{4}$ 的系数为 ( )", "options": {"A": "10", "B": "20", "C": "40", "D": "80"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "直线 $x+y+2=0$ 分别与 $x$ 轴, $y$ 轴交于 $A, B$ 两点, 点 $P$ 在圆 $(x-2)^{2}+y^{2}=2$ 上, 则 $\\triangle A B P$ 面积的取值范围是 ( )", "options": {"A": "$[2,6]$", "B": "$[4,8]$", "C": "$[\\sqrt{2}, 3 \\sqrt{2}]$", "D": "$[2 \\sqrt{2}, 3 \\sqrt{2}]$"}, "label": "A", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "某群体中的每位成员使用移动支付的概率都为 $\\mathrm{p}$, 各成员的支付方式 相互独立. 设 $X$ 为该群体的 10 位成员中使用移动支付的人数, $D X=2.4, P$ $(x=4)<p(x=6)$, 则 $p=( )", "options": {"A": "0.7", "B": "0.6", "C": "0.4", "D": "0.3"}, "label": "B", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "$\\triangle A B C$ 的内角 $A, B, C$ 的对边分别为 $a, b, c$. 若 $\\triangle A B C$ 的面积为 $\\frac{a^{2}+b^{2}-c^{2}}{4}$, 则 $C=( )", "options": {"A": "$\\frac{\\pi}{2}$", "B": "$\\frac{\\pi}{3}$", "C": "$\\frac{\\pi}{4}$", "D": "$\\frac{\\pi}{6}$"}, "label": "C", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "设 $A, B, C, D$ 是同一个半径为 4 的球的球面上四点, $\\triangle A B C$ 为等 边三角形且面积为 $9 \\sqrt{3}$, 则三棱雉 $D-A B C$ 体积的最大值为( )", "options": {"A": "$12 \\sqrt{3}$", "B": "$18 \\sqrt{3}$", "C": "$24 \\sqrt{3}$", "D": "$54 \\sqrt{3}$"}, "label": "B", "answer": null, "other": {"source": "2018年数学试卷（理科）（新课标ⅲ）"}}
|
||||
{"passage": null, "question": "设集合 $A=\\left\\{x \\mid x^{2}-4 x+3<0\\right\\}, B=\\{x \\mid 2 x-3>0\\}$, 则 $A \\cap B=( )", "options": {"A": "$\\left(-3,-\\frac{3}{2}\\right)$", "B": "$\\left(-3, \\frac{3}{2}\\right)$", "C": "$\\left(1, \\frac{3}{2}\\right)$", "D": "$\\left(\\frac{3}{2}, 3\\right)$"}, "label": "D", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "设 $(1+i) x=1+y i$, 其中 $x, y$ 是实数, 则 $|x+y i|=( )", "options": {"A": "1", "B": "$\\sqrt{2}$", "C": "$\\sqrt{3}$", "D": "2"}, "label": "B", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知等差数列 $\\left\\{a_{n}\\right\\}$ 前 9 项的和为 $27, a_{10}=8$, 则 $a_{100}=( )", "options": {"A": "100", "B": "99", "C": "98", "D": "97"}, "label": "C", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "某公司的班车在 7: 00, 8: 00, 8: 30 发车, 小明在 7:50 至 8: 30 之间到达发车站乘坐班车, 且到达发车站的时刻是随机的, 则他等车时间 不超过 10 分钟的概率是 ( )）", "options": {"A": "$\\frac{1}{3}$", "B": "$\\frac{1}{2}$", "C": "$\\frac{2}{3}$", "D": "$\\frac{3}{4}$"}, "label": "B", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "若 $a>b>1,0<c<1$, 则 ( )）", "options": {"A": "$a^{c}<b^{c}$", "B": "$a b^{c}<b a^{c}$", "C": "$a \\log _{b} c<b \\log _{a} c$", "D": "$\\log _{a} c<\\log _{b} c$"}, "label": "C", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "以抛物线 $C$ 的顶点为圆心的圆交 $C$ 于 $A$、 $B$ 两点, 交 $C$ 的准线于 $D$、 $E$ 两点. 已知 $|A B|=4 \\sqrt{2},|D E|=2 \\sqrt{5}$, 则 $C$ 的焦点到准线的距离为 ( )）", "options": {"A": "2", "B": "4", "C": "6", "D": "8"}, "label": "B", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "平面 $\\alpha$ 过正方体 $A B C D-A_{1} B_{1} C_{1} D_{1}$ 的顶点 $A, \\alpha / /$ 平面 $C B_{1} D_{1}, \\alpha \\cap$ 平 面 $A B C D=m, \\alpha \\cap$ 平面 $A B B_{1} A_{1}=n$, 则 $m$、 $n$ 所成角的正弦值为 ( )）", "options": {"A": "$\\frac{\\sqrt{3}}{2}$", "B": "$\\frac{\\sqrt{2}}{2}$", "C": "$\\frac{\\sqrt{3}}{3}$", "D": "$\\frac{1}{3}$"}, "label": "A", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
{"passage": null, "question": "已知函数 $f(x)=\\sin (\\omega x+\\phi)\\left(\\omega>0,|\\phi| \\leqslant \\frac{\\pi}{2}\\right), x=-\\frac{\\pi}{4}$ 为 $f(x)$ 的零点, $x=\\frac{\\pi}{4}$ 为 $y=f(x)$ 图象的对称轴, 且 $f(x)$ 在 $\\left(\\frac{\\pi}{18}, \\frac{5 \\pi}{36}\\right)$ 上单调, 则 $\\omega$ 的最大值为 ( )）", "options": {"A": "11", "B": "9", "C": "7", "D": "5"}, "label": "B", "answer": null, "other": {"source": "2016年数学试卷（理科）（新课标ⅰ）"}}
|
||||
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+1319
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+7473
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+5000
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+395
@@ -0,0 +1,395 @@
|
||||
{"input":"Joan found 70 seashells on the beach . she gave Sam some of her seashells . She has 27 seashell . How many seashells did she give to Sam ?","target":43.0}
|
||||
{"input":"There were 28 bales of hay in the barn . Tim stacked bales in the barn today . There are now 54 bales of hay in the barn . How many bales did he store in the barn ?","target":26.0}
|
||||
{"input":"Mary is baking a cake . The recipe wants 8 cups of flour . She already put in 2 cups . How many cups does she need to add ?","target":6.0}
|
||||
{"input":"Sara 's high school played 12 basketball games this year . The team won most of their games . They were defeated during 4 games . How many games did they win ?","target":8.0}
|
||||
{"input":"There are 22 walnut trees currently in the park . Park workers will plant walnut trees today . When the workers are finished there will be 55 walnut trees in the park . How many walnut trees did the workers plant today ?","target":33.0}
|
||||
{"input":"Mike had 34 peaches at his roadside fruit dish . He went to the orchard and picked peaches to stock up . There are now 86 peaches . how many did he pick ?","target":52.0}
|
||||
{"input":"There were 6 roses in the vase . Mary cut some roses from her flower garden . There are now 16 roses in the vase . How many roses did she cut ?","target":10.0}
|
||||
{"input":"Joan went to 4 football games this year . She went to 9 games last year . How many football games did Joan go to in all ?","target":13.0}
|
||||
{"input":"Tom has 9 yellow balloons Sara has 8 yellow balloons . How many yellow balloons do they have in total ?","target":17.0}
|
||||
{"input":"There are 4 walnut trees currently in the park . Park workers will plant 6 walnut trees today . How many walnut trees will the park have when the workers are finished ?","target":10.0}
|
||||
{"input":"Sam had 9 dimes in his bank . His dad gave him 7 dimes . How many dimes does Sam have now ?","target":16.0}
|
||||
{"input":"Alyssa 's dog had puppies . She gave 7 to her friends . She now has 5 puppies . How many puppies did she have to start with ?","target":12.0}
|
||||
{"input":"A restaurant served 9 pizzas during lunch and 6 during dinner today . How many pizzas were served today ?","target":15.0}
|
||||
{"input":"There are 2 pencils in the drawer . Tim placed 3 pencils in the drawer . How many pencils are now there in total ?","target":5.0}
|
||||
{"input":"Joan found 6 seashells and Jessica found 8 seashells on the beach . How many seashells did they find together ?","target":14.0}
|
||||
{"input":"Sandy grew 6 carrots . Sam grew 3 carrots . How many carrots did they grow in total ?","target":9.0}
|
||||
{"input":"Benny picked 2 apples and Dan picked 9 apples from the apple tree . How many apples were picked in total ?","target":11.0}
|
||||
{"input":"Sally found 9 seashells , Tom found 7 seashells , and Jessica found 5 seashells on the beach . How many seashells did they find together ?","target":21.0}
|
||||
{"input":"Tim 's cat had kittens . He gave 3 to Jessica and 6 to Sara . He now has 9 kittens . How many kittens did he have to start with ?","target":18.0}
|
||||
{"input":"Joan has 9 blue balloons , Sally has 5 blue balloons , and Jessica has 2 blue balloons . How many blue balloons do they have in total ?","target":16.0}
|
||||
{"input":"Melanie had 7 dimes in her bank . Her dad gave her 8 dimes and her mother gave her 4 dimes . How many dimes does Melanie have now ?","target":19.0}
|
||||
{"input":"A restaurant served 5 cakes during lunch and 6 during dinner today . The restaurant served 3 cakes yesterday . How many cakes were served in total ?","target":14.0}
|
||||
{"input":"Melanie picked 4 plums , Dan picked 9 plums , and Sally picked 3 plums from the plum tree . How many plums were picked in total ?","target":16.0}
|
||||
{"input":"There are 7 dogwood trees currently in the park . Park workers will plant 3 dogwood trees today and 2 dogwood trees tomorrow . How many dogwood trees will the park have when the workers are finished ?","target":12.0}
|
||||
{"input":"Sara grew 4 onions , Sally grew 5 onions , and Fred grew 9 onions . How many onions did they grow in all ?","target":18.0}
|
||||
{"input":"Jason has 43 blue and 16 red marbles . Tom has 24 blue marbles . How many blue marbles do they have in all ?","target":67.0}
|
||||
{"input":"Sam found 18 seashells and Mary found 47 seashells on the beach . How many seashells did they find together ?","target":65.0}
|
||||
{"input":"Jason grew 23 watermelons and 18 turnips . Nancy grew 28 watermelons . How many watermelons did they grow in total ?","target":51.0}
|
||||
{"input":"There are 11 rulers and 34 crayons in the drawer . Tim placed 14 rulers in the drawer . How many rulers are now there in all ?","target":25.0}
|
||||
{"input":"Sara picked 45 pears and Sally picked 11 pears from the pear tree . How many pears were picked in total ?","target":56.0}
|
||||
{"input":"Keith has 20 books . Jason has 21 books . How many books do they have together ?","target":41.0}
|
||||
{"input":"Jason had 49 quarters in his bank . His dad gave him 25 quarters . How many quarters does he have now ?","target":74.0}
|
||||
{"input":"Tom found 15 seashells and Fred found 43 seashells on the beach . When they cleaned them , they discovered that 29 were cracked . How many seashells did they find together ?","target":58.0}
|
||||
{"input":"Sara has 31 red and 15 green balloons . Sandy has 24 red balloons . How many red balloons do they have in total ?","target":55.0}
|
||||
{"input":"Joan picked 37 oranges and Sara picked 10 oranges . Alyssa picked 30 pears . How many oranges were picked in total ?","target":47.0}
|
||||
{"input":"Fred went to 36 basketball games this year , but missed 35 . He went to 11 games last year . How many basketball games did Fred go to in total ?","target":47.0}
|
||||
{"input":"There are 33 pencils and 44 crayons in the drawer . Joan placed 27 pencils in the drawer . How many pencils are now there in total ?","target":60.0}
|
||||
{"input":"Sam had 49 pennies and 24 nickels in his bank . His dad gave him 39 nickels and 31 quarters . How many nickels does he have now ?","target":63.0}
|
||||
{"input":"Dan grew 42 turnips and 38 cantelopes . Jessica grew 47 turnips . How many turnips did they grow in total ?","target":89.0}
|
||||
{"input":"There are 33 walnut trees currently in the park . Park workers will plant 44 walnut trees today . How many walnut trees will the park have when the workers are finished ?","target":77.0}
|
||||
{"input":"Sara had 21 quarters in her bank . Her dad gave her 49 quarters . How many quarters does she have now ?","target":70.0}
|
||||
{"input":"There are 41 pencils in the drawer . Mike placed 30 pencils in the drawer . How many pencils are now there in total ?","target":71.0}
|
||||
{"input":"Joan has 10 books . Tom has 38 books . How many books do they have together ?","target":48.0}
|
||||
{"input":"Joan has 40 blue balloons Melanie has 41 blue balloons . How many blue balloons do they have in total ?","target":81.0}
|
||||
{"input":"Fred grew 38 cantelopes . Tim grew 44 cantelopes . How many cantelopes did they grow in total ?","target":82.0}
|
||||
{"input":"Sam went to 14 football games this year . He went to 29 games last year . How many football games did Sam go to in all ?","target":43.0}
|
||||
{"input":"Mary found 18 seashells and Jessica found 41 seashells on the beach . How many seashells did they find together ?","target":59.0}
|
||||
{"input":"There are 39 dogwood trees currently in the park . Park workers will plant 41 dogwood trees today and 20 dogwood trees tomorrow . How many dogwood trees will the park have when the workers are finished ?","target":100.0}
|
||||
{"input":"Sandy has 10 books , Benny has 24 books , and Tim has 33 books . How many books do they have together ?","target":67.0}
|
||||
{"input":"Jason picked 46 pears , Keith picked 47 pears , and Mike picked 12 pears from the pear tree . How many pears were picked in total ?","target":105.0}
|
||||
{"input":"Keith grew 29 cantelopes , Fred grew 16 cantelopes , and Jason grew 20 cantelopes . How many cantelopes did they grow in total ?","target":65.0}
|
||||
{"input":"Melanie had 19 dimes in her bank . Her dad gave her 39 dimes and her mother gave her 25 dimes . How many dimes does Melanie have now ?","target":83.0}
|
||||
{"input":"Alyssa has 37 blue balloons , Sandy has 28 blue balloons , and Sally has 39 blue balloons . How many blue balloons do they have in all ?","target":104.0}
|
||||
{"input":"Sally had 27 Pokemon cards . Dan gave her 41 new Pokemon cards . Sally bought 20 Pokemon cards . How many Pokemon cards does Sally have now ?","target":88.0}
|
||||
{"input":"Jason went to 11 football games this month . He went to 17 games last month , and plans to go to 16 games next month . How many games will he attend in all ?","target":44.0}
|
||||
{"input":"There are 43 pencils in the drawer and 19 pencils on the desk . Dan placed 16 pencils on the desk . How many pencils are now there in total ?","target":78.0}
|
||||
{"input":"Mike has 35 books in his library . He bought several books at a yard sale over the weekend . He now has 56 books in his library . How many books did he buy at the yard sale ?","target":21.0}
|
||||
{"input":"There are 53 maple trees currently in the park . Park workers will plant maple trees today . When the workers are finished there will be 64 maple trees in the park . How many maple trees did the workers plant today ?","target":11.0}
|
||||
{"input":"Dan found 56 seashells on the beach , he gave Jessica some of his seashells . He has 22 seashell . How many seashells did he give to Jessica ?","target":34.0}
|
||||
{"input":"Sally had 13 peaches at her roadside fruit dish . She went to the orchard and picked peaches to stock up . There are now 55 peaches . how many did she pick ?","target":42.0}
|
||||
{"input":"Benny received 67 dollars for his birthday . He went to a sporting goods store and bought a baseball glove , baseball , and bat . He had 33 dollars over , how much did he spent on the baseball gear ?","target":34.0}
|
||||
{"input":"There were 3 roses in the vase . Alyssa cut some roses from her flower garden . There are now 14 roses in the vase . How many roses did she cut ?","target":11.0}
|
||||
{"input":"Last week Tom had 74 dollars . He washed cars over the weekend and now has 86 dollars . How much money did he make washing cars ?","target":12.0}
|
||||
{"input":"There were 73 bales of hay in the barn . Jason stacked bales in the barn today . There are now 96 bales of hay in the barn . How many bales did he store in the barn ?","target":23.0}
|
||||
{"input":"Nancy grew 6 potatoes . Sandy grew 7 potatoes . How many potatoes did they grow in total ?","target":13.0}
|
||||
{"input":"There are 9 crayons in the drawer . Benny placed 3 crayons in the drawer . How many crayons are now there in total ?","target":12.0}
|
||||
{"input":"There are 5 oak trees currently in the park . Park workers will plant 4 oak trees today . How many oak trees will the park have when the workers are finished ?","target":9.0}
|
||||
{"input":"Tom found 7 seashells but 4 were broken . How many unbroken seashells did Tom find ?","target":3.0}
|
||||
{"input":"Sally picked 7 lemons and Mary picked 9 lemons from the lemon tree . How many lemons were picked in total ?","target":16.0}
|
||||
{"input":"A restaurant served 6 cakes during lunch and 9 during dinner today . How many cakes were served today ?","target":15.0}
|
||||
{"input":"Joan has 8 orange balloons but lost 2 of them . How many orange balloons does Joan have now ?","target":6.0}
|
||||
{"input":"Fred had 7 dimes in his bank . His sister borrowed 3 of his dimes . How many dimes does Fred have now ?","target":4.0}
|
||||
{"input":"Joan 's cat had 8 kittens . She gave 2 to her friends . How many kittens does she have now ?","target":6.0}
|
||||
{"input":"There are 34 dogwood trees currently in the park . Park workers will plant 49 dogwood trees today . How many dogwood trees will the park have when the workers are finished ?","target":83.0}
|
||||
{"input":"There are 27 pencils in the drawer . Nancy placed 45 pencils in the drawer . How many pencils are now there in total ?","target":72.0}
|
||||
{"input":"Sam found 35 seashells on the beach , he gave Joan 18 of the seashells . How many seashells does he now have ?","target":17.0}
|
||||
{"input":"Tim has 22 books . Mike has 20 books . How many books do they have together ?","target":42.0}
|
||||
{"input":"Mike has 87 baseball cards . Sam bought 13 of Mike 's baseball cards . How many baseball cards does Mike have now ?","target":74.0}
|
||||
{"input":"Sandy grew 51 pumpkins . Mike grew 23 pumpkins . How many pumpkins did they grow in total ?","target":74.0}
|
||||
{"input":"Tim has 44 books . Sam has 52 books . How many books do they have together ?","target":96.0}
|
||||
{"input":"Dan has 64 violet marbles , he gave Mary 14 of the marbles . How many violet marbles does he now have ?","target":50.0}
|
||||
{"input":"There are 25 popular trees currently in the park . Park workers will plant 73 popular trees today . How many popular trees will the park have when the workers are finished ?","target":98.0}
|
||||
{"input":"There are 54 scissors in the drawer . Keith placed 22 scissors in the drawer . How many scissors are now there in all ?","target":76.0}
|
||||
{"input":"Alyssa picked 42 pears and Nancy picked 17 pears from the pear tree . How many pears were picked in all ?","target":59.0}
|
||||
{"input":"Sam had 98 pennies in his bank . He spent 93 of his pennies . How many pennies does he have now ?","target":5.0}
|
||||
{"input":"Joan found 79 seashells on the beach , she gave Mike 63 of the seashells . How many seashells does she now have ?","target":16.0}
|
||||
{"input":"Sam has 110 books . Joan has 102 books . How many books do they have together ?","target":212.0}
|
||||
{"input":"Mary picked 122 oranges and Jason picked 105 oranges from the orange tree . How many oranges were picked in total ?","target":227.0}
|
||||
{"input":"Sally had 760 quarters in her bank . She spent 418 of her quarters . How many quarters does she have now ?","target":342.0}
|
||||
{"input":"Melanie grew 139 turnips . Benny grew 113 turnips . How many turnips did they grow in all ?","target":252.0}
|
||||
{"input":"Jason has 676 Pokemon cards . Alyssa bought 224 of Jason 's Pokemon cards . How many Pokemon cards does Jason have now ?","target":452.0}
|
||||
{"input":"There are 107 walnut trees currently in the park . Park workers will plant 104 walnut trees today . How many walnut trees will the park have when the workers are finished ?","target":211.0}
|
||||
{"input":"Fred has 709 green balloons , he gave Sandy 221 of the balloons . How many green balloons does he now have ?","target":488.0}
|
||||
{"input":"There are 115 pencils in the drawer . Sara placed 100 pencils in the drawer . How many pencils are now there in all ?","target":215.0}
|
||||
{"input":"Jason has 3 Pokemon cards . Benny bought 2 of Jason 's Pokemon cards . How many Pokemon cards does Jason have now ?","target":1.0}
|
||||
{"input":"Mike has 8 orange marbles , he gave Sam 4 of the marbles . How many orange marbles does he now have ?","target":4.0}
|
||||
{"input":"Joan had 5 dimes in her bank . She spent 2 of her dimes . How many dimes does she have now ?","target":3.0}
|
||||
{"input":"There are 2 orchid bushes currently in the park . Park workers will plant 4 orchid bushes today . How many orchid bushes will the park have when the workers are finished ?","target":6.0}
|
||||
{"input":"Sara picked 6 pears and Tim picked 5 pears from the pear tree . How many pears were picked in total ?","target":11.0}
|
||||
{"input":"Sam grew 4 watermelons , but the rabbits ate 3 watermelons . How many watermelons does Sam have ?","target":1.0}
|
||||
{"input":"Jason had Pokemon cards . He gave 9 to his friends . He now has 4 Pokemon cards . How many Pokemon cards did he have to start with ?","target":13.0}
|
||||
{"input":"Mary had 7 nickels in her bank . Her dad gave her 5 nickels . How many nickels does Mary have now ?","target":12.0}
|
||||
{"input":"Keith grew 6 turnips . Alyssa grew 9 turnips . How many turnips did they grow in all ?","target":15.0}
|
||||
{"input":"Mary has 9 yellow marbles Joan has 3 yellow marbles . How many yellow marbles do they have in all ?","target":12.0}
|
||||
{"input":"Sally grew 6 carrots . Fred grew 4 carrots . How many carrots did they grow in all ?","target":10.0}
|
||||
{"input":"Tom found 5 seashells on the beach . he gave Jessica 2 of the seashells . How many seashells does he now have ?","target":3.0}
|
||||
{"input":"Fred has 5 baseball cards . Melanie bought 3 of Fred 's baseball cards . How many baseball cards does Fred have now ?","target":2.0}
|
||||
{"input":"Mary had 8 potatoes in the garden . The rabbits ate 3 of the potatoes . How many potatoes does Mary now have ?","target":5.0}
|
||||
{"input":"There are 9 oak trees currently in the park . Park workers had to cut down 2 oak trees that were damaged . How many oak trees will the park have when the workers are finished ?","target":7.0}
|
||||
{"input":"Jessica had 8 quarters in her bank . Her sister borrowed 3 of her quarters . How many quarters does Jessica have now ?","target":5.0}
|
||||
{"input":"A restaurant made 9 hamburgers to serve during lunch . Only 3 were actually served . How many hamburgers were over from lunch ?","target":6.0}
|
||||
{"input":"There are 7 crayons in the drawer . Mary took 3 crayons out of the drawer . How many crayons are there now ?","target":4.0}
|
||||
{"input":"Dan picked 9 limes and gave Sara 4 of the limes . How many limes does Dan have now ?","target":5.0}
|
||||
{"input":"Joan has 9 blue balloons but lost 2 of them . How many blue balloons does Joan have now ?","target":7.0}
|
||||
{"input":"Joan picked 43 apples from the orchard , and gave 27 apples to Melanie . How many apples does Joan have now ?","target":16.0}
|
||||
{"input":"Tom has 30 violet balloons , he gave Fred 16 of the balloons . How many violet balloons does he now have ?","target":14.0}
|
||||
{"input":"Fred has 40 baseball cards . Keith bought 22 of Fred 's baseball cards . How many baseball cards does Fred have now ?","target":18.0}
|
||||
{"input":"Fred found 47 seashells on the beach , he gave Jessica 25 of the seashells . How many seashells does he now have ?","target":22.0}
|
||||
{"input":"Sara grew 43 pumpkins , but the rabbits ate 23 pumpkins . How many pumpkins does Sara have ?","target":20.0}
|
||||
{"input":"Joan decided to sell all of her old books . She gathered up 33 books to sell . She sold 26 books in a yard sale . How many books does Joan now have ?","target":7.0}
|
||||
{"input":"There are 46 rulers in the drawer . Tim took 25 rulers from the drawer . How many rulers are now in the drawer ?","target":21.0}
|
||||
{"input":"There are 33 oak trees currently in the park . Park workers had to cut down 18 oak trees that were damaged . How many oak trees will be in the park when the workers are finished ?","target":15.0}
|
||||
{"input":"Joan purchased a basketball game for $ 5.20 , and a racing game for $ 4.23 . How much did Joan spend on video games ?","target":9.43}
|
||||
{"input":"Mike joined his school 's band . He bought a trumpet for $ 145.16 , and a song book which was $ 5.84 . How much did Mike spend at the music store ?","target":151.0}
|
||||
{"input":"Alyssa bought some toys . She bought a football for $ 5.71 , and spent $ 6.59 on marbles . In total , how much did Alyssa spend on toys ?","target":12.3}
|
||||
{"input":"Jessica spent $ 10.22 on a cat toy , and a cage cost her $ 11.73 . What was the total cost of Jessica 's purchases ?","target":21.95}
|
||||
{"input":"Sara got fast food for lunch . Sara spent $ 5.36 on a hotdog and $ 5.10 on a salad . What was the total of the lunch bill ?","target":10.46}
|
||||
{"input":"Jason went to the mall on Saturday to buy clothes . He spent $ 14.28 on shorts and $ 4.74 on a jacket . In total , how much money did Jason spend on clothing ?","target":19.02}
|
||||
{"input":"Alyssa loves eating fruits . Alyssa paid $ 12.08 for grapes , and $ 9.85 for cherries . In total , how much money did Alyssa spend ?","target":21.93}
|
||||
{"input":"Mary loves eating fruits . Mary paid $ 11.08 for berries , $ 14.33 for apples , and $ 9.31 for peaches . In total , how much money did she spend ?","target":34.72}
|
||||
{"input":"Sandy went to the mall to buy clothes . She spent $ 13.99 on shorts , $ 12.14 on a shirt , and $ 7.43 on a jacket . How much money did Sandy spend on clothes ?","target":33.56}
|
||||
{"input":"Jason joined his school 's band . He bought a flute for $ 142.46 , a music tool for $ 8.89 , and a song book for $ 7 . How much did Jason spend at the music store ?","target":158.35}
|
||||
{"input":"Tom purchased a football game for $ 14.02 , a strategy game for $ 9.46 , and a Batman game for $ 12.04 . How much did Tom spend on video games ?","target":35.52}
|
||||
{"input":"Mike bought some toys . He bought marbles for $ 9.05 , a football for $ 4.95 , and spent $ 6.52 on a baseball . In total , how much did Mike spend on toys ?","target":20.52}
|
||||
{"input":"A ship is filled with 5973 tons of cargo . It stops in the Bahamas , where sailors load 8723 tons of cargo onboard . How many tons of cargo does the ship hold now ?","target":14696.0}
|
||||
{"input":"Before December , customers buy 1346 ear muffs from the mall . During December , they buy 6444 , and there are none . In all , how many ear muffs do the customers buy ?","target":7790.0}
|
||||
{"input":"Diane is a beekeeper . Last year , she harvested 2479 pounds of honey . This year , she bought some new hives and increased her honey harvest by 6085 pounds . How many pounds of honey did Diane harvest this year ?","target":8564.0}
|
||||
{"input":"An oil pipe in the sea broke . Before engineers started to fix the pipe , 6522 liters of oil leaked into the water . While the engineers worked , the pipe leaked 5165 liters of oil . In all , how many liters of oil leaked into the water ?","target":11687.0}
|
||||
{"input":"A car company produced 3884 cars in North America and 2871 cars in Europe . How many cars is that in all ?","target":6755.0}
|
||||
{"input":"Abe 's family moved from the Bahamas to Japan , so they had convert their money into Japanese yen . Their checking account now has 6359 yen and their savings account now has 3485 yen . How many yen do they have ?","target":9844.0}
|
||||
{"input":"There are 1986 books in Oak Grove 's public library . In addition , there are 5106 books in its school libraries . How many books do the libraries in Oak Grove have overall ?","target":7092.0}
|
||||
{"input":"There were originally 20817 houses in Lincoln County . During a housing boom , developers built 97741 . How many houses are there now in Lincoln County ?","target":118558.0}
|
||||
{"input":"A farmer estimates that he will harvest 48097 bushels of wheat . The weather is perfect during the growing season , so he harvests 684 bushels of wheat than expected . How many bushels of wheat does the farmer harvest ?","target":48781.0}
|
||||
{"input":"Christina just transferred $ 69 out of her bank account . As a result , the account now has $ 26935 in it . How much money was in the account before the transfer ?","target":27004.0}
|
||||
{"input":"Last year at Newberg 's airport , 14507 passengers landed on time . Unfortunately , 213 passengers landed late . In all , how many passengers landed in Newberg last year ?","target":14720.0}
|
||||
{"input":"A dust storm sweeps across the prairie . It covers 64535 acres of the prairie in dust , but leaves 522 acres untouched . How many acres does the prairie cover ?","target":64013.0}
|
||||
{"input":"Some insects called aphids attack a large farm . In response , the farmer releases ladybugs onto the fields . There are 12170 ladybugs with spots and 54912 ladybugs without spots . How many ladybugs are there in all ?","target":67082.0}
|
||||
{"input":"Last year , 90171 people were born in a country , and 16320 people immigrated to it . How many new people began living in the country last year ?","target":106491.0}
|
||||
{"input":"A ship full of grain crashes into a coral reef . By the time the ship is fixed , 49952 tons of grain have spilled into the water . Only 918 tons of grain remain onboard . How many tons of grain did the ship originally contain ?","target":50870.0}
|
||||
{"input":"To fill an order , the factory dyed 61921 yards of silk green and 49500 yards pink . How many yards of silk did it dye for that order ?","target":111421.0}
|
||||
{"input":"A multi-national corporation has 2041 part-time employees and 63093 full-time employees . How many employees work for the corporation ?","target":65134.0}
|
||||
{"input":"Each year , salmon travel upstream , going from the ocean to the rivers where they were born . This year , 712261 male and 259378 female salmon returned to their rivers . How many salmon made the trip ?","target":971639.0}
|
||||
{"input":"A bathing suit manufacturer has a supply of 14797 bathing suits for men . In addition , it has 4969 bathing suits for women . How many bathing suits are available overall ?","target":19766.0}
|
||||
{"input":"Before the recent housing boom , there were 1426 houses in Lawrence County . Now , there are 2000 houses . How many houses did developers build during the housing boom ?","target":574.0}
|
||||
{"input":"A worker at a medical lab is studying blood samples . 2 samples contained a total of 7341 blood cells . The first sample contained 4221 blood cells . How many blood cells were in the second sample ?","target":3120.0}
|
||||
{"input":"So far , an orchard has sold a combined total of 9792 pounds of fresh and frozen fruit this season . If they have sold 3513 pounds of frozen fruit , how many pounds of fresh fruit have been sold so far ?","target":6279.0}
|
||||
{"input":"Recently , the value of Kate 's retirement fund decreased by $ 12 . If her fund was worth $ 1472 before , how much is it worth now ?","target":1460.0}
|
||||
{"input":"The Richmond Tigers sold a total of 9570 tickets last season . If they sold 3867 tickets in the first half of the season , how many tickets did they sell in the second half ?","target":5703.0}
|
||||
{"input":"A petri dish originally contained 600 bacteria . A scientist let the bacteria grow and now there are 8917 of them . How many more bacteria are there now ?","target":8317.0}
|
||||
{"input":"Tori is a school janitor . Last week , she picked up a total of 1576 pieces of trash . If she picked up 344 pieces of trash in the classrooms , how many pieces of trash did Tori pick up outside the classrooms ?","target":1232.0}
|
||||
{"input":"Molly owns the Wafting Pie Company . This morning , her employees used 816 eggs to bake pumpkin pies . If her employees used a total of 1339 eggs today , how many eggs did they use in the afternoon ?","target":523.0}
|
||||
{"input":"Each of farmer Cunningham 's 6048 lambs is either black or white . There are 193 white ones . How many of Farmer Cunningham 's lambs are black ?","target":5855.0}
|
||||
{"input":"Students at Arcadia schools are participating in a coat drive . 9437 coats have been collected so far . 6922 coats were collected from the high schools , and the rest from the elementary schools . How many coats were collected at the elementary schools ?","target":2515.0}
|
||||
{"input":"A company painted some houses in Hancock County white and blue using a total of 6689 gallons of paint . If they used 660 gallons of white paint , how many gallons of blue paint did the company use ?","target":6029.0}
|
||||
{"input":"The Silvergrove Public Library used a grant to purchase 2647 books . Now the library has a total of 8582 books . How many books did the library have before the grant ?","target":5935.0}
|
||||
{"input":"A cell phone company has a total of 7422 customers across the world . If 723 of its customers live in the United States , how many of its customers live in other countries ?","target":6699.0}
|
||||
{"input":"Last year , egg producers in Douglas County produced 1416 eggs . This year , those same farms produced 4636 eggs . How many more eggs did the farms produce this year ?","target":3220.0}
|
||||
{"input":"An oil pipe in the sea broke . Before engineers started to fix the pipe , 2475 gallons of oil leaked into the water . A total of 6206 gallons of oil leaked before the pipe was fixed . How many gallons of oil leaked while the engineers were fixing the pipe ?","target":3731.0}
|
||||
{"input":"A treasure hunter discovered a buried treasure chest filled with a total of 5155 gems . 45 of the gems were diamonds , and the rest were rubies . How many of the gems were rubies ?","target":5110.0}
|
||||
{"input":"Shannon and her family use up a lot of strawberry and blueberry jelly , since they eat toast every morning . At the moment , they have a combined total of 6310 grams of jelly . If they have 4518 grams of blueberry jelly , how many grams of strawberry jelly do they have ?","target":1792.0}
|
||||
{"input":"While playing a video game , Paul scored 3103 points . He and his cousin together have a total of 5816 points . How many points does Paul 's cousin have ?","target":2713.0}
|
||||
{"input":"A construction company is repaving a damaged road . So far , they have repaved a total of 4938 inches of the road . Today , they repaved 805 inches of the road . How many inches of the road had they repaved before today ?","target":4133.0}
|
||||
{"input":"Milford Lake was originally blue because it only had 809 algae plants . Now there are 3263 algae plants , and the lake has turned green . How many more algae plants are in Milford Lake now ?","target":2454.0}
|
||||
{"input":"Oscar 's bus ride to school is 0.75 mile and Charlie 's bus ride is 0.25 mile . How much longer is Oscar 's bus ride than Charlie 's ?","target":0.5}
|
||||
{"input":"Karen added 0.25 cup of walnuts to a batch of trail mix . Later , she added 0.25 cup of almonds . How many cups of nuts did Karen put in the trail mix in all ?","target":0.5}
|
||||
{"input":"Kendall is learning to drive , so this weekend she practiced driving 0.16666666666666666 mile with her mother and another 0.5 mile with her father . How far did Kendall drive in all ?","target":0.6666666667}
|
||||
{"input":"At a pie-eating contest , Erik got through 0.6666666666666666 pie before time was called ; Frank finished just 0.3333333333333333 pie . How much more pie did Erik eat than Frank ?","target":0.3333333333}
|
||||
{"input":"A tailor cut 0.75 inch off a skirt and 0.5 inch off a pair of pants . How much more did the tailor cut off the skirt than the pants ?","target":0.25}
|
||||
{"input":"At Lindsey 's Vacation Wear , 0.375 the garments are bikinis and 0.25 are trunks . What fraction of the garments are either bikinis or trunks ?","target":0.625}
|
||||
{"input":"Darnel sprinted 0.875 lap and then took a break by jogging 0.75 lap . How much farther did Darnel sprint than jog ?","target":0.125}
|
||||
{"input":"A marine biologist measured 1 fish that was 0.3 foot long and a second fish that was 0.2 foot long . How much longer was the first fish ?","target":0.1}
|
||||
{"input":"Hannah 's Vegetarian Restaurant bought 0.3333333333333333 pound of green peppers and 0.3333333333333333 pound of red peppers . How many pounds of peppers did Hannah 's Vegetarian Restaurant buy in all ?","target":0.6666666667}
|
||||
{"input":"Ella owns 2 dogs . Each day , 1 dog eats 0.125 scoop of dog food and the other dog eats 0.125 scoop . Together , how much dog food do the 2 dogs eat each day ?","target":0.25}
|
||||
{"input":"Blake filled a bucket with 0.8 gallon of water . Later , he poured out 0.2 gallon of the water . How much water is in the bucket ?","target":0.6}
|
||||
{"input":"Mandy made an apple pie . She used 0.6666666666666666 tablespoon of cinnamon and 0.5 tablespoon of nutmeg . How much more cinnamon than nutmeg did Mandy use ?","target":0.1666666667}
|
||||
{"input":"The Montoya family spends 0.6 their budget on groceries and another 0.2 going out to eat . Altogether , what fraction of their budget does the Montoya family spend on food ?","target":0.8}
|
||||
{"input":"In Mr. Olsen 's mathematics class , 0.7 the students received A 's and 0.2 received B 's . What fraction of the students received either A 's or B 's ?","target":0.9}
|
||||
{"input":"There is 0.16666666666666666 cup of oil in Scarlett 's measuring cup . If Scarlett adds 0.6666666666666666 cup more , how much oil will be in the measuring cup ?","target":0.8333333333}
|
||||
{"input":"1 evening , a restaurant served a total of 0.2 loaf of wheat bread and 0.4 loaf of white bread . How many loaves were served in all ?","target":0.6}
|
||||
{"input":"Stanley ran 0.4 mile and walked 0.2 mile . How much farther did Stanley run than walk ?","target":0.2}
|
||||
{"input":"Dina made cookies . She used 0.625 cup of flour and 0.25 cup of sugar . How much more flour than sugar did Dina use ?","target":0.375}
|
||||
{"input":"Each day , the polar bear at Richmond 's zoo eats 0.2 bucket of trout and 0.4 bucket of salmon . How many buckets of fish does the polar bear eat daily ?","target":0.6}
|
||||
{"input":"Jenny ran 0.6 mile and walked 0.4 mile . How much farther did Jenny run than walk ?","target":0.2}
|
||||
{"input":"0.5 the students in the band are in the trumpet section . 0.125 the students in the band are in the trombone section . What fraction of the students in the band are in either the trumpet section or the trombone section ?","target":0.625}
|
||||
{"input":"Suzie found 2 worms in the yard and measured them with a ruler . 1 worm was 0.8 inch long . The other worm was 0.1 inch long . How much longer was the longer worm ?","target":0.7}
|
||||
{"input":"Scarlett made a fruit salad with 0.25 pound of melon and 0.375 pound of berries . How many pounds of fruit did Scarlett use in all ?","target":0.625}
|
||||
{"input":"Vince 's bus ride to school is 0.625 mile and Zachary 's bus ride is 0.5 mile . How much longer is Vince 's bus ride than Zachary 's ?","target":0.125}
|
||||
{"input":"In 1 week , Mitch 's family drank 0.5 carton of regular milk and 0.1 carton of soy milk . How much milk did they drink in all ?","target":0.6}
|
||||
{"input":"Elizabeth went to the salon and had 0.375 inch of hair cut off . The next day she went back and asked for another 0.5 inch to be cut off . How much hair did she have cut off in all ?","target":0.875}
|
||||
{"input":"In Shannon 's apartment complex , 0.16666666666666666 the apartments are one-bedroom apartments and 0.3333333333333333 are two-bedroom apartments . What fraction of the apartments are either 1 - or two-bedroom apartments ?","target":0.5}
|
||||
{"input":"At the beach , Miki and her sister both built sandcastles and then measured their heights . Miki 's sandcastle was 0.8333333333333334 foot tall and her sister 's was 0.5 foot tall . How much taller was Miki 's sandcastle than her sister 's ?","target":0.3333333333}
|
||||
{"input":"Michelle began her pizza delivery route with 0.5 tank of gas in her car . When she made it back to the pizzeria , 0.16666666666666666 tank of gas was . How much gas did Michelle use ?","target":0.3333333333}
|
||||
{"input":"While taking inventory at her pastry shop , Kelly realizes that she had 0.4 box of baking powder yesterday , but the supply is now down to 0.3 box . How much more baking powder did Kelly have yesterday ?","target":0.1}
|
||||
{"input":"Craig walked 0.2 mile from school to David 's house and 0.7 mile from David 's house to his own house . How many miles did Craig walk in all ?","target":0.9}
|
||||
{"input":"Greg and Sharon own neighboring cornfields . Greg harvested 0.4 acre of corn on Monday and Sharon harvested 0.1 acre . How many more acres did Greg harvest than Sharon ?","target":0.3}
|
||||
{"input":"At the hardware store , 0.25 the nails are size 2d and 0.5 the nails are size 4d . What fraction of the nails are either size 2d or 4d ?","target":0.75}
|
||||
{"input":"While making desserts for a bake sale , Victor used 0.625 scoop of brown sugar as well as 0.25 scoop of white sugar . How much more brown sugar did Victor use ?","target":0.375}
|
||||
{"input":"Eve ran 0.7 mile and walked 0.6 mile . How much farther did Eve run than walk ?","target":0.1}
|
||||
{"input":"Jonah added 0.3 cup of yellow raisins and 0.4 cup of black raisins to a batch of trail mix . How many cups of raisins did Jonah add in all ?","target":0.7}
|
||||
{"input":"When Jake had 1 cat , he needed to serve 0.5 can of cat food each day . Now that Jake has adopted a second cat , he needs to serve a total of 0.9 can each day . How much extra food is needed to feed the second cat ?","target":0.4}
|
||||
{"input":"In Yardley it snowed 0.125 inch in the morning and 0.5 inch in the afternoon . What was the total amount of snowfall ?","target":0.625}
|
||||
{"input":"While making pastries , a bakery used 0.2 bag of wheat flour and 0.1 bag of white flour . How many bags of flour did the bakery use in all ?","target":0.3}
|
||||
{"input":"Kaleen filled a bucket with 0.75 gallon of water . A few minutes later , she realized only 0.5 gallon of water remained . How much water had leaked out of the bucket ?","target":0.25}
|
||||
{"input":"Irwin 's family went on a camping trip in the mountains . On the first day , they hiked from their car to the campsite . First , they hiked 0.2 mile from the car to a stream , and 0.4 mile from the stream to a meadow . Then they hiked 0.1 mile from the meadow to the campsite . How many miles did Irwin 's family hike in all ?","target":0.7}
|
||||
{"input":"During a visit to an orchard , Charlie picked 0.16666666666666666 bag of Golden Delicious apples , 0.16666666666666666 bag of Macintosh apples , and 0.3333333333333333 bag of Cortland apples . How many bags of fruit did Charlie pick in total ?","target":0.6666666667}
|
||||
{"input":"Before starting her shift , a waitress checks to make sure there is enough mustard for her customers . She finds 0.25 bottle at the first table , 0.25 bottle at the second table , and 0.375 bottle at the third table . Altogether , how many bottles of mustard does the waitress find ?","target":0.875}
|
||||
{"input":"A waitress put leftover tarts into the fridge on Thursday night . She noticed that the restaurant had 0.08333333333333333 tart filled with cherries , 0.75 tart filled with blueberries , and 0.08333333333333333 tart filled with peaches . How many leftover tarts did the restaurant have in all ?","target":0.9166666667}
|
||||
{"input":"On her vacation last summer , Trisha walked all over New York City to buy souvenirs . First , she walked 0.1111111111111111 mile from her hotel to a postcard shop . Then she walked 0.1111111111111111 mile from the postcard shop to a T-shirt shop and 0.6666666666666666 mile from the T-shirt shop back to the hotel . How many miles did Trisha walk in all ?","target":0.8888888889}
|
||||
{"input":"Brandy made trail mix for a backpacking trip . She used 0.16666666666666666 pound of peanuts , 0.16666666666666666 pound of chocolate chips , and 0.08333333333333333 pound of raisins . How many pounds of trail mix did Brandy make ?","target":0.4166666667}
|
||||
{"input":"Allie counted the leftover ice cream after a sundae party . She had 0.3333333333333333 carton of rocky road ice cream , 0.3333333333333333 carton of cookie dough ice cream , and 0.16666666666666666 carton of strawberry cheesecake ice cream . How many cartons of ice cream did Allie have in all ?","target":0.8333333333}
|
||||
{"input":"During a canned food drive , items were sorted into bins . The drive resulted in 0.125 bin of soup , 0.125 bin of vegetables , and 0.5 bin of pasta . Altogether , how many bins would the canned food take up ?","target":0.75}
|
||||
{"input":"Paco 's Countertop Company purchased pieces of marble from a quarry . The weights of the pieces they purchased were 0.3333333333333333 ton , 0.3333333333333333 ton , and 0.08333333333333333 ton . How many tons of marble did Paco 's Countertop Company purchase in all ?","target":0.75}
|
||||
{"input":"Nina did a running drill to get in shape for soccer season . First , Nina ran 0.08333333333333333 mile . Then she ran 0.08333333333333333 mile and 0.6666666666666666 mile . How many miles did Nina run in total ?","target":0.8333333333}
|
||||
{"input":"Bonnie 's science class recorded the rainfall each day . They recorded 0.16666666666666666 centimeter of rain on Monday , 0.4166666666666667 centimeter of rain on Tuesday , and 0.08333333333333333 centimeter of rain on Wednesday . How many centimeters of rain did the class record in all ?","target":0.6666666667}
|
||||
{"input":"Last Saturday , Spencer walked all over town running errands . First , he walked 0.3 mile from his house to the library and 0.1 mile from the library to the post office . Then he walked 0.4 mile from the post office back home . How many miles did Spencer walk in all ?","target":0.8}
|
||||
{"input":"A construction company ordered 0.16666666666666666 ton of concrete , 0.16666666666666666 ton of bricks , and 0.5 ton of stone . How many tons of material did the company order in all ?","target":0.8333333333}
|
||||
{"input":"Kendra made punch for her friend 's birthday party . She used 0.25 gallon of grape juice , 0.375 gallon of cranberry juice , and 0.125 gallon of club soda . How many gallons of punch did Kendra make ?","target":0.75}
|
||||
{"input":"A spaceship traveled 0.5 light-year from Earth to Planet X and 0.1 light-year from Planet X to Planet Y. Then it traveled 0.1 light-year from Planet Y back to Earth . How many light-years did the spaceship travel in all ?","target":0.7}
|
||||
{"input":"Logan recorded the snowfall every day during a snowstorm . He recorded 0.3333333333333333 centimeter on Wednesday , 0.3333333333333333 centimeter on Thursday , and 0.2222222222222222 centimeter on Friday . How many total centimeters of snow did Logan record ?","target":0.8888888889}
|
||||
{"input":"Ellen made smoothies in the blender . She used 0.2 cup of strawberries , 0.1 cup of yogurt , and 0.2 cup of orange juice . How many cups of ingredients did Ellen use for the smoothies ?","target":0.5}
|
||||
{"input":"During a school play , Jonah staffed the snack bar . He served 0.25 pitcher of lemonade during the first intermission , 0.4166666666666667 pitcher during the second , and 0.25 pitcher during the third . How many pitchers of lemonade did Jonah pour in all ?","target":0.9166666667}
|
||||
{"input":"Heather went to the county fair last weekend . When she got there , she had to walk 0.3333333333333333 mile from the car to the entrance . Then she walked 0.3333333333333333 mile to the carnival rides and 0.08333333333333333 mile from the carnival rides back to the car . How many miles did Heather walk in all ?","target":0.75}
|
||||
{"input":"A renovation project required 0.16666666666666666 truck-load of sand , 0.3333333333333333 truck-load of dirt , and 0.16666666666666666 truck-load of cement . How many truck-loads of material were needed in all ?","target":0.6666666667}
|
||||
{"input":"Karin 's science class weighed plastic rings for an experiment . They found that the orange ring weighed 0.08333333333333333 ounce , the purple ring weighed 0.3333333333333333 ounce , and the white ring weighed 0.4166666666666667 ounce . What was the total weight of the plastic rings ?","target":0.8333333333}
|
||||
{"input":"Carefully following a recipe , Kenny used exactly 0.16666666666666666 cup of oil and 1.1666666666666667 cups of water . How many cups of liquid did Kenny use in all ?","target":1.3333333333}
|
||||
{"input":"Dale 's Vegetarian Restaurant bought 2.8333333333333335 pounds of green peppers and 2.8333333333333335 pounds of red peppers . How many pounds of peppers did Dale 's Vegetarian Restaurant buy in all ?","target":5.6666666667}
|
||||
{"input":"This afternoon Craig left school , rode the bus 3.8333333333333335 miles , and then walked 0.16666666666666666 mile to get home . How much farther did Craig ride than walk ?","target":3.6666666667}
|
||||
{"input":"Kelly 's chemistry textbook weighs 7.125 pounds and her geometry textbook weighs 0.625 pound . How much more does the chemistry textbook weigh than the geometry textbook ?","target":6.5}
|
||||
{"input":"Roadster 's Paving Company used 10 tons of cement to pave Lexi 's street and 5.1 tons of cement to pave Tess 's street . How much cement did Roadster 's Paving Company use in all ?","target":15.1}
|
||||
{"input":"On a hot day , Sam poured 1 bucket of water into a plastic wading pool . A few minutes later he added another 8.8 buckets . How much water did Sam pour into the pool ?","target":9.8}
|
||||
{"input":"Kyle jogged 1.125 laps in P.E. class and 2.125 laps during track practice . How many laps did Kyle jog in all ?","target":3.25}
|
||||
{"input":"A bucket contains 3 gallons of water . If Derek adds 6.8 gallons more , how many gallons will there be in all ?","target":9.8}
|
||||
{"input":"At a pizza party , Mason and his friends drank 2.6666666666666665 bottles of lemon-lime soda and 2.6666666666666665 bottles of cola . How much soda did they drink in all ?","target":5.3333333333}
|
||||
{"input":"Professor Ellison weighed 2 pieces of metal for an experiment . The piece of iron weighed 11.166666666666666 pounds and the piece of aluminum weighed 0.8333333333333334 pound . How much more did the piece of iron weigh than the piece of aluminum ?","target":10.3333333333}
|
||||
{"input":"As part of a lesson on earthquakes , a science class is researching the movement of a nearby fault line . The fault line moved 1.25 inches during the past year and 5.25 inches the year before . How far did the fault line move in all ?","target":6.5}
|
||||
{"input":"Hoping to be named Salesperson of the Month , Rosa called the names from 10.2 pages of the phone book last week . This week , she called the people listed on another 8.6 pages of the same phone book . How many pages worth of people did Rosa call in all ?","target":18.8}
|
||||
{"input":"At the beach , Janet and her sister both built sandcastles and then measured their heights . Janet 's sandcastle was 3.6666666666666665 feet tall and her sister 's was 2.3333333333333335 feet tall . How much taller was Janet 's sandcastle than her sister 's ?","target":1.3333333333}
|
||||
{"input":"Nicole found an orange caterpillar and a green caterpillar in her backyard . The green caterpillar was 3 inches long and the orange caterpillar was 1.1666666666666667 inches long . How much longer was the green caterpillar than the orange caterpillar ?","target":1.8333333333}
|
||||
{"input":"Alec and his roommates ate 3.25 pints of ice cream on Friday night and 0.25 pint of ice cream on Saturday night . How many pints did they eat in all ?","target":3.5}
|
||||
{"input":"A farmer started the day with 8.75 buckets of seeds . After spending the morning sowing seeds , she now has 6 buckets . How many buckets of seeds did the farmer sow ?","target":2.75}
|
||||
{"input":"Irene just bought a new lamp for her bedside table . The old lamp was 1 foot tall and the new lamp is 2.3333333333333335 feet tall . How much taller is the new lamp than the old lamp ?","target":1.3333333333}
|
||||
{"input":"Ezra drew a white line that was 7.666666666666667 inches long . Then he drew a blue line that was 3.3333333333333335 inches long . How much longer was the white line than the blue line ?","target":4.3333333333}
|
||||
{"input":"There are 7.75 gallons of water in Becky 's fish tank . If Becky adds 7 gallons more , how many gallons will there be in all ?","target":14.75}
|
||||
{"input":"Wendy ran 19.833333333333332 miles and walked 9.166666666666666 miles . How much farther did Wendy run than walk ?","target":10.6666666667}
|
||||
{"input":"Kenji and his classmates placed colored blocks on a scale during a science lab . The yellow block weighed 0.6 pounds and the green block weighed 0.4 pounds . How much more did the yellow block weigh than the green block ?","target":0.2}
|
||||
{"input":"Terrell hiked 8.2 miles on Saturday . Then , on Sunday , he hiked another 1.6 miles . How far did Terrell hike all together ?","target":9.8}
|
||||
{"input":"A carpenter bought a piece of wood that was 0.41 meters long . Then she sawed 0.33 meters off the end . How long is the piece of wood now ?","target":0.08}
|
||||
{"input":"Kelly bought 0.1 pounds of peanuts and 0.4 pounds of raisins . How many pounds of snacks did she buy in all ?","target":0.5}
|
||||
{"input":"Kevin bought 2 watermelons . The first watermelon was 9.91 pounds , and the second watermelon was 4.11 pounds . How many pounds of watermelon did Kevin buy ?","target":14.02}
|
||||
{"input":"In March it rained 0.81 inches . It rained 0.35 inches less in April than in March . How much did it rain in April ?","target":0.46}
|
||||
{"input":"Ron weighed 2 colored metal balls during a science class . The blue ball weighed 6 pounds and the brown ball weighed 3.12 pounds . If Ron places both balls on the scale at the same time , what will the scale read ?","target":9.12}
|
||||
{"input":"A bee colony produced 0.36 pounds of honey , but bears ate 0.05 pounds of it . How much honey remains ?","target":0.31}
|
||||
{"input":"It rained 0.9 inches on Monday . On Tuesday , it rained 0.7 inches less than on Monday . How much did it rain on Tuesday ?","target":0.2}
|
||||
{"input":"It snowed 0.32 inches on Monday and 0.21 inches on Tuesday . How much did it snow on Monday and Tuesday combined ?","target":0.53}
|
||||
{"input":"Brennan had 0.25 grams of pepper . Then he used 0.16 grams of the pepper to make some scrambled eggs . How much pepper does Brennan have ?","target":0.09}
|
||||
{"input":"A construction company bought 5.91 tons of gravel and 8.11 tons of sand . How many tons of material did the company buy in all ?","target":14.02}
|
||||
{"input":"It rained 0.2 inches on Saturday and 0.4 inches on Sunday . How much did it rain on Saturday and Sunday combined ?","target":0.6}
|
||||
{"input":"Pamela bought 9.8 ounces of sugar , and she spilled 5.2 ounces of it on the floor . How much is ?","target":4.6}
|
||||
{"input":"Gordon bought 3.42 pounds of fruit for a class party . The class ate 2.2 pounds of the fruit . How much fruit is ?","target":1.22}
|
||||
{"input":"A chef bought 0.14 kilograms of almonds and 0.38 kilograms of pecans . How many kilograms of nuts did the chef buy in all ?","target":0.52}
|
||||
{"input":"Marta picked 2 pumpkins . The first pumpkin weighed 4 pounds , and the second pumpkin weighed 8.7 pounds . How much did the 2 pumpkins weigh all together ?","target":12.7}
|
||||
{"input":"A truck carrying 4.1 pounds of sand travels to a construction yard and loses 2.4 pounds of sand along the way . How much sand does the truck have when it arrives at the yard ?","target":1.7}
|
||||
{"input":"Tori was 4.4 feet tall . Then she grew 2.86 feet taller . How tall is Tori now ?","target":7.26}
|
||||
{"input":"A carpenter bought a piece of wood that was 8.9 centimeters long . Then he sawed 2.3 centimeters off the end . How long is the piece of wood now ?","target":6.6}
|
||||
{"input":"Jason found 49 seashells and 48 starfish on the beach . He gave 13 of the seashells to Tim . How many seashells does Jason now have ?","target":36.0}
|
||||
{"input":"Joan grew 29 carrots and 14 watermelons . Jessica grew 11 carrots . How many carrots did they grow in all ?","target":40.0}
|
||||
{"input":"Sally had 39 baseball cards , and 9 were torn . Sara bought 24 of Sally 's baseball cards . How many baseball cards does Sally have now ?","target":15.0}
|
||||
{"input":"Dan has 32 green and 38 violet marbles . Mike took 23 of Dan 's green marbles . How many green marbles does Dan now have ?","target":9.0}
|
||||
{"input":"Jason has 18 books and he has read 9 of them . Mary has 42 books . How many books do they have together ?","target":60.0}
|
||||
{"input":"Jason grew 32 watermelons and 22 cantelopes . Dan grew 31 watermelons . How many watermelons did they grow in total ?","target":63.0}
|
||||
{"input":"Benny found 66 seashells and 49 starfish on the beach . He gave 52 of the seashells to Jason . How many seashells does Benny now have ?","target":14.0}
|
||||
{"input":"There are 41 crayons and 26 pencils in the drawer . Sam placed 12 crayons in the drawer . How many crayons are now there in total ?","target":53.0}
|
||||
{"input":"Mike had 33 quarters and 87 nickels in his bank . His dad borrowed 75 nickels from Mike . How many nickels does he have now ?","target":12.0}
|
||||
{"input":"Sam has 86 yellow and 20 green marbles . Joan took 25 of Sam 's yellow marbles . How many yellow marbles does Sam now have ?","target":61.0}
|
||||
{"input":"Dan had 97 baseball cards , and 8 were torn . Sam bought 15 of Dan 's baseball cards . How many baseball cards does Dan have now ?","target":82.0}
|
||||
{"input":"There are 41 short trees and 44 tall trees currently in the park . Park workers will plant 57 short trees today . How many short trees will the park have when the workers are finished ?","target":98.0}
|
||||
{"input":"Alyssa picked 25 limes and Mike picked 32 limes . Tom picked 12 plums . How many limes were picked in all ?","target":57.0}
|
||||
{"input":"Tim found 679 seashells and 110 starfish on the beach . He gave 172 of the seashells to Sara . How many seashells does Tim now have ?","target":507.0}
|
||||
{"input":"There are 139 erasers and 118 scissors in the drawer . Jason placed 131 erasers in the drawer . How many erasers are now there in total ?","target":270.0}
|
||||
{"input":"Mike picked 123 oranges and Melanie picked 104 oranges . Fred picked 130 apples . How many oranges were picked in total ?","target":227.0}
|
||||
{"input":"Joan had 695 Pokemon cards , and 6 were torn . Sara bought 133 of Joan 's Pokemon cards . How many Pokemon cards does Joan have now ?","target":562.0}
|
||||
{"input":"Sally grew 113 turnips and 118 pumpkins . Mary grew 129 turnips . How many turnips did they grow in total ?","target":242.0}
|
||||
{"input":"Sara has 792 black and 122 red marbles . Fred took 233 of Sara 's black marbles . How many black marbles does Sara now have ?","target":559.0}
|
||||
{"input":"Joan 's high school played 864 baseball games this year , 128 of the games were played at night . She attended 395 games . How many baseball games did Joan miss ?","target":469.0}
|
||||
{"input":"There are 112 short trees and 119 tall trees currently in the park . Park workers will plant 105 short trees today . How many short trees will the park have when the workers are finished ?","target":217.0}
|
||||
{"input":"Sara had 100 pennies and 783 quarters in her bank . Her dad borrowed 271 quarters from Sara . How many quarters does she have now ?","target":512.0}
|
||||
{"input":"A restaurant made 9 hamburgers and 4 hot dogs to serve during lunch . Only 3 hamburgers were actually served . How many hamburgers were over ?","target":6.0}
|
||||
{"input":"Melanie picked 7 plums and 4 oranges from the orchard . She gave 3 plums to Sam . How many plums does she have now ?","target":4.0}
|
||||
{"input":"There are 6 short bushes and 4 tall trees currently in the park . Park workers had to cut down 2 short bushes that were damaged . How many short bushes will the park have when the workers are finished ?","target":4.0}
|
||||
{"input":"There were a total of 8 football games this year , 4 are played at night . Keith missed 4 of the games . How many football games did Keith go to in total ?","target":4.0}
|
||||
{"input":"There are 9 pencils and 4 rulers in the drawer . Sally took 4 pencils out of the drawer . How many pencils are there now ?","target":5.0}
|
||||
{"input":"Jason has 7 violet balloons and 4 red balloons . He lost 3 of the violet balloons . How many violet balloons does Jason have now ?","target":4.0}
|
||||
{"input":"Dan had 7 potatoes and 4 cantelopes in the garden . The rabbits ate 4 of the potatoes . How many potatoes does Dan now have ?","target":3.0}
|
||||
{"input":"Sara had 4 quarters and 8 dimes in her bank . Her sister borrowed 4 dimes . How many dimes does Sara have now ?","target":4.0}
|
||||
{"input":"Sandy 's dog had 8 puppies and 4 had spots . She gave 4 to her friends . How many puppies does she now have ?","target":4.0}
|
||||
{"input":"Mike found 6 seashells and 4 starfish , but 4 of the seashells were broken . How many unbroken seashells did Mike find ?","target":2.0}
|
||||
{"input":"Melanie had 30 baseball cards , and 9 were torn . Sara bought 18 of Melanie 's baseball cards . How many baseball cards does Melanie have now ?","target":12.0}
|
||||
{"input":"Sara picked 35 pears and 27 apples from the orchard . She gave 28 pears to Dan . How many pears does Sara have ?","target":7.0}
|
||||
{"input":"Jessica grew 35 watermelons and 30 carrots , but the rabbits ate 27 watermelons . How many watermelons does Jessica have ?","target":8.0}
|
||||
{"input":"Nancy found 35 seashells and 25 starfish on the beach . She gave 17 of the seashells to Jason . How many seashells does Nancy now have ?","target":18.0}
|
||||
{"input":"Joan has 47 green and 48 red marbles . Fred took 24 of Joan 's green marbles . How many green marbles does Joan now have ?","target":23.0}
|
||||
{"input":"There are 42 walnut trees and 12 orange trees currently in the park . Park workers had to cut down 13 walnut trees that were damaged . How many walnut trees will be in the park when the workers are finished ?","target":29.0}
|
||||
{"input":"Sandy had 36 pennies and 31 nickels in her bank . Her dad borrowed 20 nickels from Sandy . How many nickels does she have now ?","target":11.0}
|
||||
{"input":"There are 34 pencils and 49 crayons in the drawer . Dan took 22 pencils from the drawer . How many pencils are now in the drawer ?","target":12.0}
|
||||
{"input":"Sally paid $ 12.32 total for peaches , after a 3 dollar coupon , and $ 11.54 for cherries . In total , how much money did Sally spend ?","target":23.86}
|
||||
{"input":"Dan spent $ 11.76 on a snake toy , and a cage cost him $ 14.54 . Dan also found a dollar bill on the ground . What was the total cost of Dan 's purchases ?","target":26.3}
|
||||
{"input":"Mary went to the mall . She spent $ 13.04 on a shirt and $ 12.27 on a jacket . She went to 2 shops . In total , how much money did Mary spend on clothing ?","target":25.31}
|
||||
{"input":"Dan joined his school 's band . He bought a clarinet for $ 130.30 , and a song book which was $ 11.24 . Dan found $ 12.32 in his pocket . How much did Dan spend at the music store ?","target":141.54}
|
||||
{"input":"Tom bought a skateboard for $ 9.46 , and spent $ 9.56 on marbles . Tom also spent $ 14.50 on shorts . In total , how much did Tom spend on toys ?","target":19.02}
|
||||
{"input":"Mary got fast food for lunch . Mary spent $ 1.08 on soup and $ 4.80 on a salad . Mary paid with a 20 dollar bill . What was the total of the lunch bill ?","target":5.88}
|
||||
{"input":"For his car , Mike spent $ 118.54 on speakers and $ 106.33 on new tires . Mike wanted 3 CD 's for $ 4.58 but decided not to . In total , how much did Mike spend on car parts ?","target":224.87}
|
||||
{"input":"Tom purchased a Batman game for $ 13.60 , and a Superman game for $ 5.06 . Tom already owns 2 games . How much did Tom spend on video games ?","target":18.66}
|
||||
{"input":"Joan spent $ 15 on shorts and $ 14.82 on a jacket , and $ 12.51 on a shirt . She went to 3 shops . In total , how much money did Joan spend on clothing ?","target":42.33}
|
||||
{"input":"Joan joined her school 's band . She bought a trumpet for $ 149.16 , a music tool for $ 9.98 , and a song book which was $ 4.14 . Joan found $ 8.65 in her pocket . How much did Joan spend at the music store ?","target":163.28}
|
||||
{"input":"Melanie bought a Batman game for $ 6.95 , a strategy game for $ 7.90 , and a Superman game for $ 7.73 . Melanie already owns 4 games . How much did Melanie spend on video games ?","target":22.58}
|
||||
{"input":"Keith spent $ 6.51 on a rabbit toy , $ 5.79 on pet food , and a cage cost him $ 12.51 . He found a dollar bill on the ground . What was the total cost of Keith 's purchases ?","target":24.81}
|
||||
{"input":"Keith spent $ 136.01 on speakers , $ 139.38 on a CD player , and $ 112.46 on new tires . He wanted 3 CD 's for $ 6.16 , but did n't buy them . In total , how much did he spend ?","target":387.85}
|
||||
{"input":"Joan bought toy cars for $ 14.88 , a skateboard for $ 4.88 , and got toy trucks for $ 5.86 . She spent $ 14.55 on pants . In total , how much did Joan spend on toys ?","target":25.62}
|
||||
{"input":"After paying 6 dollars for the pie , Mary has 52 dollars , her friend has 43 dollars . How much money did she have before buying the pie ?","target":58.0}
|
||||
{"input":"There were 46 bales of hay in the barn and 32 bales in the shed . Tom stacked bales in the barn today . There are now 60 bales of hay in the barn . How many bales did he store in the barn ?","target":14.0}
|
||||
{"input":"Mary is baking a cake . The recipe calls for 7 cups of flour and 3 cups of sugar . She already put in 2 cups of flour . How many cups of flour does she need to add ?","target":5.0}
|
||||
{"input":"Last week Fred had 23 dollars and Jason had 46 dollars . Fred washed cars over the weekend and now has 86 dollars . How much money did Fred make washing cars ?","target":63.0}
|
||||
{"input":"There are 31 short trees and 32 tall trees currently in the park . Park workers will plant short trees today . When the workers are finished there will be 95 short trees in the park . How many short trees did the workers plant today ?","target":64.0}
|
||||
{"input":"There were 9 red orchids and 3 white orchids in the vase . Sally cut some red orchids from her flower garden . There are now 15 red orchids in the vase . How many red orchids did she cut ?","target":6.0}
|
||||
{"input":"Sara had 24 peaches and 37 pears at her fruit dish . She went to the orchard and picked peaches . There are now 61 peaches . how many did she pick ?","target":37.0}
|
||||
{"input":"Joan found 72 seashells and 12 starfishes on the beach . She gave Alyssa some of her seashells . She has 28 seashell . How many seashells did she give to Alyssa ?","target":44.0}
|
||||
{"input":"There are 6 pencils and 7 rulers in the drawer . Benny placed 3 pencils in the drawer . How many pencils are now there in total ?","target":9.0}
|
||||
{"input":"Sara has 3 green and 5 red marbles . Tom has 4 green marbles . How many green marbles do they have in total ?","target":7.0}
|
||||
{"input":"Joan grew 8 watermelons and 4 turnips . Tom grew 9 watermelons . How many watermelons did they grow in total ?","target":17.0}
|
||||
{"input":"There are 2 maple trees and 5 popular trees currently in the park . Park workers will plant 9 maple trees today . How many maple trees will the park have when the workers are finished ?","target":11.0}
|
||||
{"input":"Mary found 2 seashells and Keith found 5 seashells on the beach . When they cleaned them , they discovered that 9 were cracked . How many seashells did they find together ?","target":7.0}
|
||||
{"input":"Mike picked 8 pears and Jason picked 7 pears from the pear tree . Fred picked 6 apples from the apple tree . How many pears were picked in total ?","target":15.0}
|
||||
{"input":"Tim had 7 quarters and 9 nickels in his bank . His dad gave him 3 nickels and 5 pennies . How many nickels does Tim have now ?","target":12.0}
|
||||
{"input":"A restaurant served 7 slices of pie during lunch and 5 during dinner today . It served 8 of them yesterday . How many slices of pie were served today ?","target":12.0}
|
||||
{"input":"Tom went to 4 hockey games this year , but missed 7 . He went to 9 games last year . How many hockey games did Tom go to in all ?","target":13.0}
|
||||
{"input":"Sam 's dog had puppies and 8 had spots . He gave 2 to his friends . He now has 6 puppies . How many puppies did he have to start with ?","target":8.0}
|
||||
{"input":"Mike picked 7 apples , Nancy picked 3 apples , and Keith picked 6 apples and 4 pears , at the farm . How many apples were picked in total ?","target":16.0}
|
||||
{"input":"There are 7 dogwood trees currently in the park . Park workers will plant 5 dogwood trees today and 4 dogwood trees tomorrow . It took 8 workers to finish the work . How many dogwood trees will the park have when the workers are finished ?","target":16.0}
|
||||
{"input":"Dan 's cat had kittens and 5 had spots . He gave 7 to Tim and 4 to Jason . He now has 5 kittens . How many kittens did he have to start with ?","target":16.0}
|
||||
{"input":"There are 7 crayons in the drawer and 6 crayons on the desk . Sam placed 4 crayons and 8 scissors on the desk . How many crayons are now there in total ?","target":17.0}
|
||||
{"input":"Sally had 8 pennies and 7 nickels in her bank . Her dad gave her 9 nickels and her mother gave her 2 nickels . How many nickels does Sally have now ?","target":18.0}
|
||||
{"input":"Nancy went to 9 football games this month . She went to 8 games last month , and plans to go to 7 games next month . She paid 3 dollars for the tickets . How many games will she attend in all ?","target":24.0}
|
||||
{"input":"Keith found 6 seashells , Jessica found 8 seashells , and Tim found 7 seashells on the beach . When they cleaned them , they discovered that 3 were cracked . How many seashells did they find together ?","target":21.0}
|
||||
{"input":"Nancy grew 2 onions , Dan grew 9 onions , and Mike grew 4 onions . They worked for 6 days on the farm . How many onions did they grow in total ?","target":15.0}
|
||||
{"input":"Fred has 5 yellow balloons , Sam has 6 yellow balloons , and Mary has 7 yellow balloons . The balloons cost 5 dollars . How many yellow balloons do they have in total ?","target":18.0}
|
||||
{"input":"A restaurant served 4 pies during lunch and 9 during dinner today . The restaurant served 7 pies and 2 pizzas yesterday . How many pies were served in total ?","target":20.0}
|
||||
{"input":"Tim found 37 seashells and Sally found 13 seashells on the beach . When they cleaned them , they discovered that 25 were cracked . How many seashells did they find together ?","target":50.0}
|
||||
{"input":"Tom had 27 pennies and 15 dimes in his bank . His dad gave him 33 dimes and 49 nickels . How many dimes does he have now ?","target":48.0}
|
||||
{"input":"There are 39 scissors and 22 pencils in the drawer . Dan placed 13 scissors in the drawer . How many scissors are now there in total ?","target":52.0}
|
||||
{"input":"Sam has 16 blue and 25 green balloons . Alyssa has 21 blue balloons . How many blue balloons do they have in all ?","target":37.0}
|
||||
{"input":"There are 22 orchid bushes and 40 orange trees currently in the park . Park workers will plant 13 orchid bushes today . How many orchid bushes will the park have when the workers are finished ?","target":35.0}
|
||||
{"input":"Mary had 33 Pokemon cards , and 6 were torn . Sam gave Mary 23 new Pokemon cards . How many Pokemon cards does Mary have now ?","target":56.0}
|
||||
{"input":"Mary picked 14 oranges and Jason picked 41 oranges . Keith picked 38 apples . How many oranges were picked in all ?","target":55.0}
|
||||
{"input":"Jason grew 37 watermelons and 30 pumpkins . Sandy grew 11 watermelons . How many watermelons did they grow in total ?","target":48.0}
|
||||
{"input":"Mike went to 15 basketball games this year , but missed 41 . He went to 39 games last year . How many basketball games did Mike go to in total ?","target":54.0}
|
||||
{"input":"Melanie had 10 quarters and 17 pennies in her bank . Her dad gave her 27 pennies and her mother gave her 19 pennies . How many pennies does Melanie have now ?","target":63.0}
|
||||
{"input":"Joan grew 24 pumpkins , Keith grew 42 pumpkins , and Alyssa grew 13 pumpkins . They worked for 34 days on the farm . How many pumpkins did they grow in all ?","target":79.0}
|
||||
{"input":"Mary had 18 baseball cards , and 8 were torn . Fred gave Mary 26 new baseball cards . Mary bought 40 baseball cards . How many baseball cards does Mary have now ?","target":84.0}
|
||||
{"input":"Alyssa went to 11 soccer games this year , but missed 12 . She went to 13 games last year and plans to go to 15 games next year . How many soccer games will Alyssa go to in all ?","target":39.0}
|
||||
{"input":"There are 48 erasers in the drawer and 30 erasers on the desk . Alyssa placed 39 erasers and 45 rulers on the desk . How many erasers are now there in total ?","target":117.0}
|
||||
{"input":"There are 47 orchid bushes currently in the park . Park workers will plant 37 orchid bushes today and 25 orchid bushes tomorrow . It took 35 workers to finish the work . How many orchid bushes will the park have when the workers are finished ?","target":109.0}
|
||||
{"input":"Fred has 10 red balloons , Sam has 46 red balloons , and Dan has 16 red balloons . The balloons cost 10 dollars . How many red balloons do they have in all ?","target":72.0}
|
||||
{"input":"Mary found 45 seashells , Melanie found 23 seashells , and Jason found 32 seashells on the beach . When they cleaned them , they discovered that 31 were cracked . How many seashells did they find together ?","target":100.0}
|
||||
{"input":"Fred picked 36 limes , Alyssa picked 32 limes , and Nancy picked 35 limes and 18 pears , at the farm . How many limes were picked in total ?","target":103.0}
|
||||
{"input":"Benny received 79 dollars and 9 movie tickets for his birthday . He went to a sporting goods store and bought a baseball glove , baseball , and bat . He had 32 dollars over , how much did he spent on the baseball gear ?","target":47.0}
|
||||
{"input":"There are 43 maple trees and 22 orange trees currently in the park . Park workers will plant maple trees today . When the workers are finished there will be 54 maple trees in the park . How many maple trees did the workers plant today ?","target":11.0}
|
||||
{"input":"Dan had 14 peaches and 10 pears at his roadside fruit dish . He went to the orchard and picked peaches to stock up . There are now 85 peaches . how many did he pick ?","target":71.0}
|
||||
{"input":"Melanie has 41 books and 31 magazines in her library . She bought several books at a yard sale over the weekend . She now has 87 books in her library . How many books did she buy at the yard sale ?","target":46.0}
|
||||
{"input":"There were 2 red orchids and 4 white orchids in the vase . Jessica cut some red orchids from her flower garden . There are now 18 red orchids in the vase . How many red orchids did she cut ?","target":16.0}
|
||||
{"input":"Last week Tim had 12 dollars and Keith had 36 dollars . Tim washed cars over the weekend and now has 75 dollars . How much money did Tim make washing cars ?","target":63.0}
|
||||
{"input":"Joan found 75 seashells and 14 starfishes on the beach . She gave Tim some of her seashells . She has 62 seashell . How many seashells did she give to Tim ?","target":13.0}
|
||||
{"input":"There were 32 bales of hay in the barn and 26 bales in the shed . Jason stacked bales in the barn today . There are now 98 bales of hay in the barn . How many bales did he store in the barn ?","target":66.0}
|
||||
{"input":"Jessica is baking a cake . The recipe calls for 8 cups of flour and 2 cups of sugar . She already put in 4 cups of flour . How many cups of flour does she need to add ?","target":4.0}
|
||||
{"input":"Keith picked 3 pears and Jason picked 2 pears from the pear tree . Joan picked 5 apples from the apple tree . How many pears were picked in total ?","target":5.0}
|
||||
{"input":"Sally has 9 orange balloons and 4 blue balloons . She lost 2 of the orange balloons . How many orange balloons does Sally have now ?","target":7.0}
|
||||
{"input":"Sandy grew 8 carrots and 7 turnips . Mary grew 6 carrots . How many carrots did they grow in all ?","target":14.0}
|
||||
{"input":"There are 5 scissors and 3 pencils in the drawer . Jason placed 4 scissors in the drawer . How many scissors are now there in total ?","target":9.0}
|
||||
{"input":"Alyssa 's cat had 8 kittens and 8 had spots . She gave 4 to her friends . How many kittens does she now have ?","target":4.0}
|
||||
{"input":"Sam had 7 pennies and 8 dimes in his bank . His sister borrowed 4 dimes . How many dimes does Sam have now ?","target":4.0}
|
||||
{"input":"A restaurant served 9 hot dogs during lunch and 2 during dinner today . It served 5 of them yesterday . How many hot dogs were served today ?","target":11.0}
|
||||
{"input":"There are 3 short trees and 6 tall trees currently in the park . Park workers will plant 9 short trees today . How many short trees will the park have when the workers are finished ?","target":12.0}
|
||||
{"input":"Mary had 21 dimes and 38 pennies in her bank . Her dad borrowed 18 pennies from Mary . How many pennies does she have now ?","target":20.0}
|
||||
{"input":"Alyssa picked 17 plums and Jason picked 10 plums . Melanie picked 35 pears . How many plums were picked in all ?","target":27.0}
|
||||
{"input":"There are 48 pencils and 40 scissors in the drawer . Joan placed 29 pencils in the drawer . How many pencils are now there in all ?","target":77.0}
|
||||
{"input":"There are 37 short bushes and 30 tall trees currently in the park . Park workers will plant 20 short bushes today . How many short bushes will the park have when the workers are finished ?","target":57.0}
|
||||
Executable
+600
@@ -0,0 +1,600 @@
|
||||
{"input":"For Halloween Debby and her sister combined the candy they received. Debby had 32 pieces of candy while her sister had 42. If they ate 35 pieces the first night, how many pieces do they have left?","target":39}
|
||||
{"input":"A pet store had 13 siamese cats and 5 house cats. During a sale they sold 10 cats. How many cats do they have left?","target":8}
|
||||
{"input":"Luke was trying to expand his game collection. He bought 2 games from a friend and bought 2 more at a garage sale. If 2 of the games didn't work, how many good games did he end up with?","target":2}
|
||||
{"input":"The school cafeteria ordered 42 red apples and 7 green apples for students lunches. But, if only 9 students wanted fruit, how many extra did the cafeteria end up with?","target":40}
|
||||
{"input":"Lana picked 36 tulips and 37 roses to make flower bouquets. If she only used 70 of the flowers though, how many extra flowers did Lana pick?","target":3}
|
||||
{"input":"Carol and her mom were picking carrots from their garden. Carol picked 29 and her mother picked 16. If only 38 of the carrots were good, how many bad carrots did they have?","target":7}
|
||||
{"input":"Roger had 16 dollars. For his birthday he got 28 more dollars but spent 25 on a new game. How much money does he have now?","target":19}
|
||||
{"input":"While on vacation, Megan took 15 pictures at the zoo and 18 at the museum. If she later deleted 31 of the pictures, how many pictures from her vacation did she still have?","target":2}
|
||||
{"input":"Rachel bought two coloring books. One had 23 pictures and the other had 32. After one week she had already colored 44 of the pictures. How many pictures does she still have to color?","target":11}
|
||||
{"input":"Ned had to wash 9 short sleeve shirts and 21 long sleeve shirts before school. If he had only washed 29 of them by the time school started, how many did he not wash?","target":1}
|
||||
{"input":"Oliver had to wash 39 short sleeve shirts and 47 long sleeve shirts before school. If he had only washed 20 of them by the time school started, how many did he not wash?","target":66}
|
||||
{"input":"For the school bake sale Wendy made pastries. She baked 4 cupcakes and 29 cookies. After the sale she had 24 to take back home. How many pastries did she sell?","target":9}
|
||||
{"input":"While on vacation, Debby took 24 pictures at the zoo and 12 at the museum. If she later deleted 14 of the pictures, how many pictures from her vacation did she still have?","target":22}
|
||||
{"input":"Katie picked 3 tulips and 9 roses to make flower bouquets. If she only used 10 of the flowers though, how many extra flowers did Katie pick?","target":2}
|
||||
{"input":"Faye had 46 math problems and 9 science problems for homework. If she finished 40 of the problems at school, how many problems did she have to do for homework?","target":15}
|
||||
{"input":"Amy had 4 music files and 21 video files on her flash drive. If she deleted 23 of the files, how many files were still on her flash drive?","target":2}
|
||||
{"input":"Ned was trying to expand his game collection. He bought 11 games from a friend and bought 22 more at a garage sale. If 19 of the games didn't work, how many good games did he end up with?","target":14}
|
||||
{"input":"Chloe was playing a trivia game. In the first round she scored 40 points and in the second round she scored 50 points. In the last round she lost 4 points. How many points did she have at the end of the game?","target":86}
|
||||
{"input":"At the arcade, Tom won 32 tickets playing 'whack a mole' and 25 tickets playing 'skee ball'. If he spent 7 of his tickets on a hat, how many tickets does Tom have left?","target":50}
|
||||
{"input":"Bianca and her mom were picking carrots from their garden. Bianca picked 26 and her mother picked 15. If only 16 of the carrots were good, how many bad carrots did they have?","target":25}
|
||||
{"input":"For the school bake sale Katie made pastries. She baked 7 cupcakes and 5 cookies. After the sale she had 8 to take back home. How many pastries did she sell?","target":4}
|
||||
{"input":"Zoe bought two coloring books. One had 44 pictures and the other had 44. After one week she had already colored 20 of the pictures. How many pictures does she still have to color?","target":68}
|
||||
{"input":"John was trying to expand his game collection. He bought 21 games from a friend and bought 8 more at a garage sale. If 23 of the games didn't work, how many good games did he end up with?","target":6}
|
||||
{"input":"Henry had 11 dollars. For his birthday he got 18 more dollars but spent 10 on a new game. How much money does he have now?","target":19}
|
||||
{"input":"While on vacation, Gwen took 41 pictures at the zoo and 29 at the museum. If she later deleted 15 of the pictures, how many pictures from her vacation did she still have?","target":55}
|
||||
{"input":"Sam had to wash 40 short sleeve shirts and 23 long sleeve shirts before school. If he had only washed 29 of them by the time school started, how many did he not wash?","target":34}
|
||||
{"input":"A pet store had 12 siamese cats and 20 house cats. During a sale they sold 20 cats. How many cats do they have left?","target":12}
|
||||
{"input":"Faye and her mom were picking carrots from their garden. Faye picked 23 and her mother picked 5. If only 12 of the carrots were good, how many bad carrots did they have?","target":16}
|
||||
{"input":"The school cafeteria ordered 43 red apples and 32 green apples for students lunches. But, if only 2 students wanted fruit, how many extra did the cafeteria end up with?","target":73}
|
||||
{"input":"There were 39 girls and 4 boys trying out for the schools basketball team. If only 26 of them got called back, how many students didn't make the cut?","target":17}
|
||||
{"input":"A pet store had 38 siamese cats and 25 house cats. During a sale they sold 45 cats. How many cats do they have left?","target":18}
|
||||
{"input":"For the school bake sale Paige made pastries. She baked 36 cupcakes and 9 cookies. After the sale she had 4 to take back home. How many pastries did she sell?","target":41}
|
||||
{"input":"Carol was playing a trivia game. In the first round she scored 17 points and in the second round she scored 6 points. In the last round she lost 16 points. How many points did she have at the end of the game?","target":7}
|
||||
{"input":"Gwen bought two coloring books. One had 10 pictures and the other had 39. After one week she had already colored 13 of the pictures. How many pictures does she still have to color?","target":36}
|
||||
{"input":"For Halloween Janet and her sister combined the candy they received. Janet had 34 pieces of candy while her sister had 33. If they ate 4 pieces the first night, how many pieces do they have left?","target":63}
|
||||
{"input":"Haley and her mom were picking carrots from their garden. Haley picked 39 and her mother picked 38. If only 64 of the carrots were good, how many bad carrots did they have?","target":13}
|
||||
{"input":"There were 30 girls and 36 boys trying out for the schools basketball team. If only 10 of them got called back, how many students didn't make the cut?","target":56}
|
||||
{"input":"Roger had 29 dollars. For his birthday he got 20 more dollars but spent 34 on a new game. How much money does he have now?","target":15}
|
||||
{"input":"Dave had to wash 29 short sleeve shirts and 11 long sleeve shirts before school. If he had only washed 35 of them by the time school started, how many did he not wash?","target":5}
|
||||
{"input":"Edward was trying to expand his game collection. He bought 41 games from a friend and bought 14 more at a garage sale. If 31 of the games didn't work, how many good games did he end up with?","target":24}
|
||||
{"input":"At the schools book fair Sam bought 13 adventure books and 17 mystery books. If 15 of the books were used, how many new books did he buy?","target":15}
|
||||
{"input":"Amy had 26 music files and 36 video files on her flash drive. If she deleted 48 of the files, how many files were still on her flash drive?","target":14}
|
||||
{"input":"Oliver had 35 dollars. For his birthday he got 50 more dollars but spent 84 on a new game. How much money does he have now?","target":1}
|
||||
{"input":"Emily was playing a trivia game. In the first round she scored 16 points and in the second round she scored 33 points. In the last round she lost 48 points. How many points did she have at the end of the game?","target":1}
|
||||
{"input":"Paige had 43 math problems and 12 science problems for homework. If she finished 44 of the problems at school, how many problems did she have to do for homework?","target":11}
|
||||
{"input":"The school cafeteria ordered 33 red apples and 23 green apples for students lunches. But, if only 21 students wanted fruit, how many extra did the cafeteria end up with?","target":35}
|
||||
{"input":"Dave had to wash 9 short sleeve shirts and 27 long sleeve shirts before school. If he had only washed 20 of them by the time school started, how many did he not wash?","target":16}
|
||||
{"input":"While on vacation, Rachel took 6 pictures at the zoo and 9 at the museum. If she later deleted 11 of the pictures, how many pictures from her vacation did she still have?","target":4}
|
||||
{"input":"Vanessa and her mom were picking carrots from their garden. Vanessa picked 17 and her mother picked 14. If only 24 of the carrots were good, how many bad carrots did they have?","target":7}
|
||||
{"input":"A pet store had 41 siamese cats and 28 house cats. During a sale they sold 15 cats. How many cats do they have left?","target":54}
|
||||
{"input":"At the schools book fair Sam bought 13 adventure books and 17 mystery books. If 15 of the books were used, how many new books did he buy?","target":15}
|
||||
{"input":"Amy had 26 music files and 36 video files on her flash drive. If she deleted 48 of the files, how many files were still on her flash drive?","target":14}
|
||||
{"input":"Oliver had 35 dollars. For his birthday he got 50 more dollars but spent 84 on a new game. How much money does he have now?","target":1}
|
||||
{"input":"Emily was playing a trivia game. In the first round she scored 16 points and in the second round she scored 33 points. In the last round she lost 48 points. How many points did she have at the end of the game?","target":1}
|
||||
{"input":"Paige had 43 math problems and 12 science problems for homework. If she finished 44 of the problems at school, how many problems did she have to do for homework?","target":11}
|
||||
{"input":"The school cafeteria ordered 33 red apples and 23 green apples for students lunches. But, if only 21 students wanted fruit, how many extra did the cafeteria end up with?","target":35}
|
||||
{"input":"Dave had to wash 9 short sleeve shirts and 27 long sleeve shirts before school. If he had only washed 20 of them by the time school started, how many did he not wash?","target":16}
|
||||
{"input":"While on vacation, Rachel took 6 pictures at the zoo and 9 at the museum. If she later deleted 11 of the pictures, how many pictures from her vacation did she still have?","target":4}
|
||||
{"input":"Vanessa and her mom were picking carrots from their garden. Vanessa picked 17 and her mother picked 14. If only 24 of the carrots were good, how many bad carrots did they have?","target":7}
|
||||
{"input":"A pet store had 41 siamese cats and 28 house cats. During a sale they sold 15 cats. How many cats do they have left?","target":54}
|
||||
{"input":"Janet picked 4 tulips and 11 roses to make flower bouquets. If she only used 11 of the flowers though, how many extra flowers did Janet pick?","target":4}
|
||||
{"input":"Vanessa had 13 music files and 30 video files on her flash drive. If she deleted 10 of the files, how many files were still on her flash drive?","target":33}
|
||||
{"input":"Debby bought two coloring books. One had 16 pictures and the other had 40. After one week she had already colored 33 of the pictures. How many pictures does she still have to color?","target":23}
|
||||
{"input":"The school cafeteria ordered 8 red apples and 43 green apples for students lunches. But, if only 42 students wanted fruit, how many extra did the cafeteria end up with?","target":9}
|
||||
{"input":"Edward started his own lawn mowing business. In the spring he made 2 dollars mowing lawns and in the summer he made 27 dollars. If he had to spend 5 dollars buying supplies, how much money did he end up with?","target":24}
|
||||
{"input":"A pet store had 36 siamese cats and 18 house cats. During a sale they sold 26 cats. How many cats do they have left?","target":28}
|
||||
{"input":"Olivia and her mom were picking carrots from their garden. Olivia picked 20 and her mother picked 14. If only 19 of the carrots were good, how many bad carrots did they have?","target":15}
|
||||
{"input":"George had 30 dollars. For his birthday he got 16 more dollars but spent 38 on a new game. How much money does he have now?","target":8}
|
||||
{"input":"There were 6 girls and 48 boys trying out for the schools basketball team. If only 7 of them got called back, how many students didn't make the cut?","target":47}
|
||||
{"input":"For the school bake sale Amy made pastries. She baked 15 cupcakes and 48 cookies. After the sale she had 12 to take back home. How many pastries did she sell?","target":51}
|
||||
{"input":"At the schools book fair Victor bought 32 adventure books and 37 mystery books. If 16 of the books were used, how many new books did he buy?","target":53}
|
||||
{"input":"While on vacation, Haley took 50 pictures at the zoo and 8 at the museum. If she later deleted 38 of the pictures, how many pictures from her vacation did she still have?","target":20}
|
||||
{"input":"The school cafeteria ordered 37 red apples and 45 green apples for students lunches. But, if only 51 students wanted fruit, how many extra did the cafeteria end up with?","target":31}
|
||||
{"input":"While shopping, Maria bought 35 green towels and 21 white towels. If she gave her mother 34 of them, how many towels did Maria end up with?","target":22}
|
||||
{"input":"There were 17 girls and 32 boys trying out for the schools basketball team. If only 10 of them got called back, how many students didn't make the cut?","target":39}
|
||||
{"input":"Nancy and her mom were picking carrots from their garden. Nancy picked 38 and her mother picked 47. If only 71 of the carrots were good, how many bad carrots did they have?","target":14}
|
||||
{"input":"A waiter at 'The Greasy Spoon' restaurant had 29 customers to wait on. During the lunch rush he added another 20 customers. If 34 of the customers didn't leave him a tip, how many customers did leave a tip?","target":15}
|
||||
{"input":"Tom had to wash 10 short sleeve shirts and 25 long sleeve shirts before school. If he had only washed 5 of them by the time school started, how many did he not wash?","target":30}
|
||||
{"input":"Vanessa had 16 music files and 48 video files on her flash drive. If she deleted 30 of the files, how many files were still on her flash drive?","target":34}
|
||||
{"input":"Kaleb started his own lawn mowing business. In the spring he made 4 dollars mowing lawns and in the summer he made 50 dollars. If he had to spend 4 dollars buying supplies, how much money did he end up with?","target":50}
|
||||
{"input":"The school cafeteria ordered 6 red apples and 15 green apples for students lunches. But, if only 5 students wanted fruit, how many extra did the cafeteria end up with?","target":16}
|
||||
{"input":"Bianca picked 39 tulips and 49 roses to make flower bouquets. If she only used 81 of the flowers though, how many extra flowers did Bianca pick?","target":7}
|
||||
{"input":"While on vacation, Nancy took 49 pictures at the zoo and 8 at the museum. If she later deleted 38 of the pictures, how many pictures from her vacation did she still have?","target":19}
|
||||
{"input":"Gwen had 18 math problems and 11 science problems for homework. If she finished 24 of the problems at school, how many problems did she have to do for homework?","target":5}
|
||||
{"input":"For Halloween Katie and her sister combined the candy they received. Katie had 10 pieces of candy while her sister had 6. If they ate 9 pieces the first night, how many pieces do they have left?","target":7}
|
||||
{"input":"A pet store had 15 siamese cats and 49 house cats. During a sale they sold 19 cats. How many cats do they have left?","target":45}
|
||||
{"input":"There were 9 girls and 14 boys trying out for the schools basketball team. If only 2 of them got called back, how many students didn't make the cut?","target":21}
|
||||
{"input":"Haley had 27 music files and 42 video files on her flash drive. If she deleted 11 of the files, how many files were still on her flash drive?","target":58}
|
||||
{"input":"While shopping, Maria bought 40 green towels and 44 white towels. If she gave her mother 65 of them, how many towels did Maria end up with?","target":19}
|
||||
{"input":"A waiter at 'The Greasy Spoon' restaurant had 39 customers to wait on. During the lunch rush he added another 12 customers. If 49 of the customers didn't leave him a tip, how many customers did leave a tip?","target":2}
|
||||
{"input":"A pet store had 19 siamese cats and 45 house cats. During a sale they sold 56 cats. How many cats do they have left?","target":8}
|
||||
{"input":"A waiter at 'The Greasy Spoon' restaurant had 26 customers to wait on. During the lunch rush he added another 27 customers. If 27 of the customers didn't leave him a tip, how many customers did leave a tip?","target":26}
|
||||
{"input":"Rachel bought two coloring books. One had 24 pictures and the other had 39. After one week she had already colored 4 of the pictures. How many pictures does she still have to color?","target":59}
|
||||
{"input":"While shopping, Emily bought 5 green towels and 30 white towels. If she gave her mother 26 of them, how many towels did Emily end up with?","target":9}
|
||||
{"input":"For Halloween Katie and her sister combined the candy they received. Katie had 8 pieces of candy while her sister had 23. If they ate 8 pieces the first night, how many pieces do they have left?","target":23}
|
||||
{"input":"Ned was trying to expand his game collection. He bought 50 games from a friend and bought 27 more at a garage sale. If 74 of the games didn't work, how many good games did he end up with?","target":3}
|
||||
{"input":"For the school bake sale Wendy made pastries. She baked 41 cupcakes and 31 cookies. After the sale she had 32 to take back home. How many pastries did she sell?","target":40}
|
||||
{"input":"The school cafeteria ordered 25 red apples and 17 green apples for students lunches. But, if only 10 students wanted fruit, how many extra did the cafeteria end up with?","target":32}
|
||||
{"input":"At the arcade, Jerry won 29 tickets playing 'whack a mole' and 17 tickets playing 'skee ball'. If he spent 12 of his tickets on a hat, how many tickets does Jerry have left?","target":34}
|
||||
{"input":"Cody had 45 dollars. For his birthday he got 9 more dollars but spent 19 on a new game. How much money does he have now?","target":35}
|
||||
{"input":"For the school bake sale Carol made 30 cupcakes. If she sold 9 of them and then made 28 more, how many cupcakes would she have?","target":49}
|
||||
{"input":"For Halloween Faye scored 47 pieces of candy. She ate 25 pieces the first night and then her sister gave her 40 more pieces. How many pieces of candy does Faye have now?","target":62}
|
||||
{"input":"Kaleb had 34 books. If he sold 17 of them and used the money he earned to buy 7 new books, how many books would Kaleb have?","target":24}
|
||||
{"input":"In fourth grade there were 10 students at the start of the year. During the year 4 students left and 42 new students came to school. How many students were in fourth grade at the end?","target":48}
|
||||
{"input":"Oliver had 33 dollars in January. By March he had spent 4 dollars. If he got another 32 dollars from his mom, how much money would he have?","target":61}
|
||||
{"input":"A florist had 37 roses. If she sold 16 of them and then later picked 19 more, how many roses would she have?","target":40}
|
||||
{"input":"A teacher had 6 worksheets to grade. If she graded 4, but then another 18 were turned in, how many worksheets would she have to grade?","target":20}
|
||||
{"input":"A book store had 41 books in the bargin bin. If they sold 33 books, but then put 2 more in the bin, how many books would be in the bin?","target":10}
|
||||
{"input":"A waiter had 19 customers to wait on. If 14 customers left and he got another 36 customers, how many customers would he have?","target":41}
|
||||
{"input":"At the fair there were 9 people in line for the bumper cars. If 6 of them got tired of waiting and left and 3 more got in line, how many people would be in line?","target":6}
|
||||
{"input":"A teacher had 7 worksheets to grade. If she graded 2, but then another 46 were turned in, how many worksheets would she have to grade?","target":51}
|
||||
{"input":"Paige had 8 songs on her mp3 player. If she deleted 5 old songs from it and then added 30 new songs, how many songs does she have on her mp3 player?","target":33}
|
||||
{"input":"Maria picked 48 carrots from her garden. If she threw out 11 of them and then picked 15 more the next day, how many carrots would she have total?","target":52}
|
||||
{"input":"A store had 34 oranges in a bin. If they threw away 20 of the old ones and put 13 new ones in the bin how many would be in the bin?","target":27}
|
||||
{"input":"A waiter had 47 customers to wait on. If 41 customers left and he got another 20 customers, how many customers would he have?","target":26}
|
||||
{"input":"In fourth grade there were 33 students at the start of the year. During the year 18 students left and 14 new students came to school. How many students were in fourth grade at the end?","target":29}
|
||||
{"input":"For the school bake sale Katie made 26 cupcakes. If she sold 20 of them and then made 20 more, how many cupcakes would she have?","target":26}
|
||||
{"input":"Adam had 5 dollars. At the store he spent $2 on a new game. If he got another 5 dollars for his allowance, how much money does he have now?","target":8}
|
||||
{"input":"Tiffany was playing a video game and had 43 lives. In a hard part of the game she lost 14 lives. If she got 27 more lives in the next level, how many lives would she have?","target":56}
|
||||
{"input":"At the fair there were 12 people in line for the bumper cars. If 10 of them got tired of waiting and left and 15 more got in line, how many people would be in line?","target":17}
|
||||
{"input":"A waiter had 14 customers to wait on. If 3 customers left and he got another 39 customers, how many customers would he have?","target":50}
|
||||
{"input":"At the fair there were 30 people in line for the bumper cars. If 10 of them got tired of waiting and left and 5 more got in line, how many people would be in line?","target":25}
|
||||
{"input":"Faye had 34 coloring books. If she gave away 3 of them, but then bought 48 more, how many would she have total?","target":79}
|
||||
{"input":"Paige had 11 songs on her mp3 player. If she deleted 9 old songs from it and then added 8 new songs, how many songs does she have on her mp3 player?","target":10}
|
||||
{"input":"A teacher had 38 worksheets to grade. If she graded 4, but then another 15 were turned in, how many worksheets would she have to grade?","target":49}
|
||||
{"input":"In fourth grade there were 40 students at the start of the year. During the year 14 students left and 26 new students came to school. How many students were in fourth grade at the end?","target":52}
|
||||
{"input":"Wendy was playing a video game and had 43 lives. In a hard part of the game she lost 8 lives. If she got 39 more lives in the next level, how many lives would she have?","target":74}
|
||||
{"input":"A book store had 4 books in the bargin bin. If they sold 3 books, but then put 10 more in the bin, how many books would be in the bin?","target":11}
|
||||
{"input":"At the arcade Cody won 49 tickets. If he spent 25 tickets on a beanie and later won 6 more tickets, how many would he have?","target":30}
|
||||
{"input":"For the school bake sale Bianca made 14 cupcakes. If she sold 6 of them and then made 17 more, how many cupcakes would she have?","target":25}
|
||||
{"input":"Bianca had 45 coloring books. If she gave away 6 of them, but then bought 20 more, how many would she have total?","target":59}
|
||||
{"input":"Zoe had 15 songs on her mp3 player. If she deleted 8 old songs from it and then added 50 new songs, how many songs does she have on her mp3 player?","target":57}
|
||||
{"input":"At the arcade Jerry won 4 tickets. If he spent 2 tickets on a beanie and later won 47 more tickets, how many would he have?","target":49}
|
||||
{"input":"A store had 31 oranges in a bin. If they threw away 9 of the old ones and put 38 new ones in the bin how many would be in the bin?","target":60}
|
||||
{"input":"Adam had 33 books. If he sold 11 of them and used the money he earned to buy 23 new books, how many books would Adam have?","target":45}
|
||||
{"input":"At the fair there were 10 people in line for the bumper cars. If 2 of them got tired of waiting and left and 2 more got in line, how many people would be in line?","target":10}
|
||||
{"input":"The school cafeteria had 38 apples. If they used 20 to make lunch for the students and then bought 28 more, how many apples would they have?","target":46}
|
||||
{"input":"George had 28 socks. If he threw away 4 old ones that didn't fit and bought 36 new ones, how many socks would he have?","target":60}
|
||||
{"input":"For Halloween Robin scored 23 pieces of candy. She ate 7 pieces the first night and then her sister gave her 21 more pieces. How many pieces of candy does Robin have now?","target":37}
|
||||
{"input":"Rachel was playing a video game and had 10 lives. In a hard part of the game she lost 4 lives. If she got 26 more lives in the next level, how many lives would she have?","target":32}
|
||||
{"input":"A florist had 6 roses. If she sold 5 of them and then later picked 12 more, how many roses would she have?","target":13}
|
||||
{"input":"For Halloween Haley scored 33 pieces of candy. She ate 17 pieces the first night and then her sister gave her 19 more pieces. How many pieces of candy does Haley have now?","target":35}
|
||||
{"input":"The school cafeteria had 14 apples. If they used 13 to make lunch for the students and then bought 49 more, how many apples would they have?","target":50}
|
||||
{"input":"Edward had 43 books. If he sold 19 of them and used the money he earned to buy 14 new books, how many books would Edward have?","target":38}
|
||||
{"input":"A teacher had 29 worksheets to grade. If she graded 25, but then another 29 were turned in, how many worksheets would she have to grade?","target":33}
|
||||
{"input":"In fourth grade there were 35 students at the start of the year. During the year 10 students left and 10 new students came to school. How many students were in fourth grade at the end?","target":35}
|
||||
{"input":"For the school bake sale Carol made 19 cupcakes. If she sold 6 of them and then made 27 more, how many cupcakes would she have?","target":40}
|
||||
{"input":"A store had 40 oranges in a bin. If they threw away 37 of the old ones and put 7 new ones in the bin how many would be in the bin?","target":10}
|
||||
{"input":"At the arcade Victor won 46 tickets. If he spent 27 tickets on a beanie and later won 39 more tickets, how many would he have?","target":58}
|
||||
{"input":"Chloe picked 48 carrots from her garden. If she threw out 45 of them and then picked 42 more the next day, how many carrots would she have total?","target":45}
|
||||
{"input":"Sam had 10 socks. If he threw away 3 old ones that didn't fit and bought 36 new ones, how many socks would he have?","target":43}
|
||||
{"input":"Maria had 14 bottles of water in her fridge. If she drank 8 of them and then bought 45 more, how many bottles would she have?","target":51}
|
||||
{"input":"A teacher had 34 worksheets to grade. If she graded 7, but then another 36 were turned in, how many worksheets would she have to grade?","target":63}
|
||||
{"input":"At the fair there were 7 people in line for the bumper cars. If 4 of them got tired of waiting and left and 8 more got in line, how many people would be in line?","target":11}
|
||||
{"input":"Emily had 7 coloring books. If she gave away 2 of them, but then bought 14 more, how many would she have total?","target":19}
|
||||
{"input":"At the arcade Dave won 25 tickets. If he spent 22 tickets on a beanie and later won 15 more tickets, how many would he have?","target":18}
|
||||
{"input":"Robin had 30 songs on her mp3 player. If she deleted 8 old songs from it and then added 10 new songs, how many songs does she have on her mp3 player?","target":32}
|
||||
{"input":"The school cafeteria had 23 apples. If they used 20 to make lunch for the students and then bought 6 more, how many apples would they have?","target":9}
|
||||
{"input":"Janet was playing a video game and had 38 lives. In a hard part of the game she lost 16 lives. If she got 32 more lives in the next level, how many lives would she have?","target":54}
|
||||
{"input":"Megan picked 19 carrots from her garden. If she threw out 4 of them and then picked 46 more the next day, how many carrots would she have total?","target":61}
|
||||
{"input":"Roger had 45 dollars in January. By March he had spent 20 dollars. If he got another 46 dollars from his mom, how much money would he have?","target":71}
|
||||
{"input":"The school cafeteria had 17 apples. If they used 2 to make lunch for the students and then bought 23 more, how many apples would they have?","target":38}
|
||||
{"input":"Janet was playing a video game and had 47 lives. In a hard part of the game she lost 23 lives. If she got 46 more lives in the next level, how many lives would she have?","target":70}
|
||||
{"input":"Nancy picked 12 carrots from her garden. If she threw out 2 of them and then picked 21 more the next day, how many carrots would she have total?","target":31}
|
||||
{"input":"Bianca had 34 songs on her mp3 player. If she deleted 14 old songs from it and then added 44 new songs, how many songs does she have on her mp3 player?","target":64}
|
||||
{"input":"Tom had 5 books. If he sold 4 of them and used the money he earned to buy 38 new books, how many books would Tom have?","target":39}
|
||||
{"input":"John had 33 socks. If he threw away 19 old ones that didn't fit and bought 13 new ones, how many socks would he have?","target":27}
|
||||
{"input":"For the school bake sale Maria made 19 cupcakes. If she sold 5 of them and then made 10 more, how many cupcakes would she have?","target":24}
|
||||
{"input":"A store had 50 oranges in a bin. If they threw away 40 of the old ones and put 24 new ones in the bin how many would be in the bin?","target":34}
|
||||
{"input":"In fourth grade there were 8 students at the start of the year. During the year 5 students left and 8 new students came to school. How many students were in fourth grade at the end?","target":11}
|
||||
{"input":"Bianca picked 23 carrots from her garden. If she threw out 10 of them and then picked 47 more the next day, how many carrots would she have total?","target":60}
|
||||
{"input":"Zoe had 42 bottles of water in her fridge. If she drank 25 of them and then bought 30 more, how many bottles would she have?","target":47}
|
||||
{"input":"Katie had 11 songs on her mp3 player. If she deleted 7 old songs from it and then added 24 new songs, how many songs does she have on her mp3 player?","target":28}
|
||||
{"input":"A store had 5 oranges in a bin. If they threw away 2 of the old ones and put 28 new ones in the bin how many would be in the bin?","target":31}
|
||||
{"input":"Adam had 48 books. If he sold 19 of them and used the money he earned to buy 38 new books, how many books would Adam have?","target":67}
|
||||
{"input":"In fourth grade there were 31 students at the start of the year. During the year 5 students left and 11 new students came to school. How many students were in fourth grade at the end?","target":37}
|
||||
{"input":"For the school bake sale Robin made 42 cupcakes. If she sold 22 of them and then made 39 more, how many cupcakes would she have?","target":59}
|
||||
{"input":"A florist had 11 roses. If she sold 2 of them and then later picked 32 more, how many roses would she have?","target":41}
|
||||
{"input":"Frank had 11 dollars. At the store he spent $3 on a new game. If he got another 14 dollars for his allowance, how much money does he have now?","target":22}
|
||||
{"input":"Haley was playing a video game and had 14 lives. In a hard part of the game she lost 4 lives. If she got 36 more lives in the next level, how many lives would she have?","target":46}
|
||||
{"input":"Luke had 48 dollars in January. By March he had spent 11 dollars. If he got another 21 dollars from his mom, how much money would he have?","target":58}
|
||||
{"input":"Emily was playing a video game and had 42 lives. In a hard part of the game she lost 25 lives. If she got 24 more lives in the next level, how many lives would she have?","target":41}
|
||||
{"input":"Debby had 30 coloring books. If she gave away 7 of them, but then bought 35 more, how many would she have total?","target":58}
|
||||
{"input":"The school cafeteria had 12 apples. If they used 8 to make lunch for the students and then bought 19 more, how many apples would they have?","target":23}
|
||||
{"input":"At the fair there were 31 people in line for the bumper cars. If 25 of them got tired of waiting and left and 25 more got in line, how many people would be in line?","target":31}
|
||||
{"input":"A florist had 50 roses. If she sold 15 of them and then later picked 21 more, how many roses would she have?","target":56}
|
||||
{"input":"Zoe had 6 songs on her mp3 player. If she deleted 3 old songs from it and then added 20 new songs, how many songs does she have on her mp3 player?","target":23}
|
||||
{"input":"At the arcade Dave won 11 tickets. If he spent 5 tickets on a beanie and later won 10 more tickets, how many would he have?","target":16}
|
||||
{"input":"A waiter had 33 customers to wait on. If 31 customers left and he got another 26 customers, how many customers would he have?","target":28}
|
||||
{"input":"In fourth grade there were 11 students at the start of the year. During the year 6 students left and 42 new students came to school. How many students were in fourth grade at the end?","target":47}
|
||||
{"input":"Oliver had 11 socks. If he threw away 4 old ones that didn't fit and bought 26 new ones, how many socks would he have?","target":33}
|
||||
{"input":"A store had 40 oranges in a bin. If they threw away 25 of the old ones and put 21 new ones in the bin how many would be in the bin?","target":36}
|
||||
{"input":"At the arcade Edward won 9 tickets. If he spent 4 tickets on a beanie and later won 4 more tickets, how many would he have?","target":9}
|
||||
{"input":"In fourth grade there were 4 students at the start of the year. During the year 3 students left and 42 new students came to school. How many students were in fourth grade at the end?","target":43}
|
||||
{"input":"Haley picked 28 carrots from her garden. If she threw out 11 of them and then picked 9 more the next day, how many carrots would she have total?","target":26}
|
||||
{"input":"Roger had 25 books. If he sold 21 of them and used the money he earned to buy 30 new books, how many books would Roger have?","target":34}
|
||||
{"input":"A florist had 5 roses. If she sold 3 of them and then later picked 34 more, how many roses would she have?","target":36}
|
||||
{"input":"Wendy was playing a video game and had 10 lives. In a hard part of the game she lost 6 lives. If she got 37 more lives in the next level, how many lives would she have?","target":41}
|
||||
{"input":"John had 5 dollars. At the store he spent $2 on a new game. If he got another 26 dollars for his allowance, how much money does he have now?","target":29}
|
||||
{"input":"For the school bake sale Chloe made 28 cupcakes. If she sold 25 of them and then made 8 more, how many cupcakes would she have?","target":11}
|
||||
{"input":"At the town carnival Billy rode the ferris wheel 7 times and the bumper cars 3 times. If each ride cost 5 tickets, how many tickets did he use?","target":50}
|
||||
{"input":"Chloe was unboxing some of her old winter clothes. She found 4 boxes of clothing and inside each box there were 2 scarves and 6 mittens. How many pieces of winter clothing did Chloe have total?","target":32}
|
||||
{"input":"A waiter had 9 tables he was waiting on, with 7 women and 3 men at each table. How many customers total did the waiter have?","target":90}
|
||||
{"input":"April's discount flowers was having a sale where each flower was 3 dollars. If Emily bought 2 roses and 2 daisies, how much did she spend?","target":12}
|
||||
{"input":"Isabel had 2 pages of math homework and 4 pages of reading homework. If each page had 5 problems on it, how many problems did she have to complete total?","target":30}
|
||||
{"input":"Wendy was playing a video game where she scores 5 points for each treasure she finds. If she found 4 treasures on the first level and 3 on the second, what would her score be?","target":35}
|
||||
{"input":"There were 7 friends playing a video game online when 2 more players joined the game. If each player had 7 lives, how many lives did they have total?","target":63}
|
||||
{"input":"Paul bought 6 boxes of chocolate candy and 4 boxes of caramel candy. If each box has 9 pieces inside it, how much candy did he have total?","target":90}
|
||||
{"input":"A pet store has 6 bird cages. If each cage has 6 parrots and 2 parakeets in it, how many birds does the pet store have total?","target":48}
|
||||
{"input":"Rachel was organizing her book case making sure each of the shelves had exactly 9 books on it. If she had 6 shelves of mystery books and 2 shelves of picture books, how many books did she have total?","target":72}
|
||||
{"input":"Rachel was unboxing some of her old winter clothes. She found 7 boxes of clothing and inside each box there were 3 scarves and 4 mittens. How many pieces of winter clothing did Rachel have total?","target":49}
|
||||
{"input":"A pet store has 9 bird cages. If each cage has 2 parrots and 2 parakeets in it, how many birds does the pet store have total?","target":36}
|
||||
{"input":"Cody bought 7 boxes of chocolate candy and 3 boxes of caramel candy. If each box has 8 pieces inside it, how much candy did he have total?","target":80}
|
||||
{"input":"At Billy's Restaurant a group with 2 adults and 5 children came in to eat. If each meal cost 3 dollars, how much was the bill?","target":21}
|
||||
{"input":"Paul was collecting cans for recycling. On Saturday he filled 6 bags up and on Sunday he filled 3 more bags. If each bag had 8 cans in it, how many cans did he pick up total?","target":72}
|
||||
{"input":"While playing a trivia game, Adam answered 5 questions correct in the first half and 5 questions correct in the second half. If each question was worth 5 points, what was his final score?","target":50}
|
||||
{"input":"Haley's favorite band was holding a concert where tickets were 4 dollars each. Haley bought 3 tickets for herself and her friends and 5 extra tickets in case anyone else wanted to go. How much did she spend?","target":32}
|
||||
{"input":"Luke was putting his spare change into piles. He had 5 piles of quarters and 5 piles of dimes. If each pile had 3 coins in it, how many coins did he have total?","target":30}
|
||||
{"input":"Victor and his friend were buying trick decks from the magic shop for 8 dollars each. How much did they spend if Victor bought 6 decks and his friend bought 2 decks?","target":64}
|
||||
{"input":"Katie had 7 pages of math homework and 3 pages of reading homework. If each page had 9 problems on it, how many problems did she have to complete total?","target":90}
|
||||
{"input":"Faye was playing a video game where she scores 7 points for each treasure she finds. If she found 2 treasures on the first level and 6 on the second, what would her score be?","target":56}
|
||||
{"input":"Gwen was organizing her book case making sure each of the shelves had exactly 9 books on it. If she had 3 shelves of mystery books and 5 shelves of picture books, how many books did she have total?","target":72}
|
||||
{"input":"There were 2 friends playing a video game online when 2 more players joined the game. If each player had 6 lives, how many lives did they have total?","target":24}
|
||||
{"input":"Wendy bought 4 new chairs and 4 new tables for her house. If she spent 6 minutes on each piece furniture putting it together, how many minutes did it take her to finish?","target":48}
|
||||
{"input":"April's discount flowers was having a sale where each flower was 3 dollars. If Zoe bought 8 roses and 2 daisies, how much did she spend?","target":30}
|
||||
{"input":"Paige was unboxing some of her old winter clothes. She found 6 boxes of clothing and inside each box there were 5 scarves and 5 mittens. How many pieces of winter clothing did Paige have total?","target":60}
|
||||
{"input":"Sam was collecting cans for recycling. On Saturday he filled 4 bags up and on Sunday he filled 3 more bags. If each bag had 6 cans in it, how many cans did he pick up total?","target":42}
|
||||
{"input":"George was working as a sacker at a grocery store where he made 5 dollars an hour. On Monday he worked 7 hours and on Tuesday he worked 2 hours. How much money did George make in those two days?","target":45}
|
||||
{"input":"While playing a trivia game, Frank answered 3 questions correct in the first half and 2 questions correct in the second half. If each question was worth 3 points, what was his final score?","target":15}
|
||||
{"input":"Edward and his friend were buying trick decks from the magic shop for 9 dollars each. How much did they spend if Edward bought 4 decks and his friend bought 4 decks?","target":72}
|
||||
{"input":"Tom and his friend were buying trick decks from the magic shop for 8 dollars each. How much did they spend if Tom bought 3 decks and his friend bought 5 decks?","target":64}
|
||||
{"input":"While shopping for music online, Megan bought 2 country albums and 8 pop albums. Each album came with a lyric sheet and had 7 songs. How many songs did Megan buy total?","target":70}
|
||||
{"input":"Victor was working as a sacker at a grocery store where he made 6 dollars an hour. On Monday he worked 5 hours and on Tuesday he worked 5 hours. How much money did Victor make in those two days?","target":60}
|
||||
{"input":"A pet store has 6 bird cages. If each cage has 2 parrots and 7 parakeets in it, how many birds does the pet store have total?","target":54}
|
||||
{"input":"Maria bought 2 new chairs and 2 new tables for her house. If she spent 8 minutes on each piece furniture putting it together, how many minutes did it take her to finish?","target":32}
|
||||
{"input":"Olivia was playing a video game where she scores 8 points for each treasure she finds. If she found 2 treasures on the first level and 2 on the second, what would her score be?","target":32}
|
||||
{"input":"A waiter had 6 tables he was waiting on, with 3 women and 5 men at each table. How many customers total did the waiter have?","target":48}
|
||||
{"input":"Faye was selling her necklaces at a garage sale. She sold 3 bead necklaces and 7 gem stone necklaces. If each necklace cost 7 dollars, how much money did she earn?","target":70}
|
||||
{"input":"At the town carnival Oliver rode the ferris wheel 5 times and the bumper cars 4 times. If each ride cost 7 tickets, how many tickets did he use?","target":63}
|
||||
{"input":"There were 5 friends playing a video game online when 2 more players joined the game. If each player had 8 lives, how many lives did they have total?","target":56}
|
||||
{"input":"Katie was selling her necklaces at a garage sale. She sold 4 bead necklaces and 3 gem stone necklaces. If each necklace cost 3 dollars, how much money did she earn?","target":21}
|
||||
{"input":"Edward was working as a sacker at a grocery store where he made 6 dollars an hour. On Monday he worked 3 hours and on Tuesday he worked 5 hours. How much money did Edward make in those two days?","target":48}
|
||||
{"input":"While playing a trivia game, Mike answered 3 questions correct in the first half and 5 questions correct in the second half. If each question was worth 3 points, what was his final score?","target":24}
|
||||
{"input":"Amy was playing a video game where she scores 4 points for each treasure she finds. If she found 6 treasures on the first level and 2 on the second, what would her score be?","target":32}
|
||||
{"input":"A waiter had 9 tables he was waiting on, with 4 women and 3 men at each table. How many customers total did the waiter have?","target":63}
|
||||
{"input":"Chloe was organizing her book case making sure each of the shelves had exactly 6 books on it. If she had 5 shelves of mystery books and 4 shelves of picture books, how many books did she have total?","target":54}
|
||||
{"input":"While shopping for music online, Isabel bought 6 country albums and 2 pop albums. Each album came with a lyric sheet and had 9 songs. How many songs did Isabel buy total?","target":72}
|
||||
{"input":"At Oliver's Restaurant a group with 2 adults and 4 children came in to eat. If each meal cost 3 dollars, how much was the bill?","target":18}
|
||||
{"input":"Tiffany had 6 pages of math homework and 4 pages of reading homework. If each page had 3 problems on it, how many problems did she have to complete total?","target":30}
|
||||
{"input":"April's discount flowers was having a sale where each flower was 3 dollars. If Robin bought 5 roses and 4 daisies, how much did she spend?","target":27}
|
||||
{"input":"Zoe was unboxing some of her old winter clothes. She found 8 boxes of clothing and inside each box there were 4 scarves and 6 mittens. How many pieces of winter clothing did Zoe have total?","target":80}
|
||||
{"input":"There were 2 friends playing a video game online when 2 more players joined the game. If each player had 3 lives, how many lives did they have total?","target":12}
|
||||
{"input":"Adam bought 2 boxes of chocolate candy and 5 boxes of caramel candy. If each box has 4 pieces inside it, how much candy did he have total?","target":28}
|
||||
{"input":"A pet store has 8 bird cages. If each cage has 2 parrots and 7 parakeets in it, how many birds does the pet store have total?","target":72}
|
||||
{"input":"Lana's favorite band was holding a concert where tickets were 6 dollars each. Lana bought 8 tickets for herself and her friends and 2 extra tickets in case anyone else wanted to go. How much did she spend?","target":60}
|
||||
{"input":"At Tom's Restaurant a group with 2 adults and 5 children came in to eat. If each meal cost 8 dollars, how much was the bill?","target":56}
|
||||
{"input":"Gwen was organizing her book case making sure each of the shelves had exactly 4 books on it. If she had 5 shelves of mystery books and 3 shelves of picture books, how many books did she have total?","target":32}
|
||||
{"input":"While shopping for music online, Isabel bought 4 country albums and 5 pop albums. Each album came with a lyric sheet and had 8 songs. How many songs did Isabel buy total?","target":72}
|
||||
{"input":"Megan was selling her necklaces at a garage sale. She sold 7 bead necklaces and 3 gem stone necklaces. If each necklace cost 9 dollars, how much money did she earn?","target":90}
|
||||
{"input":"Rachel was playing a video game where she scores 9 points for each treasure she finds. If she found 5 treasures on the first level and 2 on the second, what would her score be?","target":63}
|
||||
{"input":"April's discount flowers was having a sale where each flower was 6 dollars. If Katie bought 5 roses and 5 daisies, how much did she spend?","target":60}
|
||||
{"input":"A pet store has 9 bird cages. If each cage has 2 parrots and 6 parakeets in it, how many birds does the pet store have total?","target":72}
|
||||
{"input":"Roger was putting his spare change into piles. He had 3 piles of quarters and 3 piles of dimes. If each pile had 7 coins in it, how many coins did he have total?","target":42}
|
||||
{"input":"There were 2 friends playing a video game online when 2 more players joined the game. If each player had 8 lives, how many lives did they have total?","target":32}
|
||||
{"input":"A waiter had 7 tables he was waiting on, with 7 women and 2 men at each table. How many customers total did the waiter have?","target":63}
|
||||
{"input":"Tiffany was playing a video game where she scores 6 points for each treasure she finds. If she found 3 treasures on the first level and 5 on the second, what would her score be?","target":48}
|
||||
{"input":"While shopping for music online, Zoe bought 3 country albums and 5 pop albums. Each album came with a lyric sheet and had 3 songs. How many songs did Zoe buy total?","target":24}
|
||||
{"input":"Megan was organizing her book case making sure each of the shelves had exactly 7 books on it. If she had 8 shelves of mystery books and 2 shelves of picture books, how many books did she have total?","target":70}
|
||||
{"input":"Isabel was selling her necklaces at a garage sale. She sold 3 bead necklaces and 3 gem stone necklaces. If each necklace cost 6 dollars, how much money did she earn?","target":36}
|
||||
{"input":"Frank was collecting cans for recycling. On Saturday he filled 5 bags up and on Sunday he filled 3 more bags. If each bag had 5 cans in it, how many cans did he pick up total?","target":40}
|
||||
{"input":"April's discount flowers was having a sale where each flower was 8 dollars. If Vanessa bought 3 roses and 3 daisies, how much did she spend?","target":48}
|
||||
{"input":"Will was working as a sacker at a grocery store where he made 8 dollars an hour. On Monday he worked 8 hours and on Tuesday he worked 2 hours. How much money did Will make in those two days?","target":80}
|
||||
{"input":"Rachel bought 7 new chairs and 3 new tables for her house. If she spent 4 minutes on each piece furniture putting it together, how many minutes did it take her to finish?","target":40}
|
||||
{"input":"While playing a trivia game, Adam answered 8 questions correct in the first half and 2 questions correct in the second half. If each question was worth 8 points, what was his final score?","target":80}
|
||||
{"input":"A pet store has 4 bird cages. If each cage has 8 parrots and 2 parakeets in it, how many birds does the pet store have total?","target":40}
|
||||
{"input":"Kaleb was collecting cans for recycling. On Saturday he filled 5 bags up and on Sunday he filled 5 more bags. If each bag had 4 cans in it, how many cans did he pick up total?","target":40}
|
||||
{"input":"Bianca was organizing her book case making sure each of the shelves had exactly 8 books on it. If she had 5 shelves of mystery books and 4 shelves of picture books, how many books did she have total?","target":72}
|
||||
{"input":"Billy was putting his spare change into piles. He had 2 piles of quarters and 3 piles of dimes. If each pile had 4 coins in it, how many coins did he have total?","target":20}
|
||||
{"input":"Chloe was playing a video game where she scores 9 points for each treasure she finds. If she found 6 treasures on the first level and 3 on the second, what would her score be?","target":81}
|
||||
{"input":"While shopping for music online, Nancy bought 3 country albums and 5 pop albums. Each album came with a lyric sheet and had 8 songs. How many songs did Nancy buy total?","target":64}
|
||||
{"input":"Lana was unboxing some of her old winter clothes. She found 3 boxes of clothing and inside each box there were 3 scarves and 4 mittens. How many pieces of winter clothing did Lana have total?","target":21}
|
||||
{"input":"Frank and his friend were buying trick decks from the magic shop for 7 dollars each. How much did they spend if Frank bought 3 decks and his friend bought 2 decks?","target":35}
|
||||
{"input":"A waiter had 5 tables he was waiting on, with 5 women and 3 men at each table. How many customers total did the waiter have?","target":40}
|
||||
{"input":"There were 8 friends playing a video game online when 2 more players joined the game. If each player had 6 lives, how many lives did they have total?","target":60}
|
||||
{"input":"Sam was collecting cans for recycling. On Saturday he filled 3 bags up and on Sunday he filled 4 more bags. If each bag had 9 cans in it, how many cans did he pick up total?","target":63}
|
||||
{"input":"April's discount flowers was having a sale where each flower was 6 dollars. If Maria bought 7 roses and 3 daisies, how much did she spend?","target":60}
|
||||
{"input":"Dave was working as a sacker at a grocery store where he made 6 dollars an hour. On Monday he worked 6 hours and on Tuesday he worked 2 hours. How much money did Dave make in those two days?","target":48}
|
||||
{"input":"Nancy was organizing her book case making sure each of the shelves had exactly 6 books on it. If she had 2 shelves of mystery books and 6 shelves of picture books, how many books did she have total?","target":48}
|
||||
{"input":"While shopping for music online, Faye bought 2 country albums and 3 pop albums. Each album came with a lyric sheet and had 6 songs. How many songs did Faye buy total?","target":30}
|
||||
{"input":"Tom was putting his spare change into piles. He had 2 piles of quarters and 4 piles of dimes. If each pile had 5 coins in it, how many coins did he have total?","target":30}
|
||||
{"input":"While playing a trivia game, George answered 6 questions correct in the first half and 4 questions correct in the second half. If each question was worth 3 points, what was his final score?","target":30}
|
||||
{"input":"Lana was unboxing some of her old winter clothes. She found 4 boxes of clothing and inside each box there were 2 scarves and 6 mittens. How many pieces of winter clothing did Lana have total?","target":32}
|
||||
{"input":"At the town carnival Oliver rode the ferris wheel 7 times and the bumper cars 3 times. If each ride cost 3 tickets, how many tickets did he use?","target":30}
|
||||
{"input":"Sarah had 4 pages of math homework and 6 pages of reading homework. If each page had 4 problems on it, how many problems did she have to complete total?","target":40}
|
||||
{"input":"Kaleb was collecting cans for recycling. On Saturday he filled 4 bags up and on Sunday he filled 6 more bags. If each bag had 4 cans in it, how many cans did he pick up total?","target":40}
|
||||
{"input":"Edward and his friend were buying trick decks from the magic shop for 6 dollars each. How much did they spend if Edward bought 3 decks and his friend bought 6 decks?","target":54}
|
||||
{"input":"Emily bought 4 new chairs and 2 new tables for her house. If she spent 8 minutes on each piece furniture putting it together, how many minutes did it take her to finish?","target":48}
|
||||
{"input":"A waiter had 9 tables he was waiting on, with 2 women and 6 men at each table. How many customers total did the waiter have?","target":72}
|
||||
{"input":"There were 4 friends playing a video game online when 5 more players joined the game. If each player had 3 lives, how many lives did they have total?","target":27}
|
||||
{"input":"While shopping for music online, Janet bought 6 country albums and 2 pop albums. Each album came with a lyric sheet and had 9 songs. How many songs did Janet buy total?","target":72}
|
||||
{"input":"The schools debate team had 26 boys and 46 girls on it. If they were split into groups of 9 how many groups could they make?","target":8}
|
||||
{"input":"Robin uploaded 35 pictures from her phone and 5 from her camera to facebook. If she sorted the pics into 5 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":8}
|
||||
{"input":"Kaleb had saved up 21 dollars. If he received another 15 dollars for his allowance, how many 6 dollar toys could he buy?","target":6}
|
||||
{"input":"Olivia was making baggies of cookies with 9 cookies in each bag. If she had 13 chocolate chip cookies and 41 oatmeal cookies, how many baggies could she make?","target":6}
|
||||
{"input":"Luke made 9 dollars mowing lawns and 18 dollars weed eating. If he only spent 3 dollar a week, how long would the money last him?","target":9}
|
||||
{"input":"There school's baseball team had 48 new players and 6 returning players. If the coach put them into groups with 6 players in each group, how many groups would there be?","target":9}
|
||||
{"input":"For a birthday party Jerry bought 41 regular sodas and 22 diet sodas. If his fridge would only hold 9 on each shelf, how many shelves would he fill up?","target":7}
|
||||
{"input":"Paige and her friends were recycling paper for their class. For every 4 pounds they recycled they earned one point. If Paige recycled 14 pounds and her friends recycled 2 pounds, how many points did they earn?","target":4}
|
||||
{"input":"For homework Nancy had 17 math problems and 15 spelling problems. If she can finish 8 problems in an hour how long will it take her to finish all the problems?","target":4}
|
||||
{"input":"Will was organizing his baseball cards in a binder with 3 on each page. If he had 8 new cards and 10 old cards to put in the binder, how many pages would he use?","target":6}
|
||||
{"input":"The schools debate team had 11 boys and 45 girls on it. If they were split into groups of 7 how many groups could they make?","target":8}
|
||||
{"input":"A group of 4 friends went into a restaurant. The chef already had 9 chicken wings cooked but cooked 7 more for the group. If they each got the same amount how many would each person get?","target":4}
|
||||
{"input":"A vase can hold 6 flowers. If you had 7 carnations and 47 roses, how many vases would you need to hold the flowers?","target":9}
|
||||
{"input":"A pet shelter had 9 puppies when another 12 were brought in. If 3 puppies a day are adopted, how long would it take for all of them to be adopted?","target":7}
|
||||
{"input":"Roger was helping the cafeteria workers pick up lunch trays, but he could only carry 4 trays at a time. If he had to pick up 10 trays from one table and 2 trays from another, how many trips will he make?","target":3}
|
||||
{"input":"A toy store had 4 giant stuffed bears in stock when they got another shipment with 10 bears in it. The put the bears onto shelves with 7 on each shelf. How many shelves did they use?","target":2}
|
||||
{"input":"John made 6 dollars mowing lawns and 18 dollars weed eating. If he only spent 3 dollar a week, how long would the money last him?","target":8}
|
||||
{"input":"Wendy's old washing machine could only wash 8 pieces of clothing at a time. If she had to wash 39 shirts and 33 sweaters how many loads would she have to do?","target":9}
|
||||
{"input":"Debby's class is going on a field trip to the zoo. If each van can hold 4 people and there are 2 students and 6 adults going, how many vans will they need?","target":2}
|
||||
{"input":"Isabel uploaded 2 pictures from her phone and 4 from her camera to facebook. If she sorted the pics into 3 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":2}
|
||||
{"input":"While playing at the arcade, Frank won 33 tickets playing 'whack a mole' and 9 tickets playing 'skee ball'. If he was trying to buy candy that cost 6 tickets a piece, how many could he buy?","target":7}
|
||||
{"input":"Mike made 14 dollars mowing lawns and 26 dollars weed eating. If he only spent 5 dollar a week, how long would the money last him?","target":8}
|
||||
{"input":"Gwen and her friends were recycling paper for their class. For every 3 pounds they recycled they earned one point. If Gwen recycled 5 pounds and her friends recycled 13 pounds, how many points did they earn?","target":6}
|
||||
{"input":"Robin uploaded 31 pictures from her phone and 5 from her camera to facebook. If she sorted the pics into 9 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":4}
|
||||
{"input":"For Halloween Megan received 11 pieces of candy from neighbors and 5 pieces from her older sister. If she only ate 8 pieces a day, how long would the candy last her?","target":2}
|
||||
{"input":"Maria was making baggies of cookies with 5 cookies in each bag. If she had 33 chocolate chip cookies and 2 oatmeal cookies, how many baggies could she make?","target":7}
|
||||
{"input":"There school's baseball team had 4 new players and 6 returning players. If the coach put them into groups with 5 players in each group, how many groups would there be?","target":2}
|
||||
{"input":"A pet shelter had 5 puppies when another 35 were brought in. If 8 puppies a day are adopted, how long would it take for all of them to be adopted?","target":5}
|
||||
{"input":"Nancy's class is going on a field trip to the zoo. If each van can hold 5 people and there are 12 students and 3 adults going, how many vans will they need?","target":3}
|
||||
{"input":"Dave was helping the cafeteria workers pick up lunch trays, but he could only carry 9 trays at a time. If he had to pick up 17 trays from one table and 55 trays from another, how many trips will he make?","target":8}
|
||||
{"input":"For a birthday party George bought 10 regular sodas and 22 diet sodas. If his fridge would only hold 4 on each shelf, how many shelves would he fill up?","target":8}
|
||||
{"input":"The schools debate team had 28 boys and 4 girls on it. If they were split into groups of 4 how many groups could they make?","target":8}
|
||||
{"input":"Maria was making baggies of cookies with 8 cookies in each bag. If she had 5 chocolate chip cookies and 19 oatmeal cookies, how many baggies could she make?","target":3}
|
||||
{"input":"For Halloween Sarah received 66 pieces of candy from neighbors and 15 pieces from her older sister. If she only ate 9 pieces a day, how long would the candy last her?","target":9}
|
||||
{"input":"Henry was helping the cafeteria workers pick up lunch trays, but he could only carry 9 trays at a time. If he had to pick up 29 trays from one table and 52 trays from another, how many trips will he make?","target":9}
|
||||
{"input":"Tiffany uploaded 7 pictures from her phone and 13 from her camera to facebook. If she sorted the pics into 5 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":4}
|
||||
{"input":"A toy store had 5 giant stuffed bears in stock when they got another shipment with 7 bears in it. The put the bears onto shelves with 6 on each shelf. How many shelves did they use?","target":2}
|
||||
{"input":"Paul made 68 dollars mowing lawns and 13 dollars weed eating. If he only spent 9 dollar a week, how long would the money last him?","target":9}
|
||||
{"input":"While playing at the arcade, Edward won 3 tickets playing 'whack a mole' and 5 tickets playing 'skee ball'. If he was trying to buy candy that cost 4 tickets a piece, how many could he buy?","target":2}
|
||||
{"input":"Megan's class is going on a field trip to the zoo. If each van can hold 5 people and there are 25 students and 5 adults going, how many vans will they need?","target":6}
|
||||
{"input":"Luke was organizing his baseball cards in a binder with 3 on each page. If he had 3 new cards and 9 old cards to put in the binder, how many pages would he use?","target":4}
|
||||
{"input":"A group of 3 friends went into a restaurant. The chef already had 6 chicken wings cooked but cooked 3 more for the group. If they each got the same amount how many would each person get?","target":3}
|
||||
{"input":"Olivia uploaded 5 pictures from her phone and 35 from her camera to facebook. If she sorted the pics into 8 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":5}
|
||||
{"input":"A toy store had 6 giant stuffed bears in stock when they got another shipment with 18 bears in it. The put the bears onto shelves with 6 on each shelf. How many shelves did they use?","target":4}
|
||||
{"input":"Jerry was helping the cafeteria workers pick up lunch trays, but he could only carry 8 trays at a time. If he had to pick up 9 trays from one table and 7 trays from another, how many trips will he make?","target":2}
|
||||
{"input":"A vase can hold 5 flowers. If you had 6 carnations and 19 roses, how many vases would you need to hold the flowers?","target":5}
|
||||
{"input":"Haley and her friends were recycling paper for their class. For every 3 pounds they recycled they earned one point. If Haley recycled 11 pounds and her friends recycled 16 pounds, how many points did they earn?","target":9}
|
||||
{"input":"A pet shelter had 17 puppies when another 55 were brought in. If 8 puppies a day are adopted, how long would it take for all of them to be adopted?","target":9}
|
||||
{"input":"There school's baseball team had 12 new players and 44 returning players. If the coach put them into groups with 8 players in each group, how many groups would there be?","target":7}
|
||||
{"input":"Frank had saved up 3 dollars. If he received another 37 dollars for his allowance, how many 8 dollar toys could he buy?","target":5}
|
||||
{"input":"Luke was organizing his baseball cards in a binder with 3 on each page. If he had 3 new cards and 9 old cards to put in the binder, how many pages would he use?","target":4}
|
||||
{"input":"A group of 3 friends went into a restaurant. The chef already had 6 chicken wings cooked but cooked 3 more for the group. If they each got the same amount how many would each person get?","target":3}
|
||||
{"input":"Olivia uploaded 5 pictures from her phone and 35 from her camera to facebook. If she sorted the pics into 8 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":5}
|
||||
{"input":"A toy store had 6 giant stuffed bears in stock when they got another shipment with 18 bears in it. The put the bears onto shelves with 6 on each shelf. How many shelves did they use?","target":4}
|
||||
{"input":"Jerry was helping the cafeteria workers pick up lunch trays, but he could only carry 8 trays at a time. If he had to pick up 9 trays from one table and 7 trays from another, how many trips will he make?","target":2}
|
||||
{"input":"A vase can hold 5 flowers. If you had 6 carnations and 19 roses, how many vases would you need to hold the flowers?","target":5}
|
||||
{"input":"Haley and her friends were recycling paper for their class. For every 3 pounds they recycled they earned one point. If Haley recycled 11 pounds and her friends recycled 16 pounds, how many points did they earn?","target":9}
|
||||
{"input":"A pet shelter had 17 puppies when another 55 were brought in. If 8 puppies a day are adopted, how long would it take for all of them to be adopted?","target":9}
|
||||
{"input":"There school's baseball team had 12 new players and 44 returning players. If the coach put them into groups with 8 players in each group, how many groups would there be?","target":7}
|
||||
{"input":"Frank had saved up 3 dollars. If he received another 37 dollars for his allowance, how many 8 dollar toys could he buy?","target":5}
|
||||
{"input":"A pet shelter had 8 puppies when another 19 were brought in. If 3 puppies a day are adopted, how long would it take for all of them to be adopted?","target":9}
|
||||
{"input":"For homework Faye had 13 math problems and 2 spelling problems. If she can finish 3 problems in an hour how long will it take her to finish all the problems?","target":5}
|
||||
{"input":"Sam had saved up 8 dollars. If he received another 7 dollars for his allowance, how many 3 dollar toys could he buy?","target":5}
|
||||
{"input":"Bianca and her friends were recycling paper for their class. For every 3 pounds they recycled they earned one point. If Bianca recycled 24 pounds and her friends recycled 3 pounds, how many points did they earn?","target":9}
|
||||
{"input":"There school's baseball team had 31 new players and 4 returning players. If the coach put them into groups with 7 players in each group, how many groups would there be?","target":5}
|
||||
{"input":"Oliver was organizing his baseball cards in a binder with 3 on each page. If he had 2 new cards and 10 old cards to put in the binder, how many pages would he use?","target":4}
|
||||
{"input":"Paul made 3 dollars mowing lawns and 3 dollars weed eating. If he only spent 3 dollar a week, how long would the money last him?","target":2}
|
||||
{"input":"Robin was making baggies of cookies with 6 cookies in each bag. If she had 23 chocolate chip cookies and 25 oatmeal cookies, how many baggies could she make?","target":8}
|
||||
{"input":"The schools debate team had 5 boys and 40 girls on it. If they were split into groups of 9 how many groups could they make?","target":5}
|
||||
{"input":"At a company picnic 23 managers and 7 employees decided to start a game of volleyball. If they split into 6 teams how many people would be on each team?","target":5}
|
||||
{"input":"For a birthday party Cody bought 4 regular sodas and 44 diet sodas. If his fridge would only hold 6 on each shelf, how many shelves would he fill up?","target":8}
|
||||
{"input":"Robin's class is going on a field trip to the zoo. If each van can hold 8 people and there are 22 students and 2 adults going, how many vans will they need?","target":3}
|
||||
{"input":"Ned was helping the cafeteria workers pick up lunch trays, but he could only carry 5 trays at a time. If he had to pick up 5 trays from one table and 5 trays from another, how many trips will he make?","target":2}
|
||||
{"input":"There school's baseball team had 2 new players and 6 returning players. If the coach put them into groups with 4 players in each group, how many groups would there be?","target":2}
|
||||
{"input":"For homework Amy had 18 math problems and 6 spelling problems. If she can finish 4 problems in an hour how long will it take her to finish all the problems?","target":6}
|
||||
{"input":"Maria was making baggies of cookies with 3 cookies in each bag. If she had 2 chocolate chip cookies and 16 oatmeal cookies, how many baggies could she make?","target":6}
|
||||
{"input":"Sarah's old washing machine could only wash 5 pieces of clothing at a time. If she had to wash 43 shirts and 2 sweaters how many loads would she have to do?","target":9}
|
||||
{"input":"A pet shelter had 3 puppies when another 3 were brought in. If 3 puppies a day are adopted, how long would it take for all of them to be adopted?","target":2}
|
||||
{"input":"While playing at the arcade, Luke won 2 tickets playing 'whack a mole' and 13 tickets playing 'skee ball'. If he was trying to buy candy that cost 3 tickets a piece, how many could he buy?","target":5}
|
||||
{"input":"A group of 3 friends went into a restaurant. The chef already had 8 chicken wings cooked but cooked 10 more for the group. If they each got the same amount how many would each person get?","target":6}
|
||||
{"input":"Nancy's old washing machine could only wash 9 pieces of clothing at a time. If she had to wash 19 shirts and 8 sweaters how many loads would she have to do?","target":3}
|
||||
{"input":"Paul had saved up 3 dollars. If he received another 7 dollars for his allowance, how many 5 dollar toys could he buy?","target":2}
|
||||
{"input":"Jerry made 14 dollars mowing lawns and 31 dollars weed eating. If he only spent 5 dollar a week, how long would the money last him?","target":9}
|
||||
{"input":"While playing at the arcade, Ned won 26 tickets playing 'whack a mole' and 19 tickets playing 'skee ball'. If he was trying to buy candy that cost 9 tickets a piece, how many could he buy?","target":5}
|
||||
{"input":"Luke was helping the cafeteria workers pick up lunch trays, but he could only carry 4 trays at a time. If he had to pick up 20 trays from one table and 16 trays from another, how many trips will he make?","target":9}
|
||||
{"input":"Vanessa and her friends were recycling paper for their class. For every 9 pounds they recycled they earned one point. If Vanessa recycled 20 pounds and her friends recycled 16 pounds, how many points did they earn?","target":4}
|
||||
{"input":"A pet shelter had 2 puppies when another 34 were brought in. If 4 puppies a day are adopted, how long would it take for all of them to be adopted?","target":9}
|
||||
{"input":"At a company picnic 9 managers and 15 employees decided to start a game of volleyball. If they split into 8 teams how many people would be on each team?","target":3}
|
||||
{"input":"John was organizing his baseball cards in a binder with 3 on each page. If he had 8 new cards and 16 old cards to put in the binder, how many pages would he use?","target":8}
|
||||
{"input":"Wendy uploaded 22 pictures from her phone and 2 from her camera to facebook. If she sorted the pics into 4 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":6}
|
||||
{"input":"A toy store had 17 giant stuffed bears in stock when they got another shipment with 10 bears in it. The put the bears onto shelves with 9 on each shelf. How many shelves did they use?","target":3}
|
||||
{"input":"Zoe and her friends were recycling paper for their class. For every 8 pounds they recycled they earned one point. If Zoe recycled 25 pounds and her friends recycled 23 pounds, how many points did they earn?","target":6}
|
||||
{"input":"Billy was organizing his baseball cards in a binder with 5 on each page. If he had 3 new cards and 42 old cards to put in the binder, how many pages would he use?","target":9}
|
||||
{"input":"For a birthday party Tom bought 4 regular sodas and 52 diet sodas. If his fridge would only hold 7 on each shelf, how many shelves would he fill up?","target":8}
|
||||
{"input":"Victor was helping the cafeteria workers pick up lunch trays, but he could only carry 7 trays at a time. If he had to pick up 23 trays from one table and 5 trays from another, how many trips will he make?","target":4}
|
||||
{"input":"Paul had saved up 4 dollars. If he received another 11 dollars for his allowance, how many 5 dollar toys could he buy?","target":3}
|
||||
{"input":"Katie uploaded 30 pictures from her phone and 51 from her camera to facebook. If she sorted the pics into 9 different albums with the same amount of pics in each album, how many pictures were in each of the albums?","target":9}
|
||||
{"input":"Debby's class is going on a field trip to the zoo. If each van can hold 9 people and there are 40 students and 14 adults going, how many vans will they need?","target":6}
|
||||
{"input":"For Halloween Emily received 5 pieces of candy from neighbors and 13 pieces from her older sister. If she only ate 9 pieces a day, how long would the candy last her?","target":2}
|
||||
{"input":"Frank made 5 dollars mowing lawns and 58 dollars weed eating. If he only spent 7 dollar a week, how long would the money last him?","target":9}
|
||||
{"input":"A new building needed 14 windows. The builder had already installed 5 of them. If it takes 4 hours to install each window, how long will it take him to install the rest?","target":36}
|
||||
{"input":"A chef needs to cook 16 potatoes. He has already cooked 7. If each potato takes 5 minutes to cook, how long will it take him to cook the rest?","target":45}
|
||||
{"input":"Ned bought 14 boxes of chocolate candy and gave 7 to his little brother. If each box has 6 pieces inside it, how many pieces did Ned still have?","target":42}
|
||||
{"input":"There were 11 friends playing a video game online when 5 players quit. If each player left had 5 lives, how many lives did they have total?","target":30}
|
||||
{"input":"Henry earned 5 dollars for each lawn he mowed. If he had 12 lawns to mow, but forgot to mow 7 of them, how much money did he actually earn?","target":25}
|
||||
{"input":"A trivia team had 5 members total, but during a game 2 members didn't show up. If each member that did show up scored 6 points, how many points were scored total?","target":18}
|
||||
{"input":"A painter needed to paint 10 rooms in a building. Each room takes 8 hours to paint. If he already painted 8 rooms, how much longer will he take to paint the rest?","target":16}
|
||||
{"input":"In a video game, each enemy defeated gives you 3 points. If a level has 6 enemies total and you destroy all but 2 of them, how many points would you earn?","target":12}
|
||||
{"input":"Wendy earned 5 points for each bag of cans she recycled. If she had 11 bags, but didn't recycle 2 of them, how many points would she have earned?","target":45}
|
||||
{"input":"Each chocolate bar in a box cost $3. If a box had 7 bars total and Olivia sold all but 4 bars, how much money would she have made?","target":9}
|
||||
{"input":"Kaleb bought 14 boxes of chocolate candy and gave 5 to his little brother. If each box has 6 pieces inside it, how many pieces did Kaleb still have?","target":54}
|
||||
{"input":"At a restaurant each adult meal costs $3 and kids eat free. If a group of 12 people came in and 7 were kids, how much would it cost for the group to eat?","target":15}
|
||||
{"input":"Jerry had 7 action figures, but needed 16 total for a complete collection. If each one costs $8, how much money would he need to finish his collection?","target":72}
|
||||
{"input":"In a video game, each enemy defeated gives you 9 points. If a level has 11 enemies total and you destroy all but 3 of them, how many points would you earn?","target":72}
|
||||
{"input":"There were 16 friends playing a video game online when 7 players quit. If each player left had 8 lives, how many lives did they have total?","target":72}
|
||||
{"input":"A new building needed 10 windows. The builder had already installed 6 of them. If it takes 5 hours to install each window, how long will it take him to install the rest?","target":20}
|
||||
{"input":"Adam earned 9 dollars for each lawn he mowed. If he had 12 lawns to mow, but forgot to mow 8 of them, how much money did he actually earn?","target":36}
|
||||
{"input":"A trivia team had 12 members total, but during a game 4 members didn't show up. If each member that did show up scored 8 points, how many points were scored total?","target":64}
|
||||
{"input":"Mike had 16 video games but 8 of them weren't working. If he wanted to sell the working games for $7 each, how much money could he earn?","target":56}
|
||||
{"input":"A painter needed to paint 12 rooms in a building. Each room takes 7 hours to paint. If he already painted 5 rooms, how much longer will he take to paint the rest?","target":49}
|
||||
{"input":"There were 8 friends playing a video game online when 5 players quit. If each player left had 5 lives, how many lives did they have total?","target":15}
|
||||
{"input":"A trivia team had 15 members total, but during a game 6 members didn't show up. If each member that did show up scored 3 points, how many points were scored total?","target":27}
|
||||
{"input":"Ned had 15 video games but 6 of them weren't working. If he wanted to sell the working games for $7 each, how much money could he earn?","target":63}
|
||||
{"input":"Each chocolate bar in a box cost $6. If a box had 13 bars total and Zoe sold all but 6 bars, how much money would she have made?","target":42}
|
||||
{"input":"Will bought 7 boxes of chocolate candy and gave 3 to his little brother. If each box has 4 pieces inside it, how many pieces did Will still have?","target":16}
|
||||
{"input":"Mike invited 13 friends to a birthday party, but 7 couldn't come. If he wanted to buy enough cupcakes so each person could have exactly 4, how many should he buy?","target":24}
|
||||
{"input":"In a video game, each enemy defeated gives you 5 points. If a level has 8 enemies total and you destroy all but 6 of them, how many points would you earn?","target":10}
|
||||
{"input":"Roger earned 9 dollars for each lawn he mowed. If he had 14 lawns to mow, but forgot to mow 8 of them, how much money did he actually earn?","target":54}
|
||||
{"input":"A magician was selling magic card decks for 2 dollars each. If he started with 5 decks and by the end of the day he had 3 left, how much money did he earn?","target":4}
|
||||
{"input":"A chef needs to cook 12 potatoes. He has already cooked 6. If each potato takes 6 minutes to cook, how long will it take him to cook the rest?","target":36}
|
||||
{"input":"A new building needed 9 windows. The builder had already installed 6 of them. If it takes 6 hours to install each window, how long will it take him to install the rest?","target":18}
|
||||
{"input":"At the fair Adam bought 13 tickets. After riding the ferris wheel he had 4 tickets left. If each ticket cost 9 dollars, how much money did Adam spend riding the ferris wheel?","target":81}
|
||||
{"input":"Dave bought 12 boxes of chocolate candy and gave 5 to his little brother. If each box has 3 pieces inside it, how many pieces did Dave still have?","target":21}
|
||||
{"input":"John had 5 action figures, but needed 7 total for a complete collection. If each one costs $5, how much money would he need to finish his collection?","target":10}
|
||||
{"input":"A painter needed to paint 9 rooms in a building. Each room takes 8 hours to paint. If he already painted 5 rooms, how much longer will he take to paint the rest?","target":32}
|
||||
{"input":"Each chocolate bar in a box cost $4. If a box had 8 bars total and Emily sold all but 3 bars, how much money would she have made?","target":20}
|
||||
{"input":"At lunch a waiter had 10 customers and 5 of them didn't leave a tip. If he got $3 each from the ones who did tip, how much money did he earn?","target":15}
|
||||
{"input":"A worksheet had 4 problems on it. If a teacher had 9 worksheets to grade and had already graded 5 of them, how many more problems does she have to grade?","target":16}
|
||||
{"input":"Will invited 9 friends to a birthday party, but 4 couldn't come. If he wanted to buy enough cupcakes so each person could have exactly 8, how many should he buy?","target":40}
|
||||
{"input":"A magician was selling magic card decks for 9 dollars each. If he started with 12 decks and by the end of the day he had 7 left, how much money did he earn?","target":45}
|
||||
{"input":"A chef needs to cook 9 potatoes. He has already cooked 7. If each potato takes 3 minutes to cook, how long will it take him to cook the rest?","target":6}
|
||||
{"input":"A magician was selling magic card decks for 5 dollars each. If he started with 14 decks and by the end of the day he had 5 left, how much money did he earn?","target":45}
|
||||
{"input":"Each chocolate bar in a box cost $3. If a box had 9 bars total and Wendy sold all but 3 bars, how much money would she have made?","target":18}
|
||||
{"input":"A painter needed to paint 12 rooms in a building. Each room takes 3 hours to paint. If he already painted 4 rooms, how much longer will he take to paint the rest?","target":24}
|
||||
{"input":"Adam bought 13 boxes of chocolate candy and gave 7 to his little brother. If each box has 6 pieces inside it, how many pieces did Adam still have?","target":36}
|
||||
{"input":"At a restaurant each adult meal costs $7 and kids eat free. If a group of 13 people came in and 9 were kids, how much would it cost for the group to eat?","target":28}
|
||||
{"input":"A trivia team had 7 members total, but during a game 2 members didn't show up. If each member that did show up scored 4 points, how many points were scored total?","target":20}
|
||||
{"input":"Kaleb had 10 video games but 8 of them weren't working. If he wanted to sell the working games for $6 each, how much money could he earn?","target":12}
|
||||
{"input":"April's discount flowers was having a sale where each rose was 7 dollars. If April started with 9 roses and had 4 roses left, how much money did she earn?","target":35}
|
||||
{"input":"At lunch a waiter had 9 customers and 5 of them didn't leave a tip. If he got $8 each from the ones who did tip, how much money did he earn?","target":32}
|
||||
{"input":"A new building needed 14 windows. The builder had already installed 8 of them. If it takes 8 hours to install each window, how long will it take him to install the rest?","target":48}
|
||||
{"input":"Bianca earned 5 points for each bag of cans she recycled. If she had 17 bags, but didn't recycle 8 of them, how many points would she have earned?","target":45}
|
||||
{"input":"A chef needs to cook 15 potatoes. He has already cooked 6. If each potato takes 8 minutes to cook, how long will it take him to cook the rest?","target":72}
|
||||
{"input":"Henry had 3 action figures, but needed 8 total for a complete collection. If each one costs $6, how much money would he need to finish his collection?","target":30}
|
||||
{"input":"John earned 8 dollars for each lawn he mowed. If he had 15 lawns to mow, but forgot to mow 7 of them, how much money did he actually earn?","target":64}
|
||||
{"input":"There were 10 friends playing a video game online when 7 players quit. If each player left had 8 lives, how many lives did they have total?","target":24}
|
||||
{"input":"In a video game, each enemy defeated gives you 8 points. If a level has 7 enemies total and you destroy all but 2 of them, how many points would you earn?","target":40}
|
||||
{"input":"A painter needed to paint 11 rooms in a building. Each room takes 7 hours to paint. If he already painted 2 rooms, how much longer will he take to paint the rest?","target":63}
|
||||
{"input":"Dave had 10 video games but 2 of them weren't working. If he wanted to sell the working games for $4 each, how much money could he earn?","target":32}
|
||||
{"input":"A worksheet had 4 problems on it. If a teacher had 16 worksheets to grade and had already graded 8 of them, how many more problems does she have to grade?","target":32}
|
||||
{"input":"A worksheet had 2 problems on it. If a teacher had 14 worksheets to grade and had already graded 7 of them, how many more problems does she have to grade?","target":14}
|
||||
{"input":"April's discount flowers was having a sale where each rose was 4 dollars. If April started with 13 roses and had 4 roses left, how much money did she earn?","target":36}
|
||||
{"input":"At a restaurant each adult meal costs $5 and kids eat free. If a group of 15 people came in and 8 were kids, how much would it cost for the group to eat?","target":35}
|
||||
{"input":"Zoe baked 5 brownies, but needed 11 total for her party. If she used 7 cups of flour on each one, how much cups of flour does she still need?","target":42}
|
||||
{"input":"A painter needed to paint 6 rooms in a building. Each room takes 5 hours to paint. If he already painted 2 rooms, how much longer will he take to paint the rest?","target":20}
|
||||
{"input":"Each chocolate bar in a box cost $4. If a box had 11 bars total and Vanessa sold all but 7 bars, how much money would she have made?","target":16}
|
||||
{"input":"Gwen earned 8 points for each bag of cans she recycled. If she had 4 bags, but didn't recycle 2 of them, how many points would she have earned?","target":16}
|
||||
{"input":"A new building needed 12 windows. The builder had already installed 6 of them. If it takes 4 hours to install each window, how long will it take him to install the rest?","target":24}
|
||||
{"input":"A chef needs to cook 15 potatoes. He has already cooked 8. If each potato takes 9 minutes to cook, how long will it take him to cook the rest?","target":63}
|
||||
{"input":"A trivia team had 14 members total, but during a game 7 members didn't show up. If each member that did show up scored 5 points, how many points were scored total?","target":35}
|
||||
{"input":"A chef needs to cook 13 potatoes. He has already cooked 5. If each potato takes 6 minutes to cook, how long will it take him to cook the rest?","target":48}
|
||||
{"input":"A trivia team had 11 members total, but during a game 6 members didn't show up. If each member that did show up scored 9 points, how many points were scored total?","target":45}
|
||||
{"input":"Each chocolate bar in a box cost $2. If a box had 13 bars total and Rachel sold all but 4 bars, how much money would she have made?","target":18}
|
||||
{"input":"Isabel baked 3 brownies, but needed 5 total for her party. If she used 5 cups of flour on each one, how much cups of flour does she still need?","target":10}
|
||||
{"input":"A worksheet had 3 problems on it. If a teacher had 15 worksheets to grade and had already graded 7 of them, how many more problems does she have to grade?","target":24}
|
||||
{"input":"At lunch a waiter had 7 customers and 4 of them didn't leave a tip. If he got $9 each from the ones who did tip, how much money did he earn?","target":27}
|
||||
{"input":"April's discount flowers was having a sale where each rose was 9 dollars. If April started with 11 roses and had 8 roses left, how much money did she earn?","target":27}
|
||||
{"input":"Tom bought 14 boxes of chocolate candy and gave 8 to his little brother. If each box has 3 pieces inside it, how many pieces did Tom still have?","target":18}
|
||||
{"input":"At a restaurant each adult meal costs $2 and kids eat free. If a group of 15 people came in and 9 were kids, how much would it cost for the group to eat?","target":12}
|
||||
{"input":"Edward earned 4 dollars for each lawn he mowed. If he had 17 lawns to mow, but forgot to mow 9 of them, how much money did he actually earn?","target":32}
|
||||
{"input":"A new building needed 11 windows. The builder had already installed 4 of them. If it takes 8 hours to install each window, how long will it take him to install the rest?","target":56}
|
||||
{"input":"Tom bought 12 boxes of chocolate candy and gave 7 to his little brother. If each box has 6 pieces inside it, how many pieces did Tom still have?","target":30}
|
||||
{"input":"John had 6 action figures, but needed 11 total for a complete collection. If each one costs $6, how much money would he need to finish his collection?","target":30}
|
||||
{"input":"At a restaurant each adult meal costs $8 and kids eat free. If a group of 11 people came in and 2 were kids, how much would it cost for the group to eat?","target":72}
|
||||
{"input":"At the fair Kaleb bought 6 tickets. After riding the ferris wheel he had 3 tickets left. If each ticket cost 9 dollars, how much money did Kaleb spend riding the ferris wheel?","target":27}
|
||||
{"input":"Tiffany baked 8 brownies, but needed 17 total for her party. If she used 8 cups of flour on each one, how much cups of flour does she still need?","target":72}
|
||||
{"input":"In a video game, each enemy defeated gives you 7 points. If a level has 11 enemies total and you destroy all but 8 of them, how many points would you earn?","target":21}
|
||||
{"input":"A magician was selling magic card decks for 7 dollars each. If he started with 16 decks and by the end of the day he had 8 left, how much money did he earn?","target":56}
|
||||
{"input":"There were 13 friends playing a video game online when 8 players quit. If each player left had 6 lives, how many lives did they have total?","target":30}
|
||||
{"input":"At lunch a waiter had 7 customers and 5 of them didn't leave a tip. If he got $3 each from the ones who did tip, how much money did he earn?","target":6}
|
||||
{"input":"At the fair Adam bought 10 tickets. After riding the ferris wheel he had 3 tickets left. If each ticket cost 9 dollars, how much money did Adam spend riding the ferris wheel?","target":63}
|
||||
{"input":"A worksheet had 7 problems on it. If a teacher had 17 worksheets to grade and had already graded 8 of them, how many more problems does she have to grade?","target":63}
|
||||
{"input":"A trivia team had 9 members total, but during a game 3 members didn't show up. If each member that did show up scored 2 points, how many points were scored total?","target":12}
|
||||
{"input":"Mike had 15 video games but 9 of them weren't working. If he wanted to sell the working games for $5 each, how much money could he earn?","target":30}
|
||||
{"input":"Edward earned 9 dollars for each lawn he mowed. If he had 6 lawns to mow, but forgot to mow 2 of them, how much money did he actually earn?","target":36}
|
||||
{"input":"At a restaurant each adult meal costs $2 and kids eat free. If a group of 9 people came in and 2 were kids, how much would it cost for the group to eat?","target":14}
|
||||
{"input":"Chloe baked 7 brownies, but needed 16 total for her party. If she used 6 cups of flour on each one, how much cups of flour does she still need?","target":54}
|
||||
{"input":"There were 8 friends playing a video game online when 3 players quit. If each player left had 3 lives, how many lives did they have total?","target":15}
|
||||
{"input":"Sam invited 9 friends to a birthday party, but 6 couldn't come. If he wanted to buy enough cupcakes so each person could have exactly 2, how many should he buy?","target":6}
|
||||
{"input":"Megan earned 8 points for each bag of cans she recycled. If she had 14 bags, but didn't recycle 5 of them, how many points would she have earned?","target":72}
|
||||
{"input":"Will had 57 dollars. If he spent 27 bucks on a new game, how many 6 dollar toys could he buy with the money he had left?","target":5}
|
||||
{"input":"A pet store had 18 puppies. In one day they sold 3 of them and put the rest into cages with 5 in each cage. How many cages did they use?","target":3}
|
||||
{"input":"Luke had 47 pieces of clothing to wash. He put 17 of them in one load, but decided to split the rest into 5 equal loads. How many pieces of clothing could go in each of the small loads?","target":6}
|
||||
{"input":"Edward bought 79 tickets at the state fair. He spent 23 tickets at the 'dunk a clown' booth and decided to use the rest on rides. If each ride cost 7 tickets, how many rides could he go on?","target":8}
|
||||
{"input":"Katie baked 18 cupcakes for her school's bake sale. If her brother, Todd, ate 8 of them how many packages could she make if she put 2 cupcake in each package?","target":5}
|
||||
{"input":"Megan had 93 files on her computer. She deleted 21 of them and put the rest into folders with 8 files in each one. How many folders did Megan end up with?","target":9}
|
||||
{"input":"Wendy uploaded 79 pictures to Facebook. She put 44 pics into one album and put the rest into 5 different albums. How many pictures were in each album?","target":7}
|
||||
{"input":"Isabel had 72 homework problems. She finished 32 of them but still had 5 pages of problems to do. If each page has the same number of problems on it, how many problems are on each page?","target":8}
|
||||
{"input":"The cafeteria had 62 apples. For lunch they handed out 8 to students and decided to use the rest to make pies. If each pie takes 9 apples, how many pies could they make?","target":6}
|
||||
{"input":"Mike made 101 dollars mowing lawns over the summer. If he spent 47 dollars buying new mower blades, how many 6 dollar games could he buy with the money he had left?","target":9}
|
||||
{"input":"Paige's team won their dodgeball game and scored 41 points total. If Paige scored 11 of the points and everyone else scored 6 points each, how many players were on her team?","target":5}
|
||||
{"input":"Oliver is at the library helping put away books. There are 46 book to put away total but a librarian takes 10 of them and leaves Oliver with the rest. If he can fit 4 books on a shelf, how many shelves will he need?","target":9}
|
||||
{"input":"Emily was planting vegetables in her garden. She started with 41 seeds and planted 29 of them in the big garden and in each of her small gardens put 4 seeds each. How many small gardens did Emily have?","target":3}
|
||||
{"input":"Edward made 37 dollars mowing lawns over the summer. If he spent 21 dollars buying new mower blades, how many 2 dollar games could he buy with the money he had left?","target":8}
|
||||
{"input":"A pet store had 81 puppies. In one day they sold 41 of them and put the rest into cages with 8 in each cage. How many cages did they use?","target":5}
|
||||
{"input":"There are 65 students trying out for the school's trivia teams. If 17 of them didn't get picked for the team and the rest were put into 8 groups, how many students would be in each group?","target":6}
|
||||
{"input":"Tom had 57 dollars. If he spent 49 bucks on a new game, how many 4 dollar toys could he buy with the money he had left?","target":2}
|
||||
{"input":"Isabel picked 66 flowers for her friend\u2019s wedding. She was making bouquets with 8 flowers in each one. If 10 of the flowers wilted before the wedding, how many bouquets could she still make?","target":7}
|
||||
{"input":"For Halloween Sarah received 108 pieces of candy. She ate 36 pieces then placed the rest into piles with 9 in each pile. How many piles could she make?","target":8}
|
||||
{"input":"Nancy had 80 files on her computer. She deleted 31 of them and put the rest into folders with 7 files in each one. How many folders did Nancy end up with?","target":7}
|
||||
{"input":"There are 36 students trying out for the school's trivia teams. If 9 of them didn't get picked for the team and the rest were put into 3 groups, how many students would be in each group?","target":9}
|
||||
{"input":"The cafeteria had 96 apples. For lunch they handed out 42 to students and decided to use the rest to make pies. If each pie takes 6 apples, how many pies could they make?","target":9}
|
||||
{"input":"A pet store had 64 puppies. In one day they sold 28 of them and put the rest into cages with 4 in each cage. How many cages did they use?","target":9}
|
||||
{"input":"Edward was selling his old games. He started out with 35 but sold 19 of them. He packed the rest up putting 8 games into each box. How many boxes did he have to use?","target":2}
|
||||
{"input":"Frank made 19 dollars mowing lawns over the summer. If he spent 11 dollars buying new mower blades, how many 2 dollar games could he buy with the money he had left?","target":4}
|
||||
{"input":"Megan had 40 homework problems. She finished 26 of them but still had 2 pages of problems to do. If each page has the same number of problems on it, how many problems are on each page?","target":7}
|
||||
{"input":"Wendy picked 103 flowers for her friend\u2019s wedding. She was making bouquets with 8 flowers in each one. If 47 of the flowers wilted before the wedding, how many bouquets could she still make?","target":7}
|
||||
{"input":"Kaleb had 39 pieces of clothing to wash. He put 19 of them in one load, but decided to split the rest into 5 equal loads. How many pieces of clothing could go in each of the small loads?","target":4}
|
||||
{"input":"A company invited 18 people to a luncheon, but 12 of them didn't show up. If the tables they had held 3 people each, how many tables do they need?","target":2}
|
||||
{"input":"Mike is at the library helping put away books. There are 82 book to put away total but a librarian takes 10 of them and leaves Mike with the rest. If he can fit 9 books on a shelf, how many shelves will he need?","target":8}
|
||||
{"input":"A pet store had 78 puppies. In one day they sold 30 of them and put the rest into cages with 8 in each cage. How many cages did they use?","target":6}
|
||||
{"input":"Bianca uploaded 33 pictures to Facebook. She put 27 pics into one album and put the rest into 3 different albums. How many pictures were in each album?","target":2}
|
||||
{"input":"Emily's team won their dodgeball game and scored 39 points total. If Emily scored 23 of the points and everyone else scored 2 points each, how many players were on her team?","target":8}
|
||||
{"input":"A company invited 47 people to a luncheon, but 7 of them didn't show up. If the tables they had held 5 people each, how many tables do they need?","target":8}
|
||||
{"input":"A waiter had 44 customers in his section. If 12 of them left and the rest of his tables had 8 people at each table, how many tables did he have?","target":4}
|
||||
{"input":"A store had 40 coloring books in stock. They ended up putting them on sale and getting rid of 20 of them. The put the ones they still had onto shelves with 4 on each shelf. How many shelves did they use?","target":5}
|
||||
{"input":"There are 64 students trying out for the school's trivia teams. If 36 of them didn't get picked for the team and the rest were put into 4 groups, how many students would be in each group?","target":7}
|
||||
{"input":"Kaleb was selling his old games. He started out with 76 but sold 46 of them. He packed the rest up putting 5 games into each box. How many boxes did he have to use?","target":6}
|
||||
{"input":"Jerry is at the library helping put away books. There are 34 book to put away total but a librarian takes 7 of them and leaves Jerry with the rest. If he can fit 3 books on a shelf, how many shelves will he need?","target":9}
|
||||
{"input":"Nancy was planting vegetables in her garden. She started with 52 seeds and planted 28 of them in the big garden and in each of her small gardens put 4 seeds each. How many small gardens did Nancy have?","target":6}
|
||||
{"input":"Will had 83 dollars. If he spent 47 bucks on a new game, how many 4 dollar toys could he buy with the money he had left?","target":9}
|
||||
{"input":"Bianca was planting vegetables in her garden. She started with 52 seeds and planted 40 of them in the big garden and in each of her small gardens put 2 seeds each. How many small gardens did Bianca have?","target":6}
|
||||
{"input":"A waiter had 21 customers in his section. If 12 of them left and the rest of his tables had 3 people at each table, how many tables did he have?","target":3}
|
||||
{"input":"There are 25 students trying out for the school's trivia teams. If 15 of them didn't get picked for the team and the rest were put into 2 groups, how many students would be in each group?","target":5}
|
||||
{"input":"The cafeteria had 47 apples. For lunch they handed out 27 to students and decided to use the rest to make pies. If each pie takes 4 apples, how many pies could they make?","target":5}
|
||||
{"input":"Megan baked 68 cupcakes for her school's bake sale. If her brother, Todd, ate 32 of them how many packages could she make if she put 6 cupcake in each package?","target":6}
|
||||
{"input":"A store had 120 coloring books in stock. They ended up putting them on sale and getting rid of 39 of them. The put the ones they still had onto shelves with 9 on each shelf. How many shelves did they use?","target":9}
|
||||
{"input":"Isabel uploaded 25 pictures to Facebook. She put 10 pics into one album and put the rest into 5 different albums. How many pictures were in each album?","target":3}
|
||||
{"input":"Sarah had 60 homework problems. She finished 20 of them but still had 5 pages of problems to do. If each page has the same number of problems on it, how many problems are on each page?","target":8}
|
||||
{"input":"Mike made 42 dollars mowing lawns over the summer. If he spent 10 dollars buying new mower blades, how many 8 dollar games could he buy with the money he had left?","target":4}
|
||||
{"input":"Will had 59 pieces of clothing to wash. He put 32 of them in one load, but decided to split the rest into 9 equal loads. How many pieces of clothing could go in each of the small loads?","target":3}
|
||||
{"input":"Oliver made 35 dollars mowing lawns over the summer. If he spent 7 dollars buying new mower blades, how many 4 dollar games could he buy with the money he had left?","target":7}
|
||||
{"input":"Emily was planting vegetables in her garden. She started with 42 seeds and planted 36 of them in the big garden and in each of her small gardens put 2 seeds each. How many small gardens did Emily have?","target":3}
|
||||
{"input":"There are 17 students trying out for the school's trivia teams. If 5 of them didn't get picked for the team and the rest were put into 3 groups, how many students would be in each group?","target":4}
|
||||
{"input":"Katie's team won their dodgeball game and scored 12 points total. If Katie scored 4 of the points and everyone else scored 4 points each, how many players were on her team?","target":2}
|
||||
{"input":"Megan baked 71 cupcakes for her school's bake sale. If her brother, Todd, ate 43 of them how many packages could she make if she put 7 cupcake in each package?","target":4}
|
||||
{"input":"Wendy uploaded 45 pictures to Facebook. She put 27 pics into one album and put the rest into 9 different albums. How many pictures were in each album?","target":2}
|
||||
{"input":"A store had 86 coloring books in stock. They ended up putting them on sale and getting rid of 37 of them. The put the ones they still had onto shelves with 7 on each shelf. How many shelves did they use?","target":7}
|
||||
{"input":"The cafeteria had 50 apples. For lunch they handed out 5 to students and decided to use the rest to make pies. If each pie takes 5 apples, how many pies could they make?","target":9}
|
||||
{"input":"Nancy had 101 homework problems. She finished 47 of them but still had 6 pages of problems to do. If each page has the same number of problems on it, how many problems are on each page?","target":9}
|
||||
{"input":"Paige had 27 files on her computer. She deleted 9 of them and put the rest into folders with 6 files in each one. How many folders did Paige end up with?","target":3}
|
||||
{"input":"For Halloween Bianca received 78 pieces of candy. She ate 30 pieces then placed the rest into piles with 8 in each pile. How many piles could she make?","target":6}
|
||||
{"input":"A company invited 45 people to a luncheon, but 35 of them didn't show up. If the tables they had held 2 people each, how many tables do they need?","target":5}
|
||||
{"input":"Haley uploaded 65 pictures to Facebook. She put 17 pics into one album and put the rest into 6 different albums. How many pictures were in each album?","target":8}
|
||||
{"input":"A pet store had 102 puppies. In one day they sold 21 of them and put the rest into cages with 9 in each cage. How many cages did they use?","target":9}
|
||||
{"input":"Roger had 120 pieces of clothing to wash. He put 48 of them in one load, but decided to split the rest into 9 equal loads. How many pieces of clothing could go in each of the small loads?","target":8}
|
||||
{"input":"Wendy picked 45 flowers for her friend\u2019s wedding. She was making bouquets with 5 flowers in each one. If 35 of the flowers wilted before the wedding, how many bouquets could she still make?","target":2}
|
||||
{"input":"Isabel baked 39 cupcakes for her school's bake sale. If her brother, Todd, ate 21 of them how many packages could she make if she put 3 cupcake in each package?","target":6}
|
||||
{"input":"Sarah had 55 homework problems. She finished 6 of them but still had 7 pages of problems to do. If each page has the same number of problems on it, how many problems are on each page?","target":7}
|
||||
{"input":"There are 58 students trying out for the school's trivia teams. If 10 of them didn't get picked for the team and the rest were put into 8 groups, how many students would be in each group?","target":6}
|
||||
{"input":"Paige picked 53 flowers for her friend\u2019s wedding. She was making bouquets with 7 flowers in each one. If 18 of the flowers wilted before the wedding, how many bouquets could she still make?","target":5}
|
||||
{"input":"A waiter had 22 customers in his section. If 14 of them left and the rest of his tables had 4 people at each table, how many tables did he have?","target":2}
|
||||
{"input":"Luke was selling his old games. He started out with 39 but sold 19 of them. He packed the rest up putting 4 games into each box. How many boxes did he have to use?","target":5}
|
||||
{"input":"Haley baked 20 cupcakes for her school's bake sale. If her brother, Todd, ate 11 of them how many packages could she make if she put 3 cupcake in each package?","target":3}
|
||||
{"input":"A pet store had 13 puppies. In one day they sold 7 of them and put the rest into cages with 2 in each cage. How many cages did they use?","target":3}
|
||||
{"input":"A store had 27 coloring books in stock. They ended up putting them on sale and getting rid of 6 of them. The put the ones they still had onto shelves with 7 on each shelf. How many shelves did they use?","target":3}
|
||||
{"input":"Wendy had 82 files on her computer. She deleted 37 of them and put the rest into folders with 5 files in each one. How many folders did Wendy end up with?","target":9}
|
||||
{"input":"Kaleb had 12 dollars. If he spent 8 bucks on a new game, how many 2 dollar toys could he buy with the money he had left?","target":2}
|
||||
{"input":"Sarah was planting vegetables in her garden. She started with 21 seeds and planted 12 of them in the big garden and in each of her small gardens put 3 seeds each. How many small gardens did Sarah have?","target":3}
|
||||
{"input":"Nancy uploaded 51 pictures to Facebook. She put 11 pics into one album and put the rest into 8 different albums. How many pictures were in each album?","target":5}
|
||||
{"input":"A company invited 24 people to a luncheon, but 10 of them didn't show up. If the tables they had held 7 people each, how many tables do they need?","target":2}
|
||||
{"input":"Bianca's team won their dodgeball game and scored 75 points total. If Bianca scored 45 of the points and everyone else scored 6 points each, how many players were on her team?","target":5}
|
||||
{"input":"For Halloween Emily received 54 pieces of candy. She ate 33 pieces then placed the rest into piles with 7 in each pile. How many piles could she make?","target":3}
|
||||
{"input":"Haley was planting vegetables in her garden. She started with 56 seeds and planted 35 of them in the big garden and in each of her small gardens put 3 seeds each. How many small gardens did Haley have?","target":7}
|
||||
{"input":"A store had 48 coloring books in stock. They ended up putting them on sale and getting rid of 38 of them. The put the ones they still had onto shelves with 5 on each shelf. How many shelves did they use?","target":2}
|
||||
{"input":"Roger is at the library helping put away books. There are 14 book to put away total but a librarian takes 2 of them and leaves Roger with the rest. If he can fit 3 books on a shelf, how many shelves will he need?","target":4}
|
||||
{"input":"A pet store had 56 puppies. In one day they sold 24 of them and put the rest into cages with 4 in each cage. How many cages did they use?","target":8}
|
||||
{"input":"The cafeteria had 86 apples. For lunch they handed out 30 to students and decided to use the rest to make pies. If each pie takes 8 apples, how many pies could they make?","target":7}
|
||||
{"input":"Jerry had 60 pieces of clothing to wash. He put 40 of them in one load, but decided to split the rest into 5 equal loads. How many pieces of clothing could go in each of the small loads?","target":4}
|
||||
{"input":"Mike made 69 dollars mowing lawns over the summer. If he spent 24 dollars buying new mower blades, how many 5 dollar games could he buy with the money he had left?","target":9}
|
||||
{"input":"Will made 104 dollars mowing lawns over the summer. If he spent 41 dollars buying new mower blades, how many 9 dollar games could he buy with the money he had left?","target":7}
|
||||
{"input":"For Halloween Bianca received 32 pieces of candy. She ate 12 pieces then placed the rest into piles with 5 in each pile. How many piles could she make?","target":4}
|
||||
{"input":"Luke was selling his old games. He started out with 57 but sold 39 of them. He packed the rest up putting 2 games into each box. How many boxes did he have to use?","target":9}
|
||||
{"input":"A pet store had 88 puppies. In one day they sold 34 of them and put the rest into cages with 6 in each cage. How many cages did they use?","target":9}
|
||||
{"input":"Katie had 85 files on her computer. She deleted 40 of them and put the rest into folders with 5 files in each one. How many folders did Katie end up with?","target":9}
|
||||
{"input":"Roger had 68 dollars. If he spent 47 bucks on a new game, how many 7 dollar toys could he buy with the money he had left?","target":3}
|
||||
{"input":"Tom bought 40 tickets at the state fair. He spent 28 tickets at the 'dunk a clown' booth and decided to use the rest on rides. If each ride cost 4 tickets, how many rides could he go on?","target":3}
|
||||
{"input":"The cafeteria had 75 apples. For lunch they handed out 19 to students and decided to use the rest to make pies. If each pie takes 8 apples, how many pies could they make?","target":7}
|
||||
{"input":"Sarah baked 38 cupcakes for her school's bake sale. If her brother, Todd, ate 14 of them how many packages could she make if she put 8 cupcake in each package?","target":3}
|
||||
{"input":"Nancy uploaded 41 pictures to Facebook. She put 37 pics into one album and put the rest into 2 different albums. How many pictures were in each album?","target":2}
|
||||
Executable
+508
@@ -0,0 +1,508 @@
|
||||
{"input":"Joan found 70 seashells on the beach. she gave Sam some of her seashells. She has 27 seashell left. How many seashells did she give to Sam ?","target":43.0}
|
||||
{"input":"There were 28 bales of hay in the barn. Tim stacked more bales in the barn today. There are now 54 bales of hay in the barn. How many bales did he store in the barn ?","target":26.0}
|
||||
{"input":"After eating at the restaurant, Sally, Sam, and Alyssa decided to divide the bill evenly. If each person paid 45 dollars, what was the total of the bill ?","target":135.0}
|
||||
{"input":"Mary is baking a cake. The recipe wants 8 cups of flour. She already put in 2 cups. How many more cups does she need to add ?","target":6.0}
|
||||
{"input":"Sara's high school played 12 basketball games this year. The team won most of their games. They were defeated during 4 games. How many games did they win ?","target":8.0}
|
||||
{"input":"There are 22 walnut trees currently in the park. Park workers will plant more walnut trees today. When the workers are finished there will be 55 walnut trees in the park. How many walnut trees did the workers plant today ?","target":33.0}
|
||||
{"input":"Mike had 34 peaches left at his roadside fruit stand. He went to the orchard and picked more peaches to stock up the stand. There are now 86 peaches at the stand, how many did he pick ?","target":52.0}
|
||||
{"input":"There were 6 roses in the vase. Mary cut some more roses from her flower garden. There are now 16 roses in the vase. How many roses did she cut ?","target":10.0}
|
||||
{"input":"Benny bought a soft drink for 2 dollars and 5 candy bars. He spent a total of 27 dollars. How much did each candy bar cost ?","target":5.0}
|
||||
{"input":"Benny bought 4 new baseball trading cards to add to his collection. The next day his dog ate half of his collection. There are now only 34 cards left. How many cards did Benny start with ?","target":64.0}
|
||||
{"input":"Alyssa spent half of her allowance going to the movies. She washed the family car and earned 8 dollars. What is her weekly allowance if she ended with 12 dollars ?","target":8.0}
|
||||
{"input":"The sum of three consecutive even numbers is 162. What is the smallest of the three numbers ?","target":52.0}
|
||||
{"input":"Sam had 79 dollars to spend on 9 books. After buying them he had 16 dollars. How much did each book cost ?","target":7.0}
|
||||
{"input":"Oceanside Bike Rental Shop charges 17 dollars plus 7 dollars an hour for renting a bike. Tom paid 80 dollars to rent a bike. How many hours did he pay to have the bike checked out ?","target":9.0}
|
||||
{"input":"On Monday, 375 students went on a trip to the zoo. All 7 buses were filled and 4 students had to travel in cars. How many students were in each bus ?","target":53.0}
|
||||
{"input":"Sandy sold half of her comic books and then bought 6 more. She now has 13. How many did she begin with ?","target":14.0}
|
||||
{"input":"The sum of three consecutive odd numbers is 69. What is the smallest of the three numbers ?","target":21.0}
|
||||
{"input":"Joan went to 4 football games this year. She went to 9 football games last year. How many football games did Joan go to in all ?","target":13.0}
|
||||
{"input":"Tom has 9 yellow balloons. Sara has 8 yellow balloons. How many yellow balloons do they have in total ?","target":17.0}
|
||||
{"input":"There are 4 walnut trees currently in the park. Park workers will plant 6 more walnut trees today. How many walnut trees will the park have when the workers are finished ?","target":10.0}
|
||||
{"input":"Sam had 9 dimes in his bank. His dad gave him 7 more dimes. How many dimes does Sam have now ?","target":16.0}
|
||||
{"input":"Alyssa's dog had puppies. She gave 7 to her friends. She now has 5 puppies left. How many puppies did she have to start with ?","target":12.0}
|
||||
{"input":"A restaurant served 9 pizzas during lunch and 6 during dinner today. How many pizzas were served today ?","target":15.0}
|
||||
{"input":"There are 2 pencils in the drawer. Tim placed 3 more pencils in the drawer. How many pencils are now there in total ?","target":5.0}
|
||||
{"input":"Joan found 6 seashells and Jessica found 8 seashells on the beach. How many seashells did they find together ?","target":14.0}
|
||||
{"input":"Sandy grew 6 carrots. Sam grew 3 carrots. How many carrots did they grow in total ?","target":9.0}
|
||||
{"input":"Benny picked 2 apples and Dan picked 9 apples from the apple tree. How many apples were picked in total ?","target":11.0}
|
||||
{"input":"Sally found 9 seashells, Tom found 7 seashells, and Jessica found 5 seashells on the beach. How many seashells did they find together ?","target":21.0}
|
||||
{"input":"Tim's cat had kittens. He gave 3 to Jessica and 6 to Sara. He now has 9 kittens left. How many kittens did he have to start with ?","target":18.0}
|
||||
{"input":"Joan has 9 blue balloons, Sally has 5 blue balloons, and Jessica has 2 blue balloons. How many blue balloons do they have in total ?","target":16.0}
|
||||
{"input":"Melanie had 7 dimes in her bank. Her dad gave her 8 dimes and her mother gave her 4 dimes. How many dimes does Melanie have now ?","target":19.0}
|
||||
{"input":"A restaurant served 5 cakes during lunch and 6 during dinner today. The restaurant served 3 cakes yesterday. How many cakes were served in total ?","target":14.0}
|
||||
{"input":"Melanie picked 4 plums, Dan picked 9 plums, and Sally picked 3 plums from the plum tree. How many plums were picked in total ?","target":16.0}
|
||||
{"input":"There are 7 dogwood trees currently in the park. Park workers will plant 3 more dogwood trees today and 2 more dogwood trees tomorrow. How many dogwood trees will the park have when the workers are finished ?","target":12.0}
|
||||
{"input":"Sara grew 4 onions, Sally grew 5 onions, and Fred grew 9 onions. How many onions did they grow in all ?","target":18.0}
|
||||
{"input":"Sam found 18 seashells and Mary found 47 seashells on the beach. How many seashells did they find together ?","target":65.0}
|
||||
{"input":"Sara picked 45 pears and Sally picked 11 pears from the pear tree. How many pears were picked in total ?","target":56.0}
|
||||
{"input":"Keith has 20 books. Jason has 21 books. How many books do they have together ?","target":41.0}
|
||||
{"input":"Jason had 49 quarters in his bank. His dad gave him 25 more quarters. How many quarters does he have now ?","target":74.0}
|
||||
{"input":"There are 33 walnut trees currently in the park. Park workers will plant 44 more walnut trees today. How many walnut trees will the park have when the workers are finished ?","target":77.0}
|
||||
{"input":"Sara had 21 quarters in her bank. Her dad gave her 49 more quarters. How many quarters does she have now ?","target":70.0}
|
||||
{"input":"There are 41 pencils in the drawer. Mike placed 30 more pencils in the drawer. How many pencils are now there in total ?","target":71.0}
|
||||
{"input":"Joan has 10 books. Tom has 38 books. How many books do they have together ?","target":48.0}
|
||||
{"input":"Joan has 40 blue balloons Melanie has 41 blue balloons. How many blue balloons do they have in total ?","target":81.0}
|
||||
{"input":"Fred grew 38 cantelopes. Tim grew 44 cantelopes. How many cantelopes did they grow in total ?","target":82.0}
|
||||
{"input":"Sam went to 14 football games this year. He went to 29 games last year. How many football games did Sam go to in all ?","target":43.0}
|
||||
{"input":"Mary found 18 seashells and Jessica found 41 seashells on the beach. How many seashells did they find together ?","target":59.0}
|
||||
{"input":"There are 39 dogwood trees currently in the park. Park workers will plant 41 more dogwood trees today and 20 more dogwood trees tomorrow. How many dogwood trees will the park have when the workers are finished ?","target":100.0}
|
||||
{"input":"Sandy has 10 books, Benny has 24 books, and Tim has 33 books. How many books do they have together ?","target":67.0}
|
||||
{"input":"Jason picked 46 pears, Keith picked 47 pears, and Mike picked 12 pears from the pear tree. How many pears were picked in total ?","target":105.0}
|
||||
{"input":"Keith grew 29 cantelopes, Fred grew 16 cantelopes, and Jason grew 20 cantelopes. How many cantelopes did they grow in total ?","target":65.0}
|
||||
{"input":"Melanie had 19 dimes in her bank. Her dad gave her 39 dimes and her mother gave her 25 dimes. How many dimes does Melanie have now ?","target":83.0}
|
||||
{"input":"Alyssa has 37 blue balloons, Sandy has 28 blue balloons, and Sally has 39 blue balloons. How many blue balloons do they have in all ?","target":104.0}
|
||||
{"input":"Sally had 27 cards. Dan gave her 41 new cards. Sally bought 20 cards. How many cards does Sally have now ?","target":88.0}
|
||||
{"input":"Jason went to 11 football games this month. He went to 17 games last month, and plans to go to 16 games next month. How many games will he attend in all ?","target":44.0}
|
||||
{"input":"There are 43 pencils in the drawer and 19 pencils on the desk. Dan placed 16 more pencils on the desk. How many pencils are now there in total ?","target":78.0}
|
||||
{"input":"Mike has 35 books in his library. He bought several books at a yard sale over the weekend. He now has 56 books in his library. How many books did he buy at the yard sale ?","target":21.0}
|
||||
{"input":"Sandy is baking a cake. The recipe wants 7 cups of flour. She already put in 4 cups. How many more cups does she need to add ?","target":3.0}
|
||||
{"input":"There are 53 maple trees currently in the park. Park workers will plant more maple trees today. When the workers are finished there will be 64 maple trees in the park. How many maple trees did the workers plant today ?","target":11.0}
|
||||
{"input":"Dan found 56 seashells on the beach, he gave Jessica some of his seashells. He has 22 seashell left. How many seashells did he give to Jessica ?","target":34.0}
|
||||
{"input":"Sally had 13 peaches left at her roadside fruit stand. She went to the orchard and picked more peaches to stock up the stand. There are now 55 peaches at the stand, how many did she pick ?","target":42.0}
|
||||
{"input":"Benny received 67 dollars for his birthday. He went to a sporting goods store and bought a baseball glove, baseball, and bat. He had 33 dollars left over. How much did he spent on the baseball gear ?","target":34.0}
|
||||
{"input":"There were 3 roses in the vase. Alyssa cut some more roses from her flower garden. There are now 14 roses in the vase. How many roses did she cut ?","target":11.0}
|
||||
{"input":"Last week Tom had 74 dollars. He washed cars over the weekend and now has 86 dollars. How much money did he make washing cars ?","target":12.0}
|
||||
{"input":"There were 73 bales of hay in the barn. Jason stacked more bales in the barn today. There are now 96 bales of hay in the barn. How many bales did he store in the barn ?","target":23.0}
|
||||
{"input":"Nancy grew 6 potatoes. Sandy grew 7 potatoes. How many potatoes did they grow in total ?","target":13.0}
|
||||
{"input":"There are 9 crayons in the drawer. Benny placed 3 more crayons in the drawer. How many crayons are now there in total ?","target":12.0}
|
||||
{"input":"There are 5 oak trees currently in the park. Park workers will plant 4 more oak trees today. How many oak trees will the park have when the workers are finished ?","target":9.0}
|
||||
{"input":"There were a total of 7 football games this year. Melanie missed 4 of the games. How many football games did Melanie go to in all ?","target":3.0}
|
||||
{"input":"Tom found 7 seashells but 4 were broken. How many unbroken seashells did Tom find ?","target":3.0}
|
||||
{"input":"Sally picked 7 lemons and Mary picked 9 lemons from the lemon tree. How many lemons were picked in total ?","target":16.0}
|
||||
{"input":"A restaurant served 6 cakes during lunch and 9 during dinner today. How many cakes were served today ?","target":15.0}
|
||||
{"input":"Joan has 8 orange balloons but lost 2 of them. How many orange balloons does Joan have now ?","target":6.0}
|
||||
{"input":"Fred had 7 dimes in his bank. His sister borrowed 3 of his dimes. How many dimes does Fred have now ?","target":4.0}
|
||||
{"input":"Joan's cat had 8 kittens. She gave 2 to her friends. How many kittens does she have now ?","target":6.0}
|
||||
{"input":"There are 34 dogwood trees currently in the park. Park workers will plant 49 more dogwood trees today. How many dogwood trees will the park have when the workers are finished ?","target":83.0}
|
||||
{"input":"There are 27 pencils in the drawer. Nancy placed 45 more pencils in the drawer. How many pencils are now there in total ?","target":72.0}
|
||||
{"input":"Benny's high school played 39 baseball games this year. He attended 14 games. How many baseball games did Benny miss ?","target":25.0}
|
||||
{"input":"Sam found 35 seashells on the beach, he gave Joan 18 of the seashells. How many seashells does he now have ?","target":17.0}
|
||||
{"input":"Tim has 22 books. Mike has 20 books. How many books do they have together ?","target":42.0}
|
||||
{"input":"Mike has 87 baseball cards. Sam bought 13 of Mike's baseball cards. How many baseball cards does Mike have now ?","target":74.0}
|
||||
{"input":"Sandy grew 51 pumpkins. Mike grew 23 pumpkins. How many pumpkins did they grow in total ?","target":74.0}
|
||||
{"input":"Tim has 44 books. Sam has 52 books. How many books do they have together ?","target":96.0}
|
||||
{"input":"Dan has 64 violet marbles, he gave Mary 14 of the marbles. How many violet marbles does he now have ?","target":50.0}
|
||||
{"input":"There are 25 popular trees currently in the park. Park workers will plant 73 more popular trees today. How many popular trees will the park have when the workers are finished ?","target":98.0}
|
||||
{"input":"There are 54 scissors in the drawer. Keith placed 22 more scissors in the drawer. How many scissors are now there in all ?","target":76.0}
|
||||
{"input":"Alyssa picked 42 pears and Nancy picked 17 pears from the pear tree. How many pears were picked in all ?","target":59.0}
|
||||
{"input":"Sam had 98 pennies in his bank. He spent 93 of his pennies. How many pennies does he have now ?","target":5.0}
|
||||
{"input":"Joan found 79 seashells on the beach, she gave Mike 63 of the seashells. How many seashells does she now have ?","target":16.0}
|
||||
{"input":"Melanie's high school played 64 soccer games this year. She attended 32 games. How many soccer games did Melanie miss ?","target":32.0}
|
||||
{"input":"Sam has 110 books. Joan has 102 books. How many books do they have together ?","target":212.0}
|
||||
{"input":"Mary picked 122 oranges and Jason picked 105 oranges from the orange tree. How many oranges were picked in total ?","target":227.0}
|
||||
{"input":"Sally had 760 quarters in her bank. She spent 418 of her quarters. How many quarters does she have now ?","target":342.0}
|
||||
{"input":"Fred's high school played 616 baseball games this year. He attended 147 games. How many baseball games did Fred miss ?","target":469.0}
|
||||
{"input":"Melanie grew 139 turnips. Benny grew 113 turnips. How many turnips did they grow in all ?","target":252.0}
|
||||
{"input":"Jason has 676 Pokemon cards. Alyssa bought 224 of Jason's Pokemon cards. How many Pokemon cards does Jason have now ?","target":452.0}
|
||||
{"input":"There are 107 walnut trees currently in the park. Park workers will plant 104 more walnut trees today. How many walnut trees will the park have when the workers are finished ?","target":211.0}
|
||||
{"input":"Fred has 709 green balloons, he gave Sandy 221 of the balloons. How many green balloons does he now have ?","target":488.0}
|
||||
{"input":"There are 115 pencils in the drawer. Sara placed 100 more pencils in the drawer. How many pencils are now there in all ?","target":215.0}
|
||||
{"input":"Mike has 45 dollars in 5 dollar bills. How many five dollars bills does he have ?","target":9.0}
|
||||
{"input":"Nancy goes fishing with Joan. They catch 18 trout. If they equally split up the trout, how many will each one get ?","target":9.0}
|
||||
{"input":"A restaurant sold 63 hamburgers last week. How many hamburgers on average were sold each day ?","target":9.0}
|
||||
{"input":"Sandy worked 45 hours in the last 5 days. Assuming that she worked the same amount of hours each day, how long did she work each day ?","target":9.0}
|
||||
{"input":"There were a total of 27 soccer games during the 3 month season. If the games are equally divided, how many soccer games are played a month ?","target":9.0}
|
||||
{"input":"Fred has 90 cents in his bank. How many dimes does Fred have ?","target":9.0}
|
||||
{"input":"Fred has 110 blue marbles. Fred has 22 times more blue marbles than Tim. How many blue marbles does Tim have?","target":5.0}
|
||||
{"input":"There are 390 students at a school. If each classroom holds 30 students, how many classrooms are needed at the school?","target":13.0}
|
||||
{"input":"Mike has 96 muffins, which he needs to box up into dozens. How many boxes does he need?","target":8.0}
|
||||
{"input":"Tim has saved 2000 cents from selling lemonade. How many dollars does Tim have?","target":20.0}
|
||||
{"input":"Jason's shelves hold 45 books each. How many shelves will Jason need if Jason has 315 books?","target":7.0}
|
||||
{"input":"Nancy bought 615 crayons that came in packs of 15. How many packs of crayons did Nancy buy?","target":41.0}
|
||||
{"input":"Mary earns $46 cleaning a home. How many homes did she clean, if she made 276 dollars?","target":6.0}
|
||||
{"input":"There were a total of 323 football games in the season. The season is played for 17 months. How many football games were played each month, if each month has the same number of games?","target":19.0}
|
||||
{"input":"Nancy, Melanie, Mary, and Alyssa each bought 540 baseball cards, which come in packs of 20. How many packs of baseball cards do they have in all?","target":108.0}
|
||||
{"input":"A teacher has 344 pieces of candy. If there are 43 students, and the candy is divided evenly, How many pieces will each student get?","target":8.0}
|
||||
{"input":"Sally earns $25.00 for each house she cleans. If she cleans 96 houses, how many dozens of dollars will she make?","target":200.0}
|
||||
{"input":"There are 60 calories in a candy bar. How many dozen calories are there in 45 candy bars?","target":225.0}
|
||||
{"input":"Sara saw 96 birds in a tree. How many dozen birds did Sara see?","target":8.0}
|
||||
{"input":"Melanie has 84 blue marbles. Sandy has 8 times more blue marbles than Melanie. How many dozen blue marbles does Sandy have?","target":56.0}
|
||||
{"input":"Sara has 192 golf balls. How many dozen golf balls does she have?","target":16.0}
|
||||
{"input":"Keith bought 72 eggs from the store to bake some cakes. How many dozen eggs did Keith buy?","target":6.0}
|
||||
{"input":"Mike has 48 books. Alyssa has 8 times more books than Mike. How many dozen books does Alyssa have?","target":32.0}
|
||||
{"input":"Sara has saved 9 dollars from washing cars. How many dozen quarters does Sara have?","target":3.0}
|
||||
{"input":"Sara, Keith, Benny, and Alyssa each have 96 baseball cards. How many dozen baseball cards do they have in all?","target":32.0}
|
||||
{"input":"Jason has 3 Pokemon cards. Benny bought 2 of Jason's Pokemon cards. How many Pokemon cards does Jason have now ?","target":1.0}
|
||||
{"input":"Mike has 8 orange marbles, he gave Sam 4 of the marbles. How many orange marbles does he now have ?","target":4.0}
|
||||
{"input":"Joan had 5 dimes in her bank. She spent 2 of her dimes. How many dimes does she have now ?","target":3.0}
|
||||
{"input":"There are 2 rose bushes currently in the park. Park workers will plant 4 more rose bushes today. How many rose bushes will the park have when the workers are finished ?","target":6.0}
|
||||
{"input":"Sara picked 6 pears and Tim picked 5 pears from the pear tree. How many pears were picked in total ?","target":11.0}
|
||||
{"input":"Sam grew 4 watermelons, but the rabbits ate 3 watermelons. How many watermelons does Sam have left ?","target":1.0}
|
||||
{"input":"There are 3 calories in a candy bar. How many calories are there in 5 candy bars ?","target":15.0}
|
||||
{"input":"Dan has 5 blue marbles. Mary has 2 times more blue marbles than Dan. How many blue marbles does Mary have ?","target":10.0}
|
||||
{"input":"Sara has 9 dozen golf balls. How many golf balls does she have ?","target":108.0}
|
||||
{"input":"Jason had Pokemon cards. He gave 9 Pokemon cards to his friends. He now has 4 Pokemon cards left. How many Pokemon cards did he have to start with ?","target":13.0}
|
||||
{"input":"Mary had 7 nickels in her bank. Her dad gave her 5 more nickels. How many nickels does Mary have now ?","target":12.0}
|
||||
{"input":"Melanie, Benny, Sally, and Jessica each have 3 baseball cards. How many baseball cards do they have in all ?","target":12.0}
|
||||
{"input":"Keith grew 6 turnips. Alyssa grew 9 turnips. How many turnips did they grow in all ?","target":15.0}
|
||||
{"input":"There are 4 children in the classroom, each student will get 2 pencils. How many pencils will the teacher have to give out ?","target":8.0}
|
||||
{"input":"Mary has 9 yellow marbles Joan has 3 yellow marbles. How many yellow marbles do they have in all ?","target":12.0}
|
||||
{"input":"There are 8 calories in a candy bar. How many calories are there in 3 candy bars ?","target":24.0}
|
||||
{"input":"Joan has saved 6 quarters from washing cars. How many cents does Joan have ?","target":150.0}
|
||||
{"input":"There are 960 students at a school. If each classroom holds 30 students, how many classrooms are needed at the school?","target":32.0}
|
||||
{"input":"Sally earns $12.50 an hour cleaning houses. If she works for 12 hours, how much money will she make ?","target":150.0}
|
||||
{"input":"There were a total of 12 basketball games in the season. The season is played for 2 months. How many basketball games were played each month, if each month has the same number of games?","target":6.0}
|
||||
{"input":"Joan bought 6 dozen eggs from the grocery store to bake some cakes. How many eggs did Joan buy ?","target":72.0}
|
||||
{"input":"Mary, Sam, Keith, and Alyssa each have 6 baseball cards. How many baseball cards do they have in all ?","target":24.0}
|
||||
{"input":"Sally bought 4 dozen eggs from the grocery store to bake some cakes. How many eggs did Sally buy ?","target":48.0}
|
||||
{"input":"There are 1110 students at a school. If each classroom holds 30 students, how many classrooms are needed at the school?","target":37.0}
|
||||
{"input":"Sally grew 6 carrots. Fred grew 4 carrots. How many carrots did they grow in all ?","target":10.0}
|
||||
{"input":"Benny has 6 blue marbles. Keith has 5 times more blue marbles than Benny. How many blue marbles does Keith have ?","target":30.0}
|
||||
{"input":"Fred earns $12.50 an hour cleaning houses. If he works for 8 hours, how much money will he make ?","target":100.0}
|
||||
{"input":"Tom found 5 seashells on the beach. he gave Jessica 2 of the seashells. How many seashells does he now have ?","target":3.0}
|
||||
{"input":"Fred has 5 baseball cards. Melanie bought 3 of Fred's baseball cards. How many baseball cards does Fred have now ?","target":2.0}
|
||||
{"input":"Nancy has saved 4900 cents from selling lemonade. How many dollars does Nancy have?","target":49.0}
|
||||
{"input":"There were a total of 10 soccer games in the season. The season is played for 5 months. How many soccer games were played each month, if each month has the same number of games?","target":2.0}
|
||||
{"input":"Benny goes out to lunch with Sara and Tom. Each person orders the $8 lunch special. Benny agrees to pay the bill. How much will he have to pay ?","target":24.0}
|
||||
{"input":"Melanie goes fishing with Tom. Melanie catches 8 trout. Tom catches 2 times as many trout as Melanie. How many trout did Tom catch ?","target":16.0}
|
||||
{"input":"Tom was at the beach for 5 days and found 7 seashells every day. How many seashells did Tom find during the beach trip ?","target":35.0}
|
||||
{"input":"Benny worked 3 hours for 6 days. How many hours did he work in total ?","target":18.0}
|
||||
{"input":"A restaurant sold 8 pies every day for a week. How many pies were sold during the week ?","target":56.0}
|
||||
{"input":"Nancy has 9 five dollars bills. How much money does she have ?","target":45.0}
|
||||
{"input":"Nancy has 7 black balloons. Mary has 4 times more black balloons than Nancy. How many black balloons does Mary have now ?","target":28.0}
|
||||
{"input":"Jessica, Sandy, and Jason each have 8 pencils. How many pencils do they have have in all ?","target":24.0}
|
||||
{"input":"Sam, Dan, Tom, and Keith each have 14 Pokemon cards. How many Pokemon cards do they have in all ?","target":56.0}
|
||||
{"input":"Alyssa has 36 books. Nancy has 7 times more books than Alyssa. How many books does Nancy have ?","target":252.0}
|
||||
{"input":"Tim has 13 dozen golf balls. How many golf balls does he have ?","target":156.0}
|
||||
{"input":"There were a total of 13 hockey games a month. The season is played for 14 months. How many hockey games are in the seasons ?","target":182.0}
|
||||
{"input":"Benny bought 7 dozen eggs from the grocery store to bake some cakes. How many eggs did Benny buy ?","target":84.0}
|
||||
{"input":"Dan has 29 violet balloons. Tim has 7 times more violet balloons than Dan. How many violet balloons does Tim have ?","target":203.0}
|
||||
{"input":"There are 31 calories in a candy bar. How many calories are there in 11 candy bars ?","target":341.0}
|
||||
{"input":"Sara has saved 11 quarters from washing cars. How many cents does Sara have ?","target":275.0}
|
||||
{"input":"There are 46 children in the classroom, each student will get 4 dozen pencils. How many pencils will the teacher have to give out ?","target":2208.0}
|
||||
{"input":"Nancy has saved 1 dozen quarters from washing cars. How much money does Nancy have ?","target":3.0}
|
||||
{"input":"Dan bought 9 dozen eggs from the grocery store to bake some cakes. How many eggs did Dan buy ?","target":108.0}
|
||||
{"input":"There are 4 dozen calories in a candy bar. How many calories are there in 42 candy bars ?","target":2016.0}
|
||||
{"input":"Sandy has 8 dozen books. Fred has 5 times more books than Sandy. How many books does Fred have ?","target":480.0}
|
||||
{"input":"Jessica has 3 dozen red marbles. Sandy has 4 times more red marbles than Jessica. How many red marbles does Sandy have ?","target":144.0}
|
||||
{"input":"Melanie, Benny, Sandy, and Jessica each have 9 dozen Pokemon cards. How many Pokemon cards do they have in all ?","target":432.0}
|
||||
{"input":"Sally saw 1 dozen birds in a tree. How many birds did Sally see ?","target":12.0}
|
||||
{"input":"Fred, Benny, and Jason have 24 crayons all together. If the crayons are equally divided, how many will each person get ?","target":8.0}
|
||||
{"input":"Sara goes fishing with Melanie. Sara catches 5 trout. Melanie catches 2 times as many trout as Sara. How many trout did Melanie catch ?","target":10.0}
|
||||
{"input":"Benny goes to lunch with Sally and Sandy. The total bill came to 15 dollars. They decided to equally split up the bill, how much will each person have to pay ?","target":5.0}
|
||||
{"input":"A restaurant sold 49 hamburgers last week. How many hamburgers on average were sold each day ?","target":7.0}
|
||||
{"input":"Sally has 6 blue balloons. Fred has 3 times more blue balloons than Sally. How many blue balloons does Fred have now ?","target":18.0}
|
||||
{"input":"Mike worked 3 hours, each day, for 5 days. How many hours did he work in total ?","target":15.0}
|
||||
{"input":"There were a total of 7 baseball games a month. The season is played for 2 months. How many baseball games are in a season ?","target":14.0}
|
||||
{"input":"Melanie is selling 4 gumballs for eight cents each. How much money can Melanie get from selling the gumballs?","target":32.0}
|
||||
{"input":"Mary had 8 potatoes in the garden. The rabbits ate 3 of the potatoes. How many potatoes does Mary now have ?","target":5.0}
|
||||
{"input":"There were a total of 6 soccer games this year. Jessica missed 4 of the games. How many soccer games did Jessica go to in all ?","target":2.0}
|
||||
{"input":"There are 9 oak trees currently in the park. Park workers had to cut down 2 oak trees that were damaged. How many oak trees will the park have when the workers are finished ?","target":7.0}
|
||||
{"input":"Jessica had 8 quarters in her bank. Her sister borrowed 3 of her quarters. How many quarters does Jessica have now ?","target":5.0}
|
||||
{"input":"A restaurant made 9 hamburgers to serve during lunch. Only 3 were actually served. How many hamburgers were left over from lunch ?","target":6.0}
|
||||
{"input":"There are 7 crayons in the drawer. Mary took 3 crayons out of the drawer. How many crayons are there now ?","target":4.0}
|
||||
{"input":"Dan picked 9 limes and gave Sara 4 of the limes. How many limes does Dan have now ?","target":5.0}
|
||||
{"input":"Dan found 7 seashells but 3 were broken. How many unbroken seashells did Dan find ?","target":4.0}
|
||||
{"input":"Joan has 9 blue balloons but lost 2 of them. How many blue balloons does Joan have now ?","target":7.0}
|
||||
{"input":"Joan picked 43 apples from the orchard, and gave 27 apples to Melanie. How many apples does Joan have now ?","target":16.0}
|
||||
{"input":"Alyssa's high school played 31 hockey games this year. She attended 13 games. How many hockey games did Alyssa miss ?","target":18.0}
|
||||
{"input":"Tom has 30 violet balloons, he gave Fred 16 of the balloons. How many violet balloons does he now have ?","target":14.0}
|
||||
{"input":"Fred has 40 baseball cards. Keith bought 22 of Fred's baseball cards. How many baseball cards does Fred have now ?","target":18.0}
|
||||
{"input":"Fred found 47 seashells on the beach, he gave Jessica 25 of the seashells. How many seashells does he now have ?","target":22.0}
|
||||
{"input":"Sara grew 43 pumpkins, but the rabbits ate 23 pumpkins. How many pumpkins does Sara have left ?","target":20.0}
|
||||
{"input":"Joan decided to sell all of her old books. She gathered up 33 books to sell. She sold 26 books in a yard sale. How many books does Joan now have ?","target":7.0}
|
||||
{"input":"There are 46 rulers in the drawer. Tim took 25 rulers from the drawer. How many rulers are now in the drawer ?","target":21.0}
|
||||
{"input":"There are 33 oak trees currently in the park. Park workers had to cut down 18 oak trees that were damaged. How many oak trees will be in the park when the workers are finished ?","target":15.0}
|
||||
{"input":"Sandy sold lemonade in her neighborhood. She got 17 half-dollars on Saturday and 6 half-dollars on Sunday. What amount of money did Sandy receive?","target":11.5}
|
||||
{"input":"Sam got 9 pennies for washing clothes, and 7 quarters for mowing lawns. How much money does Sam have?","target":1.84}
|
||||
{"input":"When Joan was visited by the toothfairy, she received 14 each of quarters, half-dollars, and dimes. How much money did the toothfairy leave Joan?","target":11.9}
|
||||
{"input":"As Alyssa was searching through her couch cushions, she found 12 quarters, and 7 pennies in the couch. How much money in total does Alyssa have?","target":3.07}
|
||||
{"input":"On Wednesday, Joan spent 4 half-dollars playing pinball. The next day, she spent 14 half-dollars on pinball. What was the total amount Joan spent playing pinball?","target":9.0}
|
||||
{"input":"Tim got 3 nickels and 13 dimes for shining shoes, and in his tip jar found 7 dimes and 9 half-dollars. How much money did Tim get?","target":6.65}
|
||||
{"input":"While digging through her clothes for ice cream money, Joan found 15 dimes in her jacket, and 4 dimes in her shorts. How much money did Joan find?","target":1.9}
|
||||
{"input":"On Friday, Sam spent 2 pennies on ice cream. The next day, Sam spent 12 dimes on baseball cards. All in all, how much money did Sam spend?","target":1.22}
|
||||
{"input":"Joan purchased a basketball game for $5.20, and a racing game for $4.23. How much did Joan spend on video games?","target":9.43}
|
||||
{"input":"Mike joined his school's band. He bought a trumpet for $145.16, and a song book which was $5.84. How much did Mike spend at the music store?","target":151.0}
|
||||
{"input":"Alyssa bought some toys. She bought a football for $5.71, and spent $6.59 on marbles. In total, how much did Alyssa spend on toys?","target":12.3}
|
||||
{"input":"Keith loves trading cards. She bought 4 packs of Digimon cards for $4.45 each, and a deck of baseball cards for $6.06. How much did Keith spend on cards?","target":23.86}
|
||||
{"input":"Jessica spent $10.22 on a cat toy, and a cage cost her $11.73. What was the total cost of Jessica's purchases?","target":21.95}
|
||||
{"input":"Sara got fast food for lunch. Sara spent $5.36 on a hotdog and $5.10 on a salad. What was the total of the lunch bill?","target":10.46}
|
||||
{"input":"Jason went to the mall on Saturday to buy clothes. He spent $14.28 on shorts and $4.74 on a jacket. In total, how much money did Jason spend on clothing?","target":19.02}
|
||||
{"input":"Alyssa loves eating fruits. Alyssa paid $12.08 for grapes, and $9.85 for cherries. In total, how much money did Alyssa spend?","target":21.93}
|
||||
{"input":"Mary loves eating fruits. Mary paid $11.08 for berries, $14.33 for apples, and $9.31 for peaches. In total, how much money did she spend?","target":34.72}
|
||||
{"input":"Sandy went to the mall to buy clothes. She spent $13.99 on shorts, $12.14 on a shirt, and $7.43 on a jacket. How much money did Sandy spend on clothes?","target":33.56}
|
||||
{"input":"Jason joined his school's band. He bought a flute for $142.46, a music stand for $8.89, and a song book for $7. How much did Jason spend at the music store?","target":158.35}
|
||||
{"input":"Tom purchased a football game for $14.02, a strategy game for $9.46, and a Batman game for $12.04. How much did Tom spend on video games?","target":35.52}
|
||||
{"input":"Fred loves trading cards. He bought 2 packs of football cards for $2.73 each, a pack of Pokemon cards for $4.01, and a deck of baseball cards for $8.95. How much did Fred spend on cards?","target":18.42}
|
||||
{"input":"On Saturday, Sara spent $10.62 each on 2 tickets to a movie theater. She also rented a movie for $1.59, and bought a movie for $13.95. How much money in total did Sara spend on movies?","target":36.78}
|
||||
{"input":"Mike bought some toys. He bought marbles for $9.05, a football for $4.95, and spent $6.52 on a baseball. In total, how much did Mike spend on toys?","target":20.52}
|
||||
{"input":"Mary loves eating fruits. Mary paid $7.19 for berries, and $6.83 for peaches with a $20 bill. How much change did Mary receive?","target":5.98}
|
||||
{"input":"Sandy went to the mall on Saturday to buy clothes. She paid $9.24 on pants and $8.25 on a shirt with a $20 bill. How much money did Sandy get in change?","target":2.51}
|
||||
{"input":"Joan paid $8.77 on a cat toy, and a cage cost her $10.97 with a $20 bill. How much change did Joan receive?","target":0.26}
|
||||
{"input":"On Friday, Fred paid $5.92 each on 2 tickets to a movie theater. He also borrowed a movie for $6.79. Fred paid with a $20 bill. How much change did Fred receive?","target":1.37}
|
||||
{"input":"Sandy bought some toys. She bought a football for $9.14, and paid $6.81 on a baseball with a $20 bill. How much change did he receive from the purchase?","target":4.05}
|
||||
{"input":"A ship is filled with 5,973 tons of cargo. It stops in the Bahamas, where sailors load 8,723 more tons of cargo onboard. How many tons of cargo does the ship hold now?","target":14696.0}
|
||||
{"input":"Before December, customers buy 1,346 ear muffs from the mall. During December, they buy 6,444 more, and there are none left. In all, how many ear muffs do the customers buy?","target":7790.0}
|
||||
{"input":"Diane is a beekeeper. Last year, she harvested 2,479 pounds of honey. This year, she bought some new hives and increased her honey harvest by 6,085 pounds. How many pounds of honey did Diane harvest this year?","target":8564.0}
|
||||
{"input":"An oil pipe in the sea broke. Before engineers started to fix the pipe, 6,522 liters of oil leaked into the water. While the engineers worked, the pipe leaked 5,165 more liters of oil. In all, how many liters of oil leaked into the water?","target":11687.0}
|
||||
{"input":"A car company produced 3,884 cars in North America and 2,871 cars in Europe. How many cars is that in all?","target":6755.0}
|
||||
{"input":"Abe's family moved from the Bahamas to Japan, so they had convert their money into Japanese yen. Their checking account now has 6,359 yen and their savings account now has 3,485 yen. How many yen do they have?","target":9844.0}
|
||||
{"input":"There are 1,986 books in Oak Grove's public library. In addition, there are 5,106 books in its school libraries. How many books do the libraries in Oak Grove have overall?","target":7092.0}
|
||||
{"input":"There were originally 20,817 houses in Lincoln County. During a housing boom, developers built 97,741 more. How many houses are there now in Lincoln County?","target":118558.0}
|
||||
{"input":"A farmer estimates that he will harvest 48,097 bushels of wheat. The weather is perfect during the growing season, so he harvests 684 more bushels of wheat than expected. How many bushels of wheat does the farmer harvest?","target":48781.0}
|
||||
{"input":"Christina just transferred $69 out of her bank account. As a result, the account now has $26,935 left in it. How much money was in the account before the transfer?","target":27004.0}
|
||||
{"input":"Last year at Newberg's airport, 14,507 passengers landed on time. Unfortunately, 213 passengers landed late. In all, how many passengers landed in Newberg last year?","target":14720.0}
|
||||
{"input":"A dust storm sweeps across the prairie. It covers 64,535 acres of the prairie in dust, but leaves 522 acres untouched. How many acres does the prairie cover?","target":65057.0}
|
||||
{"input":"Some insects called aphids attack a large farm. In response, the farmer releases ladybugs onto the fields. There are 12,170 ladybugs with spots and 54,912 ladybugs without spots. How many ladybugs are there in all?","target":67082.0}
|
||||
{"input":"Last year, 90,171 people were born in a country, and 16,320 people immigrated to it. How many new people began living in the country last year?","target":106491.0}
|
||||
{"input":"A ship full of grain crashes into a coral reef. By the time the ship is fixed, 49,952 tons of grain have spilled into the water. Only 918 tons of grain remain onboard. How many tons of grain did the ship originally contain?","target":50870.0}
|
||||
{"input":"To fill an order, the factory dyed 61,921 yards of silk green and 49,500 yards pink. How many yards of silk did it dye for that order?","target":111421.0}
|
||||
{"input":"A multi-national corporation has 2,041 part-time employees and 63,093 full-time employees. How many employees work for the corporation?","target":65134.0}
|
||||
{"input":"Each year, salmon travel upstream, going from the ocean to the rivers where they were born. This year, 712,261 male and 259,378 female salmon returned to their rivers. How many salmon made the trip?","target":971639.0}
|
||||
{"input":"A bathing suit manufacturer has a supply of 14,797 bathing suits for men. In addition, it has 4,969 bathing suits for women. How many bathing suits are available overall?","target":19766.0}
|
||||
{"input":"Before the recent housing boom, there were 1,426 houses in Lawrence County. Now, there are 2,000 houses. How many houses did developers build during the housing boom?","target":574.0}
|
||||
{"input":"A worker at a medical lab is studying blood samples. Two samples contained a total of 7,341 blood cells. The first sample contained 4,221 blood cells. How many blood cells were in the second sample?","target":3120.0}
|
||||
{"input":"So far, an orchard has sold a combined total of 9,792 pounds of fresh and frozen fruit this season. If they have sold 3,513 pounds of frozen fruit, how many pounds of fresh fruit have been sold so far?","target":6279.0}
|
||||
{"input":"On Saturday, Sara spent $10.62 each on 2 tickets to a movie theater. Sara also rented a movie for $1.59, and bought a movie for $13.95. How much money in total did Sara spend on movies?","target":36.78}
|
||||
{"input":"Jonathan wants to buy a dictionary that costs $11, a dinosaur book that costs $19, and a children's cookbook that costs $7. He has saved $8 from his allowance. How much more money does Jonathan need to buy all three books?","target":29.0}
|
||||
{"input":"Dana earns $13 per hour. She worked 9 hours on Friday, 10 hours on Saturday, and 3 hours on Sunday. How much money did Dana earn in all?","target":286.0}
|
||||
{"input":"Tanner saved $17 in September. He saved $48 in October and $25 in November. Then Tanner spent $49 on a video game. How much money does Tanner have left?","target":41.0}
|
||||
{"input":"Mika had 20 stickers. She bought 26 stickers from a store in the mall and got 20 stickers for her birthday. Then Mika gave 6 of the stickers to her sister and used 58 to decorate a greeting card. How many stickers does Mika have left?","target":2.0}
|
||||
{"input":"A group of science students went on a field trip. They took 6 vans and 8 buses. There were 6 people in each van and 18 people in each bus. How many people went on the field trip?","target":180.0}
|
||||
{"input":"Carrie's mom gave her $91 to go shopping. She bought a sweater for $24, a T-shirt for $6, and a pair of shoes for $11. How much money does Carrie have left?","target":50.0}
|
||||
{"input":"Kylie was collecting coins. She got 15 coins from her piggy bank and 13 coins from her brother. Her father gave Kylie 8 coins. Kylie gave 21 of the coins to her friend Laura. How many coins did Kylie have left?","target":15.0}
|
||||
{"input":"Justin needs 61 paper plates for a birthday party. He already has 26 blue plates and 7 red plates. How many more plates should Justin buy?","target":28.0}
|
||||
{"input":"Mark's father gave him $85. Mark bought 10 books, each of which cost $5. How much money does Mark have left?","target":35.0}
|
||||
{"input":"Elise had $8. Then she saved $13 from her allowance and spent $2 on a comic book and $18 on a puzzle. How much money does Elise have left?","target":1.0}
|
||||
{"input":"Luke had 20 stickers. He bought 12 stickers from a store in the mall and got 20 stickers for his birthday. Then Luke gave 5 of the stickers to his sister and used 8 to decorate a greeting card. How many stickers does Luke have left?","target":39.0}
|
||||
{"input":"Johnny saved $30 in September. He saved $49 in October and $46 in November. Then Johnny spent $58 on a video game. How much money does Johnny have left?","target":67.0}
|
||||
{"input":"Seth bought 20 cartons of ice cream and 2 cartons of yogurt. Each carton of ice cream cost $6 and each carton of yogurt cost $1. How much more did Seth spend on ice cream than on yogurt?","target":118.0}
|
||||
{"input":"Dalton wants to buy a jump rope that costs $7, a board game that costs $12, and a playground ball that costs $4. He has saved $6 from his allowance, and his uncle gave him $13. How much more money does Dalton need to buy the jump rope, the game, and the ball?","target":4.0}
|
||||
{"input":"Vincent bought 10 books about animals, 1 book about outer space, and 3 books about trains. Each book cost $16. How much did Vincent spend on the books?","target":224.0}
|
||||
{"input":"Priya needs 54 cupcakes for a birthday party. She already has 15 chocolate cupcakes and 25 vanilla cupcakes. How many more cupcakes should Priya buy?","target":14.0}
|
||||
{"input":"Maria needs 21 cartons of berries to make a berry cobbler. She already has 4 cartons of strawberries and 8 cartons of blueberries. How many more cartons of berries should Maria buy?","target":9.0}
|
||||
{"input":"There are 24 bicycles and 14 tricycles in the storage area at Danny's apartment building. Each bicycle has 2 wheels and each tricycle has 3 wheels. How many wheels are there in all?","target":90.0}
|
||||
{"input":"Paula's aunt gave her $109 to spend on clothes at the mall. She bought 2 shirts that cost $11 each and a pair of pants that cost $13. How much money does Paula have left to buy more clothes?","target":74.0}
|
||||
{"input":"Mary has 9 yellow marbles. Joan has 3 yellow marbles. How many yellow marbles do they have in all ?","target":12.0}
|
||||
{"input":"Jason had Pokemon cards. Jason gave 9 to his friends. Jason now has 4 Pokemon cards left. How many Pokemon cards did Jason have to start with ?","target":13.0}
|
||||
{"input":"Adam bought 9 packages of cat food and 7 packages of dog food. Each package of cat food contained 10 cans, and each package of dog food contained 5 cans. How many more cans of cat food than dog food did Adam buy?","target":55.0}
|
||||
{"input":"Zach wants to ride the Ferris wheel, the roller coaster, and the log ride. The Ferris wheel costs 2 tickets, the roller coaster costs 7 tickets and the log ride costs 1 ticket. Zach has 1 ticket. How many more tickets should Zach buy?","target":9.0}
|
||||
{"input":"Randy needs 53 cupcakes for a birthday party. He already has 7 chocolate cupcakes and 19 vanilla cupcakes. How many more cupcakes should Randy buy?","target":32.0}
|
||||
{"input":"Maggie bought 4 packs of red bouncy balls, 8 packs of yellow bouncy balls, and 4 packs of green bouncy balls. There were 10 bouncy balls in each package. How many bouncy balls did Maggie buy in all?","target":160.0}
|
||||
{"input":"Your class is having a pizza party. You buy 5 pizzas. Each pizza has 4 slices. How many slices is that altogether?","target":20.0}
|
||||
{"input":"My car gets 20 miles per gallon of gas. How many miles can I drive on 5 gallons of gas?","target":100.0}
|
||||
{"input":"58 children are taking a bus to the zoo. They sit 2 children in every seat. How many seats will the children need in all?","target":29.0}
|
||||
{"input":"Marlee has 12 guests coming to her Halloween party. Each table will hold 3 guests. How many tables will Marlee need?","target":4.0}
|
||||
{"input":"Ellen went to a garage sale to buy chairs. Each chair is 15 dollars. How much did Ellen spend for the 12 chairs she bought?","target":180.0}
|
||||
{"input":"Oscar's bus ride to school is 0.75 of a mile and Charlie's bus ride is 0.25 of a mile. How much longer is Oscar's bus ride than Charlie's?","target":0.5}
|
||||
{"input":"Karen added 0.25 of a cup of walnuts to a batch of trail mix. Later, she added 0.25 of a cup of almonds. How many cups of nuts did Karen put in the trail mix in all?","target":0.5}
|
||||
{"input":"Kendall is learning to drive, so this weekend she practiced driving 0.16666666666666666 of a mile with her mother and another 0.5 of a mile with her father. How far did Kendall drive in all?","target":0.6666666667}
|
||||
{"input":"At a pie-eating contest, Erik got through 0.6666666666666666 of a pie before time was called; Frank finished just 0.3333333333333333 of a pie. How much more pie did Erik eat than Frank?","target":0.3333333333}
|
||||
{"input":"A tailor cut 0.75 of an inch off a skirt and 0.5 of an inch off a pair of pants. How much more did the tailor cut off the skirt than the pants?","target":0.25}
|
||||
{"input":"At Lindsey's Vacation Wear, 0.375 of the garments are bikinis and 0.25 are trunks. What fraction of the garments are either bikinis or trunks?","target":0.625}
|
||||
{"input":"Darnel sprinted 0.875 of a lap and then took a break by jogging 0.75 of a lap. How much farther did Darnel sprint than jog?","target":0.125}
|
||||
{"input":"A marine biologist measured one fish that was 0.3 of a foot long and a second fish that was 0.2 of a foot long. How much longer was the first fish?","target":0.1}
|
||||
{"input":"Hannah's Vegetarian Restaurant bought 0.3333333333333333 of a pound of green peppers and 0.3333333333333333 of a pound of red peppers. How many pounds of peppers did Hannah's Vegetarian Restaurant buy in all?","target":0.6666666667}
|
||||
{"input":"Ella owns two dogs. Each day, one dog eats 0.125 of a scoop of dog food and the other dog eats 0.125 of a scoop. Together, how much dog food do the two dogs eat each day?","target":0.25}
|
||||
{"input":"Blake filled a bucket with 0.8 of a gallon of water. Later, he poured out 0.2 of a gallon of the water. How much water is left in the bucket?","target":0.6}
|
||||
{"input":"Mandy made an apple pie. She used 0.6666666666666666 of a tablespoon of cinnamon and 0.5 of a tablespoon of nutmeg. How much more cinnamon than nutmeg did Mandy use?","target":0.1666666667}
|
||||
{"input":"The Montoya family spends 0.6 of their budget on groceries and another 0.2 going out to eat. Altogether, what fraction of their budget does the Montoya family spend on food?","target":0.8}
|
||||
{"input":"In Mr. Olsen's mathematics class, 0.7 of the students received A's and 0.2 received B's. What fraction of the students received either A's or B's?","target":0.9}
|
||||
{"input":"There is 0.16666666666666666 of a cup of oil in Scarlett's measuring cup. If Scarlett adds 0.6666666666666666 of a cup more, how much oil will be in the measuring cup?","target":0.8333333333}
|
||||
{"input":"One evening, a restaurant served a total of 0.2 of a loaf of wheat bread and 0.4 of a loaf of white bread. How many loaves were served in all?","target":0.6}
|
||||
{"input":"Stanley ran 0.4 of a mile and walked 0.2 of a mile. How much farther did Stanley run than walk?","target":0.2}
|
||||
{"input":"Dina made cookies. She used 0.625 of a cup of flour and 0.25 of a cup of sugar. How much more flour than sugar did Dina use?","target":0.375}
|
||||
{"input":"Each day, the polar bear at Richmond's zoo eats 0.2 of a bucket of trout and 0.4 of a bucket of salmon. How many buckets of fish does the polar bear eat daily?","target":0.6}
|
||||
{"input":"Jenny ran 0.6 of a mile and walked 0.4 of a mile. How much farther did Jenny run than walk?","target":0.2}
|
||||
{"input":"0.5 of the students in the band are in the trumpet section. 0.125 of the students in the band are in the trombone section. What fraction of the students in the band are in either the trumpet section or the trombone section?","target":0.625}
|
||||
{"input":"Suzie found two worms in the yard and measured them with a ruler. One worm was 0.8 of an inch long. The other worm was 0.1 of an inch long. How much longer was the longer worm?","target":0.7}
|
||||
{"input":"Scarlett made a fruit salad with 0.25 of a pound of melon and 0.375 of a pound of berries. How many pounds of fruit did Scarlett use in all?","target":0.625}
|
||||
{"input":"Vince's bus ride to school is 0.625 of a mile and Zachary's bus ride is 0.5 of a mile. How much longer is Vince's bus ride than Zachary's?","target":0.125}
|
||||
{"input":"In one week, Mitch's family drank 0.5 of a carton of regular milk and 0.1 of a carton of soy milk. How much milk did they drink in all?","target":0.6}
|
||||
{"input":"Elizabeth went to the salon and had 0.375 of an inch of hair cut off. The next day she went back and asked for another 0.5 of an inch to be cut off. How much hair did she have cut off in all?","target":0.875}
|
||||
{"input":"In Shannon's apartment complex, 0.16666666666666666 of the apartments are one-bedroom apartments and 0.3333333333333333 are two-bedroom apartments. What fraction of the apartments are either one- or two-bedroom apartments?","target":0.5}
|
||||
{"input":"At the beach, Miki and her sister both built sandcastles and then measured their heights. Miki's sandcastle was 0.8333333333333334 of a foot tall and her sister's was 0.5 of a foot tall. How much taller was Miki's sandcastle than her sister's?","target":0.3333333333}
|
||||
{"input":"Michelle began her pizza delivery route with 0.5 of a tank of gas in her car. When she made it back to the pizzeria, 0.16666666666666666 of a tank of gas was left. How much gas did Michelle use?","target":0.3333333333}
|
||||
{"input":"While taking inventory at her pastry shop, Kelly realizes that she had 0.4 of a box of baking powder yesterday, but the supply is now down to 0.3 of a box. How much more baking powder did Kelly have yesterday?","target":0.1}
|
||||
{"input":"Craig walked 0.2 of a mile from school to David's house and 0.7 of a mile from David's house to his own house. How many miles did Craig walk in all?","target":0.9}
|
||||
{"input":"Greg and Sharon own neighboring cornfields. Greg harvested 0.4 of an acre of corn on Monday and Sharon harvested 0.1 of an acre. How many more acres did Greg harvest than Sharon?","target":0.3}
|
||||
{"input":"At the hardware store, 0.25 of the nails are size 2d and 0.5 of the nails are size 4d. What fraction of the nails are either size 2d or 4d?","target":0.75}
|
||||
{"input":"While making desserts for a bake sale, Victor used 0.625 of a scoop of brown sugar as well as 0.25 of a scoop of white sugar. How much more brown sugar did Victor use?","target":0.375}
|
||||
{"input":"Eve ran 0.7 of a mile and walked 0.6 of a mile. How much farther did Eve run than walk?","target":0.1}
|
||||
{"input":"Jonah added 0.3 of a cup of yellow raisins and 0.4 of a cup of black raisins to a batch of trail mix. How many cups of raisins did Jonah add in all?","target":0.7}
|
||||
{"input":"When Jake had one cat, he needed to serve 0.5 of a can of cat food each day. Now that Jake has adopted a second cat, he needs to serve a total of 0.9 of a can each day. How much extra food is needed to feed the second cat?","target":0.4}
|
||||
{"input":"In Yardley it snowed 0.125 of an inch in the morning and 0.5 of an inch in the afternoon. What was the total amount of snowfall?","target":0.625}
|
||||
{"input":"While making pastries, a bakery used 0.2 of a bag of wheat flour and 0.1 of a bag of white flour. How many bags of flour did the bakery use in all?","target":0.3}
|
||||
{"input":"Kaleen filled a bucket with 0.75 of a gallon of water. A few minutes later, she realized only 0.5 of a gallon of water remained. How much water had leaked out of the bucket?","target":0.25}
|
||||
{"input":"Irwin's family went on a camping trip in the mountains. On the first day, they hiked from their car to the campsite. First, they hiked 0.2 of a mile from the car to a stream, and 0.4 of a mile from the stream to a meadow. Then they hiked 0.1 of a mile from the meadow to the campsite. How many miles did Irwin's family hike in all?","target":0.7}
|
||||
{"input":"During a visit to an orchard, Charlie picked 0.16666666666666666 of a bag of Golden Delicious apples, 0.16666666666666666 of a bag of Macintosh apples, and 0.3333333333333333 of a bag of Cortland apples. How many bags of fruit did Charlie pick in total?","target":0.6666666667}
|
||||
{"input":"Before starting her shift, a waitress checks to make sure there is enough mustard for her customers. She finds 0.25 of a bottle at the first table, 0.25 of a bottle at the second table, and 0.375 of a bottle at the third table. Altogether, how many bottles of mustard does the waitress find?","target":0.875}
|
||||
{"input":"A waitress put leftover tarts into the fridge on Thursday night. She noticed that the restaurant had 0.08333333333333333 of a tart filled with cherries, 0.75 of a tart filled with blueberries, and 0.08333333333333333 of a tart filled with peaches. How many leftover tarts did the restaurant have in all?","target":0.9166666667}
|
||||
{"input":"On her vacation last summer, Trisha walked all over New York City to buy souvenirs. First, she walked 0.1111111111111111 of a mile from her hotel to a postcard shop. Then she walked 0.1111111111111111 of a mile from the postcard shop to a T-shirt shop and 0.6666666666666666 of a mile from the T-shirt shop back to the hotel. How many miles did Trisha walk in all?","target":0.8888888889}
|
||||
{"input":"Brandy made trail mix for a backpacking trip. She used 0.16666666666666666 of a pound of peanuts, 0.16666666666666666 of a pound of chocolate chips, and 0.08333333333333333 of a pound of raisins. How many pounds of trail mix did Brandy make?","target":0.4166666667}
|
||||
{"input":"Allie counted the leftover ice cream after a sundae party. She had 0.3333333333333333 of a carton of rocky road ice cream, 0.3333333333333333 of a carton of cookie dough ice cream, and 0.16666666666666666 of a carton of strawberry cheesecake ice cream. How many cartons of ice cream did Allie have in all?","target":0.8333333333}
|
||||
{"input":"During a canned food drive, items were sorted into bins. The drive resulted in 0.125 of a bin of soup, 0.125 of a bin of vegetables, and 0.5 of a bin of pasta. Altogether, how many bins would the canned food take up?","target":0.75}
|
||||
{"input":"Paco's Countertop Company purchased pieces of marble from a quarry. The weights of the pieces they purchased were 0.3333333333333333 of a ton, 0.3333333333333333 of a ton, and 0.08333333333333333 of a ton. How many tons of marble did Paco's Countertop Company purchase in all?","target":0.75}
|
||||
{"input":"Nina did a running drill to get in shape for soccer season. First, Nina ran 0.08333333333333333 of a mile. Then she ran 0.08333333333333333 of a mile and 0.6666666666666666 of a mile more. How many miles did Nina run in total?","target":0.8333333333}
|
||||
{"input":"Bonnie's science class recorded the rainfall each day. They recorded 0.16666666666666666 of a centimeter of rain on Monday, 0.4166666666666667 of a centimeter of rain on Tuesday, and 0.08333333333333333 of a centimeter of rain on Wednesday. How many centimeters of rain did the class record in all?","target":0.6666666667}
|
||||
{"input":"Last Saturday, Spencer walked all over town running errands. First, he walked 0.3 of a mile from his house to the library and 0.1 of a mile from the library to the post office. Then he walked 0.4 of a mile from the post office back home. How many miles did Spencer walk in all?","target":0.8}
|
||||
{"input":"A construction company ordered 0.16666666666666666 of a ton of concrete, 0.16666666666666666 of a ton of bricks, and 0.5 of a ton of stone. How many tons of material did the company order in all?","target":0.8333333333}
|
||||
{"input":"Kendra made punch for her friend's birthday party. She used 0.25 of a gallon of grape juice, 0.375 of a gallon of cranberry juice, and 0.125 of a gallon of club soda. How many gallons of punch did Kendra make?","target":0.75}
|
||||
{"input":"A spaceship traveled 0.5 of a light-year from Earth to Planet X and 0.1 of a light-year from Planet X to Planet Y. Then it traveled 0.1 of a light-year from Planet Y back to Earth. How many light-years did the spaceship travel in all?","target":0.7}
|
||||
{"input":"Logan recorded the snowfall every day during a snowstorm. He recorded 0.3333333333333333 of a centimeter on Wednesday, 0.3333333333333333 of a centimeter on Thursday, and 0.2222222222222222 of a centimeter on Friday. How many total centimeters of snow did Logan record?","target":0.8888888889}
|
||||
{"input":"Ellen made smoothies in the blender. She used 0.2 of a cup of strawberries, 0.1 of a cup of yogurt, and 0.2 of a cup of orange juice. How many cups of ingredients did Ellen use for the smoothies?","target":0.5}
|
||||
{"input":"During a school play, Jonah staffed the snack bar. He served 0.25 of a pitcher of lemonade during the first intermission, 0.4166666666666667 of a pitcher during the second, and 0.25 of a pitcher during the third. How many pitchers of lemonade did Jonah pour in all?","target":0.9166666667}
|
||||
{"input":"Heather went to the county fair last weekend. When she got there, she had to walk 0.3333333333333333 of a mile from the car to the entrance. Then she walked 0.3333333333333333 of a mile to the carnival rides and 0.08333333333333333 of a mile from the carnival rides back to the car. How many miles did Heather walk in all?","target":0.75}
|
||||
{"input":"A renovation project required 0.16666666666666666 of a truck-load of sand, 0.3333333333333333 of a truck-load of dirt, and 0.16666666666666666 of a truck-load of cement. How many truck-loads of material were needed in all?","target":0.6666666667}
|
||||
{"input":"Karin's science class weighed plastic rings for an experiment. They found that the orange ring weighed 0.08333333333333333 of an ounce, the purple ring weighed 0.3333333333333333 of an ounce, and the white ring weighed 0.4166666666666667 of an ounce. What was the total weight of the plastic rings?","target":0.8333333333}
|
||||
{"input":"Carefully following a recipe, Kenny used exactly 0.16666666666666666 of a cup of oil and 1.1666666666666667 cups of water. How many cups of liquid did Kenny use in all?","target":1.3333333333}
|
||||
{"input":"Dale's Vegetarian Restaurant bought 2.8333333333333335 pounds of green peppers and 2.8333333333333335 pounds of red peppers. How many pounds of peppers did Dale's Vegetarian Restaurant buy in all?","target":5.6666666667}
|
||||
{"input":"This afternoon Craig left school, rode the bus 3.8333333333333335 miles, and then walked 0.16666666666666666 of a mile to get home. How much farther did Craig ride than walk?","target":3.6666666667}
|
||||
{"input":"Kelly's chemistry textbook weighs 7.125 pounds and her geometry textbook weighs 0.625 of a pound. How much more does the chemistry textbook weigh than the geometry textbook?","target":6.5}
|
||||
{"input":"Roadster's Paving Company used 10 tons of cement to pave Lexi's street and 5.1 tons of cement to pave Tess's street. How much cement did Roadster's Paving Company use in all?","target":15.1}
|
||||
{"input":"On a hot day, Sam poured 1 bucket of water into a plastic wading pool. A few minutes later he added another 8.8 buckets. How much water did Sam pour into the pool?","target":9.8}
|
||||
{"input":"Kyle jogged 1.125 laps in P.E. class and 2.125 laps during track practice. How many laps did Kyle jog in all?","target":3.25}
|
||||
{"input":"A bucket contains 3 gallons of water. If Derek adds 6.8 gallons more, how many gallons will there be in all?","target":9.8}
|
||||
{"input":"At a pizza party, Mason and his friends drank 2.6666666666666665 bottles of lemon-lime soda and 2.6666666666666665 bottles of cola. How much soda did they drink in all?","target":5.3333333333}
|
||||
{"input":"Professor Ellison weighed two pieces of metal for an experiment. The piece of iron weighed 11.166666666666666 pounds and the piece of aluminum weighed 0.8333333333333334 of a pound. How much more did the piece of iron weigh than the piece of aluminum?","target":10.3333333333}
|
||||
{"input":"As part of a lesson on earthquakes, a science class is researching the movement of a nearby fault line. The fault line moved 1.25 inches during the past year and 5.25 inches the year before. How far did the fault line move in all?","target":6.5}
|
||||
{"input":"Hoping to be named Salesperson of the Month, Rosa called the names from 10.2 pages of the phone book last week. This week, she called the people listed on another 8.6 pages of the same phone book. How many pages worth of people did Rosa call in all?","target":18.8}
|
||||
{"input":"At the beach, Janet and her sister both built sandcastles and then measured their heights. Janet's sandcastle was 3.6666666666666665 feet tall and her sister's was 2.3333333333333335 feet tall. How much taller was Janet's sandcastle than her sister's?","target":1.3333333333}
|
||||
{"input":"Nicole found an orange caterpillar and a green caterpillar in her backyard. The green caterpillar was 3 inches long and the orange caterpillar was 1.1666666666666667 inches long. How much longer was the green caterpillar than the orange caterpillar?","target":1.8333333333}
|
||||
{"input":"Alec and his roommates ate 3.25 pints of ice cream on Friday night and 0.25 of a pint of ice cream on Saturday night. How many pints did they eat in all?","target":3.5}
|
||||
{"input":"A farmer started the day with 8.75 buckets of seeds. After spending the morning sowing seeds, she now has 6 buckets left. How many buckets of seeds did the farmer sow?","target":2.75}
|
||||
{"input":"Irene just bought a new lamp for her bedside table. The old lamp was 1 foot tall and the new lamp is 2.3333333333333335 feet tall. How much taller is the new lamp than the old lamp?","target":1.3333333333}
|
||||
{"input":"Ezra drew a white line that was 7.666666666666667 inches long. Then he drew a blue line that was 3.3333333333333335 inches long. How much longer was the white line than the blue line?","target":4.3333333333}
|
||||
{"input":"There are 7.75 gallons of water in Becky's fish tank. If Becky adds 7 gallons more, how many gallons will there be in all?","target":14.75}
|
||||
{"input":"Wendy ran 19.833333333333332 miles and walked 9.166666666666666 miles. How much farther did Wendy run than walk?","target":10.6666666667}
|
||||
{"input":"Kenji and his classmates placed colored blocks on a scale during a science lab. The yellow block weighed 0.6 pounds and the green block weighed 0.4 pounds. How much more did the yellow block weigh than the green block?","target":0.2}
|
||||
{"input":"Terrell hiked 8.2 miles on Saturday. Then, on Sunday, he hiked another 1.6 miles. How far did Terrell hike all together?","target":9.8}
|
||||
{"input":"A carpenter bought a piece of wood that was 0.41 meters long. Then she sawed 0.33 meters off the end. How long is the piece of wood now?","target":0.08}
|
||||
{"input":"Kelly bought 0.1 pounds of peanuts and 0.4 pounds of raisins. How many pounds of snacks did she buy in all?","target":0.5}
|
||||
{"input":"Kevin bought two watermelons. The first watermelon was 9.91 pounds, and the second watermelon was 4.11 pounds. How many pounds of watermelon did Kevin buy?","target":14.02}
|
||||
{"input":"In March it rained 0.81 inches. It rained 0.35 inches less in April than in March. How much did it rain in April?","target":0.46}
|
||||
{"input":"Ron weighed two colored metal balls during a science class. The blue ball weighed 6 pounds and the brown ball weighed 3.12 pounds. If Ron places both balls on the scale at the same time, what will the scale read?","target":9.12}
|
||||
{"input":"A bee colony produced 0.36 pounds of honey, but bears ate 0.05 pounds of it. How much honey remains?","target":0.31}
|
||||
{"input":"It rained 0.9 inches on Monday. On Tuesday, it rained 0.7 inches less than on Monday. How much did it rain on Tuesday?","target":0.2}
|
||||
{"input":"It snowed 0.32 inches on Monday and 0.21 inches on Tuesday. How much did it snow on Monday and Tuesday combined?","target":0.53}
|
||||
{"input":"Brennan had 0.25 grams of pepper. Then he used 0.16 grams of the pepper to make some scrambled eggs. How much pepper does Brennan have left?","target":0.09}
|
||||
{"input":"A construction company bought 5.91 tons of gravel and 8.11 tons of sand. How many tons of material did the company buy in all?","target":14.02}
|
||||
{"input":"It rained 0.2 inches on Saturday and 0.4 inches on Sunday. How much did it rain on Saturday and Sunday combined?","target":0.6}
|
||||
{"input":"Pamela bought 9.8 ounces of sugar, and she spilled 5.2 ounces of it on the floor. How much is left?","target":4.6}
|
||||
{"input":"Gordon bought 3.42 pounds of fruit for a class party. The class ate 2.2 pounds of the fruit. How much fruit is left?","target":1.22}
|
||||
{"input":"A chef bought 0.14 kilograms of almonds and 0.38 kilograms of pecans. How many kilograms of nuts did the chef buy in all?","target":0.52}
|
||||
{"input":"Marta picked two pumpkins. The first pumpkin weighed 4 pounds, and the second pumpkin weighed 8.7 pounds. How much did the two pumpkins weigh all together?","target":12.7}
|
||||
{"input":"A truck carrying 4.1 pounds of sand travels to a construction yard and loses 2.4 pounds of sand along the way. How much sand does the truck have when it arrives at the yard?","target":1.7}
|
||||
{"input":"Tori was 4.4 feet tall. Then she grew 2.86 feet taller. How tall is Tori now?","target":7.26}
|
||||
{"input":"A carpenter bought a piece of wood that was 8.9 centimeters long. Then he sawed 2.3 centimeters off the end. How long is the piece of wood now?","target":6.6}
|
||||
{"input":"How many cookies would you have if you had 37 bags of cookies with 19 cookies in each bag?","target":703.0}
|
||||
{"input":"The Ferris wheel in Paradise Park has 14 seats. Each seat can hold 6 people. How many people can ride the Ferris wheel at the same time?","target":84.0}
|
||||
{"input":"There are 261 fishbowls. Each fishbowl has 23 fish. How many fish are there?","target":6003.0}
|
||||
{"input":"We ordered 21 pizzas. Each pizza has 8 slices. How many slices of pizza are there altogether?","target":168.0}
|
||||
{"input":"Emily collected eggs from the hen and put them into 303 baskets. She put 28 eggs into each basket. How many eggs did Emily collect?","target":8484.0}
|
||||
{"input":"There are 37 baskets. There are 17 apples in each basket. How many apples are there in all?","target":629.0}
|
||||
{"input":"Bryan took a look at his books as well. If Bryan has 56 books in each of his 9 bookshelves, how many books does he have in total?","target":504.0}
|
||||
{"input":"If books came from all the 4 continents that Bryan had been into and he collected 122 books per continent, how many books does he have from all 4 continents combined?","target":488.0}
|
||||
{"input":"Bryan had 8 precious stones in his collection which he sold to his friend from the jewelry store. If the stones were sold at 1785 dollar each, how much money did Bryan get in total?","target":14280.0}
|
||||
{"input":"The farmers reported that they harvest 45 sacks of apples from each of the 8 sections of the orchard daily. How many sacks are harvested every day?","target":360.0}
|
||||
{"input":"Lewis then went to see the oranges being harvested. Lewis found out that they harvest 83 sacks per day. How many sacks of oranges will they have after 6 days of harvest?","target":498.0}
|
||||
{"input":"Lewis saved checking on the grapevines for his last stop. He was told by one of the pickers that they fill 324 drums of grapes per day. How many drums of grapes would be filled in 9 days?","target":2916.0}
|
||||
{"input":"Upon arriving at the circus, they went to the ticket booth and asked how much each ticket cost. If each ticket costs 44 dollars and they bought 7 tickets, how much money did they spend on tickets?","target":308.0}
|
||||
{"input":"They entered the circus tent and saw that there are 4 sections for the audience. If each section can accommodate 246 people, how many people can the tent accommodate in total?","target":984.0}
|
||||
{"input":"The first act included 5 clown mobiles, each stuffed with 28 clowns. How many clowns are inside all the clown mobiles combined?","target":140.0}
|
||||
{"input":"The next act involved several jugglers. If each juggler is juggling 6 balls at a time, how many balls are needed if there are 378 jugglers putting a show at the same time?","target":2268.0}
|
||||
{"input":"For the final act, the circus brought out dancing animals wearing crowns. If each crown is made with 7 different colored feathers, how many feathers are needed for 934 crowns?","target":6538.0}
|
||||
{"input":"The library is divided into different sections for different type of books. The science fiction section has 8 books. If each book has 478 pages, how many pages do all the books have in total?","target":3824.0}
|
||||
{"input":"Jack has a section filled with short story booklets. If each booklet has 9 pages and there are 49 booklets in the short story section, how many pages will Jack need to go through if he plans to read them all?","target":441.0}
|
||||
{"input":"The mini library also has a section for the classics. If Jack has a collection of 6 classic authors, with each author having 33 books, how many books does he have in the classics section?","target":198.0}
|
||||
{"input":"She counted her crayons and found out that she has 80 crayons which she will place in crayon boxes. Every box can contain 8 crayons. How many boxes does she need?","target":10.0}
|
||||
{"input":"There was a stack of 700 sheets of used paper. Lexie wants to place it in boxes for recycling. If every box can contain 100 sheets, how many boxes does Lexie need?","target":7.0}
|
||||
{"input":"Lexie\u2019s mom gathered all her watercolor paintings and thought of placing an equal number of paintings in 4 rooms in the house. If Lexie has 32 watercolor paintings, how many paintings will be placed in each room?","target":8.0}
|
||||
{"input":"Lexie\u2019s younger brother helped pick up all the paper clips in Lexie\u2019s room. He was able to collect 81 paper clips. If he wants to distribute the paper clips in 9 boxes, how many paper clips will each box contain?","target":9.0}
|
||||
{"input":"The junior ranger asked Christian to help him place 420 seedlings in packets. If every packet needs to contain 7 seeds, how many packets do they need?","target":60.0}
|
||||
{"input":"Christian\u2019s mother prepared lemonade. Every pitcher of lemonade can serve 5 glasses. If she was able to serve 30 glasses of lemonade, how many pitchers of lemonade did she prepare?","target":6.0}
|
||||
{"input":"Christian\u2019s father and the senior ranger gathered firewood as they walked towards the lake in the park and brought with them sacks. If every sack can contain around 20 pieces of wood, how many sacks were they able to fill if they gathered 80 pieces of wood?","target":4.0}
|
||||
{"input":"Christian and the junior ranger brought a bag of 140 nails as they visited every station assigned to the junior ranger. If they left exactly 7 nails in every station they visited, how many stations did Joline and the junior ranger visit?","target":20.0}
|
||||
{"input":"Sunday morning was spent for making wood carvings which can be sold as souvenir for tourists. They were placed in shelves that can contain 8 wood carvings at a time. If 56 wood carvings were displayed, how many shelves were filled with carvings?","target":7.0}
|
||||
{"input":"Shiela has 6 neighbors who like to collect animal drawings. Shiela drew 54 animals on small pieces of paper. If she plans to give the same number of animal drawings to her neighbors, how many will each of them receive?","target":9.0}
|
||||
{"input":"Allen, Shiela\u2019s brother, likes to play with blocks. Shiela repainted Allen\u2019s old blocks in different colors. If Allen has 49 identical blocks and there are 7 blocks for every color of paint used, how many colors did Shiela use?","target":7.0}
|
||||
{"input":"It has been tradition in Shiela\u2019s home to hang a sock above the fireplace for each member of the family. This year, she placed a cinnamon ball every day in the socks for 5 of her family members. How long can she do this if she bought 50 cinnamon balls?","target":10.0}
|
||||
{"input":"9 of Hayley\u2019s closest friends like stickers. If she plans to give all of them an equal number of stickers, how many will each receive if she has 72 stickers?","target":8.0}
|
||||
{"input":"In Haley\u2019s class, 5 are boys who love to play marbles. If Haley has 35 marbles, how many will each of the boys receive?","target":7.0}
|
||||
{"input":"When relatives visit Haley and her family, she and her cousins do origami. If she has 48 origami papers to give away to her 6 cousins, how many will each receive if she gives everyone the same number of origami papers?","target":8.0}
|
||||
{"input":"A large bag of balls was kept under Haley\u2019s bed. Her mom placed the balls in bags for children in foster homes. If every bag can contain 4 balls and Haley has 36 balls, how many bags will be used?","target":9.0}
|
||||
{"input":"Haley has 63 magazines in her cabinet. She plans to send it to the recycling office in their area. If she places it in boxes which can contain 9 magazines, how many boxes will Hayley use?","target":7.0}
|
||||
{"input":"Betty bought 88 pink flower stones and wanted to make 8 bracelets out of these stones. How many pink flower stones will each bracelet have if she used the same number of stones in each bracelet?","target":11.0}
|
||||
{"input":"Betty also bought 140 shiny blue round stones. If 14 pieces of this stone is in each bracelet, how many bracelets of blue shiny round stones will there be?","target":10.0}
|
||||
{"input":"Brenda, Betty\u2019s sister, wanted to have 3 bracelets with star-shaped stones. She also bought 36 star-shaped stones from the local store and gave it to Betty. How many star-shaped stones will there be in each of the bracelet Betty makes for Brenda?","target":12.0}
|
||||
{"input":"Shannon, Brenda\u2019s neighbor, joined Brenda in making bracelets. She brought 48 heart-shaped stones and wanted to have 8 of this type of stone in each of the bracelet she makes. How many bracelets with heart-shaped stones can Shannon make?","target":6.0}
|
||||
{"input":"Brenda\u2019s mother made cookies for 5. If she prepared 35 cookies and each of them had the same number of cookies, how many did each of them have?","target":7.0}
|
||||
{"input":"Jane had been saving large empty cans to serve as pots for sunflowers. If she has 54 sunflower seeds and there are 9 cans, how many seeds will be placed in each can if she places an equal number of seeds in each can?","target":6.0}
|
||||
{"input":"Jane\u2019s mom picked cherry tomatoes from their backyard. If she gathered 56 cherry tomatoes and is about to place them in small jars which can contain 8 cherry tomatoes at a time, how many jars will she need?","target":7.0}
|
||||
{"input":"Jane helped her mom prepare fresh lemonade. If each glass needs 2 lemons, how many glasses of fresh lemonade can she make if they have 18 lemons?","target":9.0}
|
||||
{"input":"Jane\u2019s dad brought home 24 marble potatoes. If Jane\u2019s mom made potato salad for lunch and served an equal amount of potatoes to Jane, herself and her husband, how many potatoes did each of them have?","target":8.0}
|
||||
{"input":"For dessert, Jane\u2019s mom prepared 12 pieces of bite-size cinnamon swirls. If the 3 of them ate an equal number of pieces of cinnamon swirls, how many pieces did Jane eat?","target":4.0}
|
||||
{"input":"Ellen had 380 legos, but Ellen lost 57 of them. How many legos does Ellen have now?","target":323.0}
|
||||
{"input":"Arthur baked 35 muffins. How many more muffins does Arthur have to bake to have 83 muffins?","target":48.0}
|
||||
{"input":"Willy has 1400 crayons. Lucy has 290 crayons. How many more crayons does Willy have then Lucy?","target":1110.0}
|
||||
{"input":"There are 10 stickers on a page. If you have 22 pages of stickers, how many stickers do you have?","target":220.0}
|
||||
{"input":"There are 96 cupcakes for 8 children to share. How much will each person get if they share the cupcakes equally?","target":12.0}
|
||||
{"input":"Karen has 639 crayons. Cindy has 504 crayons. How many more crayons does Karen have than Cindy?","target":135.0}
|
||||
{"input":"Jose has 85 peanuts. Kenya has 48 more than Jose. How many peanuts does Kenya have?","target":133.0}
|
||||
{"input":"Lizette has 813 stamps. Lizette has 125 more stamps than Minerva. How many stamps does Minerva have?","target":688.0}
|
||||
{"input":"White t-shirts can be purchased in packages of 6. If Mom buys 71 packages, how many white t-shirts will Mom have?","target":426.0}
|
||||
{"input":"I have 648 pencils. If I put 4 pencils in each pencil box, how many pencil boxes will I fill?","target":162.0}
|
||||
{"input":"Uncle Dave bought 143 ice cream sandwiches. If he wants to give them to his 11 hungry nieces, how many can each niece get?","target":13.0}
|
||||
{"input":"Megan had 217 markers. Robert gave her 109 more markers. How many markers does Megan have altogether?","target":326.0}
|
||||
{"input":"A DVD book holds 126 DVDs. There are 81 DVDs already in the book. How many more DVDs can be put in the book?","target":45.0}
|
||||
{"input":"Carla has some marbles. Carla bought 489 marbles. Now Calra has 2778 marbles all together. How many did Carla start with?","target":2289.0}
|
||||
{"input":"Paco had 93 cookies. Paco ate 15 of them. How many cookies did Paco have left?","target":78.0}
|
||||
{"input":"Kelly has 121 Nintendo games. How many does Kelly need to give away so that Kelly will have 22 games left?","target":99.0}
|
||||
{"input":"Connie had some marbles. Connie gave 183 to Juan. Now Connie has 593 marbles left. How many did Connie have to start with?","target":776.0}
|
||||
{"input":"Connie has 2315 red markers and 1028 blue markers. How many markers does Connie have altogether?","target":3343.0}
|
||||
{"input":"Iesha has 344 books. 136 are about school and the rest are about sports. How many books about sports does Iesha have?","target":208.0}
|
||||
{"input":"James has 1222 balloons. Amy has 513 balloons. How many more balloons does James have than Amy?","target":208.0}
|
||||
{"input":"Sean has 223 whistles. Sean has 95 more whistles than Charles. How many whistles does Charles have?","target":128.0}
|
||||
{"input":"Connie has 323 marbles. Juan has 175 more marbles than Connie. How many marbles does Juan have?","target":498.0}
|
||||
{"input":"Robin has 27 packages of gum. There are 18 pieces in each package. How many pieces of gum does Robin have?","target":486.0}
|
||||
{"input":"Adam has $5.00 to buy an airplane that costs $4.28. How much change will Adam get?","target":0.72}
|
||||
{"input":"If each ball costs $1.54, how much must Kyoko pay for 3 balls?","target":4.62}
|
||||
{"input":"Tony had $20. Tony paid $8 for a ticket to a baseball game. At the game, Tony bought a hot dog for $3. What amount of money did Tony have then?","target":9.0}
|
||||
{"input":"Mr. Guzman bought 48 doughnuts packed equally into 4 boxes. How many doughnuts were in each box?","target":12.0}
|
||||
{"input":"The Sumata family took a 5 day vacation by car. Each day they drove 250 miles. How many total miles did they drive?","target":1250.0}
|
||||
{"input":"One stamp costs 34 cents. If the cost of each stamp remains the same, how much would 4 stamps cost?","target":134.0}
|
||||
{"input":"The town of Milburg has 5256 grown-ups and 2987 children. How many people live in Milburg?","target":8243.0}
|
||||
{"input":"Lisa rented 4 DVDs for $4.80. How much did each DVD cost to rent?","target":1.2}
|
||||
{"input":"There were 3409 pieces of candy in a jar. If 145 pieces were red and the rest were blue, how many were blue?","target":3264.0}
|
||||
{"input":"On Friday, 1250 people visited the zoo. 3 times as many people visited on Saturday than on Friday. How many people visited the zoo on Saturday?","target":3750.0}
|
||||
{"input":"Students went to a concert in 8 buses. Each bus took 45 students. How many students went to the concert?","target":360.0}
|
||||
{"input":"There are 124 students making 3 stars each for the school wall. How many stars will they make all together?","target":372.0}
|
||||
{"input":"Marcia spent 300 minutes working on her science project. How many hours did Marcia spend on her science project?","target":5.0}
|
||||
{"input":"In one week, an airplane pilot flew 1134 miles on Tuesday and 1475 miles on Thursday. If the pilot flies the same number of miles 3 weeks in a row, how many miles does the pilot fly in all?","target":7827.0}
|
||||
{"input":"6 students were sitting at each table in the lunchroom. There are 34 tables. How many students were sitting in the lunchroom?","target":204.0}
|
||||
{"input":"Tyler had 15 dogs. Each dog had 5 puppies. How many puppies does Tyler now have?","target":75.0}
|
||||
{"input":"The farmer had 127 apples. The farmer gave 88 apples to his neighbor. How many apples does the farmer have now?","target":39.0}
|
||||
{"input":"John can read 4 books a day. John reads every Monday and Tuesday. How many books would John read in 6 weeks?","target":48.0}
|
||||
{"input":"Jill invited 37 people to her birthday party. They each ate 8 pieces of pizza. How many pieces of pizza did they eat?","target":296.0}
|
||||
{"input":"The Spurs basketball team has 22 players. Each player has 11 basketballs. How many basketballs do they have in all?","target":242.0}
|
||||
{"input":"Mrs. Hilt uses 2 ounces of detergent to wash a pound of clothes. How many ounces of soap will Mrs. Hilt use to wash 9 pounds of clothes?","target":18.0}
|
||||
{"input":"Mrs. Hilt went to a concert. A total of 65899 people attended the concert. The next week, Mrs. Hilt went to a second concert, which had 119 more people in attendance. How many people were at the second concert?","target":66018.0}
|
||||
{"input":"Mrs. Hilt baked pies last weekend for a holiday dinner. She baked 16 pecan pies and 14 apples pies. If she wants to arrange all of the pies in rows of 5 pies each, how many rows will she have?","target":6.0}
|
||||
{"input":"Mrs. Hilt needs to share $3.75 equally among 3 total people. How much money will each person get?","target":1.25}
|
||||
{"input":"Mrs. Hilt spent 74 cents at the school store. She bought a notebook for 35 cents, a ruler for 18 cents, and 3 pencils. What is the cost of one pencil?","target":7.0}
|
||||
{"input":"Mrs. Hilt has 40 markers. They are divided equally into 7 packages. Mrs. Hilt wants to know how many markers are in each package?","target":5.0}
|
||||
{"input":"Mrs. Hilt read 4 books. Each book had 17 chapters in it. How many chapters did Mrs. Hilt read?","target":68.0}
|
||||
{"input":"Mrs. Hilt is baking bread. She needs 5 cups of flour to bake 2 loaves of bread. How much flour will she need to make one loaf of bread?","target":2.5}
|
||||
{"input":"Mrs. Hilt bought a notebook for $1.30. She paid with nickels. How many nickels did she use to buy the notebook?","target":26.0}
|
||||
{"input":"Mrs. Hilt saw a rollercoaster. 7 students rode the rollercoaster every 5 minutes. How many students rode the rollercoaster in 15 minutes?","target":21.0}
|
||||
{"input":"Mrs. Hilt saw 144 bees in the hive. The next day she saw 3 times that many. How many bees did she see on the second day?","target":432.0}
|
||||
{"input":"Mrs. Hilt measured the distance from her desk to the water fountain. It was 30 feet. How many feet will Mrs. Hilt walk on her trips to the fountain if she goes to the water fountain 4 times today?","target":120.0}
|
||||
{"input":"Lucy has an aquarium with 212 fish. Lucy wants to buy 68 more fish. How many fish would Lucy have then?","target":280.0}
|
||||
{"input":"Lucy has 212 fish. How many more fish does Lucy need to buy to have 280 fish?","target":68.0}
|
||||
{"input":"Chris has been saving his allowance to buy a new pair of soccer cleats and a ball. His grandmother gave Chris $25 for his birthday. His aunt and uncle gave Chris $20 and his parents gave him $75. Now Chris had $279. How much money did Chris have before his birthday?","target":159.0}
|
||||
{"input":"Harry Hound had a terrible earache yesterday. When I peered into his ears yesterday, I found 36 frisky fleas having a party in his right ear and 85 baby fleas sleeping peacefully in his left ear. I cleaned out Harry Hound's ears. How many fleas perished?","target":121.0}
|
||||
{"input":"A pet supply store has 600 bags of dog food and 327 bags of cat food. How many more bags of dog food are there than cat food?","target":273.0}
|
||||
{"input":"David has 7 boxes of stuffed toy dogs. Each box has 4 dogs in it. How many dogs are there in all?","target":28.0}
|
||||
{"input":"532 people are watching a movie in a theater. The theater has 750 seats. How many seats are empty in the theater?","target":218.0}
|
||||
{"input":"Each bag contains 23 pounds of oranges. How many pounds of oranges are in 45 bags?","target":1035.0}
|
||||
{"input":"Bert runs 2 miles every day. How many miles will Bert run in 3 weeks?","target":42.0}
|
||||
{"input":"For his long distance phone service Milan pays a 2 dollars monthly fee plus 12 cents per minute. Last month , Milan 's long distance bill was 23.36 dollars. For how many minutes was Milan billed for ?","target":178.0}
|
||||
{"input":"The value of a sport utility vehicle this year is 16,000 dollars , which is 0.8 of what its value was last year. How much is the value of the vehicle last year?","target":20000.0}
|
||||
{"input":"Trenton sells electronic supplies. Each week he earns 190 dollars plus commission equal to 0.04 of his sales. This week his goal is to earn no less than 500 dollars. How much sales he must make to reach his goal?","target":7750.0}
|
||||
{"input":"Tony planted a 4 foot tree. The tree grows at a rate of 5 feet every year. How many years will it take to be 29 feet?","target":5.0}
|
||||
{"input":"The tallest player on the basketball team is 77.75 inches tall. This is 9.5 inches taller than the shortest player. How tall is the shortest player , in inches?","target":68.25}
|
||||
{"input":"0.20 of a number decreased by 4 is equal to 6. Find the number.","target":50.0}
|
||||
{"input":"Freeport McMoran projects the world grain supply will be 1800000 metric tons and the supply will be only 0.75 of the world grain demand. What will the world grain demand be?","target":2400000.0}
|
||||
{"input":"A 12 ounce can of cranberry juice sells for 84 cents. What is the cost in cents per ounce.","target":7.0}
|
||||
{"input":"Manuel opened a savings account with an initial deposit of 177 dollars. If he wants to save 500 dollars during the next 19 weeks , how much must he save each week , in dollars?","target":17.0}
|
||||
{"input":"A mechanic charged 45 dollars an hour , plus 225 dollars for the parts. If the total bill was 450 dollars , how many hours did the job take?","target":5.0}
|
||||
Executable
+562
@@ -0,0 +1,562 @@
|
||||
{"input":"If there are 7 bottle caps in a box and Linda puts 7 more bottle caps inside, how many bottle caps are in the box?","target":14.0}
|
||||
{"input":"Jose starts with 7 bottle caps. He gets 2 more from Rebecca. How many bottle caps does Jose end with?","target":9.0}
|
||||
{"input":"Bridget has 4 Skittles. Henry has 4 Skittles. If Henry gives all of his Skittles to Bridget, how many Skittles will Bridget have?","target":8.0}
|
||||
{"input":"Brenda starts with 7 Skittles. She buys 8 more. How many Skittles does Brenda end with?","target":15.0}
|
||||
{"input":"There are 4 cards. 3 cards more are added. How many are there total?","target":7.0}
|
||||
{"input":"There are 4 marbles. 7 marbles more are added. How many are there total?","target":11.0}
|
||||
{"input":"There are 8 oranges. 8 oranges more are added. How many are there total?","target":16.0}
|
||||
{"input":"Carol collects 2 peanuts. Carol's father gives Carol 5 more. How many peanuts does Carol have?","target":7.0}
|
||||
{"input":"There are 9 cards. 4 cards more are added. How many are there total?","target":13.0}
|
||||
{"input":"If there are 7 bottle caps in a box and Linda puts 7 more bottle caps inside, how many bottle caps are in the box?","target":14.0}
|
||||
{"input":"Bobby collects 2 blocks. Bobby's father gives Bobby 6 more. How many blocks does Bobby have?","target":8.0}
|
||||
{"input":"Peter starts with 8 erasers. Bridget gives Peter 3 more. How many erasers does Peter end with?","target":11.0}
|
||||
{"input":"If there are 7 crayons in a box and Gerald puts 6 more crayons inside, how many crayons are in the box?","target":13.0}
|
||||
{"input":"Michelle has 2 crayons. Janet has 2 crayons. If Janet gives all of her crayons to Michelle, how many crayons will Michelle have?","target":4.0}
|
||||
{"input":"Anna collects 6 blocks. Anna's father gives Anna 8 more. How many blocks does Anna have?","target":14.0}
|
||||
{"input":"Donald has 4 oranges. He finds another 5. How many oranges does Donald have in all?","target":9.0}
|
||||
{"input":"If there are 7 eggs in a box and Daniel puts 4 more eggs inside, how many eggs are in the box?","target":11.0}
|
||||
{"input":"Wayne collects 9 blocks. Wayne's father gives Wayne 6 more. How many blocks does Wayne have?","target":15.0}
|
||||
{"input":"Bridget has 4 Skittles. Henry has 4 Skittles. If Henry gives all of his Skittles to Bridget, how many Skittles will Bridget have?","target":8.0}
|
||||
{"input":"Ralph collects 4 cards. Ralph's father gives Ralph 8 more. How many cards does Ralph have?","target":12.0}
|
||||
{"input":"Martha starts with 3 cards. She gets 76 more from Emily. How many cards does Martha end with?","target":79.0}
|
||||
{"input":"There are 86 blocks. 9 blocks more are added. How many are there total?","target":95.0}
|
||||
{"input":"Cheryl starts with 8 Skittles. Kathryn gives Cheryl 89 more. How many Skittles does Cheryl end with?","target":97.0}
|
||||
{"input":"If there are 76 erasers in a box and Patrick puts 9 more erasers inside, how many erasers are in the box?","target":85.0}
|
||||
{"input":"Amy starts with 7 peanuts. Gerald gives Amy 55 more. How many peanuts does Amy end with?","target":62.0}
|
||||
{"input":"Teresa starts with 52 bananas. She gets 2 more from Rachel. How many bananas does Teresa end with?","target":54.0}
|
||||
{"input":"Gloria has 2 pencils. Lisa has 99 pencils. If Lisa gives all of her pencils to Gloria, how many pencils will Gloria have?","target":101.0}
|
||||
{"input":"Mildred collects 77 oranges. Mildred's father gives Mildred 2 more. How many oranges does Mildred have?","target":79.0}
|
||||
{"input":"Anna starts with 5 candies. She gets 86 more from Larry. How many candies does Anna end with?","target":91.0}
|
||||
{"input":"Martha has 2 peanuts. Joyce has 26 peanuts. If Joyce gives all of her peanuts to Martha, how many peanuts will Martha have?","target":28.0}
|
||||
{"input":"Kevin starts with 7 cards. He finds another 47. How many cards does Kevin end with?","target":54.0}
|
||||
{"input":"Joshua has 40 bottle caps. He buys 7 more. How many bottle caps does Joshua have in all?","target":47.0}
|
||||
{"input":"Nicholas starts with 8 bottle caps. He gets 85 more from Catherine. How many bottle caps does Nicholas end with?","target":93.0}
|
||||
{"input":"Mary starts with 27 eggs. She finds another 4. How many eggs does Mary end with?","target":31.0}
|
||||
{"input":"Shirley starts with 98 eggs. She buys 8 more. How many eggs does Shirley end with?","target":106.0}
|
||||
{"input":"William starts with 15 tickets. He buys 3 more. How many tickets does William end with?","target":18.0}
|
||||
{"input":"Harry starts with 79 apples. He buys 5 more. How many apples does Harry end with?","target":84.0}
|
||||
{"input":"Barbara has 9 candies. She buys 18 more. How many candies does Barbara have in all?","target":27.0}
|
||||
{"input":"Eugene has 51 pencils. He gets 6 more from Joyce. How many pencils does Eugene have in all?","target":57.0}
|
||||
{"input":"Anthony has 9 pencils. Kathryn gives Anthony 56 more. How many pencils does Anthony have in all?","target":65.0}
|
||||
{"input":"Andrew starts with 8 eggs. He buys 62 more. How many eggs does Andrew end with?","target":70.0}
|
||||
{"input":"Annie starts with 4 crayons. Matthew gives Annie 36 more. How many crayons does Annie end with?","target":40.0}
|
||||
{"input":"Angela has 11 tickets. Annie gives Angela 4 more. How many tickets does Angela have in all?","target":15.0}
|
||||
{"input":"Charles has 4 apples. Jessica gives Charles 39 more. How many apples does Charles have in all?","target":43.0}
|
||||
{"input":"Lisa starts with 91 bananas. Maria gives Lisa 8 more. How many bananas does Lisa end with?","target":99.0}
|
||||
{"input":"Evelyn starts with 18 bottle caps. She finds another 63. How many bottle caps does Evelyn end with?","target":81.0}
|
||||
{"input":"Brenda has 34 apples. She gets 37 more from Willie. How many apples does Brenda have in all?","target":71.0}
|
||||
{"input":"Joseph has 67 candies. Kathy gives Joseph 38 more. How many candies does Joseph have in all?","target":105.0}
|
||||
{"input":"Janet has 57 apples. She finds another 95. How many apples does Janet have in all?","target":152.0}
|
||||
{"input":"Nancy starts with 91 bottle caps. She finds another 88. How many bottle caps does Nancy end with?","target":179.0}
|
||||
{"input":"Harold has 53 marbles. He gets 16 more from Steve. How many marbles does Harold have in all?","target":69.0}
|
||||
{"input":"Denise removes 5 bananas from a jar. There were originally 46 bananas in the jar. How many bananas are left in the jar?","target":41.0}
|
||||
{"input":"Arthur removes 4 pencils from a jar. There were originally 87 pencils in the jar. How many pencils are left in the jar?","target":83.0}
|
||||
{"input":"Joyce starts with 75 apples. She gives 52 to Larry. How many apples does Joyce end with?","target":23.0}
|
||||
{"input":"Rachel removes 47 bottle caps from a jar. There were originally 87 bottle caps in the jar. How many bottle caps are left in the jar?","target":40.0}
|
||||
{"input":"Anne weighs 67 pounds. Douglas weighs 52 pounds. How much heavier is Anne than Douglas?","target":15.0}
|
||||
{"input":"Jessica weighs 49 pounds. Thomas weighs 44 pounds. How much heavier is Jessica than Thomas?","target":5.0}
|
||||
{"input":"Debra removes 22 apples from a jar. There were originally 57 apples in the jar. How many apples are left in the jar?","target":35.0}
|
||||
{"input":"Virginia starts with 96 eggs. Amy takes 3 away. How many eggs does Virginia end with?","target":93.0}
|
||||
{"input":"Donna weighs 69 pounds. Willie weighs 51 pounds. How much heavier is Donna than Willie?","target":18.0}
|
||||
{"input":"There are 47 eggs in a box. Harry takes 5 eggs. How many are left?","target":42.0}
|
||||
{"input":"Edward starts with 92 eggs. He gives 40 to Phillip. How many eggs does Edward end with?","target":52.0}
|
||||
{"input":"David removes 7 eggs from a jar. There were originally 27 eggs in the jar. How many eggs are left in the jar?","target":20.0}
|
||||
{"input":"There are 79 pencils in a box. Eric takes 4 pencils. How many are left?","target":75.0}
|
||||
{"input":"Jane starts with 87 crayons. 7 are eaten by a hippopotamus. How many crayons does Jane end with?","target":80.0}
|
||||
{"input":"Roger has 95 candies. He gives 3 to Stephanie. How many candies will Roger have?","target":92.0}
|
||||
{"input":"There are 96 oranges in a box. Jonathan takes 45 oranges. How many are left?","target":51.0}
|
||||
{"input":"Marie starts with 95 erasers. She loses 42. How many erasers does Marie end with?","target":53.0}
|
||||
{"input":"There are 22 oranges in a box. Paula takes 7 oranges. How many are left?","target":15.0}
|
||||
{"input":"Melissa has 88 bananas. She shares 4 with Joshua. How many bananas will Melissa have?","target":84.0}
|
||||
{"input":"Larry has 67 cards. Dennis takes 9 away. How many cards will Larry have?","target":58.0}
|
||||
{"input":"John weighs 81 pounds. Roy weighs 4 pounds. How much heavier is John than Roy?","target":77.0}
|
||||
{"input":"Larry starts with 93 stickers. He loses 6. How many stickers does Larry end with?","target":87.0}
|
||||
{"input":"Carlos starts with 39 bananas. 3 are eaten by a hippopotamus. How many bananas does Carlos end with?","target":36.0}
|
||||
{"input":"Patricia starts with 76 candies. Albert takes 5 away. How many candies does Patricia end with?","target":71.0}
|
||||
{"input":"There are 16 bottle caps in a box. Marvin takes 6 bottle caps. How many are left?","target":10.0}
|
||||
{"input":"Ernest starts with 17 crayons. Jennifer takes 6 away. How many crayons does Ernest end with?","target":11.0}
|
||||
{"input":"Mildred weighs 59 pounds. Carol weighs 9 pounds. How much heavier is Mildred than Carol?","target":50.0}
|
||||
{"input":"Diana starts with 65 bottle caps. 4 are eaten by a hippopotamus. How many bottle caps does Diana end with?","target":61.0}
|
||||
{"input":"Ruby has 63 apples. Emily takes 55 away. How many apples will Ruby have?","target":8.0}
|
||||
{"input":"Craig starts with 5 Skittles. 2 are eaten by a hippopotamus. How many Skittles does Craig end with?","target":3.0}
|
||||
{"input":"Steve has 46 oranges. He shares 4 with Patrick. How many oranges will Steve have?","target":42.0}
|
||||
{"input":"Bridget weighs 39 pounds. Martha weighs 2 pounds. How much heavier is Bridget than Martha?","target":37.0}
|
||||
{"input":"Brian has 76 cards. Wayne takes 59 away. How many cards will Brian have?","target":17.0}
|
||||
{"input":"Billy has 62 crayons. 52 are eaten by a hippopotamus. How many crayons will Billy have?","target":10.0}
|
||||
{"input":"Pamela starts with 30 bottle caps. Jean takes 26 away. How many bottle caps does Pamela end with?","target":4.0}
|
||||
{"input":"Katherine has 34 bottle caps. 8 are eaten by a hippopotamus. How many bottle caps will Katherine have?","target":26.0}
|
||||
{"input":"Melissa has 70 oranges. John takes 19 away. How many oranges will Melissa have?","target":51.0}
|
||||
{"input":"Norma starts with 47 bananas. She loses 45. How many bananas does Norma end with?","target":2.0}
|
||||
{"input":"Willie starts with 36 stickers. He gives 7 to Emily. How many stickers does Willie end with?","target":29.0}
|
||||
{"input":"Aaron starts with 81 erasers. He gives 34 to Doris. How many erasers does Aaron end with?","target":47.0}
|
||||
{"input":"Charles has 25 Skittles. Diana takes 7 away. How many Skittles will Charles have?","target":18.0}
|
||||
{"input":"Norma has 88 cards. She loses 70. How many cards will Norma have?","target":18.0}
|
||||
{"input":"Pamela has 50 Skittles. She gives 7 to Karen. How many Skittles will Pamela have?","target":43.0}
|
||||
{"input":"Evelyn starts with 76 Skittles. She shares 72 with Christine. How many Skittles does Evelyn end with?","target":4.0}
|
||||
{"input":"Helen starts with 82 cards. Gary takes 9 away. How many cards does Helen end with?","target":73.0}
|
||||
{"input":"Ruth starts with 89 apples. She shares 5 with Peter. How many apples does Ruth end with?","target":84.0}
|
||||
{"input":"Carlos has 58 blocks. He gives 21 to Rachel. How many blocks will Carlos have?","target":37.0}
|
||||
{"input":"Jesse starts with 78 pencils. He gives 44 to Joshua. How many pencils does Jesse end with?","target":34.0}
|
||||
{"input":"Roger has 92 blocks. 2 are eaten by a hippopotamus. How many blocks will Roger have?","target":90.0}
|
||||
{"input":"Heather starts with 86 blocks. She shares 41 with Jose. How many blocks does Heather end with?","target":45.0}
|
||||
{"input":"Ralph starts with 74 peanuts. He loses 59. How many peanuts does Ralph end with?","target":15.0}
|
||||
{"input":"Jack starts with 62 marbles. He shares 33 with Rebecca. How many marbles does Jack end with?","target":29.0}
|
||||
{"input":"Clarence has 99 tickets. He gives 7 to Eugene. How many tickets will Clarence have?","target":92.0}
|
||||
{"input":"Sean has 55 blocks. 29 are eaten by a hippopotamus. How many blocks will Sean have?","target":26.0}
|
||||
{"input":"Tina has 82 bananas. She gets 3 more from Joyce. How many bananas does Tina have in all?","target":85.0}
|
||||
{"input":"Janet has 3 stickers. She gets 53 more from Ruth. How many stickers does Janet have in all?","target":56.0}
|
||||
{"input":"Jennifer starts with 7 apples. She finds another 74. How many apples does Jennifer end with?","target":81.0}
|
||||
{"input":"Heather has 60 oranges. Russell takes 35 away. How many oranges will Heather have?","target":25.0}
|
||||
{"input":"Helen starts with 9 oranges. She gets 29 more from Ann. How many oranges does Helen end with?","target":38.0}
|
||||
{"input":"William has 2 bottle caps. He buys 41 more. How many bottle caps does William have in all?","target":43.0}
|
||||
{"input":"Dorothy has 72 cards. She gives 5 to Harold. How many cards will Dorothy have?","target":67.0}
|
||||
{"input":"Sara has 10 marbles. She loses 7. How many marbles will Sara have?","target":3.0}
|
||||
{"input":"Jennifer has 72 cards. 61 are eaten by a hippopotamus. How many cards will Jennifer have?","target":11.0}
|
||||
{"input":"Wanda has 4 blocks. Theresa gives Wanda 79 more. How many blocks does Wanda have in all?","target":83.0}
|
||||
{"input":"William has 10 marbles. He shares 3 with Theresa. How many marbles will William have?","target":7.0}
|
||||
{"input":"Larry starts with 91 cards. 17 are eaten by a hippopotamus. How many cards does Larry end with?","target":74.0}
|
||||
{"input":"Sarah has 26 bottle caps. She buys 3 more. How many bottle caps does Sarah have in all?","target":29.0}
|
||||
{"input":"There are 3 eggs in each box. How many eggs are in 2 boxes?","target":6.0}
|
||||
{"input":"Each bottle cap costs $2.00. How much do 6 bottle caps cost?","target":12.0}
|
||||
{"input":"Each child has 3 oranges. If there are 4 children, how many oranges are there in total?","target":12.0}
|
||||
{"input":"Each banana costs $5.00. How much do 4 bananas cost?","target":20.0}
|
||||
{"input":"Each ticket costs $9.00. How much do 4 tickets cost?","target":36.0}
|
||||
{"input":"Each bottle cap costs $5.00. How much do 5 bottle caps cost?","target":25.0}
|
||||
{"input":"Michelle has 7 boxes of crayons. Each box holds 5 crayons. How many crayons does Michelle have?","target":35.0}
|
||||
{"input":"There are 8 marbles in each box. How many marbles are in 6 boxes?","target":48.0}
|
||||
{"input":"Each child has 2 candies. If there are 9 children, how many candies are there in total?","target":18.0}
|
||||
{"input":"Each child has 5 bottle caps. If there are 9 children, how many bottle caps are there in total?","target":45.0}
|
||||
{"input":"Each child has 6 pencils. If there are 2 children, how many pencils are there in total?","target":12.0}
|
||||
{"input":"Each child has 5 bottle caps. If there are 9 children, how many bottle caps are there in total?","target":45.0}
|
||||
{"input":"Each ticket costs $9.00. How much do 4 tickets cost?","target":36.0}
|
||||
{"input":"Stephanie went to the store 8 times last month. She buys 2 oranges each time she goes to the store. How many oranges did Stephanie buy last month?","target":16.0}
|
||||
{"input":"Sarah has 7 boxes of apples. Each box holds 7 apples. How many apples does Sarah have?","target":49.0}
|
||||
{"input":"Martin has 8 boxes of crayons. Each box holds 7 crayons. How many crayons does Martin have?","target":56.0}
|
||||
{"input":"There are 9 bottle caps in each box. How many bottle caps are in 6 boxes?","target":54.0}
|
||||
{"input":"Paula has 6 boxes of peanuts. Each box holds 4 peanuts. How many peanuts does Paula have?","target":24.0}
|
||||
{"input":"Eric has 3 boxes of pencils. Each box holds 9 pencils. How many pencils does Eric have?","target":27.0}
|
||||
{"input":"There are 6 marbles in each box. How many marbles are in 3 boxes?","target":18.0}
|
||||
{"input":"Kimberly went to the store 3 times last month. She buys 7 peanuts each time she goes to the store. How many peanuts did Kimberly buy last month?","target":21.0}
|
||||
{"input":"Ronald went to the store 2 times last month. He buys 10 bananas each time he goes to the store. How many bananas did Ronald buy last month?","target":20.0}
|
||||
{"input":"Kathleen went to the store 5 times last month. She buys 5 bottle caps each time she goes to the store. How many bottle caps did Kathleen buy last month?","target":25.0}
|
||||
{"input":"There are 4 crayons in each box. How many crayons are in 3 boxes?","target":12.0}
|
||||
{"input":"Kimberly went to the store 3 times last month. She buys 7 peanuts each time she goes to the store. How many peanuts did Kimberly buy last month?","target":21.0}
|
||||
{"input":"The school is planning a field trip. There are 14 students and 2 seats on each school bus. How many buses are needed to take the trip?","target":7.0}
|
||||
{"input":"Betty has 24 oranges stored in boxes. If there are 3 boxes, how many oranges must go in each box?","target":8.0}
|
||||
{"input":"The school is planning a field trip. There are 9 students and 3 seats on each school bus. How many buses are needed to take the trip?","target":3.0}
|
||||
{"input":"There are 2 students in the class and 18 pencils. If the pencils are divided equally among the students, how many does each student get?","target":9.0}
|
||||
{"input":"There are 9 students in the class and 81 tickets. If the tickets are divided equally among the students, how many does each student get?","target":9.0}
|
||||
{"input":"There are 35 bottle caps in Beverly's bottle cap collection. If the bottle caps are organized into 7 groups, how big is each group?","target":5.0}
|
||||
{"input":"Jeffrey wants to split a collection of bottle caps into groups of 2. Jeffrey has 12 bottle caps. How many groups will be created?","target":6.0}
|
||||
{"input":"There are 9 students in the class and 81 tickets. If the tickets are divided equally among the students, how many does each student get?","target":9.0}
|
||||
{"input":"Peter has 3 candies stored in boxes. If there are 3 boxes, how many candies must go in each box?","target":1.0}
|
||||
{"input":"There are 7 students in the class and 42 apples. If the apples are divided equally among the students, how many does each student get?","target":6.0}
|
||||
{"input":"Martha is inviting 4 friends to a party. She has 12 cookies. How many cookies will each friend get?","target":3.0}
|
||||
{"input":"Jesse has 21 bananas. If he shares them among 3 friends, how many bananas does each friend get?","target":7.0}
|
||||
{"input":"There are 7 students in the class and 56 eggs. If the eggs are divided equally among the students, how many does each student get?","target":8.0}
|
||||
{"input":"Elizabeth is inviting 5 friends to a party. She has 5 cookies. How many cookies will each friend get?","target":1.0}
|
||||
{"input":"There are 72 cards in Jean's card collection. If the cards are organized into 9 groups, how big is each group?","target":8.0}
|
||||
{"input":"Eric wants to split a collection of peanuts into groups of 8. Eric has 64 peanuts. How many groups will be created?","target":8.0}
|
||||
{"input":"Henry wants to split a collection of crayons into groups of 6. Henry has 42 crayons. How many groups will be created?","target":7.0}
|
||||
{"input":"Jack is inviting 4 friends to a party. He has 4 cookies. How many cookies will each friend get?","target":1.0}
|
||||
{"input":"Harry is inviting 3 friends to a party. He has 12 cookies. How many cookies will each friend get?","target":4.0}
|
||||
{"input":"There are 30 candies in Kathy's candy collection. If the candies are organized into 10 groups, how big is each group?","target":3.0}
|
||||
{"input":"Deborah is inviting 10 friends to a party. She has 80 cookies. How many cookies will each friend get?","target":8.0}
|
||||
{"input":"The school is planning a field trip. There are 28 students and 7 seats on each school bus. How many buses are needed to take the trip?","target":4.0}
|
||||
{"input":"There are 36 bananas in Diane's banana collection. If the bananas are organized into 6 groups, how big is each group?","target":6.0}
|
||||
{"input":"There are 36 bananas in Diane's banana collection. If the bananas are organized into 6 groups, how big is each group?","target":6.0}
|
||||
{"input":"Louise has 42 oranges stored in boxes. If there are 7 boxes, how many oranges must go in each box?","target":6.0}
|
||||
{"input":"The school is planning a field trip. There are 45 students and 9 seats on each school bus. How many buses are needed to take the trip?","target":5.0}
|
||||
{"input":"Michael has 16 blocks stored in boxes. If there are 8 boxes, how many blocks must go in each box?","target":2.0}
|
||||
{"input":"Albert has 35 oranges stored in boxes. If there are 7 boxes, how many oranges must go in each box?","target":5.0}
|
||||
{"input":"Lori has 30 marbles. If she shares them among 5 friends, how many marbles does each friend get?","target":6.0}
|
||||
{"input":"Brenda wants to split a collection of erasers into groups of 90. Brenda has 270 erasers. How many groups will be created?","target":3.0}
|
||||
{"input":"The school is planning a field trip. There are 180 students and 60 seats on each school bus. How many buses are needed to take the trip?","target":3.0}
|
||||
{"input":"Bonnie has 10 tickets. If she shares them among 2 friends, how many tickets does each friend get?","target":5.0}
|
||||
{"input":"Steven wants to split a collection of cards into groups of 6. Steven has 30 cards. How many groups will be created?","target":5.0}
|
||||
{"input":"Eric has 9306 erasers. If he shares them among 99 friends, how many erasers does each friend get?","target":94.0}
|
||||
{"input":"Diana has 3840 erasers. If she shares them among 48 friends, how many erasers does each friend get?","target":80.0}
|
||||
{"input":"Ashley sold 12 boxes of Samoas. How many cases of 12 boxes, plus extra boxes does Ashley need?","target":1.0}
|
||||
{"input":"Ashley sold 12 boxes of Samoas. How many cases of 12 boxes, plus extra boxes does Ashley need?","target":1.0}
|
||||
{"input":"Ashley sold 12 boxes of Samoas. How many cases of 12 boxes, plus extra boxes does Ashley need?","target":1.0}
|
||||
{"input":"Ashley sold 12 boxes of Samoas. How many cases of 12 boxes, plus extra boxes does Ashley need?","target":1.0}
|
||||
{"input":"Sarah sold 24 boxes of Lemon Chalet Cremes. How many cases of 12 boxes, plus extra boxes does Sarah need?","target":2.0}
|
||||
{"input":"If Karen sold 36 boxes of Tagalongs, how many cases of 12 boxes does Karen pickup from the cookie mom?","target":3.0}
|
||||
{"input":"If Karen sold 36 boxes of Tagalongs, how many cases of 12 boxes does Karen pickup from the cookie mom?","target":3.0}
|
||||
{"input":"If Karen sold 36 boxes of Tagalongs, how many cases of 12 boxes does Karen pickup from the cookie mom?","target":3.0}
|
||||
{"input":"If Madison sold 24 boxes of Tagalongs, how many cases of 12 boxes does Madison pickup from the cookie mom?","target":2.0}
|
||||
{"input":"If Karen sold 36 boxes of Tagalongs, how many cases of 12 boxes does Karen pickup from the cookie mom?","target":3.0}
|
||||
{"input":"If Harold split 15 apples between 3 people in her class and kept the left overs, how many apples did each classmate get?","target":5.0}
|
||||
{"input":"If Victor split 25 Skittles between 5 people in her class and kept the left overs, how many Skittles did each classmate get?","target":5.0}
|
||||
{"input":"Jenny sold 24 boxes of Trefoils. How many cases of 8 boxes, plus extra boxes does Jenny need to deliver?","target":3.0}
|
||||
{"input":"Gumballs come in packages of 5. Nathan ate 20 gumballs. How many whole boxes did he eat and how many gumballs does he have left?","target":4.0}
|
||||
{"input":"Shirley sold 20 boxes of Do-Si-Dos. How many cases of 4 boxes, plus extra boxes does Shirley need to deliver?","target":5.0}
|
||||
{"input":"Lemon heads come in packages of 6. Louis ate 54 Lemon Heads. How many whole boxes did he eat and how many Lemon Heads does he have left?","target":9.0}
|
||||
{"input":"Lemon heads come in packages of 6. Louis ate 54 Lemon Heads. How many whole boxes did he eat and how many Lemon Heads does he have left?","target":9.0}
|
||||
{"input":"Lemon heads come in packages of 6. Louis ate 54 Lemon Heads. How many whole boxes did he eat and how many Lemon Heads does he have left?","target":9.0}
|
||||
{"input":"Marilyn starts with 51 bottle caps. She shares 36 with Nancy. How many bottle caps does Marilyn end with?","target":15.0}
|
||||
{"input":"William has 2 bottle caps. He buys 41 more. How many bottle caps does William have in all?","target":43.0}
|
||||
{"input":"Carol starts with 42 blocks. She loses 25. How many blocks does Carol end with?","target":17.0}
|
||||
{"input":"Lillian collects 88 candies. Lillian's father gives Lillian 5 more. How many candies does Lillian have?","target":93.0}
|
||||
{"input":"William has 2 bottle caps. He buys 41 more. How many bottle caps does William have in all?","target":43.0}
|
||||
{"input":"There are 8 apples in a pile on the desk. Each apple comes in a package of 11. 5 apples are added to the pile. How many apples are there in the pile?","target":13.0}
|
||||
{"input":"Sean has 9 apples. Susan gives Sean 8 more. Later, Sean buys 18 tickets at the store. How many apples does Sean have in all?","target":17.0}
|
||||
{"input":"There are 2 bananas in a pile on the desk. Each banana comes in a package of 17. 7 bananas are added to the pile. How many bananas are there in the pile?","target":9.0}
|
||||
{"input":"There are 6 candies in a pile on the desk. Each candy comes in a package of 15. 4 candies are added to the pile. How many candies are there in the pile?","target":10.0}
|
||||
{"input":"Kimberly has 5 Skittles. She buys 7 more. Later, Kimberly buys 18 oranges at the store. How many Skittles does Kimberly have in all?","target":12.0}
|
||||
{"input":"Clarence has 5 oranges. He gets 3 more from Joyce. Later, Clarence buys 9 Skittles at the store. How many oranges does Clarence have in all?","target":8.0}
|
||||
{"input":"There are 2 apples in a pile on the desk. Each apple comes in a package of 7. 4 apples are added to the pile. How many apples are there in the pile?","target":6.0}
|
||||
{"input":"Kimberly has 5 Skittles. She buys 7 more. Later, Kimberly buys 18 oranges at the store. How many Skittles does Kimberly have in all?","target":12.0}
|
||||
{"input":"Annie has 6 apples. She gets 6 more from Nathan. Later, Annie buys 10 crayons at the store. How many apples does Annie have in all?","target":12.0}
|
||||
{"input":"Joyce has 8 eggs. Marie gives Joyce 6 more. Later, Joyce buys 13 erasers at the store. How many eggs does Joyce have in all?","target":14.0}
|
||||
{"input":"There are 2 bananas in a pile on the desk. Each banana comes in a package of 13. 7 bananas are added to the pile. How many bananas are there in the pile?","target":9.0}
|
||||
{"input":"Kenneth has 3 marbles. He buys 6 more. Later, Kenneth buys 15 apples at the store. How many marbles does Kenneth have in all?","target":9.0}
|
||||
{"input":"Arthur has 6 cards. Aaron has with 5 cards. Aaron finds another 62. How many cards does Aaron end with?","target":67.0}
|
||||
{"input":"Scott has 4 tickets. Ernest has with 9 tickets. Ernest finds another 72. How many tickets does Ernest end with?","target":81.0}
|
||||
{"input":"Diane has 37 crayons. She gets 5 more from Frances. Later, Diane buys 16 cards at the store. How many crayons does Diane have in all?","target":42.0}
|
||||
{"input":"Adam has 13 blocks. Ann has with 9 blocks. Ann finds another 44. How many blocks does Ann end with?","target":53.0}
|
||||
{"input":"Ronald has 13 blocks. Martha has with 4 blocks. Martha finds another 80. How many blocks does Martha end with?","target":84.0}
|
||||
{"input":"Albert has 9 bottle caps. Anne has with 10 bottle caps. Anne finds another 5. How many bottle caps does Anne end with?","target":15.0}
|
||||
{"input":"Evelyn has 95 marbles. She gets 9 more from Henry. Later, Evelyn buys 6 cards at the store. How many marbles does Evelyn have in all?","target":104.0}
|
||||
{"input":"David has 8 bananas. He gets 50 more from Christopher. Later, David buys 9 cards at the store. How many bananas does David have in all?","target":58.0}
|
||||
{"input":"Emily collects 63 cards. Emily's father gives Emily 7 more. Bruce has 13 apples. How many cards does Emily have?","target":70.0}
|
||||
{"input":"Carolyn starts with 47 marbles and 6 oranges. She shares 42 with Diana. How many marbles does Carolyn end with?","target":5.0}
|
||||
{"input":"Willie has 48 bananas. Charles has 14 bananas. He loses 35. How many bananas will Willie have?","target":13.0}
|
||||
{"input":"There are 55 oranges in a box. Deborah has 11 oranges in a bag. Susan takes 35 oranges out of the box. How many oranges are left in the box?","target":20.0}
|
||||
{"input":"There are 77 apples in a box. Louise has 9 apples in a bag. Scott takes 75 apples out of the box. How many apples are left in the box?","target":2.0}
|
||||
{"input":"Bruce has 75 eggs. Kimberly has 17 eggs. He loses 70. How many eggs will Bruce have?","target":5.0}
|
||||
{"input":"There are 69 erasers in a box. Jane has 15 erasers in a bag. Doris takes 54 erasers out of the box. How many erasers are left in the box?","target":15.0}
|
||||
{"input":"There are 97 erasers in a box. Ernest has 17 erasers in a bag. Mark takes 97 erasers out of the box. How many erasers are left in the box?","target":0.0}
|
||||
{"input":"There are 88 candies in a box. Lisa has 18 candies in a bag. Diana takes 6 candies out of the box. How many candies are left in the box?","target":82.0}
|
||||
{"input":"Brandon has 96 Skittles. Bonnie has 4 Skittles. He loses 9. How many Skittles will Brandon have?","target":87.0}
|
||||
{"input":"Theresa has 32 crayons. Janice has 12 crayons. She shares 13 with Nancy. How many crayons will Theresa have?","target":19.0}
|
||||
{"input":"Craig has 20 apples. Judy has 11 apples. He shares 7 with Eugene. How many apples will Craig have?","target":13.0}
|
||||
{"input":"Heather weighs 87 pounds. Emily weighs 9 pounds. Elizabeth weighs 8 pounds. How much heavier is Heather than Emily?","target":78.0}
|
||||
{"input":"George has 2 boxes of blocks. Each box holds 6 blocks and there are 5 boxes in a case. How many blocks does George have?","target":12.0}
|
||||
{"input":"Maria has 3 boxes of eggs. Each box holds 7 eggs and there are 8 boxes in a case. How many eggs does Maria have?","target":21.0}
|
||||
{"input":"Gloria has 9 boxes of tickets. Each box holds 5 tickets and there are 10 boxes in a case. How many tickets does Gloria have?","target":45.0}
|
||||
{"input":"George has 5 boxes of eggs. Each box holds 3 eggs and there are 9 boxes in a case. How many eggs does George have?","target":15.0}
|
||||
{"input":"Jacqueline has 4 boxes of erasers. Each box holds 10 erasers and there are 9 boxes in a case. How many erasers does Jacqueline have?","target":40.0}
|
||||
{"input":"Marilyn has 40 bananas that must be put away in boxes. Daniel comes to help and brings 10 cookies to share with Marilyn. If there are 8 boxes, how many bananas must go in each box?","target":5.0}
|
||||
{"input":"Rose has 9 apples and 12 erasers. If she shares the apples among 3 friends, how many apples does each friend get?","target":3.0}
|
||||
{"input":"Joshua has 40 Skittles and 6 eggs. If he shares the Skittles among 5 friends, how many Skittles does each friend get?","target":8.0}
|
||||
{"input":"Ryan has 72 marbles and 17 blocks. If he shares the marbles among 9 friends, how many marbles does each friend get?","target":8.0}
|
||||
{"input":"Ruby has 36 candies and 6 bananas. If she shares the candies among 9 friends, how many candies does each friend get?","target":4.0}
|
||||
{"input":"Scott has 56 oranges that must be put away in boxes. Terry comes to help and brings 11 cookies to share with Scott. If there are 8 boxes, how many oranges must go in each box?","target":7.0}
|
||||
{"input":"Joyce has 40 bananas that must be put away in boxes. Fred comes to help and brings 18 cookies to share with Joyce. If there are 10 boxes, how many bananas must go in each box?","target":4.0}
|
||||
{"input":"Joe has 45 oranges that must be put away in boxes. Daniel comes to help and brings 16 cookies to share with Joe. If there are 9 boxes, how many oranges must go in each box?","target":5.0}
|
||||
{"input":"Ernest has 45 bananas that must be put away in boxes. Julie comes to help and brings 3 cookies to share with Ernest. If there are 5 boxes, how many bananas must go in each box?","target":9.0}
|
||||
{"input":"Virginia has 16 eggs and 8 Skittles. If she shares the eggs among 4 friends, how many eggs does each friend get?","target":4.0}
|
||||
{"input":"Shawn has 13 blocks. Mildred has with 2 blocks. Mildred finds another 84. How many blocks does Mildred end with?","target":86.0}
|
||||
{"input":"If Anne wandered for 3 hours at 2 miles per hour. How far did Anne go?","target":6.0}
|
||||
{"input":"It took Amanda 3 hours to stroll to Kimberly's house at 2 miles per hour. How far is it between Amanda's house and Kimberly's house?","target":6.0}
|
||||
{"input":"If Stephanie ran for 3 hours at 5 miles per hour. How far did Stephanie go?","target":15.0}
|
||||
{"input":"If Charles strolled 6 miles at 3 miles per hour, how long was Charles travelling?","target":2.0}
|
||||
{"input":"It took Carl 5 hours to ride to Ralph's house at 2 miles per hour. How far is it between Carl's house and Ralph's house?","target":10.0}
|
||||
{"input":"Emily sprinted to Timothy's house. It is 10 miles from Emily's house to Timothy's house. It took Emily 2 hours to get there. How fast did Emily go?","target":5.0}
|
||||
{"input":"If Norma wandered for 5 hours at 3 miles per hour. How far did Norma go?","target":15.0}
|
||||
{"input":"If Charles strolled 6 miles at 3 miles per hour, how long was Charles travelling?","target":2.0}
|
||||
{"input":"If Anne wandered for 3 hours at 2 miles per hour. How far did Anne go?","target":6.0}
|
||||
{"input":"If Rose strolled for 4 hours at 2 miles per hour. How far did Rose go?","target":8.0}
|
||||
{"input":"It took Amanda 2 hours to stroll to Kimberly's house at 4 miles per hour. How far is it between Amanda's house and Kimberly's house?","target":8.0}
|
||||
{"input":"Martin strolled to Lawrence's house. It is 12 miles from Martin's house to Lawrence's house. It took Martin 6 hours to get there. How fast did Martin go?","target":2.0}
|
||||
{"input":"It took Amanda 5 hours to stroll to Kimberly's house at 2 miles per hour. How far is it between Amanda's house and Kimberly's house?","target":10.0}
|
||||
{"input":"Martin strolled to Lawrence's house. It is 12 miles from Martin's house to Lawrence's house. It took Martin 6 hours to get there. How fast did Martin go?","target":2.0}
|
||||
{"input":"Emily sprinted to Timothy's house. It is 10 miles from Emily's house to Timothy's house. It took Emily 2 hours to get there. How fast did Emily go?","target":5.0}
|
||||
{"input":"Jose strolled to Jane's house. It is 24 miles from Jose's house to Jane's house. It took Jose 12 hours to get there. How fast did Jose go?","target":2.0}
|
||||
{"input":"It took Katherine 3 hours to run to Louis's house at 8 miles per hour. How far is it between Katherine's house and Louis's house?","target":24.0}
|
||||
{"input":"Patrick jogged to Aaron's house. It is 14 miles from Patrick's house to Aaron's house. It took Patrick 2 hours to get there. How fast did Patrick go?","target":7.0}
|
||||
{"input":"Rachel strolled to Nicholas's house. It is 10 miles from Rachel's house to Nicholas's house. It took Rachel 5 hours to get there. How fast did Rachel go?","target":2.0}
|
||||
{"input":"If Joan bicycled 25 miles at 5 miles per hour, how long was Joan travelling?","target":5.0}
|
||||
{"input":"Mark sprinted 24 miles at 6 miles per hour. How long did Mark sprint?","target":4.0}
|
||||
{"input":"If Benjamin strolled 14 miles at 2 miles per hour, how long was Benjamin travelling?","target":7.0}
|
||||
{"input":"Mark sprinted 24 miles at 6 miles per hour. How long did Mark sprint?","target":4.0}
|
||||
{"input":"If Joan bicycled 25 miles at 5 miles per hour, how long was Joan travelling?","target":5.0}
|
||||
{"input":"Christine wandered 20 miles at 4 miles per hour. How long did Christine wander?","target":5.0}
|
||||
{"input":"Christine wandered 20 miles at 4 miles per hour. How long did Christine wander?","target":5.0}
|
||||
{"input":"James rode 80 miles at 16 miles per hour. How long did James ride?","target":5.0}
|
||||
{"input":"Juan ran 80 miles at 10 miles per hour. How long did Juan run?","target":8.0}
|
||||
{"input":"James rode 80 miles at 16 miles per hour. How long did James ride?","target":5.0}
|
||||
{"input":"Lisa flew 256 miles at 32 miles per hour. How long did Lisa fly?","target":8.0}
|
||||
{"input":"Lisa flew 256 miles at 32 miles per hour. How long did Lisa fly?","target":8.0}
|
||||
{"input":"Christopher strolled 5 miles at 4 miles per hour. How long did Christopher stroll?","target":1.25}
|
||||
{"input":"Jose wandered 4 kilometers at 2 kilometers per hour. How long did Jose wander?","target":2.0}
|
||||
{"input":"If Teresa jogged 25 kilometers at 5 kilometers per hour, how long was Teresa jogging?","target":5.0}
|
||||
{"input":"If Teresa jogged 25 kilometers at 5 kilometers per hour, how long was Teresa jogging?","target":5.0}
|
||||
{"input":"If Teresa jogged 25 kilometers at 5 kilometers per hour, how long was Teresa jogging?","target":5.0}
|
||||
{"input":"If Benjamin skated 80 kilometers at 10 kilometers per hour, how long was Benjamin skating?","target":8.0}
|
||||
{"input":"If Heather bicycled 40 kilometers at 8 kilometers per hour, how long was Heather bicycling?","target":5.0}
|
||||
{"input":"If Benjamin skated 80 kilometers at 10 kilometers per hour, how long was Benjamin skating?","target":8.0}
|
||||
{"input":"Jeremy strolled 20 kilometers at 2 kilometers per hour. How long did Jeremy stroll?","target":10.0}
|
||||
{"input":"Jeremy strolled 20 kilometers at 2 kilometers per hour. How long did Jeremy stroll?","target":10.0}
|
||||
{"input":"If Lawrence walked 4 kilometers at 3 kilometers per hour, how long was Lawrence walking?","target":1.0}
|
||||
{"input":"You have 7 balloons and your friend has 5 balloons. How many more balloons do you have than your friend?","target":2.0}
|
||||
{"input":"2 birds were sitting on the fence. 4 more birds came to join them. How many birds are sitting on the fence?","target":6.0}
|
||||
{"input":"You have collected 7 crickets. How many more crickets do you need to collect to have 11 crickets?","target":4.0}
|
||||
{"input":"A bee has 6 legs. How many legs do 2 bees have?","target":12.0}
|
||||
{"input":"There are 6 birds and 3 nests. How many more birds are there than nests?","target":3.0}
|
||||
{"input":"There are 5 flowers and 3 bees. How many fewer bees are there than flowers?","target":2.0}
|
||||
{"input":"3 owls were sitting on the fence. 2 more owls joined them. How many owls are on the fence now?","target":5.0}
|
||||
{"input":"2 beavers were working on their home. 1 went for a swim. How many beavers are still working on their home?","target":1.0}
|
||||
{"input":"2 toucans are sitting on a tree limb. 1 more toucan joins them. How many toucans in all?","target":3.0}
|
||||
{"input":"There are 4 squirrels in a tree with 2 nuts. How many more squirrels are there than nuts?","target":2.0}
|
||||
{"input":"Mrs. Hilt bought a yoyo for 24 cents and a whistle for 14 cents. How much did she spend in all for the two toys?","target":38.0}
|
||||
{"input":"At Mrs. Hilt's house, there was 29 inches of snow, and Brecknock Elementary School received 17 inches of snow. How much more snow did Mrs. Hilt's house have?","target":12.0}
|
||||
{"input":"Mrs. Hilt reads 5 books a day. How many books does she read in 3 days?","target":15.0}
|
||||
{"input":"Mrs. Hilt saw 3 bugs eat 2 flowers each. How many flowers total did the bugs eat?","target":6.0}
|
||||
{"input":"Mrs. Hilt had 15 cents. She bought a pencil for 11 cents. How much money did she have left?","target":4.0}
|
||||
{"input":"Mrs. Hilt bought 2 pizzas. Each pizza had 8 slices. How many total slices of pizza did she have?","target":16.0}
|
||||
{"input":"Mrs. Hilt ate 5 apples every hour. How many apples had she eaten at the end of 3 hours?","target":15.0}
|
||||
{"input":"Mary\u2019s mom is getting ready for Mary\u2019s birthday party. She blew up 6 balloons this morning and 5 balloons this afternoon. How many balloons did she blow up in all?","target":11.0}
|
||||
{"input":"Shelby\u2019s teacher gives out gold stars for great math work. Yesterday, Shelby earned 4 gold stars. Today, she earned 3 more. How many gold stars did Shelby earn in all?","target":7.0}
|
||||
{"input":"The Litter Patrol picked up 10 glass bottles and 8 aluminum cans on Saturday. How many pieces of litter did they pick up altogether?","target":18.0}
|
||||
{"input":"Carson\u2019s teacher gives out gold stars for great math work. Yesterday, Carson earned 6 gold stars. Today, he earned 9 more. How many gold stars did Carson earn in all?","target":15.0}
|
||||
{"input":"There were 10 students riding on the school bus. At the first stop, 3 students got off of the bus. How many students are left on the bus?","target":7.0}
|
||||
{"input":"Lucy went to the grocery store. She bought 12 packs of cookies and 16 packs of noodles. How many packs of groceries did she buy in all?","target":28.0}
|
||||
{"input":"Roden went to a pet shop. He bought 15 gold fish and 7 blue fish. How many fish did he buy?","target":22.0}
|
||||
{"input":"I read 21 pages of my English book yesterday. Today, I read 17 pages. What is the total number of pages did I read?","target":38.0}
|
||||
{"input":"In a school, there are 542 girls and 387 boys. How many pupils are there in that school?","target":929.0}
|
||||
{"input":"Linda has 34 candies. Chloe has 28. How many candies do they have in all?","target":62.0}
|
||||
{"input":"Gino has 63 popsicle sticks. I have 50 popsicle sticks. What is the sum of our popsicle sticks?","target":113.0}
|
||||
{"input":"Lino picked up 292 shells at the seashore in the morning and 324 shells in the afternoon. How many shells did he pick up in all?","target":616.0}
|
||||
{"input":"There were 105 parents in the program and 698 pupils, too. How many people were present in the program?","target":803.0}
|
||||
{"input":"Last Saturday, Marie sold 425 magazines and 275 newspapers. What is the total number of reading materials she sold?","target":700.0}
|
||||
{"input":"There are 12 birds on the fence. 8 more birds land on the fence. How many birds are on the fence?","target":20.0}
|
||||
{"input":"22 boys went down the slide. 13 more boys went down the slide. How many boys went down the slide?","target":35.0}
|
||||
{"input":"13 ducks are swimming in a lake. 20 more ducks come to join them. How many ducks are swimming in the lake?","target":33.0}
|
||||
{"input":"30 dogs are barking. 10 more dogs start to bark. How many dogs are barking?","target":40.0}
|
||||
{"input":"Bobby ate 26 pieces of candy. Then, he ate 17 more. How many pieces of candy did Bobby eat?","target":43.0}
|
||||
{"input":"Sandy had 26 pet fish. She bought 6 more fish. How many pet fish does Sandy have now?","target":32.0}
|
||||
{"input":"The clown blew up 47 balloons. Then he blew up 13 more balloons. How many balloons does the clown have now?","target":60.0}
|
||||
{"input":"Our class got 54 books from the library. Then we got 23 more books from the library. How many books did our class get from the library?","target":77.0}
|
||||
{"input":"Tessa has 4 apples. Anita gave her 5 more. She needs 10 apples to make a pie. Does she have enough to make a pie?","target":9.0}
|
||||
{"input":"Molly had 14 candles on her birthday cake. She grew older and got 6 more on her birthday cake. How old is Molly now?","target":20.0}
|
||||
{"input":"James ate 22 carrot sticks before dinner and 15 more after dinner. How many carrot sticks did he eat?","target":37.0}
|
||||
{"input":"Jovana filled her bucket with 5 pounds of shells. If she adds 12 more pounds of shell to fill her bucket, how many pounds does she have?","target":17.0}
|
||||
{"input":"Isha\u2019s pencil is 12 cubes long. If she gets another pencil that is 12 cubes long, how many cubes long are both pencils?","target":24.0}
|
||||
{"input":"Isabella\u2019s hair is 18 cubes long. If her hair grows 4 more inches, how long will it be?","target":22.0}
|
||||
{"input":"Mrs. Sheridan has 17 cats. Mr. Sheridan gave her 14 more cats. How many cats does Mrs. Sheridan have altogether?","target":31.0}
|
||||
{"input":"Mrs. Sheridan has 22 fish. Her sister gave her 47 more fish. How many fish does she have now?","target":69.0}
|
||||
{"input":"Mrs. Heine is buying Valentine\u2019s Day treats for her 2 dogs. If she wants to buy them 3 heart biscuits each, how many biscuits does she need to buy?","target":6.0}
|
||||
{"input":"Cade had 87 marbles. He gave 8 to Dylan. How many does he have left?","target":79.0}
|
||||
{"input":"Michael has some fish in his fish tank. Ben gave him 18 more fish. Now he has 49. How many fish did he have to begin with?","target":31.0}
|
||||
{"input":"Alyssa had 129 cookies. Aiyanna has 140. How many more cookies does Aiyanna have than Alyssa?","target":11.0}
|
||||
{"input":"Daniel had some noodles. He gave 12 noodles to William. Now Daniel only has 54 noodles. How many noodles did Daniel have to begin with?","target":66.0}
|
||||
{"input":"Hayley had 25 meatballs on her plate. Kirsten stole some of her meatballs. Now she has 11 meatballs on her plate. How many meatballs did Kirsten steal?","target":14.0}
|
||||
{"input":"Isabella\u2019s hair is 18 inches long. If she gets a haircut and now her hair is 9 inches long, how much of Isabella\u2019s hair got cut off?","target":9.0}
|
||||
{"input":"Isabella\u2019s hair is 18 inches long. By the end of the year her hair is 24 inches long. How much hair did she grow?","target":6.0}
|
||||
{"input":"Jovana had 5 pounds of shells in her bucket. She added some shells and now has 28 pounds of shells. How many pounds did she add?","target":23.0}
|
||||
{"input":"Isha\u2019s pencil is 31 inches long. If she sharpens it, now her pencil is 14 inches long. How much did she sharpen off of her pencil?","target":17.0}
|
||||
{"input":"Mrs. Sheridan has 11 cats. How many more cats does Mrs. Sheridan need to have 43 cats?","target":32.0}
|
||||
{"input":"Mrs. Sheridan has 11 cats. Mrs. Garrett has 24 cats. How many more cats does Mrs. Garrett have than Mrs. Sheridan?","target":13.0}
|
||||
{"input":"Mrs. Wong had 30 Valentines. She gave 8 Valentines to her children. How many does she have left?","target":22.0}
|
||||
{"input":"Mrs. Franklin had 58 Valentines. Mrs. Franklin gave some to her students. Now she has 16. How many Valentines did Mrs. Franklin give to her students?","target":42.0}
|
||||
{"input":"Mrs. Snyder made 86 heart cookies. She made 36 red cookies and the rest are pink. How many pink cookies did she make?","target":50.0}
|
||||
{"input":"Mrs. Santiago has 58 red roses. Mrs. Garrett has 24. How many more red roses does Mrs. Santiago have than Mrs. Garrett?","target":34.0}
|
||||
{"input":"14 birds were sitting in a tree. 21 more birds flew up to the tree. How many birds were there altogether in the tree?","target":35.0}
|
||||
{"input":"29 birds were sitting in a tree. Some more fly up to the tree. Then there were 42 birds in the tree. How many more flew up to the tree?","target":13.0}
|
||||
{"input":"Cindy\u2019s mom baked 41 cookies. Paul\u2019s dad baked 38 cookies. They both brought them to school for a party. How many cookies did they have altogether?","target":79.0}
|
||||
{"input":"18 children were riding on the bus. At the bus stop, some more children got on the bus. Then there were 25 children altogether on the bus. How many children got on the bus at the bus stop?","target":7.0}
|
||||
{"input":"Misha has 34 dollars. How many more dollars does she have to earn to have 47 dollars?","target":13.0}
|
||||
{"input":"There were 28 girls and 35 boys on the playground at recess. How many children were there in all?","target":63.0}
|
||||
{"input":"There were 44 boys and 53 girls on the playground at recess. How many children were on the playground in all?","target":97.0}
|
||||
{"input":"There were 58 geese and 37 ducks in the marsh. How many birds were there in all?","target":95.0}
|
||||
{"input":"Paul had 42 strawberries in his basket. He picked 78 more strawberries. How many strawberries did he have then?","target":120.0}
|
||||
{"input":"Robin had 18 pieces of gum. Her brother gave her some more pieces. Now Robin has 44 pieces in all. How many pieces of gum did Robin's brother give her?","target":26.0}
|
||||
{"input":"Bobby has 142 books. Kristi has 78 books. How many more books does Bobby have than Kristi?","target":64.0}
|
||||
{"input":"Tommy had some balloons. His mom gave him 34 more balloons for his birthday. Then, Tommy had 60 balloons. How many balloons did Tommy have to start with?","target":26.0}
|
||||
{"input":"There were 14 kids on the soccer field. 22 kids decided to join in. Now how many kids are on the soccer field?","target":36.0}
|
||||
{"input":"Jane has 28 old, brown sheets of drawing paper and 27 old, yellow sheets of drawing paper. How many pieces of drawing paper does she have?","target":55.0}
|
||||
{"input":"Mikey had 356 leaves. Some of her leaves blew away. Now she has 112 leaves left. How many of her leaves blew away?","target":244.0}
|
||||
{"input":"Marcus has 210 baseball cards. He has 58 more than Carter. How many baseball cards does Carter have?","target":152.0}
|
||||
{"input":"Gavin has 23 shirts. 6 are blue the rest are green. How many green shirts does Gavin have?","target":17.0}
|
||||
{"input":"Ethan has 31 presents. Alissa has 22 more than Ethan. How many presents does Alissa have?","target":53.0}
|
||||
{"input":"On the first day of the week Pat had 39 stickers. Pat earned 22 more during the week. How many stickers did Pat have at the end of the week?","target":61.0}
|
||||
{"input":"Kelly had 56 apples. How many more apples does Kelly need to pick to have 105 apples altogether?","target":49.0}
|
||||
{"input":"Todd has some gum. Steve gave him 16 more pieces of gum. Now Todd has 54 pieces of gum. How many pieces did Todd have to start with?","target":38.0}
|
||||
{"input":"Josh had 142 pencils. He gave 31 pencils to Dorothy. How many pencils does Josh have left?","target":111.0}
|
||||
{"input":"Nell collects baseball cards. She had 304 cards. She gave some of her cards to Jeff and now has 276 cards left. How many cards did Nell give to Jeff?","target":28.0}
|
||||
{"input":"Sarah had some trucks. She gave 13 to Jeff, and now she has 38 trucks left. How many trucks did Sarah have to start with?","target":51.0}
|
||||
{"input":"There are 40 boys and some girls on the playground. There are 117 children altogether. How many girls are on the playground?","target":77.0}
|
||||
{"input":"Megan had 49 markers. Robert gave her 39 more markers. How many markers does Megan have altogether?","target":88.0}
|
||||
{"input":"Each CD rack holds 8 CDs. A shelf can hold 4 racks. How many total CDs can fit on the shelf?","target":32.0}
|
||||
{"input":"Carla has some marbles. She bought 134 marbles. Now she has 187 marbles. How many did she start with?","target":53.0}
|
||||
{"input":"Kelly has 50 Nintendo games. How many does she need to give away so that she will have 35 games left?","target":15.0}
|
||||
{"input":"Connie had some marbles. She gave 73 to Juan. Now she has 70 marbles left. How many did she have to start with?","target":143.0}
|
||||
{"input":"Connie has 41 red markers and 64 blue markers. How many markers does she have altogether?","target":105.0}
|
||||
{"input":"Iesha has 58 books. 19 are about school and the rest are about sports. How many books about sports does Iesha have?","target":39.0}
|
||||
{"input":"James has 232 balloons. Amy has 101 balloons. How many more balloons does James have than Amy?","target":131.0}
|
||||
{"input":"Sean has 45 whistles. He has 32 more whistles that Charles. How many whistles does Charles have?","target":13.0}
|
||||
{"input":"Connie has 39 marbles. Juan has 25 more marbles than Connie. How many marbles does Juan have?","target":64.0}
|
||||
{"input":"Joe had 50 toy cars. If he gets 12 more cars, how many cars will he have then?","target":62.0}
|
||||
{"input":"There are 64 pigs in the barn. Some more come to join them. Now there are 86 pigs. How many pigs came to join them?","target":22.0}
|
||||
{"input":"Rosa had 67 flowers. Andre gave her some more flowers. Now, Rosa has 90 flowers. How many flowers did Andre give to Rosa?","target":23.0}
|
||||
{"input":"Adolfo made a tower with 35 blocks. He added some more blocks and now he has 65 blocks. How many did he have to add?","target":30.0}
|
||||
{"input":"Josh had 16 marbles in his collection. He lost 7 marbles. How many marbles does he have now?","target":9.0}
|
||||
{"input":"Megan has 19 seashells. How many more seashells does she need to find to have 25 seashells in her collection?","target":6.0}
|
||||
{"input":"Brad has 17 balloons. 8 balloons are red and the rest are green. How many green balloons does Brad have?","target":9.0}
|
||||
{"input":"There are 38 books on the shelf. Marta put 10 more books on the shelf. How many books are on the shelf now?","target":48.0}
|
||||
{"input":"A bee has 6 legs. How many legs do 8 bees have?","target":48.0}
|
||||
{"input":"Mrs. Hilt bought an ice cream cone for 99 cents. How much would 2 ice cream cones cost?","target":198.0}
|
||||
{"input":"Mrs. Hilt wants to make a border around her garden. She needs 125 rocks to complete the border. She has 64 rocks. How many more rocks does she need to complete the border?","target":61.0}
|
||||
{"input":"Mrs. Hilt and her sister drove to a concert 78 miles away. They drove 32 miles and then stopped for gas. Her sister put 28 gallons of gas in the car. How many miles did they have left to drive?","target":46.0}
|
||||
{"input":"Mrs. Hilt bought 6 hot dogs. Each hot dog cost 50 cents. How much money did she pay for all of the hot dogs?","target":300.0}
|
||||
{"input":"Mrs. Hilt has 50 cents. A pencil costs 5 cents. How many pencils can she buy with the money she has?","target":10.0}
|
||||
{"input":"Mrs. Hilt bought 3 pizzas for $8 each. What was the total amount she paid for the three pizzas?","target":24.0}
|
||||
{"input":"The Hawks scored 3 touchdowns worth 7 points each. How many points do they have?","target":21.0}
|
||||
{"input":"Zach scored 42 points in the football game. Ben scored 21 points. How many more points did Zach score?","target":21.0}
|
||||
{"input":"Kate has 223 pennies. John has 388 pennies. How many more pennies does John have?","target":165.0}
|
||||
{"input":"Tim had 50 cents. He paid 45 cents for a candy bar. How much change will he get?","target":5.0}
|
||||
{"input":"Mark has 13 trees in his backyard. If he plants 12 more, how many trees will he have?","target":25.0}
|
||||
{"input":"Kim has 4 cousins. She wants to give each cousin 5 pieces of gum. How much gum will she need?","target":20.0}
|
||||
{"input":"Dan has $3.00. He bought a candy bar for $1.00. How much money is left?","target":2.0}
|
||||
{"input":"5 boats are in the lake. Each boat has 3 people. How many people are on boats in the lake?","target":15.0}
|
||||
{"input":"Charlie has 31 more snowballs than Lucy. She has 19 snowballs. How many does Charlie have?","target":50.0}
|
||||
{"input":"Randy has 78 blocks. He uses 19 blocks to build a tower. How many blocks are left?","target":59.0}
|
||||
{"input":"Bryan has 50 skittles. Ben has 20 M&M\u2019s. Who has more? How many more does he have?","target":30.0}
|
||||
{"input":"Paul got a box of 479 crayons for his birthday. At the end of the school year, he only had 134 left. How many crayons had been lost or given away?","target":345.0}
|
||||
{"input":"179 birds were sitting in a tree. 38 more birds flew up to the tree. How many birds were there altogether in the tree?","target":217.0}
|
||||
{"input":"Cindy\u2019s mom baked 1215 cookies. Paul\u2019s dad baked 1112 cookies. They both brought them to school for a party. How many cookies did they have altogether?","target":2327.0}
|
||||
{"input":"231 birds were sitting in a tree. Some more fly up to the tree. Then there were 312 birds in the tree. How many more fly up to the tree?","target":81.0}
|
||||
{"input":"64 children were riding on the bus. At the bus stop, some more children got on the bus. Then there were 78 children altogether on the bus. How many children got on the bus at the bus stop?","target":14.0}
|
||||
{"input":"Misha has 34 dollars. How many dollars does she have to earn to have 47 dollars to buy a dog?","target":13.0}
|
||||
{"input":"Elisa has 37 dollars. How many more dollars does she have to earn to have 53 dollars?","target":16.0}
|
||||
{"input":"James had 39 stickers. He got some more stickers for his birthday. Then he had 61 stickers. How many stickers did James get for his birthday?","target":22.0}
|
||||
{"input":"There were 27 boys and 35 girls on the playground at recess. How many children were on the playground at recess?","target":62.0}
|
||||
{"input":"There were 58 geese and 37 ducks in the marsh. How many birds were in the marsh?","target":95.0}
|
||||
{"input":"Paul had 28 strawberries in his basket. He picked 35 more strawberries. How many strawberries did he have then?","target":63.0}
|
||||
{"input":"Gary had 73 dollars. He spent 55 dollars on a pet snake. How many dollars did Gary have left?","target":18.0}
|
||||
{"input":"There are 397 butterflies. Each butterfly has 12 black dots and 17 yellow dots. How many black dots are there in all?","target":4764.0}
|
||||
{"input":"How many cookies would you have if you had 37 bags of cookies with 19 cookies in each bag?","target":703.0}
|
||||
{"input":"How much would 136 pieces of bubble gum cost if each piece costs 18 cents?","target":2448.0}
|
||||
{"input":"For the fifth grade play, the chairs have been put into 27 rows with 16 chairs in each row. How many chairs have been put out for the play?","target":432.0}
|
||||
{"input":"The Ferris wheel in Paradise Park has 14 seats. Each seat can hold 6 people. How many people can ride the Ferris wheel at the same time?","target":84.0}
|
||||
{"input":"A garden has 52 rows and 15 columns of bean plans. How many plants are there in all?","target":780.0}
|
||||
{"input":"A sandbox is 312 centimeters long and 146 centimeters wide. How many square centimeters of ground does the sandbox cover?","target":45552.0}
|
||||
{"input":"Sarah picked 45 apples. Her brother picked 9 apples. How many times as many apples did Sarah pick?","target":5.0}
|
||||
{"input":"There are 261 fishbowls. Each fishbowl has 23 fish. How many fish are there?","target":6003.0}
|
||||
{"input":"We ordered 21 pizzas. Each pizza has 8 slices. How many slices of pizza are there altogether?","target":168.0}
|
||||
{"input":"Emily collected eggs from the hen and put them into 303 baskets. She put 28 eggs into each basket. How many eggs did Emily collect?","target":8484.0}
|
||||
{"input":"There were 150 book shelves. Each book shelf had 15 books. How many books were on the shelves?","target":2250.0}
|
||||
{"input":"There are 84 leaves. There are 139 ladybugs on each leaf. How many ladybugs are there in all?","target":11676.0}
|
||||
{"input":"There are 37 baskets. There are 17 apples in each basket. How many apples are there in all?","target":629.0}
|
||||
{"input":"There are 544 pots. Each pot has 32 flowers in it. How many flowers are there in all?","target":17408.0}
|
||||
{"input":"Your class is having a pizza party. You buy 5 pizzas. Each pizza has 4 slices. How many slices is that altogether?","target":20.0}
|
||||
{"input":"Beth has 4 packs of crayons. Each pack has 10 crayons in it. She also has 6 extra crayons. How many crayons does Beth have altogether?","target":40.0}
|
||||
{"input":"Ted has 15 candy bars. He wants to put them into 5 bags so there are the same number of candy bars in each bag. How many candy bars should go in each bag?","target":3.0}
|
||||
{"input":"A candy store has 6 boxes of chocolates. Each box has 500 pieces. How many pieces are there altogether in the boxes?","target":3000.0}
|
||||
{"input":"Martha bought 18 small cakes. She has 3 children. She would like to divide the cakes among her children so that each child gets the same amount. How many cakes would each child get?","target":6.0}
|
||||
{"input":"Julian is writing a comic book. His story has 143 frames in all. If he wants to put exactly 11 frames on each page, how many pages would he have?","target":13.0}
|
||||
{"input":"Jesse\u2019s room is 12 feet long and 8 feet wide. How much carpet does she need to cover the whole floor?","target":96.0}
|
||||
{"input":"There are 5 people on the Green Bay High track team. If a relay race is 150 meters long, how far will each team member have to run?","target":30.0}
|
||||
{"input":"It takes 4 feet of cotton to make a tee-shirt. How many tee-shirts can be made with 60 feet of material?","target":15.0}
|
||||
{"input":"Lukas averages 12 points per game in basketball. How many points would he score in 5 games?","target":60.0}
|
||||
{"input":"My car gets 20 miles per gallon. How many miles can I drive on 5 gallons of gas?","target":100.0}
|
||||
{"input":"It takes 7 minutes to bake one pan of cookies. How long will it take to bake 4 pans of cookies?","target":28.0}
|
||||
{"input":"My car gets 20 miles per gallon of gas. If Grandma\u2019s house is 100 miles away, how many gallons of gas would it take to get to her house?","target":5.0}
|
||||
{"input":"Melissa scored 120 points in each game. How many points did she score in 10 games?","target":1200.0}
|
||||
{"input":"Johnny practiced for the track team and ran 3 laps per minute. How many minutes did it take Johnny to run 10 laps?","target":3.33333}
|
||||
{"input":"Our watermelons have 100 seeds each. If we have 4 watermelons, how many seeds should there be when all seeds are taken out of the watermelons?","target":400.0}
|
||||
{"input":"Robin has 9 packages of gum. There are 15 pieces in each package. How many pieces of gum does Robin have?","target":135.0}
|
||||
{"input":"63 people are going to the zoo. There are 3 cars to take people to the zoo. How many will go in each car if the same number go in each car?","target":21.0}
|
||||
{"input":"I have 80 cents to buy candy. If each gumdrop costs 4 cents, how many gumdrops can I buy?","target":20.0}
|
||||
{"input":"Maggi had 3 packages of cupcakes. There are 4 cupcakes in each package. She ate 5 cupcakes. How many are left?","target":12.0}
|
||||
{"input":"58 children are taking a bus to the zoo. They sit 2 children in every seat. How many seats will the children need in all?","target":29.0}
|
||||
{"input":"Marlee has 12 guests coming to her Halloween party. Each table will hold 3 guests. How many tables will she need?","target":4.0}
|
||||
{"input":"Warren has 252 guests coming to his party. Each table will hold 4 guests. How many tables will he need?","target":63.0}
|
||||
{"input":"Arthur baked 35 muffins. How many more muffins does Arthur have to bake to have 83 muffins?","target":48.0}
|
||||
{"input":"There are 10 stickers on a page. If you have 22 pages of stickers, how many stickers do you have?","target":220.0}
|
||||
{"input":"There are 96 cupcakes for 8 children to share. How much will each person get if they share the cupcakes equally?","target":12.0}
|
||||
{"input":"Jose has 85 peanuts. Kenya has 48 more than Jose. How many peanuts does Kenya have?","target":133.0}
|
||||
{"input":"White t-shirts can be purchased in packages of 6. If Mom buys 71 packages, how many white t-shirts will she have?","target":426.0}
|
||||
{"input":"I have 648 pencils. If I put 4 pencils in each pencil box, how many pencil boxes will I fill?","target":162.0}
|
||||
{"input":"Uncle Dave bought 143 ice cream sandwiches. If he wants to give them to his 11 hungry nieces, how many can each niece get?","target":13.0}
|
||||
{"input":"Megan had 217 markers. Robert gave her 109 more markers. How many markers does Megan have altogether?","target":326.0}
|
||||
{"input":"A DVD book holds 126 DVDs. There are 81 DVDs already in the book. How many more DVDs can be put in the book?","target":45.0}
|
||||
{"input":"Carla has some marbles. She bought 489 marbles. Now she has 2778 marbles all together. How many did she start with?","target":2289.0}
|
||||
{"input":"Kelly has 121 Nintendo games. How many does she need to give away so that she will have 22 games left?","target":99.0}
|
||||
{"input":"Connie had some marbles. She gave 183 to Juan. Now she has 593 marbles left. How many did she have to start with?","target":776.0}
|
||||
{"input":"Connie has 2315 red markers and 1028 blue markers. How many markers does she have altogether?","target":3343.0}
|
||||
{"input":"Iesha has 344 books. 136 are about school and the rest are about sports. How many books about sports does Iesha have?","target":208.0}
|
||||
{"input":"Sean has 223 whistles. He has 95 more whistles that Charles. How many whistles does Charles have?","target":128.0}
|
||||
{"input":"Robin has 27 packages of gum. There are 18 pieces in each package. How many pieces of gum does Robin have?","target":486.0}
|
||||
{"input":"Adam has $5.00 to buy an airplane that costs $4.28. How much change will he get?","target":0.72}
|
||||
{"input":"If each ball costs $1.54, how much must Kyoko pay for 3 balls?","target":4.62}
|
||||
{"input":"Mr. Guzman bought 48 doughnuts packed equally into 4 boxes. How many doughnuts were in each box?","target":12.0}
|
||||
{"input":"The Sumata family took a 5-day vacation by car. Each day they drove 250 miles. How many total miles did they drive?","target":1250.0}
|
||||
{"input":"The town of Milburg has 5256 grown-ups and 2987 children. How many people live in Milburg?","target":8243.0}
|
||||
{"input":"Lisa rented 4 DVDs for $4.80. How much did each DVD cost to rent?","target":1.2}
|
||||
{"input":"There were 3409 pieces of candy in a jar. If 145 pieces were red and the rest were blue, how many were blue?","target":3264.0}
|
||||
{"input":"Third-grade students went to a concert in 8 buses. Each bus took 45 students. How many students went to the concert?","target":360.0}
|
||||
{"input":"There are 124 students making 3 stars each for the school wall. How many stars will they make all together?","target":372.0}
|
||||
{"input":"6 students were sitting at each table in the lunchroom. There are 34 tables. How many students were sitting in the lunchroom?","target":204.0}
|
||||
{"input":"Tyler had 15 dogs. Each dog had 5 puppies. How many puppies does Tyler now have?","target":75.0}
|
||||
{"input":"The farmer had 127 apples. He gave 88 apples to his neighbor. How many apples does he have now?","target":39.0}
|
||||
{"input":"Jill invited 37 people to her birthday party. They each ate 8 pieces of pizza. How many pieces of pizza did they eat?","target":296.0}
|
||||
{"input":"Mrs. Hilt uses 2 ounces of detergent to wash a pound of clothes. How many ounces of soap will she use to wash 9 pounds of clothes?","target":18.0}
|
||||
{"input":"Mrs. Hilt went to a concert. A total of 65899 people attended the concert. The next week, she went to a second concert, which had 119 more people in attendance. How many people were at the second concert?","target":66018.0}
|
||||
{"input":"Mrs. Hilt baked pies last weekend for a holiday dinner. She baked 16 pecan pies and 14 apples pies. If she wants to arrange all of the pies in rows of 5 pies each, how many rows will she have?","target":30.0}
|
||||
{"input":"Mrs. Hilt needs to share $3.75 equally among 3 total people. How much money will each person get?","target":1.25}
|
||||
{"input":"Mrs. Hilt read 4 books. Each book had 17 chapters in it. How many chapters did Mrs. Hilt read?","target":68.0}
|
||||
{"input":"Mrs. Hilt saw 144 bees in the hive. The next day she saw 3 times that many. How many bees did she see on the second day?","target":432.0}
|
||||
{"input":"Mrs. Hilt measured the distance from her desk to the water fountain. It was 30 feet. How many feet will Mrs. Hilt walk on her trips to the fountain if she goes to the water fountain 4 times today?","target":120.0}
|
||||
{"input":"Lucy has an aquarium with 212 fish. She wants to buy 68 more fish. How many fish would Lucy have then?","target":280.0}
|
||||
{"input":"Lucy has 212 fish. How many more fish does she need to buy to have 280 fish?","target":68.0}
|
||||
{"input":"Rupert and Ronald aced their math test. So their mother bought for them a wonderful trampoline yesterday. Ronald jumped 157 times on the trampoline. Rupert jumped 86 more times than Ronald. How many times did they jump altogether?","target":243.0}
|
||||
{"input":"Harry Hound had a terrible earache yesterday. When I peered into his ears yesterday, I found 36 frisky fleas having a party in his right ear and 85 baby fleas sleeping peacefully in his left ear. I cleaned out Harry Hound's ears. How many fleas perished?","target":121.0}
|
||||
{"input":"A pet supply store has 600 bags of dog food and 327 bags of cat food. How many more bags of dog food are there than cat food?","target":273.0}
|
||||
{"input":"532 people are watching a movie in a theater. The theater has 750 seats. How many seats are empty in the theater?","target":218.0}
|
||||
{"input":"Each bag contains 23 pounds of oranges. How many pounds of oranges are in 45 bags?","target":1035.0}
|
||||
{"input":"Ellen went to a garage sale to buy chairs. Each chair cost 15 dollars. How much money did Ellen spend for the 12 chairs she bought?","target":180.0}
|
||||
{"input":"Albert has two snakes. The garden snake is 10 inches long. The boa constrictor is 7 times longer than the garden snake. How long is the boa constrictor?","target":70.0}
|
||||
{"input":"Albert\u2019s cabbage patch has 12 rows of cabbage. In each row, there are 15 heads of cabbage. How many heads of cabbage does Albert have in all?","target":180.0}
|
||||
{"input":"Marie can bike at a speed of 12 miles an hour. How far can she bike in 31 hours?","target":372.0}
|
||||
{"input":"Tammy drove 55 miles in one hour. At that rate, how far can she drive in 36 hours?","target":1980.0}
|
||||
{"input":"It takes 4 apples to make 1 pie. How many apples does it take to make 504 pies?","target":2016.0}
|
||||
{"input":"You have 24 cookies and want to share them equally with 6 people. How many cookies would each person get?","target":4.0}
|
||||
{"input":"You are reading a book with 120 pages. If you want to read the same number of pages each night, how many would you have to read each night to finish in 10 days?","target":12.0}
|
||||
{"input":"A cereal box holds 18 cups of cereal. Each serving is 2 cups. How many servings are in the whole box?","target":9.0}
|
||||
{"input":"A box of books weighs 42 pounds. Each book weighs 3 pounds. How many books are there in the box?","target":14.0}
|
||||
{"input":"Sue\u2019s mother made 75 cookies. She put the cookies in bags, with 3 cookies in each bag. How many bags could she fill up?","target":25.0}
|
||||
{"input":"Frank worked 8 hours on the first 4 days of the week. How many hours did he work in all?","target":32.0}
|
||||
{"input":"Sue\u2019s family went on vacation. Her mom drove the car at 60 mph. They camped at a campground after traveling for 5 hours. How far was the campground from their home?","target":300.0}
|
||||
{"input":"Brett drove 55 miles every hour. How many miles would he drive in 8 hours?","target":440.0}
|
||||
{"input":"A perfect score is 21 points. How many points would you have after 3 perfect games in a row?","target":63.0}
|
||||
{"input":"Brian\u2019s car gets 20 miles per gallon. On his last trip, he used 3 gallons of gas. How many miles did he travel on his last trip?","target":60.0}
|
||||
{"input":"Bob\u2019s car gets 10 kilometers per gallon. How far can he drive on 10 gallons of gas?","target":100.0}
|
||||
{"input":"Each cup contains 8 ounces. How many ounces are in 33 cups?","target":264.0}
|
||||
{"input":"A chocolate chip cookie recipe calls for 2 cups of chocolate chips. You want to make 23 recipes for a bake sale. How many cups of chocolate chips will be needed to make all the cookie recipes?","target":46.0}
|
||||
{"input":"I have a pet golden retriever. Each year he gains 11 pounds. He is 8 years old. How many pounds does he weigh?","target":88.0}
|
||||
{"input":"Mary\u2019s car gets 20 miles per gallon of gas. How far can she drive on 14 gallons of gas?","target":280.0}
|
||||
{"input":"I walked 2 miles in 1 hour for Relay for Life. If I maintained this pace for the 8 hours I walk, how many miles total will I walk?","target":16.0}
|
||||
{"input":"I walk 1 mile every 15 minutes. I walked 3 miles. How many minutes did it take me ?","target":45.0}
|
||||
{"input":"Lansing has 25 elementary schools. There are 247 students in each school. How many elementary students are there altogether in Lansing?","target":6175.0}
|
||||
{"input":"256 students are going to the zoo. They have to be divided into groups so that each teacher has one group. There are 8 teachers. How many students will be in each group?","target":32.0}
|
||||
{"input":"A fruit farm packs oranges in boxes that hold 10 each. One day it packs 2650 oranges. How many boxes did they use?","target":265.0}
|
||||
{"input":"You want to give your baseball cards to your 5 best friends. You have 455 baseball cards. How many would each get, if you share them equally?","target":91.0}
|
||||
{"input":"Ellen had 2080 Legos, but she lost 17 Legos. How many Legos does she have now?","target":2063.0}
|
||||
{"input":"Arthur baked 115 muffins. James baked 12 times as many. How many muffins did James bake?","target":1380.0}
|
||||
{"input":"There are 14240 books in a library. They are arranged on shelves that hold 8 books each. How many shelves are in the library?","target":1780.0}
|
||||
{"input":"219 people are going to the zoo, and there are 3 buses to take people. How many will go in each bus if the same number go in each one and all of the people go to the zoo?","target":73.0}
|
||||
{"input":"I have 224 cents to buy candy. If each piece of bulk candy costs 8 cents, how many gumdrops can I buy?","target":28.0}
|
||||
{"input":"My dog had some bones. Then, he dug up 367 bones. Now he has 860 bones. How many bones did he start with?","target":493.0}
|
||||
{"input":"Jazmin had 1209 dolls and Geraldine had 2186 dolls. If they put their dolls together, how many would they have?","target":3395.0}
|
||||
{"input":"Mariela was in the hospital and she got 403 get well cards from around the country. When she got home she got 287 more cards from friends and family. How many get well cards did Mariela get?","target":690.0}
|
||||
{"input":"Cody is 14 years old. His grandmother is 6 times as old as he is. How old is Cody\u2019s grandmother?","target":84.0}
|
||||
{"input":"Jorge scored 156 goals playing soccer last season. This season he scored 187 goals. What is the total number of goals Jorge scored?","target":343.0}
|
||||
{"input":"Ceasar needs to read a 563 page book for school. He has already read 147 pages. How many pages does he have left to read?","target":416.0}
|
||||
{"input":"Omar and Karen made egg rolls to share at the school potluck. Omar rolled 219 egg rolls. Karen rolled 229 egg rolls. What is the total number of egg rolls Omar and Karen rolled?","target":448.0}
|
||||
{"input":"Beka flew 873 miles to visit her aunt. Jackson flew 563 miles to visit his aunt. How many more miles did Beka fly than Jackson?","target":310.0}
|
||||
{"input":"Darius drove 679 miles to visit his grandmother. Julia drove 998 miles to visit her grandmother. What is the total number of miles Darius and Julia drove?","target":1677.0}
|
||||
{"input":"Isabel bought 900 pieces of paper. She used 156 pieces of the paper. How many pieces of paper does she have left?","target":744.0}
|
||||
{"input":"Mrs. Hilt looked at her car's odometer before a trip. The odometer showed that she had traveled 212.3 miles. When she stopped for lunch, the odometer read 372.0. How many miles had she traveled?","target":159.7}
|
||||
{"input":"Mrs. Hilt impressed 2436 fans at the basketball game on Friday. If the fans were seated in equal groups on 3 sets of bleachers, how many fans were on each set?","target":812.0}
|
||||
{"input":"Your class had a pizza party. 0.375 of a pizza was left over, and 0.5 of another pizza was left over. You put them both into one box. How much pizza do you have altogether?","target":0.875}
|
||||
{"input":"A cake recipe requires 0.6 cup of sugar for the frosting and 0.2 cup of sugar for the cake. How much sugar is that altogether?","target":0.8}
|
||||
{"input":"After a party, 0.625 of the cake is left over. That night, big brother eats 0.25 of the cake. How much is left over after that?","target":0.375}
|
||||
{"input":"You go out for a long walk. You walk 0.75 mile and then sit down to take a rest. Then you walk 0.25 of a mile. How far did you walk altogether?","target":1.0}
|
||||
{"input":"John needs $2.50. He has $0.75. How much more money does he need?","target":1.75}
|
||||
{"input":"Jane buys an apple for $0.75 and pays with a $5.00 bill. How much change will she get?","target":4.25}
|
||||
{"input":"John walks 0.7 miles to school and Nina walks 0.4 miles to school. How much farther does John walk than Nina?","target":0.3}
|
||||
{"input":"One pencil weighs 28.3 grams. How much do 5 pencils weigh?","target":141.5}
|
||||
{"input":"Andrew spent 3 day working on his Science report. He worked for 2.5 hours each day. How many hours did he work?","target":7.5}
|
||||
Executable
+2065
File diff suppressed because it is too large
Load Diff
Executable
+2
@@ -0,0 +1,2 @@
|
||||
MIT OpenCourseWare:
|
||||
- Solving Quantitative Reasoning Problems with Language Models. https://openreview.net/forum?id=IFXTZERXdM7
|
||||
Executable
+272
@@ -0,0 +1,272 @@
|
||||
{"problem": "Each of the two Magellan telescopes has a diameter of $6.5 \\mathrm{~m}$. In one configuration the effective focal length is $72 \\mathrm{~m}$. Find the diameter of the image of a planet (in $\\mathrm{cm}$ ) at this focus if the angular diameter of the planet at the time of the observation is $45^{\\prime \\prime}$.", "solution": "Start with:\n\\[\ns=\\alpha f \\text {, }\n\\]\nwhere $s$ is the diameter of the image, $f$ the focal length, and $\\alpha$ the angular diameter of the planet. For the values given in the problem:\n\\[\ns=\\frac{45}{3600} \\frac{\\pi}{180} 7200=\\boxed{1.6} \\mathrm{~cm}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 0}
|
||||
{"problem": "A white dwarf star has an effective temperature, $T_{e}=50,000$ degrees Kelvin, but its radius, $R_{\\mathrm{WD}}$, is comparable to that of the Earth. Take $R_{\\mathrm{WD}}=10^{4} \\mathrm{~km}\\left(10^{7} \\mathrm{~m}\\right.$ or $\\left.10^{9} \\mathrm{~cm}\\right)$. Compute the luminosity (power output) of the white dwarf. Treat the white dwarf as a blackbody radiator. Give your answer in units of ergs per second, to two significant figures.", "solution": "\\[\n\\begin{aligned}\nL=4 \\pi R^{2} \\sigma T_{e}^{4} &=4 \\pi\\left(10^{9}\\right)^{2}\\left(5.7 \\times 10^{-5}\\right)(50,000)^{4} \\operatorname{ergs~s}^{-1} \\\\\nL & \\simeq \\boxed{4.5e33} \\mathrm{ergs} \\mathrm{s}^{-1} \\simeq 1 L_{\\odot}\n\\end{aligned}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 1}
|
||||
{"problem": "Preamble: A prism is constructed from glass and has sides that form a right triangle with the other two angles equal to $45^{\\circ}$. The sides are $L, L$, and $H$, where $L$ is a leg and $H$ is the hypotenuse. A parallel light beam enters side $L$ normal to the surface, passes into the glass, and then strikes $H$ internally. The index of refraction of the glass is $n=1.5$.\n\nCompute the critical angle for the light to be internally reflected at $H$. Give your answer in degrees to 3 significant figures.", "solution": "From Snell's law we have:\n\\[\n\\begin{gathered}\nn_{g} \\sin \\left(\\theta_{g}\\right)=n_{\\text {air }} \\sin \\left(\\theta_{\\text {air }}\\right) \\\\\n\\sin \\left(\\theta_{\\text {crit }}\\right)=\\frac{1}{1.5} \\sin \\left(90^{\\circ}\\right) \\Rightarrow \\theta_{\\text {crit }}=\\boxed{41.8}^{\\circ}\n\\end{gathered}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 2}
|
||||
{"problem": "A particular star has an absolute magnitude $M=-7$. If this star is observed in a galaxy that is at a distance of $3 \\mathrm{Mpc}$, what will its apparent magnitude be?", "solution": "\\[\n\\text { Given: } M=-7 \\text { and } d=3 \\mathrm{Mpc}\n\\]\n\\[\n\\begin{aligned}\n & \\text { Apparent Magnitude: } m=M+5 \\log \\left[\\frac{d}{10 \\mathrm{pc}}\\right]=-7+5 \\log \\left[\\frac{3 \\times 10^{6}}{10}\\right]=\\boxed{20.39} \\\\\n\\end{aligned}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 3}
|
||||
{"problem": "Find the gravitational acceleration due to the Sun at the location of the Earth's orbit (i.e., at a distance of $1 \\mathrm{AU}$ ). Give your answer in meters per second squared, and express it to one significant figure.", "solution": "\\begin{equation}\nF = ma = \\frac{GM_{\\odot}m}{r^2},\n\\end{equation}\nso \n\\begin{equation}\na = \\frac{GM_{\\odot}{r^2}}\n\\end{equation}\n\nPlugging in values for $G$, $M_{\\odot}$, and $r$ gives $a = \\boxed{0.006}$ meters per second squared.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 4}
|
||||
{"problem": "Preamble: A collimated light beam propagating in water is incident on the surface (air/water interface) at an angle $\\theta_w$ with respect to the surface normal.\n\nSubproblem 0: If the index of refraction of water is $n=1.3$, find an expression for the angle of the light once it emerges from the water into the air, $\\theta_a$, in terms of $\\theta_w$.\n\n\nSolution: Using Snell's law, $1.3 \\sin{\\theta_w} = \\sin{\\theta_a}$. So $\\theta_a = \\boxed{\\arcsin{1.3 \\sin{\\theta_w}}}$.\n\nFinal answer: The final answer is \\arcsin{1.3 \\sin{\\theta_w}}. I hope it is correct.\n\nSubproblem 1: What is the critical angle, i.e., the critical value of $\\theta_w$ such that the light will not emerge from the water? Leave your answer in terms of inverse trigonometric functions; i.e., do not evaluate the function.", "solution": "The relation derived in the previous problem is $\\theta_a = \\arcsin{1.3 \\sin{\\theta_w}}$. The critical angle thus occurs when $1.3 \\sin{\\theta_w}$ exceeds unity, because then there is no corresponding solution for $\\theta_a$. So the answer is $\\boxed{np.arcsin(10/13)}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 5}
|
||||
{"problem": "Find the theoretical limiting angular resolution (in arcsec) of a commercial 8-inch (diameter) optical telescope being used in the visible spectrum (at $\\lambda=5000 \\AA=500 \\mathrm{~nm}=5 \\times 10^{-5} \\mathrm{~cm}=5 \\times 10^{-7} \\mathrm{~m}$). Answer in arcseconds to two significant figures.", "solution": "\\[\n\\theta=1.22 \\frac{\\lambda}{D}=1.22 \\frac{5 \\times 10^{-5} \\mathrm{~cm}}{8 \\times 2.54 \\mathrm{~cm}}=2.46 \\times 10^{-6} \\text { radians }=\\boxed{0.49} \\operatorname{arcsecs}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 6}
|
||||
{"problem": "A star has a measured parallax of $0.01^{\\prime \\prime}$, that is, $0.01$ arcseconds. How far away is it, in parsecs?", "solution": "Almost by definition, it is $\\boxed{100}$ parsecs away.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 7}
|
||||
{"problem": "An extrasolar planet has been observed which passes in front of (i.e., transits) its parent star. If the planet is dark (i.e., contributes essentially no light of its own) and has a surface area that is $2 \\%$ of that of its parent star, find the decrease in magnitude of the system during transits.", "solution": "The flux goes from a maximum of $F_{0}$, when the planet is not blocking any light, to $0.98 F_{0}$ when the planet is in front of the stellar disk. So, the uneclipsed magnitude is:\n\\[\nm_{0}=-2.5 \\log \\left(F_{0} / F_{\\text {ref }}\\right) \\quad .\n\\]\nWhen the planet blocks $2 \\%$ of the stellar disk, the magnitude increases to:\n\\[\nm=-2.5 \\log \\left(F / F_{\\text {ref }}\\right)=-2.5 \\log \\left(0.98 F_{0} / F_{\\text {ref }}\\right) \\quad .\n\\]\nThus, the change in magnitude is:\n\\[\n\\Delta m=m-m_{0}=-2.5 \\log (0.98) \\simeq \\boxed{0.022} \\quad \\text { magnitudes }\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 8}
|
||||
{"problem": "If the Bohr energy levels scale as $Z^{2}$, where $Z$ is the atomic number of the atom (i.e., the charge on the nucleus), estimate the wavelength of a photon that results from a transition from $n=3$ to $n=2$ in Fe, which has $Z=26$. Assume that the Fe atom is completely stripped of all its electrons except for one. Give your answer in Angstroms, to two significant figures.", "solution": "\\[\n\\begin{gathered}\nh \\nu=13.6 Z^{2}\\left[\\frac{1}{n_{f}^{2}}-\\frac{1}{n_{i}^{2}}\\right] \\mathrm{eV} \\\\\nh \\nu=13.6 \\times 26^{2}\\left[\\frac{1}{2^{2}}-\\frac{1}{3^{2}}\\right] \\mathrm{eV} \\\\\nh \\nu=1280 \\mathrm{eV}=1.28 \\mathrm{keV} \\Rightarrow \\boxed{9.6} \\AA\n\\end{gathered}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 9}
|
||||
{"problem": "If the Sun's absolute magnitude is $+5$, find the luminosity of a star of magnitude $0$ in ergs/s. A useful constant: the luminosity of the sun is $3.83 \\times 10^{33}$ ergs/s.", "solution": "The relation between luminosity and absolute magnitude is: $m - n = 2.5 \\log (f_n/f_m)$; note the numerator and denominator: brighter objects have numericallly smaller magnitudes. If a star has magnitude $0$, then since the difference in magnitudes from the sun is $5$, it must have $100$ times the sun's luminosity. Therefore, the answer is $\\boxed{3.83e35}$ ergs/s.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 10}
|
||||
{"problem": "Preamble: A spectrum is taken of a single star (i.e., one not in a binary). Among the observed spectral lines is one from oxygen whose rest wavelength is $5007 \\AA$. The Doppler shifted oxygen line from this star is observed to be at a wavelength of $5012 \\AA$. The star is also observed to have a proper motion, $\\mu$, of 1 arc second per year (which corresponds to $\\sim 1.5 \\times 10^{-13}$ radians per second of time). It is located at a distance of $60 \\mathrm{pc}$ from the Earth. Take the speed of light to be $3 \\times 10^8$ meters per second.\n\nWhat is the component of the star's velocity parallel to its vector to the Earth (in kilometers per second)?", "solution": "To find this longitudinal velocity component, we use the Doppler shift, finding $V_{r}=\\frac{\\Delta \\lambda}{\\lambda} c=\\frac{5}{5000} c=\\boxed{300} \\mathrm{~km} / \\mathrm{s}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 11}
|
||||
{"problem": "The differential luminosity from a star, $\\Delta L$, with an approximate blackbody spectrum, is given by:\n\\[\n\\Delta L=\\frac{8 \\pi^{2} c^{2} R^{2}}{\\lambda^{5}\\left[e^{h c /(\\lambda k T)}-1\\right]} \\Delta \\lambda\n\\]\nwhere $R$ is the radius of the star, $T$ is its effective surface temperature, and $\\lambda$ is the wavelength. $\\Delta L$ is the power emitted by the star between wavelengths $\\lambda$ and $\\lambda+\\Delta \\lambda$ (assume $\\Delta \\lambda \\ll \\lambda)$. The star is at distance $d$. Find the star's spectral intensity $I(\\lambda)$ at the Earth, where $I(\\lambda)$ is defined as the power per unit area per unit wavelength interval.", "solution": "\\[\nI(\\lambda)=\\frac{1}{4 \\pi d^{2}} \\frac{\\Delta L}{\\Delta \\lambda}=\\boxed{\\frac{2 \\pi c^{2} R^{2}}{\\lambda^{5}\\left[e^{h c /(\\lambda k T)}-1\\right] d^{2}}}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 12}
|
||||
{"problem": "Preamble: A very hot star is detected in the galaxy M31 located at a distance of $800 \\mathrm{kpc}$. The star has a temperature $T = 6 \\times 10^{5} K$ and produces a flux of $10^{-12} \\mathrm{erg} \\cdot \\mathrm{s}^{-1} \\mathrm{cm}^{-2}$ at the Earth. Treat the star's surface as a blackbody radiator.\n\nSubproblem 0: Find the luminosity of the star (in units of $\\mathrm{erg} \\cdot \\mathrm{s}^{-1}$).\n\n\nSolution: \\[\n L=4 \\pi D^{2} \\text { Flux }_{\\text {Earth }}=10^{-12} 4 \\pi\\left(800 \\times 3 \\times 10^{21}\\right)^{2}=\\boxed{7e37} \\mathrm{erg} \\cdot \\mathrm{s}^{-1}\n\\]\n\nFinal answer: The final answer is 7e37. I hope it is correct.\n\nSubproblem 1: Compute the star's radius in centimeters.", "solution": "\\[\n R=\\left(L / 4 \\pi \\sigma T^{4}\\right)^{1 / 2}=\\boxed{8.7e8} \\mathrm{~cm}=0.012 R_{\\odot}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 13}
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{"problem": "A star is at a distance from the Earth of $300 \\mathrm{pc}$. Find its parallax angle, $\\pi$, in arcseconds to one significant figure.", "solution": "\\[\n\\begin{aligned}\nD &=1 \\mathrm{pc} / \\pi^{\\prime \\prime} \\\\\n\\pi^{\\prime \\prime} &=1 \\mathrm{pc} / 300 \\mathrm{pc} \\\\\n\\pi^{\\prime \\prime} &=\\boxed{0.003}^{\\prime \\prime}\n\\end{aligned}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 14}
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{"problem": "The Sun's effective temperature, $T_{e}$, is 5800 Kelvin, and its radius is $7 \\times 10^{10} \\mathrm{~cm}\\left(7 \\times 10^{8}\\right.$ m). Compute the luminosity (power output) of the Sun in erg/s. Treat the Sun as a blackbody radiator, and give your answer to one significant figure.", "solution": "Using the standard formula for power output of a blackbody radiator gives $P = \\sigma A T^4$, where the area in this case is $4\\piR_{sun}^2$. Plugging in the numbers given in the problem yields that the sun's power output is (to one significant figure) $\\boxed{4e33}$ ergs.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 15}
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{"problem": "Use the Bohr model of the atom to compute the wavelength of the transition from the $n=100$ to $n=99$ levels, in centimeters. [Uscful relation: the wavelength of $L \\alpha$ ( $\\mathrm{n}=2$ to $\\mathrm{n}=1$ transition) is $1216 \\AA$.]", "solution": "The inverse wavelength of radiation is proportional to the energy difference between the initial and final energy levels. So for our transition of interest, we have \n\\begin{equation}\n \\lambda^{-1} = R(\\frac{1}{99^2} - \\frac{1}{100^2}).\n\\end{equation}\nUsing the information given in the problem for the $L \\alpha$ transition, we get\n\\begin{equation}\n (1216 \\AA)^{-1} = R(\\frac{1}{1^2} - \\frac{1}{2^2}).\n\\end{equation}\nCombining the above two relations yields $\\lambda = \\boxed{4.49}$ cm.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 16}
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{"problem": "Preamble: A radio interferometer, operating at a wavelength of $1 \\mathrm{~cm}$, consists of 100 small dishes, each $1 \\mathrm{~m}$ in diameter, distributed randomly within a $1 \\mathrm{~km}$ diameter circle. \n\nWhat is the angular resolution of a single dish, in radians?", "solution": "The angular resolution of a single dish is roughly given by the wavelength over its radius, in this case $\\boxed{0.01}$ radians.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 17}
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{"problem": "Preamble: Orbital Dynamics: A binary system consists of two stars in circular orbit about a common center of mass, with an orbital period, $P_{\\text {orb }}=10$ days. Star 1 is observed in the visible band, and Doppler measurements show that its orbital speed is $v_{1}=20 \\mathrm{~km} \\mathrm{~s}^{-1}$. Star 2 is an X-ray pulsar and its orbital radius about the center of mass is $r_{2}=3 \\times 10^{12} \\mathrm{~cm}=3 \\times 10^{10} \\mathrm{~m}$.\n\nSubproblem 0: Find the orbital radius, $r_{1}$, of the optical star (Star 1) about the center of mass, in centimeters.\n\n\nSolution: \\[\n\\begin{gathered}\nv_{1}=\\frac{2 \\pi r_{1}}{P_{\\text {orb }}} \\\\\nr_{1}=\\frac{P_{\\text {orb }} v_{1}}{2 \\pi}=\\boxed{2.75e11} \\mathrm{~cm}\n\\end{gathered}\n\\]\n\nFinal answer: The final answer is 2.75e11. I hope it is correct.\n\nSubproblem 1: What is the total orbital separation between the two stars, $r=r_{1}+r_{2}$ (in centimeters)?", "solution": "\\[\n r=r_{1}+r_{2}=2.75 \\times 10^{11}+3 \\times 10^{12}=\\boxed{3.3e12} \\quad \\mathrm{~cm}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 18}
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{"problem": "If a star cluster is made up of $10^{4}$ stars, each of whose absolute magnitude is $-5$, compute the combined apparent magnitude of the cluster if it is located at a distance of $1 \\mathrm{Mpc}$.", "solution": "The absolute magnitude of one of the stars is given by:\n\\[\nM=-2.5 \\log \\left(L / L_{\\mathrm{ref}}\\right)=-5\n\\]\nwhere $L$ is the stellar luminosity, and $L_{\\text {ref }}$ is the luminosity of a zero magnitude star. This equation implies that $L=100 L_{\\text {ref }}$. Armed with this fact, we can now compute the combined magnitude of the collection of $10^{4}$ stars:\n\\[\nM_{\\text {TOT }}=-2.5 \\log \\left[\\left(10^{4} \\times 100 L_{\\text {ref }}\\right) / L_{\\text {ref }}\\right]=-2.5 \\log \\left(10^{6}\\right)=-15\n\\]\nFinally, the distance modulus corresponding to $1 \\mathrm{Mpc}$ is $5 \\log \\left(10^{6} / 10\\right)=25$. Therefore, the apparent magnitude of the star cluster at this distance is:\n\\[\nm=M+\\text { distance modulus } \\Rightarrow m=-15+25=+\\boxed{10} .\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 19}
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{"problem": "A galaxy moves directly away from us with a speed of $3000 \\mathrm{~km} \\mathrm{~s}^{-1}$. Find the wavelength of the $\\mathrm{H} \\alpha$ line observed at the Earth, in Angstroms. The rest wavelength of $\\mathrm{H} \\alpha$ is $6565 \\AA$. Take the speed of light to be $3\\times 10^8$ meters per second.", "solution": "We have that the velocity of the galaxy is $0.01$ times $c$, the speed of light. So, using Doppler effect formulas,\n\\begin{equation}\n\\lambda_{obs} = (6565 \\AA)(1 + v/c) = (6565 \\AA)(1.01)\n\\end{equation}\nSo the answer is $\\boxed{6630}$ Angstroms.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 20}
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{"problem": "The Spitzer Space Telescope has an effective diameter of $85 \\mathrm{cm}$, and a typical wavelength used for observation of $5 \\mu \\mathrm{m}$, or 5 microns. Based on this information, compute an estimate for the angular resolution of the Spitzer Space telescope in arcseconds.", "solution": "Using the formula for angular resolution $\\theta$ in terms of the effective size $d$ and the wavelength $\\lambda$, namely $\\theta = \\lambda/d$, gives \\boxed{1.2} arcseconds.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 21}
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{"problem": "It has long been suspected that there is a massive black hole near the center of our Galaxy. Recently, a group of astronmers determined the parameters of a star that is orbiting the suspected black hole. The orbital period is 15 years, and the orbital radius is $0.12$ seconds of arc (as seen from the Earth). Take the distance to the Galactic center to be $8 \\mathrm{kpc}$. Compute the mass of the black hole, starting from $F=m a$. Express your answer in units of the Sun's mass; i.e., answer the question `what is the ratio of masses between this black hole and our Sun'? Give your answer to 1 significant figure. (Assume that Newton's law of gravity is applicable for orbits sufficiently far from a black hole, and that the orbiting star satisfies this condition.)", "solution": "The force of gravitational attraction between the black hole (of mass $M_{BH}$) and the star (of mass $M_s$) is given by\n\\begin{equation}\nF = \\frac{G M_{BH} M_s}{R^2},\n\\end{equation}\nwhere $R$ is the distance between the star and black hole (assuming a circular orbit). Equating this to the centripetal force gives\n\\begin{equation}\nF = \\frac{G M_{BH} M_s}{R^2} = \\frac{M_s v^2}{R},\n\\end{equation}\nwhere $v$, the (linear) orbital velocity, is related to the orbital period $P$ by\n\\begin{equation}\nv = \\frac{2\\pi R}{P}.\n\\end{equation}\nCombining the above equations, we get\n\\begin{equation}\n\\frac{G M_{BH} M_s}{R^2} = \\frac{M_s 4 \\pi^2 R^2}{RP^2},\n\\end{equation}\nor \n\\begin{equation}\nG M_{BH} = 4 \\pi^2 R^3 / P^2\n\\end{equation}\nSince this equation should also be valid for Earth's orbit around the Sun, if we replace $M_{BH}$ by the Sun's mass, $R$ by the Earth-sun distance, and $P$ by the orbital period of 1 year, we find that the ratio of masses between the black hole and our Sun is given by $(R / 1 \\mathrm{year})^3 / (P / 1 \\mathrm{a.u.})^2$.\nTo evaluate the above expression, we need to find $R$ from the information given in the problem; since we know the angle its orbital radius subtends ($0.12$ arcseconds) at a distance of $8 \\mathrm{kpc}$, we simply multiply these two quantities to find that $R = 900~\\mathrm{a.u.}$. So $M_{BH}/M_{sun} = (900)^3/(15)^2$, or $\\boxed{3e6}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 22}
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{"problem": "Preamble: A very hot star is detected in the galaxy M31 located at a distance of $800 \\mathrm{kpc}$. The star has a temperature $T = 6 \\times 10^{5} K$ and produces a flux of $10^{-12} \\mathrm{erg} \\cdot \\mathrm{s}^{-1} \\mathrm{cm}^{-2}$ at the Earth. Treat the star's surface as a blackbody radiator.\n\nFind the luminosity of the star (in units of $\\mathrm{erg} \\cdot \\mathrm{s}^{-1}$).", "solution": "\\[\n L=4 \\pi D^{2} \\text { Flux }_{\\text {Earth }}=10^{-12} 4 \\pi\\left(800 \\times 3 \\times 10^{21}\\right)^{2}=\\boxed{7e37} \\mathrm{erg} \\cdot \\mathrm{s}^{-1}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 23}
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{"problem": "A large ground-based telescope has an effective focal length of 10 meters. Two astronomical objects are separated by 1 arc second in the sky. How far apart will the two corresponding images be in the focal plane, in microns?", "solution": "\\[\ns=f \\theta=1000 \\mathrm{~cm} \\times \\frac{1}{2 \\times 10^{5}} \\text { radians }=0.005 \\mathrm{~cm}=\\boxed{50} \\mu \\mathrm{m}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 24}
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{"problem": "The equation of state for cold (non-relativistic) matter may be approximated as:\n\\[\nP=a \\rho^{5 / 3}-b \\rho^{4 / 3}\n\\]\nwhere $P$ is the pressure, $\\rho$ the density, and $a$ and $b$ are fixed constants. Use a dimensional analysis of the equation of hydrostatic equilibrium to estimate the ``radius-mass'' relation for planets and low-mass white dwarfs whose material follows this equation of state. Specifically, find $R(M)$ in terms of $G$ and the constants $a$ and $b$. You should set all constants of order unity (e.g., $4, \\pi, 3$, etc.) to $1.0$. [Hint: solve for $R(M)$ rather than $M(R)$ ]. You can check your answer by showing that for higher masses, $R \\propto M^{-1 / 3}$, while for the lower-masses $R \\propto M^{+1 / 3}$.", "solution": "\\[\n\\begin{gathered}\n\\frac{d P}{d r}=-g \\rho \\\\\n\\frac{a \\rho^{5 / 3}-b \\rho^{4 / 3}}{R} \\sim\\left(\\frac{G M}{R^{2}}\\right)\\left(\\frac{M}{R^{3}}\\right) \\\\\n\\frac{a M^{5 / 3}}{R^{6}}-\\frac{b M^{4 / 3}}{R^{5}} \\sim\\left(\\frac{G M^{2}}{R^{5}}\\right) \\\\\nG M^{2} \\sim \\frac{a M^{5 / 3}}{R}-b M^{4 / 3} \\\\\nR \\frac{a M^{5 / 3}}{G M^{2}+b M^{4 / 3}} \\simeq \\boxed{\\frac{a M^{1 / 3}}{G M^{2 / 3}+b}}\n\\end{gathered}\n\\]\nFor small masses, $R \\propto M^{1 / 3}$ as for rocky planets, while for larger masses, $R \\propto M^{-1 / 3}$ as for white dwarfs where the degenerate electrons are not yet relativistic.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 25}
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{"problem": "Take the total energy (potential plus thermal) of the Sun to be given by the simple expression:\n\\[\nE \\simeq-\\frac{G M^{2}}{R}\n\\]\nwhere $M$ and $R$ are the mass and radius, respectively. Suppose that the energy generation in the Sun were suddenly turned off and the Sun began to slowly contract. During this contraction its mass, $M$, would remain constant and, to a fair approximation, its surface temperature would also remain constant at $\\sim 5800 \\mathrm{~K}$. Assume that the total energy of the Sun is always given by the above expression, even as $R$ gets smaller. By writing down a simple (differential) equation relating the power radiated at Sun's surface with the change in its total energy (using the above expression), integrate this equation to find the time (in years) for the Sun to shrink to $1 / 2$ its present radius. Answer in units of years.", "solution": "\\[\n\\begin{gathered}\nL=4 \\pi \\sigma R^{2} T^{4}=d E / d t=\\left(\\frac{G M^{2}}{R^{2}}\\right) \\frac{d R}{d t} \\\\\n\\int_{R}^{0.5 R} \\frac{d R}{R^{4}}=-\\int_{0}^{t} \\frac{4 \\pi \\sigma T^{4}}{G M^{2}} d t \\\\\n-\\frac{1}{3(R / 2)^{3}}+\\frac{1}{3 R^{3}}=-\\left(\\frac{4 \\pi \\sigma T^{4}}{G M^{2}}\\right) t \\\\\nt=\\frac{G M^{2}}{12 \\pi \\sigma T^{4}}\\left(\\frac{8}{R^{3}}-\\frac{1}{R^{3}}\\right) \\\\\nt=\\frac{7 G M^{2}}{12 \\pi \\sigma T^{4} R^{3}}=2.2 \\times 10^{15} \\mathrm{sec}=75 \\text { million years }\n\\end{gathered}\n\\]\nSo the answer is $\\boxed{7.5e7}$ years.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 26}
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{"problem": "Preamble: Once a star like the Sun starts to ascend the giant branch its luminosity, to a good approximation, is given by:\n\\[\nL=\\frac{10^{5} L_{\\odot}}{M_{\\odot}^{6}} M_{\\text {core }}^{6}\n\\]\nwhere the symbol $\\odot$ stands for the solar value, and $M_{\\text {core }}$ is the mass of the He core of the star. Further, assume that as more hydrogen is burned to helium - and becomes added to the core - the conversion efficiency between rest mass and energy is:\n\\[\n\\Delta E=0.007 \\Delta M_{\\text {core }} c^{2} .\n\\]\n\nUse these two expressions to write down a differential equation, in time, for $M_{\\text {core }}$. For ease of writing, simply use the variable $M$ to stand for $M_{\\text {core }}$. Leave your answer in terms of $c$, $M_{\\odot}$, and $L_{\\odot}$.", "solution": "\\[\nL \\equiv \\frac{\\Delta E}{\\Delta t}=\\frac{0.007 \\Delta M c^{2}}{\\Delta t}=\\frac{10^{5} L_{\\odot}}{M_{\\odot}^{6}} M^{6}.\n\\]\nConverting these to differentials, we get\n\\begin{equation}\n\\frac{0.007 dM c^{2}}{dt}=\\frac{10^{5} L_{\\odot}}{M_{\\odot}^{6}} M^{6}, or\n\\end{equation}\n\\begin{equation}\n\\boxed{\\frac{dM}{dt}=\\frac{10^{5} L_{\\odot}}{0.007 c^{2} M_{\\odot}^{6}} M^{6}}\n\\end{equation}", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 27}
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{"problem": "A star of radius, $R$, and mass, $M$, has an atmosphere that obeys a polytropic equation of state:\n\\[\nP=K \\rho^{5 / 3} \\text {, }\n\\]\nwhere $P$ is the gas pressure, $\\rho$ is the gas density (mass per unit volume), and $K$ is a constant throughout the atmosphere. Assume that the atmosphere is sufficiently thin (compared to $R$ ) that the gravitational acceleration can be taken to be a constant.\nUse the equation of hydrostatic equilibrium to derive the pressure as a function of height $z$ above the surface of the planet. Take the pressure at the surface to be $P_{0}$.", "solution": "Start with the equation of hydrostatic equilibrium:\n\\[\n\\frac{d P}{d z}=-g \\rho\n\\]\nwhere $g$ is approximately constant through the atmosphere, and is given by $G M / R^{2}$. We can use the polytropic equation of state to eliminate $\\rho$ from the equation of hydrostatic equilibrium:\n\\[\n\\frac{d P}{d z}=-g\\left(\\frac{P}{K}\\right)^{3 / 5}\n\\]\nSeparating variables, we find:\n\\[\nP^{-3 / 5} d P=-g\\left(\\frac{1}{K}\\right)^{3 / 5} d z\n\\]\nWe then integrate the left-hand side from $P_{0}$ to $P$ and the right hand side from 0 to $z$ to find:\n\\[\n\\frac{5}{2}\\left(P^{2 / 5}-P_{0}^{2 / 5}\\right)=-g K^{-3 / 5} z\n\\]\nSolving for $P(z)$ we have:\n\\[\n P(z)=\\boxed{\\left[P_{0}^{2 / 5}-\\frac{2}{5} g K^{-3 / 5} z\\right]^{5 / 2}}=P_{0}\\left[1-\\frac{2}{5} \\frac{g}{P_{0}^{2 / 5} K^{3 / 5}} z\\right]^{5 / 2}\n\\]\nThe pressure therefore, goes to zero at a finite height $z_{\\max }$, where:\n\\[\nz_{\\max }=\\frac{5 P_{0}^{2 / 5} K^{3 / 5}}{2 g}=\\frac{5 K \\rho_{0}^{2 / 3}}{2 g}=\\frac{5 P_{0}}{2 g \\rho_{0}}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 28}
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{"problem": "An eclipsing binary consists of two stars of different radii and effective temperatures. Star 1 has radius $R_{1}$ and $T_{1}$, and Star 2 has $R_{2}=0.5 R_{1}$ and $T_{2}=2 T_{1}$. Find the change in bolometric magnitude of the binary, $\\Delta m_{\\text {bol }}$, when the smaller star is behind the larger star. (Consider only bolometric magnitudes so you don't have to worry about color differences.)", "solution": "\\[\n\\begin{gathered}\n\\mathcal{F}_{1 \\& 2}=4 \\pi \\sigma\\left(T_{1}^{4} R_{1}^{2}+T_{2}^{4} R_{2}^{2}\\right) \\\\\n\\mathcal{F}_{\\text {eclipse }}=4 \\pi \\sigma T_{1}^{4} R_{1}^{2} \\\\\n\\Delta m=-2.5 \\log \\left(\\frac{\\mathcal{F}_{1 \\& 2}}{\\mathcal{F}_{\\text {eclipse }}}\\right) \\\\\n\\Delta m=-2.5 \\log \\left(1+\\frac{T_{2}^{4} R_{2}^{2}}{T_{1}^{4} R_{1}^{2}}\\right) \\\\\n\\Delta m=-2.5 \\log \\left(1+\\frac{16}{4}\\right)=-1.75\n\\end{gathered}\n\\]\nSo, the binary is $\\boxed{1.75}$ magnitudes brighter out of eclipse than when star 2 is behind star 1 .", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 29}
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{"problem": "Preamble: It has been suggested that our Galaxy has a spherically symmetric dark-matter halo with a density distribution, $\\rho_{\\text {dark }}(r)$, given by:\n\\[\n\\rho_{\\text {dark }}(r)=\\rho_{0}\\left(\\frac{r_{0}}{r}\\right)^{2},\n\\]\nwhere $\\rho_{0}$ and $r_{0}$ are constants, and $r$ is the radial distance from the center of the galaxy. For star orbits far out in the halo you can ignore the gravitational contribution of the ordinary matter in the Galaxy.\n\nCompute the rotation curve of the Galaxy (at large distances), i.e., find $v(r)$ for circular orbits.", "solution": "\\[\n\\begin{gathered}\n-\\frac{G M(<r)}{r^{2}}=-\\frac{v^{2}}{r} \\quad(\\text { from } F=m a) \\\\\nM(<r)=\\int_{0}^{r} \\rho_{0}\\left(\\frac{r_{0}}{r}\\right)^{2} 4 \\pi r^{2} d r=4 \\pi \\rho_{0} r_{0}^{2} r\n\\end{gathered}\n\\]\nNote that, in general, $M \\neq \\rho \\times$ volume! You must integrate over $\\rho(r)$. From these expressions we find:\n\\[\nv(r)=\\boxed{\\sqrt{4 \\pi G \\rho_{0} r_{0}^{2}}}=\\text { constant }\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 30}
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{"problem": "The Very Large Array (VLA) telescope has an effective diameter of $36 \\mathrm{~km}$, and a typical wavelength used for observation at this facility might be $6 \\mathrm{~cm}$. Based on this information, compute an estimate for the angular resolution of the VLA in arcseconds", "solution": "Using the formula for angular resolution $\\theta$ in terms of the effective size $d$ and the wavelength $\\lambda$, namely $\\theta = \\lambda/d$, gives \\boxed{0.33} arcseconds.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 31}
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{"problem": "Subproblem 0: A particular star has an absolute magnitude $M=-7$. If this star is observed in a galaxy that is at a distance of $3 \\mathrm{Mpc}$, what will its apparent magnitude be? \n\n\nSolution: \\[\n\\text { Given: } M=-7 \\text { and } d=3 \\mathrm{Mpc}\n\\]\n\\[\n\\begin{aligned}\n & \\text { Apparent Magnitude: } m=M+5 \\log \\left[\\frac{d}{10 \\mathrm{pc}}\\right]=-7+5 \\log \\left[\\frac{3 \\times 10^{6}}{10}\\right]=\\boxed{20.39} \\\\\n\\end{aligned}\n\\]\n\nFinal answer: The final answer is 20.39. I hope it is correct.\n\nSubproblem 1: What is the distance modulus to this galaxy?", "solution": "Distance Modulus: $DM=m-M=20.39+7=\\boxed{27.39}$\n\\end{aligned}", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 32}
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{"problem": "Find the distance modulus to the Andromeda galaxy (M31). Take the distance to Andromeda to be $750 \\mathrm{kpc}$, and answer to three significant figures.", "solution": "\\[\n\\mathrm{DM}=5 \\log \\left(\\frac{d}{10 \\mathrm{pc}}\\right)=5 \\log (75,000)=\\boxed{24.4}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 33}
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{"problem": "The Hubble Space telescope has an effective diameter of $2.5 \\mathrm{~m}$, and a typical wavelength used for observation by the Hubble might be $0.6 \\mu \\mathrm{m}$, or 600 nanometers (typical optical wavelength). Based on this information, compute an estimate for the angular resolution of the Hubble Space telescope in arcseconds.", "solution": "Using the formula for angular resolution $\\theta$ in terms of the effective size $d$ and the wavelength $\\lambda$, namely $\\theta = \\lambda/d$, gives \\boxed{0.05} arcseconds.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 34}
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{"problem": "Preamble: A collimated light beam propagating in water is incident on the surface (air/water interface) at an angle $\\theta_w$ with respect to the surface normal.\n\nIf the index of refraction of water is $n=1.3$, find an expression for the angle of the light once it emerges from the water into the air, $\\theta_a$, in terms of $\\theta_w$.", "solution": "Using Snell's law, $1.3 \\sin{\\theta_w} = \\sin{\\theta_a}$. So $\\theta_a = \\boxed{\\arcsin{1.3 \\sin{\\theta_w}}}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 35}
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{"problem": "What fraction of the rest mass energy is released (in the form of radiation) when a mass $\\Delta M$ is dropped from infinity onto the surface of a neutron star with $M=1 M_{\\odot}$ and $R=10$ $\\mathrm{km}$ ?", "solution": "\\[\n\\Delta E=\\frac{G M \\Delta m}{R}\n\\]\nThe fractional rest energy lost is $\\Delta E / \\Delta m c^{2}$, or\n\\[\n\\frac{\\Delta E}{\\Delta m c^{2}}=\\frac{G M}{R c^{2}} \\simeq \\boxed{0.15}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 36}
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{"problem": "Preamble: The density of stars in a particular globular star cluster is $10^{6} \\mathrm{pc}^{-3}$. Take the stars to have the same radius as the Sun, and to have an average speed of $10 \\mathrm{~km} \\mathrm{sec}^{-1}$.\n\nFind the mean free path for collisions among stars. Express your answer in centimeters, to a single significant figure.", "solution": "\\[\n\\begin{gathered}\n\\ell \\simeq \\frac{1}{n \\sigma}=\\frac{1}{10^{6} \\mathrm{pc}^{-3} \\pi R^{2}} \\\\\n\\ell \\simeq \\frac{1}{3 \\times 10^{-50} \\mathrm{~cm}^{-3} \\times 1.5 \\times 10^{22} \\mathrm{~cm}^{2}} \\simeq \\boxed{2e27} \\mathrm{~cm}\n\\end{gathered}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 37}
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{"problem": "For a gas supported by degenerate electron pressure, the pressure is given by:\n\\[\nP=K \\rho^{5 / 3}\n\\]\nwhere $K$ is a constant and $\\rho$ is the mass density. If a star is totally supported by degenerate electron pressure, use a dimensional analysis of the equation of hydrostatic equilibrium:\n\\[\n\\frac{d P}{d r}=-g \\rho\n\\]\nto determine how the radius of such a star depends on its mass, $M$. Specifically, you will find that $R$ is proportional to some power of $M$; what is that power?", "solution": "\\[\n\\begin{gathered}\n\\frac{K \\rho^{5 / 3}}{R} \\simeq\\left(\\frac{G M}{R^{2}}\\right)\\left(\\frac{M}{R^{3}}\\right) \\\\\n\\rho \\sim \\frac{M}{R^{3}} \\\\\n\\frac{K M^{5 / 3}}{R R^{5}} \\simeq \\frac{G M^{2}}{R^{5}} \\\\\nR \\simeq \\frac{K}{G M^{1 / 3}}\n\\end{gathered}\n\\]\nSo the answer is $\\boxed{-1./3}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 38}
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{"problem": "A galaxy moves directly away from us with speed $v$, and the wavelength of its $\\mathrm{H} \\alpha$ line is observed to be $6784 \\AA$. The rest wavelength of $\\mathrm{H} \\alpha$ is $6565 \\AA$. Find $v/c$.", "solution": "\\[\n\\lambda \\simeq \\lambda_{0}(1+v / c)\n\\]\nwhere $\\lambda=6784 \\AA$ and $\\lambda_{0}=6565 \\AA$. Rearranging,\n\\[\n\\frac{v}{c} \\simeq \\frac{\\lambda-\\lambda_{0}}{\\lambda_{0}} \\simeq \\frac{6784-6565}{6565} \\Rightarrow v \\simeq 0.033 c\n\\]\nSo $v/c \\simeq \\boxed{0.033}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 39}
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{"problem": "A candle has a power in the visual band of roughly $3$ Watts. When this candle is placed at a distance of $3 \\mathrm{~km}$ it has the same apparent brightness as a certain star. Assume that this star has the same luminosity as the Sun in the visual band $\\left(\\sim 10^{26}\\right.$ Watts $)$. How far away is the star (in pc)?", "solution": "The fact that the two sources have the same apparent brightness implies that the flux at the respective distances is the same; since flux varies with distance as $1/d^2$, we find that (with distances in km) $\\frac{3}{3^2} = \\frac{10^{26}}{d^2}$, so $d = 10^{13}\\times\\frac{3}{\\sqrt{3}}$, or roughly $1.7\\times 10^{13}$ kilometers. In parsecs, this is $\\boxed{0.5613}$ parsecs.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 40}
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{"problem": "Preamble: A galaxy is found to have a rotation curve, $v(r)$, given by\n\\[\nv(r)=\\frac{\\left(\\frac{r}{r_{0}}\\right)}{\\left(1+\\frac{r}{r_{0}}\\right)^{3 / 2}} v_{0}\n\\]\nwhere $r$ is the radial distance from the center of the galaxy, $r_{0}$ is a constant with the dimension of length, and $v_{0}$ is another constant with the dimension of speed. The rotation curve is defined as the orbital speed of test stars in circular orbit at radius $r$.\n\nFind an expression for $\\omega(r)$, where $\\omega$ is the angular velocity. The constants $v_{0}$ and $r_{0}$ will appear in your answer.", "solution": "$\\omega=v / r & \\Rightarrow \\omega(r)=\\boxed{\\frac{v_{0}}{r_{0}} \\frac{1}{\\left(1+r / r_{0}\\right)^{3 / 2}}}$", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 41}
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{"problem": "Preamble: Orbital Dynamics: A binary system consists of two stars in circular orbit about a common center of mass, with an orbital period, $P_{\\text {orb }}=10$ days. Star 1 is observed in the visible band, and Doppler measurements show that its orbital speed is $v_{1}=20 \\mathrm{~km} \\mathrm{~s}^{-1}$. Star 2 is an X-ray pulsar and its orbital radius about the center of mass is $r_{2}=3 \\times 10^{12} \\mathrm{~cm}=3 \\times 10^{10} \\mathrm{~m}$.\n\nFind the orbital radius, $r_{1}$, of the optical star (Star 1) about the center of mass, in centimeters.", "solution": "\\[\n\\begin{gathered}\nv_{1}=\\frac{2 \\pi r_{1}}{P_{\\text {orb }}} \\\\\nr_{1}=\\frac{P_{\\text {orb }} v_{1}}{2 \\pi}=\\boxed{2.75e11} \\mathrm{~cm}\n\\end{gathered}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 42}
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{"problem": "Preamble: The density of stars in a particular globular star cluster is $10^{6} \\mathrm{pc}^{-3}$. Take the stars to have the same radius as the Sun, and to have an average speed of $10 \\mathrm{~km} \\mathrm{sec}^{-1}$.\n\nSubproblem 0: Find the mean free path for collisions among stars. Express your answer in centimeters, to a single significant figure.\n\n\nSolution: \\[\n\\begin{gathered}\n\\ell \\simeq \\frac{1}{n \\sigma}=\\frac{1}{10^{6} \\mathrm{pc}^{-3} \\pi R^{2}} \\\\\n\\ell \\simeq \\frac{1}{3 \\times 10^{-50} \\mathrm{~cm}^{-3} \\times 1.5 \\times 10^{22} \\mathrm{~cm}^{2}} \\simeq \\boxed{2e27} \\mathrm{~cm}\n\\end{gathered}\n\\]\n\nFinal answer: The final answer is 2e27. I hope it is correct.\n\nSubproblem 1: Find the corresponding mean time between collisions. (Assume that the stars move in straight-line paths, i.e., are not deflected by gravitational interactions.) Answer in units of years, to a single significant figure.", "solution": "$\\tau_{\\text {coll }} \\simeq \\frac{2 \\times 10^{27} \\mathrm{~cm}}{10^{6} \\mathrm{~cm} / \\mathrm{sec}} \\simeq 2 \\times 10^{21} \\mathrm{sec} \\simeq \\boxed{6e13} \\text { years }$", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 43}
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{"problem": "Preamble: A radio interferometer, operating at a wavelength of $1 \\mathrm{~cm}$, consists of 100 small dishes, each $1 \\mathrm{~m}$ in diameter, distributed randomly within a $1 \\mathrm{~km}$ diameter circle. \n\nSubproblem 0: What is the angular resolution of a single dish, in radians?\n\n\nSolution: The angular resolution of a single dish is roughly given by the wavelength over its radius, in this case $\\boxed{0.01}$ radians.\n\nFinal answer: The final answer is 0.01. I hope it is correct.\n\nSubproblem 1: What is the angular resolution of the interferometer array for a source directly overhead, in radians?", "solution": "The angular resolution of the full array is given by the wavelength over the dimension of the array, in this case $\\boxed{1e-5}$ radians.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 44}
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{"problem": "If a star cluster is made up of $10^{6}$ stars whose absolute magnitude is the same as that of the Sun (+5), compute the combined magnitude of the cluster if it is located at a distance of $10 \\mathrm{pc}$.", "solution": "At $10 \\mathrm{pc}$, the magnitude is (by definition) just the absolute magnitude of the cluster. Since the total luminosity of the cluster is $10^{6}$ times the luminosity of the Sun, we have that \n\\begin{equation}\n\\delta m = 2.5 \\log \\left( \\frac{L_{TOT}}{L_{sun}} \\right) = 2.5 \\log 10^6 = 15.\n\\end{equation}\nSince the Sun has absolute magnitude +5, the magnitude of the cluser is $\\boxed{-10}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 45}
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{"problem": "A certain red giant has a radius that is 500 times that of the Sun, and a temperature that is $1 / 2$ that of the Sun's temperature. Find its bolometric (total) luminosity in units of the bolometric luminosity of the Sun.", "solution": "Power output goes as $T^4r^2$, so the power output of this star is $\\boxed{15625}$ times that of the Sun.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 46}
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{"problem": "Suppose air molecules have a collision cross section of $10^{-16} \\mathrm{~cm}^{2}$. If the (number) density of air molecules is $10^{19} \\mathrm{~cm}^{-3}$, what is the collision mean free path in cm? Answer to one significant figure.", "solution": "\\[\n\\ell=\\frac{1}{n \\sigma}=\\frac{1}{10^{19} 10^{-16}}=\\boxed{1e-3} \\mathrm{~cm}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 47}
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{"problem": "Two stars have the same surface temperature. Star 1 has a radius that is $2.5$ times larger than the radius of star 2. Star 1 is ten times farther away than star 2. What is the absolute value of the difference in apparent magnitude between the two stars, rounded to the nearest integer?", "solution": "Total power output goes as $r^2 T^4$, where $r$ is the star's radius, and $T$ is its temperature. Flux, at a distance $R$ away thus goes as $r^2 T^4 / R^2$. In our case, the ratio of flux from star 1 to star 2 is $1/16$ (i.e., star 2 is greater in apparent magnitude). Using the relation between apparent magnitude and flux, we find that that the absolute value of the difference in apparent magnitudes is $2.5 \\log{16}$, which rounded to the nearest integer is $\\boxed{3}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 48}
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{"problem": "What is the slope of a $\\log N(>F)$ vs. $\\log F$ curve for a homogeneous distribution of objects, each of luminosity, $L$, where $F$ is the flux at the observer, and $N$ is the number of objects observed per square degree on the sky?", "solution": "The number of objects detected goes as the cube of the distance for objects with flux greater than a certain minimum flux. At the same time the flux falls off with the inverse square of the distance. Thus, the slope of the $\\log N(>F)$ vs. $\\log F$ curve is $\\boxed{-3./2}$.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 49}
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{"problem": "Preamble: Comparison of Radio and Optical Telescopes.\n\nThe Very Large Array (VLA) is used to make an interferometric map of the Orion Nebula at a wavelength of $10 \\mathrm{~cm}$. What is the best angular resolution of the radio image that can be produced, in radians? Note that the maximum separation of two antennae in the VLA is $36 \\mathrm{~km}$.", "solution": "The best angular resolution will occur at the maximum separation, and is simply the ratio of wavelength to this separation $p$: $\\theta = \\frac{\\lambda}{p}$, or $\\frac{0.1}{36\\times 10^3}$, which is $\\boxed{2.7778e-6}$ radians.", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 50}
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{"problem": "A globular cluster has $10^{6}$ stars each of apparent magnitude $+8$. What is the combined apparent magnitude of the entire cluster?", "solution": "\\[\n\\begin{gathered}\n+8=-2.5 \\log \\left(F / F_{0}\\right) \\\\\nF=6.3 \\times 10^{-4} F_{0} \\\\\nF_{\\text {cluster }}=10^{6} \\times 6.3 \\times 10^{-4} F_{0}=630 F_{0} \\\\\nm_{\\text {cluster }}=-2.5 \\log (630)=\\boxed{-7}\n\\end{gathered}\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 51}
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{"problem": "Preamble: A very hot star is detected in the galaxy M31 located at a distance of $800 \\mathrm{kpc}$. The star has a temperature $T = 6 \\times 10^{5} K$ and produces a flux of $10^{-12} \\mathrm{erg} \\cdot \\mathrm{s}^{-1} \\mathrm{cm}^{-2}$ at the Earth. Treat the star's surface as a blackbody radiator.\n\nSubproblem 0: Find the luminosity of the star (in units of $\\mathrm{erg} \\cdot \\mathrm{s}^{-1}$).\n\n\nSolution: \\[\n L=4 \\pi D^{2} \\text { Flux }_{\\text {Earth }}=10^{-12} 4 \\pi\\left(800 \\times 3 \\times 10^{21}\\right)^{2}=\\boxed{7e37} \\mathrm{erg} \\cdot \\mathrm{s}^{-1}\n\\]\n\nFinal answer: The final answer is 7e37. I hope it is correct.\n\nSubproblem 1: Compute the star's radius in centimeters.\n\n\nSolution: \\[\n R=\\left(L / 4 \\pi \\sigma T^{4}\\right)^{1 / 2}=\\boxed{8.7e8} \\mathrm{~cm}=0.012 R_{\\odot}\n\\]\n\nFinal answer: The final answer is 8.7e8. I hope it is correct.\n\nSubproblem 2: At what wavelength is the peak of the emitted radiation? Answer in $\\AA$.", "solution": "Using the Wien displacement law:\n\\[\n \\lambda_{\\max }=0.29 / T \\mathrm{~cm}=\\boxed{48} \\AA\n\\]", "type": "Introduction to Astronomy (8.282J Spring 2006)", "idx": 52}
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{"problem": "A Boolean function $F(A, B)$ is said to be universal if any arbitrary boolean function can be constructed by using nested $F(A, B)$ functions. A universal function is useful, since using it we can build any function we wish out of a single part. For example, when implementing boolean logic on a computer chip a universal function (called a 'gate' in logic-speak) can simplify design enormously. We would like to find a universal boolean function. In this problem we will denote the two boolean inputs $A$ and $B$ and the one boolean output as $C$. \nFirst, to help us organize our thoughts, let's enumerate all of the functions we'd like to be able to construct. How many different possible one-output boolean functions of two variables are there? I.e., how many functions are there of the form $F(A, B)=C ?$", "solution": "This particular definition of universality only treats arbitrary functions of two Boolean variables, but with any number of outputs. It appears to be an onerous task to prove universality for an arbitrary number of outputs. However, since each individual output of a multi-output function can be considered a separate one-ouput function, it is sufficient to prove the case of only one-output functions. This is why we begin by listing all one-output functions of one variable.\nEach variable $A$ and $B$ has two possible values, making four different combinations of inputs $(A, B)$. Each combination of inputs (four possible) can cause one of two output values. Therefore the number of possible one-output binary functions of two binary variables is $2^{4}$, or \\boxed{16}. They are enumerated in the table below.\n\\begin{tabular}{cc|ccccccccccccccccccc}\n$A$ & $B$ & $b_{0}$ & $b_{1}$ & $b_{2}$ & $b_{3}$ & $b_{4}$ & $b_{5}$ & $b_{6}$ & $b_{7}$ & $b_{8}$ & $b_{9}$ & $b_{10}$ & $b_{11}$ & $b_{12}$ & $b_{13}$ & $b_{14}$ & $b_{15}$ & \\\\\n\\hline\n0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 & \\\\\n0 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & \\\\\n1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & \\\\\n1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & \\\\\n\\end{tabular}", "type": "Information and Entropy (6.050J Spring 2008)", "idx": 53}
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{"problem": "Unfortunately, a mutant gene can turn box people into triangles late in life. A laboratory test has been developed which can spot the gene early so that the dreaded triangle transformation can be prevented by medications. This test is 95 percent accurate at spotting the gene when it is there. However, the test gives a \"false positive\" $0.4$ percent of the time, falsely indicating that a healthy box person has the mutant gene. If $0.1$ percent (be careful - that's one-tenth of one percent) of the box people have the mutant gene, what's the probability that a box person actually has the mutant gene if the test indicates that he or she does?", "solution": "We see that the probability that a person has the disease given that the test is positive, is:\n\\[\n\\frac{0.001 \\times 0.95}{0.001 \\times 0.95+0.999 \\times 0.004}=19.2 \\%\n\\]\n$\\begin{array}{ccccc}\\text { Have Disease? } & \\text { Percent } & \\text { Test Results } & \\text { Percent } & \\text { Total } \\\\ \\text { Yes } & 0.001 & \\text { Positive } & 0.95 & 0.00095 \\\\ & & \\text { Negative } & 0.05 & 0.00005 \\\\ \\text { No } & 0.999 & \\text { Positive } & 0.004 & 0.003996 \\\\ & & \\text { Negative } & 0.996 & 0.95504\\end{array}$\nAnswer: \\boxed{0.192}.", "type": "Information and Entropy (6.050J Spring 2008)", "idx": 54}
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{"problem": "Buzz, the hot new dining spot on campus, emphasizes simplicity. It only has two items on the menu, burgers and zucchini. Customers make a choice as they enter (they are not allowed to order both), and inform the cooks in the back room by shouting out either \"B\" or \"Z\". Unfortunately the two letters sound similar so $8 \\%$ of the time the cooks misinterpret what was said. The marketing experts who designed the restaurant guess that $90 \\%$ of the orders will be for burgers and $10 \\%$ for zucchini.\nThe cooks can hear one order per second. The customers arrive at the rate of one per second. One of the chefs says that this system will never work because customers can only send one bit per second, the rate at which orders can be accepted, so you could barely keep up even if there were no noise in the channel. You are hired as an outside consultant to deal with the problem.\nWhat is the channel capacity $\\mathrm{C}$ of this communication channel in bits per second?", "solution": "This is a noisy channel with the same probabilities for mixing up $Z$ and $B$. Channel capacity is defined as the maximum mutual information (for any possible input probability) times the rate $W$. The rate of error is $\\epsilon=0.08$. So the channel capacity for this channel is given by:\n\\[\n\\begin{aligned}\nC &=M_{\\max } W \\\\\n&=1-\\epsilon \\log _{2}\\left(\\frac{1}{\\epsilon}\\right)-(1-\\epsilon) \\log _{2}\\left(\\frac{1}{(1-\\epsilon)}\\right) \\\\\n&=1-0.08 \\log _{2}\\left(\\frac{1}{0.08}\\right)-(0.92) \\log _{2}\\left(\\frac{1}{0.92}\\right) \\\\\n&=0.5978 \\mathrm{bits} / \\mathrm{second}\n\\end{aligned}\n\\]\nSo the final answer is \\boxed{0.5978} bits/s.", "type": "Information and Entropy (6.050J Spring 2008)", "idx": 55}
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{"problem": "Preamble: Given the following data from an Experimental Forest, answer the following questions. Show your work and units.\n$\\begin{array}{ll}\\text { Total vegetative biomass } & 80,000 \\mathrm{kcal} \\mathrm{m}^{-2} \\\\ \\text { Detritus and organic matter in soil } & 120,000 \\mathrm{kcal } \\mathrm{m}^{-2} \\\\ \\text { Total Gross Primary Productivity } & 20,000 \\mathrm{kcal } \\mathrm{m}^{-2} \\mathrm{yr}^{-1} \\\\ \\text { Total Plant Respiration } & 5,000 \\mathrm{kcal} \\mathrm{m}^{-2} \\mathrm{yr}^{-1} \\\\ \\text { Total Community Respiration } & 9,000 \\mathrm{kcal} \\mathrm{m}^{-2} \\mathrm{yr}^{-1}\\end{array}$\n\nSubproblem 0: What is the net primary productivity of the forest?\n\n\nSolution: NPP $=$ GPP $-R_{A}=20,000-5,000=\\boxed{15000} \\mathrm{kcal} \\mathrm{m}^{-2} \\mathrm{yr}^{-1}$\n\nFinal answer: The final answer is 15000. I hope it is correct.\n\nSubproblem 1: What is the net community production?", "solution": "$\\mathrm{NCP}=\\mathrm{GPP}-\\mathrm{R}_{\\mathrm{A}}-\\mathrm{R}_{\\mathrm{H}}=20,000-9000=\\boxed{11000} \\mathrm{kcal} \\mathrm{m}^{-2} \\mathrm{yr}^{-1}$", "type": "Ecology I (1.018J Fall 2009)", "idx": 56}
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{"problem": "Preamble: A population of 100 ferrets is introduced to a large island in the beginning of 1990 . Ferrets have an intrinsic growth rate, $r_{\\max }$ of $1.3 \\mathrm{yr}^{-1}$.\n\nSubproblem 0: Assuming unlimited resources-i.e., there are enough resources on this island to last the ferrets for hundreds of years-how many ferrets will there be on the island in the year 2000? (Show your work!)\n\n\nSolution: $N_o = 100$ (in 1990)\n\\\\\n$N = ?$ (in 2000)\n\\\\\n$t = 10$ yr\n\\\\\n$r = 1.3 \\text{yr}^{-1}$\n\\\\\n$N = N_{o}e^{rt} = 100*e^{(1.3/\\text{yr})(10 \\text{yr})} = 4.4 x 10^7$ ferrets\n\\\\\nThere will be \\boxed{4.4e7} ferrets on the island in the year 2000. \n\nFinal answer: The final answer is 4.4e7. I hope it is correct.\n\nSubproblem 1: What is the doubling time of the ferret population? (Show your work!)", "solution": "$N_o = 100$ (in 1990)\n\\\\\n$t = 10$ yr\n\\\\\n$r = 1.3 \\text{yr}^{-1}$\n\\\\\n$t_d = (ln(2))/r = 0.693/(1.3 \\text{yr}^{-1}) = 0.53$ years\n\\\\\nThe doubling time of the ferret population is \\boxed{0.53} years.", "type": "Ecology I (1.018J Fall 2009)", "idx": 57}
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{"problem": "Preamble: Given the following data from an Experimental Forest, answer the following questions. Show your work and units.\n$\\begin{array}{ll}\\text { Total vegetative biomass } & 80,000 \\mathrm{kcal} \\mathrm{m}^{-2} \\\\ \\text { Detritus and organic matter in soil } & 120,000 \\mathrm{kcal } \\mathrm{m}^{-2} \\\\ \\text { Total Gross Primary Productivity } & 20,000 \\mathrm{kcal } \\mathrm{m}^{-2} \\mathrm{yr}^{-1} \\\\ \\text { Total Plant Respiration } & 5,000 \\mathrm{kcal} \\mathrm{m}^{-2} \\mathrm{yr}^{-1} \\\\ \\text { Total Community Respiration } & 9,000 \\mathrm{kcal} \\mathrm{m}^{-2} \\mathrm{yr}^{-1}\\end{array}$\n\nWhat is the net primary productivity of the forest?", "solution": "NPP $=$ GPP $-R_{A}=20,000-5,000=\\boxed{15000} \\mathrm{kcal} \\mathrm{m}^{-2} \\mathrm{yr}^{-1}$", "type": "Ecology I (1.018J Fall 2009)", "idx": 58}
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{"problem": "Preamble: The Peak District Moorlands in the United Kingdom store 20 million tonnes of carbon, almost half of the carbon stored in the soils of the entire United Kingdom (the Moorlands are only $8 \\%$ of the land area). In pristine condition, these peatlands can store an additional 13,000 tonnes of carbon per year.\n\nGiven this rate of productivity, how long did it take for the Peatlands to sequester this much carbon?", "solution": "$20,000,000$ tonnes $C / 13,000$ tonnes $C y^{-1}=\\boxed{1538}$ years", "type": "Ecology I (1.018J Fall 2009)", "idx": 59}
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{"problem": "Preamble: A population of 100 ferrets is introduced to a large island in the beginning of 1990 . Ferrets have an intrinsic growth rate, $r_{\\max }$ of $1.3 \\mathrm{yr}^{-1}$.\n\nAssuming unlimited resources-i.e., there are enough resources on this island to last the ferrets for hundreds of years-how many ferrets will there be on the island in the year 2000? (Show your work!)", "solution": "$N_o = 100$ (in 1990)\n\\\\\n$N = ?$ (in 2000)\n\\\\\n$t = 10$ yr\n\\\\\n$r = 1.3 \\text{yr}^{-1}$\n\\\\\n$N = N_{o}e^{rt} = 100*e^{(1.3/\\text{yr})(10 \\text{yr})} = 4.4 x 10^7$ ferrets\n\\\\\nThere will be \\boxed{4.4e7} ferrets on the island in the year 2000.", "type": "Ecology I (1.018J Fall 2009)", "idx": 60}
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{"problem": "Preamble: The following subproblems refer to a circuit with the following parameters. Denote by $I(t)$ the current (where the positive direction is, say, clockwise) in the circuit and by $V(t)$ the voltage increase across the voltage source, at time $t$. Denote by $R$ the resistance of the resistor and $C$ the capacitance of the capacitor (in units which we will not specify)-both positive numbers. Then\n\\[\nR \\dot{I}+\\frac{1}{C} I=\\dot{V}\n\\]\n\nSubproblem 0: Suppose that $V$ is constant, $V(t)=V_{0}$. Solve for $I(t)$, with initial condition $I(0)$.\n\n\nSolution: When $V$ is constant, the equation becomes $R \\dot{I}+\\frac{1}{C} I=0$, which is separable. Solving gives us\n\\[\nI(t)=\\boxed{I(0) e^{-\\frac{t}{R C}}\n}\\]. \n\nFinal answer: The final answer is I(0) e^{-\\frac{t}{R C}}\n. I hope it is correct.\n\nSubproblem 1: It is common to write the solution to the previous subproblem in the form $c e^{-t / \\tau}$. What is $c$ in this case?", "solution": "$c=\\boxed{I(0)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 61}
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{"problem": "Consider the following \"mixing problem.\" A tank holds $V$ liters of salt water. Suppose that a saline solution with concentration of $c \\mathrm{gm} /$ liter is added at the rate of $r$ liters/minute. A mixer keeps the salt essentially uniformly distributed in the tank. A pipe lets solution out of the tank at the same rate of $r$ liters/minute. The differential equation for the amount of salt in the tank is given by \n\\[\nx^{\\prime}+\\frac{r}{V} x-r c=0 .\n\\]\nSuppose that the out-flow from this tank leads into another tank, also of volume 1 , and that at time $t=1$ the water in it has no salt in it. Again there is a mixer and an outflow. Write down a differential equation for the amount of salt in this second tank, as a function of time, assuming the amount of salt in the second tank at moment $t$ is given by $y(t)$, and the amount of salt in the first tank at moment $t$ is given by $x(t)$.", "solution": "The differential equation for $y(t)$ is $\\boxed{y^{\\prime}+r y-r x(t)=0}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 62}
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{"problem": "Find the general solution of $x^{2} y^{\\prime}+2 x y=\\sin (2 x)$, solving for $y$. Note that a general solution to a differential equation has the form $x=x_{p}+c x_{h}$ where $x_{h}$ is a nonzero solution of the homogeneous equation $\\dot{x}+p x=0$. Additionally, note that the left hand side is the derivative of a product.", "solution": "We see that $\\left(x^{2} y\\right)^{\\prime}=x^{2} y^{\\prime}+2 x y$. Thus, $x^{2} y=-\\frac{1}{2} \\cos (2 x)+c$, and $y=\\boxed{c x^{-2}-\\frac{\\cos (2 x)}{2 x^{2}}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 63}
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{"problem": "An African government is trying to come up with good policy regarding the hunting of oryx. They are using the following model: the oryx population has a natural growth rate of $k$, and we suppose a constant harvesting rate of $a$ oryxes per year.\nWrite down an ordinary differential equation describing the evolution of the oryx population given the dynamics above, using $x(t)$ to denote the oryx population (the number of individual oryx(es)) at time $t$, measured in years.", "solution": "The natural growth rate is $k$, meaning that after some short time $\\Delta t$ year(s) passes, we expect $k x(t) \\Delta t$ new oryxes to appear. However, meanwhile the population is reduced by $a \\Delta t$ oryxes due to the harvesting. Therefore, we are led to\n\\[\nx(t+\\Delta t) \\simeq x(t)+k x(t) \\Delta t-a \\Delta t,\n\\]\nand the unit on both sides is oryx $(\\mathrm{es})$. If we let $\\Delta t$ approach 0 , then we get the differential equation\n\\[\n\\boxed{\\frac{d x}{d t}=k x-a} .\n\\]", "type": "Differential Equations (18.03 Spring 2010)", "idx": 64}
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{"problem": "If the complex number $z$ is given by $z = 1+\\sqrt{3} i$, what is the magnitude of $z^2$?", "solution": "$z^{2}$ has argument $2 \\pi / 3$ and radius 4, so by Euler's formula, $z^{2}=4 e^{i 2 \\pi / 3}$. Thus $A=4, \\theta=\\frac{2\\pi}{3}$, so our answer is $\\boxed{4}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 65}
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{"problem": "In the polar representation $(r, \\theta)$ of the complex number $z=1+\\sqrt{3} i$, what is $r$?", "solution": "For z, $r=2$ and $\\theta=\\pi / 3$, so its polar coordinates are $\\left(2, \\frac{\\pi}{3}\\right)$. So $r=\\boxed{2}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 66}
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{"problem": "Preamble: In the following problems, take $a = \\ln 2$ and $b = \\pi / 3$. \n\nGiven $a = \\ln 2$ and $b = \\pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers.", "solution": "Using Euler's formula, we find that the answer is $\\boxed{1+\\sqrt{3} i}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 67}
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{"problem": "Subproblem 0: Find the general solution of the differential equation $y^{\\prime}=x-2 y$ analytically using integrating factors, solving for $y$. Note that a function $u(t)$ such that $u \\dot{x}+u p x=\\frac{d}{d t}(u x)$ is an integrating factor. Additionally, note that a general solution to a differential equation has the form $x=x_{p}+c x_{h}$ where $x_{h}$ is a nonzero solution of the homogeneous equation $\\dot{x}+p x=0$.\n\n\nSolution: In standard form, $y^{\\prime}+2 y=x$, so $u=C e^{2 x}$. Then $y=u^{-1} \\int u x d x=e^{-2 x} \\int x e^{2 x} d x$. Integrating by parts yields $\\int x e^{2 x} d x=$ $\\frac{x}{2} e^{2 x}-\\frac{1}{2} \\int e^{2 x} d x=\\frac{x}{2} e^{2 x}-\\frac{1}{4} e^{2 x}+c$. Therefore, $y=\\boxed{x / 2-1 / 4+c e^{-2 x}}$.\n\nFinal answer: The final answer is x / 2-1 / 4+c e^{-2 x}. I hope it is correct.\n\nSubproblem 1: For what value of $c$ does the straight line solution occur?", "solution": "The straight line solution occurs when $c=\\boxed{0}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 68}
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{"problem": "Preamble: The following subproblems relate to applying Euler's Method (a first-order numerical procedure for solving ordinary differential equations with a given initial value) onto $y^{\\prime}=y^{2}-x^{2}=F(x, y)$ at $y(0)=-1$, with $h=0.5$. Recall the notation \\[x_{0}=0, y_{0}=-1, x_{n+1}=x_{h}+h, y_{n+1}=y_{n}+m_{n} h, m_{n}=F\\left(x_{n}, y_{n}\\right)\\]. \n\nUse Euler's method to estimate the value at $x=1.5$.", "solution": "$y_3 = \\boxed{-0.875}$", "type": "Differential Equations (18.03 Spring 2010)", "idx": 69}
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{"problem": "Rewrite the function $f(t) = \\cos (2 t)+\\sin (2 t)$ in the form $A \\cos (\\omega t-\\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$.", "solution": "Here, our right triangle has hypotenuse $\\sqrt{2}$, so $A=\\sqrt{2}$. Both summands have \"circular frequency\" 2, so $\\omega=2 . \\phi$ is the argument of the hypotenuse, which is $\\pi / 4$, so $f(t)=\\boxed{\\sqrt{2} \\cos (2 t-\\pi / 4)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 70}
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{"problem": "Given the ordinary differential equation $\\ddot{x}-a^{2} x=0$, where $a$ is a nonzero real-valued constant, find a solution $x(t)$ to this equation such that $x(0) = 0$ and $\\dot{x}(0)=1$.", "solution": "First, notice that both $x(t)=e^{a t}$ and $x(t)=e^{-a t}$ are solutions to $\\ddot{x}-a^{2} x=0$. Then for any constants $c_{1}$ and $c_{2}$, $x(t)=c_{1} e^{a t}+c_{2} e^{-a t}$ are also solutions to $\\ddot{x}-a^{2} x=0$. Moreover, $x(0)=c_{1}+c_{2}$, and $\\dot{x}(0)=a\\left(c_{1}-c_{2}\\right)$. Assuming $a \\neq 0$, to satisfy the given conditions, we need $c_{1}+c_{2}=0$ and $a\\left(c_{1}-c_{2}\\right)=1$, which implies $c_{1}=-c_{2}=\\frac{1}{2 a}$. So $x(t)=\\boxed{\\frac{1}{2a}(\\exp{a*t} - \\exp{-a*t})}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 71}
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{"problem": "Find a solution to the differential equation $\\ddot{x}+\\omega^{2} x=0$ satisfying the initial conditions $x(0)=x_{0}$ and $\\dot{x}(0)=\\dot{x}_{0}$.", "solution": "Suppose \\[x(t)=a \\cos (\\omega t)+b \\sin (\\omega t)\\] $x(0)=a$, therefore $a=x_{0}$. Then \\[x^{\\prime}(0)=-a \\omega \\sin 0+b \\omega \\cos 0=b \\omega=\\dot{x}_{0}\\] Then $b=\\dot{x}_{0} / \\omega$. The solution is then $x=\\boxed{x_{0} \\cos (\\omega t)+$ $\\dot{x}_{0} \\sin (\\omega t) / \\omega}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 72}
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{"problem": "Find the complex number $a+b i$ with the smallest possible positive $b$ such that $e^{a+b i}=1+\\sqrt{3} i$.", "solution": "$1+\\sqrt{3} i$ has modulus 2 and argument $\\pi / 3+2 k \\pi$ for all integers k, so $1+\\sqrt{3} i$ can be expressed as a complex exponential of the form $2 e^{i(\\pi / 3+2 k \\pi)}$. Taking logs gives us the equation $a+b i=\\ln 2+i(\\pi / 3+2 k \\pi)$. The smallest positive value of $b$ is $\\pi / 3$. Thus we have $\\boxed{\\ln 2 + i\\pi / 3}$", "type": "Differential Equations (18.03 Spring 2010)", "idx": 73}
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{"problem": "Subproblem 0: Find the general solution of the differential equation $\\dot{x}+2 x=e^{t}$, using $c$ for the arbitrary constant of integration which will occur.\n\n\nSolution: We can use integrating factors to get $(u x)^{\\prime}=u e^{t}$ for $u=e^{2 t}$. Integrating yields $e^{2 t} x=e^{3 t} / 3+c$, or $x=\\boxed{\\frac{e^{t}} {3}+c e^{-2 t}}$. \n\nFinal answer: The final answer is \\frac{e^{t}} {3}+c e^{-2 t}. I hope it is correct.\n\nSubproblem 1: Find a solution of the differential equation $\\dot{x}+2 x=e^{t}$ of the form $w e^{t}$, where $w$ is a constant (which you should find).", "solution": "When $c=0, x=\\boxed{e^{t} / 3}$ is the solution of the required form.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 74}
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{"problem": "Subproblem 0: For $\\omega \\geq 0$, find $A$ such that $A \\cos (\\omega t)$ is a solution of $\\ddot{x}+4 x=\\cos (\\omega t)$.\n\n\nSolution: If $x=A \\cos (\\omega t)$, then taking derivatives gives us $\\ddot{x}=-\\omega^{2} A \\cos (\\omega t)$, and $\\ddot{x}+4 x=\\left(4-\\omega^{2}\\right) A \\cos (\\omega t)$. Then $A=\\boxed{\\frac{1}{4-\\omega^{2}}}$. \n\nFinal answer: The final answer is \\frac{1}{4-\\omega^{2}}. I hope it is correct.\n\nSubproblem 1: For what value of $\\omega$ does resonance occur?", "solution": "Resonance occurs when $\\omega=\\boxed{2}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 75}
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{"problem": "Subproblem 0: Find a purely sinusoidal solution of $\\frac{d^{4} x}{d t^{4}}-x=\\cos (2 t)$.\n\n\nSolution: We choose an exponential input function whose real part is $\\cos (2 t)$, namely $e^{2 i t}$. Since $p(s)=s^{4}-1$ and $p(2 i)=15 \\neq 0$, the exponential response formula yields the solution $\\frac{e^{2 i t}}{15}$. A sinusoidal solution to the original equation is given by the real part: $\\boxed{\\frac{\\cos (2 t)}{15}}$. \n\nFinal answer: The final answer is \\frac{\\cos (2 t)}{15}. I hope it is correct.\n\nSubproblem 1: Find the general solution to $\\frac{d^{4} x}{d t^{4}}-x=\\cos (2 t)$, denoting constants as $C_{1}, C_{2}, C_{3}, C_{4}$.", "solution": "To get the general solution, we take the sum of the general solution to the homogeneous equation and the particular solution to the original equation. The homogeneous equation is $\\frac{d^{4} x}{d t^{4}}-x=0$. The characteristic polynomial $p(s)=s^{4}-1$ has 4 roots: $\\pm 1, \\pm i$. So the general solution to $\\frac{d^{4} x}{d t^{4}}-x=0$ is given by $C_{1} e^{t}+C_{2} e^{-t}+C_{3} \\cos (t)+C_{4} \\sin (t)$ for arbitrary real constants $C_{1}, C_{2}, C_{3}, C_{4}$.\nThe solution to the equation is $\\boxed{\\frac{\\cos (2 t)}{15}+C_{1} e^{t}+C_{2} e^{-t}+C_{3} \\cos (t)+C_{4} \\sin (t)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 76}
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{"problem": "For $\\omega \\geq 0$, find $A$ such that $A \\cos (\\omega t)$ is a solution of $\\ddot{x}+4 x=\\cos (\\omega t)$.", "solution": "If $x=A \\cos (\\omega t)$, then taking derivatives gives us $\\ddot{x}=-\\omega^{2} A \\cos (\\omega t)$, and $\\ddot{x}+4 x=\\left(4-\\omega^{2}\\right) A \\cos (\\omega t)$. Then $A=\\boxed{\\frac{1}{4-\\omega^{2}}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 77}
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{"problem": "Find a solution to $\\dot{x}+2 x=\\cos (2 t)$ in the form $k_0\\left[f(k_1t) + g(k_2t)\\right]$, where $f, g$ are trigonometric functions. Do not include homogeneous solutions to this ODE in your solution.", "solution": "$\\cos (2 t)=\\operatorname{Re}\\left(e^{2 i t}\\right)$, so $x$ can be the real part of any solution $z$ to $\\dot{z}+2 z=e^{2 i t}$. One solution is given by $x=\\operatorname{Re}\\left(e^{2 i t} /(2+2 i)\\right)=\\boxed{\\frac{\\cos (2 t)+\\sin (2 t)}{4}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 78}
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{"problem": "Preamble: The following subproblems refer to the differential equation. $\\ddot{x}+4 x=\\sin (3 t)$\n\nFind $A$ so that $A \\sin (3 t)$ is a solution of $\\ddot{x}+4 x=\\sin (3 t)$.", "solution": "We can find this by brute force. If $x=A \\sin (3 t)$, then $\\ddot{x}=-9 A \\sin (3 t)$, so $\\ddot{x}+4 x=-5 A \\sin (3 t)$. Therefore, when $A=\\boxed{-0.2}, x_{p}(t)=-\\sin (3 t) / 5$ is a solution of the given equation.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 79}
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{"problem": "Find the general solution of the differential equation $y^{\\prime}=x-2 y$ analytically using integrating factors, solving for $y$. Note that a function $u(t)$ such that $u \\dot{x}+u p x=\\frac{d}{d t}(u x)$ is an integrating factor. Additionally, note that a general solution to a differential equation has the form $x=x_{p}+c x_{h}$ where $x_{h}$ is a nonzero solution of the homogeneous equation $\\dot{x}+p x=0$.", "solution": "In standard form, $y^{\\prime}+2 y=x$, so $u=C e^{2 x}$. Then $y=u^{-1} \\int u x d x=e^{-2 x} \\int x e^{2 x} d x$. Integrating by parts yields $\\int x e^{2 x} d x=$ $\\frac{x}{2} e^{2 x}-\\frac{1}{2} \\int e^{2 x} d x=\\frac{x}{2} e^{2 x}-\\frac{1}{4} e^{2 x}+c$. Therefore, $y=\\boxed{x / 2-1 / 4+c e^{-2 x}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 80}
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{"problem": "Subproblem 0: Find a purely exponential solution of $\\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$.\n\n\nSolution: The characteristic polynomial of the homogeneous equation is given by $p(s)=$ $s^{4}-1$. Since $p(-2)=15 \\neq 0$, the exponential response formula gives the solution $\\frac{e^{-2 t}}{p(-2)}=\\boxed{\\frac{e^{-2 t}}{15}}$.\n\nFinal answer: The final answer is \\frac{e^{-2 t}}{15}. I hope it is correct.\n\nSubproblem 1: Find the general solution to $\\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$, denoting constants as $C_{1}, C_{2}, C_{3}, C_{4}$.", "solution": "To get the general solution, we take the sum of the general solution to the homogeneous equation and the particular solution to the original equation. The homogeneous equation is $\\frac{d^{4} x}{d t^{4}}-x=0$. The characteristic polynomial $p(s)=s^{4}-1$ has 4 roots: $\\pm 1, \\pm i$. So the general solution to $\\frac{d^{4} x}{d t^{4}}-x=0$ is given by $C_{1} e^{t}+C_{2} e^{-t}+C_{3} \\cos (t)+C_{4} \\sin (t)$ for arbitrary real constants $C_{1}, C_{2}, C_{3}, C_{4}$.\nTherefore, the general solution to the equation is $\\boxed{\\frac{e^{-2 t}}{15}+C_{1} e^{t}+C_{2} e^{-t}+ C_{3} \\cos (t)+C_{4} \\sin (t)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 81}
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{"problem": "Preamble: Consider the differential equation $\\ddot{x}+\\omega^{2} x=0$. \\\\\n\nA differential equation $m \\ddot{x}+b \\dot{x}+k x=0$ (where $m, b$, and $k$ are real constants, and $m \\neq 0$ ) has corresponding characteristic polynomial $p(s)=m s^{2}+b s+k$.\\\\\nWhat is the characteristic polynomial $p(s)$ of $\\ddot{x}+\\omega^{2} x=0$?", "solution": "The characteristic polynomial $p(s)$ is $p(s)=\\boxed{s^{2}+\\omega^{2}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 82}
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{"problem": "Rewrite the function $\\cos (\\pi t)-\\sqrt{3} \\sin (\\pi t)$ in the form $A \\cos (\\omega t-\\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$.", "solution": "The right triangle has hypotenuse of length $\\sqrt{1^{2}+(-\\sqrt{3})^{2}}=2$. The circular frequency of both summands is $\\pi$, so $\\omega=\\pi$. The argument of the hypotenuse is $-\\pi / 3$, so $f(t)=\\boxed{2 \\cos (\\pi t+\\pi / 3)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 83}
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{"problem": "Preamble: The following subproblems refer to the damped sinusoid $x(t)=A e^{-a t} \\cos (\\omega t)$.\n\nWhat is the spacing between successive maxima of $x(t)$? Assume that $\\omega \\neq 0$.", "solution": "The extrema of $x(t)=A e^{-a t} \\cos (\\omega t)$ occur when $\\dot{x}(t)=0$, i.e., $-a \\cos (\\omega t)=\\omega \\sin (\\omega t)$. When $\\omega \\neq 0$, the extrema are achieved at $t$ where $\\tan (\\omega t)=-a / \\omega$. Since minima and maxima of $x(t)$ are alternating, the maxima occur at every other $t \\operatorname{such}$ that $\\tan (\\omega t)=-a / \\omega$. If $t_{0}$ and $t_{1}$ are successive maxima, then $t_{1}-t_{0}=$ twice the period of $\\tan (\\omega t)=\\boxed{2 \\pi / \\omega}$,", "type": "Differential Equations (18.03 Spring 2010)", "idx": 84}
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{"problem": "Preamble: The following subproblems refer to a spring/mass/dashpot system driven through the spring modeled by the equation $m \\ddot{x}+b \\dot{x}+k x=k y$. Here $x$ measures the position of the mass, $y$ measures the position of the other end of the spring, and $x=y$ when the spring is relaxed.\n\nIn this system, regard $y(t)$ as the input signal and $x(t)$ as the system response. Take $m=1, b=3, k=4, y(t)=A \\cos t$. Replace the input signal by a complex exponential $y_{c x}(t)$ of which it is the real part, and compute the exponential (\"steady state\") system response $z_p(t)$; leave your answer in terms of complex exponentials, i.e. do not take the real part.", "solution": "The equation is $\\ddot{x}+3 \\dot{x}+4 x=4 A \\cos t$, with the characteristic polynomial $p(s)=s^{2}+3 s+4$. The complex exponential corresponding to the input signal is $y_{c x}=A e^{i t}$ and $p(i)=3+3 i \\neq 0$. By the Exponential Response Formula, $z_{p}=\\frac{4 A}{p(i)} e^{i t}=\\boxed{\\frac{4 A}{3+3 i} e^{i t}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 85}
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{"problem": "Preamble: The following subproblems refer to a circuit with the following parameters. Denote by $I(t)$ the current (where the positive direction is, say, clockwise) in the circuit and by $V(t)$ the voltage increase across the voltage source, at time $t$. Denote by $R$ the resistance of the resistor and $C$ the capacitance of the capacitor (in units which we will not specify)-both positive numbers. Then\n\\[\nR \\dot{I}+\\frac{1}{C} I=\\dot{V}\n\\]\n\nSuppose that $V$ is constant, $V(t)=V_{0}$. Solve for $I(t)$, with initial condition $I(0)$.", "solution": "When $V$ is constant, the equation becomes $R \\dot{I}+\\frac{1}{C} I=0$, which is separable. Solving gives us\n\\[\nI(t)=\\boxed{I(0) e^{-\\frac{t}{R C}}\n}\\].", "type": "Differential Equations (18.03 Spring 2010)", "idx": 86}
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{"problem": "Subproblem 0: Find the general (complex-valued) solution of the differential equation $\\dot{z}+2 z=e^{2 i t}$, using $C$ to stand for any complex-valued integration constants which may arise.\n\n\nSolution: Using integrating factors, we get $e^{2 t} z=e^{(2+2 i) t} /(2+2 i)+C$, or $z=\\boxed{\\frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}}$, where $C$ is any complex number.\n\nFinal answer: The final answer is \\frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}. I hope it is correct.\n\nSubproblem 1: Find a solution of the differential equation $\\dot{z}+2 z=e^{2 i t}$ in the form $w e^{t}$, where $w$ is a constant (which you should find).", "solution": "When $C=0, z=\\boxed{\\frac{e^{2 i t}}{(2+2 i)}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 87}
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{"problem": "Preamble: The following subproblems consider a second order mass/spring/dashpot system driven by a force $F_{\\text {ext }}$ acting directly on the mass: $m \\ddot{x}+b \\dot{x}+k x=F_{\\text {ext }}$. So the input signal is $F_{\\text {ext }}$ and the system response is $x$. We're interested in sinusoidal input signal, $F_{\\text {ext }}(t)=A \\cos (\\omega t)$, and in the steady state, sinusoidal system response, $x_{p}(t)=g A \\cos (\\omega t-\\phi)$. Here $g$ is the gain of the system and $\\phi$ is the phase lag. Both depend upon $\\omega$, and we will consider how that is the case. \\\\\nTake $A=1$, so the amplitude of the system response equals the gain, and take $m=1, b=\\frac{1}{4}$, and $k=2$.\\\\\n\nCompute the complex gain $H(\\omega)$ of this system. (This means: make the complex replacement $F_{\\mathrm{cx}}=e^{i \\omega t}$, and express the exponential system response $z_{p}$ as a complex multiple of $F_{\\mathrm{cx}}, i.e. z_{p}=H(\\omega) F_{\\mathrm{cx}}$).", "solution": "Set $F_{\\mathrm{cx}}=e^{i \\omega t}$. The complex replacement of the equation is $\\ddot{z}+\\frac{1}{4} \\dot{z}+2 z=e^{i \\omega t}$, with the characteristic polynomial $p(s)=s^{2}+\\frac{1}{4} s+2.$ Given that $p(i \\omega)=-\\omega^{2}+\\frac{\\omega}{4} i+2 \\neq 0$, so by the exponential response formula, $z_{p}=e^{i \\omega t} / p(i \\omega)=F_{\\mathrm{cx}} / p(i \\omega)$, and $H(\\omega)=z_{p} / F_{\\mathrm{cx}}=1 / p(i \\omega)=$ $\\frac{2-\\omega^{2}-\\omega i / 4}{\\left(2-\\omega^{2}\\right)^{2}+(\\omega / 4)^{2}}=\\boxed{\\frac{2-\\omega^{2}-\\omega i / 4}{\\omega^{4}-\\frac{63}{16} \\omega^{2}+4}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 88}
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{"problem": "Preamble: The following subproblems refer to the following \"mixing problem\": A tank holds $V$ liters of salt water. Suppose that a saline solution with concentration of $c \\mathrm{gm} /$ liter is added at the rate of $r$ liters/minute. A mixer keeps the salt essentially uniformly distributed in the tank. A pipe lets solution out of the tank at the same rate of $r$ liters/minute. \n\nWrite down the differential equation for the amount of salt in the tank in standard linear form. [Not the concentration!] Use the notation $x(t)$ for the number of grams of salt in the tank at time $t$.", "solution": "The concentration of salt at any given time is $x(t) / V \\mathrm{gm} /$ liter, so for small $\\Delta t$, we lose $r x(t) \\Delta t / V$ gm from the exit pipe, and we gain $r c \\Delta t \\mathrm{gm}$ from the input pipe. The equation is $x^{\\prime}(t)=r c-\\frac{r x(t)}{V}$, and in standard linear form, it is\n$\\boxed{x^{\\prime}+\\frac{r}{V} x-r c=0}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 89}
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{"problem": "Find the polynomial solution of $\\ddot{x}-x=t^{2}+t+1$, solving for $x(t)$.", "solution": "Since the constant term of the right-hand side is nonzero, the undetermined coefficients theorem asserts that there is a unique quadratic polynomial $a t^{2}+b t+c$ satisfying this equation. Substituting this form into the left side of the equation, we see that $a=-1,-b=1$, and $2 a-c=1$, so $b=-1$ and $c=-3$. Finally, $x(t) = \\boxed{-t^2 - t - 3}$", "type": "Differential Equations (18.03 Spring 2010)", "idx": 90}
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{"problem": "Preamble: In the following problems, take $a = \\ln 2$ and $b = \\pi / 3$. \n\nSubproblem 0: Given $a = \\ln 2$ and $b = \\pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers. \n\n\nSolution: Using Euler's formula, we find that the answer is $\\boxed{1+\\sqrt{3} i}$.\n\nFinal answer: The final answer is 1+\\sqrt{3} i. I hope it is correct.\n\nSubproblem 1: Given $a = \\ln 2$ and $b = \\pi / 3$, rewrite $e^{2(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.\n\n\nSolution: $e^{n(a+b i)}=(1+\\sqrt{3} i)^{n}$, so the answer is $\\boxed{-2+2 \\sqrt{3} i}$.\n\nFinal answer: The final answer is -2+2 \\sqrt{3} i. I hope it is correct.\n\nSubproblem 2: Rewrite $e^{3(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.", "solution": "$e^{n(a+b i)}=(1+\\sqrt{3} i)^{n}$, so the answer is $\\boxed{-8}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 91}
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{"problem": "Find a purely sinusoidal solution of $\\frac{d^{4} x}{d t^{4}}-x=\\cos (2 t)$.", "solution": "We choose an exponential input function whose real part is $\\cos (2 t)$, namely $e^{2 i t}$. Since $p(s)=s^{4}-1$ and $p(2 i)=15 \\neq 0$, the exponential response formula yields the solution $\\frac{e^{2 i t}}{15}$. A sinusoidal solution to the original equation is given by the real part: $\\boxed{\\frac{\\cos (2 t)}{15}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 92}
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{"problem": "Preamble: In the following problems, take $a = \\ln 2$ and $b = \\pi / 3$. \n\nSubproblem 0: Given $a = \\ln 2$ and $b = \\pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers. \n\n\nSolution: Using Euler's formula, we find that the answer is $\\boxed{1+\\sqrt{3} i}$.\n\nFinal answer: The final answer is 1+\\sqrt{3} i. I hope it is correct.\n\nSubproblem 1: Given $a = \\ln 2$ and $b = \\pi / 3$, rewrite $e^{2(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.", "solution": "$e^{n(a+b i)}=(1+\\sqrt{3} i)^{n}$, so the answer is $\\boxed{-2+2 \\sqrt{3} i}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 93}
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{"problem": "Find a solution of $\\ddot{x}+4 x=\\cos (2 t)$, solving for $x(t)$, by using the ERF on a complex replacement. The ERF (Exponential Response Formula) states that a solution to $p(D) x=A e^{r t}$ is given by $x_{p}=A \\frac{e^{r t}}{p(r)}$, as long as $\\left.p (r\\right) \\neq 0$). The ERF with resonance assumes that $p(r)=0$ and states that a solution to $p(D) x=A e^{r t}$ is given by $x_{p}=A \\frac{t e^{r t}}{p^{\\prime}(r)}$, as long as $\\left.p^{\\prime} ( r\\right) \\neq 0$.", "solution": "The complex replacement of the equation is $\\ddot{z}+4 z=e^{2 i t}$, with the characteristic polynomial $p(s)=s^{2}+4$. Because $p(2 i)=0$ and $p^{\\prime}(2 i)=4 i \\neq 0$, we need to use the Resonant ERF, which leads to $z_{p}=\\frac{t e^{2 i t}}{4 i}$. A solution of the original equation is given by $x_{p}=\\operatorname{Re}\\left(z_{p}\\right)=\\boxed{\\frac{t}{4} \\sin (2 t)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 94}
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{"problem": "Given the ordinary differential equation $\\ddot{x}-a^{2} x=0$, where $a$ is a nonzero real-valued constant, find a solution $x(t)$ to this equation such that $x(0) = 1$ and $\\dot{x}(0)=0$.", "solution": "First, notice that both $x(t)=e^{a t}$ and $x(t)=e^{-a t}$ are solutions to $\\ddot{x}-a^{2} x=0$. Then for any constants $c_{1}$ and $c_{2}$, $x(t)=c_{1} e^{a t}+c_{2} e^{-a t}$ are also solutions to $\\ddot{x}-a^{2} x=0$. Moreover, $x(0)=c_{1}+c_{2}$, and $\\dot{x}(0)=a\\left(c_{1}-c_{2}\\right)$. Assuming $a \\neq 0$, to satisfy the given conditions, we need $c_{1}+c_{2}=1$ and $a\\left(c_{1}-c_{2}\\right)=0$, which implies $c_{1}=c_{2}=1 / 2$. So $x(t)=\\boxed{\\frac{1}{2}(\\exp{a*t} + \\exp{-a*t})}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 95}
|
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{"problem": "Find the general solution of the differential equation $\\dot{x}+2 x=e^{t}$, using $c$ for the arbitrary constant of integration which will occur.", "solution": "We can use integrating factors to get $(u x)^{\\prime}=u e^{t}$ for $u=e^{2 t}$. Integrating yields $e^{2 t} x=e^{3 t} / 3+c$, or $x=\\boxed{\\frac{e^{t}} {3}+c e^{-2 t}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 96}
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{"problem": "Find a solution of $\\ddot{x}+3 \\dot{x}+2 x=t e^{-t}$ in the form $x(t)=u(t) e^{-t}$ for some function $u(t)$. Use $C$ for an arbitrary constant, should it arise.", "solution": "$\\dot{x}=\\dot{u} e^{-t}-u e^{-t}$ and $\\ddot{x}=\\ddot{u} e^{-t}-2 \\dot{u} e^{-t}+u e^{-t}$. Plugging into the equation leads to $e^{-t}(\\ddot{u}+\\dot{u})=t e^{-t}$. Cancelling off $e^{-t}$ from both sides, we get $\\ddot{u}+\\dot{u}=t$. To solve this equation for $u$, we use the undetermined coefficient method. However, the corresponding characteristic polynomial $p(s)=s^{2}+s$ has zero as its constant term. So set $w=\\dot{u}$, then the equation can be rewritten as $\\dot{w}+w=t$. This can be solved and one solution is $w=t-1$, and hence $\\dot{u}=t-1$, and one solution for $u$ is $u=\\frac{t^{2}}{2}-t+C$. Back to the original equation, one solution is given by $x=\\boxed{\\left(\\frac{t^{2}}{2}-t+C\\right) e^{-t}}$", "type": "Differential Equations (18.03 Spring 2010)", "idx": 97}
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{"problem": "If the complex number $z$ is given by $z = 1+\\sqrt{3} i$, what is the real part of $z^2$?", "solution": "$z^{2}$ has argument $2 \\pi / 3$ and radius 4 , so by Euler's formula, $z^{2}=4 e^{i 2 \\pi / 3}=-2+2 \\sqrt{3} i$. Thus $a = -2, b = 2\\sqrt 3$, so our answer is \\boxed{-2}.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 98}
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{"problem": "Find a purely exponential solution of $\\frac{d^{4} x}{d t^{4}}-x=e^{-2 t}$.", "solution": "The characteristic polynomial of the homogeneous equation is given by $p(s)=$ $s^{4}-1$. Since $p(-2)=15 \\neq 0$, the exponential response formula gives the solution $\\frac{e^{-2 t}}{p(-2)}=\\boxed{\\frac{e^{-2 t}}{15}}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 99}
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{"problem": "Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \\cos (3 t)$, which we will assume is a solution of the differential equation $m \\ddot{x}+b \\dot{x}+k x=0$. \n\nWhat is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$.", "solution": "We can write $e^{-t / 2} \\cos (3 t)=\\operatorname{Re} e^{(-1 / 2 \\pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\\frac{1}{2} \\pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\\left(s^{2}+s+\\frac{37}{4}\\right)$. Then $b=\\boxed{m}$,", "type": "Differential Equations (18.03 Spring 2010)", "idx": 100}
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{"problem": "Preamble: The following subproblems refer to the differential equation. $\\ddot{x}+4 x=\\sin (3 t)$\n\nSubproblem 0: Find $A$ so that $A \\sin (3 t)$ is a solution of $\\ddot{x}+4 x=\\sin (3 t)$.\n\n\nSolution: We can find this by brute force. If $x=A \\sin (3 t)$, then $\\ddot{x}=-9 A \\sin (3 t)$, so $\\ddot{x}+4 x=-5 A \\sin (3 t)$. Therefore, when $A=\\boxed{-0.2}, x_{p}(t)=-\\sin (3 t) / 5$ is a solution of the given equation.\n\nFinal answer: The final answer is -0.2. I hope it is correct.\n\nSubproblem 1: What is the general solution, in the form $f_0(t) + C_1f_1(t) + C_2f_2(t)$, where $C_1, C_2$ denote arbitrary constants?", "solution": "To find the general solution, we add to $x_{p}$ the general solution to the homogeneous equation $\\ddot{x}+4 x=0$. The characteristic polynomial is $p(s)=s^{2}+4$, with roots $\\pm 2 i$, so the general solution to $\\ddot{x}+4 x=0$ is $C_{1} \\sin (2 t)+C_{2} \\cos (2 t)$. Therefore, the general solution to $\\ddot{x}+4 x=\\sin (3 t)$ is given by $\\boxed{-\\sin (3 t) / 5+ C_{1} \\sin (2 t)+C_{2} \\cos (2 t)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 101}
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{"problem": "What is the smallest possible positive $k$ such that all functions $x(t)=A \\cos (\\omega t-\\phi)$---where $\\phi$ is an odd multiple of $k$---satisfy $x(0)=0$? \\\\", "solution": "$x(0)=A \\cos \\phi$. When $A=0$, then $x(t)=0$ for every $t$; when $A \\neq 0$, $x(0)=0$ implies $\\cos \\phi=0$, and hence $\\phi$ can be any odd multiple of $\\pi / 2$, i.e., $\\phi=\\pm \\pi / 2, \\pm 3 \\pi / 2, \\pm 5 \\pi / 2, \\ldots$ this means $k=\\boxed{\\frac{\\pi}{2}}$", "type": "Differential Equations (18.03 Spring 2010)", "idx": 102}
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{"problem": "Preamble: The following subproblems refer to the differential equation $\\ddot{x}+b \\dot{x}+x=0$.\\\\\n\nWhat is the characteristic polynomial $p(s)$ of $\\ddot{x}+b \\dot{x}+x=0$?", "solution": "The characteristic polynomial is $p(s)=\\boxed{s^{2}+b s+1}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 103}
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{"problem": "Preamble: The following subproblems refer to the exponential function $e^{-t / 2} \\cos (3 t)$, which we will assume is a solution of the differential equation $m \\ddot{x}+b \\dot{x}+k x=0$. \n\nSubproblem 0: What is $b$ in terms of $m$? Write $b$ as a constant times a function of $m$.\n\n\nSolution: We can write $e^{-t / 2} \\cos (3 t)=\\operatorname{Re} e^{(-1 / 2 \\pm 3 i) t}$, so $p(s)=m s^{2}+b s+k$ has solutions $-\\frac{1}{2} \\pm 3 i$. This means $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\\left(s^{2}+s+\\frac{37}{4}\\right)$. Then $b=\\boxed{m}$, \n\nFinal answer: The final answer is m. I hope it is correct.\n\nSubproblem 1: What is $k$ in terms of $m$? Write $k$ as a constant times a function of $m$.", "solution": "Having found that $p(s)=m(s+1 / 2-3 i)(s+1 / 2+3 i)=m\\left(s^{2}+s+\\frac{37}{4}\\right)$ in the previous subproblem, $k=\\boxed{\\frac{37}{4} m}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 104}
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{"problem": "Preamble: In the following problems, take $a = \\ln 2$ and $b = \\pi / 3$. \n\nSubproblem 0: Given $a = \\ln 2$ and $b = \\pi / 3$, rewrite $e^{a+b i}$ in the form $x + yi$, where $x, y$ are real numbers. \n\n\nSolution: Using Euler's formula, we find that the answer is $\\boxed{1+\\sqrt{3} i}$.\n\nFinal answer: The final answer is 1+\\sqrt{3} i. I hope it is correct.\n\nSubproblem 1: Given $a = \\ln 2$ and $b = \\pi / 3$, rewrite $e^{2(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.\n\n\nSolution: $e^{n(a+b i)}=(1+\\sqrt{3} i)^{n}$, so the answer is $\\boxed{-2+2 \\sqrt{3} i}$.\n\nFinal answer: The final answer is -2+2 \\sqrt{3} i. I hope it is correct.\n\nSubproblem 2: Rewrite $e^{3(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers. \n\n\nSolution: $e^{n(a+b i)}=(1+\\sqrt{3} i)^{n}$, so the answer is $\\boxed{-8}$.\n\nFinal answer: The final answer is -8. I hope it is correct.\n\nSubproblem 3: Rewrite $e^{4(a+b i)}$ in the form $x + yi$, where $x, y$ are real numbers.", "solution": "$e^{n(a+b i)}=(1+\\sqrt{3} i)^{n}$, so the answer is $\\boxed{-8-8 \\sqrt{3} i}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 105}
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{"problem": "Rewrite the function $\\operatorname{Re} \\frac{e^{i t}}{2+2 i}$ in the form $A \\cos (\\omega t-\\phi)$. It may help to begin by drawing a right triangle with sides $a$ and $b$.", "solution": "$e^{i t}=\\cos (t)+i \\sin (t)$, and $\\frac{1}{2+2 i}=\\frac{1-i}{4}$. the real part is then $\\frac{1}{4} \\cos (t)+$ $\\frac{1}{4} \\sin (t)$. The right triangle here has hypotenuse $\\frac{\\sqrt{2}}{4}$ and argument $\\pi / 4$, so $f(t)=\\boxed{\\frac{\\sqrt{2}}{4} \\cos (t-\\pi / 4)}$.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 106}
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{"problem": "Preamble: The following subproblems refer to the differential equation $\\ddot{x}+b \\dot{x}+x=0$.\\\\\n\nSubproblem 0: What is the characteristic polynomial $p(s)$ of $\\ddot{x}+b \\dot{x}+x=0$?\n\n\nSolution: The characteristic polynomial is $p(s)=\\boxed{s^{2}+b s+1}$.\n\nFinal answer: The final answer is s^{2}+b s+1. I hope it is correct.\n\nSubproblem 1: For what value of $b$ does $\\ddot{x}+b \\dot{x}+x=0$ exhibit critical damping?", "solution": "To exhibit critical damping, the characteristic polynomial $s^{2}+b s+1$ must be a square, i.e., $(s-k)^{2}$ for some $k$. Multiplying and comparing yields $-2 k=b$ and $k^{2}=1$. Therefore, $b$ could be either one of $=-2, 2$. When $b=-2, e^{t}$ is a solution, and it exhibits exponential growth instead of damping, so we reject that value of $b$. Therefore, the value of $b$ for which $\\ddot{x}+b \\dot{x}+x=0$ exhibits critical damping is $b=\\boxed{2}$", "type": "Differential Equations (18.03 Spring 2010)", "idx": 107}
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{"problem": "Find the general (complex-valued) solution of the differential equation $\\dot{z}+2 z=e^{2 i t}$, using $C$ to stand for any complex-valued integration constants which may arise.", "solution": "Using integrating factors, we get $e^{2 t} z=e^{(2+2 i) t} /(2+2 i)+C$, or $z=\\boxed{\\frac{e^{2 i t}}{(2+2 i)}+C e^{-2 t}}$, where $C$ is any complex number.", "type": "Differential Equations (18.03 Spring 2010)", "idx": 108}
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{"problem": "Preamble: Consider the first-order system\n\\[\n\\tau \\dot{y}+y=u\n\\]\ndriven with a unit step from zero initial conditions. The input to this system is \\(u\\) and the output is \\(y\\). \n\nDerive and expression for the settling time \\(t_{s}\\), where the settling is to within an error \\(\\pm \\Delta\\) from the final value of 1.", "solution": "Rise and Settling Times. We are given the first-order transfer function\n\\[\nH(s)=\\frac{1}{\\tau s+1}\n\\]\nThe response to a unit step with zero initial conditions will be \\(y(t)=1-e^{-t / \\tau}\\). To determine the amount of time it take \\(y\\) to settle to within \\(\\Delta\\) of its final value, we want to find the time \\(t_{s}\\) such that \\(y\\left(t_{s}\\right)=1-\\Delta\\). Thus, we obtain\n\\[\n\\begin{aligned}\n&\\Delta=e^{-t_{s} / \\tau} \\\\\n&t_{s}=\\boxed{-\\tau \\ln \\Delta}\n\\end{aligned}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 109}
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{"problem": "Preamble: Consider the first-order system\n\\[\n\\tau \\dot{y}+y=u\n\\]\ndriven with a unit step from zero initial conditions. The input to this system is \\(u\\) and the output is \\(y\\). \n\nSubproblem 0: Derive and expression for the settling time \\(t_{s}\\), where the settling is to within an error \\(\\pm \\Delta\\) from the final value of 1.\n\n\nSolution: Rise and Settling Times. We are given the first-order transfer function\n\\[\nH(s)=\\frac{1}{\\tau s+1}\n\\]\nThe response to a unit step with zero initial conditions will be \\(y(t)=1-e^{-t / \\tau}\\). To determine the amount of time it take \\(y\\) to settle to within \\(\\Delta\\) of its final value, we want to find the time \\(t_{s}\\) such that \\(y\\left(t_{s}\\right)=1-\\Delta\\). Thus, we obtain\n\\[\n\\begin{aligned}\n&\\Delta=e^{-t_{s} / \\tau} \\\\\n&t_{s}=\\boxed{-\\tau \\ln \\Delta}\n\\end{aligned}\n\\]\n\nFinal answer: The final answer is -\\tau \\ln \\Delta. I hope it is correct.\n\nSubproblem 1: Derive an expression for the \\(10-90 \\%\\) rise time \\(t_{r}\\) in terms of $\\tau$.", "solution": "The \\(10-90 \\%\\) rise time \\(t_{r}\\) may be thought of as the difference between the \\(90 \\%\\) settling time \\((\\Delta=0.1)\\) and the \\(10 \\%\\) settling time \\((\\Delta=0.9)\\).\n\\[\nt_{r}=t_{\\Delta=0.1}-t_{\\Delta=0.9}\n\\]\nTherefore, we find \\(t_{r}=\\boxed{2.2 \\tau}\\).", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 110}
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{"problem": "Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :\n\n$y(t)=e^{-a t}$", "solution": "This function is one of the most widely used in dynamic systems, so we memorize its transform!\n\\[\nY(s)=\\boxed{\\frac{1}{s+a}}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 111}
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{"problem": "Preamble: For each Laplace Transform \\(Y(s)\\), find the function \\(y(t)\\) :\n\nSubproblem 0: \\[\nY(s)=\\boxed{\\frac{1}{(s+a)(s+b)}}\n\\]\n\n\nSolution: We can simplify with partial fractions:\n\\[\nY(s)=\\frac{1}{(s+a)(s+b)}=\\frac{C}{s+a}+\\frac{D}{s+b}\n\\]\nfind the constants \\(C\\) and \\(D\\) by setting \\(s=-a\\) and \\(s=-b\\)\n\\[\n\\begin{aligned}\n\\frac{1}{(s+a)(s+b)} &=\\frac{C}{s+a}+\\frac{D}{s+b} \\\\\n1 &=C(s+b)+D(s+a) \\\\\nC &=\\frac{1}{b-a} \\\\\nD &=\\frac{1}{a-b}\n\\end{aligned}\n\\]\ntherefore\n\\[\nY(s)=\\frac{1}{b-a} \\frac{1}{s+a}-\\frac{1}{b-a} \\frac{1}{s+b}\n\\]\nBy looking up the inverse Laplace Transform of \\(\\frac{1}{s+b}\\), we find the total solution \\(y(t)\\)\n\\[\ny(t)=\\boxed{\\frac{1}{b-a}\\left(e^{-a t}-e^{-b t}\\right)}\n\\]\n\nFinal answer: The final answer is \\frac{1}{b-a}\\left(e^{-a t}-e^{-b t}\\right). I hope it is correct.\n\nSubproblem 1: \\[\nY(s)=\\frac{s}{\\frac{s^{2}}{\\omega_{n}^{2}}+\\frac{2 \\zeta}{\\omega_{n}} s+1}\n\\]\nYou may assume that $\\zeta < 1$.", "solution": "First, note that the transform is\n\\[\n\\begin{aligned}\nY(s) &=\\frac{s}{\\frac{s^{2}}{\\omega_{n}^{2}}+\\frac{2 \\zeta}{\\omega_{n}} s+1} \\\\\n&=s \\cdot \\frac{\\omega_{n}^{2}}{s^{2}+2 \\zeta \\omega_{n} s+\\omega_{n}^{2}}\n\\end{aligned}\n\\]\nWe will solve this problem using the property\n\\[\n\\frac{d f}{d t}=s F(s)-f(0)\n\\]\ntherefore\n\\[\n\\begin{aligned}\ny(t) &=\\frac{d}{d t}\\left(\\frac{\\omega_{n}}{\\sqrt{1-\\zeta^{2}}} e^{-\\zeta \\omega_{n} t} \\sin \\left(\\omega_{n} \\sqrt{1-\\zeta^{2}} t\\right)\\right) \\\\\n&=\\boxed{\\omega_{n}^{2} e^{-\\zeta \\omega_{n} t} \\cos \\left(\\omega_{n} \\sqrt{1-\\zeta^{2}} t\\right)-\\frac{\\zeta \\omega_{n}^{2}}{\\sqrt{1-\\zeta^{2}}} e^{-\\zeta \\omega_{n} t} \\sin \\left(\\omega_{n} \\sqrt{1-\\zeta^{2}} t\\right)}\n\\end{aligned}\n\\]\nremember that for this form to be correct, \\(\\zeta\\) must be less than 1 .", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 112}
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{"problem": "A signal \\(x(t)\\) is given by\n\\[\nx(t)=\\left(e^{-t}-e^{-1}\\right)\\left(u_{s}(t)-u_{s}(t-1)\\right)\n\\]\nCalculate its Laplace transform \\(X(s)\\). Make sure to clearly show the steps in your calculation.", "solution": "Simplify the expression in to a sum of terms,\n\\[\nx(t)=e^{-t} u_{s}(t)-e^{-1} u_{s}(t)-e^{-t} u_{s}(t-1)+e^{-1} u_{s}(t-1)\n\\]\nNow take the Laplace transform of the first, second and fourth terms,\n\\[\nX(s)=\\frac{1}{s+1}-\\frac{e^{-1}}{s}-\\mathcal{L} e^{-t} u_{s}(t-1)+\\frac{e^{-1} e^{-s}}{s}\n\\]\nThe third term requires some massaging to get it in a form available on the table. The term can be modified into the form of a time delay, by factoring out \\(e^{-1}\\).\n\\[\n\\mathcal{L}\\left\\{e^{-t} u_{s}(t-1)\\right\\}=e^{-1} \\mathcal{L}\\left\\{e^{-(t-1)} u_{s}(t-1)\\right\\}\n\\]\nNow applying the Laplace Transform for a time delay from the table\n\\[\ne^{-1} \\mathcal{L}\\left\\{e^{-(t-1)} u_{s}(t-1)\\right\\}=\\frac{e^{-1} e^{-s}}{s+1}\n\\]\nSubstituting this piece back into the expression above gives the solution\n\\[\nX(s)=\\boxed{\\frac{1}{s+1}-\\frac{e^{-1}}{s}-\\frac{e^{-1} e^{-s}}{s+1}+\\frac{e^{-1} e^{-s}}{s}}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 113}
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{"problem": "Preamble: You are given an equation of motion of the form:\n\\[\n\\dot{y}+5 y=10 u\n\\]\n\nSubproblem 0: What is the time constant for this system?\n\n\nSolution: We find the homogenous solution, solving:\n\\[\n\\dot{y}+5 y=0\n\\]\nby trying a solution of the form $y=A \\cdot e^{s, t}$.\nCalculation:\n\\[\n\\dot{y}=A \\cdot s \\cdot e^{s \\cdot t} \\mid \\Rightarrow A \\cdot s \\cdot e^{s t}+5 A \\cdot e^{s t}=0\n\\]\nyields that $s=-5$, meaning the solution is $y=A \\cdot e^{-5 \\cdot t}=A \\cdot e^{-t / \\tau}$, meaning $\\tau = \\boxed{0.2}$.\n\nFinal answer: The final answer is 0.2. I hope it is correct.\n\nSubproblem 1: If \\(u=10\\), what is the final or steady-state value for \\(y(t)\\)?", "solution": "Steady state implies $\\dot{y} = 0$, so in the case when $u=10$, we get $y=\\boxed{20}$.", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 114}
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{"problem": "A signal \\(w(t)\\) is defined as\n\\[\nw(t)=u_{s}(t)-u_{s}(t-T)\n\\]\nwhere \\(T\\) is a fixed time in seconds and \\(u_{s}(t)\\) is the unit step. Compute the Laplace transform \\(W(s)\\) of \\(w(t)\\). Show your work.", "solution": "The Laplace Transform of \\(x(t)\\) is defined as\n\\[\n\\mathcal{L}[x(t)]=X(s)=\\int_{0}^{\\infty} x(t) e^{-s t} d t\n\\]\ntherefore\n\\[\n\\begin{aligned}\nW(s) &=\\int_{0}^{\\infty} e^{-s t} d t-\\left(\\int_{0}^{T} 0 d t+\\int_{T}^{\\infty} e^{-s t} d t\\right) \\\\\n&=-\\left.\\frac{1}{s} e^{-s t}\\right|_{0} ^{\\infty}-\\left(0+-\\left.\\frac{1}{s} e^{-s t}\\right|_{T} ^{\\infty}\\right) \\\\\n&=\\boxed{\\frac{1}{s}-\\frac{1}{s} e^{-s T}}\n\\end{aligned}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 115}
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{"problem": "Preamble: Assume that we apply a unit step in force separately to a mass \\(m\\), a dashpot \\(c\\), and a spring \\(k\\). The mass moves in inertial space. The spring and dashpot have one end connected to inertial space (reference velocity \\(=0\\) ), and the force is applied to the other end. Assume zero initial velocity and position for the elements.\nRecall that the unit step function \\(u_{S}(t)\\) is defined as \\(u_{S}(t)=0 ; t<0\\) and \\(u_{S}(t)=1 ; t \\geq 0\\). We will also find it useful to introduce the unit impulse function \\(\\delta(t)\\) which can be defined via\n\\[\nu_{S}(t)=\\int_{-\\infty}^{t} \\delta(\\tau) d \\tau\n\\]\nThis means that we can also view the unit impulse as the derivative of the unit step:\n\\[\n\\delta(t)=\\frac{d u_{S}(t)}{d t}\n\\]\n\nSolve for the resulting velocity of the mass.", "solution": "\\[\n\\begin{aligned}\nm \\ddot{x}_{m} &=u_{s}(t) \\\\\n\\dot{x}_{m}=v_{m} &=\\int_{-\\infty}^{t} \\frac{1}{m} u_{s}(t) d t=\\boxed{\\frac{1}{m} t} \\\\\n\\end{aligned}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 116}
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{"problem": "Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :\n\nSubproblem 0: $y(t)=e^{-a t}$\n\n\nSolution: This function is one of the most widely used in dynamic systems, so we memorize its transform!\n\\[\nY(s)=\\boxed{\\frac{1}{s+a}}\n\\]\n\nFinal answer: The final answer is \\frac{1}{s+a}. I hope it is correct.\n\nSubproblem 1: $y(t)=e^{-\\sigma t} \\sin \\omega_{d} t$\n\n\nSolution: \\[\nY(s)=\\boxed{\\frac{\\omega_{d}}{(s+\\sigma)^{2}+\\omega_{d}^{2}}}\n\\]\n\nFinal answer: The final answer is \\frac{\\omega_{d}}{(s+\\sigma)^{2}+\\omega_{d}^{2}}. I hope it is correct.\n\nSubproblem 2: $y(t)=e^{-\\sigma t} \\cos \\omega_{d} t$", "solution": "\\[\nY(s)=\\boxed{\\frac{s+\\sigma}{(s+\\sigma)^{2}+\\omega_{d}^{2}}}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 117}
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{"problem": "Preamble: For each of the functions $y(t)$, find the Laplace Transform $Y(s)$ :\n\nSubproblem 0: $y(t)=e^{-a t}$\n\n\nSolution: This function is one of the most widely used in dynamic systems, so we memorize its transform!\n\\[\nY(s)=\\boxed{\\frac{1}{s+a}}\n\\]\n\nFinal answer: The final answer is \\frac{1}{s+a}. I hope it is correct.\n\nSubproblem 1: $y(t)=e^{-\\sigma t} \\sin \\omega_{d} t$", "solution": "\\[\nY(s)=\\boxed{\\frac{\\omega_{d}}{(s+\\sigma)^{2}+\\omega_{d}^{2}}}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 118}
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{"problem": "Preamble: Consider the mass \\(m\\) sliding horizontally under the influence of the applied force \\(f\\) and a friction force which can be approximated by a linear friction element with coefficient \\(b\\). \n\nFormulate the state-determined equation of motion for the velocity \\(v\\) as output and the force \\(f\\) as input.", "solution": "The equation of motion is\n\\[\n\\boxed{m \\frac{d v}{d t}+b v=f} \\quad \\text { or } \\quad \\frac{d v}{d t}=-\\frac{b}{m} v+\\frac{1}{m} f\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 119}
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{"problem": "Preamble: Consider the rotor with moment of inertia \\(I\\) rotating under the influence of an applied torque \\(T\\) and the frictional torques from two bearings, each of which can be approximated by a linear frictional element with coefficient \\(B\\).\n\nSubproblem 0: Formulate the state-determined equation of motion for the angular velocity $\\omega$ as output and the torque $T$ as input.\n\n\nSolution: The equation of motion is\n\\[\n\\boxed{I \\frac{d \\omega}{d t}+2 B \\omega=T} \\quad \\text { or } \\quad \\frac{d \\omega}{d t}=-\\frac{2 B}{I} \\omega+\\frac{1}{I} T\n\\]\n\nFinal answer: The final answer is I \\frac{d \\omega}{d t}+2 B \\omega=T. I hope it is correct.\n\nSubproblem 1: Consider the case where:\n\\[\n\\begin{aligned}\nI &=0.001 \\mathrm{~kg}-\\mathrm{m}^{2} \\\\\nB &=0.005 \\mathrm{~N}-\\mathrm{m} / \\mathrm{r} / \\mathrm{s}\n\\end{aligned}\n\\]\nWhat is the steady-state velocity \\(\\omega_{s s}\\), in radians per second, when the input is a constant torque of 10 Newton-meters?", "solution": "The steady-state angular velocity, when \\(T=10\\) Newton-meters, and \\(I=0.001 \\mathrm{~kg}-\\mathrm{m}^{2}\\), and \\(B=0.005 \\mathrm{~N}-\\mathrm{m} / \\mathrm{r} / \\mathrm{s}\\) is\n\\[\n\\omega_{s s}=\\frac{T}{2 B}=\\frac{10}{2(0.005)}=\\boxed{1000} \\mathrm{r} / \\mathrm{s}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 120}
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{"problem": "Preamble: Consider the mass \\(m\\) sliding horizontally under the influence of the applied force \\(f\\) and a friction force which can be approximated by a linear friction element with coefficient \\(b\\). \n\nSubproblem 0: Formulate the state-determined equation of motion for the velocity \\(v\\) as output and the force \\(f\\) as input.\n\n\nSolution: The equation of motion is\n\\[\n\\boxed{m \\frac{d v}{d t}+b v=f} \\quad \\text { or } \\quad \\frac{d v}{d t}=-\\frac{b}{m} v+\\frac{1}{m} f\n\\]\n\nFinal answer: The final answer is m \\frac{d v}{d t}+b v=f. I hope it is correct.\n\nSubproblem 1: Consider the case where:\n\\[\n\\begin{aligned}\nm &=1000 \\mathrm{~kg} \\\\\nb &=100 \\mathrm{~N} / \\mathrm{m} / \\mathrm{s}\n\\end{aligned}\n\\]\nWhat is the steady-state velocity \\(v_{s s}\\) when the input is a constant force of 10 Newtons? Answer in meters per second.", "solution": "The steady-state velocity, when \\(f=10\\) Newtons, and \\(m=1000 \\mathrm{~kg}\\), and \\(b=100 \\mathrm{~N} / \\mathrm{m} / \\mathrm{s}\\) is\n\\[\nv_{s s}=\\frac{f}{b}=\\frac{10}{100}=\\boxed{0.10} \\mathrm{~m} / \\mathrm{s}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 121}
|
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{"problem": "Obtain the inverse Laplace transform of the following frequency-domain expression: $F(s) = -\\frac{(4 s-10)}{s(s+2)(s+5)}$.\nUse $u(t)$ to denote the unit step function.", "solution": "Using partial fraction expansion, the above can be rewritten as \n\\[\nF(s) = \\frac{1}{s} - \\frac{3}{s+2} + \\frac{2}{s+5}\n\\]\nApply the inverse Laplace transform, then we end up with\n\\[\nf(t) = \\boxed{(1 - 3e^{-2t} + 2e^{-5t}) u(t)}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 122}
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{"problem": "A signal has a Laplace transform\n\\[\nX(s)=b+\\frac{a}{s(s+a)}\n\\]\nwhere \\(a, b>0\\), and with a region of convergence of \\(|s|>0\\). Find \\(x(t), t>0\\).", "solution": "Each term of \\(X(s)\\) can be evaluated directly using a table of Laplace Transforms:\n\\[\n\\mathcal{L}^{-1}\\{b\\}=b \\delta(t)\n\\]\nand\n\\[\n\\mathcal{L}^{-1}\\left\\{\\frac{a}{s(s+a)}\\right\\}=1-e^{-a t}\n\\]\nThe final result is then\n\\[\n\\mathcal{L}^{-1}\\{X(s)\\}=\\boxed{b \\delta(t)+1-e^{-a t}}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 123}
|
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{"problem": "Preamble: For each Laplace Transform \\(Y(s)\\), find the function \\(y(t)\\) :\n\n\\[\nY(s)=\\boxed{\\frac{1}{(s+a)(s+b)}}\n\\]", "solution": "We can simplify with partial fractions:\n\\[\nY(s)=\\frac{1}{(s+a)(s+b)}=\\frac{C}{s+a}+\\frac{D}{s+b}\n\\]\nfind the constants \\(C\\) and \\(D\\) by setting \\(s=-a\\) and \\(s=-b\\)\n\\[\n\\begin{aligned}\n\\frac{1}{(s+a)(s+b)} &=\\frac{C}{s+a}+\\frac{D}{s+b} \\\\\n1 &=C(s+b)+D(s+a) \\\\\nC &=\\frac{1}{b-a} \\\\\nD &=\\frac{1}{a-b}\n\\end{aligned}\n\\]\ntherefore\n\\[\nY(s)=\\frac{1}{b-a} \\frac{1}{s+a}-\\frac{1}{b-a} \\frac{1}{s+b}\n\\]\nBy looking up the inverse Laplace Transform of \\(\\frac{1}{s+b}\\), we find the total solution \\(y(t)\\)\n\\[\ny(t)=\\boxed{\\frac{1}{b-a}\\left(e^{-a t}-e^{-b t}\\right)}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 124}
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{"problem": "Preamble: Consider the rotor with moment of inertia \\(I\\) rotating under the influence of an applied torque \\(T\\) and the frictional torques from two bearings, each of which can be approximated by a linear frictional element with coefficient \\(B\\).\n\nFormulate the state-determined equation of motion for the angular velocity $\\omega$ as output and the torque $T$ as input.", "solution": "The equation of motion is\n\\[\n\\boxed{I \\frac{d \\omega}{d t}+2 B \\omega=T} \\quad \\text { or } \\quad \\frac{d \\omega}{d t}=-\\frac{2 B}{I} \\omega+\\frac{1}{I} T\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 125}
|
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{"problem": "Obtain the inverse Laplace transform of the following frequency-domain expression: $F(s) = \\frac{4}{s^2(s^2+4)}$.\nUse $u(t)$ to denote the unit step function.", "solution": "Since $F(s) = \\frac{1}{s^2} + \\frac{-1}{s^2+4}$, its inverse Laplace transform is \n\\[\nf(t) = \\boxed{(t + \\frac{1}{2} \\sin{2t}) u(t)}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 126}
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{"problem": "Preamble: This problem considers the simple RLC circuit, in which a voltage source $v_{i}$ is in series with a resistor $R$, inductor $L$, and capacitor $C$. We measure the voltage $v_{o}$ across the capacitor. $v_{i}$ and $v_{o}$ share a ground reference.\n\nCalculate the transfer function \\(V_{o}(s) / V_{i}(s)\\).", "solution": "Using the voltage divider relationship:\n\\[\n\\begin{aligned}\nV_{o}(s) &=\\frac{Z_{e q}}{Z_{\\text {total }}}V_{i}(s)=\\frac{\\frac{1}{C s}}{R+L s+\\frac{1}{C s}} V_{i}(s) \\\\\n\\frac{V_{o}(s)}{V_{i}(s)} &=\\boxed{\\frac{1}{L C s^{2}+R C s+1}}\n\\end{aligned}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 127}
|
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{"problem": "Preamble: You are given an equation of motion of the form:\n\\[\n\\dot{y}+5 y=10 u\n\\]\n\nWhat is the time constant for this system?", "solution": "We find the homogenous solution, solving:\n\\[\n\\dot{y}+5 y=0\n\\]\nby trying a solution of the form $y=A \\cdot e^{s, t}$.\nCalculation:\n\\[\n\\dot{y}=A \\cdot s \\cdot e^{s \\cdot t} \\mid \\Rightarrow A \\cdot s \\cdot e^{s t}+5 A \\cdot e^{s t}=0\n\\]\nyields that $s=-5$, meaning the solution is $y=A \\cdot e^{-5 \\cdot t}=A \\cdot e^{-t / \\tau}$, meaning $\\tau = \\boxed{0.2}$.", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 128}
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{"problem": "Preamble: This problem considers the simple RLC circuit, in which a voltage source $v_{i}$ is in series with a resistor $R$, inductor $L$, and capacitor $C$. We measure the voltage $v_{o}$ across the capacitor. $v_{i}$ and $v_{o}$ share a ground reference.\n\nSubproblem 0: Calculate the transfer function \\(V_{o}(s) / V_{i}(s)\\).\n\n\nSolution: Using the voltage divider relationship:\n\\[\n\\begin{aligned}\nV_{o}(s) &=\\frac{Z_{e q}}{Z_{\\text {total }}}V_{i}(s)=\\frac{\\frac{1}{C s}}{R+L s+\\frac{1}{C s}} V_{i}(s) \\\\\n\\frac{V_{o}(s)}{V_{i}(s)} &=\\boxed{\\frac{1}{L C s^{2}+R C s+1}}\n\\end{aligned}\n\\]\n\nFinal answer: The final answer is \\frac{1}{L C s^{2}+R C s+1}. I hope it is correct.\n\nSubproblem 1: Let \\(L=0.01 \\mathrm{H}\\). Choose the value of $C$ such that \\(\\omega_{n}=10^{5}\\) and \\(\\zeta=0.05\\). Give your answer in Farads.", "solution": "$C=\\frac{1}{\\omega_{n}^{2}L}=\\boxed{1e-8}[\\mathrm{~F}]$", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 129}
|
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{"problem": "Preamble: Here we consider a system described by the differential equation\n\\[\n\\ddot{y}+10 \\dot{y}+10000 y=0 .\n\\]\n\nWhat is the value of the natural frequency \\(\\omega_{n}\\) in radians per second?", "solution": "$\\omega_{n}=\\sqrt{\\frac{k}{m}}$\nSo\n$\\omega_{n} =\\boxed{100} \\mathrm{rad} / \\mathrm{s}$", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 130}
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{"problem": "Preamble: Consider a circuit in which a voltage source of voltage in $v_{i}(t)$ is connected in series with an inductor $L$ and capacitor $C$. We consider the voltage across the capacitor $v_{o}(t)$ to be the output of the system.\nBoth $v_{i}(t)$ and $v_{o}(t)$ share ground reference.\n\nWrite the governing differential equation for this circuit.", "solution": "Using Kirchoff Current Law at the node between the inductor and capacitor with the assumed currents both positive into the node gives the following:\n\\[\n\\begin{gathered}\ni_{L}+i_{C}=0 \\\\\ni_{L}=\\frac{1}{L} \\int v_{L} d t \\\\\ni_{C}=C \\frac{d v_{c}}{d t}\n\\end{gathered}\n\\]\nThe above equation must be differentiated before substituting for the currents and from the direction of our assumed currents, \\(v_{L}=v_{i}-v_{o}\\) and \\(v_{C}=0-v_{o}\\). The governing differential equation is then\n\\[\n\\boxed{\\frac{d^{2} v_{o}}{d t^{2}}+\\frac{v_{o}}{L C}=\\frac{v_{i}}{L C}}\n\\]", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 131}
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{"problem": "Write (but don't solve) the equation of motion for a pendulum consisting of a mass $m$ attached to a rigid massless rod, suspended from the ceiling and free to rotate in a single vertical plane. Let the rod (of length $l$) make an angle of $\\theta$ with the vertical. Gravity ($mg$) acts directly downward, the system input is a horizontal external force $f(t)$, and the system output is the angle $\\theta(t)$. \nNote: Do NOT make the small-angle approximation in your equation.", "solution": "From force balance, we can derive the equation of motion. Choosing the system variable system variable $\\theta(t)$ with polar coordinates, we don't need to care about tension on the rod and centrifugal force.\nWe can use the relation between torque and angular momentum to immediately write down the equation for $\\theta(t)$:\n\\[\nm l^{2} \\ddot{\\theta}(t)-m g l \\sin \\theta(t)=f(t) l \\cos \\theta(t) .\n\\]\nDividing both sides by $l$ gives:\n\\[\n\\boxed{m l \\ddot{\\theta}(t)-m g \\sin \\theta(t)=f(t) \\cos \\theta(t)} .\n\\]\nNote that inertia of the mass with respect to the rotation axis is $m l^{2}$. It is a non linear differential equation because it has $\\sin \\theta(t)$ term.", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 132}
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{"problem": "Preamble: Here we consider a system described by the differential equation\n\\[\n\\ddot{y}+10 \\dot{y}+10000 y=0 .\n\\]\n\nSubproblem 0: What is the value of the natural frequency \\(\\omega_{n}\\) in radians per second?\n\n\nSolution: $\\omega_{n}=\\sqrt{\\frac{k}{m}}$\nSo\n$\\omega_{n} =\\boxed{100} \\mathrm{rad} / \\mathrm{s}$\n\nFinal answer: The final answer is 100. I hope it is correct.\n\nSubproblem 1: What is the value of the damping ratio \\(\\zeta\\)? \n\n\nSolution: $\\zeta=\\frac{b}{2 \\sqrt{k m}}$\nSo\n$\\zeta =\\boxed{0.05}$\n\nFinal answer: The final answer is 0.05. I hope it is correct.\n\nSubproblem 2: What is the value of the damped natural frequency \\(\\omega_{d}\\) in radians per second? Give your answer to three significant figures.", "solution": "$\\omega_{d}=\\omega_{n} \\sqrt{1-\\zeta^{2}}$\nSo\n$\\omega_{d}=\\boxed{99.9} \\mathrm{rad} / \\mathrm{s}$", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 133}
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{"problem": "Preamble: Here we consider a system described by the differential equation\n\\[\n\\ddot{y}+10 \\dot{y}+10000 y=0 .\n\\]\n\nSubproblem 0: What is the value of the natural frequency \\(\\omega_{n}\\) in radians per second?\n\n\nSolution: $\\omega_{n}=\\sqrt{\\frac{k}{m}}$\nSo\n$\\omega_{n} =\\boxed{100} \\mathrm{rad} / \\mathrm{s}$\n\nFinal answer: The final answer is 100. I hope it is correct.\n\nSubproblem 1: What is the value of the damping ratio \\(\\zeta\\)?", "solution": "$\\zeta=\\frac{b}{2 \\sqrt{k m}}$\nSo\n$\\zeta =\\boxed{0.05}$", "type": "Dynamics and Control (2.003 Spring 2005)", "idx": 134}
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{"problem": "What is the speed of light in meters/second to 1 significant figure? Use the format $a \\times 10^{b}$ where a and b are numbers.", "solution": "$\\boxed{3e8}$ m/s.", "type": "Relativity (8.033 Fall 2006)", "idx": 135}
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{"problem": "Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested. \n\nSubproblem 0: Age of our universe when most He nuclei were formed in minutes: \n\n\nSolution: \\boxed{1} minute.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: Age of our universe when hydrogen atoms formed in years:\n\n\nSolution: \\boxed{400000} years.\n\nFinal answer: The final answer is 400000. I hope it is correct.\n\nSubproblem 2: Age of our universe today in Gyr:\n\n\nSolution: \\boxed{10} Gyr.\n\nFinal answer: The final answer is 10. I hope it is correct.\n\nSubproblem 3: Number of stars in our Galaxy: (Please format your answer as 'xen' representing $x * 10^n$)", "solution": "\\boxed{1e11}.", "type": "Relativity (8.033 Fall 2006)", "idx": 136}
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{"problem": "Preamble: In a parallel universe, the Boston baseball team made the playoffs.\n\nManny Relativirez hits the ball and starts running towards first base at speed $\\beta$. How fast is he running, given that he sees third base $45^{\\circ}$ to his left (as opposed to straight to his left before he started running)? Assume that he is still very close to home plate. Give your answer in terms of the speed of light, $c$.", "solution": "Using the aberration formula with $\\cos \\theta^{\\prime}=-1 / \\sqrt{2}, \\beta=1 / \\sqrt{2}$, so $v=\\boxed{\\frac{1}{\\sqrt{2}}c}$.", "type": "Relativity (8.033 Fall 2006)", "idx": 137}
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{"problem": "Preamble: In the Sun, one of the processes in the He fusion chain is $p+p+e^{-} \\rightarrow d+\\nu$, where $d$ is a deuteron. Make the approximations that the deuteron rest mass is $2 m_{p}$, and that $m_{e} \\approx 0$ and $m_{\\nu} \\approx 0$, since both the electron and the neutrino have negligible rest mass compared with the proton rest mass $m_{p}$.\n\nIn the lab frame, the two protons have the same energy $\\gamma m_{p}$ and impact angle $\\theta$, and the electron is at rest. Calculate the energy $E_{\\nu}$ of the neutrino in the rest frame of the deuteron in terms of $\\theta, m_{p}$ and $\\gamma$.", "solution": "Use the fact that the quantity $E^{2}-p^{2} c^{2}$ is invariant. In the deutron's rest frame, after the collison:\n\\[\n\\begin{aligned}\nE^{2}-p^{2} c^{2} &=\\left(2 m_{p} c^{2}+E_{\\nu}\\right)^{2}-E_{\\nu}^{2} \\\\\n&=4 m_{p}^{2} c^{4}+4 m_{p} c^{2} E_{\\nu}=4 m_{p} c^{2}\\left(m_{p} c^{2}+E_{\\nu}\\right)\n\\end{aligned}\n\\]\nIn the lab frame, before collison:\n\\[\n\\begin{aligned}\nE^{2}-p^{2} c^{2} &=\\left(2 E_{p}\\right)^{2}-\\left(2 p_{p} \\cos \\theta c\\right)^{2} \\\\\n&=\\left(2 \\gamma m_{p} c^{2}\\right)^{2}-\\left(2 \\gamma \\beta m_{p} \\cos \\theta c^{2}\\right)^{2}\n\\end{aligned}\n\\]\nUse $\\gamma^{2} \\beta^{2}=\\left(\\gamma^{2}-1\\right)$ in the second term and simplify the algebra to find\n\\[\nE^{2}-p^{2} c^{2}=4 m_{p}^{2} c^{4}\\left(\\gamma^{2}-\\left(\\gamma^{2}-1\\right) \\cos ^{2} \\theta\\right)\n\\]\nEquating the invariants in the two frames, we have\n\\[\n\\begin{aligned}\n4 m_{p} c^{2}\\left(m_{p} c^{2}+E_{\\nu}\\right) &=4 m_{p}^{2} c^{4}\\left(\\gamma^{2}-\\left(\\gamma^{2}-1\\right) \\cos ^{2} \\theta\\right) \\\\\n\\Rightarrow E_{\\nu} &= \\boxed{m_{p} c^{2}\\left(\\gamma^{2}-1\\right) \\sin ^{2} \\theta}\n\\end{aligned}\n\\]", "type": "Relativity (8.033 Fall 2006)", "idx": 138}
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{"problem": "Preamble: In a parallel universe, the Boston baseball team made the playoffs.\n\nSubproblem 0: Manny Relativirez hits the ball and starts running towards first base at speed $\\beta$. How fast is he running, given that he sees third base $45^{\\circ}$ to his left (as opposed to straight to his left before he started running)? Assume that he is still very close to home plate. Give your answer in terms of the speed of light, $c$.\n\n\nSolution: Using the aberration formula with $\\cos \\theta^{\\prime}=-1 / \\sqrt{2}, \\beta=1 / \\sqrt{2}$, so $v=\\boxed{\\frac{1}{\\sqrt{2}}c}$.\n\nFinal answer: The final answer is \\frac{1}{\\sqrt{2}}c. I hope it is correct.\n\nSubproblem 1: A player standing on third base is wearing red socks emitting light of wavelength $\\lambda_{\\text {red}}$. What wavelength does Manny see in terms of $\\lambda_{\\text {red}}$?", "solution": "Using the doppler shift formula, $\\lambda^{\\prime}= \\boxed{\\lambda_{\\text {red}} / \\sqrt{2}}$.", "type": "Relativity (8.033 Fall 2006)", "idx": 139}
|
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{"problem": "Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested. \n\nSubproblem 0: Age of our universe when most He nuclei were formed in minutes: \n\n\nSolution: \\boxed{1} minute.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: Age of our universe when hydrogen atoms formed in years:\n\n\nSolution: \\boxed{400000} years.\n\nFinal answer: The final answer is 400000. I hope it is correct.\n\nSubproblem 2: Age of our universe today in Gyr:", "solution": "\\boxed{10} Gyr.", "type": "Relativity (8.033 Fall 2006)", "idx": 140}
|
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{"problem": "How many down quarks does a tritium ($H^3$) nucleus contain?", "solution": "\\boxed{5}.", "type": "Relativity (8.033 Fall 2006)", "idx": 141}
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{"problem": "How many up quarks does a tritium ($H^3$) nucleus contain?", "solution": "\\boxed{4}.", "type": "Relativity (8.033 Fall 2006)", "idx": 142}
|
||||
{"problem": "Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested. \n\nAge of our universe when most He nuclei were formed in minutes:", "solution": "\\boxed{1} minute.", "type": "Relativity (8.033 Fall 2006)", "idx": 143}
|
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{"problem": "Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested. \n\nSubproblem 0: Age of our universe when most He nuclei were formed in minutes: \n\n\nSolution: \\boxed{1} minute.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: Age of our universe when hydrogen atoms formed in years:\n\n\nSolution: \\boxed{400000} years.\n\nFinal answer: The final answer is 400000. I hope it is correct.\n\nSubproblem 2: Age of our universe today in Gyr:\n\n\nSolution: \\boxed{10} Gyr.\n\nFinal answer: The final answer is 10. I hope it is correct.\n\nSubproblem 3: Number of stars in our Galaxy: (Please format your answer as 'xen' representing $x * 10^n$)\n\n\nSolution: \\boxed{1e11}.\n\nFinal answer: The final answer is 1e11. I hope it is correct.\n\nSubproblem 4: Light travel time to closest star (Sun!:) in minutes. (Please format your answer as an integer.)", "solution": "\\boxed{8} minutes.", "type": "Relativity (8.033 Fall 2006)", "idx": 144}
|
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{"problem": "Preamble: Give each of the following quantities to the nearest power of 10 and in the units requested. \n\nSubproblem 0: Age of our universe when most He nuclei were formed in minutes: \n\n\nSolution: \\boxed{1} minute.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: Age of our universe when hydrogen atoms formed in years:", "solution": "\\boxed{400000} years.", "type": "Relativity (8.033 Fall 2006)", "idx": 145}
|
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{"problem": "Potassium metal can be used as the active surface in a photodiode because electrons are relatively easily removed from a potassium surface. The energy needed is $2.15 \\times 10^{5} J$ per mole of electrons removed ( 1 mole $=6.02 \\times 10^{23}$ electrons). What is the longest wavelength light (in nm) with quanta of sufficient energy to eject electrons from a potassium photodiode surface?", "solution": "\\includegraphics[scale=0.5]{set_02_img_00.jpg}\n\\nonessentialimage\n$I_{p}$, the photocurrent, is proportional to the intensity of incident radiation, i.e. the number of incident photons capable of generating a photoelectron.\nThis device should be called a phototube rather than a photodiode - a solar cell is a photodiode. \nRequired: $1 eV=1.6 \\times 10^{-19} J$\n\\[\nE_{\\text {rad }}=h v=(hc) / \\lambda\n\\]\nThe question is: below what threshold energy (hv) will a photon no longer be able to generate a photoelectron?\\\\\n$2.15 x 10^{5}$ J/mole photoelectrons $\\times \\frac{1 \\text{mole}}{6.02 \\times 10^{23} \\text{photoelectrons}} = 3.57 \\times 10^{-19}$ J/photoelectron\\\\\n$\\lambda_{\\text {threshold }}=\\frac{hc}{3.57 \\times 10^{-19}}=\\frac{6.62 \\times 10^{-34} \\times 3 \\times 10^{8}}{3.57 \\times 10^{-19}}=5.6 \\times 10^{-7} m= \\boxed{560} nm$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 146}
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{"problem": "Preamble: For red light of wavelength $(\\lambda) 6.7102 \\times 10^{-5} cm$, emitted by excited lithium atoms, calculate:\n\nSubproblem 0: the frequency $(v)$ in Hz, to 4 decimal places. \n\n\nSolution: $c=\\lambda v$ and $v=c / \\lambda$ where $v$ is the frequency of radiation (number of waves/s).\nFor: $\\quad \\lambda=6.7102 \\times 10^{-5} cm=6.7102 \\times 10^{-7} m$\n\\[\nv=\\frac{2.9979 \\times 10^{8} {ms}^{-1}}{6.7102 \\times 10^{-7} m}=4.4677 \\times 10^{14} {s}^{-1}= \\boxed{4.4677} Hz\n\\]\n\nFinal answer: The final answer is 4.4677. I hope it is correct.\n\nSubproblem 1: the wave number $(\\bar{v})$ in ${cm}^{-1}$. Please format your answer as $n \\times 10^x$, where $n$ is to 4 decimal places. \n\n\nSolution: $\\bar{v}=\\frac{1}{\\lambda}=\\frac{1}{6.7102 \\times 10^{-7} m}=1.4903 \\times 10^{6} m^{-1}= \\boxed{1.4903e4} {cm}^{-1}$\n\nFinal answer: The final answer is 1.4903e4. I hope it is correct.\n\nSubproblem 2: the wavelength $(\\lambda)$ in nm, to 2 decimal places.", "solution": "$\\lambda=6.7102 \\times 10^{-5} cm \\times \\frac{1 nm}{10^{-7} cm}= \\boxed{671.02} cm$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 147}
|
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{"problem": "What is the net charge of arginine in a solution of $\\mathrm{pH} \\mathrm{} 1.0$ ? Please format your answer as +n or -n.", "solution": "\\boxed{+2}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 148}
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{"problem": "Preamble: For red light of wavelength $(\\lambda) 6.7102 \\times 10^{-5} cm$, emitted by excited lithium atoms, calculate:\n\nSubproblem 0: the frequency $(v)$ in Hz, to 4 decimal places. \n\n\nSolution: $c=\\lambda v$ and $v=c / \\lambda$ where $v$ is the frequency of radiation (number of waves/s).\nFor: $\\quad \\lambda=6.7102 \\times 10^{-5} cm=6.7102 \\times 10^{-7} m$\n\\[\nv=\\frac{2.9979 \\times 10^{8} {ms}^{-1}}{6.7102 \\times 10^{-7} m}=4.4677 \\times 10^{14} {s}^{-1}= \\boxed{4.4677} Hz\n\\]\n\nFinal answer: The final answer is 4.4677. I hope it is correct.\n\nSubproblem 1: the wave number $(\\bar{v})$ in ${cm}^{-1}$. Please format your answer as $n \\times 10^x$, where $n$ is to 4 decimal places.", "solution": "$\\bar{v}=\\frac{1}{\\lambda}=\\frac{1}{6.7102 \\times 10^{-7} m}=1.4903 \\times 10^{6} m^{-1}= \\boxed{1.4903e4} {cm}^{-1}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 149}
|
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{"problem": "Determine the atomic weight of ${He}^{++}$ in amu to 5 decimal places from the values of its constituents.", "solution": "The mass of the constituents $(2 p+2 n)$ is given as:\n\\[\n\\begin{array}{ll}\n2 p= & 2 \\times 1.6726485 \\times 10^{-24} g \\\\\n2 n= & 2 \\times 16749543 \\times 10^{-24} g\n\\end{array}\n\\]\nThe atomic weight (calculated) in amu is given as:\n\\[\n\\begin{aligned}\n&\\frac{6.6952056 \\times 10^{-24} g}{1.660565 \\times 10^{-24} g} / amu \\\\\n&{He}=\\boxed{4.03188} amu\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 150}
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{"problem": "Preamble: Determine the following values from a standard radio dial. \n\nSubproblem 0: What is the minimum wavelength in m for broadcasts on the AM band? Format your answer as an integer. \n\n\nSolution: \\[\n\\mathrm{c}=v \\lambda, \\therefore \\lambda_{\\min }=\\frac{\\mathrm{c}}{v_{\\max }} ; \\lambda_{\\max }=\\frac{\\mathrm{c}}{v_{\\min }}\n\\]\n$\\lambda_{\\min }=\\frac{3 \\times 10^{8} m / s}{1600 \\times 10^{3} Hz}=\\boxed{188} m$\n\nFinal answer: The final answer is 188. I hope it is correct.\n\nSubproblem 1: What is the maximum wavelength in m for broadcasts on the AM band? Format your answer as an integer.", "solution": "\\[\n\\mathrm{c}=v \\lambda, \\therefore \\lambda_{\\min }=\\frac{\\mathrm{c}}{v_{\\max }} ; \\lambda_{\\max }=\\frac{\\mathrm{c}}{v_{\\min }}\n\\]\n\\[\n\\lambda_{\\max }=\\frac{3 \\times 10^{8}}{530 \\times 10^{3}}=\\boxed{566} m\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 151}
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{"problem": "Determine the wavelength of radiation emitted by hydrogen atoms in angstroms upon electron transitions from $n=6$ to $n=2$.", "solution": "From the Rydberg relationship we obtain:\n\\[\n\\begin{aligned}\n&\\frac{1}{\\lambda}=\\bar{v}=R\\left(\\frac{1}{n_{i}^{2}}-\\frac{1}{n_{f}^{2}}\\right)=1.097 \\times 10^{7}\\left(\\frac{1}{36}-\\frac{1}{4}\\right)=(-) 2.44 \\times 10^{6} \\\\\n&\\lambda=\\frac{1}{v}=\\frac{1}{2.44 \\times 10^{6}}=4.1 \\times 10^{-7} {~m}=0.41 \\mu {m}=\\boxed{4100} \\text{angstroms}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 152}
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{"problem": "Preamble: Determine the following values from a standard radio dial. \n\nSubproblem 0: What is the minimum wavelength in m for broadcasts on the AM band? Format your answer as an integer. \n\n\nSolution: \\[\n\\mathrm{c}=v \\lambda, \\therefore \\lambda_{\\min }=\\frac{\\mathrm{c}}{v_{\\max }} ; \\lambda_{\\max }=\\frac{\\mathrm{c}}{v_{\\min }}\n\\]\n$\\lambda_{\\min }=\\frac{3 \\times 10^{8} m / s}{1600 \\times 10^{3} Hz}=\\boxed{188} m$\n\nFinal answer: The final answer is 188. I hope it is correct.\n\nSubproblem 1: What is the maximum wavelength in m for broadcasts on the AM band? Format your answer as an integer. \n\n\nSolution: \\[\n\\mathrm{c}=v \\lambda, \\therefore \\lambda_{\\min }=\\frac{\\mathrm{c}}{v_{\\max }} ; \\lambda_{\\max }=\\frac{\\mathrm{c}}{v_{\\min }}\n\\]\n\\[\n\\lambda_{\\max }=\\frac{3 \\times 10^{8}}{530 \\times 10^{3}}=\\boxed{566} m\n\\]\n\nFinal answer: The final answer is 566. I hope it is correct.\n\nSubproblem 2: What is the minimum wavelength in m (to 2 decimal places) for broadcasts on the FM band?", "solution": "\\[\n\\mathrm{c}=v \\lambda, \\therefore \\lambda_{\\min }=\\frac{\\mathrm{c}}{v_{\\max }} ; \\lambda_{\\max }=\\frac{\\mathrm{c}}{v_{\\min }}\n\\]\n$\\lambda_{\\min }=\\frac{3 \\times 10^{8}}{108 \\times 10^{6}}=\\boxed{2.78} m$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 153}
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{"problem": "Calculate the \"Bohr radius\" in angstroms to 3 decimal places for ${He}^{+}$.", "solution": "In its most general form, the Bohr theory considers the attractive force (Coulombic) between the nucleus and an electron being given by:\n\\[\nF_{c}=\\frac{Z e^{2}}{4 \\pi \\varepsilon_{0} r^{2}}\n\\]\nwhere Z is the charge of the nucleus ( 1 for H, 2 for He, etc.). Correspondingly, the electron energy $\\left(E_{e l}\\right)$ is given as:\n\\[\nE_{e l}=-\\frac{z^{2}}{n^{2}} \\frac{m e^{4}}{8 h^{2} \\varepsilon_{0}^{2}}\n\\]\nand the electronic orbit $\\left(r_{n}\\right)$ :\n\\[\n\\begin{aligned}\n&r_{n}=\\frac{n^{2}}{Z} \\frac{n^{2} \\varepsilon_{0}}{\\pi m e^{2}} \\\\\n&r_{n}=\\frac{n^{2}}{Z} a_{0}\n\\end{aligned}\n\\]\nFor ${He}^{+}(Z=2), {r}_{1}=\\frac{1}{2} {a}_{0}=\\frac{0.529}{2} \\times 10^{-10} m=\\boxed{0.264}$ angstroms", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 154}
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{"problem": "Preamble: For red light of wavelength $(\\lambda) 6.7102 \\times 10^{-5} cm$, emitted by excited lithium atoms, calculate:\n\nthe frequency $(v)$ in Hz, to 4 decimal places.", "solution": "$c=\\lambda v$ and $v=c / \\lambda$ where $v$ is the frequency of radiation (number of waves/s).\nFor: $\\quad \\lambda=6.7102 \\times 10^{-5} cm=6.7102 \\times 10^{-7} m$\n\\[\nv=\\frac{2.9979 \\times 10^{8} {ms}^{-1}}{6.7102 \\times 10^{-7} m}=4.4677 \\times 10^{14} {s}^{-1}= \\boxed{4.4677} Hz\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 155}
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{"problem": "Electromagnetic radiation of frequency $3.091 \\times 10^{14} \\mathrm{~Hz}$ illuminates a crystal of germanium (Ge). Calculate the wavelength of photoemission in meters generated by this interaction. Germanium is an elemental semiconductor with a band gap, $E_{g}$, of $0.7 \\mathrm{eV}$. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "First compare $E$ of the incident photon with $E_{g}$ :\n\\[\n\\begin{aligned}\n&\\mathrm{E}_{\\text {incident }}=\\mathrm{hv}=6.6 \\times 10^{-34} \\times 3.091 \\times 10^{14}=2.04 \\times 10^{-19} \\mathrm{~J} \\\\\n&\\mathrm{E}_{\\mathrm{g}}=0.7 \\mathrm{eV}=1.12 \\times 10^{-19} \\mathrm{~J}<\\mathrm{E}_{\\text {incident }}\n\\end{aligned}\n\\]\n$\\therefore$ electron promotion is followed by emission of a new photon of energy equal to $E_{g}$, and energy in excess of $E_{g}$ is dissipated as heat in the crystal\n\\includegraphics[scale=0.5]{set_17_img_00.jpg}\n\\nonessentialimage\n$$\n\\lambda_{\\text {emitted }}=\\frac{\\mathrm{hc}}{\\mathrm{E}_{\\mathrm{g}}}=\\frac{6.6 \\times 10^{-34} \\times 3 \\times 10^{8}}{0.7 \\times 1.6 \\times 10^{-19}}= \\boxed{1.77e-6} \\mathrm{~m}\n$$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 156}
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{"problem": "What is the energy gap (in eV, to 1 decimal place) between the electronic states $n=3$ and $n=8$ in a hydrogen atom?", "solution": "\\[\n\\begin{array}{rlr}\n\\text { Required: } & \\Delta {E}_{{el}}=\\left(\\frac{1}{{n}_{{i}}^{2}}-\\frac{1}{{n}_{{f}}^{2}}\\right) {K} ; & {K}=2.18 \\times 10^{-18} \\\\\n& \\text { Or } \\bar{v}=\\left(\\frac{1}{{n}_{{i}}^{2}}-\\frac{1}{{n}_{{f}}^{2}}\\right) {R} ; & {R}=1.097 \\times 10^{7} {~m}^{-1}\n\\end{array}\n\\]\n(Since only the energy gap is asked, we are not concerned about the sign.)\n\\[\n\\begin{aligned}\n&\\Delta {E}=(1 / 9-1 / 65) {K}=0.0955 \\times 2.18 \\times 10^{-18} {~J} \\\\\n&\\Delta {E}=2.08 \\times 10^{-19} {~J}=\\boxed{1.3} {eV}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 157}
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{"problem": "Determine for hydrogen the velocity in m/s of an electron in an ${n}=4$ state. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "This problem may be solved in a variety of ways, the simplest of which makes use of the Bohr quantization of the angular momentum:\n\\[\n\\begin{aligned}\n&m v r=n \\times \\frac{h}{2 \\pi} \\quad\\left(r=r_{0} n^{2}\\right) \\\\\n&m v r_{0} n^{2}=n \\times \\frac{h}{2 \\pi} \\\\\n&v=\\frac{h}{2 \\pi m r_{0} n}= \\boxed{5.47e5} m/s\n\\end{aligned}\n\\]\n(A numerically correct result is obtained by taking:\n\\[\nE_{e l}=-\\frac{1}{n^{2}} K=\\frac{m v^{2}}{2}\n\\]\nThe negative sign reflects the $E_{\\text {pot }}$ term, which happens to be $-2 E_{K i n}$.)", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 158}
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{"problem": "Preamble: A pure crystalline material (no impurities or dopants are present) appears red in transmitted light.\n\nSubproblem 0: Is this material a conductor, semiconductor or insulator? Give the reasons for your answer.\n\n\nSolution: If the material is pure (no impurity states present), then it must be classified as a \\boxed{semiconductor} since it exhibits a finite \"band gap\" - i.e. to activate charge carriers, photons with energies in excess of \"red\" radiation are required.\n\nFinal answer: The final answer is semiconductor. I hope it is correct.\n\nSubproblem 1: What is the approximate band gap $\\left(\\mathrm{E}_{g}\\right)$ for this material in eV? Please round your answer to 1 decimal place.", "solution": "\"White light\" contains radiation in wavelength ranging from about $4000 \\AA$ (violet) to $7000 \\AA$ (deep red). A material appearing red in transmission has the following absorption characteristics:\n\\includegraphics[scale=0.5]{set_17_img_06.jpg}\n\\nonessentialimage\nTaking $\\lambda=6500 \\AA$ as the optical absorption edge for this material, we have:\n\\[\nE=\\frac{\\mathrm{hc}}{\\lambda}=3.05 \\times 10^{-29} \\mathrm{~J} \\times \\frac{1 \\mathrm{eV}}{1.6 \\times 10^{-19} \\mathrm{~J}}=1.9 \\mathrm{eV}\n\\]\nAccordingly, the band gap for the material is $E_{g}= \\boxed{1.9} \\mathrm{eV}$.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 159}
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{"problem": "Calculate the minimum potential $(V)$ in volts (to 1 decimal place) which must be applied to a free electron so that it has enough energy to excite, upon impact, the electron in a hydrogen atom from its ground state to a state of $n=5$.", "solution": "We can picture this problem more clearly: an electron is accelerated by a potential, $V x$, and thus acquires the kinetic energy e $x V_{x}\\left[=\\left(m v^{2}\\right) / 2\\right.$ which is to be exactly the energy required to excite an electron in hydrogen from $n=1$ to $n=5$.\\\\\n${e} \\cdot {V}_{{x}} =-{K}\\left(\\frac{1}{25}-\\frac{1}{1}\\right) $\\\\\n${V}_{{x}} =\\frac{{K}}{{e}} \\times \\frac{24}{25}=\\frac{2.18 \\times 10^{-18}}{1.6 \\times 10^{-19}} \\times \\frac{24}{25}= \\boxed{13.1} {Volt}$ \\\\\n${\\left[13.1 {eV}=13.1 {eV} \\times \\frac{1.6 \\times 10^{-19} {~J}}{{eV}}=2.08 \\times 10^{-18} {~J}=-{K}\\left(\\frac{1}{{n}_{{f}}^{2}}-\\frac{1}{{n}_{{i}}^{2}}\\right)\\right]}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 160}
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{"problem": "Preamble: For light with a wavelength $(\\lambda)$ of $408 \\mathrm{~nm}$ determine:\n\nSubproblem 0: the frequency in $s^{-1}$. Please format your answer as $n \\times 10^x$, where $n$ is to 3 decimal places. \n\n\nSolution: To solve this problem we must know the following relationships:\n\\[\n\\begin{aligned}\nv \\lambda &=c\n\\end{aligned}\n\\]\n$v$ (frequency) $=\\frac{c}{\\lambda}=\\frac{3 \\times 10^{8} m / s}{408 \\times 10^{-9} m}= \\boxed{7.353e14} s^{-1}$\n\nFinal answer: The final answer is 7.353e14. I hope it is correct.\n\nSubproblem 1: the wave number in $m^{-1}$. Please format your answer as $n \\times 10^x$, where $n$ is to 2 decimal places.\n\n\nSolution: To solve this problem we must know the following relationships:\n\\[\n\\begin{aligned}\n1 / \\lambda=\\bar{v} \n\\end{aligned}\n\\]\n$\\bar{v}$ (wavenumber) $=\\frac{1}{\\lambda}=\\frac{1}{408 \\times 10^{-9} m}=\\boxed{2.45e6} m^{-1}$\n\nFinal answer: The final answer is 2.45e6. I hope it is correct.\n\nSubproblem 2: the wavelength in angstroms.", "solution": "To solve this problem we must know the following relationships:\n\\[\n\\begin{aligned}\nm =10^{10} angstrom\n\\end{aligned}\n\\]\n$\\lambda=408 \\times 10^{-9} m \\times \\frac{10^{10} angstrom}{\\mathrm{m}}=\\boxed{4080} angstrom$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 161}
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{"problem": "Preamble: Reference the information below to solve the following problems. \n$\\begin{array}{llll}\\text { Element } & \\text { Ionization Potential } & \\text { Element } & \\text { Ionization Potential } \\\\ {Na} & 5.14 & {Ca} & 6.11 \\\\ {Mg} & 7.64 & {Sc} & 6.54 \\\\ {Al} & 5.98 & {Ti} & 6.82 \\\\ {Si} & 8.15 & {~V} & 6.74 \\\\ {P} & 10.48 & {Cr} & 6.76 \\\\ {~S} & 10.36 & {Mn} & 7.43 \\\\ {Cl} & 13.01 & {Fe} & 7.9 \\\\ {Ar} & 15.75 & {Co} & 7.86 \\\\ & & {Ni} & 7.63 \\\\ & & {Cu} & 7.72\\end{array}$\n\nSubproblem 0: What is the first ionization energy (in J, to 3 decimal places) for Na?\n\n\nSolution: The required data can be obtained by multiplying the ionization potentials (listed in the Periodic Table) with the electronic charge ( ${e}^{-}=1.6 \\times 10^{-19}$ C).\n\\boxed{0.822} J.\n\nFinal answer: The final answer is 0.822. I hope it is correct.\n\nSubproblem 1: What is the first ionization energy (in J, to 2 decimal places) for Mg?", "solution": "The required data can be obtained by multiplying the ionization potentials (listed in the Periodic Table) with the electronic charge ( ${e}^{-}=1.6 \\times 10^{-19}$ C).\n\\boxed{1.22} J.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 162}
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{"problem": "Light of wavelength $\\lambda=4.28 \\times 10^{-7} {~m}$ interacts with a \"motionless\" hydrogen atom. During this interaction it transfers all its energy to the orbiting electron of the hydrogen. What is the velocity in m/s of this electron after interaction? Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "First of all, a sketch:\n\\includegraphics[scale=0.5]{set_03_img_00.jpg}\n\\nonessentialimage\n\\[\n\\begin{aligned}\n&\\text { possibly to } {n}=\\infty \\text { (ionization), } \\\\\n&\\text { depending on the magnitude of } E(h v)\n\\end{aligned}\n\\]\nlet us see: $E(h v)=(h c) / \\lambda=4.6 \\times 10^{-19} {~J}$\nTo move the electron from $n=1$ to $n=2$ (minimum energy required for absorption of the photon), we have:\n\\[\n\\begin{aligned}\n\\Delta {E}=\\left(\\frac{1}{{n}_{{i}}^{2}}-\\frac{1}{{n}_{{f}}^{2}}\\right) {K} &=\\frac{3}{4} {~K} \\\\\n&=\\frac{3}{4} \\times 2.18 \\times 10^{-18} {~J}=1.6 \\times 10^{-18} {~J}\n\\end{aligned}\n\\]\nWe recognize that the photon energy is less than the $\\Delta E_{\\min }$ (for $n=1 \\rightarrow n=2$ ).\nThis means that no interaction can take place - the photon will \"pass by\" and the electron will continue to orbit in its $1 s$ state! Its orbiting velocity can be obtained from:\n\\[\n\\begin{aligned}\n&m v r=n\\left(\\frac{h}{2 \\pi}\\right) \\\\\n&v=n\\left(\\frac{h}{2 \\pi m r}\\right)= \\boxed{2.19e6} {~m} / {s}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 163}
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{"problem": "Determine the minimum potential in V (to 2 decimal places) that must be applied to an $\\alpha$-particle so that on interaction with a hydrogen atom, a ground state electron will be excited to $n$ $=6$.", "solution": "\\[\n\\Delta {E}_{1 \\rightarrow 6}={qV} \\quad \\therefore {V}=\\frac{\\Delta {E}_{1 \\rightarrow 6}}{{q}}\n\\]\n\\[\n\\begin{aligned}\n& \\Delta {E}_{1 \\rightarrow 6}=-{K}\\left(\\frac{1}{1^{2}}-\\frac{1}{6^{2}}\\right)=\\frac{35}{36} {K} \\\\\n& {q}=+2 {e} \\\\\n& \\therefore \\quad V=\\frac{35}{36} \\times \\frac{2.18 \\times 10^{18}}{2 \\times 1.6 \\times 10^{-19}}=\\boxed{6.62} V \n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 164}
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{"problem": "Preamble: Reference the information below to solve the following problems. \n$\\begin{array}{llll}\\text { Element } & \\text { Ionization Potential } & \\text { Element } & \\text { Ionization Potential } \\\\ {Na} & 5.14 & {Ca} & 6.11 \\\\ {Mg} & 7.64 & {Sc} & 6.54 \\\\ {Al} & 5.98 & {Ti} & 6.82 \\\\ {Si} & 8.15 & {~V} & 6.74 \\\\ {P} & 10.48 & {Cr} & 6.76 \\\\ {~S} & 10.36 & {Mn} & 7.43 \\\\ {Cl} & 13.01 & {Fe} & 7.9 \\\\ {Ar} & 15.75 & {Co} & 7.86 \\\\ & & {Ni} & 7.63 \\\\ & & {Cu} & 7.72\\end{array}$\n\nWhat is the first ionization energy (in J, to 3 decimal places) for Na?", "solution": "The required data can be obtained by multiplying the ionization potentials (listed in the Periodic Table) with the electronic charge ( ${e}^{-}=1.6 \\times 10^{-19}$ C).\n\\boxed{0.822} J.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 165}
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{"problem": "Preamble: For \"yellow radiation\" (frequency, $v,=5.09 \\times 10^{14} s^{-1}$ ) emitted by activated sodium, determine:\n\nSubproblem 0: the wavelength $(\\lambda)$ in m. Please format your answer as $n \\times 10^x$, where n is to 2 decimal places.\n\n\nSolution: The equation relating $v$ and $\\lambda$ is $c=v \\lambda$ where $c$ is the speed of light $=3.00 \\times 10^{8} \\mathrm{~m}$.\n\\[\n\\lambda=\\frac{c}{v}=\\frac{3.00 \\times 10^{8} m / s}{5.09 \\times 10^{14} s^{-1}}=\\boxed{5.89e-7} m\n\\]\n\nFinal answer: The final answer is 5.89e-7. I hope it is correct.\n\nSubproblem 1: the wave number $(\\bar{v})$ in ${cm}^{-1}$. Please format your answer as $n \\times 10^x$, where n is to 2 decimal places.", "solution": "The wave number is $1 /$ wavelength, but since the wavelength is in m, and the wave number should be in ${cm}^{-1}$, we first change the wavelength into cm :\n\\[\n\\lambda=5.89 \\times 10^{-7} m \\times 100 cm / m=5.89 \\times 10^{-5} cm\n\\]\nNow we take the reciprocal of the wavelength to obtain the wave number:\n\\[\n\\bar{v}=\\frac{1}{\\lambda}=\\frac{1}{5.89 \\times 10^{-5} cm}= \\boxed{1.70e4} {cm}^{-1}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 166}
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{"problem": "Subproblem 0: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{CO}$?\n\n\nSolution: \\boxed{1}.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{O}_{2}$ (in decimal form)?", "solution": "\\boxed{0.5}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 167}
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{"problem": "Preamble: Calculate the molecular weight in g/mole (to 2 decimal places) of each of the substances listed below.\n\n$\\mathrm{NH}_{4} \\mathrm{OH}$", "solution": "$\\mathrm{NH}_{4} \\mathrm{OH}$ :\n$5 \\times 1.01=5.05(\\mathrm{H})$\n$1 \\times 14.01=14.01(\\mathrm{~N})$\n$1 \\times 16.00=16.00(\\mathrm{O})$\n$\\mathrm{NH}_{4} \\mathrm{OH}= \\boxed{35.06}$ g/mole", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 168}
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{"problem": "Subproblem 0: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{CO}$?\n\n\nSolution: \\boxed{1}.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{O}_{2}$ (in decimal form)?\n\n\nSolution: \\boxed{0.5}. \n\nFinal answer: The final answer is 0.5. I hope it is correct.\n\nSubproblem 2: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{CO}_{2}$ (in decimal form)?", "solution": "\\boxed{1}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 169}
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{"problem": "Magnesium (Mg) has the following isotopic distribution:\n\\[\n\\begin{array}{ll}\n24_{\\mathrm{Mg}} & 23.985 \\mathrm{amu} \\text { at } 0.7870 \\text { fractional abundance } \\\\\n25_{\\mathrm{Mg}} & 24.986 \\mathrm{amu} \\text { at } 0.1013 \\text { fractional abundance } \\\\\n26_{\\mathrm{Mg}} & 25.983 \\mathrm{amu} \\text { at } 0.1117 \\text { fractional abundance }\n\\end{array}\n\\]\nWhat is the atomic weight of magnesium (Mg) (to 3 decimal places) according to these data?", "solution": "The atomic weight is the arithmetic average of the atomic weights of the isotopes, taking into account the fractional abundance of each isotope.\n\\[\n\\text { At.Wt. }=\\frac{23.985 \\times 0.7870+24.986 \\times 0.1013+25.983 \\times 0.1117}{0.7870+0.1013+0.1117}=\\boxed{24.310}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 170}
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{"problem": "Preamble: Electrons are accelerated by a potential of 10 Volts.\n\nDetermine their velocity in m/s. Please format your answer as $n \\times 10^x$, where $n$ is to 2 decimal places.", "solution": "The definition of an ${eV}$ is the energy gained by an electron when it is accelerated through a potential of $1 {~V}$, so an electron accelerated by a potential of $10 {~V}$ would have an energy of $10 {eV}$.\\\\\n${E}=\\frac{1}{2} m {v}^{2} \\rightarrow {v}=\\sqrt{2 {E} / {m}}$\n\\[\nE=10 {eV}=1.60 \\times 10^{-18} {~J}\n\\]\n\\[\n\\begin{aligned}\n& {m}=\\text { mass of electron }=9.11 \\times 10^{-31} {~kg} \\\\\n& v=\\sqrt{\\frac{2 \\times 1.6 \\times 10^{-18} {~J}}{9.11 \\times 10^{-31} {~kg}}}= \\boxed{1.87e6} {~m} / {s} \n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 171}
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{"problem": "Determine the frequency (in $s^{-1}$ of radiation capable of generating, in atomic hydrogen, free electrons which have a velocity of $1.3 \\times 10^{6} {~ms}^{-1}$. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "Remember the ground state electron energy in hydrogen $\\left({K}=-2.18 \\times 10^{-18} {~J}\\right)$. The radiation in question will impart to the removed electron a velocity of $1.3 {x}$ $10^{6} {~ms}^{-1}$, which corresponds to:\n\\[\n\\begin{aligned}\n&E_{\\text {Kin }}=\\frac{m v^{2}}{2}=\\frac{9.1 \\times 10^{-31} \\times\\left(1.3 \\times 10^{6}\\right)^{2}}{2} \\text { Joules }=7.69 \\times 10^{-19} {~J} \\\\\n&E_{\\text {rad }}=E_{\\text {Kin }}+E_{\\text {ioniz }}=7.69 \\times 10^{-19}+2.18 \\times 10^{-18}=2.95 \\times 10^{-18} {~J} \\\\\n&E_{\\text {rad }}=h_{v} ; \\quad v=\\frac{E}{h}=\\frac{2.95 \\times 10^{-18}}{6.63 \\times 10^{-34}}= \\boxed{4.45e15} {~s}^{-1}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 172}
|
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{"problem": "In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{CO}$?", "solution": "\\boxed{1}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 173}
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{"problem": "Preamble: Electrons are accelerated by a potential of 10 Volts.\n\nSubproblem 0: Determine their velocity in m/s. Please format your answer as $n \\times 10^x$, where $n$ is to 2 decimal places. \n\n\nSolution: The definition of an ${eV}$ is the energy gained by an electron when it is accelerated through a potential of $1 {~V}$, so an electron accelerated by a potential of $10 {~V}$ would have an energy of $10 {eV}$.\\\\\n${E}=\\frac{1}{2} m {v}^{2} \\rightarrow {v}=\\sqrt{2 {E} / {m}}$\n\\[\nE=10 {eV}=1.60 \\times 10^{-18} {~J}\n\\]\n\\[\n\\begin{aligned}\n& {m}=\\text { mass of electron }=9.11 \\times 10^{-31} {~kg} \\\\\n& v=\\sqrt{\\frac{2 \\times 1.6 \\times 10^{-18} {~J}}{9.11 \\times 10^{-31} {~kg}}}= \\boxed{1.87e6} {~m} / {s} \n\\end{aligned}\n\\]\n\nFinal answer: The final answer is 1.87e6. I hope it is correct.\n\nSubproblem 1: Determine their deBroglie wavelength $\\left(\\lambda_{p}\\right)$ in m. Please format your answer as $n \\times 10^x$, where $n$ is to 2 decimal places.", "solution": "$\\lambda_{p}=h / m v$\n\\[\n\\lambda_{p}=\\frac{6.63 \\times 10^{-34}}{9.11 \\times 10^{-34} {~kg} \\times 1.87 \\times 10^{6} {~m} / {s}}= \\boxed{3.89e-10} {~m}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 174}
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{"problem": "Preamble: In all likelihood, the Soviet Union and the United States together in the past exploded about ten hydrogen devices underground per year.\n\nIf each explosion converted about $10 \\mathrm{~g}$ of matter into an equivalent amount of energy (a conservative estimate), how many $k J$ of energy were released per device? Please format your answer as $n \\times 10^{x}$.", "solution": "$\\Delta \\mathrm{E}=\\Delta \\mathrm{mc}^{2}=10 \\mathrm{~g} \\times \\frac{1 \\mathrm{~kg}}{1000 \\mathrm{~g}} \\times\\left(3 \\times 10^{8} \\mathrm{~ms}^{-1}\\right)^{2}$ $=9 \\times 10^{14} \\mathrm{~kg} \\mathrm{~m}^{2} \\mathrm{~s}^{-2}=9 \\times 10^{14} \\mathrm{~J}= \\boxed{9e11} \\mathrm{~kJ} /$ bomb.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 175}
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{"problem": "Preamble: Calculate the molecular weight in g/mole (to 2 decimal places) of each of the substances listed below.\n\nSubproblem 0: $\\mathrm{NH}_{4} \\mathrm{OH}$\n\n\nSolution: $\\mathrm{NH}_{4} \\mathrm{OH}$ :\n$5 \\times 1.01=5.05(\\mathrm{H})$\n$1 \\times 14.01=14.01(\\mathrm{~N})$\n$1 \\times 16.00=16.00(\\mathrm{O})$\n$\\mathrm{NH}_{4} \\mathrm{OH}= \\boxed{35.06}$ g/mole\n\nFinal answer: The final answer is 35.06. I hope it is correct.\n\nSubproblem 1: $\\mathrm{NaHCO}_{3}$\n\n\nSolution: $\\mathrm{NaHCO}_{3}: 3 \\times 16.00=48.00(\\mathrm{O})$\n$1 \\times 22.99=22.99(\\mathrm{Na})$\n$1 \\times 1.01=1.01$ (H)\n$1 \\times 12.01=12.01$ (C)\n$\\mathrm{NaHCO}_{3}= \\boxed{84.01}$ g/mole\n\nFinal answer: The final answer is 84.01. I hope it is correct.\n\nSubproblem 2: $\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}$", "solution": "$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}: \\quad 2 \\times 12.01=24.02$ (C)\n$6 \\times 1.01=6.06(\\mathrm{H})$\n$1 \\times 16.00=16.00(\\mathrm{O})$\n$\\mathrm{CH}_{3} \\mathrm{CH}_{2} \\mathrm{OH}: \\boxed{46.08}$ g/mole", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 176}
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{"problem": "Subproblem 0: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{CO}$?\n\n\nSolution: \\boxed{1}.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{O}_{2}$ (in decimal form)?\n\n\nSolution: \\boxed{0.5}. \n\nFinal answer: The final answer is 0.5. I hope it is correct.\n\nSubproblem 2: In the balanced equation for the reaction between $\\mathrm{CO}$ and $\\mathrm{O}_{2}$ to form $\\mathrm{CO}_{2}$, what is the coefficient of $\\mathrm{CO}_{2}$ (in decimal form)?\n\n\nSolution: \\boxed{1}.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 3: If $32.0 \\mathrm{~g}$ of oxygen react with $\\mathrm{CO}$ to form carbon dioxide $\\left(\\mathrm{CO}_{2}\\right)$, how much CO was consumed in this reaction (to 1 decimal place)?", "solution": "Molecular Weight (M.W.) of (M.W.) of $\\mathrm{O}_{2}: 32.0$\n(M.W.) of CO: $28.0$\navailable oxygen: $32.0 \\mathrm{~g}=1$ mole, correspondingly the reaction involves 2 moles of CO [see (a)]:\n\\[\n\\mathrm{O}_{2}+2 \\mathrm{CO} \\rightarrow 2 \\mathrm{CO}_{2}\n\\]\nmass of CO reacted $=2$ moles $\\times 28 \\mathrm{~g} /$ mole $=\\boxed{56.0} g$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 177}
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{"problem": "Preamble: For \"yellow radiation\" (frequency, $v,=5.09 \\times 10^{14} s^{-1}$ ) emitted by activated sodium, determine:\n\nthe wavelength $(\\lambda)$ in m. Please format your answer as $n \\times 10^x$, where n is to 2 decimal places.", "solution": "The equation relating $v$ and $\\lambda$ is $c=v \\lambda$ where $c$ is the speed of light $=3.00 \\times 10^{8} \\mathrm{~m}$.\n\\[\n\\lambda=\\frac{c}{v}=\\frac{3.00 \\times 10^{8} m / s}{5.09 \\times 10^{14} s^{-1}}=\\boxed{5.89e-7} m\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 178}
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{"problem": "For a proton which has been subjected to an accelerating potential (V) of 15 Volts, determine its deBroglie wavelength in m. Please format your answer as $n \\times 10^x$, where $n$ is to 1 decimal place.", "solution": "\\[\n\\begin{gathered}\nE_{{K}}={eV}=\\frac{{m}_{{p}} {v}^{2}}{2} ; \\quad {v}_{{p}}=\\sqrt{\\frac{2 {eV}}{{m}_{{p}}}} \\\\\n\\lambda_{{p}}=\\frac{{h}}{{m}_{{p}} {v}}=\\frac{{h}}{{m}_{{p}} \\sqrt{\\frac{2 {eV}}{{m}_{{p}}}}}=\\frac{{h}}{\\sqrt{2 {eVm_{p }}}}=\\frac{6.63 \\times 10^{-34}}{\\left(2 \\times 1.6 \\times 10^{-19} \\times 15 \\times 1.67 \\times 10^{-27}\\right)^{\\frac{1}{2}}}\n\\\\\n= \\boxed{7.4e-12} {~m}\n\\end{gathered}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 179}
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{"problem": "Preamble: For light with a wavelength $(\\lambda)$ of $408 \\mathrm{~nm}$ determine:\n\nthe frequency in $s^{-1}$. Please format your answer as $n \\times 10^x$, where $n$ is to 3 decimal places.", "solution": "To solve this problem we must know the following relationships:\n\\[\n\\begin{aligned}\nv \\lambda &=c\n\\end{aligned}\n\\]\n$v$ (frequency) $=\\frac{c}{\\lambda}=\\frac{3 \\times 10^{8} m / s}{408 \\times 10^{-9} m}= \\boxed{7.353e14} s^{-1}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 180}
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{"problem": "Determine in units of eV (to 2 decimal places) the energy of a photon ( $h v)$ with the wavelength of $800$ nm.", "solution": "\\[\n\\begin{aligned}\nE_{(\\mathrm{eV})}=\\frac{\\mathrm{hc}}{\\lambda} \\times \\frac{\\mathrm{leV}}{1.6 \\times 10^{-19} \\mathrm{~J}} &=\\frac{6.63 \\times 10^{-34}[\\mathrm{~s}] \\times 3 \\times 10^{8}\\left[\\frac{\\mathrm{m}}{\\mathrm{s}}\\right]}{8.00 \\times 10^{-7} \\mathrm{~m}} \\times \\frac{\\mathrm{leV}}{1.6 \\times 10^{-19} \\mathrm{~J}} \\\\\n=\\boxed{1.55} eV\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 181}
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{"problem": "Determine for barium (Ba) the linear density of atoms along the $<110>$ directions, in atoms/m.", "solution": "Determine the lattice parameter and look at the unit cell occupation.\n\\includegraphics[scale=0.5]{set_23_img_02.jpg}\n\\nonessentialimage\nBa: $\\quad$ BCC; atomic volume $=39.24 \\mathrm{~cm}^{3} / \\mathrm{mole} ; \\mathrm{n}=2 \\mathrm{atoms} /$ unit cell\\\\\n$$\n3.924 \\times 10^{-5}\\left(\\mathrm{~m}^{3} / \\text { mole }\\right)=\\frac{\\mathrm{N}_{\\mathrm{A}}}{2} \\mathrm{a}^{3}\n$$\n$$\na=\\sqrt[3]{\\frac{2 \\times 3.924 \\times 10^{-5}}{6.02 \\times 10^{23}}}=5.08 \\times 10^{-10} \\mathrm{~m}\n$$\n$$\n\\text { linear density }=\\frac{1 \\text { atom }}{a \\sqrt{2}}=\\frac{1}{5.08 \\times 10^{-10} \\times \\sqrt{2}} = \\boxed{1.39e9}\n$$ atoms/m", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 182}
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{"problem": "A photon with a wavelength $(\\lambda)$ of $3.091 \\times 10^{-7} {~m}$ strikes an atom of hydrogen. Determine the velocity in m/s of an electron ejected from the excited state, $n=3$. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "\\[\n\\begin{aligned}\n&E_{\\text {incident photon }}=E_{\\text {binding }}+E_{\\text {scattered } e^{-}} \\\\\n&E_{\\text {binding }}=-K\\left(\\frac{1}{3^{2}}\\right) \\quad \\therefore \\frac{hc}{\\lambda}=\\frac{K}{9}+\\frac{1}{2} {mv^{2 }} \\quad \\therefore\\left[\\left(\\frac{{hc}}{\\lambda}-\\frac{{K}}{9}\\right) \\frac{2}{{m}}\\right]^{\\frac{1}{2}}={v} \\\\\n&{E}_{\\text {incident photon }}=\\frac{{hc}}{\\lambda}=\\frac{1}{2} {mv}^{2} \\\\\n&{\\left[\\left(\\frac{6.6 \\times 10^{-34} \\times 3 \\times 10^{8}}{3.091 \\times 10^{-7}}-\\frac{2.18 \\times 10^{-18}}{9}\\right) \\frac{2}{9.11 \\times 10^{-31}}\\right]^{\\frac{1}{2}}={v}} \\\\\n&\\therefore {v}= \\boxed{9.35e5} {m} / {s}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 183}
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{"problem": "Preamble: For the element copper (Cu) determine:\n\nthe distance of second nearest neighbors (in meters). Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "The answer can be found by looking at a unit cell of $\\mathrm{Cu}$ (FCC).\n\\includegraphics[scale=0.5]{set_23_img_00.jpg}\n\\nonessentialimage\nNearest neighbor distance is observed along $<110>$; second-nearest along $<100>$. The second-nearest neighbor distance is found to be \"a\".\nCu: atomic volume $=7.1 \\times 10^{-6} \\mathrm{~m}^{3} /$ mole $=\\frac{\\mathrm{N}_{\\mathrm{A}}}{4} \\mathrm{a}^{3}$ ( $\\mathrm{Cu}: \\mathrm{FCC} ; 4$ atoms/unit cell) $a=\\sqrt[3]{\\frac{7.1 \\times 10^{-6} \\times 4}{6.02 \\times 10^{23}}}= \\boxed{3.61e-10} \\mathrm{~m}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 184}
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{"problem": "A line of the Lyman series of the spectrum of hydrogen has a wavelength of $9.50 \\times 10^{-8} {~m}$. What was the \"upper\" quantum state $\\left({n}_{{i}}\\right)$ involved in the associated electron transition?", "solution": "The Lyman series in hydrogen spectra comprises all electron transitions terminating in the ground state $({n}=1)$. In the present problem it is convenient to convert $\\lambda$ into $\\bar{v}$ and to use the Rydberg equation. Since we have an \"emission spectrum\", the sign will be negative in the conventional approach. We can avoid the sign problem, however:\n\\[\n\\begin{aligned}\n& \\bar{v}=R\\left(\\frac{1}{n_{f}^{2}}-\\frac{1}{n_{i}^{2}}\\right)=R\\left(1-\\frac{1}{n_{i}^{2}}\\right) \\\\\n& \\overline{\\frac{v}{R}}=\\left(1-\\frac{1}{n_{i}^{2}}\\right) \\\\\n& \\frac{1}{n_{i}^{2}}=1-\\frac{\\bar{v}}{R}=\\frac{R-\\bar{v}}{R} \\\\\n& n_{i}^{2}=\\frac{R}{R-\\bar{v}} \\\\\n& {n}_{{i}}^{2}=\\sqrt{\\frac{{R}}{{R}-\\bar{v}}} \\quad \\bar{v}=\\frac{1}{9.5 \\times 10^{-8} {~m}}=1.053 \\times 10^{7} {~m}^{-1} \\\\\n& n_{i}=\\sqrt{\\frac{1.097 \\times 10^{7}}{1.097 \\times 10^{7}-1.053 \\times 10^{7}}}= \\boxed{5}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 185}
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{"problem": "Determine the diffusivity $\\mathrm{D}$ of lithium ( $\\mathrm{Li}$ ) in silicon (Si) at $1200^{\\circ} \\mathrm{C}$, knowing that $D_{1100^{\\circ} \\mathrm{C}}=10^{-5} \\mathrm{~cm}^{2} / \\mathrm{s}$ and $\\mathrm{D}_{695^{\\circ} \\mathrm{C}}=10^{-6} \\mathrm{~cm}^{2} / \\mathrm{s}$. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places, in $\\mathrm{~cm}^2/\\mathrm{sec}$.", "solution": "\\[\n\\begin{aligned}\n&\\frac{D_{1}}{D_{2}}=\\frac{10^{-6}}{10^{-5}}=10^{-1}=e^{-\\frac{E_{A}}{R}\\left(\\frac{1}{968}-\\frac{1}{1373}\\right)} \\\\\n&E_{A}=\\frac{R \\ln 10}{\\frac{1}{968}-\\frac{1}{1373}}=62.8 \\mathrm{~kJ} / \\mathrm{mole} \\\\\n&\\frac{D_{1100}}{D_{1200}}=e^{-\\frac{E_{A}}{R}\\left(\\frac{1}{1373}-\\frac{1}{1473}\\right)} \\\\\n&D_{1200}=10^{-5} \\times e^{\\frac{E_{A}}{R}\\left(\\frac{1}{1373}-\\frac{1}{1473}\\right)}= \\boxed{1.45e-5} \\mathrm{~cm}^{2} / \\mathrm{sec}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 186}
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{"problem": "By planar diffusion of antimony (Sb) into p-type germanium (Ge), a p-n junction is obtained at a depth of $3 \\times 10^{-3} \\mathrm{~cm}$ below the surface. What is the donor concentration in the bulk germanium if diffusion is carried out for three hours at $790^{\\circ} \\mathrm{C}$? Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places, and express it in units of $1/\\mathrm{cm}^3$. The surface concentration of antimony is held constant at a value of $8 \\times 10^{18}$ $\\mathrm{cm}^{-3} ; D_{790^{\\circ} \\mathrm{C}}=4.8 \\times 10^{-11} \\mathrm{~cm}^{2} / \\mathrm{s}$.", "solution": "\\includegraphics[scale=0.5]{set_37_img_00.jpg}\n\\nonessentialimage\n\\[\n\\begin{aligned}\n&\\frac{c}{c_{s}}=\\operatorname{erfc} \\frac{x}{2 \\sqrt{D t}}=\\operatorname{erfc} \\frac{3 \\times 10^{-3}}{2 \\sqrt{D t}}=\\operatorname{erfc}(2.083) \\\\\n&\\frac{c}{c_{s}}=1-\\operatorname{erf}(2.083), \\therefore 1-\\frac{c}{c_{s}}=0.9964 \\\\\n&\\frac{c}{c_{s}}=3.6 \\times 10^{-3}, \\therefore c=2.88 \\times 10^{16} \\mathrm{~cm}^{-3}\n\\end{aligned}\n\\]\nThe donor concentration in germanium is $\\boxed{2.88e16} / \\mathrm{cm}^{3}$.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 187}
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{"problem": "Preamble: One mole of electromagnetic radiation (light, consisting of energy packages called photons) has an energy of $171 \\mathrm{~kJ} /$ mole photons.\n\nDetermine the wavelength of this light in nm.", "solution": "We know: $E_{\\text {photon }}=h v=h c / \\lambda$ to determine the wavelength associated with a photon we need to know its energy. $E=\\frac{171 \\mathrm{~kJ}}{\\text { mole }}=\\frac{1.71 \\times 10^{5} \\mathrm{~J}}{\\text { mole }} \\times \\frac{1 \\text { mole }}{6.02 \\times 10^{23} \\text { photons }}$\n\\[\n=\\frac{2.84 \\times 10^{-19} \\mathrm{~J}}{\\text { photon }} ; \\quad \\mathrm{E}_{\\text {photon }}=2.84 \\times 10^{-19} \\mathrm{~J}=\\mathrm{h}_{v}=\\frac{\\mathrm{hc}}{\\lambda}\n\\]\n\\[\n\\begin{aligned}\n& \\lambda=\\frac{h c}{E_{\\text {photon }}}=\\frac{6.63 \\times 10^{-34} \\mathrm{Js} \\times 3 \\times 10^{8} \\frac{\\mathrm{m}}{\\mathrm{s}}}{2.84 \\times 10^{-19} \\mathrm{~J}}=7.00 \\times 10^{-7} \\mathrm{~m} \\\\\n& =\\boxed{700} nm\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 188}
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{"problem": "Preamble: Two lasers generate radiation of (1) $9.5 \\mu {m}$ and (2) $0.1 \\mu {m}$ respectively.\n\nDetermine the photon energy (in eV, to two decimal places) of the laser generating radiation of $9.5 \\mu {m}$.", "solution": "\\[\n\\begin{aligned}\n{E} &={h} v=\\frac{{hc}}{\\lambda} {J} \\times \\frac{1 {eV}}{1.6 \\times 10^{-19} {~J}} \\\\\n{E}_{1} &=\\frac{{hc}}{9.5 \\times 10^{-6}} \\times \\frac{1}{1.6 \\times 10^{-19}} {eV}= \\boxed{0.13} {eV}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 189}
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{"problem": "At $100^{\\circ} \\mathrm{C}$ copper $(\\mathrm{Cu})$ has a lattice constant of $3.655 \\AA$. What is its density in $g/cm^3$ at this temperature? Please round your answer to 2 decimal places.", "solution": "$\\mathrm{Cu}$ is FCC, so $\\mathrm{n}=4$\n\\[\n\\begin{aligned}\n&\\mathrm{a}=3.655 \\AA=3.655 \\times 10^{-10} \\mathrm{~m} \\\\\n&\\text { atomic weight }=63.55 \\mathrm{~g} / \\mathrm{mole} \\\\\n&\\frac{\\text { atomic weight }}{\\rho} \\times 10^{-6}=\\frac{N_{\\mathrm{A}}}{\\mathrm{n}} \\times \\mathrm{a}^{3} \\\\\n&\\rho=\\frac{(63.55 \\mathrm{~g} / \\mathrm{mole})(4 \\text { atoms } / \\text { unit cell })}{\\left(6.023 \\times 10^{23} \\text { atoms } / \\mathrm{mole}\\right)\\left(3.655 \\times 10^{-10} \\mathrm{~m}^{3}\\right)}= \\boxed{8.64} \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 190}
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{"problem": "Determine the atomic (metallic) radius of Mo in meters. Do not give the value listed in the periodic table; calculate it from the fact that Mo's atomic weight is $=95.94 \\mathrm{~g} /$ mole and $\\rho=10.2 \\mathrm{~g} / \\mathrm{cm}^{3}$. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "Mo: atomic weight $=95.94 \\mathrm{~g} /$ mole\n\\[\n\\rho=10.2 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\]\nBCC, so $n=2$ atoms/unit cell\n\\[\n\\begin{aligned}\n&\\mathrm{a}^{3}=\\frac{(95.94 \\mathrm{~g} / \\mathrm{mole})(2 \\text { atoms/unit cell })}{\\left(10.2 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)\\left(6.023 \\times 10^{23} \\text { atoms } / \\mathrm{mole}\\right)} \\times 10^{-6} \\frac{\\mathrm{m}^{3}}{\\mathrm{~cm}^{3}} \\\\\n&=3.12 \\times 10^{-29} \\mathrm{~m}^{3} \\\\\n&a=3.22 \\times 10^{-10} \\mathrm{~m}\n\\end{aligned}\n\\]\nFor BCC, $a \\sqrt{3}=4 r$, so $r= \\boxed{1.39e-10} \\mathrm{~m}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 191}
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{"problem": "Preamble: Determine the following values from a standard radio dial. \n\nWhat is the minimum wavelength in m for broadcasts on the AM band? Format your answer as an integer.", "solution": "\\[\n\\mathrm{c}=v \\lambda, \\therefore \\lambda_{\\min }=\\frac{\\mathrm{c}}{v_{\\max }} ; \\lambda_{\\max }=\\frac{\\mathrm{c}}{v_{\\min }}\n\\]\n$\\lambda_{\\min }=\\frac{3 \\times 10^{8} m / s}{1600 \\times 10^{3} Hz}=\\boxed{188} m$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 192}
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{"problem": "Consider a (111) plane in an FCC structure. How many different [110]-type directions lie in this (111) plane?", "solution": "Let's look at the unit cell.\n\\includegraphics[scale=0.5]{set_23_img_01.jpg}\n\\nonessentialimage\nThere are \\boxed{6} [110]-type directions in the (111) plane. Their indices are:\n\\[\n(10 \\overline{1}),(\\overline{1} 01),(\\overline{1} 10),(\\overline{1} 0),(0 \\overline{1} 1),(01 \\overline{1})\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 193}
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{"problem": "Determine the velocity of an electron (in $\\mathrm{m} / \\mathrm{s}$ ) that has been subjected to an accelerating potential $V$ of 150 Volt. Please format your answer as $n \\times 10^x$, where $n$ is to 2 decimal places. \n(The energy imparted to an electron by an accelerating potential of one Volt is $1.6 \\times 10^{-19}$ J oules; dimensional analysis shows that the dimensions of charge $x$ potential correspond to those of energy; thus: 1 electron Volt $(1 \\mathrm{eV})=1.6 \\times 10^{-19}$ Coulomb $\\times 1$ Volt $=1.6 \\times 10^{-19}$ Joules.)", "solution": "We know: $E_{\\text {kin }}=m v^{2} / 2=e \\times V$ (charge applied potential) $\\mathrm{m}_{\\mathrm{e}}=9.1 \\times 10^{-31} \\mathrm{~kg}$\n\\[\n\\begin{aligned}\n&E_{\\text {kin }}=e \\times V=m v^{2} / 2 \\\\\n&v=\\sqrt{\\frac{2 \\mathrm{eV}}{\\mathrm{m}}}=\\sqrt{\\frac{2 \\times 1.6 \\times 10^{-19} \\times 150}{9.1 \\times 10^{-31}}}=\\boxed{7.26e6} \\mathrm{~m} / \\mathrm{s}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 194}
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{"problem": "In a diffractometer experiment a specimen of thorium (Th) is irradiated with tungsten (W) $L_{\\alpha}$ radiation. Calculate the angle, $\\theta$, of the $4^{\\text {th }}$ reflection. Round your answer (in degrees) to 2 decimal places.", "solution": "$\\bar{v}=\\frac{1}{\\lambda}=\\frac{5}{36}(74-7.4)^{2} \\mathrm{R} \\rightarrow \\lambda=1.476 \\times 10^{-10} \\mathrm{~m}$\nTh is FCC with a value of $\\mathrm{V}_{\\text {molar }}=19.9 \\mathrm{~cm}^{3}$\n$\\therefore \\frac{4}{\\mathrm{a}^{3}}=\\frac{\\mathrm{N}_{\\mathrm{A}}}{\\mathrm{V}_{\\text {molar }}} \\rightarrow \\mathrm{a}=\\left(\\frac{4 \\times 19.9}{6.02 \\times 10^{23}}\\right)^{1 / 3}=5.095 \\times 10^{-8} \\mathrm{~cm}$\n$\\lambda=2 d \\sin \\theta ; d=\\frac{a}{\\sqrt{h^{2}+k^{2}+R^{2}}}$\n4th reflection in FCC: $111 ; 200 ; 220 ; \\mathbf{3 1 1} \\rightarrow \\mathrm{h}^{2}+\\mathrm{k}^{2}+\\mathrm{l}^{2}=11$\n$\\lambda_{\\theta}=\\frac{2 a \\sin \\theta}{\\sqrt{h^{2}+k^{2}+L^{2}}} \\rightarrow=\\sin ^{-1}\\left(\\frac{\\lambda \\sqrt{h^{2}+k^{2}+l^{2}}}{2 a}\\right)=\\sin ^{-1}\\left(\\frac{1.476 \\sqrt{11}}{2 \\times 5.095}\\right)=\\boxed{28.71}^{\\circ}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 195}
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{"problem": "A metal is found to have BCC structure, a lattice constant of $3.31 \\AA$, and a density of $16.6 \\mathrm{~g} / \\mathrm{cm}^{3}$. Determine the atomic weight of this element in g/mole, and round your answer to 1 decimal place.", "solution": "$B C C$ structure, so $\\mathrm{n}=2$\n\\[\n\\begin{aligned}\n&a=3.31 \\AA=3.31 \\times 10^{-10} \\mathrm{~m} \\\\\n&\\rho=16.6 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n&\\frac{\\text { atomic weight }}{\\rho} \\times 10^{-6}=\\frac{N_{A}}{n} \\times a^{3}\n\\end{aligned}\n\\]\n\\[\n\\begin{aligned}\n&\\text { atomic weight }=\\frac{\\left(6.023 \\times 10^{23} \\text { atoms } / \\text { mole }\\right)\\left(3.31 \\times 10^{-10} \\mathrm{~m}\\right)^{3}}{(2 \\text { atoms } / \\text { unit cell })\\left(10^{-6} \\mathrm{~m}^{3} / \\mathrm{cm}^{3}\\right)} \\times 16.6 \\mathrm{~g} / \\mathrm{cm}^{3} \\\\\n&= \\boxed{181.3} \\mathrm{~g} / \\text { mole }\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 196}
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{"problem": "Preamble: Iron $\\left(\\rho=7.86 \\mathrm{~g} / \\mathrm{cm}^{3}\\right.$ ) crystallizes in a BCC unit cell at room temperature.\n\nCalculate the radius in cm of an iron atom in this crystal. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "In $\\mathrm{BCC}$ there are 2 atoms per unit cell, so $\\frac{2}{\\mathrm{a}^{3}}=\\frac{\\mathrm{N}_{\\mathrm{A}}}{\\mathrm{V}_{\\text {molar }}}$, where $\\mathrm{V}_{\\text {molar }}=\\mathrm{A} / \\rho ; \\mathrm{A}$ is the atomic mass of iron.\n\\[\n\\begin{aligned}\n&\\frac{2}{a^{3}}=\\frac{N_{A} \\times p}{A} \\\\\n&\\therefore a=\\left(\\frac{2 A}{N_{A} \\times \\rho}\\right)^{\\frac{1}{3}}=\\frac{4}{\\sqrt{3}} r \\\\\n&\\therefore r= \\boxed{1.24e-8} \\mathrm{~cm}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 197}
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{"problem": "Preamble: For the element copper (Cu) determine:\n\nSubproblem 0: the distance of second nearest neighbors (in meters). Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.\n\n\nSolution: The answer can be found by looking at a unit cell of $\\mathrm{Cu}$ (FCC).\n\\includegraphics[scale=0.5]{set_23_img_00.jpg}\n\\nonessentialimage\nNearest neighbor distance is observed along $<110>$; second-nearest along $<100>$. The second-nearest neighbor distance is found to be \"a\".\nCu: atomic volume $=7.1 \\times 10^{-6} \\mathrm{~m}^{3} /$ mole $=\\frac{\\mathrm{N}_{\\mathrm{A}}}{4} \\mathrm{a}^{3}$ ( $\\mathrm{Cu}: \\mathrm{FCC} ; 4$ atoms/unit cell) $a=\\sqrt[3]{\\frac{7.1 \\times 10^{-6} \\times 4}{6.02 \\times 10^{23}}}= \\boxed{3.61e-10} \\mathrm{~m}$\n\nFinal answer: The final answer is 3.61e-10. I hope it is correct.\n\nSubproblem 1: the interplanar spacing of $\\{110\\}$ planes (in meters). Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "$d_{h k l}=\\frac{a}{\\sqrt{h^{2}+k^{2}+1^{2}}}$\n\\[\nd_{110}=\\frac{3.61 \\times 10^{-10}}{\\sqrt{2}}= \\boxed{2.55e-10} \\mathrm{~m}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 198}
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{"problem": "Subproblem 0: What is the working temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1950}.\n\nFinal answer: The final answer is 1950. I hope it is correct.\n\nSubproblem 1: What is the softening temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1700}.\n\nFinal answer: The final answer is 1700. I hope it is correct.\n\nSubproblem 2: What is the working temperature for Pyrex in Celsius?\n\n\nSolution: \\boxed{1200}.\n\nFinal answer: The final answer is 1200. I hope it is correct.\n\nSubproblem 3: What is the softening temperature for Pyrex in Celsius?\n\n\nSolution: \\boxed{800}.\n\nFinal answer: The final answer is 800. I hope it is correct.\n\nSubproblem 4: What is the working temperature for soda-lime glass in Celsius?", "solution": "\\boxed{900}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 199}
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{"problem": "Determine the wavelength of $\\lambda_{K_{\\alpha}}$ for molybdenum (Mo). Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places, in meters.", "solution": "$M o: Z=42 ; \\mathrm{K}_{\\alpha} \\rightarrow \\mathrm{n}_{\\mathrm{i}}=2 ; \\mathrm{n}_{\\mathrm{f}}=1 ; \\sigma=1$\n\\[\n\\begin{aligned}\n&\\bar{v}_{\\mathrm{K}_{\\alpha}}=R(Z-1)^{2}\\left[\\frac{1}{\\mathrm{n}_{\\mathrm{f}}^{2}}-\\frac{1}{\\mathrm{n}_{\\mathrm{i}}^{2}}\\right] \\\\\n&\\bar{v}_{\\mathrm{K}_{\\alpha}}=1.097 \\times 10^{7}\\left[\\frac{1}{\\mathrm{~m}}\\right](42-1)^{2}\\left[\\frac{1}{1^{2}}-\\frac{1}{2^{2}}\\right] \\\\\n&\\bar{v}_{\\mathrm{K}_{\\alpha}}=1.38 \\times 10^{10} \\mathrm{~m}^{-1} \\\\\n&\\lambda_{\\mathrm{K}_{\\alpha}}=\\frac{1}{\\bar{v}_{\\mathrm{K}_{\\alpha}}}= \\boxed{7.25e-11} \\mathrm{~m}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 200}
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{"problem": "Determine the second-nearest neighbor distance (in pm) for nickel (Ni) at $100^{\\circ} \\mathrm{C}$ if its density at that temperature is $8.83 \\mathrm{~g} / \\mathrm{cm}^{3}$. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "\\[\n\\begin{array}{ll}\n\\mathrm{Ni}: \\mathrm{n}=4 \\\\\n\\text { atomic weight }=58.70 \\mathrm{~g} / \\mathrm{mole} \\\\\n\\rho=8.83 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{array}\n\\]\nFor a face-centered cubic structure, the second nearest neighbor distance equals \"a\".\n\\[\n\\begin{aligned}\n& \\frac{\\text { atomic weight }}{\\rho} \\times 10^{-6}=\\frac{N_{A}}{n} \\times a^{3} \\\\\n& a^{3}=\\frac{(58.70 \\mathrm{~g} / \\mathrm{mole})\\left(10^{-6} \\mathrm{~m}^{3} / \\mathrm{cm}^{3}\\right)(4 \\text { atoms } / \\text { unit cell })}{\\left(6.023 \\times 10^{23} \\text { atoms } / \\mathrm{mole}\\right)\\left(8.83 \\mathrm{~g} / \\mathrm{cm}^{3}\\right)} \\\\\n& =4.41 \\times 10^{-29} \\mathrm{~m}^{3} \\\\\n& \\mathrm{a}=3.61 \\times 10^{-10} \\mathrm{~m} \\times \\frac{10^{12} \\mathrm{pm}}{\\mathrm{m}}= \\boxed{3.61e2} \\mathrm{pm} \n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 201}
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{"problem": "What is the working temperature for silica glass in Celsius?", "solution": "\\boxed{1950}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 202}
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{"problem": "What acceleration potential $V$ must be applied to electrons to cause electron diffraction on $\\{220\\}$ planes of gold $(\\mathrm{Au})$ at $\\theta=5^{\\circ}$ ? Format your answer as an integer, in Volts.", "solution": "We first determine the wavelength of particle waves $\\left(\\lambda_{p}\\right)$ required for diffraction and then the voltage to be applied to the electrons:\n\\[\n\\begin{aligned}\n&\\lambda=2 \\mathrm{~d}_{\\{220\\}} \\sin \\theta=2 \\frac{\\mathrm{a}}{\\sqrt{8}} \\sin 5^{\\circ} \\\\\n&\\mathrm{a}_{\\mathrm{Au}}=\\sqrt[3]{\\frac{4 \\times 10.2 \\times 10^{-6}}{6.02 \\times 10^{23}}}=4.08 \\times 10^{-10} \\mathrm{~m} \\\\\n&\\lambda=\\frac{2 \\times 4.08 \\times 10^{-10}}{\\sqrt{8}} \\sin 5^{\\circ}=\\frac{4.08 \\times 10^{-10}}{\\sqrt{2}} \\times 0.087=0.25 \\times 10^{-10} \\mathrm{~m}=\\lambda_{\\mathrm{p}} \\\\\n&\\mathrm{eV}=\\frac{\\mathrm{mv}}{2}, \\therefore \\mathrm{v}=\\sqrt{\\frac{2 \\mathrm{eV}}{\\mathrm{m}}} \\\\\n&\\lambda_{\\mathrm{P}}=\\frac{\\mathrm{h}}{\\mathrm{mv}}=\\frac{\\mathrm{h}}{\\sqrt{2 \\mathrm{meV}}}, \\therefore V=\\frac{\\mathrm{h}^{2}}{2 \\lambda^{2} \\mathrm{me}}= \\boxed{2415} \\mathrm{~V}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 203}
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{"problem": "To increase its corrosion resistance, chromium $(\\mathrm{Cr})$ is diffused into steel at $980^{\\circ} \\mathrm{C}$. If during diffusion the surface concentration of chromium remains constant at $100 \\%$, how long will it take (in days) to achieve a $\\mathrm{Cr}$ concentration of $1.8 \\%$ at a depth of $0.002 \\mathrm{~cm}$ below the steel surface? Round your answer to 1 decimal place. $\\left(D_{o}=0.54 \\mathrm{~cm}^{2} / \\mathrm{s} ; E_{A}=286 \\mathrm{~kJ} / \\mathrm{mol}\\right.$ )", "solution": "A solution to Fick's second law for the given boundary conditions is:\n$\\frac{c}{c_{s}}=1-\\operatorname{erf} \\frac{x}{2 \\sqrt{D t}}$, from which we get erf $\\frac{x}{2 \\sqrt{D t}}=1-0.018=0.982$\nFrom the error function tables, $0.982$ is the erf of $1.67$. This means that\n\\[\n\\frac{0.002}{2 \\sqrt{D t}}=\\frac{0.001}{\\sqrt{D t}}=1.67\n\\]\n\\[\n\\begin{aligned}\n& \\mathrm{D}=\\mathrm{D}_{0} \\mathrm{e}^{\\left(\\frac{-286 \\times 10^{5}}{8.314 \\times 1253}\\right)}=6.45 \\times 10^{-13} \\mathrm{~cm}^{2} / \\mathrm{s} \\\\\n& \\therefore \\mathrm{t}=\\frac{0.001^{2}}{1.67^{2} \\times 6.45 \\times 10^{-13}}=5.56 \\times 10^{5} \\mathrm{sec}=\\boxed{6.4} \\text { days }\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 204}
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{"problem": "Subproblem 0: What is the working temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1950}.\n\nFinal answer: The final answer is 1950. I hope it is correct.\n\nSubproblem 1: What is the softening temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1700}.\n\nFinal answer: The final answer is 1700. I hope it is correct.\n\nSubproblem 2: What is the working temperature for Pyrex in Celsius?", "solution": "\\boxed{1200}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 205}
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{"problem": "Preamble: Calculate the vacancy fraction in copper (Cu) in $\\mathrm{~cm}^{-3}$ at the following temperatures. Measurements have determined the values of the enthalpy of vacancy formation, $\\Delta \\mathrm{H}_{\\mathrm{V}}$, to be $1.03 \\mathrm{eV}$ and the entropic prefactor, A, to be 1.1. Please format your answers as $n \\times 10^x$ where $n$ is to 2 decimal places.\n\n$20^{\\circ} \\mathrm{C}$.", "solution": "number of sites / unit volume (also known as site density) is given by:\n\\[\n\\begin{aligned}\n\\frac{\\mathrm{N}_{\\mathrm{A}}}{\\mathrm{V}_{\\text {molar }}} & \\therefore \\text { site density }=6.02 \\times 10^{23} / 7.11 \\mathrm{~cm}^{3}=8.47 \\times 10^{22} \\\\\n& \\rightarrow \\text { vacancy density }=\\mathrm{f}_{\\mathrm{v}} \\times \\text { site density }\n\\end{aligned}\n\\]\n$f_{V}=A e^{-\\frac{\\Delta H_{V}}{k_{B} T}}=1.1 \\times e^{-\\frac{1.03 \\times 1.6 \\times 10^{-19}}{1.38 \\times 10^{-22} \\times(20+273)}}=2.19 \\times 10^{-18}$\nvacancy density at $20^{\\circ} \\mathrm{C}= \\boxed{1.85e5} \\mathrm{~cm}^{-3}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 206}
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{"problem": "Preamble: For aluminum at $300 \\mathrm{~K}$, \n\nCalculate the planar packing fraction (fractional area occupied by atoms) of the ( 110 ) plane. Please round your answer to 3 decimal places.", "solution": "Aluminum at $300 \\mathrm{~K}$ has FCC structure:\n\\includegraphics[scale=0.5]{set_23_img_03.jpg}\n\\nonessentialimage\nVolume unit of a cell:\n\\[\n\\begin{aligned}\n&V=\\frac{10 \\mathrm{~cm}^{3}}{\\text { mole }} \\times \\frac{1 \\text { mole }}{6.02 \\times 10^{23} \\text { atoms }} \\times \\frac{4 \\text { atoms }}{1 \\text { unit cell }} \\\\\n&=6.64 \\times 10^{-23} \\mathrm{~cm}^{3} / \\text { unit cell }\n\\end{aligned}\n\\]\nFor FCC: $\\sqrt{2} \\mathrm{a}=4 \\mathrm{r} \\rightarrow$ atomic radius $\\mathrm{r}=\\frac{\\sqrt{2}}{4} \\mathrm{a}=\\frac{\\sqrt{2}}{4}\\left(4.05 \\times 10^{-8} \\mathrm{~cm}\\right)$\n\\[\n=1.43 \\times 10^{-8} \\mathrm{~cm}\n\\]\nPlanar packing fraction of the $(110)$ plane:\narea of shaded plane in above unit cell $=\\sqrt{2} a^{2}$\nnumber of lattice points in the shaded area $=2\\left(\\frac{1}{2}\\right)+4\\left(\\frac{1}{4}\\right)=2$\narea occupied by 1 atom $=\\pi r^{2}$\npacking fraction $=\\frac{\\text { area occupied by atoms }}{\\text { total area }}=\\frac{2 \\pi \\mathrm{r}^{2}}{\\sqrt{2} \\mathrm{a}^{2}}$\n\\[\n=\\frac{2 \\pi\\left(1.43 \\times 10^{-8} \\mathrm{~cm}\\right)^{2}}{\\sqrt{2}\\left(4.05 \\times 10^{-8} \\mathrm{~cm}\\right)^{2}}= \\boxed{0.554}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 207}
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{"problem": "Determine the inter-ionic equilibrium distance in meters between the sodium and chlorine ions in a sodium chloride molecule knowing that the bond energy is $3.84 \\mathrm{eV}$ and that the repulsive exponent is 8. Please format your answer as $n \\times 10^x$ where $n$ is to 1 decimal place.", "solution": "$\\mathrm{E}_{\\mathrm{equ}}=-3.84 \\mathrm{eV}=-3.84 \\times 1.6 \\times 10^{-19} \\mathrm{~J}=-\\frac{\\mathrm{e}^{2}}{4 \\pi \\varepsilon_{0} r_{0}}\\left(1-\\frac{1}{\\mathrm{n}}\\right)$\n\\\\\n$r_{0}=\\frac{\\left(1.6 \\times 10^{-19}\\right)^{2}}{4 \\pi 8.85 \\times 10^{-12} \\times 6.14 \\times 10^{-19}}\\left(1-\\frac{1}{8}\\right)= \n\\boxed{3.3e-10} \\mathrm{~m}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 208}
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{"problem": "Preamble: A formation energy of $2.0 \\mathrm{eV}$ is required to create a vacancy in a particular metal. At $800^{\\circ} \\mathrm{C}$ there is one vacancy for every 10,000 atoms.\n\nAt what temperature (in Celsius) will there be one vacancy for every 1,000 atoms? Format your answer as an integer.", "solution": "We need to know the temperature dependence of the vacancy density:\n\\[\n\\frac{1}{10^{4}}=A e^{-\\frac{\\Delta H_{v}}{k T_{1}}} \\quad \\text { and } \\frac{1}{10^{3}}=A e^{-\\frac{\\Delta H_{v}}{k T_{x}}}\n\\]\nFrom the ratio: $\\frac{\\frac{1}{10^{4}}}{\\frac{1}{10^{3}}}=\\frac{10^{3}}{10^{4}}=\\frac{\\mathrm{Ae}^{-\\Delta \\mathrm{H}_{v} / \\mathrm{k} T_{1}}}{\\mathrm{Ae}^{-\\Delta \\mathrm{H}_{v} / \\mathrm{kT} \\mathrm{x}}}$ we get $-\\ln 10=-\\frac{\\Delta \\mathrm{H}_{\\mathrm{v}}}{\\mathrm{k}}\\left(\\frac{1}{\\mathrm{~T}_{1}}-\\frac{1}{\\mathrm{~T}_{\\mathrm{x}}}\\right)$\n\\[\n\\begin{aligned}\n&\\therefore \\quad\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{x}}\\right)=\\frac{k \\ln 10}{\\Delta H_{v}} \\\\\n&\\frac{1}{T_{x}}=\\frac{1}{T_{1}}-\\frac{k \\ln 10}{\\Delta H_{v}}=\\frac{1}{1073}-\\frac{1.38 \\times 10^{-23} \\times \\ln 10}{2 \\times 1.6 \\times 10^{-19}}=8.33 \\times 10^{-4} \\\\\n&T_{x}=1200 \\mathrm{~K}= \\boxed{928}^{\\circ} \\mathrm{C}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 209}
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{"problem": "For $\\mathrm{NaF}$ the repulsive (Born) exponent, $\\mathrm{n}$, is 8.7. Making use of data given in your Periodic Table, calculate the crystal energy ( $\\left.\\Delta \\mathrm{E}_{\\text {cryst }}\\right)$ in kJ/mole, to 1 decimal place.", "solution": "\\[\n\\Delta E=\\frac{e^{2} N_{A} M}{4 \\pi \\varepsilon_{0} r_{0}}\\left(1-\\frac{1}{n}\\right)\n\\]\nThe assumption must be made that the distance of separation of Na- $F$ is given by the sum of the ionic radii (that in a crystal they touch each other - a not unreasonable assumption). Thus, $r_{0}=0.95 \\times 10^{-10}+1.36 \\times 10^{-10} \\mathrm{~m}=2.31 \\AA$ and you must also assume $M$ is the same as for $\\mathrm{NaCl}=1.747$ : $\\mathrm{E}_{\\text {cryst }}=-\\frac{\\left(1.6 \\times 10^{-19}\\right)^{2} 6.02 \\times 10^{23} \\times 1.747}{4 \\pi 8.85 \\times 10^{-12} \\times 2.31 \\times 10^{-10}}\\left(1-\\frac{1}{8.7}\\right)$\n\\\\\n$\\mathrm{E}_{\\text {cryst }}=\\boxed{927.5} /$ kJ/mole", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 210}
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{"problem": "Preamble: Calculate the molecular weight in g/mole (to 2 decimal places) of each of the substances listed below.\n\nSubproblem 0: $\\mathrm{NH}_{4} \\mathrm{OH}$\n\n\nSolution: $\\mathrm{NH}_{4} \\mathrm{OH}$ :\n$5 \\times 1.01=5.05(\\mathrm{H})$\n$1 \\times 14.01=14.01(\\mathrm{~N})$\n$1 \\times 16.00=16.00(\\mathrm{O})$\n$\\mathrm{NH}_{4} \\mathrm{OH}= \\boxed{35.06}$ g/mole\n\nFinal answer: The final answer is 35.06. I hope it is correct.\n\nSubproblem 1: $\\mathrm{NaHCO}_{3}$", "solution": "$\\mathrm{NaHCO}_{3}: 3 \\times 16.00=48.00(\\mathrm{O})$\n$1 \\times 22.99=22.99(\\mathrm{Na})$\n$1 \\times 1.01=1.01$ (H)\n$1 \\times 12.01=12.01$ (C)\n$\\mathrm{NaHCO}_{3}= \\boxed{84.01}$ g/mole", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 211}
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{"problem": "In iridium (Ir), the vacancy fraction, $n_{v} / \\mathrm{N}$, is $3.091 \\times 10^{-5}$ at $12340^{\\circ} \\mathrm{C}$ and $5.26 \\times 10^{-3}$ at the melting point. Calculate the enthalpy of vacancy formation, $\\Delta \\mathrm{H}_{\\mathrm{v}}$. Round your answer to 1 decimal place.", "solution": "All we need to know is the temperature dependence of the vacancy density:\n$\\frac{n_{v}}{N}=A e^{-\\frac{\\Delta H_{v}}{R T}}$, where $T$ is in Kelvins and the melting point of $I r$ is $2446^{\\circ} \\mathrm{C}$\n$3.091 \\times 10^{-5}=\\mathrm{Ae}^{-\\frac{\\Delta \\mathrm{H}_{\\mathrm{V}}}{\\mathrm{RT}_{1}}}$, where $\\mathrm{T}_{1}=1234^{\\circ} \\mathrm{C}=1507 \\mathrm{~K}$\n$5.26 \\times 10^{-3}=A e^{-\\frac{\\Delta H_{v}}{R T_{2}}}$, where $T_{2}=2446^{\\circ} \\mathrm{C}=2719 \\mathrm{~K}$\nTaking the ratio:\n\\[\n\\begin{aligned}\n&\\frac{5.26 \\times 10^{-3}}{3.091 \\times 10^{-5}}=\\frac{A e^{-\\frac{\\Delta H_{v}}{R T_{1}}}}{A e^{-\\frac{\\Delta H_{v}}{R T_{2}}}}=e^{-\\frac{\\Delta H_{v}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)} \\\\\n&\\therefore \\ln 170.2=-\\frac{\\Delta H_{v}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right) \\\\\n&\\therefore \\Delta H_{v}=-\\frac{R \\times \\ln 170.2}{\\frac{1}{1507}-\\frac{1}{2719}}=-\\frac{8.314 \\times \\ln 170.2}{\\frac{1}{1507}-\\frac{1}{2719}}=1.44 \\times 10^{5} \\mathrm{~J} / \\mathrm{mole} \\cdot \\mathrm{vac} \\\\\n&\\therefore \\Delta \\mathrm{H}_{\\mathrm{v}}=\\frac{1.44 \\times 10^{5}}{6.02 \\times 10^{23}}=2.40 \\times 10^{-19} \\mathrm{~J} / \\mathrm{vac}= \\boxed{1.5} \\mathrm{eV} / \\mathrm{vac}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 212}
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{"problem": "If no electron-hole pairs were produced in germanium (Ge) until the temperature reached the value corresponding to the energy gap, at what temperature (Celsius) would Ge become conductive? Please format your answer as $n \\times 10^x$ where n is to 1 decimal place. $\\left(\\mathrm{E}_{\\mathrm{th}}=3 / 2 \\mathrm{kT}\\right)$", "solution": "\\[\n\\begin{aligned}\n&E_{t h}=\\frac{3 K T}{2} ; E_{g}=0.72 \\times 1.6 \\times 10^{-19} \\mathrm{~J} \\\\\n&T=\\frac{0.72 \\times 1.6 \\times 10^{-19} \\times 2}{3 \\times 1.38 \\times 10^{-23}}=5565 \\mathrm{~K}=5.3 \\times 10^{3}{ }^{\\circ} \\mathrm{C}\n\\end{aligned}\n\\]\nThe temperature would have to be $\\boxed{5.3e3}{ }^{\\circ} \\mathrm{C}$, about $4400^{\\circ} \\mathrm{C}$ above the melting point of Ge.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 213}
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{"problem": "Preamble: A first-order chemical reaction is found to have an activation energy $\\left(E_{A}\\right)$ of 250 $\\mathrm{kJ} /$ mole and a pre-exponential (A) of $1.7 \\times 10^{14} \\mathrm{~s}^{-1}$.\n\nDetermine the rate constant at $\\mathrm{T}=750^{\\circ} \\mathrm{C}$. Round your answer to 1 decimal place, in units of $\\mathrm{s}^{-1}$.", "solution": "$\\mathrm{k}=\\mathrm{Ae} \\mathrm{e}^{-\\frac{\\mathrm{E}_{\\mathrm{A}}}{\\mathrm{RT}}}=1.7 \\times 10^{14} \\times \\mathrm{e}^{-\\frac{2.5 \\times 10^{5}}{8.31 \\times 10^{23}}}= \\boxed{28.8} \\mathrm{~s}^{-1}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 214}
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{"problem": "A cubic metal $(r=0.77 \\AA$ ) exhibits plastic deformation by slip along $<111>$ directions. Determine its planar packing density (atoms $/ \\mathrm{m}^{2}$) for its densest family of planes. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "Slip along $<111>$ directions suggests a BCC system, corresponding to $\\{110\\},<111>$ slip. Therefore:\n\\[\n\\begin{aligned}\n&a \\sqrt{3}=4 r \\\\\n&a=\\frac{4 r}{\\sqrt{3}}=1.78 \\times 10^{-10} \\mathrm{~m}\n\\end{aligned}\n\\]\nDensest planes are $\\{110\\}$, so we find:\n\\[\n\\frac{2 \\text { atoms }}{a^{2} \\sqrt{2}}=\\boxed{4.46e19} \\text { atoms } / \\mathrm{m}^{2}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 215}
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{"problem": "Determine the total void volume $(\\mathrm{cm}^{3} / mole)$ for gold (Au) at $27^{\\circ} \\mathrm{C}$; make the hard-sphere approximation in your calculation. Note that the molar volume of gold (Au) is $10.3 \\mathrm{~cm}^{3} / \\mathrm{mole}$. Please round your answer to 2 decimal places.", "solution": "First determine the packing density for Au, which is $\\mathrm{FC}$; then relate it to the molar volume given in the periodic table.\n\\[\n\\begin{aligned}\n&\\text { packing density }=\\frac{\\text { volume of atoms/unit cell }}{\\text { volume of unit cell }}=\\frac{\\frac{16 \\pi \\mathrm{r}^{3}}{3}}{\\mathrm{a}^{3}}=\\frac{16 \\pi \\mathrm{r}^{3}}{3 \\mathrm{a}^{3}} \\\\\n&\\text { packing density }=\\frac{16 \\pi \\mathrm{r}^{3}}{3 \\times 16 \\sqrt{2} \\mathrm{r}^{3}}=\\frac{\\pi}{3 \\sqrt{2}}=0.74=74 \\% \\\\\n&\\text { void volume }=1-\\text { packing density }=26 \\%\n\\end{aligned}\n\\]\nFrom the packing density $(74 \\%)$ we recognize the void volume to be $26 \\%$. Given the molar volume as $10.3 \\mathrm{~cm}^{3} / \\mathrm{mole}$, the void volume is:\n\\[\n0.26 \\times 10.3 \\mathrm{~cm}^{3} / \\text { mole }= \\boxed{2.68} \\mathrm{~cm}^{3} / \\text { mole }\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 216}
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{"problem": "Subproblem 0: What is the working temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1950}.\n\nFinal answer: The final answer is 1950. I hope it is correct.\n\nSubproblem 1: What is the softening temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1700}.\n\nFinal answer: The final answer is 1700. I hope it is correct.\n\nSubproblem 2: What is the working temperature for Pyrex in Celsius?\n\n\nSolution: \\boxed{1200}.\n\nFinal answer: The final answer is 1200. I hope it is correct.\n\nSubproblem 3: What is the softening temperature for Pyrex in Celsius?\n\n\nSolution: \\boxed{800}.\n\nFinal answer: The final answer is 800. I hope it is correct.\n\nSubproblem 4: What is the working temperature for soda-lime glass in Celsius?\n\n\nSolution: \\boxed{900}.\n\nFinal answer: The final answer is 900. I hope it is correct.\n\nSubproblem 5: What is the softening temperature for soda-lime glass in Celsius?", "solution": "\\boxed{700}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 217}
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{"problem": "What is the maximum wavelength $(\\lambda)$ (in meters) of radiation capable of second order diffraction in platinum (Pt)? Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "The longest wavelength capable of $1^{\\text {st }}$ order diffraction in Pt can be identified on the basis of the Bragg equation: $\\lambda=2 \\mathrm{~d} \\sin \\theta . \\lambda_{\\max }$ will diffract on planes with maximum interplanar spacing (in compliance with the selection rules): $\\{111\\}$ at the maximum value $\\theta\\left(90^{\\circ}\\right)$. We determine the lattice constant a for $\\mathrm{Pt}$, and from it obtain $\\mathrm{d}_{\\{111\\}}$. Pt is FCC with a value of atomic volume or $V_{\\text {molar }}=9.1 \\mathrm{~cm}^{3} /$ mole.\n\\[\n\\mathrm{V}_{\\text {molar }}=\\frac{N_{\\mathrm{A}}}{4} \\mathrm{a}^{3} ; \\mathrm{a}=\\sqrt[3]{\\frac{9.1 \\times 10^{-6} \\times 4}{\\mathrm{~N}_{\\mathrm{A}}}}=3.92 \\times 10^{-10} \\mathrm{~m}\n\\]\nIf we now look at $2^{\\text {nd }}$ order diffraction, we find $2 \\lambda=2 \\mathrm{~d}_{\\{111\\}} \\sin 90^{\\circ}$\n\\[\n\\therefore \\lambda_{\\max }=\\mathrm{d}_{\\{111\\}}=\\frac{\\mathrm{a}}{\\sqrt{3}}=\\frac{3.92 \\times 10^{-10}}{\\sqrt{3}}= \\boxed{2.26e-10} \\mathrm{~m}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 218}
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{"problem": "What is the activation energy of a process which is observed to increase by a factor of three when the temperature is increased from room temperature $\\left(20^{\\circ} \\mathrm{C}\\right)$ to $40^{\\circ} \\mathrm{C}$ ? Round your answer to 1 decimal place, and express it in $\\mathrm{~kJ} / \\mathrm{mole}$.", "solution": "\\[\n\\mathrm{k}_{1}=A \\mathrm{e}^{\\frac{-E_{A}}{R T_{1}}} ; k_{2}=3 k_{1}=A e^{\\frac{-E_{A}}{R T_{2}}} \\rightarrow \\frac{1}{3}=e^{-\\frac{E_{A}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right)}\n\\]\n\\[\n\\begin{aligned}\n&\\ln 3=\\frac{E_{A}}{R}\\left(\\frac{1}{T_{1}}-\\frac{1}{T_{2}}\\right) \\\\\n&E_{A}=\\frac{R \\times \\ln 3}{\\frac{1}{293}-\\frac{1}{313}}=4.19 \\times 10^{4}= \\boxed{41.9} \\mathrm{~kJ} / \\mathrm{mole}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 219}
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{"problem": "How much oxygen (in kg, to 3 decimal places) is required to completely convert 1 mole of $\\mathrm{C}_{2} \\mathrm{H}_{6}$ into $\\mathrm{CO}_{2}$ and $\\mathrm{H}_{2} \\mathrm{O}$ ?", "solution": "To get the requested answer, let us formulate a ``stoichiometric'' equation (molar quantities) for the reaction: $\\mathrm{C}_{2} \\mathrm{H}_{6}+70 \\rightarrow 2 \\mathrm{CO}_{2}+3 \\mathrm{H}_{2} \\mathrm{O}_{\\text {. Each } \\mathrm{C}_{2} \\mathrm{H}_{6}}$ (ethane) molecule requires 7 oxygen atoms for complete combustion. In molar quantities: 1 mole of $\\mathrm{C}_{2} \\mathrm{H}_{6}=2 \\times 12.01+6 \\times 1.008=30.07 \\mathrm{~g}$ requires\n$7 \\times 15.9984 \\mathrm{~g}=1.12 \\times 10^{2}$ oxygen $=\\boxed{0.112} kg$ oxygen\nWe recognize the oxygen forms molecules, $\\mathrm{O}_{2}$, and therefore a more appropriate formulation would be: $\\mathrm{C}_{2} \\mathrm{H}_{6}+7 / 2 \\mathrm{O}_{2} \\rightarrow 2 \\mathrm{CO}_{2}+3 \\mathrm{H}_{2} \\mathrm{O}$. The result would be the same.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 220}
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{"problem": "Determine the differences in relative electronegativity $(\\Delta x$ in $e V)$ for the systems ${H}-{F}$ and ${C}-{F}$ given the following data:\n$\\begin{array}{cl}\\text { Bond Energy } & {kJ} / \\text { mole } \\\\ {H}_{2} & 436 \\\\ {~F}_{2} & 172 \\\\ {C}-{C} & 335 \\\\ {H}-{F} & 565 \\\\ {C}-{H} & 410\\end{array}$\n\\\\\nPlease format your answer to 2 decimal places.", "solution": "According to Pauling, the square of the difference in electro negativity for two elements $\\left(X_{A}-X_{B}\\right)^{2}$ is given by the following relationship: $\\left(X_{A}-X_{B}\\right)^{2}=[$ Bond Energy $(A-B)-\\sqrt{\\text { Bond Energy AA. Bond Energy } B B}] \\times \\frac{1}{96.3}$\nIf bond energies are given in ${kJ}$.\\\\\n$\\left(X_{H}-X_{F}\\right)^{2}=[565-\\sqrt{436 \\times 172}] \\frac{1}{96.3}=3.02$\n\\[\n\\begin{aligned}\n& \\left({X}_{{H}}-{X}_{{F}}\\right)=\\sqrt{3.02}=1.7 \\\\\n& \\left(X_{C}-X_{H}\\right)^{2}=[410-\\sqrt{335 \\times 436}] \\frac{1}{96.3}=0.29 \\\\\n& \\left(X_{C}-X_{H}\\right)=\\sqrt{0.29}= \\boxed{0.54}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 221}
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{"problem": "Preamble: The number of electron-hole pairs in intrinsic germanium (Ge) is given by:\n\\[\nn_{i}=9.7 \\times 10^{15} \\mathrm{~T}^{3 / 2} \\mathrm{e}^{-\\mathrm{E}_{g} / 2 \\mathrm{KT}}\\left[\\mathrm{cm}^{3}\\right] \\quad\\left(\\mathrm{E}_{\\mathrm{g}}=0.72 \\mathrm{eV}\\right)\n\\]\n\nWhat is the density of pairs at $\\mathrm{T}=20^{\\circ} \\mathrm{C}$, in inverse $\\mathrm{cm}^3$? Please format your answer as $n \\times 10^x$ where n is to 2 decimal places.", "solution": "Recall: $\\mathrm{T}$ in thermally activated processes is the absolute temperature: $\\mathrm{T}^{\\circ} \\mathrm{K}=$ $\\left(273.16+\\mathrm{t}^{\\circ} \\mathrm{C}\\right)$; Boltzmann's constant $=\\mathrm{k}=1.38 \\times 10^{-23} \\mathrm{~J} /{ }^{\\circ} \\mathrm{K}$\n$\\mathrm{T}=293.16 \\mathrm{~K}:$\n\\[\n\\begin{aligned}\n&n_{i}=9.7 \\times 10^{15} \\times 293.16^{\\frac{3}{2}} \\times e^{-\\frac{0.72 \\times 16 \\times 10^{-19}}{2 \\times 1.38 \\times 10^{-23} \\times 293.16}} \\\\\n&=9.7 \\times 10^{15} \\times 5019 \\times 6.6 \\times 10^{-7} \\\\\n&n_{i}= \\boxed{3.21e13} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 222}
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{"problem": "Preamble: For light with a wavelength $(\\lambda)$ of $408 \\mathrm{~nm}$ determine:\n\nSubproblem 0: the frequency in $s^{-1}$. Please format your answer as $n \\times 10^x$, where $n$ is to 3 decimal places. \n\n\nSolution: To solve this problem we must know the following relationships:\n\\[\n\\begin{aligned}\nv \\lambda &=c\n\\end{aligned}\n\\]\n$v$ (frequency) $=\\frac{c}{\\lambda}=\\frac{3 \\times 10^{8} m / s}{408 \\times 10^{-9} m}= \\boxed{7.353e14} s^{-1}$\n\nFinal answer: The final answer is 7.353e14. I hope it is correct.\n\nSubproblem 1: the wave number in $m^{-1}$. Please format your answer as $n \\times 10^x$, where $n$ is to 2 decimal places.", "solution": "To solve this problem we must know the following relationships:\n\\[\n\\begin{aligned}\n1 / \\lambda=\\bar{v} \n\\end{aligned}\n\\]\n$\\bar{v}$ (wavenumber) $=\\frac{1}{\\lambda}=\\frac{1}{408 \\times 10^{-9} m}=\\boxed{2.45e6} m^{-1}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 223}
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{"problem": "Calculate the volume in mL of $0.25 \\mathrm{M} \\mathrm{NaI}$ that would be needed to precipitate all the $\\mathrm{g}^{2+}$ ion from $45 \\mathrm{~mL}$ of a $0.10 \\mathrm{M} \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}$ solution according to the following reaction:\n\\[\n2 \\mathrm{NaI}(\\mathrm{aq})+\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}(\\mathrm{aq}) \\rightarrow \\mathrm{HgI}_{2}(\\mathrm{~s})+2 \\mathrm{NaNO}_{3}(\\mathrm{aq})\n\\]", "solution": "\\[\n\\begin{aligned}\n&2 \\mathrm{NaI}(\\mathrm{aq})+\\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}(\\mathrm{aq}) \\rightarrow \\mathrm{HgI}_{2}(\\mathrm{~s})+\\mathrm{NaNO}_{3}(\\mathrm{aq}) \\\\\n&\\frac{0.10 \\mathrm{~mol} \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}}{1 \\mathrm{~L}} \\times 0.045 \\mathrm{~L}=4.5 \\times 10^{-3} \\mathrm{~mol} \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2} \\\\\n&4.5 \\times 10^{-3} \\mathrm{~mol} \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2} \\times \\frac{2 \\mathrm{~mol} \\mathrm{NaI}}{1 \\mathrm{~mol} \\mathrm{Hg}\\left(\\mathrm{NO}_{3}\\right)_{2}}=9.00 \\times 10^{-3} \\mathrm{~mol} \\mathrm{NaI} \\\\\n&\\frac{9.00 \\times 10^{-3} \\mathrm{~mol} \\mathrm{NaI}}{0.25 \\frac{\\mathrm{mol} \\mathrm{NaI}}{\\mathrm{L}}}=3.6 \\times 10^{-2} \\mathrm{~L} \\times \\frac{1000 \\mathrm{ml}}{1 \\mathrm{~L}}=\\boxed{36} \\mathrm{~mL} \\mathrm{NaI}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 224}
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{"problem": "A slab of plate glass containing dissolved helium (He) is placed in a vacuum furnace at a temperature of $400^{\\circ} \\mathrm{C}$ to remove the helium from the glass. Before vacuum treatment, the concentration of helium is constant throughout the glass. After 10 minutes in vacuum at $400^{\\circ} \\mathrm{C}$, at what depth (in $\\mu \\mathrm{m}$) from the surface of the glass has the concentration of helium decreased to $1 / 3$ of its initial value? The diffusion coefficient of helium in the plate glass at the processing temperature has a value of $3.091 \\times 10^{-6} \\mathrm{~cm}^{2} / \\mathrm{s}$.", "solution": "\\includegraphics[scale=0.5]{set_37_img_01.jpg}\n\\nonessentialimage\n\\[\n\\begin{aligned}\n&c=A+B \\text { erf } \\frac{x}{2 \\sqrt{D t}} ; c(0, t)=0=A ; c(\\infty, t)=c_{0}=B \\\\\n&\\therefore c(x, t)=c_{0} \\operatorname{erf} \\frac{x}{2 \\sqrt{D t}}\n\\end{aligned}\n\\]\nWhat is $\\mathrm{x}$ when $\\mathrm{c}=\\mathrm{c}_{0} / 3$ ?\n\\[\n\\begin{gathered}\n\\frac{c_{0}}{3}=c_{0} \\operatorname{erf} \\frac{x}{2 \\sqrt{D t}} \\rightarrow 0.33=\\operatorname{erf} \\frac{x}{2 \\sqrt{D t}} ; \\operatorname{erf}(0.30)=0.3286 \\approx 0.33 \\\\\n\\therefore \\frac{x}{2 \\sqrt{D t}}=0.30 \\rightarrow x=2 \\times 0.30 \\times \\sqrt{3.091 \\times 10^{-6} \\times 10 \\times 60}=2.58 \\times 10^{-2} \\mathrm{~cm}=\\boxed{258} \\mu \\mathrm{m}\n\\end{gathered}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 225}
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{"problem": "Subproblem 0: What is the working temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1950}.\n\nFinal answer: The final answer is 1950. I hope it is correct.\n\nSubproblem 1: What is the softening temperature for silica glass in Celsius?", "solution": "\\boxed{1700}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 226}
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{"problem": "Preamble: Two lasers generate radiation of (1) $9.5 \\mu {m}$ and (2) $0.1 \\mu {m}$ respectively.\n\nSubproblem 0: Determine the photon energy (in eV, to two decimal places) of the laser generating radiation of $9.5 \\mu {m}$.\n\n\nSolution: \\[\n\\begin{aligned}\n{E} &={h} v=\\frac{{hc}}{\\lambda} {J} \\times \\frac{1 {eV}}{1.6 \\times 10^{-19} {~J}} \\\\\n{E}_{1} &=\\frac{{hc}}{9.5 \\times 10^{-6}} \\times \\frac{1}{1.6 \\times 10^{-19}} {eV}= \\boxed{0.13} {eV}\n\\end{aligned}\n\\]\n\nFinal answer: The final answer is 0.13. I hope it is correct.\n\nSubproblem 1: Determine the photon energy (in eV, to one decimal place) of the laser generating radiation of $0.1 \\mu {m}$.", "solution": "\\[\n\\begin{aligned}\n{E} &={h} v=\\frac{{hc}}{\\lambda} {J} \\times \\frac{1 {eV}}{1.6 \\times 10^{-19} {~J}} \\\\\n{E}_{2} &=\\frac{{hc}}{0.1 \\times 10^{-6}} \\times \\frac{1}{1.6 \\times 10^{-19}} {eV}= \\boxed{12.4} {eV}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 227}
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{"problem": "Preamble: $\\mathrm{Bi}_{2} \\mathrm{~S}_{3}$ dissolves in water according to the following reaction:\n\\[\n\\mathrm{Bi}_{2} \\mathrm{~S}_{3}(\\mathrm{~s}) \\Leftrightarrow 2 \\mathrm{Bi}^{3+}(\\mathrm{aq})+3 \\mathrm{~s}^{2-}(\\mathrm{aq})\n\\]\nfor which the solubility product, $\\mathrm{K}_{\\mathrm{sp}}$, has the value of $1.6 \\times 10^{-72}$ at room temperature.\n\nAt room temperature how many moles of $\\mathrm{Bi}_{2} \\mathrm{~S}_{3}$ will dissolve in $3.091 \\times 10^{6}$ liters of water? Please format your answer as $n \\times 10^x$ where $n$ is to 1 decimal place.", "solution": "$\\mathrm{Bi}_{2} \\mathrm{~S}_{3}=2 \\mathrm{Bi}^{3+}(\\mathrm{aq})+3 \\mathrm{~S}^{2-}(\\mathrm{aq})$\n\\[\n\\therefore\\left[\\mathrm{Bi}^{3+}\\right]=2 \\mathrm{C}_{\\mathrm{s}} \\text { and }\\left[\\mathrm{s}^{2}\\right]=3 \\mathrm{C}_{\\mathrm{s}}\n\\]\n\\[\n\\begin{aligned}\n& \\therefore \\mathrm{K}_{\\mathrm{sp}}=\\left(2 \\mathrm{C}_{\\mathrm{s}}\\right)^{2}\\left(3 \\mathrm{C}_{\\mathrm{s}}\\right)^{3}=4 \\mathrm{C}_{\\mathrm{s}}^{2} \\cdot 27 \\mathrm{C}_{\\mathrm{s}}^{3}=108 \\mathrm{C}_{\\mathrm{s}}^{5} \\\\\n& \\therefore C_{\\mathrm{s}}=\\left(\\frac{\\mathrm{K}_{\\mathrm{sp}}}{108}\\right)^{1 / 5}=1.715 \\times 10^{-15} \\mathrm{~mol} / \\mathrm{L} \\\\\n& \\therefore \\text { in } 3.091 \\times 10^{6} \\mathrm{~L} \\Rightarrow \\boxed{5.3e-9} \\mathrm{~mol} \\mathrm{Bi}_{2} \\mathrm{~S}_{3}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 228}
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{"problem": "Whiskey, suspected to be of the \"moonshine\" variety, is analyzed for its age by determining its amount of naturally occurring tritium (T) which is a radioactive hydrogen isotope $\\left({ }^{3} \\mathrm{H}\\right)$ with a half-life of $12.5$ years. In this \"shine\" the activity is found to be $6 \\%$ of that encountered in fresh bourbon. What is the age (in years) of the whiskey in question?", "solution": "\\[\n\\begin{aligned}\n&\\frac{c_{o}}{c}=e^{k t} ; c=0.06 c_{0} \\\\\n&\\ln \\frac{c_{0}}{0.06 c_{0}}=k t_{x} \\\\\n&\\ln 0.06=-k_{x} \\\\\n&t_{x}=-\\frac{\\ln 0.06}{\\frac{\\ln 2}{t_{1 / 2}}}=\\frac{\\ln 0.06}{\\frac{0.693}{12.5}}= \\boxed{50.7} \\text { years }\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 229}
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{"problem": "Subproblem 0: What is the working temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1950}.\n\nFinal answer: The final answer is 1950. I hope it is correct.\n\nSubproblem 1: What is the softening temperature for silica glass in Celsius?\n\n\nSolution: \\boxed{1700}.\n\nFinal answer: The final answer is 1700. I hope it is correct.\n\nSubproblem 2: What is the working temperature for Pyrex in Celsius?\n\n\nSolution: \\boxed{1200}.\n\nFinal answer: The final answer is 1200. I hope it is correct.\n\nSubproblem 3: What is the softening temperature for Pyrex in Celsius?", "solution": "\\boxed{800}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 230}
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{"problem": "Preamble: A first-order chemical reaction is found to have an activation energy $\\left(E_{A}\\right)$ of 250 $\\mathrm{kJ} /$ mole and a pre-exponential (A) of $1.7 \\times 10^{14} \\mathrm{~s}^{-1}$.\n\nSubproblem 0: Determine the rate constant at $\\mathrm{T}=750^{\\circ} \\mathrm{C}$. Round your answer to 1 decimal place, in units of $\\mathrm{s}^{-1}$.\n\n\nSolution: $\\mathrm{k}=\\mathrm{Ae} \\mathrm{e}^{-\\frac{\\mathrm{E}_{\\mathrm{A}}}{\\mathrm{RT}}}=1.7 \\times 10^{14} \\times \\mathrm{e}^{-\\frac{2.5 \\times 10^{5}}{8.31 \\times 10^{23}}}= \\boxed{28.8} \\mathrm{~s}^{-1}$\n\nFinal answer: The final answer is 28.8. I hope it is correct.\n\nSubproblem 1: What percent of the reaction will be completed at $600^{\\circ} \\mathrm{C}$ in a period of 10 minutes?", "solution": "Requires knowledge of $k_{600}$ :\n\\[\n\\begin{aligned}\n&\\mathrm{k}_{600}=1.7 \\times 10^{14} \\times \\mathrm{e}^{-\\frac{2.5 \\times 10^{5}}{8.31 \\times 873}}=0.184 \\\\\n&\\frac{\\mathrm{c}}{\\mathrm{c}_{0}}=\\mathrm{e}^{-\\mathrm{kt}}=\\mathrm{e}^{-0.184 \\times 600}=1.3 \\times 10^{-48} \\approx 0\n\\end{aligned}\n\\]\n$c=0$ means the reaction is essentially $ \\boxed{100} \\%$ complete.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 231}
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{"problem": "Determine the energy gap (in eV) between the electronic states $n=7$ and $n=8$ in hydrogen. Please format your answer as $n \\times 10^x$ where $n$ is to 1 decimal place.", "solution": "Here we need to know the \"basis\" of the Rydberg equation [ $E_{e l}=-\\left(1 / n^{2}\\right) K$ ] and $1 {eV}=1.6 \\times 10^{-19} {~J}$ :\n\\[\n\\begin{aligned}\n&\\Delta {E}_{{el}}={K}\\left(\\frac{1}{{n}_{{i}}^{2}}-\\frac{1}{{n}_{{f}}^{2}}\\right)=2.18 \\times 10^{-18}\\left(\\frac{1}{49}-\\frac{1}{64}\\right)=1.043 \\times 10^{-20} {~J} \\\\\n&\\Delta {E}_{{el}}=1.043 \\times 10^{-20} {~J} \\times \\frac{1 {eV}}{\\left(1.6 \\times 10^{-19} {~J}\\right)}= \\boxed{6.5e-2} {eV}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 232}
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{"problem": "Preamble: The decay rate of ${ }^{14} \\mathrm{C}$ in living tissue is $15.3$ disintegrations per minute per gram of carbon. Experimentally, the decay rate can be measured to $\\pm 0.1$ disintegrations per minute per gram of carbon. The half-life of ${ }^{14} \\mathrm{C}$ is 5730 years.\n\nWhat is the maximum age of a sample that can be dated, in years?", "solution": "Radioactive decay is a $1^{\\text {st }}$ order reaction which can be modeled as:\n\\[\n-\\frac{d c}{d t}=k c \\text { or } c=c_{0} e^{-k t}\n\\]\nWith a little algebra we can get an expression for the relationship between time, $\\mathrm{t}$, and the instant value of the decay rate.\nAt any time, t, we can write $\\quad-\\frac{\\mathrm{dc}}{\\mathrm{dt}}=\\mathrm{kc}=\\mathrm{kc}_{0} \\mathrm{e}^{-\\mathrm{kt}}$\nand at time zero,\n\\[\n-\\frac{d c}{d t}=k c_{0}\n\\]\nDivide eq. 1 by eq. 2 to get\nwhere to reduce clutter let $r=\\frac{d c}{d t}$\nTake the logarithm of both sides of eq. 3 and substitute $k=\\frac{\\ln 2}{t_{1 / 2}}$.\nWith rearrangement, this gives $\\quad t=-\\frac{t_{1 / 2}}{\\ln 2} \\times \\ln \\frac{r_{t}}{r_{0}}$\nSo, for the oldest specimen we would measure the minimum instant decay rate of $0.1 \\pm 0.1$ disintegrations per minute per gram. Set this equal to $r_{t}$ in eq. 4 and solve for $t$ to get $\\boxed{41585} \\pm 5730$ years.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 233}
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{"problem": "Estimate the ionic radius of ${Cs}^{+}$ in Angstroms to 2 decimal places. The lattice energy of $\\mathrm{CsCl}$ is $633 \\mathrm{~kJ} / \\mathrm{mol}$. For $\\mathrm{CsCl}$ the Madelung constant, $\\mathrm{M}$, is $1.763$, and the Born exponent, $\\mathrm{n}$, is 10.7. The ionic radius of $\\mathrm{Cl}^{-}$is known to be $1.81 \\AA$.", "solution": "\\[\n\\mathrm{E}_{\\text {lattice }}=\\frac{\\mathrm{Mq}_{1} \\mathrm{q}_{2}}{4 \\pi \\varepsilon_{0} r_{\\mathrm{o}}}\\left(1-\\frac{1}{\\mathrm{n}}\\right) \\text { and } \\mathrm{r}_{\\mathrm{o}}=\\mathrm{r}_{\\mathrm{Cs}^{+}}+\\mathrm{r}_{\\mathrm{Cl}}\n\\]\nSolve first for $r_{0}$\n\\[\n\\begin{aligned}\nr_{0} &=\\frac{M q_{1} q_{2} N_{A v}}{4 \\pi \\varepsilon_{0} E_{\\text {lattice }}}\\left(1-\\frac{1}{n}\\right)=\\frac{1.763\\left(1.6 \\times 10^{-19}\\right)^{2} 6.02 \\times 10^{23}}{4 \\pi 8.85 \\times 10^{-12} 6.33 \\times 10^{5}}\\left(1-\\frac{1}{10.7}\\right) \\\\\n&=3.50 \\times 10^{-10} \\mathrm{~m}=3.50 \\AA=r_{\\mathrm{Cs}^{+}}+r_{\\mathrm{Cr}} \\\\\n\\therefore & r_{\\mathrm{Cs}^{+}}=3.50-1.81=\\boxed{1.69} \\AA\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 234}
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{"problem": "Given the ionic radii, $\\mathrm{Cs}^{+}=1.67 \\AA, \\mathrm{Cl}^{-}=1.81 \\AA$, and the Madelung constant $\\mathrm{M}(\\mathrm{CsCl})=1.763$, determine to the best of your ability the molar Crystal energy ( $\\Delta \\mathrm{E}_{\\text {cryst }}$ ) for $\\mathrm{CsCl}$. Please format your answer as $n \\times 10^x$ where n is to 2 decimal places; answer in $\\mathrm{J} / \\text{mole}$.", "solution": "Given the radii $\\mathrm{Cs}^{+}=1.67 \\AA$ and $\\mathrm{Cl}^{-}=1.81 \\AA$, we can assume that $\\mathrm{r}_{0}$ is the sum of the two. However, we need to know the exponential constant of the repulsive term which is not provided. Considering only the attractive force:\n\\[\n\\begin{array}{ll}\n\\Delta \\mathrm{E}_{\\text {cryst }}=\\frac{-\\mathrm{e}^{2} \\mathrm{~N}_{\\mathrm{A}} \\mathrm{MQ}_{1} \\mathrm{Q}_{2}}{4 \\pi \\varepsilon_{0} r_{0}} & \\text { where: } \\mathrm{Q}_{1}=\\mathrm{Q}_{2}=1 \\\\\n& \\mathrm{M}=1.763 \\\\\n& \\mathrm{~N}_{\\mathrm{A}}=6.02 \\times 10^{23} \\text { particle/mole }\n\\end{array}\n\\]\n\\[\n\\begin{aligned}\n& \\Delta \\mathrm{E}_{\\text {cryst }}=\\frac{-\\left(1.6 \\times 10^{-19} \\mathrm{coul}\\right)^{2} \\times 6.02 \\times 10^{23} \\times 1.763 \\times 1 \\times 1}{4 \\pi 8.85 \\times 10^{-12} \\times(1.81+1.67) \\times 10^{-10} \\mathrm{~m}} \\\\\n& = \\boxed{7.02e5} \\mathrm{~J} / \\text { mole }\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 235}
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{"problem": "Determine the amount (in grams) of boron (B) that, substitutionally incorporated into $1 \\mathrm{~kg}$ of germanium (Ge), will establish a charge carrier density of $3.091 \\mathrm{x}$ $10^{17} / \\mathrm{cm}^{3}$. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "The periodic table gives the molar volume of Ge as $13.57 \\mathrm{~cm}^{3}$ and 1 mole of Ge weighs $72.61 \\mathrm{~g}$, so set up the ratio $\\frac{72.61}{13.6}=\\frac{1000 \\mathrm{~g}}{\\mathrm{x}}$ and solve for $\\mathrm{x}$ to get $187.30$ $\\mathrm{cm}^{3}$ for the total volume. The addition of boron gives 1 charge carrier/B atom.\n$\\rightarrow \\mathrm{B}$ concentration in Si must be $3.091 \\times 10^{17} \\mathrm{~B} / \\mathrm{cm}^{3}$\n$\\mathrm{N}_{\\mathrm{A}}$ of $B$ atoms weighs $10.81 \\mathrm{~g}$\n$\\therefore 3.091 \\times 10^{17} \\mathrm{~B}$ atoms weigh $\\frac{3.091 \\times 10^{17}}{6.02 \\times 10^{23}} \\times 10.81=5.55 \\times 10^{-6} \\mathrm{~g}$\n$\\therefore$ for every $1 \\mathrm{~cm}^{3}$ of Ge, add $5.55 \\times 10^{-6} \\mathrm{~g} \\mathrm{~B}$\n$\\rightarrow$ for $187.30 \\mathrm{~cm}^{3}$ of Ge, add $187.30 \\times 5.55 \\times 10^{-6}= \\boxed{1.04e-3} \\mathrm{~g} \\mathrm{~B}$", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 236}
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{"problem": "Subproblem 0: Is an energy level of $-1.362 \\times 10^{-19} {~J}$ an allowed electron energy state in atomic hydrogen?\n\n\nSolution: $E_{e l} =-\\frac{1}{n^{2}} {~K}$ \\\\\n$-1.362 \\times 10^{-19} {~J}=-\\frac{1}{{n}^{2}} \\times 2.18 \\times 10^{-18} {~J}$\\\\\n${n} &=\\sqrt{\\frac{2.18 \\times 10^{-18}}{1.362 \\times 10^{-19}}}=4.00$\\\\\nThe answer is \\boxed{Yes}.\n\nFinal answer: The final answer is Yes. I hope it is correct.\n\nSubproblem 1: If your answer is yes, determine its principal quantum number $(n)$. If your answer is no, determine ${n}$ for the \"nearest allowed state\".", "solution": "n = \\boxed{4}.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 237}
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{"problem": "Determine the highest linear density of atoms (atoms/m) encountered in vanadium (V). Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "\\[\n\\begin{aligned}\n&\\mathrm{V}: \\quad \\text { atomic weight }=50.94 \\mathrm{~g} / \\text { mole } \\\\\n&\\rho=5.8 \\mathrm{~g} / \\mathrm{cm}^{3}\n\\end{aligned}\n\\]\n$B C C$, so $n=2$\nThe highest density would be found in the [111] direction. To find \"a\":\n\\[\n\\begin{aligned}\n&\\frac{\\text { atomic weight }}{\\rho}=a^{3} \\frac{N_{A}}{n} \\rightarrow a^{3}=\\frac{50.94 \\times 2}{5.8 \\times 6.023 \\times 10^{23}} \\\\\n&a=3.08 \\times 10^{-8} \\mathrm{~cm}=3.08 \\times 10^{-10} \\mathrm{~m}\n\\end{aligned}\n\\]\nThe length in the [111] direction is $\\mathrm{a} \\sqrt{3}$, so there are:\n\\[\n\\begin{aligned}\n&2 \\text { atoms } / \\mathrm{a} \\sqrt{3}=2 \\text { atoms/ }\\left(3.08 \\times 10^{-10} \\mathrm{~m} \\times \\sqrt{3}\\right) \\\\\n&= \\boxed{3.75e9} \\text { atoms } / \\mathrm{m}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 238}
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{"problem": "Strontium fluoride, $\\mathrm{SrF}_{2}$, has a $\\mathrm{K}_{\\mathrm{sp}}$ value in water of $2.45 \\times 10^{-9}$ at room temperature.\nCalculate the solubility of $\\mathrm{SrF}_{2}$ in water. Express your answer in units of molarity. Please format your answer as $n \\times 10^x$ where $n$ is to 2 decimal places.", "solution": "\\[\n\\begin{aligned}\n&\\mathrm{SrF}_{2}=\\mathrm{Sr}^{2+}+2 \\mathrm{~F}^{-} \\quad \\mathrm{K}_{\\mathrm{sp}}=\\left[\\mathrm{Sr}^{2+}\\right]\\left[\\mathrm{F}^{-}\\right]^{2}, \\quad \\text { but }[\\mathrm{F}]=2\\left[\\mathrm{Sr}^{2+}\\right]=2 \\mathrm{c}_{\\mathrm{s}} \\\\\n&\\therefore \\mathrm{K}_{\\mathrm{sp}}=\\mathrm{c}_{\\mathrm{s}}\\left(2 \\mathrm{c}_{\\mathrm{s}}\\right)^{2}=4 \\mathrm{c}_{\\mathrm{s}}^{3} \\quad \\therefore \\quad \\mathrm{c}_{\\mathrm{s}}=\\left(\\frac{\\mathrm{K}_{\\mathrm{sp}}}{4}\\right)^{1 / 3}= \\boxed{8.49e-4} \\mathrm{M}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 239}
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{"problem": "You wish to dope a single crystal of silicon (Si) with boron (B). The specification reads $5 \\times 10^{16}$ boron atoms/ $\\mathrm{cm}^{3}$ at a depth of $25 \\mu \\mathrm{m}$ from the surface of the silicon. What must be the effective concentration of boron in units of atoms/ $\\mathrm{cm}^{3}$ if you are to meet this specification within a time of 90 minutes? Round your answer to 4 decimal places. Assume that initially the concentration of boron in the silicon crystal is zero. The diffusion coefficient of boron in silicon has a value of $7.23 \\times 10^{-9} \\mathrm{~cm}^{2} / \\mathrm{s}$ at the processing temperature.", "solution": "\\[\n\\begin{aligned}\n&c(x, t)=A+B \\text { erf } \\frac{x}{2 \\sqrt{D t}} ; c(0, t)=c_{s}=A ; c(x, 0)=c_{i}=0 \\\\\n&c(\\infty, t)=c_{i}=0=A+B \\rightarrow A=-B \\\\\n&\\therefore c(x, t)=c_{s}-c_{s} \\operatorname{erf} \\frac{x}{2 \\sqrt{D t}}=c_{s} \\operatorname{erfc} \\frac{x}{2 \\sqrt{D t}} \\rightarrow 5 \\times 10^{16}=c_{s} \\text { erfc } \\frac{25 \\times 10^{-4}}{2 \\sqrt{7.23 \\times 10^{-9} \\times 90 \\times 60}} \\\\\n&\\therefore c_{s}=\\frac{5 \\times 10^{16}}{\\operatorname{erfc} \\frac{25 \\times 10^{-4}}{2 \\sqrt{7.23 \\times 10^{-9} \\times 5400}}}=6.43 \\times 10^{16} \\mathrm{~cm}^{-3} \\\\\n&\\operatorname{erfc}(0.20)=1-\\operatorname{erf}(0.20)=1-0.2227=\\boxed{0.7773}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 240}
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{"problem": "An electron beam strikes a crystal of cadmium sulfide (CdS). Electrons scattered by the crystal move at a velocity of $4.4 \\times 10^{5} \\mathrm{~m} / \\mathrm{s}$. Calculate the energy of the incident beam. Express your result in eV, and as an integer. CdS is a semiconductor with a band gap, $E_{g}$, of $2.45$ eV.", "solution": "\\includegraphics[scale=0.5]{set_18_img_01.jpg}\n\\nonessentialimage\n\\[\n\\begin{aligned}\n&E_{\\text {incident } e^{-}}=E_{\\text {emitted } \\mathrm{v}}+E_{\\text {scattered } e^{-}}=E_{g}+\\frac{\\mathrm{mv}^{2}}{2} \\\\\n&=2.45 \\mathrm{eV}+\\frac{1}{2} \\times \\frac{9.11 \\times 10^{-31} \\mathrm{~kg} \\times\\left(4.4 \\times 10^{5} \\mathrm{~m} / \\mathrm{s}\\right)^{2}}{1.6 \\times 10^{-19} \\mathrm{eV} / \\mathrm{J}} \\\\\n&=2.45 \\mathrm{eV}+0.55 \\mathrm{eV}=\\boxed{3} \\mathrm{eV}\n\\end{aligned}\n\\]", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 241}
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{"problem": "Subproblem 0: Determine the inter-ionic equilibrium distance in meters between the sodium and chlorine ions in a sodium chloride molecule knowing that the bond energy is $3.84 \\mathrm{eV}$ and that the repulsive exponent is 8. Please format your answer as $n \\times 10^x$ where $n$ is to 1 decimal place.\n\n\nSolution: $\\mathrm{E}_{\\mathrm{equ}}=-3.84 \\mathrm{eV}=-3.84 \\times 1.6 \\times 10^{-19} \\mathrm{~J}=-\\frac{\\mathrm{e}^{2}}{4 \\pi \\varepsilon_{0} r_{0}}\\left(1-\\frac{1}{\\mathrm{n}}\\right)$\n\\\\\n$r_{0}=\\frac{\\left(1.6 \\times 10^{-19}\\right)^{2}}{4 \\pi 8.85 \\times 10^{-12} \\times 6.14 \\times 10^{-19}}\\left(1-\\frac{1}{8}\\right)= \n\\boxed{3.3e-10} \\mathrm{~m}$\n\nFinal answer: The final answer is 3.3e-10. I hope it is correct.\n\nSubproblem 1: At the equilibrium distance, how much (in percent) is the contribution to the attractive bond energy by electron shell repulsion?", "solution": "Shell \"repulsion\" obviously constitutes a \"negative\" contribution to the bond energy. Looking at the energy equation we find:\n\\[\n\\begin{array}{ll}\n\\text { the attractive term as: } & -E \\times(1)=-E \\\\\n\\text { the repulsion term as: } & -E \\times(-1 / n)=E / n=E / 8\n\\end{array}\n\\]\nThe contribution to the bond energy by the repulsion term $=1 / 8 \\times 100 = \\boxed{12.5}\\%$ Since the bond energy is negative, the $12.5 \\%$ constitute a reduction in bond strength.", "type": "Introduction to Solid State Chemistry (3.091 Fall 2010)", "idx": 242}
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{"problem": "Preamble: A consumer's preferences are representable by the following utility function:\n\\[\n u(x, y)=x^{\\frac{1}{2}}+y\n\\]\n\nObtain the marginal rate of substitution of the consumer at an arbitrary point $(X,Y)$, where $X>0$ and $Y>0$.", "solution": "\\[ M R S=-\\frac{\\frac{1}{2} x^{-\\frac{1}{2}}}{1}=\\boxed{-\\frac{1}{2} X^{-\\frac{1}{2}}} \\]", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 243}
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{"problem": "Preamble: Xiaoyu spends all her income on statistical software $(S)$ and clothes (C). Her preferences can be represented by the utility function: $U(S, C)=4 \\ln (S)+6 \\ln (C)$.\n\nCompute the marginal rate of substitution of software for clothes.", "solution": "We have that $M R S=\\frac{\\frac{4}{S}}{\\frac{6}{C}}=\\boxed{\\frac{2}{3} \\frac{C}{S}}$.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 244}
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{"problem": "What algebraic condition describes a firm that is at an output level that maximizes its profits, given its capital in the short-term? Use standard acronyms in your condition.", "solution": "The required condition is \\boxed{MR=SRMC}, or marginal revenue is equal to short-run marginal cost.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 245}
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{"problem": "Preamble: Moldavia is a small country that currently trades freely in the world barley market. Demand and supply for barley in Moldavia is governed by the following schedules:\nDemand: $Q^{D}=4-P$\nSupply: $Q^{S}=P$\nThe world price of barley is $\\$ 1 /$ bushel.\n\nSubproblem 0: Calculate the free trade equilibrium price of barley in Moldavia, in dollars per bushel. \n\n\nSolution: In free trade, Moldavia will import barley because the world price of $\\$ 1 /$ bushel is lower than the autarkic price of $\\$ 2$ /bushel. Free trade equilibrium price will be \\boxed{1} dollar per bushel.\n\nFinal answer: The final answer is 1. I hope it is correct.\n\nSubproblem 1: Calculate the free trade equilibrium quantity of barley in Moldavia (in bushels).", "solution": "In free trade, Moldavia will import barley because the world price of $\\$ 1 /$ bushel is lower than the autarkic price of $\\$ 2$ /bushel. Free trade equilibrium quantity will be \\boxed{3} bushels, of which 1 is produced at home and 2 are imported.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 246}
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{"problem": "Preamble: Consider the market for apple juice. In this market, the supply curve is given by $Q_{S}=$ $10 P_{J}-5 P_{A}$ and the demand curve is given by $Q_{D}=100-15 P_{J}+10 P_{T}$, where $J$ denotes apple juice, $A$ denotes apples, and $T$ denotes tea.\n\nSubproblem 0: Assume that $P_{A}$ is fixed at $\\$ 1$ and $P_{T}=5$. Calculate the equilibrium price in the apple juice market.\n\n\nSolution: We have the system of equations $Q=10 P_{J}-5 \\cdot 1$ and $Q=100-15 P_{J}+10 \\cdot 5$. Solving for $P_{J}$ we get that $P_{J}=\\boxed{6.2}$.\n\nFinal answer: The final answer is 6.2. I hope it is correct.\n\nSubproblem 1: Assume that $P_{A}$ is fixed at $\\$ 1$ and $P_{T}=5$. Calculate the equilibrium quantity in the apple juice market.", "solution": "We have the system of equations $Q=10 P_{J}-5 \\cdot 1$ and $Q=100-15 P_{J}+10 \\cdot 5$. Solving for $Q$ we get that $Q=\\boxed{57}$.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 247}
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{"problem": "Preamble: Suppose, in the short run, the output of widgets is supplied by 100 identical competitive firms, each having a cost function:\n\\[\nc_{s}(y)=\\frac{1}{3} y^{3}+2\n\\]\nThe demand for widgets is given by:\n\\[\ny^{d}(p)=6400 / p^{\\frac{1}{2}}\n\\]\n\nSubproblem 0: Obtain the short run industry supply function for widgets.\n\n\nSolution: Since $P=M C=y^{2}$, the supply function of each firm is given by $y_{i}^{s}=p^{\\frac{1}{2}}$. \nThe industry supply function is $y^{s}(p)=100 y_{i}^{s}(p)=\\boxed{100 p^{\\frac{1}{2}}}$.\n\nFinal answer: The final answer is 100 p^{\\frac{1}{2}}. I hope it is correct.\n\nSubproblem 1: Obtain the short run equilibrium price of widgets.\n\n\nSolution: $y^{s}=y^{d} \\longrightarrow 100 p^{\\frac{1}{2}}=\\frac{6400}{p^{\\frac{1}{2}}} \\longrightarrow p=\\boxed{64}$. \n\nFinal answer: The final answer is 64. I hope it is correct.\n\nSubproblem 2: Obtain the the output of widgets supplied by each firm.", "solution": "$y^{s}=y^{d} \\longrightarrow 100 p^{\\frac{1}{2}}=\\frac{6400}{p^{\\frac{1}{2}}} \\longrightarrow p=64$. Hence $y^{*}=100 \\cdot 8=800$ and $y_{i}=\\boxed{8}.$", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 248}
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{"problem": "Preamble: Sebastian owns a coffee factory in Argentina. His production function is:\n\\[\nF(K, L)=(K-1)^{\\frac{1}{4}} L^{\\frac{1}{4}}\n\\]\nConsider the cost of capital to be $r$ and the wage to be $w$. Both inputs are variable, and Sebastian faces no fixed costs.\n\nWhat is the marginal rate of technical substitution of labor for capital?", "solution": "\\[\nM R T S=\\frac{M P_{L}}{M P_{K}}=\\boxed{\\frac{K-1}{L}}\n\\]", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 249}
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{"problem": "Preamble: There are two algebraic conditions describing a firm that is at a capital level that minimizes its costs in the long-term.\n\nWrite the condition which involves the SRAC, or short-run average cost?", "solution": "\\boxed{SRAC=LRAC}, short-run average cost equals long-run average cost.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 250}
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{"problem": "Preamble: There are two algebraic conditions describing a firm that is at a capital level that minimizes its costs in the long-term.\n\nSubproblem 0: Write the condition which involves the SRAC, or short-run average cost?\n\n\nSolution: \\boxed{SRAC=LRAC}, short-run average cost equals long-run average cost.\n\nFinal answer: The final answer is SRAC=LRAC. I hope it is correct.\n\nSubproblem 1: Write the condition which involves SRMC, or short-run marginal cost?", "solution": "\\boxed{SRMC=LRMC}, or short-run marginal cost equals long-run levels.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 251}
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{"problem": "Preamble: Suppose, in the short run, the output of widgets is supplied by 100 identical competitive firms, each having a cost function:\n\\[\nc_{s}(y)=\\frac{1}{3} y^{3}+2\n\\]\nThe demand for widgets is given by:\n\\[\ny^{d}(p)=6400 / p^{\\frac{1}{2}}\n\\]\n\nObtain the short run industry supply function for widgets.", "solution": "Since $P=M C=y^{2}$, the supply function of each firm is given by $y_{i}^{s}=p^{\\frac{1}{2}}$. \nThe industry supply function is $y^{s}(p)=100 y_{i}^{s}(p)=\\boxed{100 p^{\\frac{1}{2}}}$.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 252}
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{"problem": "Preamble: Moldavia is a small country that currently trades freely in the world barley market. Demand and supply for barley in Moldavia is governed by the following schedules:\nDemand: $Q^{D}=4-P$\nSupply: $Q^{S}=P$\nThe world price of barley is $\\$ 1 /$ bushel.\n\nCalculate the free trade equilibrium price of barley in Moldavia, in dollars per bushel.", "solution": "In free trade, Moldavia will import barley because the world price of $\\$ 1 /$ bushel is lower than the autarkic price of $\\$ 2$ /bushel. Free trade equilibrium price will be \\boxed{1} dollar per bushel.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 253}
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{"problem": "Preamble: Suppose, in the short run, the output of widgets is supplied by 100 identical competitive firms, each having a cost function:\n\\[\nc_{s}(y)=\\frac{1}{3} y^{3}+2\n\\]\nThe demand for widgets is given by:\n\\[\ny^{d}(p)=6400 / p^{\\frac{1}{2}}\n\\]\n\nSubproblem 0: Obtain the short run industry supply function for widgets.\n\n\nSolution: Since $P=M C=y^{2}$, the supply function of each firm is given by $y_{i}^{s}=p^{\\frac{1}{2}}$. \nThe industry supply function is $y^{s}(p)=100 y_{i}^{s}(p)=\\boxed{100 p^{\\frac{1}{2}}}$.\n\nFinal answer: The final answer is 100 p^{\\frac{1}{2}}. I hope it is correct.\n\nSubproblem 1: Obtain the short run equilibrium price of widgets.", "solution": "$y^{s}=y^{d} \\longrightarrow 100 p^{\\frac{1}{2}}=\\frac{6400}{p^{\\frac{1}{2}}} \\longrightarrow p=\\boxed{64}$.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 254}
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{"problem": "Preamble: A consumer's preferences are representable by the following utility function:\n\\[\n u(x, y)=x^{\\frac{1}{2}}+y\n\\]\n\nSubproblem 0: Obtain the marginal rate of substitution of the consumer at an arbitrary point $(X,Y)$, where $X>0$ and $Y>0$.\n\n\nSolution: \\[ M R S=-\\frac{\\frac{1}{2} x^{-\\frac{1}{2}}}{1}=\\boxed{-\\frac{1}{2} X^{-\\frac{1}{2}}} \\]\n\nFinal answer: The final answer is -\\frac{1}{2} X^{-\\frac{1}{2}}. I hope it is correct.\n\nSubproblem 1: Suppose the price of the second good $(y)$ is 1 , and the price of the first good $(x)$ is denoted by $p>0$. If the consumer's income is $m>\\frac{1}{4p}$, in the optimal consumption bundle of the consumer (in terms of $m$ and $p$ ), what is the quantity of the first good $(x)$?", "solution": "The consumer solves $\\max x^{\\frac{1}{2}}+y$ so that $p x+y=m$. We look for stationary values of the Lagrangian $L=x^{\\frac{1}{2}}+y+\\lambda(m-p x-y)$. The first-order conditions for stationarity are\n\\[\n \\begin{aligned}\n &\\frac{\\partial L}{\\partial x}=\\frac{1}{2} x^{-\\frac{1}{2}}-\\lambda p=0 \\\\\n &\\frac{\\partial L}{\\partial y}=1-\\lambda=0 \\\\\n &\\frac{\\partial L}{\\partial \\lambda}=m-p x-y=0\n \\end{aligned}\n\\]\nCombining the first two equations above gives $\\frac{1}{2 x^{\\frac{1}{2}}}=p$, or $x^{*}=\\frac{1}{4 p^{2}}$. Substituting $x^{*}$ into the budget constraint gives $y=m-p x^{*}=m-\\frac{1}{4 p}$.\nCase 1) $m \\geq \\frac{1}{4 p} \\longrightarrow x^{*}=\\frac{1}{4 p^{2}}$ and $y=m-\\frac{1}{4 p} \\geq 0$.\nCase 2) $m \\leq \\frac{1}{4 p} \\longrightarrow x^{*}=\\frac{m}{p}$ and $y=0$.\nSince we know $m>\\frac{1}{4p}$, we use case 1, in which case our optimal consumption bundle $(x*,y*)$ is $(\\frac{1}{4p^2},m-\\frac{1}{4p})$. So the answer is $\\boxed{\\frac{1}{4p^2}}$.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 255}
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{"problem": "Preamble: Consider the market for apple juice. In this market, the supply curve is given by $Q_{S}=$ $10 P_{J}-5 P_{A}$ and the demand curve is given by $Q_{D}=100-15 P_{J}+10 P_{T}$, where $J$ denotes apple juice, $A$ denotes apples, and $T$ denotes tea.\n\nAssume that $P_{A}$ is fixed at $\\$ 1$ and $P_{T}=5$. Calculate the equilibrium price in the apple juice market.", "solution": "We have the system of equations $Q=10 P_{J}-5 \\cdot 1$ and $Q=100-15 P_{J}+10 \\cdot 5$. Solving for $P_{J}$ we get that $P_{J}=\\boxed{6.2}$.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 256}
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{"problem": "Preamble: In Cambridge, shoppers can buy apples from two sources: a local orchard, and a store that ships apples from out of state. The orchard can produce up to 50 apples per day at a constant marginal cost of 25 cents per apple. The store can supply any remaining apples demanded, at a constant marginal cost of 75 cents per unit. When apples cost 75 cents per apple, the residents of Cambridge buy 150 apples in a day.\n\nAssume that the city of Cambridge sets the price of apples within its borders. What price should it set, in cents?", "solution": "The city should set the price of apples to be $\\boxed{75}$ cents since that is the marginal cost when residents eat at least 50 apples a day, which they do when the price is 75 cents or less.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 257}
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{"problem": "Preamble: You manage a factory that produces cans of peanut butter. The current market price is $\\$ 10 /$ can, and you know the following about your costs (MC stands for marginal cost, and ATC stands for average total cost):\n\\[\n\\begin{array}{l}\nMC(5)=10 \\\\\nATC(5)=6 \\\\\nMC(4)=4 \\\\\nATC(4)=4\n\\end{array}\n\\]\n\nA case of food poisoning breaks out due to your peanut butter, and you lose a lawsuit against your company. As punishment, Judge Judy decides to take away all of your profits, and considers the following two options to be equivalent:\ni. Pay a lump sum in the amount of your profits.\nii. Impose a tax of $\\$\\left[P-A T C\\left(q^{*}\\right)\\right]$ per can since that is your current profit per can, where $q^{*}$ is the profit maximizing output before the lawsuit.\nHow much is the tax, in dollars per can?", "solution": "You maximize profits where $P=M C$, and since $P=10=M C(5)$ you would set $q^{*}=5$.\n\\[\n\\pi / q=(P-A T C)=(10-6)=4\n\\]\nThe tax would be $\\$ \\boxed{4} /$ can.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 258}
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{"problem": "Preamble: Suppose there are exactly two consumers (Albie and Bubbie) who demand strawberries. Suppose that Albie's demand for strawberries is given by\n\\[\nq_{a}(p)=p^{\\alpha} f_{a}\\left(I_{a}\\right)\n\\]\nand Bubbie's demand is given by\n\\[\nq_{b}(p)=p^{\\beta} f_{b}\\left(I_{b}\\right)\n\\]\nwhere $I_{a}$ and $I_{b}$ are Albie and Bubbie's incomes, and $f_{a}(\\cdot)$ and $f_{b}(\\cdot)$ are two unknown functions.\n\nFind Albie's (own-price) elasticity of demand, $\\epsilon_{q_{a}, p}$. Use the sign convention that $\\epsilon_{y, x}=\\frac{\\partial y}{\\partial x} \\frac{x}{y}$.", "solution": "\\[\n\\epsilon_{q_{a}, p}=\\frac{\\partial q_{a}}{\\partial p} \\frac{p}{q_{a}(p)}=\\left[\\alpha p^{\\alpha-1} f_{a}\\left(I_{a} s\\right)\\right] \\frac{p}{p^{\\alpha} f_{a}\\left(I_{a}\\right)}=\\boxed{\\alpha}\n\\]", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 259}
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{"problem": "Preamble: You have been asked to analyze the market for steel. From public sources, you are able to find that last year's price for steel was $\\$ 20$ per ton. At this price, 100 million tons were sold on the world market. From trade association data you are able to obtain estimates for the own price elasticities of demand and supply on the world markets as $-0.25$ for demand and $0.5$ for supply. Assume that steel has linear demand and supply curves throughout, and that the market is competitive.\n\nSolve for the equations of demand in this market. Use $P$ to represent the price of steel in dollars per ton, and $X_{d}$ to represent the demand in units of millions of tons.", "solution": "Assume that this is a competitive market and assume that demand and supply are linear. Thus, $X_{d}=a-b P$ and $X_{s}=c+d P$. We know from the equation for own-price elasticity of demand that\n\\[\nE_{Q_{X} P_{X}}=\\frac{d X_{d}}{d P_{X}} \\frac{P_{X}}{X_{d}}=-b \\frac{P_{X}}{X_{d}}=-b \\frac{20}{100}=-0.25\n\\]\nSolving for $b$, then, we have $b=1.25$. Substituting back into the equation for demand, $X_{d}=$ $a-1.25 P$ or $100=a-1.25(20)$. Solving for $a$ we have $a=125$. Hence, the equation for last year's demand is $\\boxed{X_{d}=125-1.25 P}$.", "type": "Principles of Microeconomics (14.01 Fall 2011)", "idx": 260}
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{"problem": "Harmonic Oscillator Subjected to Perturbation by an Electric Field: An electron is connected by a harmonic spring to a fixed point at $x=0$. It is subject to a field-free potential energy\n\\[\nV(x)=\\frac{1}{2} k x^{2} .\n\\]\nThe energy levels and eigenstates are those of a harmonic oscillator where\n\\[\n\\begin{aligned}\n\\omega &=\\left[k / m_{e}\\right]^{1 / 2} \\\\\nE_{v} &=\\hbar \\omega(v+1 / 2) \\\\\n\\psi_{v}(x) &=(v !)^{-1 / 2}\\left(\\hat{\\boldsymbol{a}}^{\\dagger}\\right)^{v} \\psi_{v=0}(x) .\n\\end{aligned}\n\\]\nNow a constant electric field, $E_{0}$, is applied and $V(x)$ becomes\n\\[\nV(x)=\\frac{1}{2} k x^{2}+E_{0} e x \\quad(e>0 \\text { by definition }) .\n\\]\nWrite an expression for the energy levels $E_{v}$ as a function of the strength of the electric field.", "solution": "The total potential, including the interaction with the electric field is\n\\[\nV(x)=\\frac{m \\omega^{2}}{2} x^{2}+E_{0} e x .\n\\]\nWe find its minimum to be\n\\[\n\\begin{aligned}\n\\frac{d V}{d x}=m \\omega^{2} x &+E_{0} e=0 \\\\\n\\Rightarrow x_{\\min } &=\\frac{E_{0} e}{m \\omega^{2}}, \\\\\nV\\left(x_{\\min }\\right) &=\\frac{m \\omega^{2}}{2} \\frac{E_{0}^{2} e^{2}}{m^{2} \\omega^{2}}-\\frac{E_{0}^{2} e^{2}}{m \\omega^{2}} \\\\\n&=\\frac{E_{0}^{2} e^{2}}{2 m \\omega^{2}} .\n\\end{aligned}\n\\]\nDefining the displacement from the minimum $x^{\\prime}=x-x_{\\min }$, we arrive at\n\\[\n\\begin{aligned}\nV\\left(x^{\\prime}\\right) &=\\frac{m \\omega^{2}}{2}\\left(x^{\\prime}-\\frac{E_{0} e}{m \\omega^{2}}\\right)^{2}+E_{0} e\\left(x^{\\prime}-\\frac{E_{0} e}{m \\omega^{2}}\\right) \\\\\n&=\\frac{m \\omega^{2}}{2} x^{\\prime 2}-\\frac{E_{0}^{2} e^{2}}{2 m \\omega^{2}}\n\\end{aligned}\n\\]\nThus, we see that the system is still harmonic! All we have done is to shift the minimum position and minimum energy, but the potential is still quadratic. The harmonic frequency $\\omega$ remains unchanged.\nSince the potential now is a harmonic oscillator with frequency $\\omega$ and a constant offset, we can easily write down the energy levels:\n\\[\nE_{v}=\\boxed{\\hbar \\omega(v+1 / 2)-\\frac{E_{0}^{2} e^{2}}{2 m \\omega^{2}}}\n\\]", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 261}
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{"problem": "Preamble: The following concern the independent particle model. You may find the following set of Coulomb and exchange integrals useful (energies in $\\mathrm{eV}$):\n$\\mathrm{J}_{1 s 1 s}=17.0 Z$ \n$\\mathrm{~J}_{1 s 2 s}=4.8 Z$ \n$\\mathrm{~K}_{1 s 2 s}=0.9 Z$ \n$\\mathrm{~J}_{2 s 2 s}=3.5 Z$ \n$\\mathrm{J}_{1 s 2 p}=6.6 Z$ \n$\\mathrm{~K}_{1 s 2 p}=0.5 Z$ \n$\\mathrm{~J}_{2 s 2 p}=4.4 Z$ \n$\\mathrm{~K}_{2 s 2 p}=0.8 Z$ \n$\\mathrm{J}_{2 p_{i}, 2 p_{i}}=3.9 Z$\n$\\mathrm{~J}_{2 p_{i}, 2 p_{k}}=3.5 Z$\n$\\mathrm{~K}_{2 p_{i}, 2 p_{k}}=0.2 Z i \\neq k$ \n\nUsing the independent particle model, what is the energy difference between the $1 s^{2} 2 p_{x}^{2}$ configuration and the $1 s^{2} 2 s^{2}$ configuration? Give your answer in eV, in terms of $Z$, and round to a single decimal place.", "solution": "We are asked to calculate the energy difference between a $1 s^{2} 2 p_{x}^{2}$ and a $1 s^{2} 2 s^{2}$ configuration. Let's compute the energy for each using the independent particle model\n\\[\n\\begin{aligned}\nE\\left[1 s^{2} 2 p_{x}^{2}\\right]=& \\sum_{i} E_{i}+\\sum_{i, j}^{i>j} \\widetilde{J}_{i j}-\\widetilde{K}_{i j} \\\\\n=& 2 E_{1 s}+2 E_{2 p} \\\\\n&+\\widetilde{J}_{1 s \\alpha, 1 s \\beta}+\\widetilde{J}_{1 s \\alpha, 2 p_{x} \\alpha}+\\widetilde{J}_{1 s \\alpha, 2 p_{x} \\beta}+\\widetilde{J}_{1 s \\beta, 2 p_{x} \\alpha}+\\widetilde{J}_{1 s \\beta, 2 p_{x} \\beta}+\\widetilde{J}_{2 p_{x} \\alpha, 2 p_{x} \\beta} \\\\\n&-\\widetilde{K}_{1 s \\alpha, 1 s \\beta}-\\widetilde{K}_{1 s \\alpha, 2 p_{x} \\alpha}-\\widetilde{K}_{1 s \\alpha, 2 p_{x} \\beta}-\\widetilde{K}_{1 s \\beta, 2 p_{x} \\alpha}-\\widetilde{K}_{1 s \\beta, 2 p_{x} \\beta}-\\widetilde{K}_{2 p_{x} \\alpha, 2 p_{x} \\beta} \\\\\n=& 2 E_{1 s}+2 E_{2 p}+J_{1 s, 1 s}+4 J_{1 s, 2 p}+J_{2 p_{i}, 2 p_{i}}-2 K_{1 s, 2 p} \\\\\nE\\left[1 s^{2} 2 s^{2}\\right]=& 2 E_{1 s}+2 E_{2 s}+J_{1 s, 1 s}+4 J_{1 s, 2 s}+J_{2 s, 2 s}-2 K_{1 s, 2 s} \\\\\n\\Rightarrow \\Delta E=& 4\\left(J_{1 s, 2 p}-J_{1 s, 2 s}\\right)+\\left(J_{2 p_{i}, 2 p_{i}}-J_{2 s, 2 s}\\right)-2\\left(K_{1 s, 2 p}-K_{1 s, 2 s}\\right) \\\\\n=& Z[4(6.6-4.8)-(3.9-3.5)-2(0.5-0.9)] \\\\\n=&+\\boxed{7.6 Z} \\mathrm{eV}\n\\end{aligned}\n\\]", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 262}
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{"problem": "Preamble: A pulsed Nd:YAG laser is found in many physical chemistry laboratories.\n\nFor a $2.00 \\mathrm{~mJ}$ pulse of laser light, how many photons are there at $1.06 \\mu \\mathrm{m}$ (the Nd:YAG fundamental) in the pulse? PAnswer to three significant figures.", "solution": "For $1.06 \\mu \\mathrm{m}$ Light\nEnergy of one photon $=E_{p}=h \\nu ; \\nu=c / \\lambda ; E_{p}=h c / \\lambda$\n\\[\n\\begin{aligned}\n\\lambda &=1.06 \\mu \\mathrm{m}=1.06 \\times 10^{-6} \\mathrm{~m} \\\\\nc &=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s} \\\\\nh &=\\text { Planck's constant }=6.626 \\times 10^{-34} \\mathrm{~kg} \\mathrm{} \\mathrm{m}^{2} / \\mathrm{s}\n\\end{aligned}\n\\]\n$E_{p}=1.88 \\times 10^{-19} \\mathrm{~J}$\n$1.88 \\times 10^{-19} \\mathrm{~J} /$ photon, we want photons/pulse.\n\\[\n\\frac{1}{1.88 \\times 10^{19} \\mathrm{~J} / \\text { photon }} \\times \\frac{2.00 \\times 10^{-3}}{\\text { pulse }}=\\boxed{1.07e16} \\mathrm{photons} / \\mathrm{pulse}\n\\]", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 263}
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{"problem": "Given that the work function of chromium is $4.40 \\mathrm{eV}$, calculate the kinetic energy of electrons in Joules emitted from a clean chromium surface that is irradiated with ultraviolet radiation of wavelength $200 \\mathrm{~nm}$.", "solution": "The chromium surface is irradiated with $200 \\mathrm{~nm}$ UV light. These photons have energy\n\\[\n\\begin{aligned}\nE &=\\frac{h c}{\\lambda}=\\frac{\\left(6.626 \\times 10^{34} \\mathrm{~J} \\cdot \\mathrm{s}\\right)\\left(3 \\times 10^{8} \\mathrm{~m} \\cdot \\mathrm{s}^{-1}\\right)}{200 \\times 10^{-9} \\mathrm{~m}} \\\\\n&=9.94 \\times 10^{-19} \\mathrm{~J} \\\\\n&=6.20 \\mathrm{eV}\n\\end{aligned}\n\\]\nThe photo-ejected electron has kinetic energy\n\\[\nK E=E_{\\text {photon }}-\\phi_{o}=6.20 \\mathrm{eV}-4.40 \\mathrm{eV}=1.80 \\mathrm{eV}=\\boxed{2.88e-19} \\mathrm{~J}\n\\]", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 264}
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{"problem": "Compute the momentum of one $500 \\mathrm{~nm}$ photon using $p_{\\text {photon }}=E_{\\text {photon }} / c$ where $c$ is the speed of light, $c=3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}$, and $\\nu=c / \\lambda$. Express your answer in kilogram meters per second, rounding your answer to three decimal places.", "solution": "\\[\n\\begin{aligned}\np_{\\text {proton }} &=E_{\\text {proton }} / c \\\\\np &=\\text { Momentum } \\\\\nE &=\\text { Energy }=h \\nu \\\\\nc &=\\text { Speed of light, } 3 \\times 10^{8} \\mathrm{~m} / \\mathrm{s}\n\\end{aligned}\n\\]\n\\[\n\\begin{aligned}\n& p_{\\mathrm{PH}}=\\frac{h \\nu}{c} \\quad \\nu=c / \\lambda \\\\\n& p_{\\mathrm{PH}}=h / \\lambda(\\lambda \\text { in meters }), 500 \\mathrm{~nm}=500 \\times 10^{-9} \\mathrm{~m} \\\\\n& p_{\\mathrm{PH}}=h / 500 \\times 10^{-9}=6.626 \\times 10^{-34} / 500 \\times 10^{-9}=\\boxed{1.325e-27} \\mathrm{~kg} \\mathrm{~m} / \\mathrm{s}\n\\end{aligned}\n\\]", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 265}
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{"problem": "Preamble: This problem deals with the H\\\"uckel MO theory of $\\pi$-conjugated systems.\nTo answer each question, you will need to construct the H\u00fcckel MOs for each of the molecules pictured, divide them into sets of occupied and unoccupied orbitals, and determine the relevant properties, such as ground state energy, bond order, etc.\nNOTE: For all parts we take $\\alpha=\\alpha_{\\mathrm{C}}=-11.2 \\mathrm{eV}$ and $\\beta=\\beta_{\\mathrm{CC}}=-0.7 \\mathrm{eV}$.\n\nDetermine the ionization potential of benzene (remember, ionization potential $\\left[\\mathrm{IP}=\\mathrm{E}\\left(\\mathrm{B}^{+}\\right)-\\mathrm{E}(\\mathrm{B})\\right]$), in $\\mathrm{eV}$, rounded to one decimal place. The benzene molecule is shown below:\n\\chemfig{C*6((-H)-C(-H)=C(-H)-C(-H)=C(-H)-C(-H)=)}", "solution": "Let's build the H\u00fcckel MO Hamiltonian from the 6 carbon atoms. The differences between benzene and hexatriene are only connectivity:\n\\[\nH_{\\text {benzene }}=\\left(\\begin{array}{cccccc}\n\\alpha & \\beta & 0 & 0 & 0 & \\beta \\\\\n\\beta & \\alpha & \\beta & 0 & 0 & 0 \\\\\n0 & \\beta & \\alpha & \\beta & 0 & 0 \\\\\n0 & 0 & \\beta & \\alpha & \\beta & 0 \\\\\n0 & 0 & 0 & \\beta & \\alpha & \\beta \\\\\n\\beta & 0 & 0 & 0 & \\beta & \\alpha\n\\end{array}\\right)\n\\]\nWe now substitute $\\alpha$ and $\\beta$ with the values above and find the eigenvalues of the Hamiltonian numerically. The eigenvalues of $\\mathrm{H}_{\\text {benzene }}$ (in $\\mathrm{eV}$ ) are\n\\[\nE^{\\mu}=\\{-12.6,-11.9,-11.9,-10.5,-10.5,-9.8\\}\n\\].\nThe ionization potential in this model is simply the energy of the HOMO of the ground state of each molecule (this is the orbital from which the electron is ejected). Since there are $6 \\pi$-electrons, we can fill the three lowest MOs and the HOMO will be the third lowest. Therefore, the IP of benzene is $\\boxed{11.9} \\mathrm{eV}$", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 266}
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{"problem": "A baseball has diameter $=7.4 \\mathrm{~cm}$. and a mass of $145 \\mathrm{~g}$. Suppose the baseball is moving at $v=1 \\mathrm{~nm} /$ second. What is its de Broglie wavelength\n\\[\n\\lambda=\\frac{h}{p}=\\frac{h}{m \\nu}\n\\]\n? Give answer in meters.", "solution": "\\[\n\\begin{aligned}\nD_{\\text {ball }} &=0.074 m \\\\\nm_{\\text {ball }} &=0.145 \\mathrm{~kg} \\\\\nv_{\\text {ball }} &=1 \\mathrm{~nm} / \\mathrm{s}=1 \\times 10^{-9} \\mathrm{~m} / \\mathrm{s}\n\\end{aligned}\n\\]\nUsing de Broglie:\n\\[\n\\lambda_{\\text {ball }}=\\frac{h}{p}=\\frac{h}{m \\nu}=\\frac{6.626 \\times 10^{-34} \\mathrm{~m}^{2} \\mathrm{~kg} / \\mathrm{s}}{0.145 \\mathrm{~kg} \\cdot 1 \\times 10^{-9} \\mathrm{~m} / \\mathrm{s}}=\\boxed{4.6e-24} \\mathrm{~m}=\\lambda_{\\text {ball }}\n\\]", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 267}
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{"problem": "Preamble: Consider the Particle in an Infinite Box ``superposition state'' wavefunction,\n\\[\n\\psi_{1,2}=(1 / 3)^{1 / 2} \\psi_{1}+(2 / 3)^{1 / 2} \\psi_{2}\n\\]\nwhere $E_{1}$ is the eigen-energy of $\\psi_{1}$ and $E_{2}$ is the eigen-energy of $\\psi_{2}$.\n\nSubproblem 0: Suppose you do one experiment to measure the energy of $\\psi_{1,2}$. List the possible result(s) of your measurement.\n\n\nSolution: Since the only eigenergies are $E_{1}$ and $E_{2}$, the possible outcomes of the measurement are $\\boxed{E_{1},E_{2}}$.\n\nFinal answer: The final answer is E_{1},E_{2}. I hope it is correct.\n\nSubproblem 1: Suppose you do many identical measurements to measure the energies of identical systems in state $\\psi_{1,2}$. What average energy will you observe?", "solution": "\\[\n\\langle E\\rangle =\\boxed{\\frac{1}{3} E_{1}+\\frac{2}{3} E_{2}}\n\\]\nThis value of $\\langle E\\rangle$ is between $E_{1}$ and $E_{2}$ and is the weighted average energy.", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 268}
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{"problem": "Preamble: Consider the Particle in an Infinite Box ``superposition state'' wavefunction,\n\\[\n\\psi_{1,2}=(1 / 3)^{1 / 2} \\psi_{1}+(2 / 3)^{1 / 2} \\psi_{2}\n\\]\nwhere $E_{1}$ is the eigen-energy of $\\psi_{1}$ and $E_{2}$ is the eigen-energy of $\\psi_{2}$.\n\nSuppose you do one experiment to measure the energy of $\\psi_{1,2}$. List the possible result(s) of your measurement.", "solution": "Since the only eigenergies are $E_{1}$ and $E_{2}$, the possible outcomes of the measurement are $\\boxed{E_{1},E_{2}}$.", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 269}
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{"problem": "Preamble: Evaluate the following integrals for $\\psi_{J M}$ eigenfunctions of $\\mathbf{J}^{2}$ and $\\mathbf{J}_{z}$. \n\n$\\int \\psi_{22}^{*}\\left(\\widehat{\\mathbf{J}}^{+}\\right)^{4} \\psi_{2,-2} d \\tau$", "solution": "\\[\n\\begin{gathered}\n\\int \\psi_{22}^{*}\\left(\\hat{J}_{+}\\right)^{4} \\psi_{2,-2} d \\tau=\\int \\psi_{22}^{*} \\sqrt{2(2+1)-(-2)(-2+1)}\\left(\\hat{J}_{+}\\right)^{3} \\psi_{2,-1} d \\tau \\\\\n=\\int \\psi_{22}^{*} \\sqrt{2(2+1)-(-2)(-2+1)} \\sqrt{2(2+1)-(-1)(-1+1)}\\left(\\hat{J}_{+}\\right)^{2} \\psi_{2,0} d \\tau \\\\\n=\\int \\psi_{22}^{*} \\sqrt{2(2+1)-(-2)(-2+1)} \\sqrt{2(2+1)-(-1)(-1+1)} \\\\\n\\times \\sqrt{2(2+1)-(0)(0+1)}\\left(\\hat{J}_{+}\\right) \\psi_{2,1} d \\tau \\\\\n=\\int \\psi_{22}^{*} \\sqrt{2(2+1)-(-2)(-2+1)} \\sqrt{2(2+1)-(-1)(-1+1)} \\\\\n\\times \\sqrt{2(2+1)-(0)(0+1)} \\sqrt{2(2+1)-(1)(1+1)} \\psi_{22} d \\tau \\\\\n=\\sqrt{4} \\times \\sqrt{6} \\times \\sqrt{6} \\times \\sqrt{4} \\int \\psi_{22}^{*} \\psi_{22} d \\tau \\\\\n=\\boxed{24}\n\\end{gathered}\n\\]", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 270}
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{"problem": "Preamble: Consider the 3-level $\\mathbf{H}$ matrix\n\\[\n\\mathbf{H}=\\hbar \\omega\\left(\\begin{array}{ccc}\n10 & 1 & 0 \\\\\n1 & 0 & 2 \\\\\n0 & 2 & -10\n\\end{array}\\right)\n\\]\nLabel the eigen-energies and eigen-functions according to the dominant basis state character. The $\\widetilde{10}$ state is the one dominated by the zero-order state with $E^{(0)}=10, \\tilde{0}$ by $E^{(0)}=0$, and $-\\widetilde{10}$ by $E^{(0)}=-10$ (we will work in units where $\\hbar \\omega = 1$, and can be safely ignored).\n\nUse non-degenerate perturbation theory to derive the energy $E_{\\widetilde{10}}$. Carry out your calculations to second order in the perturbing Hamiltonian, and round to one decimal place.", "solution": "$E_{\\widetilde{10}} = 10 + \\frac{1^2}{10 - 0} = \\boxed{10.1}.$", "type": "Physical Chemistry (5.61 Fall 2017)", "idx": 271}
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{"id": "0", "question": "The graph of the polynomial function $f$, where $y=f(x)$, has $x$-intercepts of $(-6,0)$ and $(6,0)$. Which of the following must be true?", "options": "A) $f(-6)=0$ B) $f(6)=-6$ C) $f(-6)=6$ D) $f(0)=-6$", "Answer": "A"}
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{"id": "1", "question": "$$\\begin{gathered} y=4 x+6 \\\\-5 x-y=21\\end{gathered}$$ What is the solution $(x, y)$ to the given system of equations?", "options": "A) $(-3,-6)$ B) $\\left(-\\frac{5}{3},-\\frac{2}{3}\\right)$ C) $(3,18)$ D) $(15,66)$", "Answer": "A"}
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{"id": "2", "question": "$\\lvert x-10 \\rvert = 0$ What are all the possible solutions to the given equation?", "options": "A) -10 B) 0 C) 10 D) -10 and 10", "Answer": "C"}
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{"id": "3", "question": "$$q=s(r-1)^2$$ The given equation relates the positive numbers $q, r$, and $s$. Which equation gives $r$ in terms of $q$ and $s$, when $r>1$?", "options": "A) $r=1+\\sqrt{\\frac{q}{s}}$ B) $r=1+\\frac{\\sqrt{q}}{s}$ C) $r=-1-\\sqrt{\\frac{q}{s}}$ D) $r=-1-\\frac{\\sqrt{q}}{s}$", "Answer": "A"}
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{"id": "4", "question": "In the relationship between variables $x$ and $y$, each increase of $1$ in the value of $x$ decreases the value of $y$ by 2. When $x=0$, $y=5$. Which equation represents this relationship?", "options": "A) $y=-\\frac{1}{2}x+5$ B) $y=-\\frac{1}{2}x-5$ C) $y=-2x-5$ D) $y=-2x+5$", "Answer": "D"}
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{"id": "5", "question": "An isosceles right triangle has a hypotenuse of length 4 inches. What is the perimeter, in inches, of this triangle?", "options": "A) $2\\sqrt{2}$ B) $4\\sqrt{2}$ C) $4+4\\sqrt{2}$ D) $4+8\\sqrt{2}$", "Answer": "C"}
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{"id": "6", "question": "How many solutions does the equation $4(x-2) = -2(x+4)$ have?", "options": "A) Zero B) Exactly one C) Exactly two D) Infinitely many", "Answer": "B"}
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{"id": "7", "question": "$R(t) = 1,830 - 790(2.71)^{-.18t}$ The function $R$ gives the predicted average rating, expressed as a number of points, in the German chess federation database for a player based on the number of years, $t$, the player has participated in professional chess tournaments. Which of the following represents the predicted average rating of a player who has just entered their first professional chess tournament?", "options": "A) $R(-0.18)$ B) $R(0)$ C) $R(790)$ D) $R(1,830)$", "Answer": "B"}
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{"id": "8", "question": "Alice took 60 minutes to complete a task on her first trial. The time it took Alice to complete the task decreased by 10% of the previous time for each additional trial. Approximately how many minutes will it take Alice to complete the task on her fifth trial?", "options": "A) 50 B) 42 C) 39 D) 35", "Answer": "C"}
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{"id": "9", "question": "$$ \\begin{aligned} & y<\\frac{2}{5} x+3 \\\\& y>\\frac{1}{2} x-6\\end{aligned}$$ In which of the following tables are all the values of $x$ and their corresponding values of $y$ solutions to the system of inequalities shown?", "options": "A) \\begin{tabular}{|r|r|} \\hline$x$ & $y$ \\\\\\hline-2 & -8 \\\\\\hline 0 & -4 \\\\\\hline 4 & 4 \\\\\\hline\\end{tabular} B) \\begin{tabular}{|c|c|}\\hline$x$ & $y$ \\\\\\hline-2 & -8 \\\\\\hline 0 & 4 \\\\\\hline 4 & 4 \\\\\\hline\\end{tabular} C) \\begin{tabular}{|r|r|}\\hline$x$ & $y$ \\\\\\hline-2 & 3 \\\\\\hline 0 & 2 \\\\\\hline 4 & -3 \\\\\\hline\\end{tabular} D) \\begin{tabular}{|r|r|}\\hline$x$ & $y$ \\\\\\hline-2 & 2 \\\\\\hline 0 & -3 \\\\\\hline 4 & 3 \\\\\\hline\\end{tabular}", "Answer": "D"}
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{"id": "10", "question": "Which of the following is equivalent to $(\\sqrt{32})(\\sqrt[5]{64})$?", "options": "A) $6\\left(\\sqrt[7]{2^5}\\right)$ B) $6\\left(\\sqrt[10]{2^7}\\right)$ C) $8\\left(\\sqrt[7]{2^5}\\right)$ D) $8\\left(\\sqrt[10]{2^7}\\right)$", "Answer": "D"}
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{"id": "11", "question": "An object has a mass of 3,300 milligrams. What is the mass of the object in grams? (1 gram = 1,000 milligrams)", "options": "A) 0.33 B) 3.30 C) 33.00 D) 330.00", "Answer": "B"}
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{"id": "12", "question": "On average, one square inch of human skin contains 650 sweat glands. A certain area of skin contains 1,170 sweat glands. Based on this information, which of the following is closest to the size of this area, in square inches?", "options": "A) 0.44 B) 0.56 C) 0.80 D) 1.80", "Answer": "D"}
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{"id": "13", "question": "The table give the heights, in feet, of 5 peaks in the Rocky Mountains and 5 peaks in the Appalachian Mountains. \\begin{tabular}{|l|l|l|l|l|} \\hline $\\begin{array}{l}\\text { Rocky } \\\\\\text { Mountain } \\\\\\text { Peak }\\end{array}$ & $\\begin{array}{l}\\text { Height } \\\\\\text { (in feet) }\\end{array}$ & $\\begin{array}{l}\\text { Appalachian } \\\\\\text { Mountain } \\\\\\text { Peak }\\end{array}$ & $\\begin{array}{l}\\text { Height } \\\\\\text { (in feet) }\\end{array}$ \\\\\\hline $\\begin{array}{l}\\text { Mount } \\\\\\text { Elbert }\\end{array}$ & 14,439 & $\\begin{array}{l}\\text { Mount } \\\\\\text { Mitchell }\\end{array}$ & 6,684 \\\\\\hline $\\begin{array}{l}\\text { Mount } \\\\\\text { Massive }\\end{array}$ & 14,429 & Mount Craig & 6,647 \\\\\\hline $\\begin{array}{l}\\text { Mount } \\\\\\text { Harvard }\\end{array}$ & 14,419 & $\\begin{array}{l}\\text { Clingman's } \\\\\\text { Dome }\\end{array}$ & 6,643 \\\\\\hline $\\begin{array}{l}\\text { Blanca } \\\\\\text { Peak }\\end{array}$ & 14,350 & $\\begin{array}{l}\\text { Mount } \\\\\\text { Guyot }\\end{array}$ & 6,621 \\\\\\hline $\\begin{array}{l}\\text { La Plata } \\\\\\text { Peak }\\end{array}$ & 14,343 & $\\begin{array}{l}\\text { Balsam } \\\\\\text { Cone }\\end{array}$ & 6,611 \\\\\\hline\\end{tabular} What is the height, in meters, of Blanca Peak? (Use 1 meter $=3.28$ feet)", "options": "A) 437.5 B) 4,375 C) 47,045 D) 47,068", "Answer": "B"}
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{"id": "14", "question": "The table give the heights, in feet, of 5 peaks in the Rocky Mountains and 5 peaks in the Appalachian Mountains. \\begin{tabular}{|l|l|l|l|l|} \\hline $\\begin{array}{l}\\text { Rocky } \\\\\\text { Mountain } \\\\\\text { Peak }\\end{array}$ & $\\begin{array}{l}\\text { Height } \\\\\\text { (in feet) }\\end{array}$ & $\\begin{array}{l}\\text { Appalachian } \\\\\\text { Mountain } \\\\\\text { Peak }\\end{array}$ & $\\begin{array}{l}\\text { Height } \\\\\\text { (in feet) }\\end{array}$ \\\\\\hline $\\begin{array}{l}\\text { Mount } \\\\\\text { Elbert }\\end{array}$ & 14,439 & $\\begin{array}{l}\\text { Mount } \\\\\\text { Mitchell }\\end{array}$ & 6,684 \\\\\\hline $\\begin{array}{l}\\text { Mount } \\\\\\text { Massive }\\end{array}$ & 14,429 & Mount Craig & 6,647 \\\\\\hline $\\begin{array}{l}\\text { Mount } \\\\\\text { Harvard }\\end{array}$ & 14,419 & $\\begin{array}{l}\\text { Clingman's } \\\\\\text { Dome }\\end{array}$ & 6,643 \\\\\\hline $\\begin{array}{l}\\text { Blanca } \\\\\\text { Peak }\\end{array}$ & 14,350 & $\\begin{array}{l}\\text { Mount } \\\\\\text { Guyot }\\end{array}$ & 6,621 \\\\\\hline $\\begin{array}{l}\\text { La Plata } \\\\\\text { Peak }\\end{array}$ & 14,343 & $\\begin{array}{l}\\text { Balsam } \\\\\\text { Cone }\\end{array}$ & 6,611 \\\\\\hline\\end{tabular} For the given Appalachian Mountain peaks, the height of the highest peak is approximately what percent greater than the height of the lowest peak?", "options": "A) $1.1 \\%$ B) $9.9 \\%$ C) $73.0 \\%$ D) $101.1 \\%$", "Answer": "A"}
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{"id": "15", "question": "Data set $A: 2,4,6,6,8,12$ Data set B: $2,4,6,6,8,12,26$ Two data sets are shown. Which statement best compares the medians of the data sets?", "options": "A) The median of data set A is greater than the median of data set $B$ B) The median of data set A is less than the median of data set B C) The medians of data sets A and B are equal D) There is not enough information to compare the medians", "Answer": "C"}
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{"id": "16", "question": "$$0.79 x+1.0 y=100$$ The mass of a solution of isopropanol and water is 100 grams. The given equation represents this situation, where $x$ is the volume of isopropanol, in cubic centimeters, and $y$ is the volume of water, in cubic centimeters. If the volume of isopropanol is 70 cubic centimeters, what is the approximate volume of water, in cubic centimeters?", "options": "A) 45 B) 55 C) 70 D) 79", "Answer": "A"}
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{"id": "17", "question": "There are 435 voting members of the US House of Representatives. If $b$ voting members are in favor of a certain bill, which expression represents the percentage of the voting members in favor of the bill?", "options": "A. $100\\left(\\frac{b}{435}\\right)$ B. $100\\left(\\frac{435}{b}\\right)$ C. $435\\left(\\frac{b}{100}\\right)$ D. $435(100 b)$", "Answer": "A"}
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{"id": "18", "question": "$$10(x+120)=120$$ Which of the following equations has the same solution as the given equation?", "options": "A) $x+120=12$ B) $x+120=130$ C) $x+12=12$ D) $x+12=120$", "Answer": "A"}
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{"id": "19", "question": "The given function $C$ models the annual soybean use in China, in millions of metric tons, between 1995 and 2014, where $x$ is the number of years after 1995. $$C(x)=4.3 x+19$$ According to the model, what is the best interpretation of 4.3 in this context?", "options": "A) Each year between 1995 and 2014, China used 4.3 million metric tons of soybeans B) Each year between 1995 and 2014, China's annual use of soybeans increased by 4.3 million metric tons C) China used 4.3 million metric tons of soybeans in 1995 D) China used a total of 4.3 million metric tons of soybeans between 1995 and 2014", "Answer": "B"}
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{"id": "20", "question": "$$ \\begin{gathered} C(x)=50,000+0.75 x \\\\ R(x)=4.75 x \\end{gathered}$$ The given function $C$ models the total cost (sum of fixed cost and variable cost), in dollars, of growing and harvesting $x$ bales of hay on a certain farm. The given function $R$ models the revenue, in dollars, earned from selling $x$ bales of hay. According to the function $R$, how many bales of hay would have to be sold to earn a revenue of $\\$1,425$?", "options": "A) 100 B) 300 C) 500 D) 1,000", "Answer": "B"}
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{"id": "21", "question": "$$ \\begin{gathered} C(x)=50,000+0.75 x \\\\ R(x)=4.75 x \\end{gathered}$$ The given function $C$ models the total cost (sum of fixed cost and variable cost), in dollars, of growing and harvesting $x$ bales of hay on a certain farm. The given function $R$ models the revenue, in dollars, earned from selling $x$ bales of hay. Which of the following inequalities models the number of bales of hay that must be sold to earn a profit of $\\$ 10,000$ or more? (profit $=$ revenue - cost)", "options": "A) $10,000 \\leq 4 x-50,000$ B) $10,000 \\geq 4 x-50,000$ C) $10,000 \\leq 4 x+50,000$ D) $10,000 \\geq 4 x+50,000$", "Answer": "A"}
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{"id": "22", "question": "Which expression is equivalent to $\\left(x^2+4\\right)^2+(x-2)(x+2) ?$", "options": "A) $x^4+x^2+20$ B) $x^4+5 x^2+16$ C) $x^4+9 x^2$ D) $x^4+9 x^2+12$", "Answer": "D"}
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{"id": "23", "question": "$$ \\begin{aligned} & y=4 x+1 \\\\ & y=4 x+3 \\end{aligned}$$ How many solutions does the given system of equations have?", "options": "A) Zero B) Exactly one C) Exactly two D) Infinitely many", "Answer": "A"}
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{"id": "24", "question": "$$ h(x)=3 x+3 $$ Which inequality represents all values of $x$ for which the graph of $y=h(x)$ in the $x y$-plane is above the $x$-axis?", "options": "A) $x<3$ B) $x<-1$ C) $x>-1$ D) $x>3$", "Answer": "C"}
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{"id": "25", "question": "Which quadratic equation has no real solutions?", "options": "A) $3 x^2-3=0$ B) $3 x^2+3 x=0$ C) $3 x^2+3 x+3=0$ D) $3 x^2-6 x+3=0$", "Answer": "C"}
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{"id": "26", "question": "In 1976, there were approximately 1,000 gray wolves in northern Minnesota. The number of gray wolves in northern Minnesota in 2008 was 190% greater than in 1976. Approximately how many gray wolves were in northern Minnesota in 2008?", "options": "A. 1,190 B. 1,900 C. 2,900 D. 19,000", "Answer": "C"}
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{"id": "27", "question": "When the quadratic function $f$ is graphed in the $x y$-plane, where $y=f(x)$, its vertex is $(-2,5)$. One of the $x$-intercepts of this graph is $\\left(-\\frac{7}{3}, 0\\right)$. What is the other $x$-intercept of the graph?", "options": "A. $\\left(-\\frac{13}{3}, 0\\right)$ B. $\\left(-\\frac{5}{3}, 0\\right)$ C. $\\left(\\frac{1}{3}, 0\\right)$ D. $\\left(\\frac{7}{3}, 0\\right)$", "Answer": "B"}
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{"id": "28", "question": "For an exponential function $g$, the value of $g(x)$ decreases by $20 \\%$ for each 1-unit increase in the value of $x$. If $g(2)=16$, which equation could define $g$ ?", "options": "A) $g(x)=16(0.8)^{x-2}$ B) $g(x)=16(0.8)^{x+2}$ C) $g(x)=16(0.2)^{x-2}$ D) $g(x)=16(0.2)^{x+2}$", "Answer": "A"}
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{"id": "29", "question": "Micha and Rana each selected a random sample of students at their school and asked how many soft drink servings each student had consumed the previous week. Micha estimated that the mean number of soft drink servings was 7.1, with an associated margin of error of 1.2. Rana estimated that the mean number of soft drink servings was 8.3, with an associated margin of error of 0.8. Assuming the margins of error were calculated in the same way, which of the following best explains why Rana obtained a smaller margin of error than Micha?", "options": "A. Rana's sample contained more students than Micha's sample contained. B. Rana's sample contained more students who drank soft drinks than Micha's sample contained. C. Rana's sample contained more students who drank exactly seven soft drink servings than Micha's sample contained. D. Rana's sample contained more students who drank exactly eight soft drink servings than Micha's sample contained.", "Answer": "A"}
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{"id": "30", "question": "A circle in the $x y$-plane has its center at $(-3,4)$ and the point $(-2,1)$ lies on the circle. Which equation represents this circle?", "options": "A) $(x-3)^2+(y+4)^2=\\sqrt{10}$ B) $(x+3)^2+(y-4)^2=\\sqrt{10}$ C) $(x-3)^2+(y+4)^2=10$ D) $(x+3)^2+(y-4)^2=10$", "Answer": "D"}
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{"id": "31", "question": "\\begin{tabular}{|c|c|} \\hline$x$ & $h(x)$ \\\\\\hline 2 & 0 \\\\\\hline 4 & 0 \\\\\\hline 6 & 8 \\\\\\hline \\end{tabular} For the quadratic function $h$, the table gives three values of $x$ and their corresponding values of $h(x)$. At what value of $x$ does $h$ reach its minimum?", "options": "A) -1 B) 0 C) 3 D) 4", "Answer": "C"}
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Reference in New Issue
Block a user