domain list | difficulty float64 | problem string | solution string | answer string | source string |
|---|---|---|---|---|---|
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | The integers \(a, b,\) and \(c\) satisfy the equations \(a + 5 = b\), \(5 + b = c\), and \(b + c = a\). What is the value of \(b\)? | Since \(a + 5 = b\), then \(a = b - 5\). Substituting \(a = b - 5\) and \(c = 5 + b\) into \(b + c = a\), we obtain \(b + (5 + b) = b - 5\). Simplifying, we get \(2b + 5 = b - 5\), which gives \(b = -10\). | -10 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | Shuxin begins with 10 red candies, 7 yellow candies, and 3 blue candies. After eating some of the candies, there are equal numbers of red, yellow, and blue candies remaining. What is the smallest possible number of candies that Shuxin ate? | For there to be equal numbers of each colour of candy, there must be at most 3 red candies and at most 3 yellow candies, since there are 3 blue candies to start. Thus, Shuxin ate at least 7 red candies and at least 4 yellow candies. This means that Shuxin ate at least $7+4=11$ candies. We note that if Shuxin eats 7 red candies, 4 yellow candies, and 0 blue candies, there will indeed be equal numbers of each colour. | 11 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1.5 | What is the quantity equivalent to '2% of 1'? | The quantity $2 \%$ is equivalent to the fraction $\frac{2}{100}$, so '2% of 1' is equal to $\frac{2}{100}$. | \frac{2}{100} | fermat |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Eugene swam on Sunday, Monday, and Tuesday. On Monday, he swam for 30 minutes. On Tuesday, he swam for 45 minutes. His average swim time over the three days was 34 minutes. For how many minutes did he swim on Sunday? | Since Eugene swam three times and had an average swim time of 34 minutes, he swam for \( 3 \times 34 = 102 \) minutes in total. Since he swam for 30 minutes and 45 minutes on Monday and Tuesday, then on Sunday, he swam for \( 102 - 30 - 45 = 27 \) minutes. | 27 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | How many of the numbers in Grace's sequence, starting from 43 and each number being 4 less than the previous one, are positive? | We write out the numbers in the sequence until we obtain a negative number: $43,39,35,31,27,23,19,15,11,7,3,-1$. Since each number is 4 less than the number before it, then once a negative number is reached, every following number will be negative. Thus, Grace writes 11 positive numbers in the sequence. | 11 | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | How many of the 200 students surveyed said that their favourite food was sandwiches, given the circle graph results? | Since the angle in the sector representing cookies is $90^{\circ}$, then this sector represents $\frac{1}{4}$ of the total circle. Therefore, 25% of the students chose cookies as their favourite food. Thus, the percentage of students who chose sandwiches was $100\%-30\%-25\%-35\%=10\%$. Since there are 200 students in total, then $200 \times \frac{10}{100}=20$ students said that their favourite food was sandwiches. | 20 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1 | If the perimeter of a square is 28, what is the side length of the square? | Since a square has four equal sides, the side length of a square equals one-quarter of the perimeter of the square. Thus, the side length of a square with perimeter 28 is $28 \div 4 = 7$. | 7 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Storage space on a computer is measured in gigabytes (GB) and megabytes (MB), where $1 \mathrm{~GB} = 1024 \mathrm{MB}$. Julia has an empty 300 GB hard drive and puts 300000 MB of data onto it. How much storage space on the hard drive remains empty? | Since $1 \mathrm{~GB} = 1024 \mathrm{MB}$, then Julia's 300 GB hard drive has $300 \times 1024 = 307200 \mathrm{MB}$ of storage space. When Julia puts 300000 MB of data on the empty hard drive, the amount of empty space remaining is $307200 - 300000 = 7200 \mathrm{MB}$. | 7200 \mathrm{MB} | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | The regular price for a bicycle is $\$320$. The bicycle is on sale for $20\%$ off. The regular price for a helmet is $\$80$. The helmet is on sale for $10\%$ off. If Sandra bought both items on sale, what is her percentage savings on the total purchase? | Since the regular price for the bicycle is $\$320$ and the savings are $20\%$, then the amount of money that Sandra saves on the bicycle is $\$320 \times 20\%=\$320 \times 0.2=\$64$. Since the regular price for the helmet is $\$80$ and the savings are $10\%$, then the amount of money that Sandra saves on the helmet is $\$80 \times 10\%=\$80 \times 0.1=\$8$. The total of the original prices for the bicycle and helmet is $\$320+\$80=\$400$. Sandra's total savings are $\$64+\$8=\$72$. Therefore, her total percentage savings is $\frac{\$72}{\$400} \times 100\%=\frac{72}{4} \times 1\%=18\%$. | 18\% | fermat |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | If points $P, Q, R$, and $S$ are arranged in order on a line segment with $P Q=1, Q R=2 P Q$, and $R S=3 Q R$, what is the length of $P S$? | Since $P Q=1$ and $Q R=2 P Q$, then $Q R=2$. Since $Q R=2$ and $R S=3 Q R$, then $R S=3(2)=6$. Therefore, $P S=P Q+Q R+R S=1+2+6=9$. | 9 | fermat |
[
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 1.5 | In $\triangle ABC$, points $D$ and $E$ lie on $AB$, as shown. If $AD=DE=EB=CD=CE$, what is the measure of $\angle ABC$? | Since $CD=DE=EC$, then $\triangle CDE$ is equilateral, which means that $\angle DEC=60^{\circ}$. Since $\angle DEB$ is a straight angle, then $\angle CEB=180^{\circ}-\angle DEC=180^{\circ}-60^{\circ}=120^{\circ}$. Since $CE=EB$, then $\triangle CEB$ is isosceles with $\angle ECB=\angle EBC$. Since $\angle ECB+\angle CEB+\angle EBC=180^{\circ}$, then $2 \times \angle EBC+120^{\circ}=180^{\circ}$, which means that $2 \times \angle EBC=60^{\circ}$ or $\angle EBC=30^{\circ}$. Therefore, $\angle ABC=\angle EBC=30^{\circ}$. | 30^{\circ} | pascal |
[
"Mathematics -> Algebra -> Algebra -> Polynomial Operations"
] | 1.5 | For every real number $x$, what is the value of the expression $(x+1)^{2} - x^{2}$? | Expanding and simplifying, $(x+1)^{2} - x^{2} = (x^{2} + 2x + 1) - x^{2} = 2x + 1$. | 2x + 1 | fermat |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1 | Calculate the value of $(3,1) \nabla (4,2)$ using the operation ' $\nabla$ ' defined by $(a, b) \nabla (c, d)=ac+bd$. | From the definition, $(3,1) \nabla (4,2)=(3)(4)+(1)(2)=12+2=14$. | 14 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Decimals"
] | 1 | If each of Bill's steps is $ rac{1}{2}$ metre long, how many steps does Bill take to walk 12 metres in a straight line? | Since each of Bill's steps is $ rac{1}{2}$ metre long, then 2 of Bill's steps measure 1 m. To walk 12 m, Bill thus takes $12 imes 2=24$ steps. | 24 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Integers",
"Mathematics -> Number Theory -> Perfect Squares -> Other"
] | 1.5 | For how many positive integers $n$, with $n \leq 100$, is $n^{3}+5n^{2}$ the square of an integer? | For $n^{3}+5n^{2}$ to be the square of an integer, $\sqrt{n^{3}+5n^{2}}$ must be an integer. We know that $\sqrt{n^{3}+5n^{2}}=\sqrt{n^{2}(n+5)}=\sqrt{n^{2}} \sqrt{n+5}=n \sqrt{n+5}$. For $n \sqrt{n+5}$ to be an integer, $\sqrt{n+5}$ must be an integer. In other words, $n+5$ must be a perfect square. Since $n$ is between 1 and 100, then $n+5$ is between 6 and 105. The perfect squares in this range are $3^{2}=9,4^{2}=16, \ldots, 10^{2}=100$. Thus, there are 8 perfect squares in this range. Therefore, there are 8 values of $n$ for which $\sqrt{n+5}$ is an integer, and thus for which $n^{3}+5n^{2}$ is the square of an integer. | 8 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | How many foonies are in a stack that has a volume of $50 \mathrm{~cm}^{3}$, given that each foonie has a volume of $2.5 \mathrm{~cm}^{3}$? | Since the face of a foonie has area $5 \mathrm{~cm}^{2}$ and its thickness is 0.5 cm, then the volume of one foonie is $5 \times 0.5=2.5 \mathrm{~cm}^{3}$. If a stack of foonies has a volume of $50 \mathrm{~cm}^{3}$ and each foonie has a volume of $2.5 \mathrm{~cm}^{3}$, then there are $50 \div 2.5=20$ foonies in the stack. | 20 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | If $x = 2y$ and $y \neq 0$, what is the value of $(x-y)(2x+y)$? | Since $x = 2y$, then $(x-y)(2x+y) = (2y-y)(2(2y)+y) = (y)(5y) = 5y^{2}$. | 5y^{2} | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | A sequence consists of 2010 terms. Each term after the first is 1 larger than the previous term. The sum of the 2010 terms is 5307. When every second term is added up, starting with the first term and ending with the second last term, what is the sum? | We label the terms $x_{1}, x_{2}, x_{3}, \ldots, x_{2009}, x_{2010}$. Suppose that $S$ is the sum of the odd-numbered terms in the sequence; that is, $S=x_{1}+x_{3}+x_{5}+\cdots+x_{2007}+x_{2009}$. We know that the sum of all of the terms is 5307; that is, $x_{1}+x_{2}+x_{3}+\cdots+x_{2009}+x_{2010}=5307$. Next, we pair up the terms: each odd-numbered term with the following even-numbered term. That is, we pair the first term with the second, the third term with the fourth, and so on, until we pair the 2009th term with the 2010th term. There are 1005 such pairs. In each pair, the even-numbered term is one bigger than the odd-numbered term. That is, $x_{2}-x_{1}=1, x_{4}-x_{3}=1$, and so on. Therefore, the sum of the even-numbered terms is 1005 greater than the sum of the oddnumbered terms. Thus, the sum of the even-numbered terms is $S+1005$. Since the sum of all of the terms equals the sum of the odd-numbered terms plus the sum of the even-numbered terms, then $S+(S+1005)=5307$ or $2 S=4302$ or $S=2151$. Thus, the required sum is 2151. | 2151 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | What is the average (mean) number of hamburgers eaten per student if 12 students ate 0 hamburgers, 14 students ate 1 hamburger, 8 students ate 2 hamburgers, 4 students ate 3 hamburgers, and 2 students ate 4 hamburgers? | The mean number of hamburgers eaten per student equals the total number of hamburgers eaten divided by the total number of students. 12 students each eat 0 hamburgers. This is a total of 0 hamburgers eaten. 14 students each eat 1 hamburger. This is a total of 14 hamburgers eaten. 8 students each eat 2 hamburgers. This is a total of 16 hamburgers eaten. 4 students each eat 3 hamburgers. This is a total of 12 hamburgers eaten. 2 students each eat 4 hamburgers. This is a total of 8 hamburgers eaten. Thus, a total of $0 + 14 + 16 + 12 + 8 = 50$ hamburgers are eaten. The total number of students is $12 + 14 + 8 + 4 + 2 = 40$. Therefore, the mean number of hamburgers eaten is $\frac{50}{40} = 1.25$. | 1.25 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | There are 20 students in a class. In total, 10 of them have black hair, 5 of them wear glasses, and 3 of them both have black hair and wear glasses. How many of the students have black hair but do not wear glasses? | Since 10 students have black hair and 3 students have black hair and wear glasses, then a total of $10-3=7$ students have black hair but do not wear glasses. | 7 | pascal |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 1 | After a fair die with faces numbered 1 to 6 is rolled, the number on the top face is $x$. What is the most likely outcome? | With a fair die that has faces numbered from 1 to 6, the probability of rolling each of 1 to 6 is $\frac{1}{6}$. We calculate the probability for each of the five choices. There are 4 values of $x$ that satisfy $x>2$, so the probability is $\frac{4}{6}=\frac{2}{3}$. There are 2 values of $x$ that satisfy $x=4$ or $x=5$, so the probability is $\frac{2}{6}=\frac{1}{3}$. There are 3 values of $x$ that are even, so the probability is $\frac{3}{6}=\frac{1}{2}$. There are 2 values of $x$ that satisfy $x<3$, so the probability is $\frac{2}{6}=\frac{1}{3}$. There is 1 value of $x$ that satisfies $x=3$, so the probability is $\frac{1}{6}$. Therefore, the most likely of the five choices is that $x$ is greater than 2. | x > 2 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | Suppose that $x$ and $y$ are positive numbers with $xy=\frac{1}{9}$, $x(y+1)=\frac{7}{9}$, and $y(x+1)=\frac{5}{18}$. What is the value of $(x+1)(y+1)$? | If we multiply the second and third equations together, we obtain $x(y+1)y(y+1)=\frac{7}{9} \cdot \frac{5}{18}$ or $xy(x+1)(y+1)=\frac{35}{162}$. From the first equation, $xy=\frac{1}{9}$. Therefore, $\frac{1}{9}(x+1)(y+1)=\frac{35}{162}$ or $(x+1)(y+1)=9\left(\frac{35}{162}\right)=\frac{35}{18}$. | \frac{35}{18} | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | When a number is tripled and then decreased by 5, the result is 16. What is the original number? | To get back to the original number, we undo the given operations. We add 5 to 16 to obtain 21 and then divide by 3 to obtain 7. These are the 'inverse' operations of decreasing by 5 and multiplying by 3. | 7 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations",
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | In her last basketball game, Jackie scored 36 points. These points raised the average number of points that she scored per game from 20 to 21. To raise this average to 22 points, how many points must Jackie score in her next game? | Suppose that Jackie had played $n$ games before her last game. Since she scored an average of 20 points per game over these $n$ games, then she scored $20n$ points over these $n$ games. In her last game, she scored 36 points and so she has now scored $20n+36$ points in total. But, after her last game, she has now played $n+1$ games and has an average of 21 points scored per game. Therefore, we can also say that her total number of points scored is $21(n+1)$. Thus, $21(n+1)=20n+36$ or $21n+21=20n+36$ and so $n=15$. This tells us that after 16 games, Jackie has scored $20(15)+36=336$ points. For her average to be 22 points per game after 17 games, she must have scored a total of $17 \cdot 22=374$ points. This would mean that she must score $374-336=38$ points in her next game. | 38 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | Lauren plays basketball with her friends. She makes 10 baskets. Each of these baskets is worth either 2 or 3 points. Lauren scores a total of 26 points. How many 3 point baskets did she make? | Suppose that Lauren makes $x$ baskets worth 3 points each. Since she makes 10 baskets, then $10-x$ baskets that she made are worth 2 points each. Since Lauren scores 26 points, then $3 x+2(10-x)=26$ and so $3 x+20-x=26$ which gives $x=6$. Therefore, Lauren makes 6 baskets worth 3 points. | 6 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1 | An integer $n$ is decreased by 2 and then multiplied by 5. If the result is 85, what is the value of $n$? | We undo each of the operations in reverse order. The final result, 85, was obtained by multiplying a number by 5. This number was $85 \div 5=17$. The number 17 was obtained by decreasing $n$ by 2. Thus, $n=17+2=19$. | 19 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Averages -> Other"
] | 1.5 | Fifty numbers have an average of 76. Forty of these numbers have an average of 80. What is the average of the other ten numbers? | If 50 numbers have an average of 76, then the sum of these 50 numbers is $50(76)=3800$. If 40 numbers have an average of 80, then the sum of these 40 numbers is $40(80)=3200$. Therefore, the sum of the 10 remaining numbers is $3800-3200=600$, and so the average of the 10 remaining numbers is $ rac{600}{10}=60$. | 60 | fermat |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | A $3 \times 3$ table starts with every entry equal to 0 and is modified using the following steps: (i) adding 1 to all three numbers in any row; (ii) adding 2 to all three numbers in any column. After step (i) has been used a total of $a$ times and step (ii) has been used a total of $b$ times, the table appears as \begin{tabular}{|l|l|l|} \hline 7 & 1 & 5 \\ \hline 9 & 3 & 7 \\ \hline 8 & 2 & 6 \\ \hline \end{tabular} shown. What is the value of $a+b$? | Since the second column includes the number 1, then step (ii) was never used on the second column, otherwise each entry would be at least 2. To generate the 1,3 and 2 in the second column, we thus need to have used step (i) 1 time on row 1,3 times on row 2, and 2 times on row 3. This gives: \begin{tabular}{|l|l|l|} \hline 1 & 1 & 1 \\ \hline 3 & 3 & 3 \\ \hline 2 & 2 & 2 \\ \hline \end{tabular} We cannot use step (i) any more times, otherwise the entries in column 2 will increase. Thus, $a=1+3+2=6$. To obtain the final grid from this current grid using only step (ii), we must increase each entry in column 1 by 6 (which means using step (ii) 3 times) and increase each entry in column 3 by 4 (which means using step (ii) 2 times). Thus, $b=3+2=5$. Therefore, $a+b=11$. | 11 | pascal |
[
"Mathematics -> Applied Mathematics -> Statistics -> Probability -> Counting Methods -> Combinations"
] | 1.5 | What is the least number of gumballs that Wally must buy to guarantee that he receives 3 gumballs of the same colour? | It is possible that after buying 7 gumballs, Wally has received 2 red, 2 blue, 1 white, and 2 green gumballs. This is the largest number of each colour that he could receive without having three gumballs of any one colour. If Wally buys another gumball, he will receive a blue or a green or a red gumball. In each of these cases, he will have at least 3 gumballs of one colour. In summary, if Wally buys 7 gumballs, he is not guaranteed to have 3 of any one colour; if Wally buys 8 gumballs, he is guaranteed to have 3 of at least one colour. Therefore, the least number that he must buy to guarantee receiving 3 of the same colour is 8. | 8 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | What is the largest possible value for $n$ if the average of the two positive integers $m$ and $n$ is 5? | Since the average of $m$ and $n$ is 5, then $\frac{m+n}{2}=5$ which means that $m+n=10$. In order for $n$ to be as large as possible, we need to make $m$ as small as possible. Since $m$ and $n$ are positive integers, then the smallest possible value of $m$ is 1, which means that the largest possible value of $n$ is $n=10-m=10-1=9$. | 9 | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | In a survey, 100 students were asked if they like lentils and were also asked if they like chickpeas. A total of 68 students like lentils. A total of 53 like chickpeas. A total of 6 like neither lentils nor chickpeas. How many of the 100 students like both lentils and chickpeas? | Suppose that $x$ students like both lentils and chickpeas. Since 68 students like lentils, these 68 students either like chickpeas or they do not. Since $x$ students like lentils and chickpeas, then $x$ of the 68 students that like lentils also like chickpeas and so $68-x$ students like lentils but do not like chickpeas. Since 53 students like chickpeas, then $53-x$ students like chickpeas but do not like lentils. We know that there are 100 students in total and that 6 like neither lentils nor chickpeas. Since there are 100 students in total, then $(68-x)+x+(53-x)+6=100$ which gives $127-x=100$ and so $x=27$. Therefore, there are 27 students that like both lentils and chickpeas. | 27 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | In the subtraction shown, $K, L, M$, and $N$ are digits. What is the value of $K+L+M+N$? | We work from right to left as we would if doing this calculation by hand. In the units column, we have $L-1$ giving 1. Thus, $L=2$. (There is no borrowing required.) In the tens column, we have $3-N$ giving 5. Since 5 is larger than 3, we must borrow from the hundreds column. Thus, $13-N$ gives 5, which means $N=8$. In the hundreds column, we have $(K-1)-4$ giving 4, which means $K=9$. In the thousands column, we have 5 (with nothing borrowed) minus $M$ giving 4. Thus, $5-M=4$ or $M=1$. Finally, $K+L+M+N=9+2+1+8=20$. | 20 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1 | If $2.4 \times 10^{8}$ is doubled, what is the result? | When $2.4 \times 10^{8}$ is doubled, the result is $2 \times 2.4 \times 10^{8}=4.8 \times 10^{8}$. | 4.8 \times 10^{8} | cayley |
[
"Mathematics -> Geometry -> Plane Geometry -> Polygons"
] | 1 | What is the perimeter of the shaded region in a \( 3 \times 3 \) grid where some \( 1 \times 1 \) squares are shaded? | The top, left and bottom unit squares each contribute 3 sides of length 1 to the perimeter. The remaining square contributes 1 side of length 1 to the perimeter. Therefore, the perimeter is \( 3 \times 3 + 1 \times 1 = 10 \). | 10 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1 | What is 25% of 60? | Expressed as a fraction, $25 \%$ is equivalent to $\frac{1}{4}$. Since $\frac{1}{4}$ of 60 is 15, then $25 \%$ of 60 is 15. | 15 | fermat |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | Solve for $2d$ if $10d + 8 = 528$. | Since $10d + 8 = 528$, then $10d = 520$ and so $\frac{10d}{5} = \frac{520}{5}$ which gives $2d = 104$. | 104 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1.5 | The product \( \left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right) \) is equal to what? | Simplifying, \( \left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)=\left(\frac{2}{3}\right)\left(\frac{3}{4}\right)\left(\frac{4}{5}\right) \). We can simplify further by dividing equal numerators and denominators to obtain a final value of \( \frac{2}{5} \). | \frac{2}{5} | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1.5 | If Kai will celebrate his 25th birthday in March 2020, in what year was Kai born? | Kai was born 25 years before 2020 and so was born in the year $2020 - 25 = 1995$. | 1995 | cayley |
[
"Mathematics -> Geometry -> Plane Geometry -> Angles"
] | 1.5 | The minute hand on a clock points at the 12. After rotating $120^{\circ}$ clockwise, which number will it point at? | Since there are 12 equally spaced numbers and the total angle in a complete circle is $360^{\circ}$, then the angle between two consecutive numbers is $360^{\circ} \div 12=30^{\circ}$. To rotate $120^{\circ}$, the minute hand must move by $120^{\circ} \div 30^{\circ}=4$ numbers clockwise from the 12. Therefore, the hand will be pointing at the 4. | 4 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1 | For what value of $k$ is the line through the points $(3,2k+1)$ and $(8,4k-5)$ parallel to the $x$-axis? | A line segment joining two points is parallel to the $x$-axis exactly when the $y$-coordinates of the two points are equal. Here, this means that $2k+1=4k-5$ and so $6=2k$ or $k=3$. (We can check that when $k=3$, the coordinates of the points are $(3,7)$ and $(8,7)$.) | 3 | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Matilda has a summer job delivering newspapers. She earns \$6.00 an hour plus \$0.25 per newspaper delivered. Matilda delivers 30 newspapers per hour. How much money will she earn during a 3-hour shift? | During a 3-hour shift, Matilda will deliver \( 3 \times 30=90 \) newspapers. Therefore, she earns a total of \( 3 \times \$6.00+90 \times \$0.25=\$18.00+\$22.50=\$40.50 \) during her 3-hour shift. | \$40.50 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1.5 | Simplify $ rac{1}{2+ rac{2}{3}}$. | Evaluating, $ rac{1}{2+ rac{2}{3}}= rac{1}{ rac{8}{3}}= rac{3}{8}$. | \frac{3}{8} | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Percentages -> Other"
] | 1 | What percentage of students did not receive a muffin, given that 38\% of students received a muffin? | Since $38\%$ of students received a muffin, then $100\% - 38\% = 62\%$ of students did not receive a muffin. | 62\% | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1.5 | What fraction of the entire wall is painted red if Matilda paints half of her section red and Ellie paints one third of her section red? | Matilda and Ellie each take $\frac{1}{2}$ of the wall. Matilda paints $\frac{1}{2}$ of her half, or $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the entire wall. Ellie paints $\frac{1}{3}$ of her half, or $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the entire wall. Therefore, $\frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12}$ of the wall is painted red. | \frac{5}{12} | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1.5 | An integer $x$ is chosen so that $3 x+1$ is an even integer. Which of the following must be an odd integer? | If $x$ is an integer for which $3 x+1$ is even, then $3 x$ is odd, since it is 1 less than an even integer. If $3 x$ is odd, then $x$ must be odd (since if $x$ is even, then $3 x$ would be even). If $x$ is odd, then $7 x$ is odd (odd times odd equals odd) and so $7 x+4$ is odd (odd plus even equals odd). Therefore, the one expression which must be odd is $7 x+4$. | 7x+4 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1.5 | A basket contains 12 apples and 15 bananas. If 3 more bananas are added to the basket, what fraction of the fruit in the basket will be bananas? | When 3 bananas are added to the basket, there are 12 apples and 18 bananas in the basket. Therefore, the fraction of the fruit in the basket that is bananas is \( \frac{18}{12+18}=\frac{18}{30}=\frac{3}{5} \). | \frac{3}{5} | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | A movie is 1 hour and 48 minutes long. A second movie is 25 minutes longer than the first. How long is the second movie? | We add 25 minutes to 1 hour and 48 minutes in two steps. First, we add 12 minutes to 1 hour and 48 minutes to get 2 hours. Then we add $25-12=13$ minutes to 2 hours to get 2 hours and 13 minutes. Alternatively, we could note that 1 hour and 48 minutes is $60+48=108$ minutes, and so the time that is 25 minutes longer is 133 minutes, which is $120+13$ minutes or 2 hours and 13 minutes. | 2 ext{ hours and } 13 ext{ minutes} | cayley |
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 1.5 | Simplify the expression $20(x+y)-19(y+x)$ for all values of $x$ and $y$. | Simplifying, we see that $20(x+y)-19(y+x)=20x+20y-19y-19x=x+y$ for all values of $x$ and $y$. | x+y | fermat |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1 | What was the range of temperatures on Monday in Fermatville, given that the minimum temperature was $-11^{\circ} \mathrm{C}$ and the maximum temperature was $14^{\circ} \mathrm{C}$? | Since the maximum temperature was $14^{\circ} \mathrm{C}$ and the minimum temperature was $-11^{\circ} \mathrm{C}$, then the range of temperatures was $14^{\circ} \mathrm{C} - (-11^{\circ} \mathrm{C}) = 25^{\circ} \mathrm{C}$. | 25^{\circ} \mathrm{C} | fermat |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1.5 | Karim has 23 candies. He eats $n$ candies and divides the remaining candies equally among his three children so that each child gets an integer number of candies. Which of the following is not a possible value of $n$? | After Karim eats $n$ candies, he has $23-n$ candies remaining. Since he divides these candies equally among his three children, the integer $23-n$ must be a multiple of 3. If $n=2,5,11,14$, we obtain $23-n=21,18,12,9$, each of which is a multiple of 3. If $n=9$, we obtain $23-n=14$, which is not a multiple of 3. Therefore, $n$ cannot equal 9. | 9 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | A factory makes chocolate bars. Five boxes, labelled $V, W, X, Y, Z$, are each packed with 20 bars. Each of the bars in three of the boxes has a mass of 100 g. Each of the bars in the other two boxes has a mass of 90 g. One bar is taken from box $V$, two bars are taken from box $W$, four bars are taken from box $X$, eight bars are taken from box $Y$, and sixteen bars are taken from box $Z$. The total mass of these bars taken from the boxes is 2920 g. Which boxes contain the 90 g bars? | The number of bars taken from the boxes is $1+2+4+8+16=31$. If these bars all had mass 100 g, their total mass would be 3100 g. Since their total mass is 2920 g, they are $3100 \mathrm{~g}-2920 \mathrm{~g}=180 \mathrm{~g}$ lighter. Since all of the bars have a mass of 100 g or of 90 g, then it must be the case that 18 of the bars are each 10 g lighter (that is, have a mass of 90 g). Thus, we want to write 18 as the sum of two of $1,2,4,8,16$ in order to determine the boxes from which the 90 g bars were taken. We note that $18=2+16$ and so the 90 g bars must have been taken from box $W$ and box $Z$. | W \text{ and } Z | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Violet has one-half of the money she needs to buy her mother a necklace. After her sister gives her $\$30$, she has three-quarters of the amount she needs. How much will Violet's father give her? | Violet starts with one-half of the money that she needed to buy the necklace. After her sister gives her money, she has three-quarters of the amount that she needs. This means that her sister gave her $\frac{3}{4}-\frac{1}{2}=\frac{1}{4}$ of the total amount that she needs. Since she now has three-quarters of the amount that she needs, then she still needs one-quarter of the total cost. In other words, her father will give her the same amount that her sister gave her, or $\$30$. | $30 | fermat |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Who is the third oldest among Dhruv, Bev, Elcim, Andy, and Cao given that Dhruv is older than Bev, Bev is older than Elcim, Elcim is younger than Andy, Andy is younger than Bev, and Bev is younger than Cao? | We use $A, B, C, D$, and $E$ to represent Andy, Bev, Cao, Dhruv, and Elcim, respectively. We use the notation $D>B$ to represent the fact "Dhruv is older than Bev". The five sentences give $D>B$ and $B>E$ and $A>E$ and $B>A$ and $C>B$. These show us that Dhruv and Cao are older than Bev, and Elcim and Andy are younger than Bev. This means that two people are older than Bev and two people are younger than Bev, which means that Bev must be the third oldest. | Bev | fermat |
[
"Mathematics -> Algebra -> Prealgebra -> Percentages -> Other"
] | 1.5 | A positive number is increased by $60\%$. By what percentage should the result be decreased to return to the original value? | Solution 1: Suppose that the original number is 100. When 100 is increased by $60\%$, the result is 160. To return to the original value of 100, 160 must be decreased by 60. This percentage is $\frac{60}{160} \times 100\%=\frac{3}{8} \times 100\%=37.5\%$. Solution 2: Suppose that the original number is $x$ for some $x>0$. When $x$ is increased by $60\%$, the result is $1.6x$. To return to the original value of $x$, $1.6x$ must be decreased by $0.6x$. This percentage is $\frac{0.6x}{1.6x} \times 100\%=\frac{3}{8} \times 100\%=37.5\%$. | 37.5\% | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1 | If $10 \%$ of $s$ is $t$, what does $s$ equal? | The percentage $10 \%$ is equivalent to the fraction $\frac{1}{10}$. Therefore, $t=\frac{1}{10} s$, or $s=10 t$. | 10t | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Megan and Hana raced their remote control cars for 100 m. The two cars started at the same time. The average speed of Megan's car was $\frac{5}{4} \mathrm{~m} / \mathrm{s}$. Hana's car finished 5 seconds before Megan's car. What was the average speed of Hana's car? | Megan's car travels 100 m at $\frac{5}{4} \mathrm{~m} / \mathrm{s}$, and so takes $\frac{100 \mathrm{~m}}{5 / 4 \mathrm{~m} / \mathrm{s}}=\frac{400}{5} \mathrm{~s}=80 \mathrm{~s}$. Hana's car completes the 100 m in 5 s fewer, and so takes 75 s. Thus, the average speed of Hana's car was $\frac{100 \mathrm{~m}}{75 \mathrm{~s}}=\frac{100}{75} \mathrm{~m} / \mathrm{s}=\frac{4}{3} \mathrm{~m} / \mathrm{s}$. | \frac{4}{3} \mathrm{~m} / \mathrm{s} | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | The gas tank in Catherine's car is $\frac{1}{8}$ full. When 30 litres of gas are added, the tank becomes $\frac{3}{4}$ full. If the gas costs Catherine $\$ 1.38$ per litre, how much will it cost her to fill the remaining quarter of the tank? | When Catherine adds 30 litres of gasoline, the tank goes from $\frac{1}{8}$ full to $\frac{3}{4}$ full. Since $\frac{3}{4}-\frac{1}{8}=\frac{6}{8}-\frac{1}{8}=\frac{5}{8}$, then $\frac{5}{8}$ of the capacity of the tank is 30 litres. Thus, $\frac{1}{8}$ of the capacity of the tank is $30 \div 5=6$ litres. Also, the full capacity of the tank is $8 \times 6=48$ litres. To fill the remaining $\frac{1}{4}$ of the tank, Catherine must add an additional $\frac{1}{4} \times 48=12$ litres of gas. Because each litre costs $\$ 1.38$, it will cost $12 \times \$ 1.38=\$ 16.56$ to fill the rest of the tank. | \$16.56 | pascal |
[
"Mathematics -> Number Theory -> Factorization"
] | 1.5 | The integer 119 is a multiple of which number? | The ones digit of 119 is not even, so 119 is not a multiple of 2. The ones digit of 119 is not 0 or 5, so 119 is not a multiple of 5. Since $120=3 \times 40$, then 119 is 1 less than a multiple of 3 so is not itself a multiple of 3. Since $110=11 \times 10$ and $121=11 \times 11$, then 119 is between two consecutive multiples of 11, so is not itself a multiple of 11. Finally, $119 \div 7=17$, so 119 is a multiple of 7. | 7 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1.5 | A string has been cut into 4 pieces, all of different lengths. The length of each piece is 2 times the length of the next smaller piece. What fraction of the original string is the longest piece? | Let \(L\) be the length of the string. If \(x\) is the length of the shortest piece, then since each of the other pieces is twice the length of the next smaller piece, then the lengths of the remaining pieces are \(2x, 4x\), and \(8x\). Since these four pieces make up the full length of the string, then \(x+2x+4x+8x=L\) or \(15x=L\) and so \(x=\frac{1}{15}L\). Thus, the longest piece has length \(8x=\frac{8}{15}L\), which is \(\frac{8}{15}\) of the length of the string. | \frac{8}{15} | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1.5 | The expression $(5 \times 5)+(5 \times 5)+(5 \times 5)+(5 \times 5)+(5 \times 5)$ is equal to what? | The given sum includes 5 terms each equal to $(5 \times 5)$. Thus, the given sum is equal to $5 \times(5 \times 5)$ which equals $5 \times 25$ or 125. | 125 | cayley |
[
"Mathematics -> Algebra -> Prealgebra -> Decimals"
] | 1.5 | Which of the following numbers is closest to 1: $ rac{11}{10}$, $ rac{111}{100}$, 1.101, $ rac{1111}{1000}$, 1.011? | When we convert each of the possible answers to a decimal, we obtain 1.1, 1.11, 1.101, 1.111, and 1.011. Since the last of these is the only one greater than 1 and less than 1.1, it is closest to 1. | 1.011 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | The integer 2014 is between which powers of 10? | Since \( 10^{0}=1,10^{1}=10,10^{2}=100,10^{3}=1000,10^{4}=10000 \), and \( 10^{5}=100000 \), then 2014 is between \( 10^{3} \) and \( 10^{4} \). | 10^{3} \text{ and } 10^{4} | fermat |
[
"Mathematics -> Applied Mathematics -> Math Word Problems",
"Mathematics -> Geometry -> Plane Geometry -> Perimeter"
] | 1 | Anna and Aaron walk along paths formed by the edges of a region of squares. How far did they walk in total? | Each square has an area of \( 400 \text{ m}^2 \) and side length 20 m. Anna's path is 400 m and Aaron's path is 240 m. Therefore, the total distance walked is \( 400 + 240 = 640 \text{ m} \). | 640 \text{ m} | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1 | What number should go in the $\square$ to make the equation $\frac{3}{4}+\frac{4}{\square}=1$ true? | For $\frac{3}{4}+\frac{4}{\square}=1$ to be true, we must have $\frac{4}{\square}=1-\frac{3}{4}=\frac{1}{4}$. Since $\frac{1}{4}=\frac{4}{16}$, we rewrite the right side using the same numerator to obtain $\frac{4}{\square}=\frac{4}{16}$. Therefore, $\square=16$ makes the equation true. | 16 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | Which of the following divisions is not equal to a whole number: $\frac{60}{12}$, $\frac{60}{8}$, $\frac{60}{5}$, $\frac{60}{4}$, $\frac{60}{3}$? | Since $\frac{60}{8}=60 \div 8=7.5$, then this choice is not equal to a whole number. Note as well that $\frac{60}{12}=5, \frac{60}{5}=12, \frac{60}{4}=15$, and $\frac{60}{3}=20$ are all whole numbers. | 7.5 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Decimals"
] | 1 | Which number is greater than 0.7? | Each of \( 0.07, -0.41, 0.35, \) and \(-0.9\) is less than 0.7. The number 0.8 is greater than 0.7. | 0.8 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Anca and Bruce drove along a highway. Bruce drove at 50 km/h and Anca at 60 km/h, but stopped to rest. How long did Anca stop? | Bruce drove 200 km in 4 hours. Anca drove the same distance in \( 3 \frac{1}{3} \) hours. The difference is \( \frac{2}{3} \) hours, or 40 minutes. | 40 \text{ minutes} | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | Simplify the expression $(\sqrt{100}+\sqrt{9}) \times(\sqrt{100}-\sqrt{9})$. | Simplifying, $(\sqrt{100}+\sqrt{9}) \times(\sqrt{100}-\sqrt{9})=(10+3) \times(10-3)=13 \times 7=91$. | 91 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1 | Anna walked at a constant rate. If she walked 600 metres in 4 minutes, how far did she walk in 6 minutes? | If Anna walked 600 metres in 4 minutes, then she walked $\frac{600}{4}=150$ metres each minute. Therefore, in 6 minutes, she walked $6 \times 150=900$ metres. | 900 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Which number from the set $\{1,2,3,4,5,6,7,8,9,10,11\}$ must be removed so that the mean (average) of the numbers remaining in the set is 6.1? | The original set contains 11 elements whose sum is 66. When one number is removed, there will be 10 elements in the set. For the average of these elements to be 6.1, their sum must be $10 \times 6.1=61$. Since the sum of the original 11 elements is 66 and the sum of the remaining 10 elements is 61, then the element that must be removed is $66-61=5$. | 5 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | Arrange the numbers $2011, \sqrt{2011}, 2011^{2}$ in increasing order. | Since $2011^{2}=4044121$ and $\sqrt{2011} \approx 44.8$, then the list of numbers in increasing order is $\sqrt{2011}, 2011, 2011^{2}$. (If $n$ is a positive integer with $n>1$, then $n^{2}>n$ and $\sqrt{n}<n$, so the list $\sqrt{n}, n, n^{2}$ is always in increasing order.) | \sqrt{2011}, 2011, 2011^{2} | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Luca mixes 50 mL of milk for every 250 mL of flour to make pizza dough. How much milk does he mix with 750 mL of flour? | We divide the 750 mL of flour into portions of 250 mL. We do this by calculating $750 \div 250 = 3$. Therefore, 750 mL is three portions of 250 mL. Since 50 mL of milk is required for each 250 mL of flour, then $3 \times 50 = 150 \text{ mL}$ of milk is required in total. | 150 \text{ mL} | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | The value of $\sqrt{3^{3}+3^{3}+3^{3}}$ is what? | Since $3^{3}=3 \times 3 \times 3=3 \times 9=27$, then $\sqrt{3^{3}+3^{3}+3^{3}}=\sqrt{27+27+27}=\sqrt{81}=9$. | 9 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1.5 | An integer $x$ is chosen so that $3x+1$ is an even integer. Which of the following must be an odd integer? (A) $x+3$ (B) $x-3$ (C) $2x$ (D) $7x+4$ (E) $5x+3$ | Solution 1: If $x=1$, then $3x+1=4$, which is an even integer. In this case, the five given choices are (A) $x+3=4$, (B) $x-3=-2$, (C) $2x=2$, (D) $7x+4=11$, (E) $5x+3=8$. Of these, the only odd integer is (D). Therefore, since $x=1$ satisfies the initial criteria, then (D) must be the correct answer as the result must be true no matter what integer value of $x$ is chosen that makes $3x+1$ even. Solution 2: If $x$ is an integer for which $3x+1$ is even, then $3x$ is odd, since it is 1 less than an even integer. If $3x$ is odd, then $x$ must be odd (since if $x$ is even, then $3x$ would be even). If $x$ is odd, then $x+3$ is even (odd plus odd equals even), so (A) cannot be correct. If $x$ is odd, then $x-3$ is even (odd minus odd equals even), so (B) cannot be correct. If $x$ is odd, then $2x$ is even (even times odd equals even), so (C) cannot be correct. If $x$ is odd, then $7x$ is odd (odd times odd equals odd) and so $7x+4$ is odd (odd plus even equals odd). If $x$ is odd, then $5x$ is odd (odd times odd equals odd) and so $5x+3$ is even (odd plus odd equals even), so (E) cannot be correct. Therefore, the one expression which must be odd is $7x+4$. | 7x+4 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | A string has been cut into 4 pieces, all of different lengths. The length of each piece is 2 times the length of the next smaller piece. What fraction of the original string is the longest piece? | Let $L$ be the length of the string. If $x$ is the length of the shortest piece, then since each of the other pieces is twice the length of the next smaller piece, then the lengths of the remaining pieces are $2x, 4x$, and $8x$. Since these four pieces make up the full length of the string, then $x+2x+4x+8x=L$ or $15x=L$ and so $x=\frac{1}{15}L$. Thus, the longest piece has length $8x=\frac{8}{15}L$, which is $\frac{8}{15}$ of the length of the string. | \frac{8}{15} | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1 | The 17th day of a month is Saturday. What was the first day of that month? | Since the 17th day of the month is a Saturday and there are 7 days in a week, then the previous Saturday was the $17-7=10$th day of the month and the Saturday before that was the $10-7=3$rd day of the month. Since the 3rd day of the month was a Saturday, then the 2nd day was a Friday and the 1st day of the month was a Thursday. | Thursday | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Decimals"
] | 1.5 | Sophie has written three tests. Her marks were $73\%$, $82\%$, and $85\%$. She still has two tests to write. All tests are equally weighted. Her goal is an average of $80\%$ or higher. With which of the following pairs of marks on the remaining tests will Sophie not reach her goal: $79\%$ and $82\%$, $70\%$ and $91\%$, $76\%$ and $86\%$, $73\%$ and $83\%$, $61\%$ and $99\%$? | For Sophie's average over 5 tests to be $80\%$, the sum of her marks on the 5 tests must be $5 \times 80\% = 400\%$. After the first 3 tests, the sum of her marks is $73\% + 82\% + 85\% = 240\%$. Therefore, she will reach her goal as long as the sum of her marks on the two remaining tests is at least $400\% - 240\% = 160\%$. The sums of the pairs of marks given are (A) $161\%$, (B) $161\%$, (C) $162\%$, (D) $156\%$, (E) $160\%$. Thus, the pair with which Sophie would not meet her goal is (D). | 73\% and 83\% | pascal |
[
"Mathematics -> Algebra -> Algebra -> Algebraic Expressions"
] | 1 | Which of the following expressions is not equivalent to $3x + 6$? | We look at each of the five choices: (A) $3(x + 2) = 3x + 6$ (B) $\frac{-9x - 18}{-3} = \frac{-9x}{-3} + \frac{-18}{-3} = 3x + 6$ (C) $\frac{1}{3}(3x) + \frac{2}{3}(9) = x + 6$ (D) $\frac{1}{3}(9x + 18) = 3x + 6$ (E) $3x - 2(-3) = 3x + (-2)(-3) = 3x + 6$ The expression that is not equivalent to $3x + 6$ is the expression from (C). | \frac{1}{3}(3x) + \frac{2}{3}(9) | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Starting at 1:00 p.m., Jorge watched three movies. The first movie was 2 hours and 20 minutes long. He took a 20 minute break and then watched the second movie, which was 1 hour and 45 minutes long. He again took a 20 minute break and then watched the last movie, which was 2 hours and 10 minutes long. At what time did the final movie end? | Starting at 1:00 p.m., Jorge watches a movie that is 2 hours and 20 minutes long. This first movie ends at 3:20 p.m. Then, Jorge takes a 20 minute break. This break ends at 3:40 p.m. Then, Jorge watches a movie that is 1 hour and 45 minutes long. After 20 minutes of this movie, it is 4:00 p.m. and there is still 1 hour and 25 minutes left in the movie. This second movie thus ends at 5:25 p.m. Then, Jorge takes a 20 minute break which ends at 5:45 p.m. Finally, Jorge watches a movie that is 2 hours and 10 minutes long. This final movie ends at 7:55 p.m. | 7:55 \text{ p.m.} | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1.5 | Which of the following expressions is equal to an odd integer for every integer $n$? | When $n=1$, the values of the five expressions are 2014, 2018, 2017, 2018, 2019.
When $n=2$, the values of the five expressions are 2011, 2019, 4034, 2021, 2021.
Only the fifth expression $(2017+2n)$ is odd for both of these choices of $n$, so this must be the correct answer.
We note further that since 2017 is an odd integer and $2n$ is always an even integer, then $2017+2n$ is always an odd integer, as required. | 2017+2n | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Decimals"
] | 1.5 | Country music songs are added to a playlist so that now $40\%$ of the songs are Country. If the ratio of Hip Hop songs to Pop songs remains the same, what percentage of the total number of songs are now Hip Hop? | Since $40\%$ of the songs on the updated playlist are Country, then the remaining $100\%-40\%$ or $60\%$ must be Hip Hop or Pop songs. Since the ratio of Hip Hop songs to Pop songs does not change, then $65\%$ of this remaining $60\%$ must be Hip Hop songs. Overall, this is $65\% \times 60\%=0.65 \times 0.6=0.39=39\%$ of the total number of songs on the playlist. | 39\% | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | Sophie has written three tests. Her marks were $73\%$, $82\%$, and $85\%$. She still has two tests to write. All tests are equally weighted. Her goal is an average of $80\%$ or higher. With which of the following pairs of marks on the remaining tests will Sophie not reach her goal: $79\%$ and $82\%$, $70\%$ and $91\%$, $76\%$ and $86\%$, $73\%$ and $83\%$, $61\%$ and $99\%$? | For Sophie's average over 5 tests to be $80\%$, the sum of her marks on the 5 tests must be $5 \times 80\% = 400\%$. After the first 3 tests, the sum of her marks is $73\% + 82\% + 85\% = 240\%$. Therefore, she will reach her goal as long as the sum of her marks on the two remaining tests is at least $400\% - 240\% = 160\%$. The sums of the pairs of marks given are (A) $161\%$, (B) $161\%$, (C) $162\%$, (D) $156\%$, (E) $160\%$. Thus, the pair with which Sophie would not meet her goal is (D). | 73\% and 83\% | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | John lists the integers from 1 to 20 in increasing order. He then erases the first half of the integers in the list and rewrites them in order at the end of the second half of the list. Which integer in the new list has exactly 12 integers to its left? | John first writes the integers from 1 to 20 in increasing order. When he erases the first half of the numbers, he erases the numbers from 1 to 10 and rewrites these at the end of the original list. Therefore, the number 1 has 10 numbers to its left. (These numbers are $11,12, \ldots, 20$.) Thus, the number 2 has 11 numbers to its left, and so the number 3 has 12 numbers to its left. | 3 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Six friends ate at a restaurant and agreed to share the bill equally. Because Luxmi forgot her money, each of her five friends paid an extra \$3 to cover her portion of the total bill. What was the total bill? | Since each of five friends paid an extra \$3 to cover Luxmi's portion of the bill, then Luxmi's share was $5 \times \$3=\$15$. Since each of the six friends had an equal share, then the total bill is $6 \times \$15=\$90$. | \$90 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1.5 | If $x$ is $20 \%$ of $y$ and $x$ is $50 \%$ of $z$, then what percentage is $z$ of $y$? | Since $x$ is $20 \%$ of $y$, then $x=\frac{20}{100} y=\frac{1}{5} y$. Since $x$ is $50 \%$ of $z$, then $x=\frac{1}{2} z$. Therefore, $\frac{1}{5} y=\frac{1}{2} z$ which gives $\frac{2}{5} y=z$. Thus, $z=\frac{40}{100} y$ and so $z$ is $40 \%$ of $y$. | 40 \% | cayley |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | At what speed does Jeff run if Jeff and Ursula each run 30 km, Ursula runs at a constant speed of $10 \mathrm{~km} / \mathrm{h}$, and Jeff's time to complete the 30 km is 1 hour less than Ursula's time? | When Ursula runs 30 km at $10 \mathrm{~km} / \mathrm{h}$, it takes her $\frac{30 \mathrm{~km}}{10 \mathrm{~km} / \mathrm{h}}=3 \mathrm{~h}$.
This means that Jeff completes the same distance in $3 \mathrm{~h}-1 \mathrm{~h}=2 \mathrm{~h}$.
Therefore, Jeff's constant speed is $\frac{30 \mathrm{~km}}{2 \mathrm{~h}}=15 \mathrm{~km} / \mathrm{h}$. | 15 \mathrm{~km} / \mathrm{h} | pascal |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Inequalities"
] | 1 | Which of the following is a possible value of $x$ if given two different numbers on a number line, the number to the right is greater than the number to the left, and the positions of $x, x^{3}$ and $x^{2}$ are marked on a number line? | From the number line shown, we see that $x<x^{3}<x^{2}$.
If $x>1$, then successive powers of $x$ are increasing (that is, $x<x^{2}<x^{3}$ ).
Since this is not the case, then it is not true that $x>1$.
If $x=0$ or $x=1$, then successive powers of $x$ are equal. This is not the case either.
If $0<x<1$, then successive powers of $x$ are decreasing (that is, $x^{3}<x^{2}<x$ ). This is not the case either.
Therefore, it must be the case that $x<0$.
If $x<-1$, we would have $x^{3}<x<0<x^{2}$. This is because when $x<-1$, then $x$ is negative and we have $x^{2}>1$ which gives $x^{3}=x^{2} \times x<1 \times x$. This is not the case here either.
Therefore, it must be the case that $-1<x<0$.
From the given possibilities, this means that $-\frac{2}{5}$ is the only possible value of $x$.
We can check that if $x=-\frac{2}{5}=-0.4$, then $x^{2}=0.16$ and $x^{3}=-0.064$, and so we have $x<x^{3}<x^{2}$. We can also check by substitution that none of the other possible answers gives the correct ordering of $x, x^{2}$ and $x^{3}$. | -\frac{2}{5} | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Elena earns $\$ 13.25$ per hour working at a store. How much does Elena earn in 4 hours? | Elena works for 4 hours and earns $\$ 13.25$ per hour. This means that she earns a total of $4 \times \$ 13.25=\$ 53.00$. | \$53.00 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1 | Which of the following is equal to $110 \%$ of 500? | Solution 1: $10 \%$ of 500 is $\frac{1}{10}$ of 500, which equals 50. Thus, $110 \%$ of 500 equals $500+50$, which equals 550. Solution 2: $110 \%$ of 500 is equal to $\frac{110}{100} \times 500=110 \times 5=550$. | 550 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Ratios and Proportions -> Other",
"Mathematics -> Number Theory -> Other"
] | 1 | The ratio of apples to bananas in a box is $3: 2$. What total number of apples and bananas in the box cannot be equal to? | Since the ratio of apples to bananas is $3: 2$, then we can let the numbers of apples and bananas equal $3n$ and $2n$, respectively, for some positive integer $n$. Therefore, the total number of apples and bananas is $3n + 2n = 5n$, which is a multiple of 5. Of the given choices, only (E) 72 is not a multiple of 5 and so cannot be the total. (Each of the other choices can be the total by picking an appropriate value of $n$.) | 72 | pascal |
[
"Mathematics -> Algebra -> Intermediate Algebra -> Inequalities"
] | 1.5 | If $x$ is a number less than -2, which of the following expressions has the least value: $x$, $x+2$, $\frac{1}{2}x$, $x-2$, or $2x$? | For any negative real number $x$, the value of $2x$ will be less than the value of $\frac{1}{2}x$. Therefore, $\frac{1}{2}x$ cannot be the least of the five values. Thus, the least of the five values is either $x-2$ or $2x$. When $x < -2$, we know that $2x - (x-2) = x + 2 < 0$. Since the difference between $2x$ and $x-2$ is negative, then $2x$ has the smaller value and so is the least of all five values. | 2x | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Integers"
] | 1 | Which number is closest to \(-3.4\) on a number line? | On a number line, \(-3.4\) is between \(-4\) and \(-3\). This means that \(-3.4\) is closer to \(-3\) than to \(-4\), and so the answer is \(-3\). | -3 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Dhruv is older than Bev. Bev is older than Elcim. Elcim is younger than Andy. Andy is younger than Bev. Bev is younger than Cao. Who is the third oldest? | The five sentences give \(D > B\) and \(B > E\) and \(A > E\) and \(B > A\) and \(C > B\). These show us that Dhruv and Cao are older than Bev, and Elcim and Andy are younger than Bev. This means that two people are older than Bev and two people are younger than Bev, which means that Bev must be the third oldest. | Bev | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | Mayar and Rosie are 90 metres apart. Starting at the same time, they run towards each other. Mayar runs twice as fast as Rosie. How far has Mayar run when they meet? | Suppose that Rosie runs \(x\) metres from the time that they start running until the time that they meet. Since Mayar runs twice as fast as Rosie, then Mayar runs \(2x\) metres in this time. When Mayar and Rosie meet, they will have run a total of 90 m, since between the two of them, they have covered the full 90 m. Therefore, \(2x + x = 90\) and so \(3x = 90\) or \(x = 30\). Since \(2x = 60\), this means that Mayar has run 60 m when they meet. | 60 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Fractions"
] | 1 | In the list 7, 9, 10, 11, 18, which number is the average (mean) of the other four numbers? | The average of the numbers \(7, 9, 10, 11\) is \(\frac{7+9+10+11}{4} = \frac{37}{4} = 9.25\), which is not equal to 18, which is the fifth number. The average of the numbers \(7, 9, 10, 18\) is \(\frac{7+9+10+18}{4} = \frac{44}{4} = 11\), which is equal to 11, the remaining fifth number. | 11 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | A sign has 31 spaces on a single line. The word RHOMBUS is written from left to right in 7 consecutive spaces. There is an equal number of empty spaces on each side of the word. Counting from the left, in what space number should the letter $R$ be put? | Since the letters of RHOMBUS take up 7 of the 31 spaces on the line, there are $31-7=24$ spaces that are empty. Since the numbers of empty spaces on each side of RHOMBUS are the same, there are $24 \div 2=12$ empty spaces on each side. Therefore, the letter R is placed in space number $12+1=13$, counting from the left. | 13 | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | Which of the following is closest in value to 7? | We note that $7=\sqrt{49}$ and that $\sqrt{40}<\sqrt{49}<\sqrt{50}<\sqrt{60}<\sqrt{70}<\sqrt{80}$.
This means that $\sqrt{40}$ or $\sqrt{50}$ is the closest to 7 of the given choices.
Since $\sqrt{40} \approx 6.32$ and $\sqrt{50} \approx 7.07$, then $\sqrt{50}$ is closest to 7. | \sqrt{50} | pascal |
[
"Mathematics -> Algebra -> Prealgebra -> Simple Equations"
] | 1.5 | What number did Janet pick if she added 7 to the number, multiplied the sum by 2, subtracted 4, and the final result was 28? | We undo Janet's steps to find the initial number.
To do this, we start with 28, add 4 (to get 32), then divide the sum by 2 (to get 16), then subtract 7 (to get 9).
Thus, Janet's initial number was 9. | 9 | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1 | Who is the tallest if Igor is shorter than Jie, Faye is taller than Goa, Jie is taller than Faye, and Han is shorter than Goa? | Since Igor is shorter than Jie, then Igor cannot be the tallest.
Since Faye is taller than Goa, then Goa cannot be the tallest.
Since Jie is taller than Faye, then Faye cannot be the tallest.
Since Han is shorter than Goa, then Han cannot be the tallest.
The only person of the five who has not been eliminated is Jie, who must thus be the tallest. | Jie | pascal |
[
"Mathematics -> Applied Mathematics -> Math Word Problems"
] | 1.5 | On February 1, it was $16.2^{\circ} \mathrm{C}$ outside Jacinta's house at 3:00 p.m. On February 2, it was $-3.6^{\circ} \mathrm{C}$ outside Jacinta's house at 2:00 a.m. If the temperature changed at a constant rate between these times, what was the rate at which the temperature decreased? | The total decrease in temperature between these times is $16.2^{\circ} \mathrm{C}-\left(-3.6^{\circ} \mathrm{C}\right)=19.8^{\circ} \mathrm{C}$. The length of time between 3:00 p.m. one day and 2:00 a.m. the next day is 11 hours, since it is 1 hour shorter than the length of time between 3:00 p.m. and 3:00 a.m. Since the temperature decreased at a constant rate over this period of time, the rate of decrease in temperature was $\frac{19.8^{\circ} \mathrm{C}}{11 \mathrm{~h}}=1.8^{\circ} \mathrm{C} / \mathrm{h}$. | 1.8^{\circ} \mathrm{C} / \mathrm{h} | pascal |
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