violetxi/omni-math-difficulty-1_2 · Datasets at Hugging Face

domain list	difficulty float64	problem string	solution string	answer string	source string
[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]	1	Morgan uses a spreadsheet to create a table of values. In the first column, she lists the positive integers from 1 to 400. She then puts integers in the second column in the following way: if the integer in the first column of a given row is $n$, the number in the second column of that row is $3 n+1$. Which of the following integers does not appear in the second column: 31, 94, 131, 331, 907?	Since $31=3 imes 10+1$ and $94=3 imes 31+1$ and $331=3 imes 110+1$ and $907=3 imes 302+1$, then each of $31,94,331$, and 907 appear in the second column of Morgan's spreadsheet. Thus, 131 must be the integer that does not appear in Morgan's spreadsheet. (We note that 131 is 2 more than $3 imes 43=129$ so is not 1 more than a multiple of 3.)	131	pascal
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1	Peyton puts 30 L of oil and 15 L of vinegar into a large empty can. He then adds 15 L of oil to create a new mixture. What percentage of the new mixture is oil?	After Peyton has added 15 L of oil, the new mixture contains $30+15=45 \mathrm{~L}$ of oil and 15 L of vinegar. Thus, the total volume of the new mixture is $45+15=60 \mathrm{~L}$. Of this, the percentage that is oil is $\frac{45}{60} \times 100 \%=\frac{3}{4} \times 100 \%=75 \%$.	75\%	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]	1.5	Anna thinks of an integer that is not a multiple of three, not a perfect square, and the sum of its digits is a prime number. What could the integer be?	Since 12 and 21 are multiples of 3 (12 = 4 \times 3 and 21 = 7 \times 3), the answer is not 12 or 21. 16 is a perfect square (16 = 4 \times 4) so the answer is not 16. The sum of the digits of 26 is 8, which is not a prime number, so the answer is not 26. Since 14 is not a multiple of three, 14 is not a perfect square, and the sum of the digits of 14 is 1 + 4 = 5 which is prime, then the answer is 14.	14	fermat
[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]	1.5	Ewan writes out a sequence where he counts by 11s starting at 3. Which number will appear in Ewan's sequence?	Ewan's sequence starts with 3 and each following number is 11 larger than the previous number. Since every number in the sequence is some number of 11s more than 3, this means that each number in the sequence is 3 more than a multiple of 11. Furthermore, every such positive integer is in Ewan's sequence. Since $110 = 11 \times 10$ is a multiple of 11, then $113 = 110 + 3$ is 3 more than a multiple of 11, and so is in Ewan's sequence.	113	fermat
[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]	1	If \( n = 7 \), which of the following expressions is equal to an even integer: \( 9n, n+8, n^2, n(n-2), 8n \)?	When \( n=7 \), we have \( 9n=63, n+8=15, n^2=49, n(n-2)=35, 8n=56 \). Therefore, \( 8n \) is even. For every integer \( n \), the expression \( 8n \) is equal to an even integer.	8n	pascal
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1	In a certain year, July 1 was a Wednesday. What day of the week was July 17 in that year?	Since July 1 is a Wednesday, then July 8 and July 15 are both Wednesdays. Since July 15 is a Wednesday, then July 17 is a Friday.	Friday	pascal
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1.5	A store sells jellybeans at a fixed price per gram. The price for 250 g of jellybeans is $\$ 7.50$. What mass of jellybeans sells for $\$ 1.80$?	The store sells 250 g of jellybeans for $\$ 7.50$, which is 750 cents. Therefore, 1 g of jellybeans costs $750 \div 250=3$ cents. This means that $\$ 1.80$, which is 180 cents, will buy $180 \div 3=60 \mathrm{~g}$ of jellybeans.	60 \mathrm{~g}	cayley
[ "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 -> Geometry -> Plane Geometry -> Area" ]	1.5	A rectangle has length 13 and width 10. The length and the width of the rectangle are each increased by 2. By how much does the area of the rectangle increase?	The area of the original rectangle is $13 imes 10=130$. When the dimensions of the original rectangle are each increased by 2, we obtain a rectangle that is 15 by 12. The area of the new rectangle is $15 imes 12=180$, and so the area increased by $180-130=50$.	50	pascal
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1	If July 3, 2030 is a Wednesday, what day of the week is July 14, 2030?	July 14 is 11 days after July 3 of the same year. Since there are 7 days in a week, then July 10 and July 3 occur on the same day of the week, namely Wednesday. July 14 is 4 days after July 10, and so is a Sunday.	Sunday	pascal
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1	A hiker is exploring a trail. The trail has three sections: the first $25 \%$ of the trail is along a river, the next $\frac{5}{8}$ of the trail is through a forest, and the remaining 3 km of the trail is up a hill. How long is the trail?	Since $25 \%$ is equivalent to $\frac{1}{4}$, then the fraction of the trail covered by the section along the river and the section through the forest is $\frac{1}{4}+\frac{5}{8}=\frac{2}{8}+\frac{5}{8}=\frac{7}{8}$. This means that the final section up a hill represents $1-\frac{7}{8}=\frac{1}{8}$ of the trail. Since $\frac{1}{8}$ of the trail is 3 km long, then the entire trail is $8 \times 3 \mathrm{~km}=24 \mathrm{~km}$ long.	24 \text{ km}	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]	1.5	At the end of which year did Steve have more money than Wayne for the first time?	Steve's and Wayne's amounts of money double and halve each year, respectively. By 2004, Steve has more money than Wayne.	2004	pascal
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1.5	How much money does Roman give Dale if Roman wins a contest with a prize of $\$ 200$, gives $30 \%$ of the prize to Jackie, and then splits $15 \%$ of what remains equally between Dale and Natalia?	To determine $30 \%$ of Roman's $\$ 200$ prize, we calculate $\$ 200 \times 30 \%=\$ 200 \times \frac{30}{100}=\$ 2 \times 30=\$ 60$. After Roman gives $\$ 60$ to Jackie, he has $\$ 200-\$ 60=\$ 140$ remaining. He splits $15 \%$ of this between Dale and Natalia. The total that he splits is $\$ 140 \times 15 \%=\$ 140 \times 0.15=\$ 21$. Since Roman splits $\$ 21$ equally between Dale and Natalia, then Roman gives Dale a total of $\$ 21 \div 2=\$ 10.50$.	\$ 10.50	cayley
[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]	1.5	In 12 years, Janice will be 8 times as old as she was 2 years ago. How old is Janice now?	Suppose that Janice is \( x \) years old now. Two years ago, Janice was \( x - 2 \) years old. In 12 years, Janice will be \( x + 12 \) years old. From the given information \( x + 12 = 8(x - 2) \) and so \( x + 12 = 8x - 16 \) which gives \( 7x = 28 \) and so \( x = 4 \).	4	cayley
[ "Mathematics -> Algebra -> Prealgebra -> Fractions" ]	1	What fraction of the pizza is left for Wally if Jovin takes $\frac{1}{3}$ of the pizza, Anna takes $\frac{1}{6}$ of the pizza, and Olivia takes $\frac{1}{4}$ of the pizza?	Since Jovin, Anna and Olivia take $\frac{1}{3}, \frac{1}{6}$ and $\frac{1}{4}$ of the pizza, respectively, then the fraction of the pizza with which Wally is left is $$ 1-\frac{1}{3}-\frac{1}{6}-\frac{1}{4}=\frac{12}{12}-\frac{4}{12}-\frac{2}{12}-\frac{3}{12}=\frac{3}{12}=\frac{1}{4} $$	\frac{1}{4}	pascal
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1.5	The first two hours of Melanie's trip were spent travelling at $100 \mathrm{~km} / \mathrm{h}$. The remaining 200 km of Melanie's trip was spent travelling at $80 \mathrm{~km} / \mathrm{h}$. What was Melanie's average speed during this trip?	In 2 hours travelling at $100 \mathrm{~km} / \mathrm{h}$, Melanie travels $2 \mathrm{~h} \times 100 \mathrm{~km} / \mathrm{h}=200 \mathrm{~km}$. When Melanie travels 200 km at $80 \mathrm{~km} / \mathrm{h}$, it takes $\frac{200 \mathrm{~km}}{80 \mathrm{~km} / \mathrm{h}}=2.5 \mathrm{~h}$. Melanie travels a total of $200 \mathrm{~km}+200 \mathrm{~km}=400 \mathrm{~km}$. Melanie travels for a total of $2 \mathrm{~h}+2.5 \mathrm{~h}=4.5 \mathrm{~h}$. Therefore, Melanie's average speed is $\frac{400 \mathrm{~km}}{4.5 \mathrm{~h}} \approx 88.89 \mathrm{~km} / \mathrm{h}$. Of the given choices, this is closest to $89 \mathrm{~km} / \mathrm{h}$.	89 \mathrm{~km} / \mathrm{h}	pascal
[ "Mathematics -> Algebra -> Intermediate Algebra -> Inequalities" ]	1.5	The value of $\frac{x}{2}$ is less than the value of $x^{2}$. The value of $x^{2}$ is less than the value of $x$. Which of the following could be a value of $x$?	Since $x^{2}<x$ and $x^{2} \geq 0$, then $x>0$ and so it cannot be the case that $x$ is negative. Thus, neither (D) nor (E) is the answer. Since $x^{2}<x$, then we cannot have $x>1$. This is because when $x>1$, we have $x^{2}>x$. Thus, (A) is not the answer and so the answer is (B) or (C). If $x=\frac{1}{3}$, then $x^{2}=\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}$ and $\frac{x}{2}=\frac{1 / 3}{2}=\frac{1}{6}$. Since $\frac{1}{6}>\frac{1}{9}$, then $(B)$ cannot be the answer. Therefore, the answer must be (C). Checking, when $x=\frac{3}{4}$, we have $x^{2}=\frac{9}{16}$ and $\frac{x}{2}=\frac{3}{8}$. Since $\frac{x}{2}=\frac{3}{8}=\frac{6}{16}<\frac{9}{16}=x^{2}$, then $\frac{x}{2}<x^{2}$. Also, $x^{2}=\frac{9}{16}<\frac{12}{16}=\frac{3}{4}=x$. This confirms that $x=\frac{3}{4}$ does satisfy the required conditions.	\frac{3}{4}	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]	1	Which graph is linear with a slope of 0?	A graph that is linear with a slope of 0 is a horizontal straight line. This is Graph Q.	Graph Q	cayley
[ "Mathematics -> Algebra -> Prealgebra -> Decimals" ]	1.5	In the decimal representation of $rac{1}{7}$, the 100th digit to the right of the decimal is?	The digits to the right of the decimal place in the decimal representation of $rac{1}{7}$ occur in blocks of 6, repeating the block of digits 142857. Since $16 imes 6=96$, then the 96th digit to the right of the decimal place is the last in one of these blocks; that is, the 96th digit is 7. This means that the 97th digit is 1, the 98th digit is 4, the 99th digit is 2, and the 100th digit is 8.	8	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Fractions" ]	1.5	Which of the following numbers is less than $\frac{1}{20}$?	If $0<a<20$, then $\frac{1}{a}>\frac{1}{20}$. Therefore, $\frac{1}{15}>\frac{1}{20}$ and $\frac{1}{10}>\frac{1}{20}$. Also, $\frac{1}{20}=0.05$ which is less than both 0.5 and 0.055. Lastly, $\frac{1}{20}>\frac{1}{25}$ since $0<20<25$. Therefore, $\frac{1}{25}$ is the only one of the choices that is less than $\frac{1}{20}$.	\frac{1}{25}	cayley
[ "Mathematics -> Algebra -> Prealgebra -> Decimals" ]	1	Convert 2 meters plus 3 centimeters plus 5 millimeters into meters.	Since there are 100 cm in 1 m, then 1 cm is 0.01 m. Thus, 3 cm equals 0.03 m. Since there are 1000 mm in 1 m, then 1 mm is 0.001 m. Thus, 5 mm equals 0.005 m. Therefore, 2 m plus 3 cm plus 5 mm equals $2+0.03+0.005=2.035 \mathrm{~m}$.	2.035 \text{ m}	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Percentages -> Other", "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]	1.5	At the start of this month, Mathilde and Salah each had 100 coins. For Mathilde, this was $25 \%$ more coins than she had at the start of last month. For Salah, this was $20 \%$ fewer coins than he had at the start of last month. What was the total number of coins that they had at the start of last month?	Suppose that Mathilde had $m$ coins at the start of last month and Salah had $s$ coins at the start of last month. From the given information, 100 is $25 \%$ more than $m$, so $100=1.25 m$ which means that $m=\frac{100}{1.25}=80$. From the given information, 100 is $20 \%$ less than $s$, so $100=0.80$ s which means that $s=\frac{100}{0.80}=125$. Therefore, at the beginning of last month, they had a total of $m+s=80+125=205$ coins.	205	pascal
[ "Mathematics -> Algebra -> Intermediate Algebra -> Inequalities" ]	1.5	For which of the following values of $x$ is $x$ greater than $x^{2}$: $x=-2$, $x=-rac{1}{2}$, $x=0$, $x=rac{1}{2}$, $x=2$?	When $x=-2$, we get $x^{2}=4$. Here, $x<x^{2}$. When $x=-rac{1}{2}$, we get $x^{2}=rac{1}{4}$. Here, $x<x^{2}$. When $x=0$, we get $x^{2}=0$. Here, $x=x^{2}$. When $x=rac{1}{2}$, we get $x^{2}=rac{1}{4}$. Here, $x>x^{2}$. When $x=2$, we get $x^{2}=4$. Here, $x<x^{2}$. This means that $x=rac{1}{2}$ is the only choice where $x>x^{2}$.	rac{1}{2}	fermat
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1.5	Owen spends $\$ 1.20$ per litre on gasoline. He uses an average of 1 L of gasoline to drive 12.5 km. How much will Owen spend on gasoline to drive 50 km?	Since Owen uses an average of 1 L to drive 12.5 km, then it costs Owen $\$ 1.20$ in gas to drive 12.5 km. To drive 50 km, he drives 12.5 km a total of $50 \div 12.5=4$ times. Therefore, it costs him $4 \times \$ 1.20=\$ 4.80$ in gas to drive 50 km.	\$ 4.80	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Percentages -> Other", "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1.5	Meg started with the number 100. She increased this number by $20\%$ and then increased the resulting number by $50\%$. What was her final result?	$20\%$ of the number 100 is 20, so when 100 is increased by $20\%$, it becomes $100 + 20 = 120$. $50\%$ of a number is half of that number, so $50\%$ of 120 is 60. Thus, when 120 is increased by $50\%$, it becomes $120 + 60 = 180$. Therefore, Meg's final result is 180.	180	cayley
[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]	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 has been removed is $66-61=5$.	5	fermat
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1	Which of the following words has the largest value, given that the first five letters of the alphabet are assigned the values $A=1, B=2, C=3, D=4, E=5$?	We calculate the value of each of the five words as follows: - The value of $B A D$ is $2+1+4=7$ - The value of $C A B$ is $3+1+2=6$ - The value of $D A D$ is $4+1+4=9$ - The value of $B E E$ is $2+5+5=12$ - The value of $B E D$ is $2+5+4=11$. Of these, the word with the largest value is $B E E$.	BEE	cayley
[ "Mathematics -> Number Theory -> Prime Numbers" ]	1.5	Which of the following integers cannot be written as a product of two integers, each greater than 1: 6, 27, 53, 39, 77?	We note that $6=2 imes 3$ and $27=3 imes 9$ and $39=3 imes 13$ and $77=7 imes 11$, which means that each of $6,27,39$, and 77 can be written as the product of two integers, each greater than 1. Thus, 53 must be the integer that cannot be written in this way. We can check that 53 is indeed a prime number.	53	fermat
[ "Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other" ]	1.5	Which combination of outcomes is not possible for a soccer team that played three games, each ending in a win, loss, or tie, if the team scored more goals than were scored against them?	If a team has 0 wins, 1 loss, and 2 ties, then it scored fewer goals than its opponent once (the 1 loss) and the same number of goals as its opponent twice (the 2 ties). Therefore, it is not possible for a team to have 0 wins, 1 loss, and 2 ties, and to have scored more goals than were scored against them.	0 wins, 1 loss, 2 ties	cayley
[ "Mathematics -> Algebra -> Algebra -> Algebraic Expressions" ]	1.5	For which value of \( x \) is \( x^3 < x^2 \)?	If \( x = 1 \), then \( x^2 = 1 \) and \( x^3 = 1 \) and so \( x^3 = x^2 \). If \( x > 1 \), then \( x^3 \) equals \( x \) times \( x^2 \); since \( x > 1 \), then \( x \) times \( x^2 \) is greater than \( x^2 \) and so \( x^3 > x^2 \). Therefore, if \( x \) is positive with \( x^3 < x^2 \), we must have \( 0 < x < 1 \). Of the given choices, only \( x = \frac{3}{4} \) satisfies \( 0 < x < 1 \), and so the answer is (B).	\frac{3}{4}	cayley
[ "Mathematics -> Algebra -> Intermediate Algebra -> Inequalities" ]	1.5	Order the numbers $3$, $\frac{5}{2}$, and $\sqrt{10}$ from smallest to largest.	Since $3 = \frac{6}{2}$ and $\frac{5}{2} < \frac{6}{2}$, then $\frac{5}{2} < 3$. Since $3 = \sqrt{9}$ and $\sqrt{9} < \sqrt{10}$, then $3 < \sqrt{10}$. Thus, $\frac{5}{2} < 3 < \sqrt{10}$, and so the list of the three numbers in order from smallest to largest is $\frac{5}{2}, 3, \sqrt{10}$.	\frac{5}{2}, 3, \sqrt{10}	cayley
[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]	1	Jurgen is travelling to Waterloo by bus. He packs for 25 minutes, walks to the bus station for 35 minutes, and arrives 60 minutes before his bus leaves at 6:45 p.m. At what time did he start packing?	Jurgen takes $25+35=60$ minutes to pack and then walk to the bus station. Since Jurgen arrives 60 minutes before the bus leaves, he began packing $60+60=120$ minutes, or 2 hours, before the bus leaves. Since the bus leaves at 6:45 p.m., Jurgen began packing at 4:45 p.m.	4:45 ext{ p.m.}	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Fractions" ]	1	The value of \( \frac{1}{2} + \frac{2}{4} + \frac{4}{8} + \frac{8}{16} \) is what?	In the given sum, each of the four fractions is equivalent to \( \frac{1}{2} \). Therefore, the given sum is equal to \( \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2 \).	2	pascal
[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]	1	Solve for $x$ in the equation $(-1)(2)(x)(4)=24$.	Since $(-1)(2)(x)(4)=24$, then $-8x=24$ or $x=\frac{24}{-8}=-3$.	-3	cayley
[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]	1.5	The average (mean) of a list of 10 numbers is 17. When one number is removed from the list, the new average is 16. What number was removed?	When 10 numbers have an average of 17, their sum is $10 \times 17=170$. When 9 numbers have an average of 16, their sum is $9 \times 16=144$. Therefore, the number that was removed was $170-144=26$.	26	pascal

[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]

1

Morgan uses a spreadsheet to create a table of values. In the first column, she lists the positive integers from 1 to 400. She then puts integers in the second column in the following way: if the integer in the first column of a given row is $n$, the number in the second column of that row is $3 n+1$. Which of the following integers does not appear in the second column: 31, 94, 131, 331, 907?

Since $31=3 imes 10+1$ and $94=3 imes 31+1$ and $331=3 imes 110+1$ and $907=3 imes 302+1$, then each of $31,94,331$, and 907 appear in the second column of Morgan's spreadsheet. Thus, 131 must be the integer that does not appear in Morgan's spreadsheet. (We note that 131 is 2 more than $3 imes 43=129$ so is not 1 more than a multiple of 3.)

131

pascal

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1

Peyton puts 30 L of oil and 15 L of vinegar into a large empty can. He then adds 15 L of oil to create a new mixture. What percentage of the new mixture is oil?

After Peyton has added 15 L of oil, the new mixture contains $30+15=45 \mathrm{~L}$ of oil and 15 L of vinegar. Thus, the total volume of the new mixture is $45+15=60 \mathrm{~L}$. Of this, the percentage that is oil is $\frac{45}{60} \times 100 \%=\frac{3}{4} \times 100 \%=75 \%$.

75\%

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]

1.5

Anna thinks of an integer that is not a multiple of three, not a perfect square, and the sum of its digits is a prime number. What could the integer be?

Since 12 and 21 are multiples of 3 (12 = 4 \times 3 and 21 = 7 \times 3), the answer is not 12 or 21. 16 is a perfect square (16 = 4 \times 4) so the answer is not 16. The sum of the digits of 26 is 8, which is not a prime number, so the answer is not 26. Since 14 is not a multiple of three, 14 is not a perfect square, and the sum of the digits of 14 is 1 + 4 = 5 which is prime, then the answer is 14.

14

fermat

[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]

1.5

Ewan writes out a sequence where he counts by 11s starting at 3. Which number will appear in Ewan's sequence?

Ewan's sequence starts with 3 and each following number is 11 larger than the previous number. Since every number in the sequence is some number of 11s more than 3, this means that each number in the sequence is 3 more than a multiple of 11. Furthermore, every such positive integer is in Ewan's sequence. Since $110 = 11 \times 10$ is a multiple of 11, then $113 = 110 + 3$ is 3 more than a multiple of 11, and so is in Ewan's sequence.

113

fermat

[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]

1

If $ n = 7 $, which of the following expressions is equal to an even integer: $ 9n, n+8, n^2, n(n-2), 8n $?

When $ n=7 $, we have $ 9n=63, n+8=15, n^2=49, n(n-2)=35, 8n=56 $. Therefore, $ 8n $ is even. For every integer $ n $, the expression $ 8n $ is equal to an even integer.

8n

pascal

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1

In a certain year, July 1 was a Wednesday. What day of the week was July 17 in that year?

Since July 1 is a Wednesday, then July 8 and July 15 are both Wednesdays. Since July 15 is a Wednesday, then July 17 is a Friday.

Friday

pascal

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1.5

A store sells jellybeans at a fixed price per gram. The price for 250 g of jellybeans is $\$ 7.50$. What mass of jellybeans sells for $\$ 1.80$?

The store sells 250 g of jellybeans for $\$ 7.50$, which is 750 cents. Therefore, 1 g of jellybeans costs $750 \div 250=3$ cents. This means that $\$ 1.80$, which is 180 cents, will buy $180 \div 3=60 \mathrm{~g}$ of jellybeans.

60 \mathrm{~g}

cayley

[ "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 -> Geometry -> Plane Geometry -> Area" ]

1.5

A rectangle has length 13 and width 10. The length and the width of the rectangle are each increased by 2. By how much does the area of the rectangle increase?

The area of the original rectangle is $13 imes 10=130$. When the dimensions of the original rectangle are each increased by 2, we obtain a rectangle that is 15 by 12. The area of the new rectangle is $15 imes 12=180$, and so the area increased by $180-130=50$.

50

pascal

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1

If July 3, 2030 is a Wednesday, what day of the week is July 14, 2030?

July 14 is 11 days after July 3 of the same year. Since there are 7 days in a week, then July 10 and July 3 occur on the same day of the week, namely Wednesday. July 14 is 4 days after July 10, and so is a Sunday.

Sunday

pascal

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1

A hiker is exploring a trail. The trail has three sections: the first $25 \%$ of the trail is along a river, the next $\frac{5}{8}$ of the trail is through a forest, and the remaining 3 km of the trail is up a hill. How long is the trail?

Since $25 \%$ is equivalent to $\frac{1}{4}$, then the fraction of the trail covered by the section along the river and the section through the forest is $\frac{1}{4}+\frac{5}{8}=\frac{2}{8}+\frac{5}{8}=\frac{7}{8}$. This means that the final section up a hill represents $1-\frac{7}{8}=\frac{1}{8}$ of the trail. Since $\frac{1}{8}$ of the trail is 3 km long, then the entire trail is $8 \times 3 \mathrm{~km}=24 \mathrm{~km}$ long.

24 \text{ km}

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]

1.5

At the end of which year did Steve have more money than Wayne for the first time?

Steve's and Wayne's amounts of money double and halve each year, respectively. By 2004, Steve has more money than Wayne.

2004

pascal

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1.5

How much money does Roman give Dale if Roman wins a contest with a prize of $\$ 200$, gives $30 \%$ of the prize to Jackie, and then splits $15 \%$ of what remains equally between Dale and Natalia?

To determine $30 \%$ of Roman's $\$ 200$ prize, we calculate $\$ 200 \times 30 \%=\$ 200 \times \frac{30}{100}=\$ 2 \times 30=\$ 60$. After Roman gives $\$ 60$ to Jackie, he has $\$ 200-\$ 60=\$ 140$ remaining. He splits $15 \%$ of this between Dale and Natalia. The total that he splits is $\$ 140 \times 15 \%=\$ 140 \times 0.15=\$ 21$. Since Roman splits $\$ 21$ equally between Dale and Natalia, then Roman gives Dale a total of $\$ 21 \div 2=\$ 10.50$.

\$ 10.50

cayley

[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]

1.5

In 12 years, Janice will be 8 times as old as she was 2 years ago. How old is Janice now?

Suppose that Janice is $ x $ years old now. Two years ago, Janice was $ x - 2 $ years old. In 12 years, Janice will be $ x + 12 $ years old. From the given information $ x + 12 = 8(x - 2) $ and so $ x + 12 = 8x - 16 $ which gives $ 7x = 28 $ and so $ x = 4 $.

4

cayley

[ "Mathematics -> Algebra -> Prealgebra -> Fractions" ]

1

What fraction of the pizza is left for Wally if Jovin takes $\frac{1}{3}$ of the pizza, Anna takes $\frac{1}{6}$ of the pizza, and Olivia takes $\frac{1}{4}$ of the pizza?

Since Jovin, Anna and Olivia take $\frac{1}{3}, \frac{1}{6}$ and $\frac{1}{4}$ of the pizza, respectively, then the fraction of the pizza with which Wally is left is $$ 1-\frac{1}{3}-\frac{1}{6}-\frac{1}{4}=\frac{12}{12}-\frac{4}{12}-\frac{2}{12}-\frac{3}{12}=\frac{3}{12}=\frac{1}{4} $$

\frac{1}{4}

pascal

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1.5

The first two hours of Melanie's trip were spent travelling at $100 \mathrm{~km} / \mathrm{h}$. The remaining 200 km of Melanie's trip was spent travelling at $80 \mathrm{~km} / \mathrm{h}$. What was Melanie's average speed during this trip?

In 2 hours travelling at $100 \mathrm{~km} / \mathrm{h}$, Melanie travels $2 \mathrm{~h} \times 100 \mathrm{~km} / \mathrm{h}=200 \mathrm{~km}$. When Melanie travels 200 km at $80 \mathrm{~km} / \mathrm{h}$, it takes $\frac{200 \mathrm{~km}}{80 \mathrm{~km} / \mathrm{h}}=2.5 \mathrm{~h}$. Melanie travels a total of $200 \mathrm{~km}+200 \mathrm{~km}=400 \mathrm{~km}$. Melanie travels for a total of $2 \mathrm{~h}+2.5 \mathrm{~h}=4.5 \mathrm{~h}$. Therefore, Melanie's average speed is $\frac{400 \mathrm{~km}}{4.5 \mathrm{~h}} \approx 88.89 \mathrm{~km} / \mathrm{h}$. Of the given choices, this is closest to $89 \mathrm{~km} / \mathrm{h}$.

89 \mathrm{~km} / \mathrm{h}

pascal

[ "Mathematics -> Algebra -> Intermediate Algebra -> Inequalities" ]

1.5

The value of $\frac{x}{2}$ is less than the value of $x^{2}$. The value of $x^{2}$ is less than the value of $x$. Which of the following could be a value of $x$?

Since $x^{2}<x$ and $x^{2} \geq 0$, then $x>0$ and so it cannot be the case that $x$ is negative. Thus, neither (D) nor (E) is the answer. Since $x^{2}<x$, then we cannot have $x>1$. This is because when $x>1$, we have $x^{2}>x$. Thus, (A) is not the answer and so the answer is (B) or (C). If $x=\frac{1}{3}$, then $x^{2}=\frac{1}{3} \times \frac{1}{3}=\frac{1}{9}$ and $\frac{x}{2}=\frac{1 / 3}{2}=\frac{1}{6}$. Since $\frac{1}{6}>\frac{1}{9}$, then $(B)$ cannot be the answer. Therefore, the answer must be (C). Checking, when $x=\frac{3}{4}$, we have $x^{2}=\frac{9}{16}$ and $\frac{x}{2}=\frac{3}{8}$. Since $\frac{x}{2}=\frac{3}{8}=\frac{6}{16}<\frac{9}{16}=x^{2}$, then $\frac{x}{2}<x^{2}$. Also, $x^{2}=\frac{9}{16}<\frac{12}{16}=\frac{3}{4}=x$. This confirms that $x=\frac{3}{4}$ does satisfy the required conditions.

\frac{3}{4}

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Integers" ]

1

Which graph is linear with a slope of 0?

A graph that is linear with a slope of 0 is a horizontal straight line. This is Graph Q.

Graph Q

cayley

[ "Mathematics -> Algebra -> Prealgebra -> Decimals" ]

1.5

In the decimal representation of $rac{1}{7}$, the 100th digit to the right of the decimal is?

The digits to the right of the decimal place in the decimal representation of $rac{1}{7}$ occur in blocks of 6, repeating the block of digits 142857. Since $16 imes 6=96$, then the 96th digit to the right of the decimal place is the last in one of these blocks; that is, the 96th digit is 7. This means that the 97th digit is 1, the 98th digit is 4, the 99th digit is 2, and the 100th digit is 8.

8

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Fractions" ]

1.5

Which of the following numbers is less than $\frac{1}{20}$?

If $0<a<20$, then $\frac{1}{a}>\frac{1}{20}$. Therefore, $\frac{1}{15}>\frac{1}{20}$ and $\frac{1}{10}>\frac{1}{20}$. Also, $\frac{1}{20}=0.05$ which is less than both 0.5 and 0.055. Lastly, $\frac{1}{20}>\frac{1}{25}$ since $0<20<25$. Therefore, $\frac{1}{25}$ is the only one of the choices that is less than $\frac{1}{20}$.

\frac{1}{25}

cayley

[ "Mathematics -> Algebra -> Prealgebra -> Decimals" ]

1

Convert 2 meters plus 3 centimeters plus 5 millimeters into meters.

Since there are 100 cm in 1 m, then 1 cm is 0.01 m. Thus, 3 cm equals 0.03 m. Since there are 1000 mm in 1 m, then 1 mm is 0.001 m. Thus, 5 mm equals 0.005 m. Therefore, 2 m plus 3 cm plus 5 mm equals $2+0.03+0.005=2.035 \mathrm{~m}$.

2.035 \text{ m}

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Percentages -> Other", "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]

1.5

At the start of this month, Mathilde and Salah each had 100 coins. For Mathilde, this was $25 \%$ more coins than she had at the start of last month. For Salah, this was $20 \%$ fewer coins than he had at the start of last month. What was the total number of coins that they had at the start of last month?

Suppose that Mathilde had $m$ coins at the start of last month and Salah had $s$ coins at the start of last month. From the given information, 100 is $25 \%$ more than $m$, so $100=1.25 m$ which means that $m=\frac{100}{1.25}=80$. From the given information, 100 is $20 \%$ less than $s$, so $100=0.80$ s which means that $s=\frac{100}{0.80}=125$. Therefore, at the beginning of last month, they had a total of $m+s=80+125=205$ coins.

205

pascal

[ "Mathematics -> Algebra -> Intermediate Algebra -> Inequalities" ]

1.5

For which of the following values of $x$ is $x$ greater than $x^{2}$: $x=-2$, $x=-rac{1}{2}$, $x=0$, $x=rac{1}{2}$, $x=2$?

When $x=-2$, we get $x^{2}=4$. Here, $x<x^{2}$. When $x=-rac{1}{2}$, we get $x^{2}=rac{1}{4}$. Here, $x<x^{2}$. When $x=0$, we get $x^{2}=0$. Here, $x=x^{2}$. When $x=rac{1}{2}$, we get $x^{2}=rac{1}{4}$. Here, $x>x^{2}$. When $x=2$, we get $x^{2}=4$. Here, $x<x^{2}$. This means that $x=rac{1}{2}$ is the only choice where $x>x^{2}$.

rac{1}{2}

fermat

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1.5

Owen spends $\$ 1.20$ per litre on gasoline. He uses an average of 1 L of gasoline to drive 12.5 km. How much will Owen spend on gasoline to drive 50 km?

Since Owen uses an average of 1 L to drive 12.5 km, then it costs Owen $\$ 1.20$ in gas to drive 12.5 km. To drive 50 km, he drives 12.5 km a total of $50 \div 12.5=4$ times. Therefore, it costs him $4 \times \$ 1.20=\$ 4.80$ in gas to drive 50 km.

\$ 4.80

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Percentages -> Other", "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1.5

Meg started with the number 100. She increased this number by $20\%$ and then increased the resulting number by $50\%$. What was her final result?

$20\%$ of the number 100 is 20, so when 100 is increased by $20\%$, it becomes $100 + 20 = 120$. $50\%$ of a number is half of that number, so $50\%$ of 120 is 60. Thus, when 120 is increased by $50\%$, it becomes $120 + 60 = 180$. Therefore, Meg's final result is 180.

180

cayley

[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]

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 has been removed is $66-61=5$.

5

fermat

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1

Which of the following words has the largest value, given that the first five letters of the alphabet are assigned the values $A=1, B=2, C=3, D=4, E=5$?

We calculate the value of each of the five words as follows: - The value of $B A D$ is $2+1+4=7$ - The value of $C A B$ is $3+1+2=6$ - The value of $D A D$ is $4+1+4=9$ - The value of $B E E$ is $2+5+5=12$ - The value of $B E D$ is $2+5+4=11$. Of these, the word with the largest value is $B E E$.

BEE

cayley

[ "Mathematics -> Number Theory -> Prime Numbers" ]

1.5

Which of the following integers cannot be written as a product of two integers, each greater than 1: 6, 27, 53, 39, 77?

We note that $6=2 imes 3$ and $27=3 imes 9$ and $39=3 imes 13$ and $77=7 imes 11$, which means that each of $6,27,39$, and 77 can be written as the product of two integers, each greater than 1. Thus, 53 must be the integer that cannot be written in this way. We can check that 53 is indeed a prime number.

53

fermat

[ "Mathematics -> Applied Mathematics -> Statistics -> Probability -> Other" ]

1.5

Which combination of outcomes is not possible for a soccer team that played three games, each ending in a win, loss, or tie, if the team scored more goals than were scored against them?

If a team has 0 wins, 1 loss, and 2 ties, then it scored fewer goals than its opponent once (the 1 loss) and the same number of goals as its opponent twice (the 2 ties). Therefore, it is not possible for a team to have 0 wins, 1 loss, and 2 ties, and to have scored more goals than were scored against them.

0 wins, 1 loss, 2 ties

cayley

[ "Mathematics -> Algebra -> Algebra -> Algebraic Expressions" ]

1.5

For which value of $ x $ is $ x^3 < x^2 $?

If $ x = 1 $, then $ x^2 = 1 $ and $ x^3 = 1 $ and so $ x^3 = x^2 $. If $ x > 1 $, then $ x^3 $ equals $ x $ times $ x^2 $; since $ x > 1 $, then $ x $ times $ x^2 $ is greater than $ x^2 $ and so $ x^3 > x^2 $. Therefore, if $ x $ is positive with $ x^3 < x^2 $, we must have $ 0 < x < 1 $. Of the given choices, only $ x = \frac{3}{4} $ satisfies $ 0 < x < 1 $, and so the answer is (B).

\frac{3}{4}

cayley

[ "Mathematics -> Algebra -> Intermediate Algebra -> Inequalities" ]

1.5

Order the numbers $3$, $\frac{5}{2}$, and $\sqrt{10}$ from smallest to largest.

Since $3 = \frac{6}{2}$ and $\frac{5}{2} < \frac{6}{2}$, then $\frac{5}{2} < 3$. Since $3 = \sqrt{9}$ and $\sqrt{9} < \sqrt{10}$, then $3 < \sqrt{10}$. Thus, $\frac{5}{2} < 3 < \sqrt{10}$, and so the list of the three numbers in order from smallest to largest is $\frac{5}{2}, 3, \sqrt{10}$.

\frac{5}{2}, 3, \sqrt{10}

cayley

[ "Mathematics -> Applied Mathematics -> Math Word Problems" ]

1

Jurgen is travelling to Waterloo by bus. He packs for 25 minutes, walks to the bus station for 35 minutes, and arrives 60 minutes before his bus leaves at 6:45 p.m. At what time did he start packing?

Jurgen takes $25+35=60$ minutes to pack and then walk to the bus station. Since Jurgen arrives 60 minutes before the bus leaves, he began packing $60+60=120$ minutes, or 2 hours, before the bus leaves. Since the bus leaves at 6:45 p.m., Jurgen began packing at 4:45 p.m.

4:45 ext{ p.m.}

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Fractions" ]

1

The value of $ \frac{1}{2} + \frac{2}{4} + \frac{4}{8} + \frac{8}{16} $ is what?

In the given sum, each of the four fractions is equivalent to $ \frac{1}{2} $. Therefore, the given sum is equal to $ \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2 $.

2

pascal

[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]

1

Solve for $x$ in the equation $(-1)(2)(x)(4)=24$.

Since $(-1)(2)(x)(4)=24$, then $-8x=24$ or $x=\frac{24}{-8}=-3$.

-3

cayley

[ "Mathematics -> Algebra -> Prealgebra -> Simple Equations" ]

1.5

The average (mean) of a list of 10 numbers is 17. When one number is removed from the list, the new average is 16. What number was removed?

When 10 numbers have an average of 17, their sum is $10 \times 17=170$. When 9 numbers have an average of 16, their sum is $9 \times 16=144$. Therefore, the number that was removed was $170-144=26$.

26

pascal