prompt_id stringlengths 16 16 | sample_idx int64 0 3 ⌀ | prompt stringlengths 10 4.31k | answer stringlengths 1 210 ⌀ | completion stringlengths 0 1.33M ⌀ | reward float64 0 1 ⌀ | advantage float64 0 1.5 ⌀ | group_size int64 4 4 ⌀ | group_reward_mean float64 0 1 ⌀ | group_reward_std float64 0 0.58 ⌀ | all_positive bool 2
classes | all_negative bool 2
classes |
|---|---|---|---|---|---|---|---|---|---|---|---|
5d8af06aba66ff86 | 0 | The harmonic mean of a set of non-zero numbers is the reciprocal of the average of the reciprocals of the numbers. What is the harmonic mean of 1, 2, and 4? | \frac{12}{7} | The harmonic mean of a set of numbers is given by \( H = \frac{n}{\sum_{i=1}^n \frac{1}{x_i}} \). For the numbers 1, 2, and 4, we have \( n = 3 \). The sum of reciprocals is \( \frac{1}{1} + \frac{1}{2} + \frac{1}{4} = 1 + 0.5 + 0.25 = 1.75 = \frac{7}{4} \). Thus, \( H = \frac{3}{\frac{7}{4}} = 3 \times \frac{4}{7} = \... | 1 | 0 | 4 | 1 | 0 | true | false |
8a2ac0e76ef7e796 | 0 | Chandler wants to buy a $500$ dollar mountain bike. For his birthday, his grandparents send him $50$ dollars, his aunt sends him $35$ dollars and his cousin gives him $15$ dollars. He earns $16$ dollars per week for his paper route. He will use all of his birthday money and all of the money he earns from his paper rout... | 25 | Chandler receives $50 + 35 + 15 = 100$ dollars from his birthday. He needs $500$ dollars total, so he must earn an additional $500 - 100 = 400$ dollars from his paper route. He earns $16$ dollars per week, so the number of weeks needed is $400 \div 16 = 25$ weeks. After 25 weeks, he will have exactly $100 + 25 \cdot 16... | 1 | 0 | 4 | 1 | 0 | true | false |
6c6c4220fe253dcc | 0 | If $991+993+995+997+999=5000-N$, then $N=$ | 25 | The sum of the five odd numbers is \(991 + 993 + 995 + 997 + 999 = 4975\).
Given \(4975 = 5000 - N\), solving for \(N\) yields \(N = 5000 - 4975 = 25\).
\boxed{25} | 1 | 0 | 4 | 1 | 0 | true | false |
59566e62ec4d6f20 | 0 | On average, for every 4 sports cars sold at the local dealership, 7 sedans are sold. The dealership predicts that it will sell 28 sports cars next month. How many sedans does it expect to sell? | 49 | The ratio of sports cars to sedans is 4:7. For 28 sports cars, the number of sedans is \( \frac{7}{4} \times 28 = 49 \).
\boxed{49} | 1 | 0 | 4 | 1 | 0 | true | false |
37d7f48d0f359839 | 0 | At Euclid Middle School the mathematics teachers are Mrs. Germain, Mr. Newton, and Mrs. Young. There are $11$ students in Mrs. Germain's class, $8$ students in Mr. Newton's class, and $9$ students in Mrs. Young's class taking the AMC $8$ this year. How many mathematics students at Euclid Middle School are taking the co... | 28 | The mathematics teachers at Euclid Middle School are Mrs. Germain, Mr. Newton, and Mrs. Young. The numbers of students taking the AMC 8 from their classes are 11, 8, and 9 respectively. Adding these gives the total number of mathematics students taking the contest: \(11 + 8 + 9 = 28\).
\boxed{28} | 1 | 0 | 4 | 1 | 0 | true | false |
e06043fb5585a184 | 0 | The Fort Worth Zoo has a number of two-legged birds and a number of four-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds? | 139 | The number of two-legged birds is 139.
Let \( b \) be the number of birds and \( m \) be the number of mammals. Each animal has one head, so \( b + m = 200 \). The total number of legs is \( 2b + 4m = 522 \). Solving the system:
\[
\begin{aligned}
b + m &= 200 \\
2b + 4m &= 522
\end{aligned}
\]
Multiply the first eq... | 1 | 0 | 4 | 1 | 0 | true | false |
03fd71fbf2db9deb | 0 | The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$, in the order shown. The value of $x$ is: | 0 | The common difference of an arithmetic progression is constant. Thus, the difference between the second and first terms equals the difference between the third and second terms:
\[
(x + 1) - (x - 1) = (2x + 3) - (x + 1).
\]
Simplify both sides:
Left: \(x + 1 - x + 1 = 2\).
Right: \(2x + 3 - x - 1 = x + 2\).
Set th... | 1 | 0 | 4 | 1 | 0 | true | false |
9add70569edc6fe0 | 0 | If $\frac{1}{x} - \frac{1}{y} = \frac{1}{z}$, then $z$ equals: | \frac{xy}{y - x} | The given equation is \(\frac{1}{x} - \frac{1}{y} = \frac{1}{z}\). Combine the left-hand side:
\[
\frac{1}{x} - \frac{1}{y} = \frac{y - x}{xy}.
\]
Thus, \(\frac{y - x}{xy} = \frac{1}{z}\). Taking reciprocals (provided \(y - x \neq 0\) and \(x, y \neq 0\)) gives
\[
z = \frac{xy}{y - x}.
\]
Therefore, \(z\) equals \(... | 1 | 0 | 4 | 1 | 0 | true | false |
9af4a98fc94fc96d | 0 | Alicia had two containers. The first was $\frac{5}{6}$ full of water and the second was empty. She poured all the water from the first container into the second container, at which point the second container was $\frac{3}{4}$ full of water. What is the ratio of the volume of the first container to the volume of the sec... | \frac{9}{10} | The first container is initially \(\frac{5}{6}\) full, so the volume of water is \(\frac{5}{6} V_1\). After pouring into the second container, it is \(\frac{3}{4}\) full, so \(\frac{5}{6} V_1 = \frac{3}{4} V_2\). Solving for the ratio \(V_1 : V_2\):
\[
\frac{V_1}{V_2} = \frac{3/4}{5/6} = \frac{3}{4} \cdot \frac{6}{5} ... | 1 | 0 | 4 | 1 | 0 | true | false |
8891dde31a85b5b1 | 0 | Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60$. | \frac{60}{13} | The minimum value of \(\sqrt{x^2+y^2}\) subject to the constraint \(5x+12y=60\) is the distance from the origin to the line \(5x+12y-60=0\). Using the distance formula:
\[
\text{Distance} = \frac{|5(0)+12(0)-60|}{\sqrt{5^2+12^2}} = \frac{60}{\sqrt{169}} = \frac{60}{13}.
\]
Thus the minimum value is \(\frac{60}{13}\).... | 1 | 0 | 4 | 1 | 0 | true | false |
236b24ac3ca2e91d | 0 | Andrea and Lauren are $20$ kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreasing at a rate of $1$ kilometer per minute. After $5$ minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from t... | 65 | Andrea and Lauren start 20 km apart and bike toward each other. Andrea's speed is three times Lauren's speed, so let Lauren's speed be \(v\) km/min and Andrea's speed be \(3v\) km/min. The distance between them decreases at a rate of \(3v + v = 4v\) km/min, which is given as 1 km/min. Thus, \(4v = 1\) so \(v = 0.25\) k... | 1 | 0 | 4 | 1 | 0 | true | false |
81becf6d55521520 | 0 | The total in-store price for an appliance is $99.99$. A television commercial advertises the same product for three easy payments of $29.98$ and a one-time shipping and handling charge of $9.98$. How many cents are saved by buying the appliance from the television advertiser? | 7 | The in-store price is $99.99$, which is $9999$ cents. The television advertiser's total is three payments of $29.98$ plus a shipping fee of $9.98$:
\[
3 \times 29.98 = 89.94 \quad \text{dollars},\qquad 89.94 + 9.98 = 99.92 \quad \text{dollars} = 9992 \text{ cents}.
\]
The savings in cents is
\[
9999 - 9992 = 7.
\... | 1 | 0 | 4 | 1 | 0 | true | false |
6d903c5cde234c34 | 0 | At $2:15$ o'clock, the hour and minute hands of a clock form an angle of: | 22\frac {1}{2}^{\circ} | The hour hand moves 0.5 degrees per minute. At 2:00, it is at 60° (since each hour is 30°). At 2:15, it has moved \(15 \times 0.5 = 7.5°\), so it is at \(60° + 7.5° = 67.5°\). The minute hand at 15 minutes is at the 3, which is \(90°\). The difference is \(|90° - 67.5°| = 22.5°\). This is the smaller angle between the ... | 1 | 0 | 4 | 1 | 0 | true | false |
e576ad2d78f60e93 | 0 | Alicia earns 20 dollars per hour, of which $1.45\%$ is deducted to pay local taxes. How many cents per hour of Alicia's wages are used to pay local taxes? | 29 | Alicia earns $20 per hour, and 1.45% of that is deducted for local taxes. To find the deduction in cents per hour:
1.45% of $20 = 0.0145 × 20 = 0.29 dollars.
Since 1 dollar = 100 cents, 0.29 dollars = 29 cents.
Thus, Alicia pays 29 cents per hour in local taxes.
\boxed{29} | 1 | 0 | 4 | 1 | 0 | true | false |
7286ee3ec8c1117a | 0 | The values of $k$ for which the equation $2x^2-kx+x+8=0$ will have real and equal roots are: | 9 and -7 | The given equation simplifies to \(2x^2 + (1 - k)x + 8 = 0\). For real and equal roots, the discriminant must be zero:
\[
(1 - k)^2 - 4 \cdot 2 \cdot 8 = (1 - k)^2 - 64 = 0.
\]
Solving \((1 - k)^2 = 64\) gives \(1 - k = \pm 8\). Thus, \(k = -7\) or \(k = 9\). Both values yield double roots.
\boxed{-7 \text{ and } 9} | 1 | 0.5 | 4 | 0.75 | 0.5 | false | false |
2d2e4d664db307cf | 0 | What is the value of $2^{0^{1^9}} + (2^0)^{1^9}$? | 2 | The expression is \(2^{0^{1^9}} + (2^0)^{1^9}\). First, evaluate \(1^9 = 1\). Then \(0^{1^9} = 0^1 = 0\). Thus \(2^{0^{1^9}} = 2^0 = 1\). Next, \(2^0 = 1\), and \(1^9 = 1\), so \((2^0)^{1^9} = 1^1 = 1\). Summing gives \(1 + 1 = 2\).
\boxed{2} | 1 | 0 | 4 | 1 | 0 | true | false |
b2dbffd92e5b7f37 | 0 | How many positive factors of 36 are also multiples of 4? | 3 | The prime factorization of 36 is \(2^2 \times 3^2\). The positive factors are of the form \(2^a \times 3^b\) with \(a = 0,1,2\) and \(b = 0,1,2\). For a factor to be a multiple of 4 (\(2^2\)), we need \(a \ge 2\), so \(a = 2\). Then \(b\) can be 0, 1, or 2, giving the factors \(4, 12, 36\). Thus there are 3 such factor... | 1 | 0 | 4 | 1 | 0 | true | false |
18350ed3676e861a | 0 | During the softball season, Judy had $35$ hits. Among her hits were $1$ home run, $1$ triple and $5$ doubles. The rest of her hits were single. What percent of her hits were single? | 80\% | Judy had 35 hits total. Among them, there were 1 home run, 1 triple, and 5 doubles, which sum to \(1+1+5=7\) hits that were not singles. Therefore, the number of singles is \(35-7=28\). The percentage of hits that were singles is \(\frac{28}{35} \times 100\% = \frac{4}{5} \times 100\% = 80\%\).
\boxed{80\%} | 1 | 0 | 4 | 1 | 0 | true | false |
73bbb1d4e278c84d | 0 | Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower? | 0.4 | The volume of the original water tower's spherical top is 100,000 liters, and the model's sphere holds 0.1 liters. The scale factor for volume is \( \frac{0.1}{100,000} = \frac{1}{1,000,000} \). Since volume scales with the cube of the linear scale factor \(k\), we have \(k^3 = \frac{1}{1,000,000}\), so \(k = \frac{1}{... | 1 | 0 | 4 | 1 | 0 | true | false |
a7fafaa840c11d23 | 0 | The points $(2,-3)$, $(4,3)$, and $(5, k/2)$ are on the same straight line. The value(s) of $k$ is (are): | 12 | The points \((2,-3)\), \((4,3)\), and \((5, k/2)\) are collinear. The slope between \((2,-3)\) and \((4,3)\) is \(\frac{3 - (-3)}{4 - 2} = \frac{6}{2} = 3\).
Using the slope between \((4,3)\) and \((5, k/2)\): \(\frac{k/2 - 3}{5 - 4} = k/2 - 3 = 3\), so \(k/2 = 6\) and \(k = 12\).
Thus the value of \(k\) is \(12\).... | 1 | 0 | 4 | 1 | 0 | true | false |
02f53822174662a8 | 0 | The sum of two natural numbers is $17402$. One of the two numbers is divisible by $10$. If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers? | 14238 | The sum of two natural numbers is 17402. One number is divisible by 10. Erasing its units digit (which is 0) gives the other number, so the larger number is ten times the smaller. Let the smaller number be \(y\) and the larger number be \(x = 10y\). Their sum is \(10y + y = 11y = 17402\), so \(y = 17402 / 11 = 1582\). ... | 1 | 0 | 4 | 1 | 0 | true | false |
022cdf2d764a0b08 | 0 | How many perfect cubes lie between $2^8+1$ and $2^{18}+1$, inclusive? | 58 | We need to count perfect cubes \( n^3 \) such that \( 2^8 + 1 \le n^3 \le 2^{18} + 1 \).
Compute the bounds:
\( 2^8 = 256 \), so lower bound \( L = 256 + 1 = 257 \).
\( 2^{18} = 262144 \), so upper bound \( U = 262144 + 1 = 262145 \).
Find the smallest integer \( m \) with \( m^3 \ge 257 \):
\( 6^3 = 216 < 2... | 1 | 0 | 4 | 1 | 0 | true | false |
43735b2a4e5955ee | 0 | A team won $40$ of its first $50$ games. How many of the remaining $40$ games must this team win so it will have won exactly $70 \%$ of its games for the season? | 23 | The team has played 50 games and won 40. The season consists of 90 games total (50 + 40). To win exactly 70% of the season's games, the team needs \(0.70 \times 90 = 63\) wins. Since they already have 40 wins, they need \(63 - 40 = 23\) more wins from the remaining 40 games.
\boxed{23} | 1 | 0 | 4 | 1 | 0 | true | false |
4664f50b0051a03a | 0 | Claire adds the degree measures of the interior angles of a convex polygon and arrives at a sum of $2017$. She then discovers that she forgot to include one angle. What is the degree measure of the forgotten angle? | 143 | The sum of the interior angles of a convex polygon with \(n\) sides is \((n-2) \cdot 180^\circ\). Claire's computed sum of \(2017^\circ\) excludes one angle, \(x\). Thus,
\[
(n-2) \cdot 180 = 2017 + x.
\]
Solving for \(x\):
\[
x = (n-2) \cdot 180 - 2017.
\]
Since the polygon is convex, \(0 < x < 180\). Let \(k = n-2 \g... | 1 | 0 | 4 | 1 | 0 | true | false |
bbb83cd5ff49cf9c | 0 | In a certain year the price of gasoline rose by $20\%$ during January, fell by $20\%$ during February, rose by $25\%$ during March, and fell by $x\%$ during April. The price of gasoline at the end of April was the same as it had been at the beginning of January. To the nearest integer, what is $x$ | 17 | The price of gasoline at the beginning of January is \(P\). After a 20% increase in January, the price becomes \(1.20P\). After a 20% decrease in February, the price becomes \(1.20P \times 0.80 = 0.96P\). After a 25% increase in March, the price becomes \(0.96P \times 1.25 = 1.20P\). After a decrease of \(x\%\) in Apri... | 1 | 0 | 4 | 1 | 0 | true | false |
62e61f9984abe880 | 0 | Mary's top book shelf holds five books with the following widths, in centimeters: $6$, $\dfrac{1}{2}$, $1$, $2.5$, and $10$. What is the average book width, in centimeters? | 4 | The sum of the book widths is \(6 + \frac{1}{2} + 1 + 2.5 + 10 = 20\) centimeters. There are 5 books, so the average width is \(20 \div 5 = 4\) centimeters.
\boxed{4} | 1 | 0 | 4 | 1 | 0 | true | false |
b9993b2eb98e1bc0 | 0 | Everyday at school, Jo climbs a flight of $6$ stairs. Jo can take the stairs $1$, $2$, or $3$ at a time. For example, Jo could climb $3$, then $1$, then $2$. In how many ways can Jo climb the stairs? | 24 | The number of ways to climb 6 stairs using steps of 1, 2, or 3 at a time is given by the recurrence \(a_n = a_{n-1} + a_{n-2} + a_{n-3}\) with \(a_0 = 1\), \(a_1 = 1\), \(a_2 = 2\). Computing successively:
- \(a_3 = a_2 + a_1 + a_0 = 2 + 1 + 1 = 4\)
- \(a_4 = a_3 + a_2 + a_1 = 4 + 2 + 1 = 7\)
- \(a_5 = a_4 + a_3 + a_2... | 1 | 0 | 4 | 1 | 0 | true | false |
e5d6d177acc28d19 | 0 | If $a$ and $b$ are digits for which
$\begin{array}{ccc}& 2 & a\ \times & b & 3\ \hline & 6 & 9\ 9 & 2 & \ \hline 9 & 8 & 9\end{array}$
then $a+b =$ | 7 | The given multiplication is:
\[
\begin{array}{c@{\,}c@{\,}c}
& 2 & a \\
\times & b & 3 \\
\hline
& 6 & 9 \\
9 & 2 & \\
\hline
9 & 8 & 9
\end{array}
\]
The first partial product is \( (2a) \times 3 = 69 \). Since \( 2a = 20 + a \), we have \( 3(20 + a) = 60 + 3a = 69 \). Solving gives \( 3a = 9 \), so \( a = 3 \).
Th... | 1 | 0 | 4 | 1 | 0 | true | false |
9ee2ec5295d18d53 | 0 | If the following instructions are carried out by a computer, what value of \(X\) will be printed because of instruction \(5\)?
1. START \(X\) AT \(3\) AND \(S\) AT \(0\).
2. INCREASE THE VALUE OF \(X\) BY \(2\).
3. INCREASE THE VALUE OF \(S\) BY THE VALUE OF \(X\).
4. IF \(S\) IS AT LEAST \(10000\),
TH... | 23 | The process starts with \(X = 3\) and \(S = 0\). Each iteration increases \(X\) by 2 and then adds the new \(X\) to \(S\). After \(n\) iterations, \(X = 3 + 2n\) and \(S = n^2 + 4n\). The loop continues until \(S \ge 10000\). Solving \(n^2 + 4n \ge 10000\) gives \(n = 99\) as the smallest integer satisfying the conditi... | 0 | 0 | 4 | 0 | 0 | false | true |
b61778e74e197f83 | 0 | Triangle $ABC$ has vertices $A = (3,0)$, $B = (0,3)$, and $C$, where $C$ is on the line $x + y = 7$. What is the area of $\triangle ABC$? | 6 | The area of triangle \(ABC\) is constant because \(C\) lies on the line \(x + y = 7\), which is parallel to \(AB\) (the line through \(A\) and \(B\) is \(x + y = 3\)). The distance between these parallel lines is \(\frac{|7-3|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\). The length of \(AB\) is \(\sqrt{(3-0)^2 + (0-3... | 1 | 0 | 4 | 1 | 0 | true | false |
c1dca300db22eab4 | 0 | Julie is preparing a speech for her class. Her speech must last between one-half hour and three-quarters of an hour. The ideal rate of speech is 150 words per minute. If Julie speaks at the ideal rate, what number of words would be an appropriate length for her speech? | 5650 | The speech must last between 30 minutes (half an hour) and 45 minutes (three-quarters of an hour). At 150 words per minute, the number of words for 30 minutes is \(150 \times 30 = 4500\), and for 45 minutes is \(150 \times 45 = 6750\). Therefore, any number of words between 4500 and 6750 inclusive is appropriate.
\box... | 0 | 0 | 4 | 0 | 0 | false | true |
f58b3ecf3f089ffc | 0 | Positive integers $a$ and $b$ are each less than $6$. What is the smallest possible value for $2 \cdot a - a \cdot b$? | -15 | The expression is \(2a - ab = a(2 - b)\). Since \(a\) and \(b\) are positive integers less than 6, they can be 1, 2, 3, 4, or 5. To minimize the value, we want the product to be as negative as possible. The factor \((2 - b)\) is negative when \(b > 2\), and its most negative value is \(-3\) when \(b = 5\). The factor \... | 1 | 0 | 4 | 1 | 0 | true | false |
ca5257c849b303da | 0 | Zara has a collection of $4$ marbles: an Aggie, a Bumblebee, a Steelie, and a Tiger. She wants to display them in a row on a shelf, but does not want to put the Steelie and the Tiger next to one another. In how many ways can she do this? | 12 | The total number of ways to arrange the four distinct marbles is \(4! = 24\). To count the arrangements where the Steelie (S) and Tiger (T) are adjacent, treat them as a single block. Within the block, they can be ordered in \(2\) ways (ST or TS). Together with the other two marbles (Aggie and Bumblebee), there are \(3... | 1 | 0 | 4 | 1 | 0 | true | false |
cd343d940b21dfa4 | 0 | Let $(a_n)$ and $(b_n)$ be the sequences of real numbers such that
\[ (2 + i)^n = a_n + b_ni \]for all integers $n\geq 0$, where $i = \sqrt{-1}$. What is
\[\sum_{n=0}^\infty\frac{a_nb_n}{7^n}\,?\] | \frac{7}{16} | The sequences satisfy \((2+i)^n = a_n + b_n i\) for \(n\ge 0\). Write \(2+i = \sqrt{5} e^{i\theta}\) with \(\theta = \arctan(1/2)\). Then \(a_n = 5^{n/2}\cos(n\theta)\) and \(b_n = 5^{n/2}\sin(n\theta)\), so
\[
a_n b_n = 5^n \cos(n\theta)\sin(n\theta) = \frac{5^n}{2}\sin(2n\theta).
\]
Hence the sum is
\[
S = \sum_{n=0}... | 1 | 0 | 4 | 1 | 0 | true | false |
ce8287d71434c1f9 | 0 | The operation $\otimes$ is defined for all nonzero numbers by $a \otimes b = \frac{a^{2}}{b}$. Determine $[(1 \otimes 2) \otimes 3] - [1 \otimes (2 \otimes 3)]$. | -\frac{2}{3} | The operation \(\otimes\) is defined by \(a \otimes b = \frac{a^2}{b}\) for nonzero numbers.
First, compute \(1 \otimes 2\):
\[
1 \otimes 2 = \frac{1^2}{2} = \frac{1}{2}.
\]
Then compute \((1 \otimes 2) \otimes 3 = \left(\frac{1}{2}\right) \otimes 3\):
\[
\left(\frac{1}{2}\right) \otimes 3 = \frac{\left(\frac{1}{2}\ri... | 1 | 0 | 4 | 1 | 0 | true | false |
718c022c49eebe40 | 0 | Carrie has a rectangular garden that measures $6$ feet by $8$ feet. She plants the entire garden with strawberry plants. Carrie is able to plant $4$ strawberry plants per square foot, and she harvests an average of $10$ strawberries per plant. How many strawberries can she expect to harvest? | 1920 | The garden area is \(6 \times 8 = 48\) square feet. With 4 plants per square foot, there are \(48 \times 4 = 192\) plants. Each plant yields 10 strawberries, so total strawberries are \(192 \times 10 = 1920\).
\boxed{1920} | 1 | 0 | 4 | 1 | 0 | true | false |
757a1077030ac8b9 | 0 | The sum of the greatest integer less than or equal to $x$ and the least integer greater than or equal to $x$ is $5$. The solution set for $x$ is | \{x \mid 2 < x < 3\} | The sum of the greatest integer less than or equal to \(x\) (floor) and the least integer greater than or equal to \(x\) (ceil) is 5. Let \(n = \lfloor x \rfloor\). If \(x\) is an integer, then \(\lceil x \rceil = x\) and the sum is \(2x = 5\), so \(x = 2.5\), which is not an integer. Hence no integer solution. If \(x\... | 0 | 0 | 4 | 0 | 0 | false | true |
41ec838326aba29a | 0 | The number $2.5252525\ldots$ can be written as a fraction.
When reduced to lowest terms the sum of the numerator and denominator of this fraction is: | 349 | The number \(2.5252525\ldots\) can be expressed as a repeating decimal with the repeating block "52". Let \(x = 2.\overline{52}\). Then \(100x = 252.\overline{52}\). Subtracting, \(100x - x = 252.\overline{52} - 2.\overline{52} = 250\), so \(99x = 250\) and \(x = \frac{250}{99}\). The fraction is already in lowest term... | 1 | 0 | 4 | 1 | 0 | true | false |
d0342b0575e7ca41 | 0 | The graph, $G$ of $y=\log_{10}x$ is rotated $90^{\circ}$ counter-clockwise about the origin to obtain a new graph $G'$. What is the equation for $G'$? | 10^{-x} | The original graph is given by \(y = \log_{10} x\). Rotating a point \((x, y)\) by \(90^\circ\) counterclockwise about the origin transforms it to \((-y, x)\). Let \((X, Y)\) be coordinates on the rotated graph. Then there exists \((x_0, y_0)\) on the original graph such that \(X = -y_0\) and \(Y = x_0\). Solving, \(x_... | 1 | 0 | 4 | 1 | 0 | true | false |
9dd0529036c196e3 | 0 | A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $15$ and $25$ meters. What fraction of the yard is occupied by the flower beds?
[asy]
unitsize(2mm); defaultpen(... | \frac{1}{5} | We are given a rectangular yard with two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths 15 and 25 meters. We need to find what fraction of the yard is occupied by the flower beds.
First, inte... | 1 | 0 | 4 | 1 | 0 | true | false |
6dedd193b2338f2b | 0 | The expression $\frac{1^{4y-1}}{5^{-1}+3^{-1}}$ is equal to: | \frac{15}{8} | The numerator is \(1^{4y-1}\), which equals 1 for any real exponent. The denominator is \(5^{-1} + 3^{-1} = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}\). Thus, the expression simplifies to \(\frac{1}{\frac{8}{15}} = \frac{15}{8}\).
\boxed{\frac{15}{8}} | 1 | 0 | 4 | 1 | 0 | true | false |
88a00b02a1540e34 | 0 | At noon on a certain day, Minneapolis is $N$ degrees warmer than St. Louis. At $4{:}00$ the temperature in Minneapolis has fallen by $5$ degrees while the temperature in St. Louis has risen by $3$ degrees, at which time the temperatures in the two cities differ by $2$ degrees. What is the product of all possible values... | 60 | The temperatures at noon are \(T_M = T_S + N\). At 4:00, \(T_M' = T_M - 5 = T_S + N - 5\) and \(T_S' = T_S + 3\). The difference at 4:00 is \((T_S + N - 5) - (T_S + 3) = N - 8\). The absolute difference is 2, so \(|N - 8| = 2\), giving \(N = 10\) or \(N = 6\). The product of these values is \(10 \times 6 = 60\).
\boxe... | 1 | 0 | 4 | 1 | 0 | true | false |
83da7981583f00e5 | 0 | The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y=3x$ has equation $y=kx.$ What is $k?$ | \frac{1+\sqrt{5}}{2} | The angle bisectors of the lines \(y = x\) and \(y = 3x\) are given by
\[
\frac{y - x}{\sqrt{2}} = \pm \frac{y - 3x}{\sqrt{10}}.
\]
Substituting \(y = kx\) yields
\[
\frac{k - 1}{\sqrt{2}} = \pm \frac{k - 3}{\sqrt{10}}.
\]
The positive sign gives \(k = \frac{\sqrt{5} - 3}{\sqrt{5} - 1} \approx -0.618\), which lies outs... | 1 | 0 | 4 | 1 | 0 | true | false |
45aaf5b2d4204dab | 0 | A rectangular floor measures $a$ by $b$ feet, where $a$ and $b$ are positive integers with $b > a$. An artist paints a rectangle on the floor with the sides of the rectangle parallel to the sides of the floor. The unpainted part of the floor forms a border of width $1$ foot around the painted rectangle and occupies hal... | 2 | The equation from the given conditions is \(ab = 2(a-2)(b-2)\), which simplifies to \((a-4)(b-4)=8\). Since \(a\) and \(b\) are positive integers with \(b > a\) and the border requires \(a>2\) and \(b>2\), the only factor pairs of 8 with both factors positive and \(b>a\) are \((1,8)\) and \((2,4)\). These give \((a,b) ... | 1 | 0 | 4 | 1 | 0 | true | false |
243749518f54c572 | 0 | A square and an equilateral triangle have the same perimeter. Let $A$ be the area of the circle circumscribed about the square and $B$ the area of the circle circumscribed around the triangle. Find $A/B$. | \frac{27}{32} | The square and equilateral triangle have the same perimeter \(P\). For the square, side length \(s = P/4\) and circumradius \(R_s = s/\sqrt{2} = P/(4\sqrt{2})\). Its circumcircle area is \(A = \pi R_s^2 = \pi P^2/32\). For the triangle, side length \(t = P/3\) and circumradius \(R_t = t/\sqrt{3} = P/(3\sqrt{3})\). Its ... | 1 | 0 | 4 | 1 | 0 | true | false |
1fdf65986afa7402 | 0 | Positive integers $a$ and $b$ are such that the graphs of $y=ax+5$ and $y=3x+b$ intersect the $x$-axis at the same point. What is the sum of all possible $x$-coordinates of these points of intersection? | -8 | The x-intercepts of the lines \(y = ax + 5\) and \(y = 3x + b\) are \(-\frac{5}{a}\) and \(-\frac{b}{3}\), respectively. Setting them equal gives \(\frac{5}{a} = \frac{b}{3}\), so \(ab = 15\). Since \(a\) and \(b\) are positive integers, the possible pairs are \((1,15)\), \((3,5)\), \((5,3)\), \((15,1)\). The correspon... | 1 | 0 | 4 | 1 | 0 | true | false |
9464d196a863fba7 | 0 | Two years ago Pete was three times as old as his cousin Claire. Two years before that, Pete was four times as old as Claire. In how many years will the ratio of their ages be $2$ : $1$ ? | 4 | Let Pete's current age be \(P\) and Claire's current age be \(C\). Two years ago, Pete was three times as old as Claire:
\[
P - 2 = 3(C - 2) \implies P = 3C - 4.
\]
Two years before that (four years ago), Pete was four times as old as Claire:
\[
P - 4 = 4(C - 4) \implies P = 4C - 12.
\]
Equating the expressions for \(P... | 1 | 0 | 4 | 1 | 0 | true | false |
6cafbee62f045b81 | 0 | In the multiplication problem below $A$, $B$, $C$, $D$ are different digits. What is $A+B$?
$\begin{array}{cccc} & A & B & A\\ \times & & C & D\\ \hline C & D & C & D\\ \end{array}$ | 1 | The multiplication is \((100A + 10B + A) \times (10C + D) = 1000C + 100D + 10C + D\).
Simplify the left side: \((101A + 10B) \times (10C + D)\).
Simplify the right side: \(1010C + 101D = 101(10C + D)\).
Thus \((101A + 10B)(10C + D) = 101(10C + D)\).
Since \(10C + D \neq 0\) (it is a two‑digit number), divide bo... | 1 | 0 | 4 | 1 | 0 | true | false |
e14813c2b66609d0 | 0 | The sides of a triangle have lengths $6.5$, $10$, and $s$, where $s$ is a whole number. What is the smallest possible value of $s$? | 4 | The triangle inequalities require that the sum of any two sides is greater than the third side. For sides \(6.5\), \(10\), and \(s\) (with \(s\) an integer), the inequalities are:
- \(6.5 + 10 > s \Rightarrow 16.5 > s \Rightarrow s \le 16\).
- \(6.5 + s > 10 \Rightarrow s > 3.5 \Rightarrow s \ge 4\).
- \(10 + s > 6.5\... | 1 | 0 | 4 | 1 | 0 | true | false |
fe9ce89df1aeb5a3 | 0 | Let $f$ be the function defined by $f(x)=ax^2-\sqrt{2}$ for some positive $a$. If $f(f(\sqrt{2}))=-\sqrt{2}$ then $a=$ | \frac{\sqrt{2}}{2} | The function is \(f(x) = a x^2 - \sqrt{2}\) with \(a > 0\). Compute \(f(\sqrt{2})\):
\[
f(\sqrt{2}) = a (\sqrt{2})^2 - \sqrt{2} = 2a - \sqrt{2}.
\]
Then compute \(f(f(\sqrt{2})) = f(2a - \sqrt{2})\):
\[
f(2a - \sqrt{2}) = a (2a - \sqrt{2})^2 - \sqrt{2}.
\]
Set this equal to \(-\sqrt{2}\):
\[
a (2a - \sqrt{2})^2 - ... | 1 | 0 | 4 | 1 | 0 | true | false |
72d2c4c8ef42abdd | 0 | Given $0 \le x_0 < 1$, let
\[x_n = \begin{cases} 2x_{n-1} & \text{ if } 2x_{n-1} < 1 \\ 2x_{n-1} - 1 & \text{ if } 2x_{n-1} \ge 1 \end{cases}\]for all integers $n > 0$. For how many $x_0$ is it true that $x_0 = x_5$? | 31 | The map \(x_n = 2x_{n-1} \mod 1\) is the doubling map. The condition \(x_0 = x_5\) means \(x_0\) is a fixed point of the fifth iterate, i.e., \(T^5(x_0)=x_0\). This equation is equivalent to \(2^5 x_0 - x_0 \in \mathbb{Z}\), or \(31 x_0 \in \mathbb{Z}\). Since \(0 \le x_0 < 1\), we have \(x_0 = k/31\) for integers \(k ... | 1 | 0 | 4 | 1 | 0 | true | false |
dbd2f7c3d993cc5f | 0 | In $\triangle PQR$, $PR=15$, $QR=20$, and $PQ=25$. Points $A$ and $B$ lie on $\overline{PQ}$, points $C$ and $D$ lie on $\overline{QR}$, and points $E$ and $F$ lie on $\overline{PR}$, with $PA=QB=QC=RD=RE=PF=5$. Find the area of hexagon $ABCDEF$. | 150 | The triangle \(PQR\) has sides \(PR=15\), \(QR=20\), and \(PQ=25\). Since \(15^2+20^2=225+400=625=25^2\), triangle \(PQR\) is right-angled at \(R\). Its area is \(\frac{1}{2}\cdot15\cdot20=150\).
Points are placed on the sides: on \(PQ\), \(A\) and \(B\) with \(PA=5\) and \(QB=5\); on \(QR\), \(C\) and \(D\) with \(QC... | 0 | 0 | 4 | 0 | 0 | false | true |
eba2f8b1db1f4b38 | 0 | If $q_1(x)$ and $r_1$ are the quotient and remainder, respectively, when the polynomial $x^8$ is divided by $x + \frac{1}{2}$, and if $q_2(x)$ and $r_2$ are the quotient and remainder, respectively, when $q_1(x)$ is divided by $x + \frac{1}{2}$, then $r_2$ equals | -\frac{1}{16} | The remainder when \(x^8\) is divided by \(x + \frac{1}{2}\) is \(r_1 = \left(-\frac{1}{2}\right)^8 = \frac{1}{256}\). The quotient \(q_1(x)\) satisfies \(x^8 = \left(x + \frac{1}{2}\right) q_1(x) + \frac{1}{256}\). By the Remainder Theorem, \(r_2 = q_1\left(-\frac{1}{2}\right)\). Since \(q_1(x)\) is the polynomial suc... | 1 | 0 | 4 | 1 | 0 | true | false |
a972f9040ef9414f | 0 | Medians $AD$ and $CE$ of $\triangle ABC$ intersect in $M$. The midpoint of $AE$ is $N$.
Let the area of $\triangle MNE$ be $k$ times the area of $\triangle ABC$. Then $k$ equals: | \frac{1}{6} | The medians \(AD\) and \(CE\) intersect at the centroid \(M\). The midpoint of \(AE\) is \(N\). Using coordinates: let \(A=(0,0)\), \(B=(2,0)\) so that \(E=(1,0)\) (midpoint of \(AB\)), and \(N=(0.5,0)\) (midpoint of \(AE\)). Let \(C=(2u,2v)\) with \(v>0\). Then area of \(\triangle ABC\) is \(\frac12\cdot AB\cdot (\tex... | 0 | 0 | 4 | 0 | 0 | false | true |
a2febbdb64113979 | 0 | A palindrome between $1000$ and $10000$ is chosen at random. What is the probability that it is divisible by $7$? | \frac{1}{5} | The four-digit palindromes between 1000 and 10000 are of the form \(abba\) with \(a = 1,\dots,9\) and \(b = 0,\dots,9\). There are \(9 \times 10 = 90\) such numbers.
A palindrome \(abba\) equals \(1001a + 110b\). Since \(1001 = 7 \times 143\), it is divisible by 7. Therefore, the palindrome is divisible by 7 if and on... | 1 | 0 | 4 | 1 | 0 | true | false |
cce0f008bb5cda7f | 0 | Adams plans a profit of $10$ % on the selling price of an article and his expenses are $15$ % of sales. The rate of markup on an article that sells for $ $5.00$ is: | 33\frac {1}{3}\% | The selling price is $5.00. Profit is 10% of the selling price, so profit = $0.50. Expenses are 15% of sales, so expenses = $0.75. The cost is the selling price minus profit and expenses:
\[
\text{cost} = 5.00 - 0.50 - 0.75 = 3.75.
\]
The markup is the difference between selling price and cost:
\[
\text{markup} =... | 1 | 0 | 4 | 1 | 0 | true | false |
d82d59b8a70f5a8f | 0 | In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB, BC$, and $CD$ are diameters of circle $O, N$, and $P$, respectively. Circles $O, N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has len... | 20 | The line \(AG\) passes through \(A(0,0)\) and is tangent to circle \(P\) (center \((75,0)\), radius \(15\)). The slope \(m\) satisfies the tangency condition: distance from \(P\) to line \(y=mx\) equals \(15\).
\[
\frac{75|m|}{\sqrt{1+m^2}} = 15 \quad\Rightarrow\quad \frac{|m|}{\sqrt{1+m^2}} = \frac{1}{5}.
\]
For c... | 0 | 0 | 4 | 0 | 0 | false | true |
35dc883e5aa2aff4 | 0 | Points $A,B,C,D,E$ and $F$ lie, in that order, on $\overline{AF}$, dividing it into five segments, each of length 1. Point $G$ is not on line $AF$. Point $H$ lies on $\overline{GD}$, and point $J$ lies on $\overline{GF}$. The line segments $\overline{HC}, \overline{JE},$ and $\overline{AG}$ are parallel. Find $HC/JE$. | \frac{5}{3} | The points \(A, B, C, D, E, F\) lie in order on \(\overline{AF}\) with each segment of length 1. Place them on the \(x\)-axis: \(A=(0,0)\), \(B=(1,0)\), \(C=(2,0)\), \(D=(3,0)\), \(E=(4,0)\), \(F=(5,0)\). Let \(G=(p,q)\) with \(q\neq 0\). Then \(AG\) has direction vector \((p,q)\). Let \(H\) be on \(\overline{GD}\) and... | 1 | 0 | 4 | 1 | 0 | true | false |
78093557a3c04ff6 | 1 | For all non-zero numbers $x$ and $y$ such that $x = 1/y$, $\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)$ equals | x^2-y^2 | Given the condition \(x = \frac{1}{y}\), we can substitute \(y = \frac{1}{x}\) into the expression:
\[
\left(x - \frac{1}{x}\right)\left(y + \frac{1}{y}\right) = \left(x - \frac{1}{x}\right)\left(\frac{1}{x} + x\right) = \left(x - \frac{1}{x}\right)\left(x + \frac{1}{x}\right) = x^2 - \frac{1}{x^2}.
\]
Alternatively,... | 1 | 0.866025 | 4 | 0.5 | 0.57735 | false | false |
f28a17200deb9373 | 0 | In square $ABCD$, points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$, respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$, with $BR = 6$ and $PR = 7$. What is the area of the square? | 117 | The area of the square is \(117\). The side length \(s\) satisfies \(s^2 + p^2 = 169\) from \(BP = 13\), and perpendicularity gives \(q = s - p\). Solving the intersection condition yields \(s = \frac{3}{2}p\), leading to \(s^2 = 117\). Thus the area is \(117\).
\boxed{117} | 1 | 0 | 4 | 1 | 0 | true | false |
202ee7b40196721f | 0 | A man on his way to dinner shortly after $6:00$ p.m. observes that the hands of his watch form an angle of $110^{\circ}$. Returning before $7:00$ p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$. The number of minutes that he has been away is: | 40 | The man leaves shortly after 6:00 p.m. and returns before 7:00 p.m. The angle between the hour and minute hands is \(110^\circ\) at both times.
Let \(t\) be the number of minutes after 6:00. The hour hand is at \(180^\circ + 0.5t\) and the minute hand is at \(6t\). The difference is \(180 - 5.5t\) for \(t < \frac{180}... | 1 | 0 | 4 | 1 | 0 | true | false |
3a164ab2f07fc4a5 | 0 | A set $S$ of points in the $xy$-plane is symmetric about the origin, both coordinate axes, and the line $y=x$. If $(2,3)$ is in $S$, what is the smallest number of points in $S$? | 8 | The set \(S\) must be invariant under reflection across the \(x\)-axis, reflection across the \(y\)-axis, reflection across the line \(y=x\), and inversion through the origin. Starting from \((2,3)\), these symmetries generate the following points:
- Reflection across \(x\)-axis: \((2,-3)\)
- Reflection across \(y\)-a... | 1 | 0 | 4 | 1 | 0 | true | false |
598860e7cace6a9a | 0 | In this diagram, not drawn to scale, Figures $I$ and $III$ are equilateral triangular regions with respective areas of $32\sqrt{3}$ and $8\sqrt{3}$ square inches. Figure $II$ is a square region with area $32$ square inches. Let the length of segment $AD$ be decreased by $12\frac{1}{2}$% of itself, while the lengths of ... | 25 | The areas of the equilateral triangles and the square give the side lengths:
- For equilateral triangle I (area \(32\sqrt{3}\)):
\(\frac{\sqrt{3}}{4}s^2 = 32\sqrt{3} \Rightarrow s^2 = 128 \Rightarrow s = 8\sqrt{2}\). So \(AB = 8\sqrt{2}\) inches.
- For equilateral triangle III (area \(8\sqrt{3}\)):
\(\frac{\s... | 0 | 0 | 4 | 0 | 0 | false | true |
7777b9fae19fb25e | 0 | Let $ABC$ be an equilateral triangle. Extend side $\overline{AB}$ beyond $B$ to a point $B'$ so that $BB'=3 \cdot AB$. Similarly, extend side $\overline{BC}$ beyond $C$ to a point $C'$ so that $CC'=3 \cdot BC$, and extend side $\overline{CA}$ beyond $A$ to a point $A'$ so that $AA'=3 \cdot CA$. What is the ratio of the... | 16 | The ratio of the area of $\triangle A'B'C'$ to the area of $\triangle ABC$ is $37$.
\boxed{37} | 0 | 0 | 4 | 0 | 0 | false | true |
7189e002e0e793cd | 0 | A shopper plans to purchase an item that has a listed price greater than $\$100$ and can use any one of the three coupons. Coupon A gives $15\%$ off the listed price, Coupon B gives $\$30$ off the listed price, and Coupon C gives $25\%$ off the amount by which the listed price exceeds
$\$100$.
Let $x$ and $y$ be the s... | 50 | The listed price \(P > 100\) satisfies the conditions that Coupon A saves at least as many dollars as Coupon B and at least as many dollars as Coupon C.
Savings:
- Coupon A: \(0.15P\)
- Coupon B: \(30\)
- Coupon C: \(0.25(P-100)\)
Inequalities:
1. \(0.15P \ge 30 \Rightarrow P \ge 200\)
2. \(0.15P \ge 0.25(P-100) \Rig... | 1 | 0 | 4 | 1 | 0 | true | false |
b70e4f2c910ef552 | 0 | An amusement park has a collection of scale models, with a ratio of $1: 20$, of buildings and other sights from around the country. The height of the United States Capitol is $289$ feet. What is the height in feet of its duplicate to the nearest whole number? | 14 | The scale ratio of 1:20 means the model is \(\frac{1}{20}\) the size of the actual building. The actual height of the United States Capitol is 289 feet. Therefore, the model's height is \(\frac{289}{20} = 14.45\) feet. Rounding to the nearest whole number gives 14 feet.
\boxed{14} | 1 | 0 | 4 | 1 | 0 | true | false |
56ff814d7548827d | 0 | Doug constructs a square window using $8$ equal-size panes of glass. The ratio of the height to width for each pane is $5 : 2$, and the borders around and between the panes are $2$ inches wide. In inches, what is the side length of the square window? | 26 | The square window is constructed from 8 equal-sized panes arranged in a rectangular grid. Let the number of rows be \(r\) and columns be \(c\), so \(rc = 8\). Each pane has height \(h\) and width \(w\) with \(h:w = 5:2\), so \(h = \frac{5}{2}w\). The borders (frame and mullions) are 2 inches wide. The total height of t... | 1 | 0 | 4 | 1 | 0 | true | false |
c37b8f16dac85444 | 0 | In $\triangle ABC$, $\angle ABC=45^\circ$. Point $D$ is on $\overline{BC}$ so that $2 \cdot BD=CD$ and $\angle DAB=15^\circ$. Find $\angle ACB.$ | 75^\circ | The correct answer is \(\angle ACB = 75^\circ\).
We are given triangle \(ABC\) with \(\angle ABC = 45^\circ\). Point \(D\) lies on \(\overline{BC}\) such that \(2 \cdot BD = CD\) (so \(BD:DC = 1:2\)) and \(\angle DAB = 15^\circ\). We need \(\angle ACB\).
In \(\triangle ABD\), \(\angle ABD = 45^\circ\) and \(\angle DA... | 1 | 0 | 4 | 1 | 0 | true | false |
f3a82ab4d2598a74 | 0 | In $\triangle ABC$ with right angle at $C$, altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$, then the area of $\triangle ABC$ is | 4K | The right triangle \( \triangle ABC \) with right angle at \( C \) has altitude \( CH \) and median \( CM \) that trisect the right angle. This forces the triangle to be a 30-60-90 triangle. Without loss of generality, assume the legs are \( AC = b \) and \( BC = a \) with \( a = \sqrt{3}b \), \( c = 2b \). Using coord... | 1 | 0 | 4 | 1 | 0 | true | false |
48f5622fe1ef2b5f | 0 | The numbers $-2, 4, 6, 9$ and $12$ are rearranged according to these rules:
1. The largest isn't first, but it is in one of the first three places.
2. The smallest isn't last, but it is in one of the last three places.
3. The median isn't first or last.
What is the average of the first and l... | 6.5 | The numbers \(-2, 4, 6, 9, 12\) are rearranged. The largest is \(12\), the smallest is \(-2\), and the median is \(6\). The rules restrict their positions:
- \(12\) cannot be first but must be among the first three positions, so \(12 \in \{2,3\}\).
- \(-2\) cannot be last but must be among the last three positions, so ... | 1 | 0 | 4 | 1 | 0 | true | false |
347410354498b3d6 | 0 | Five positive consecutive integers starting with $a$ have average $b$. What is the average of $5$ consecutive integers that start with $b$? | $a+4$ | The five consecutive integers starting with \(a\) are \(a, a+1, a+2, a+3, a+4\). Their average is
\[
b = \frac{a + (a+1) + (a+2) + (a+3) + (a+4)}{5} = \frac{5a+10}{5} = a+2.
\]
Now consider five consecutive integers starting with \(b\): \(b, b+1, b+2, b+3, b+4\). Their average is
\[
\frac{b + (b+1) + (b+2) + (b+3) + (b... | 0 | 0 | 4 | 0 | 0 | false | true |
4488e36443f85cd0 | 0 | A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\frac{7}{17}$. What number was erased? | 7 | The set consists of consecutive positive integers from 1 to \(N\). After erasing one number \(k\), the average of the remaining numbers is \(35\frac{7}{17} = \frac{602}{17}\). This gives
\[
\frac{\frac{N(N+1)}{2} - k}{N-1} = \frac{602}{17}.
\]
Multiplying both sides by \(2(N-1)\) yields
\[
N(N+1) - 2k = \frac{1204}{... | 1 | 0 | 4 | 1 | 0 | true | false |
05fbe6833e50430e | 0 | Consider all 1000-element subsets of the set $\{1, 2, 3, \dots , 2015\}$. From each such subset choose the least element. The arithmetic mean of all of these least elements is $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p + q$. | 2016 | The average minimum of all 1000-element subsets of $\{1,2,\dots,2015\}$ is $\frac{2016}{1001}$, which reduces to $\frac{288}{143}$. Therefore $p+q = 288+143 = 431$.
\boxed{431} | 0 | 0 | 4 | 0 | 0 | false | true |
91cb4ca9adfffc11 | 0 | Four circles, no two of which are congruent, have centers at $A$, $B$, $C$, and $D$, and points $P$ and $Q$ lie on all four circles. The radius of circle $A$ is $\frac{5}{8}$ times the radius of circle $B$, and the radius of circle $C$ is $\frac{5}{8}$ times the radius of circle $D$. Furthermore, $AB = CD = 39$ and $PQ... | 192 | The centers of the circles all lie on the perpendicular bisector of \(PQ\). Place \(R\), the midpoint of \(PQ\), at the origin and let \(PQ\) be vertical, so \(P=(0,24)\), \(Q=(0,-24)\). Then the centers have coordinates \((x,0)\) and the radii satisfy \(r^2 = x^2 + 24^2 = x^2 + 576\). For a pair of centers with radii ... | 1 | 0 | 4 | 1 | 0 | true | false |
e847a140e6ba9445 | 0 | If $x, 2x+2, 3x+3, \dots$ are in geometric progression, the fourth term is: | -13\frac{1}{2} | The terms are in geometric progression, so the ratio between consecutive terms is constant:
\[
\frac{2x+2}{x} = \frac{3x+3}{2x+2}.
\]
Cross-multiplying gives:
\[
(2x+2)^2 = x(3x+3).
\]
Expanding both sides:
\[
4x^2 + 8x + 4 = 3x^2 + 3x.
\]
Bring all terms to one side:
\[
x^2 + 5x + 4 = 0.
\]
Factoring:
\[
(x+1)(x+4) = ... | 1 | 0 | 4 | 1 | 0 | true | false |
55d4f310988cac0b | 0 | A line that passes through the origin intersects both the line $x = 1$ and the line $y=1+ \frac{\sqrt{3}}{3} x$. The three lines create an equilateral triangle. What is the perimeter of the triangle? | 3 + 2\sqrt{3} | The line through the origin is \(y = mx\). The fixed lines are \(x = 1\) and \(y = 1 + \frac{\sqrt{3}}{3}x\).
The angle between the fixed lines is \(60^\circ\) because the vertical line has angle \(90^\circ\) and the second line has slope \(\frac{\sqrt{3}}{3}\) giving angle \(30^\circ\); the difference is \(60^\circ\... | 1 | 0 | 4 | 1 | 0 | true | false |
01ad6c00de60af03 | 0 | Trapezoid $ABCD$ has $\overline{AB} \parallel \overline{CD}, BC=CD=43$, and $\overline{AD} \perp \overline{BD}$. Let $O$ be the intersection of the diagonals $\overline{AC}$ and $\overline{BD}$, and let $P$ be the midpoint of $\overline{BD}$. Given that $OP=11$, the length of $AD$ can be written in the form $m\sqrt{n}$... | 194 | We have trapezoid \(ABCD\) with \(\overline{AB} \parallel \overline{CD}\), \(BC = CD = 43\), and \(\overline{AD} \perp \overline{BD}\).
Place \(D\) at the origin, let \(\overline{AD}\) lie on the positive \(x\)-axis, and \(\overline{BD}\) lie on the positive \(y\)-axis.
Set \(A = (a,0)\) where \(a = AD > 0\) and \(... | 1 | 0 | 4 | 1 | 0 | true | false |
699f0909b41710ce | 0 | If each of two intersecting lines intersects a hyperbola and neither line is tangent to the hyperbola, then the possible number of points of intersection with the hyperbola is: | 2, 3, or 4 | The two intersecting lines each intersect the hyperbola and are not tangent. A line can intersect a hyperbola in either 1 point (if it is parallel to an asymptote) or 2 points (if it is a secant). Since the two lines intersect, they share exactly one point. The total number of distinct intersection points of the two li... | 0 | 0 | 4 | 0 | 0 | false | true |
265814acf3a84c1c | 0 | An architect is building a structure that will place vertical pillars at the vertices of regular hexagon $ABCDEF$, which is lying horizontally on the ground. The six pillars will hold up a flat solar panel that will not be parallel to the ground. The heights of pillars at $A$, $B$, and $C$ are $12$, $9$, and $10$ meter... | 17 | The heights at the vertices of a regular hexagon follow a linear function because the solar panel is flat. For a centrally symmetric polygon, the sum of the heights at opposite vertices is constant and equal to twice the height at the center. Let \(c\) be the height at the center. Then for opposite pairs:
\[
h_A + h_D ... | 1 | 0 | 4 | 1 | 0 | true | false |
e440b38dc9a147c6 | 0 | A finite sequence of three-digit integers has the property that the tens and units digits of each term are, respectively, the hundreds and tens digits of the next term, and the tens and units digits of the last term are, respectively, the hundreds and tens digits of the first term. For example, such a sequence might be... | 37 | The sum of all terms in the sequence is \(S = 111U\), where \(U\) is the sum of the units digits of all terms. Since each units digit is between 1 and 9, \(U\) can vary (e.g., \(U=1\) for the sequence consisting of the single term \(111\)). Thus the only prime factors guaranteed to divide \(S\) are the prime factors of... | 1 | 0 | 4 | 1 | 0 | true | false |
9d1f8b048fe98451 | 0 | A powderman set a fuse for a blast to take place in $30$ seconds. He ran away at a rate of $8$ yards per second. Sound travels at the rate of $1080$ feet per second. When the powderman heard the blast, he had run approximately: | 245 yd. | The explosion occurs 30 seconds after the fuse is lit. During this time, the powderman runs at 8 yards per second, covering a distance of \(8 \times 30 = 240\) yards. After the explosion, sound travels from the blast site to the powderman at 1080 feet per second, which is \(1080/3 = 360\) yards per second. The powderma... | 1 | 0 | 4 | 1 | 0 | true | false |
731350ee7a558dd7 | 0 | The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$, respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$, and that $L_1$ has 4 times the slope of $L_2$. If $L_1$ is not horizontal, then $mn$ is | 2 | The slopes of the lines are \(m\) and \(n\) with \(m = 4n\). Let \(\theta\) be the angle \(L_2\) makes with the positive \(x\)-axis, so \(n = \tan\theta\). Then \(L_1\) makes angle \(2\theta\) and \(m = \tan(2\theta)\). Using the double-angle formula,
\[
\tan(2\theta) = \frac{2\tan\theta}{1 - \tan^2\theta}.
\]
Substitu... | 1 | 0 | 4 | 1 | 0 | true | false |
16f219499e0f60e4 | 0 | Two fair dice, each with at least $6$ faces are rolled. On each face of each die is printed a distinct integer from $1$ to the number of faces on that die, inclusive. The probability of rolling a sum of $7$ is $\frac34$ of the probability of rolling a sum of $10,$ and the probability of rolling a sum of $12$ is $\frac{... | 17 | The dice have standard numbering, with the first die having \(m\) faces labeled \(1,\ldots,m\) and the second die having \(n\) faces labeled \(1,\ldots,n\), where \(m,n\ge 6\). For any such dice, the number of ways to roll a sum of \(7\) is \(f(7)=6\) because \(7\le \min(m,n)+1\). The given condition \(P(7)=\frac34 P(1... | 1 | 0 | 4 | 1 | 0 | true | false |
a7986c74e93a5151 | 0 | In the cube $ABCDEFGH$ with opposite vertices $C$ and $E,$ $J$ and $I$ are the midpoints of segments $\overline{FB}$ and $\overline{HD},$ respectively. Let $R$ be the ratio of the area of the cross-section $EJCI$ to the area of one of the faces of the cube. What is $R^2?$ | \frac{9}{4} | The cube has side length \(s\). Place it in a coordinate system with vertices:
\[
A = (0,0,0),\; B = (s,0,0),\; C = (s,s,0),\; D = (0,s,0),\; E = (0,0,s),\; F = (s,0,s),\; G = (s,s,s),\; H = (0,s,s).
\]
\(J\) is the midpoint of \(\overline{FB}\) and \(I\) the midpoint of \(\overline{HD}\):
\[
J = \left(s,0,\frac{s}{2}\... | 0 | 0 | 4 | 0 | 0 | false | true |
0927268611c1cab5 | 0 | Quadrilateral $ABCD$ has $AB = BC = CD$, $m\angle ABC = 70^\circ$ and $m\angle BCD = 170^\circ$. What is the degree measure of $\angle BAD$? | 85 | The quadrilateral \(ABCD\) has \(AB = BC = CD\), \(\angle ABC = 70^\circ\), and \(\angle BCD = 170^\circ\). Using the given side equalities, triangles \(ABC\) and \(BCD\) are isosceles.
In \(\triangle ABC\), \(AB = BC\) gives \(\angle BAC = \angle BCA = \frac{180^\circ-70^\circ}{2}=55^\circ\).
In \(\triangle BCD\),... | 1 | 0 | 4 | 1 | 0 | true | false |
8b13f71f5faf2c06 | 0 | What is the sum of all possible values of $t$ between $0$ and $360$ such that the triangle in the coordinate plane whose vertices are $(\cos 40^\circ,\sin 40^\circ)$, $(\cos 60^\circ,\sin 60^\circ)$, and $(\cos t^\circ,\sin t^\circ)$ is isosceles? | 380 | The triangle vertices are on the unit circle, so side lengths are proportional to the sine of half the central angle between the points. The given points are at \(40^\circ\) and \(60^\circ\); the side \(AB\) corresponds to a central angle of \(20^\circ\) (since \(\min(|40-60|,360-|40-60|)=20\)). For the triangle to be ... | 1 | 0 | 4 | 1 | 0 | true | false |
4b5e38006d9dd6ae | 0 | What is the greatest three-digit positive integer $n$ for which the sum of the first $n$ positive integers is not a divisor of the product of the first $n$ positive integers? | 996 | The sum of the first \(n\) positive integers is \(S = \frac{n(n+1)}{2}\), and the product is \(P = n!\). The condition that \(S\) does not divide \(P\) is equivalent to \(\frac{n(n+1)}{2} \nmid n!\). Simplifying,
\[
\frac{n!}{n(n+1)/2} = \frac{2(n-1)!}{n+1},
\]
so \(S \mid P\) iff \(n+1 \mid 2(n-1)!\).
For \(n \ge... | 1 | 0 | 4 | 1 | 0 | true | false |
82f8f4c0b0563dc3 | 0 | Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC=20,$ and $CD=30.$ Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E,$ and $AE=5.$ What is the area of quadrilateral $ABCD?$ | 360 | The area of quadrilateral \(ABCD\) is the sum of the areas of right triangles \(ABC\) and \(ACD\).
Triangle \(ACD\) has legs \(AC = 20\) and \(CD = 30\), so its area is \(\frac{1}{2} \cdot 20 \cdot 30 = 300\).
Place coordinates: \(C = (0,0)\), \(A = (20,0)\) (since \(AC = 20\)), and \(D = (0,30)\) (since \(CD = 30\... | 1 | 0 | 4 | 1 | 0 | true | false |
beed5c7974a048d6 | 0 | Mary thought of a positive two-digit number. She multiplied it by $3$ and added $11$. Then she switched the digits of the result, obtaining a number between $71$ and $75$, inclusive. What was Mary's number? | 12 | The original number is denoted \(N\), a positive two-digit integer. After multiplying by 3 and adding 11, we obtain \(R = 3N + 11\). The digits of \(R\) are then switched (reversed) to produce a number \(S\) with \(71 \le S \le 75\).
Assume \(R\) is a two-digit number (the most natural interpretation of "switched the ... | 1 | 0.5 | 4 | 0.75 | 0.5 | false | false |
fddc576a2cdf2a79 | 0 | Letters $A, B, C,$ and $D$ represent four different digits selected from $0, 1, 2, \ldots ,9.$ If $(A+B)/(C+D)$ is an integer that is as large as possible, what is the value of $A+B$? | 17 | The maximum possible value of \((A+B)/(C+D)\) occurs when the numerator is as large as possible and the denominator is as small as possible, while ensuring the quotient is an integer and all four digits are distinct. The smallest positive sum of two distinct digits is \(0+1=1\). The largest sum of two distinct digits f... | 1 | 0 | 4 | 1 | 0 | true | false |
5a80792661636065 | 0 | Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle at points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle? | \frac{4}{3}-\frac{4\sqrt{3}\pi}{27} | The equilateral triangle \(ABC\) has side length \(s\). Place coordinates: \(B=(0,0)\), \(C=(s,0)\), \(A=\bigl(\frac{s}{2},\frac{\sqrt{3}}{2}s\bigr)\).
The circle is tangent to \(AB\) at \(B\) and to \(AC\) at \(C\). The center \(O\) is the intersection of the lines through \(B\) perpendicular to \(AB\) and through \... | 1 | 0 | 4 | 1 | 0 | true | false |
fd4324e620cfebf7 | 0 | Let $P(x)$ be a polynomial of degree $3n$ such that
\begin{align*} P(0) = P(3) = \dots = P(3n) &= 2, \\ P(1) = P(4) = \dots = P(3n+1-2) &= 1, \\ P(2) = P(5) = \dots = P(3n+2-2) &= 0. \end{align*}
Also, $P(3n+1) = 730$. Determine $n$. | 1 | The polynomial \(P(x)\) of degree \(3n\) satisfies
\[
P(k) = \begin{cases}
2, & k \equiv 0 \pmod{3}\\
1, & k \equiv 1 \pmod{3}\\
0, & k \equiv 2 \pmod{3}
\end{cases}
\]
for \(k = 0,1,\dots,3n\), and additionally \(P(3n+1)=730\).
Since the \((3n+1)\)-st forward difference of a polynomial of degree \(3n\) is zero,
\[
\s... | 0 | 0 | 4 | 0 | 0 | false | true |
65430ab6d0228693 | 0 | A student council must select a two-person welcoming committee and a three-person planning committee from among its members. There are exactly $10$ ways to select a two-person team for the welcoming committee. It is possible for students to serve on both committees. In how many different ways can a three-person plannin... | 10 | The number of ways to select a two-person welcoming committee is given as 10. For a council with \(n\) members, this number is \(\binom{n}{2} = \frac{n(n-1)}{2} = 10\). Solving \(n(n-1)=20\) gives \(n=5\) (since \(n\) is positive). The number of ways to select a three-person planning committee from the same 5 members i... | 1 | 0 | 4 | 1 | 0 | true | false |
4caf1fa59dc3d7ae | 0 | An $11 \times 11 \times 11$ wooden cube is formed by gluing together $11^3$ unit cubes. What is the greatest number of unit cubes that can be seen from a single point? | 331 | The greatest number of unit cubes that can be seen from a single point occurs when the viewpoint is located so that three mutually perpendicular faces of the large cube are visible. For an \(11 \times 11 \times 11\) cube, each face contains \(11^2 = 121\) unit cubes, each edge contains 11 cubes, and each corner is a si... | 1 | 0 | 4 | 1 | 0 | true | false |
cc905742599fe813 | 1 | The circumference of the circle with center $O$ is divided into $12$ equal arcs, marked the letters $A$ through $L$ as seen below. What is the number of degrees in the sum of the angles $x$ and $y$? | 90 | The circle is divided into 12 equal arcs, so each central angle is \(360^\circ/12 = 30^\circ\). Points \(A\) and \(G\) are opposite because they are 6 arcs apart, making \(AG\) a diameter. In triangle \(ADG\), \(\angle ADG\) is an inscribed angle subtending the diameter \(AG\), so \(\angle ADG = 90^\circ\). The sum of ... | 1 | 1.5 | 4 | 0.25 | 0.5 | false | false |
6ae58502ed0acb90 | 0 | A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$. What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube? | 90 | The 3×3×3 cube consists of 27 dice. Each die has opposite faces summing to 7. The visible faces are the 54 faces on the surface of the large cube. To minimize the sum of these visible numbers, we want each die to show its smallest numbers on its exposed faces.
- The central die has no exposed faces; its entire sum (21... | 1 | 0 | 4 | 1 | 0 | true | false |
1ed807025133afe6 | 0 | Two long cylindrical tubes of the same length but different diameters lie parallel to each other on a flat surface. The larger tube has radius $72$ and rolls along the surface toward the smaller tube, which has radius $24$. It rolls over the smaller tube and continues rolling along the flat surface until it comes to re... | 312 | The larger cylinder of radius \(R=72\) rolls without slipping on a flat surface, over a fixed smaller cylinder of radius \(r=24\), and continues until it completes one full revolution. The total rotation is \(2\pi\) radians. During the overpass, the center of the larger cylinder moves along a circular arc of radius \(R... | 0 | 0 | 4 | 0 | 0 | false | true |
a6aa25cbc57f9d1b | 0 | Square $EFGH$ has one vertex on each side of square $ABCD$. Point $E$ is on $AB$ with $AE=7\cdot EB$. What is the ratio of the area of $EFGH$ to the area of $ABCD$? | \frac{25}{32} | The side length of square \(ABCD\) is \(AB = AE + EB\). Given \(AE = 7 \cdot EB\), let \(EB = x\), then \(AE = 7x\) and \(AB = 8x\). Place square \(ABCD\) with coordinates \(A(0,0)\), \(B(8x,0)\), \(C(8x,8x)\), \(D(0,8x)\). Then \(E = (7x,0)\).
Let \(F\) on \(BC\) be \((8x, y_F)\), \(G\) on \(CD\) be \((x_G, 8x)\), \(... | 1 | 0 | 4 | 1 | 0 | true | false |
382ed772f39bb016 | 0 | Jose is $4$ years younger than Zack. Zack is $3$ years older than Inez. Inez is $15$ years old. How old is Jose? | 14 | Inez is 15 years old. Zack is 3 years older than Inez, so Zack is 15 + 3 = 18 years old. Jose is 4 years younger than Zack, so Jose is 18 - 4 = 14 years old.
\boxed{14} | 1 | 0 | 4 | 1 | 0 | true | false |
b1f7311d004f9d4f | 0 | The product $8 \times .25 \times 2 \times .125 =$ | $\frac{1}{2}$ | The product is computed as follows:
\[
8 \times 0.25 = 2,
\]
\[
2 \times 2 = 4,
\]
\[
4 \times 0.125 = 0.5.
\]
Alternatively, using properties: \(8 \times 0.125 = 1\) and \(0.25 \times 2 = 0.5\), so the product is \(1 \times 0.5 = 0.5\).
Thus, the result is \(0.5\).
\boxed{0.5} | 1 | 0 | 4 | 1 | 0 | true | false |
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Check out the documentation for more information.
compare-offline-grpo
Сравнение 6 методов оффлайн RL/RFT для дистилляции ризонинга.
Полный план — в CLAUDE.md. Этот файл — короткий обзор для быстрого старта.
Quickstart
# 1) deps
python -m venv .venv && source .venv/bin/activate
pip install -r requirements.txt
# 2) secrets
cp .env.example .env
# .env уже содержит OPENROUTER_API_KEY (см. CLAUDE.md). Заполни HF_TOKEN, WANDB_API_KEY.
# 3) eval-сеты + промпты для тренировки
python scripts/download_eval_sets.py
python scripts/download_prompts.py --source deepscaler
# 4) роллауты учителя (smoke на 50 промптах)
python scripts/generate_rollouts.py --limit 50 --K 4
# 5) verify
python scripts/verify_rollouts.py
# 6) verify + prepare method-specific views from raw rollouts
python scripts/prepare_rollouts.py \
--input-dir data/rollouts/raw \
--verified-output data/rollouts/verified/all.parquet \
--train-dir data/train
# 7) train (один метод)
python scripts/train.py --config configs/train_sft.yaml
python scripts/train.py --config configs/train_rft.yaml
python scripts/train.py --config configs/train_dft.yaml
python scripts/train.py --config configs/train_rift.yaml
python scripts/train.py --config configs/train_offline_grpo.yaml
python scripts/train.py --config configs/train_dpo.yaml
# 8) eval
python scripts/eval_pass_at_k.py --checkpoint results/checkpoints/sft_<id>/last
Layout
data/
prompts/ # тренировочные задачи (DeepScaleR 40K)
rollouts/raw/ # K completions/prompt от учителя
rollouts/verified # + reward + ref_logprobs
train/ # method-specific срезы (один источник)
eval/ # GSM8K, MATH-500, AMC23, AIME25
src/
losses/ # SFT, RFT, DFT, RIFT, GRPO, DPO/KTO
verifier/ # math-verify + boxed extractor + string-match
data/ # Datasets, collators
eval/ # pass@k, бенчи
scripts/ # downloaders, generate, verify, prepare, train, eval
configs/ # YAML-конфиги
results/ # checkpoints, eval-логи, таблицы, графики
Модели
- Учитель:
deepseek/deepseek-v4-pro(OpenRouter) - Ученик:
liquid/lfm-2.5-1.2b-thinking:freeдля inference / HF-аналог для тренировки
См. CLAUDE.md → раздел «Модели» (про OpenRouter ID vs HF веса).
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