Benchmark Freeform Rewrites
Collection
Evals rewritten into free-form worked answers • 9 items • Updated
id string | question string | answer string | freeform_answer string |
|---|---|---|---|
0 | Cities $A$ and $B$ are $45$ miles apart. Alicia lives in $A$ and Beth lives in $B$. Alicia bikes towards $B$ at 18 miles per hour. Leaving at the same time, Beth bikes toward $A$ at 12 miles per hour. How many miles from City $A$ will they be when they meet? | 27 | They meet after 45/(18+12) = 45/30 = 3/2 hours. Alicia travels 18*(3/2) = 27 miles from A. \boxed{27} |
1 | Positive real numbers $x$ and $y$ satisfy $y^3=x^2$ and $(y-x)^2=4y^2$. What is $x+y$? | 36 | From (y-x)^2 = 4y^2, take square roots: y-x = 2y (taking positive root since y>x would give y-x positive, but let's check both).
Case 1: y-x = 2y → -x = y → impossible (positive reals).
Case 2: y-x = -2y → y+2y = x → x = 3y.
Now y^3 = x^2 = 9y^2, so y = 9, x = 27.
x+y = 36.
\boxed{36} |
11 | The weight of $\frac{1}{3}$ of a large pizza together with $3 \frac{1}{2}$ cups of orange slices is the same as the weight of $\frac{3}{4}$ of a large pizza together with $\frac{1}{2}$ cup of orange slices. A cup of orange slices weighs $\frac{1}{4}$ of a pound. What is the weight, in pounds, of a large pizza? The answ... | 4 | Let p = weight of pizza, c = 1/4 pound per cup.
(1/3)p + 3.5(1/4) = (3/4)p + 0.5(1/4)
(1/3)p + 7/8 = (3/4)p + 1/8
7/8 - 1/8 = (3/4 - 1/3)p
6/8 = (9/12 - 4/12)p
3/4 = (5/12)p
p = (3/4)(12/5) = 9/5
m=9, n=5, m-n=4
\boxed{4} |
13 | If $A$ and $B$ are vertices of a polyhedron, define the distance $d(A,B)$ to be the minimum number of edges of the polyhedron one must traverse in order to connect $A$ and $B$. For example, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$, but if $\overline{AC}$ and $\overline{CB}$ are edges and $\ov... | 29 | A regular icosahedron has 12 vertices. Each vertex connects to 5 others at distance 1, 5 others at distance 2, and 1 vertex at distance 3 (the antipode).
For any fixed vertex R, the distances from R: 5 vertices at d=1, 5 at d=2, 1 at d=3.
We need P(d(Q,R) > d(R,S)) where Q, R, S are distinct.
Total ways to choose or... |
15 | How many ordered pairs of positive real numbers $(a,b)$ satisfy the equation
\[(1+2a)(2+2b)(2a+b) = 32ab?\] | 1 | By AM-GM on each factor:
1+2a >= 2*sqrt(2a), with equality when 1=2a, i.e., a=1/2.
2+2b >= 2*sqrt(2b)*... let me apply AM-GM directly to the product.
Apply AM-GM: (1+2a) >= 2*sqrt(2a), (2+2b) >= 2*2*sqrt(b/...
Let me try: write 1+2a >= 2sqrt(2a), 2+2b = 2(1+b) >= 2*2*sqrt(b) = 4sqrt(b), 2a+b >= 2sqrt(2ab).
Product... |
16 | Let $K$ be the number of sequences $A_1$, $A_2$, $\dots$, $A_n$ such that $n$ is a positive integer less than or equal to $10$, each $A_i$ is a subset of $\{1, 2, 3, \dots, 10\}$, and $A_{i-1}$ is a subset of $A_i$ for each $i$ between $2$ and $n$, inclusive. For example, $\{\}$, $\{5, 7\}$, $\{2, 5, 7\}$, $\{2, 5, 7\}... | 5 | For each element in {1,...,10}, independently decide its "entry pattern" across the sequence. For a sequence of length n, each element can first appear at position i (for i=1,...,n) or never appear. That gives (n+1) choices per element.
So for fixed n, the count is (n+1)^10.
K = sum_{n=1}^{10} (n+1)^10 = 2^10 + 3^10 ... |
17 | There is a unique sequence of integers $a_1, a_2, \cdots a_{2023}$ such that
\[\tan2023x = \frac{a_1 \tan x + a_3 \tan^3 x + a_5 \tan^5 x + \cdots + a_{2023} \tan^{2023} x}{1 + a_2 \tan^2 x + a_4 \tan^4 x \cdots + a_{2022} \tan^{2022} x}\]whenever $\tan 2023x$ is defined. What is $a_{2023}?$ | -1 | Using the formula tan(nx) expressed via the imaginary/real parts of (cos x + i sin x)^n divided appropriately, we know tan(2023x) = Im((1+it)^2023)/Re((1+it)^2023) where t = tan x.
The numerator contains odd powers of t and denominator even powers. The leading term of the numerator (highest odd power) is i^2022 * C(20... |
18 | How many positive perfect squares less than $2023$ are divisible by $5$? | 8 | Perfect squares divisible by 5 must be divisible by 25. So we need multiples of 25 that are perfect squares less than 2023: 25, 100, 225, 400, 625, 900, 1225, 1600, 2025... 2025 > 2023, so we stop at 1600 = (40)^2. These are (5)^2, (10)^2, ..., (40)^2, giving 40/5 = 8 values.
\boxed{8} |
19 | How many digits are in the base-ten representation of $8^5 \cdot 5^{10} \cdot 15^5$? | 18 | 8^5 · 5^10 · 15^5 = 2^15 · 5^10 · (3·5)^5 = 2^15 · 5^10 · 3^5 · 5^5 = 2^15 · 5^15 · 3^5 = 10^15 · 243 = 243 × 10^15, which has 3 + 15 = 18 digits. \boxed{18} |
2 | What is the degree measure of the acute angle formed by lines with slopes $2$ and $\frac{1}{3}$? | 45 | tan(theta) = |(2 - 1/3)/(1 + 2·(1/3))| = |(5/3)/(5/3)| = 1, so theta = 45°.
\boxed{45} |
20 | Janet rolls a standard $6$-sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3$? The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 265 | The running total equals 3 at some point if the first roll is 3, or the first two rolls sum to 3, or the first three rolls sum to 3.
Case 1: First roll = 3. Probability = 1/6.
Case 2: First roll ≠ 3, first two rolls sum to 3. First roll must be 1 or 2 (since rolls are 1-6). If first roll = 1, second = 2; if first rol... |
21 | Points $A$ and $B$ lie on the graph of $y=\log_{2}x$. The midpoint of $\overline{AB}$ is $(6, 2)$. What is the positive difference between the $x$-coordinates of $A$ and $B$? The final answer can be written in the form $m \sqrt{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 9 | Let A=(a, log2(a)) and B=(b, log2(b)). Midpoint conditions: (a+b)/2=6 and (log2(a)+log2(b))/2=2, so a+b=12 and log2(ab)=4, meaning ab=16.
Then (a-b)^2=(a+b)^2-4ab=144-64=80, so |a-b|=sqrt(80)=4sqrt(5).
m=4, n=5, gcd(4,5)=1, so m+n=9.
\boxed{9} |
23 | Maureen is keeping track of the mean of her quiz scores this semester. If Maureen scores an $11$ on the next quiz, her mean will increase by $1$. If she scores an $11$ on each of the next three quizzes, her mean will increase by $2$. What is the mean of her quiz scores currently? | 7 | Let n = number of quizzes taken, m = current mean.
From condition 1: (nm + 11)/(n+1) = m+1, so nm+11 = nm+n+m+1, giving 11 = n+m+1, so n+m = 10.
From condition 2: (nm + 33)/(n+3) = m+2, so nm+33 = nm+2n+3m+6, giving 33 = 2n+3m+6, so 2n+3m = 27.
From n = 10-m: 2(10-m)+3m = 27, so 20+m = 27, m = 7.
\boxed{7} |
25 | Mrs. Jones is pouring orange juice into four identical glasses for her four sons. She fills the first three glasses completely but runs out of juice when the fourth glass is only $\frac{1}{3}$ full. What fraction of a glass must Mrs. Jones pour from each of the first three glasses into the fourth glass so that all four... | 7 | Total juice = 3 + 1/3 = 10/3 glasses. Each son gets 10/12 = 5/6 glass.
First three glasses have 1 each, need 5/6, so pour out 1 - 5/6 = 1/6 from each.
m=1, n=6, m+n=7.
\boxed{7} |
26 | In the $xy$-plane, a circle of radius $4$ with center on the positive $x$-axis is tangent to the $y$-axis at the origin, and a circle with radius $10$ with center on the positive $y$-axis is tangent to the $x$-axis at the origin. What is the slope of the line passing through the two points at which these circles inters... | 7 | Circle 1: center (4,0), radius 4: (x-4)^2 + y^2 = 16, expands to x^2 - 8x + y^2 = 0.
Circle 2: center (0,10), radius 10: x^2 + (y-10)^2 = 100, expands to x^2 + y^2 - 20y = 0.
Subtract circle 1 from circle 2: 8x - 20y = 0, so y = (2/5)x.
Slope = 2/5, m+n = 2+5 = 7.
\boxed{7} |
27 | Calculate the maximum area of an isosceles trapezoid that has legs of length $1$ and one base twice as long as the other. The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m^2+n^2$? | 13 | Let the shorter base be a, longer base 2a, legs = 1.
Height h = sqrt(1 - (a/2)^2) since each side extends a/2 beyond the shorter base.
Area = (1/2)(a + 2a)(h) = (3a/2)*sqrt(1 - a^2/4)
Maximize f(a) = (3a/2)*sqrt(1 - a^2/4). Let u = a^2, maximize u*(1 - u/4).
d/du[u - u^2/4] = 1 - u/2 = 0, so u = 2, a = sqrt(2).
h ... |
28 | For complex number $u = a+bi$ and $v = c+di$ (where $i=\sqrt{-1}$), define the binary operation
$u \otimes v = ac + bdi$
Suppose $z$ is a complex number such that $z\otimes z = z^{2}+40$. What is $|z|^2$? | 50 | Let z = a+bi. Then z⊗z = a²+b²i² ... wait, z⊗z = a·a + b·b·i = a²+b²i.
Also z² = (a²-b²)+2abi.
Set z⊗z = z²+40:
a²+b²i = (a²-b²+40)+2abi
Real: a² = a²-b²+40 → b² = 40
Imaginary: b² = 2ab → 40 = 2ab → ab = 20 → a = 20/b
b² = 40, so b = ±2√10, a = 20/b = ±√10.
|z|² = a²+b² = 10+40 = 50.
\boxed{50} |
29 | A rectangular box $P$ has distinct edge lengths $a$, $b$, and $c$. The sum of the lengths of all $12$ edges of $P$ is $13$, the areas of all $6$ faces of $P$ is $\frac{11}{2}$, and the volume of $P$ is $\frac{1}{2}$. Find the length of the longest interior diagonal connecting two vertices of $P$. The final answer can b... | 13 | 4(a+b+c)=13, so a+b+c=13/4.
2(ab+bc+ca)=11/2, so ab+bc+ca=11/4.
abc=1/2.
Diagonal = sqrt(a^2+b^2+c^2) = sqrt((a+b+c)^2 - 2(ab+bc+ca)) = sqrt(169/16 - 22/4) = sqrt(169/16 - 88/16) = sqrt(81/16) = 9/4.
m+n = 9+4 = 13.
\boxed{13} |
3 | What is the value of
\[2^3 - 1^3 + 4^3 - 3^3 + 6^3 - 5^3 + \dots + 18^3 - 17^3?\] | 3159 | Each pair: (2k)^3 - (2k-1)^3 for k=1 to 9.
a^3 - b^3 = (a-b)(a^2+ab+b^2), with a=2k, b=2k-1:
= 1·(4k^2 + 2k(2k-1) + (2k-1)^2) = 4k^2 + 4k^2-2k + 4k^2-4k+1 = 12k^2 - 6k + 1
Sum from k=1 to 9: 12·(285) - 6·(45) + 9 = 3420 - 270 + 9 = 3159
\boxed{3159} |
30 | For how many ordered pairs $(a,b)$ of integers does the polynomial $x^3+ax^2+bx+6$ have $3$ distinct integer roots? | 5 | The roots r,s,t are distinct integers with rst = -6 and r+s+t = -a, rs+rt+st = b.
Find all sets of 3 distinct integers multiplying to -6:
{1,2,-3}: product=-6 ✓
{1,-2,3}: product=-6 ✓
{-1,2,3}: product=-6 ✓
{-1,-2,-3}: product=-6 ✓
{1,-1,6}: product=-6 ✓
{1,-1,-6}: product=6 ✗
{2,-1,-3}: same as {-1,-3,2}... wait, (-... |
33 | Triangle $ABC$ has side lengths in arithmetic progression, and the smallest side has length $6.$ If the triangle has an angle of $120^\circ,$ Find the area of $ABC$. The final answer can be simplified in the form $m \sqrt{n}$, where $m$ and $n$ are positive integers and $n$ without square factore. What is $m+n$? | 18 | Let sides be 6, 6+d, 6+2d. The 120° angle is opposite the largest side 6+2d.
By law of cosines: (6+2d)^2 = 6^2 + (6+d)^2 - 2(6)(6+d)cos(120°)
cos(120°) = -1/2, so:
(6+2d)^2 = 36 + (6+d)^2 + 6(6+d)
36+24d+4d^2 = 36 + 36+12d+d^2 + 36+6d
36+24d+4d^2 = 108+18d+d^2
3d^2+6d-72=0, d^2+2d-24=0, (d+6)(d-4)=0, d=4.
Sides:... |
36 | Carlos went to a sports store to buy running shoes. Running shoes were on sale, with prices reduced by $20\%$ on every pair of shoes. Carlos also knew that he had to pay a $7.5\%$ sales tax on the discounted price. He had $$43$ dollars. What is the original (before discount) price of the most expensive shoes he could a... | 50 | Let original price be P. Discounted price = 0.8P. After tax = 0.8P × 1.075 = 0.86P. Set 0.86P ≤ 43, so P ≤ 43/0.86 = 50. \boxed{50} |
43 | A $3-4-5$ right triangle is inscribed in circle $A$, and a $5-12-13$ right triangle is inscribed in circle $B$. Find the ratio of the area of circle $A$ to the area of circle $B$. The final answer can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 194 | For a right triangle, the hypotenuse is the diameter. Circle A has diameter 5, radius 5/2. Circle B has diameter 13, radius 13/2.
Ratio of areas = (5/2)^2 / (13/2)^2 = 25/169.
m+n = 25+169 = 194.
\boxed{194} |
44 | Jackson's paintbrush makes a narrow strip with a width of $6.5$ millimeters. Jackson has enough paint to make a strip $25$ meters long. How many square centimeters of paper could Jackson cover with paint? | 1625 | 25 meters = 2500 cm, width = 6.5 mm = 0.65 cm. Area = 2500 × 0.65 = 1625 square centimeters.
\boxed{1625} |
47 | For how many integers $n$ does the expression\[\sqrt{\frac{\log (n^2) - (\log n)^2}{\log n - 3}}\]represent a real number, where log denotes the base $10$ logarithm? | 901 | Let t = log n. The expression requires n > 0 (n ≠ 0) and the fraction inside to be ≥ 0.
The numerator: log(n²) - (log n)² = 2t - t² = t(2-t)
The denominator: t - 3
Need t(2-t)/(t-3) ≥ 0 and t ≠ 3.
t(2-t)/(t-3) ≥ 0 means -t(t-2)/(t-3) ≥ 0, i.e., t(t-2)/(t-3) ≤ 0.
Sign analysis of t(t-2)/(t-3):
- t < 0: negative·nega... |
48 | How many nonempty subsets $B$ of ${0, 1, 2, 3, \cdots, 12}$ have the property that the number of elements in $B$ is equal to the least element of $B$? For example, $B = {4, 6, 8, 11}$ satisfies the condition. | 144 | If the least element is k, then B has k elements, with the smallest being k. We need to choose the remaining k-1 elements from {k+1, k+2, ..., 12}, which has 12-k elements.
k must satisfy k >= 1 (nonempty, least element k means k elements total, so k >= 1) and k-1 <= 12-k, i.e., 2k <= 13, so k <= 6.
Count = sum from ... |
49 | What is the area of the region in the coordinate plane defined by
$| | x | - 1 | + | | y | - 1 | \le 1$? | 8 | By symmetry consider x,y≥0: ||x|-1|+||y|-1| = |x-1|+|y-1|≤1, a square with vertices (0,1),(1,0),(2,1),(1,2), area=2.
The full condition uses |x| and |y|, so reflecting into all four quadrants gives area = 4×2 = 8.
\boxed{8} |
5 | How many complex numbers satisfy the equation $z^5=\overline{z}$, where $\overline{z}$ is the conjugate of the complex number $z$? | 7 | If z=0, that works. Otherwise write z=r·e^(iθ) with r>0. Then z^5 = r^5·e^(5iθ) and conjugate z = r·e^(-iθ). Equating moduli: r^5 = r, so r^4=1, r=1. Equating arguments: 5θ = -θ + 2πk, so 6θ = 2πk, θ = πk/3 for k=0,1,2,3,4,5. That gives 6 solutions on the unit circle, plus z=0, total 7.
\boxed{7} |
7 | Consider the set of complex numbers $z$ satisfying $|1+z+z^{2}|=4$. The maximum value of the imaginary part of $z$ can be written in the form $\tfrac{\sqrt{m}}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$? | 21 | Let z = x + yi. We need |1 + z + z²| = 4.
Note 1 + z + z² = (z + 1/2)² + 3/4. Let w = z + 1/2, so |w² + 3/4| = 4.
Write w = a + bi where a = x + 1/2, b = y.
w² = a² - b² + 2abi
|w² + 3/4|² = (a² - b² + 3/4)² + 4a²b² = 16
To maximize y = b, take derivative approach. Let s = a², t = b².
(s - t + 3/4)² + 4st = 16
E... |
AMC 2023 Freeform Rewrites
DatologyAI is pleased to release a rewrite of AMC 2023 that enables language-model evaluation through language-modeling loss on verified-correct free-form answers. Part of the DatologyAI Benchmark Freeform Rewrites collection.
Motivation
Practitioners developing language models need evaluations capable of quantifying the effects of small changes, such as adjustments to the training data mix, even when evaluating the smaller models trained during research and development.
Generative evaluation with pass/fail checks is susceptible to emergence effects when applied to development-scale models. Unconditioned language-modeling loss over held-out documents provides a denser signal, but held-out documents may not map clearly onto the capabilities of interest.
Few-shot-conditioned free-form question answering, the setup this dataset enables, sits between the two. It targets specific capabilities and measures in-context learning while providing a dense, emergence-free signal.
This data
The DatologyAI Benchmark Freeform Rewrites datasets transform existing multiple-choice and short-answer benchmarks into free-form long-answer tasks. Each row pairs a self-contained question, the original short answer, and a worked continuation verified to answer the question correctly.
A technical write-up describing the rewrite and verification process and the validation results is coming soon.
Fields
| Field | Description |
|---|---|
id |
Item id from the source benchmark; unique within the dataset. |
question |
Self-contained rewritten question or problem statement. |
answer |
Short original answer from the source benchmark. |
freeform_answer |
Free-form worked answer that arrives at answer. |
Loading
from datasets import load_dataset
ds = load_dataset("DatologyAI/amc23-freeform-rewrites", split="train")
Related datasets
- Downloads last month
- 11