Printable paper
2025 AMC 10B — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2025 AMC 10B — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | B | Ratios, Percents & Averages — Each mug needs a full 20 grams. Seventeen mugs use 340 grams, and the 10 grams left over cannot make another proper mug. |
| 2 | D | Modular Arithmetic — The last digit of a square depends only on the last digit of the number, so the list repeats every ten terms, and each block of ten adds to 45. |
| 3 | D | Sequences & Series — From the second row on, each row's total is exactly double the one before, so the totals go 11, 22, 44, … and the 11th row totals 5632. |
| 4 | A | Bases & Digits — Write both numbers in ordinary place value: 7a + b = 9b + a. This simplifies to 3a = 4b, and a digit in base seven is at most 6. |
| 5 | C | Circles — Angle B is 130 degrees, so the far arc AC is 260 and the near arc is 100. Triangle AOC is isosceles with a 100-degree angle at O. |
| 6 | C | Coordinate Geometry — The line crosses both side edges, so the lower region is a trapezoid of area 8/3. The piece left of x = a must have area 4/3. |
| 7 | C | Coordinate Geometry — Measure from the locked gate. The west route is 17 + (x + 8), using the 8-15-17 right triangle; the east route is √(15² + (x+9)²) + 9. |
| 8 | C | Divisibility & Factors — The total in cents, ABBBA, is a multiple of 36. The test for 4 makes A even; the test for 9 makes A a multiple of 3. |
| 9 | C | Basic Counting — Add the inequalities in pairs to trap x, y and z between 0 and 2. Then list the few triples, sorted by their sum. |
| 10 | A | Polynomials — Subtracting gives f − g = (n−4)(n+1), a factor of both. The fraction becomes (n−2)/(n−3), but n = 4 is out because g(4) = 0. |
| 11 | B | Basic Probability — Only Tuesday matters: compare it with Monday. Choose the 2 students who repeat, then the other 4 must all get a new tutor, which happens in 9 ways. |
| 12 | C | Circles — Cut the hexagon into six equilateral triangles and the rhombus into two: each disk is the circle inside one triangle, so H = R. |
| 13 | A | Triangle Centers & Cevians — Put the right angle at the origin. The crossing point and the altitude's foot lie on one line from there; compare how far across each is. |
| 14 | C | Basic Probability — The three tallest are the captains exactly when they land in three different groups. Only their three wristbands matter: 6/8 × 3/7. |
| 15 | D | Sequences & Series — The bottom factors as k(k+2)(k+4). Each term is one quarter of the difference 1/(k(k+2)) − 1/((k+2)(k+4)), so the sum cancels down to its first two pieces. |
| 16 | D | Arrangements with Restrictions — Each color covers two non-neighboring sectors. Split the sectors into three such pairs (4 ways), then give the pairs the three colors (6 ways). |
| 17 | B | Sequences & Series — The average of the first k terms is 2028 − k, so their total is k(2028 − k). Subtracting totals shows later terms drop by 2. |
| 18 | D | Sequences & Series — The value n appears 2n + 1 times, from n² to just before (n+1)². So add n(2n + 1) for n = 1 to 44, plus 45. |
| 19 | D | Solid Geometry — Extend the slanted sides to a point: the container is a pyramid with its tip cut off, and volume below a square of side s grows like s³. |
| 20 | A | Circles — The four semicircle centers form a square of side 2r around the center, so each is r√2 from the center, and the small circle's radius is r√2 − r. |
| 21 | C | Arrangements with Restrictions — Renaming colors red to blue, blue to yellow, yellow to red keeps every rule, so count only colorings with a red center and multiply by 3. |
| 22 | A | Conditional Probability & States — A digit sum of 61 is 2 short of seven 9s, leaving only 28 numbers. The test for 11 passes only when the whole shortfall is in even places. |
| 23 | C | Diophantine Equations — Horace writes 91(r−1) + c and Vera writes 141(c−1) + r. Setting them equal gives 9r = 14c − 5, solved by c = 1, 10, 19, … |
| 24 | E | Basic Probability — Let p1, p2, p3 be the chances of reaching 4 from 1, 2, 3. Each is a quarter of the chances one step down and one step up. |
| 25 | B | Transformations & Symmetry — Flip the square instead of bouncing the path. The path becomes one straight line through a grid of squares, ending at the first grid corner it meets. |
2025 AMC 10B — answer key. Problems © Mathematical Association of America, American Mathematics Competitions (maa.org/amc), licensed CC BY-NC-SA 4.0. Typeset and annotated by AMC 10 Step by Step (amcstep.com) and shared under the same licence. Not endorsed by the MAA.