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2025 AMC 10A — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2025 AMC 10A — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | E | Linear Equations & Word Problems — Only each rider's distance from home matters, and that is speed times riding time. Remember Betsy has ridden one hour less than Andy. |
| 2 | B | Ratios, Percents & Averages — Follow only the peanuts. The peanut pounds must be 40 percent of the new total weight; that tells how much mix was added, and then count the cashews. |
| 3 | D | Triangles: Area & Pythagorean — Split by which sides are equal. Two sides of 2025 works with any third side up to 2025. Two equal shorter sides must add to more than 2025. |
| 4 | A | Ratios, Percents & Averages — Turn each average into a total. Then Ash's age is 14 plus 2 per student, and also 52 minus 3 per teacher. Both agree at 7 students. |
| 5 | E | Sequences & Series — Break the sequence into groups that each end with 1, count how many terms the first few groups hold, then walk forward to position 2025. |
| 6 | C | Angles & Polygons — Each hexagon corner is where rays from two triangle corners meet. The small triangle they form with a side has base angles 20 and 20, or 40 and 40. |
| 7 | E | Polynomials — Dividing by x - 1 leaves the value at x = 1 as the remainder; same for x - 2. That gives two easy equations in a and b. |
| 8 | B | Logic Puzzles — Every statement only talks about how many statements are true. Try each possible count from 0 to 4 and keep the one that agrees with itself. |
| 9 | C | Functions — Sliding the graph sideways does not change its heights. So count how many times the original graph reaches height 25; each such spot gives one value of a. |
| 10 | C | Circles — The small radius equals the height of the chord above AB. A right triangle with legs 8 and that height gives big radius squared minus small radius squared = 64. |
| 11 | E | Sequences & Series — The geometric sequence 1, p, q, z makes z a perfect cube. The arithmetic one makes z one more than a multiple of 3. Test cubes 8, 27, 64. |
| 12 | D | Basic Counting — Sort digits 1 to 9 into groups: 2 is even and prime, 4, 6, 8 only even, 3, 5, 7 only prime, 1 and 9 neither. Then count two cases. |
| 13 | D | Sequences & Series — Each shaded ring is a shrunken copy of the one two rings further out, so the shaded areas add up to 1/(1 + k squared). |
| 14 | B | Basic Probability — The students sit side by side with chance 6/15. The four chairs left then form a row, not a circle, so the teachers have 3 good pairs out of 6. |
| 15 | A | Similar & Congruent Triangles — Draw AE. It is the hypotenuse of both right triangles ABE and ADE, so DE = 5. Then triangles ABC and EDC are similar, with EDC five times bigger. |
| 16 | D | Expected Value — List the 27 equally likely ways to drop the coins. Just 3 put all coins together and 6 spread them out; the other 18 give a fullest jar of 2. |
| 17 | E | GCD & LCM — Take away the remainders: N divides both 273420 and 272745, so it divides 45, their greatest common factor. A remainder of 16 forces N above 16, so N = 45. |
| 18 | B | Quadratics — For every factor, the reciprocals of its two roots add to -4/3, whatever k is, and the harmonic mean only needs these reciprocals. |
| 19 | A | Sequences & Series — Each row adds up to twice the row above, so 12,288 is the sum of row 13. Then follow only the second and third numbers down to that row. |
| 20 | A | Coordinate Geometry — Put the silo's center at (0, 0). The line of sight touches the silo when its distance from the center equals the radius, 10. That gives one equation for g. |
| 21 | C | Number Properties — Look at the largest number in the set. Pair up the smaller numbers so each pair adds to it; the set can hold at most one number from each pair. |
| 22 | B | Circles — The three outer centers are 3, 4 and 5 apart, a right triangle. Put it on coordinate axes and write the small circle's distance to each outer center. |
| 23 | D | Triangles: Area & Pythagorean — Use Pythagoras twice to find where the altitude from C meets AB. The bisector from B then cuts that altitude in the ratio of the sides beside angle B. |
| 24 | C | Basic Counting — A fair number's digits climb to the largest digit, then fall. Once the digits are chosen, every digit except the largest just picks the left or right side. |
| 25 | A | Geometric Probability — Whether AP beats BP depends on the square's vertical middle line; whether AP beats AB depends on a quarter circle around A. AP is the middle length in two regions. |
2025 AMC 10A — 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.