2025 AMC 10A
All 25 problems, 75 minutes, scored 6 / 1.5 / 0. The answer key and the idea behind each problem are on this page too, folded away until you ask for them.
The problems
Show the answer key, topics and key insights
This gives away all 25 answers and the idea behind each one. Sit the paper first if you mean to.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | E | Linear Equations & Word Problems | Each rider's distance from the start is just speed times time, so set 8t = 12(t-1) with Betsy's clock running an hour behind Andy's. | |
| 2 | B | Ratios, Percents & Averages | Track only the peanuts: 5 + 0.2x pounds of peanuts must be 40 percent of the new total 10 + x, which forces x = 5. | |
| 3 | D | Triangles: Area & Pythagorean | Split on where the equal pair sits: (2025, 2025, b) needs only 1 <= b <= 2025, while (a, a, 2025) needs 2a > 2025, so a >= 1013. | |
| 4 | A | Ratios, Percents & Averages | Ash's age read two ways gives A = 14 + 2s and A = 52 - 3t; with s + t = 15 these meet at s = 7. | |
| 5 | E | Sequences & Series | The sequence runs up to n and back to 1; the first n runs use n(n-1) + 1 terms, and 45*44 + 1 = 1981 falls just short of 2025. | |
| 6 | C | Angles & Polygons | Each hexagon vertex is where two trisectors from different corners cross, and the triangle cut off has base angles 20 and 20, or 40 and 40. | |
| 7 | E | Polynomials | The Remainder Theorem turns the two divisions into P(1) = 4 and P(2) = 6, a two-by-two linear system in a and b. | |
| 8 | B | Logic Puzzles | Each statement is a condition on T, the number of true statements, so test T = 0, 1, 2, 3, 4 and keep the value that reproduces itself. | |
| 9 | C | Functions | y = f(x-a) is the graph of f slid sideways, so the condition is f(1-a) = 25: count how many times the cubic takes the value 25. | |
| 10 | C | Circles | Tangency makes the small radius equal CD's height above AB, so the half-chord 8 gives R^2 - r^2 = 64 and the area is 32pi. | |
| 11 | E | Sequences & Series | The geometric sequence starts at 1, so z = p^3; the arithmetic one forces 3 | z - 1, and p^3 = p mod 3 makes p = 4 smallest. | |
| 12 | D | Basic Counting | Sort the digits 1-9 into even only {4,6,8}, prime only {3,5,7}, both {2}, and neither {1,9}, then split on whether one digit does both jobs or two digits do. | |
| 13 | D | Sequences & Series | The shaded rings form a geometric series with first term 1 - k^2 and ratio k^4, and the sum collapses to (1 - k^2)/(1 - k^4) = 1/(1 + k^2). | |
| 14 | B | Basic Probability | Work with pairs of chairs, not seatings: the students take an adjacent pair with probability 6/15, leaving an arc of four in which 3 of 6 pairs touch. | |
| 15 | A | Similar & Congruent Triangles | Draw AE: it is the hypotenuse of both right triangles ABE and ADE, so DE = 5 and triangles ABC and EDC are similar in ratio 1 : 5. | |
| 16 | D | Expected Value | Sort the 27 placements by shape: 3 give 3-0-0, 6 give 1-1-1, 18 give 2-1-0, so the expectation is (9 + 6 + 36)/27. | |
| 17 | E | GCD & LCM | N divides 273436 - 16 = 273420 and 272760 - 15 = 272745, hence their gcd 45; a remainder of 16 forces N > 16, leaving N = 45. | |
| 18 | B | Quadratics | The harmonic mean needs only the reciprocals' sum, and Vieta on kx^2 - 4x - 3 gives 1/p + 1/q = -4/3 for every k. | |
| 19 | A | Sequences & Series | Each row sum doubles, since every entry is used twice except the fixed ends that restore it, so S_n = 3 * 2^(n-1) and 12,288 is row 13. | |
| 20 | A | Coordinate Geometry | Put the silo's centre at the origin: the line of sight is tangent exactly when its distance from the origin equals 10, one equation in g. | |
| 21 | C | Number Properties | If m is the largest element of a sum-free set S, then S minus {m} and its mirror about m are disjoint in {1, ..., m-1}, so 2(|S|-1) <= 19. | |
| 22 | B | Circles | The three outer centres lie 3, 4, 5 apart, so they form a right triangle; subtracting circle equations makes the inner centre's coordinates linear in r. | |
| 23 | D | Triangles: Area & Pythagorean | With H the foot of the altitude from C, triangle BHP is right-angled at H with angle B/2, so BP = BH / cos(B/2) and cos B = 7/18. | |
| 24 | C | Basic Counting | No interior digit may be a local minimum, so the digits rise to a peak then fall, and any chosen set of k digits admits exactly 2^(k-1) orderings. | |
| 25 | A | Geometric Probability | AP is the middle side exactly when P lies right of the bisector x = 1/2 and inside A's unit circle, or left of it and outside. |
Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic tags, difficulty ratings and key insights on this page are original to this site. No problem statements are reproduced here — each links to its own page.