2015 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 | C | Exponents, Logarithms & Radicals | Inside the parentheses 2^0 = 1 cancels the -1, leaving 25; a negative-one exponent means reciprocal, so the value is 5/25. | |
| 2 | D | Linear Equations & Word Problems | If all 25 tiles were triangles there would be 75 edges; each square adds exactly one extra edge, so 84 - 75 = 9 squares. | |
| 3 | D | Basic Counting | Count horizontal toothpicks row by row (n + n + (n-1) + ... + 1), double by symmetry: an n-step staircase uses n(n+3) toothpicks, so 40 - 18 = 22. | |
| 4 | B | Ratios, Percents & Averages | Give Mia 1 egg, Sofia 2, Pablo 6; the fair share is 3 each, so Sofia needs 1 of Pablo's 6 eggs. | |
| 5 | E | Ratios, Percents & Averages | Averages hide totals: 14 tests average 80 (sum 1120), 15 tests average 81 (sum 1215), and the difference 95 is Payton's score. | |
| 6 | B | Linear Equations & Word Problems | Write a + b = 5(a - b) and collect terms: 4a = 6b, so a/b = 3/2. | |
| 7 | B | Sequences & Series | The number of terms is (last - first)/(common difference) + 1 = (73 - 13)/3 + 1 = 21. | |
| 8 | B | Linear Equations & Word Problems | Set up both past conditions in terms of current ages, solve to get Pete 20 and Claire 8, then find x with 20 + x = 2(8 + x). | |
| 9 | D | Ratios, Percents & Averages | Equal volumes mean r^2 h is constant; radius scaled by 1.1 scales r^2 by 1.21, so the first height equals 1.21 times the second. | |
| 10 | C | Arrangements with Restrictions | Letter b may only touch d and c only a, so b and c sit at the ends beside their lone partners: only bdac and cadb. | |
| 11 | C | Triangles: Area & Pythagorean | Sides 4x and 3x make the diagonal 5x, so d = 5x and the area 12x^2 becomes 12(d/5)^2 = (12/25)d^2. | |
| 12 | C | Algebraic Manipulation | Rearranged, the curve is (y - x^2)^2 = 1, so y = x^2 plus or minus 1 and the two y-values at any x differ by exactly 2. | |
| 13 | C | Basic Counting | With at least one nickel every multiple of 5 up to the total is attainable, so the count of values is 12 + n, forcing n = 5. | |
| 14 | C | Circles | A disk rolling outside a circle of triple its radius spins 3 times per lap, so the arrow is upright again after one third of a lap: 4 o'clock. | |
| 15 | B | Diophantine Equations | Cross-multiplying gives y(10 - x) = 11x, so x is at most 9; checking x = 1 to 9 leaves 5/11 as the only fraction in lowest terms. | |
| 16 | B | Systems of Equations | Subtract the equations and cancel x - y to get x + y = 3; add them to get x^2 + y^2 = 5(x + y) = 15. | |
| 17 | D | Coordinate Geometry | Slope sqrt(3)/3 means a 30-degree line, so the third line is its mirror image y = -x/sqrt(3); they meet x = 1 at points 1 + 2/sqrt(3) apart. | |
| 18 | E | Bases & Digits | Since 1000 = 3E8 in hex, every all-numeric hex string up to 399 (decimal 921) is in range: 4*10*10 - 1 = 399 numbers, digit sum 21. | |
| 19 | D | Triangles: Area & Pythagorean | The altitude from C bisects DE by symmetry and makes 15-degree angles with CD and CE, so the area is h^2 tan 15 with h = 5/sqrt(2). | |
| 20 | B | Algebraic Manipulation | Since A + P + 4 = (x+2)(y+2), a value works exactly when adding 4 gives a product of two factors each at least 3; 106 = 2*53 fails. | |
| 21 | C | Solid Geometry | Faces ABC and ABD are congruent 3-4-5 right triangles sharing hypotenuse AB; their altitudes to AB meet at the same foot, and CD = (12/5)sqrt2 makes them perpendicular. | |
| 22 | A | Recursive Counting | Count circular strings with no two adjacent 1s: fix person 1; seated gives a line of 7 (34 ways), standing gives a line of 5 (13 ways), total 47. | |
| 23 | C | Quadratics | By Vieta rs = 2(r + s), hence (r - 2)(s - 2) = 4; the factor pairs of 4, negatives included, give a = 9, 8, 0, -1. | |
| 24 | B | Diophantine Equations | Pythagoras gives BC^2 = 4(AD - 1): BC = 2k, AD = k^2 + 1, p = 2k^2 + 2k + 4, below 2015 for k = 1..31. | |
| 25 | A | Geometric Probability | Condition on which sides hold the points: same side (prob 1/4, success 1/4), adjacent (1/2, success 1 - pi/16), opposite (1/4, always); total (26 - pi)/32. |
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