2021 AMC Fall 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 | Algebraic Manipulation | Since 2112 - 2021 = 91 = 7 * 13 and 169 = 13^2, the fraction is (91/13)^2 = 7^2 = 49. | |
| 2 | E | Linear Equations & Word Problems | Only shortening the 4-inch side gives 3 * 6 = 18, so the other side is the 6-inch one, and 4 * 5 = 20. | |
| 3 | D | Solid Geometry | Reshaping means only volume matters: cube volume 216 divided by ball volume 32pi/3 is about 6.45, so 6 balls fit. | |
| 4 | B | Linear Equations & Word Problems | Convert each leg to minutes: Route A takes 12 minutes, Route B takes 6.75 + 1.5 = 8.25 minutes, so the difference is 3.75. | |
| 5 | E | Primes | Divisibility rules kill every digit: even digits and 5 fail, digit sum rules out 1 and 7, and 202103 has alternating sum 0, so 11 divides it. | |
| 6 | B | Linear Equations & Word Problems | There are 40 gaps, not 41, so each gap is 5280/40 = 132 feet; a stride is 132/44 = 3 feet and a leap 132/12 = 11 feet. | |
| 7 | D | Angles & Polygons | At D the square's 90 degrees and the given 110 degrees leave angle FDE = 160, so the isosceles triangle DEF has base angles 10 and angle AFE = 170. | |
| 8 | B | Bases & Digits | The condition 10a + b = a + b^2 becomes 9a = b(b-1); checking digits b, only b = 9 gives a valid tens digit a = 8. | |
| 9 | E | Basic Probability | P(even) = 3/4 and P(odd) = 1/4 per roll; an even sum needs both even or both odd, so (3/4)^2 + (1/4)^2 = 10/16 = 5/8. | |
| 10 | B | Statistics & Data | t is the plain average 20; s is size-weighted, (50^2 + 2*20^2 + 2*5^2)/100 = 33.5, so t - s = -13.5. | |
| 11 | A | Linear Equations & Word Problems | Measure in Emily's steps: walking with the ship she gains 210 - 210k, against it 42 + 42k, where k is the ship's speed ratio; both equal the length. | |
| 12 | D | Modular Arithmetic | Because 9 = -1 (mod 5), N is congruent to its alternating base-nine digit sum: 2 - 5 + 6 - 7 + 2 = -2, i.e. remainder 3. | |
| 13 | D | Basic Probability | A black ball needs at least 3 white balls and a white ball needs at least 3 black, so the split must be exactly 3-3: C(6,3)/2^6 = 20/64. | |
| 14 | D | Absolute Value & Inequalities | The parabola y = 3 - x^2/3 meets the diamond |x|+|y| = 3 only at its three vertices and cuts the lower half of |x|+|y| = 5 twice. | |
| 15 | C | Circles | Tangent radii give right angles at B and C, so the circumcircle of ABC has diameter AO, and AO^2 = 54 + 50 = 104 gives radius sqrt(26). | |
| 16 | D | Functions | Replacing x by 1 - x swaps the two floor terms, so f(1 - x) = -f(x): the graph is symmetric about the point (1/2, 0). | |
| 17 | D | Solid Geometry | Heights on a plane are linear in position: parallelogram OABC gives h(O) = h(A) + h(C) - h(B) = 13, and O bisects BE, so h(E) = 17. | |
| 18 | C | Arrangements with Restrictions | Diagonal cells never touch: fix them first; then each of the other two cells has 3 options if the diagonal pair matches and 2 otherwise: 36 + 48. | |
| 19 | A | Circles | Inside, the disk misses each corner by 1 - pi/4, so A = 8s - 20 + pi; outside it sweeps 8s + 4pi, which must equal 2A. | |
| 20 | B | Quadratics | Both discriminants must be nonpositive: b^2 <= 4c and c^2 <= 4b, which force b, c <= 4; checking the 16 small pairs leaves 6. | |
| 21 | E | Basic Counting | Both probabilities share the 5^20 denominator, so p/q is a ratio of multinomials: 5 * 4 arrangements of bins times 4!4!/(3!5!) = 20 * 4/5 = 16. | |
| 22 | B | Solid Geometry | The three centers form an equilateral triangle of side 2r at height r, so each center is 2r/sqrt(3) from the axis; then the slant-line distance condition gives r. | |
| 23 | D | Divisibility & Factors | Since f_1(n) = 2d(n) <= 20 here, iterating small even numbers shows only 12, 18, 20 reach 12; so count n <= 50 with d(n) in {6, 9, 10}. | |
| 24 | E | Arrangements with Restrictions | Six edges are labeled 1 and the top and bottom faces take 4, so exactly 2 vertical edges are 1: diagonal pairs give 12 labelings, adjacent pairs give 8. | |
| 25 | A | Quadratics | Three solutions of p(p(x)) = 0 force one root of p to equal its minimum value, so (r - s)^2 = -4r, and the root sum peaks at 1/2. |
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.