2022 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 | D | Fractions & Decimals | Evaluate the nested fraction from the inside out: 3+1/3 = 10/3, then 3+3/10 = 33/10, then 3+10/33 = 109/33. | |
| 2 | B | Ratios, Percents & Averages | Laps are proportional to time at constant speed: 15 laps times 27/57 is 135/19, slightly more than 7. | |
| 3 | E | Linear Equations & Word Problems | Write everything in terms of the third number c: 6c + (c + 40) + c = 96 gives c = 7, so the numbers are 42, 47, 7. | |
| 4 | E | Ratios, Percents & Averages | Convert 100 kilometers to 100m miles, divide by x miles per gallon to get gallons used, then multiply by l liters per gallon. | |
| 5 | C | Triangles: Area & Pythagorean | AP = s leaves PB = 1 - s, and PQ is the hypotenuse of an isosceles right triangle at corner B, so s = (1 - s) sqrt(2). | |
| 6 | A | Absolute Value & Inequalities | For a < 0, sqrt((a-1)^2) = |a-1| = 1 - a, so the inside becomes 2a - 3, which is negative; its absolute value is 3 - 2a. | |
| 7 | B | GCD & LCM | Compare prime exponents: lcm(n, 18) = 180 = 2^2 3^2 5 forces 2^2 and 5 into n; gcd(n, 45) = 15 pins the exponent of 3 at exactly 1. | |
| 8 | D | Statistics & Data | The mean (20 + X)/6 must equal 1, 2, 5, 7, or X itself; only 5, 7, and X yield a positive integer X, namely 10, 22, and 4. | |
| 9 | D | Arrangements with Restrictions | The wide bottom-middle region touches all four others; the remaining four form a path (bottom-left, top-left, top-right, bottom-right), giving 5*4*3*3*3. | |
| 10 | E | Quadrilaterals & Polygon Areas | With sides a, b: a^2 + b^2 = 64 and (a-2)^2 + (b-2)^2 = 32; subtracting gives a + b = 10, then 2ab = 10^2 - 64 = 36. | |
| 11 | C | Exponents, Logarithms & Radicals | Write both sides as powers of 2 using 4096 = 2^12: the exponents give m - 6 = 1 - 12/m, a quadratic with roots 3 and 4. | |
| 12 | A | Logic Puzzles | Tabulate each type's yes/no pattern: truth-tellers Y N N, liars Y Y N, alternaters N N N or Y Y Y; then peel the counts from question 3 backward. | |
| 13 | C | Triangle Centers & Cevians | Reflect B across bisector AP: it lands at E on AC with AE = AB = 2k, EC = k, and triangles EAD, ECB are similar with ratio 2. | |
| 14 | E | Arrangements with Restrictions | 8-14 must be the larger members, so 1-7 are the smaller; pair 7, 6, 5, 4 first (1, 2, 3, 4 options), then 3, 2, 1 freely. | |
| 15 | D | Circles | 7-24-25 and 15-20-25 share the hypotenuse 25, so diagonal AC = 25 is a diameter: circle area 625 pi / 4 minus triangle areas 84 + 150. | |
| 16 | D | Polynomials | The new volume is (r+2)(s+2)(t+2): expand it as rst + 2(rs+st+tr) + 4(r+s+t) + 8 and read every symmetric sum from Vieta. | |
| 17 | D | Fractions & Decimals | Repeating 0.abc is abc/999 and repeating 0.a is a/9, so the condition becomes 7a = 3b + 4c, i.e. b = a - 4t and c = a + 3t. | |
| 18 | A | Transformations & Symmetry | Track the angle theta of the point: T_k sends theta to 180 - theta - k, so each consecutive pair T_{2j-1}, T_{2j} nets -1 degree; check odd n too. | |
| 19 | C | Modular Arithmetic | h is the sum of L_17/k; every term except k = 17 keeps the factor 17, so h is congruent to L_17/17 = lcm(1..16) mod 17, which is 5. | |
| 20 | E | Sequences & Series | Subtracting consecutive terms removes the arithmetic part's constant: differences 3 and 31 give b(r-1) + d = 3 and br(r-1) + d = 31, so b(r-1)^2 = 28. | |
| 21 | B | Solid Geometry | The shared slanted edge shows each hexagon's top edge sits horizontally 1 unit outside the square, so the rim is a 3 by 3 square with four corner triangles removed. | |
| 22 | D | Basic Counting | If the first pass collects 1 through k, the arrangement is determined by which k positions hold them: C(13,k) choices, minus the one sorted arrangement done in a single pass. | |
| 23 | B | Quadrilaterals & Polygon Areas | With the symmetry axis as the y-axis: PD^2 - PA^2 = -4ax and PC^2 - PB^2 = -4bx where 2a = AD, 2b = BC, so BC/AD = 5/15. | |
| 24 | E | Recursive Counting | In a bad string, the smallest failing j has exactly j-1 digits below j-1 (a good string) and the rest at least j; recursing gives 1, 3, 16, 125, 1296. | |
| 25 | B | Coordinate Geometry | Side n has (n+1)^2 lattice points; set r+1 = 3k, s+1 = 2k, and the 27-times condition forces t+1 congruent to -1 mod 13, then 4(t+1)^2 = 13k^2 - 2k. |
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.