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2018 AMC 10A — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2018 AMC 10A — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | B | Exponents, Logarithms & Radicals — A power of -1 just means flip the fraction upside down. Start at the innermost brackets and repeat: flip, then add 1. |
| 2 | A | Ratios, Percents & Averages — Pick an easy amount for Jacqueline, like 100. Then Liliane has 150 and Alice has 125, and you just compare 150 with 125. |
| 3 | E | Clocks, Calendars & Time — One day is 60 times 60 times 24 seconds. Split 10! into those same factors and the leftover is exactly 42 days, which lands on noon of February 12. |
| 4 | E | Arrangements with Restrictions — First choose which three periods hold math, with gaps between them; only four choices work. Then put the three different courses into those periods in 6 orders. |
| 5 | D | Absolute Value & Inequalities — Every statement is false, so flip each one: less than 6, more than 5, more than 4. The numbers that pass all three lie strictly between 5 and 6. |
| 6 | B | Ratios, Percents & Averages — Out of every 100 votes, 65 are likes and 35 are dislikes, so each 100 votes adds 30 to the score. A score of 90 needs 300 votes. |
| 7 | E | Exponents, Logarithms & Radicals — 4000 has five 2s and three 5s. Positive n uses up a 5 each step, negative n uses up a 2. Count both directions and n = 0. |
| 8 | C | Linear Equations & Word Problems — Call the number of 5-cent coins x. Write the other two counts using x, and the total value becomes one equation in x. |
| 9 | E | Similar & Congruent Triangles — Four small triangles fit side by side along DE, so triangle ADE is a small one scaled up 4 times, with area 4 times 4 = 16. |
| 10 | A | Algebraic Manipulation — The two numbers under the roots differ by 24. Big minus small, times big plus small, equals that difference, so 3 times the sum is 24. |
| 11 | E | Distributions & Stars and Bars — Each die shows at least 1, so seven dice already give 7. Only 3 extra points are left to hand out, and there are just three ways to split 3. |
| 12 | C | Absolute Value & Inequalities — Write x using y from the line. The line then splits into three stretches where x and y keep their signs, and each stretch gives simple equations. |
| 13 | D | Transformations & Symmetry — The crease is the line through the middle of AB at a right angle to it. It forms a small right triangle shaped like the paper. |
| 14 | A | Exponents, Logarithms & Radicals — The 3-powers dominate, so the value is near 3 to the fourth, 81. The small 2-power terms pull it just below 81, not above. |
| 15 | D | Circles — A touching point lies on the line through both centers. Stretching the center segments from 8 to 13 turns triangle OPQ into triangle OAB. |
| 16 | D | Triangles: Area & Pythagorean — Sliding along the hypotenuse, the segment shrinks from 20 to the height (about 14.5), then grows to 21, hitting each whole number once per side. |
| 17 | C | Divisibility & Factors — A small smallest number bans all its multiples, and leftover pairs like 4 and 8 still clash. Smallest 2 or 3 allows only 5 numbers. |
| 18 | D | Bases & Digits — Digits -1, 0, 1 with powers of 3 make every whole number from -3280 to 3280 exactly once; the nonnegative ones are 0 through 3280. |
| 19 | E | Modular Arithmetic — Only the last digit of m matters: 1 always works, 5 never, 9 needs an even exponent, 3 and 7 need a multiple of 4. |
| 20 | B | Paths & Grids — All the rotations and flips together fold the grid onto one triangular slice, one-eighth of the square, holding 10 cells. Color those freely: 2^10 ways, minus all-black and all-white. |
| 21 | E | Coordinate Geometry — Put y + a in place of x squared. The equation in y factors: one root is the parabola's bottom point, the other adds two points when a > 1/2. |
| 22 | D | GCD & LCM — Follow the 2s and 3s along the chain. The gcd of d and a has exactly one 2 and one 3, which forces it to be 78. |
| 23 | D | Triangles: Area & Pythagorean — Join the square's outer corner to the far corners. The pieces are the square, a height-2 triangle on the hypotenuse, and two thin triangles. |
| 24 | D | Triangle Centers & Cevians — Since AB is 5 times AC, the bisector splits both triangle ABC and half-size triangle ADE into areas 5 to 1. |
| 25 | D | Bases & Digits — Using R = 11...1, the equation becomes (9c minus a squared) times R = b minus 2c. Two values of n force both sides to be 0. |
2018 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.