2019 AMC 10B
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 | Ratios, Percents & Averages | The same amount of water is 5/6 of the first container and 3/4 of the second, so the volume ratio is (3/4) divided by (5/6). | |
| 2 | E | Logic Puzzles | A counterexample must make the hypothesis true and the conclusion false: n composite and n-2 composite, which only 27 (with 25) does. | |
| 3 | B | Ratios, Percents & Averages | Count non-players: 60% of seniors plus 30% of non-seniors equals 46.8% of 500, which pins down the split and then 70% of non-seniors play. | |
| 4 | A | Coordinate Geometry | Two specific progressions such as (1,1,1) and (1,2,3) give two lines whose intersection (-1,2) must be the common point. | |
| 5 | E | Coordinate Geometry | Reflecting over y = x swaps coordinates, so a segment of slope m becomes one of slope 1/m; the product is 1, not -1, so they are not perpendicular. | |
| 6 | C | Algebraic Manipulation | Factor n! out of both factorials: (n+1)! + (n+2)! = n!(n+1)(n+3), so (n+1)(n+3) = 440 = 20 times 22. | |
| 7 | B | GCD & LCM | Casper's money is a common multiple of 12, 14 and 15, so at least lcm = 420 cents, which buys 420/20 = 21 purple candies. | |
| 8 | B | Quadrilaterals & Polygon Areas | The triangles' apexes meet at the center, so each triangle's height sqrt(3) is half the square's side; the square has area 12 and the triangles 4 sqrt(3). | |
| 9 | A | Functions | For x at least 0 both terms agree, giving 0; for a negative non-integer the floor rounds away from zero, so the difference is exactly -1. | |
| 10 | A | Geometric Optimization | Area 100 on base 10 needs height 20, but with AC + BC = 40 the height is largest when AC = BC = 20, giving sqrt(375) < 20. | |
| 11 | A | Ratios, Percents & Averages | Each jar holds 90k marbles, so the greens are 9k + 10k = 19k = 95 and the blue difference is 81k - 80k = k = 5. | |
| 12 | C | Bases & Digits | Below 2019 there are at most four base-7 digits; digit sum 23 needs three 6s and a 5, but 5666 base 7 = 2057 is too big. | |
| 13 | A | Statistics & Data | The median is 6, x, or 8 depending on where x falls; set each equal to (35 + x)/5 and keep only solutions consistent with their own case. | |
| 14 | C | Divisibility & Factors | 19! ends in three zeros, so H = 0; then the digit-sum test for 9 and the alternating-sum test for 11 pin down T and M. | |
| 15 | A | Triangles: Area & Pythagorean | Two shared legs would force equal areas, so the hypotenuse of the small triangle must be a leg of the big one, doubling one leg and forcing a 30-60-90 shape. | |
| 16 | A | Triangles: Area & Pythagorean | The isosceles triangles force angle CDE = 90, so CDE is 3-4-5, BC = 8, and perpendiculars from C and E bisect AD and DB. | |
| 17 | C | Basic Probability | Red higher and green higher are equally likely; the only other outcome is a tie, with probability the geometric sum of 4^(-k), which is 1/3. | |
| 18 | C | Sequences & Series | At the limit the two turning points reproduce each other: A = B/4 and B = A + (3/4)(2 - A); solving gives A = 2/5, B = 8/5. | |
| 19 | C | Divisibility & Factors | Products of two divisors of 2^5 5^5 are the 121 numbers 2^x 5^y, exponents 0 to 10; only four corners force equal divisors. | |
| 20 | E | Circles | Slice the disk by lines EG and AD: a half-disk above, two rectangle-minus-quarter-circles gaps between the semicircles, and a 120-degree segment below. | |
| 21 | B | Conditional Probability & States | The first four flips are forced to be THTH (probability 1/16); from there, the chance of eventually ending with HH rather than TT is 2/3. | |
| 22 | B | Conditional Probability & States | Only two kinds of holdings occur, (1,1,1) and a permutation of (2,1,0), and from either one the next state is (1,1,1) with probability exactly 1/4. | |
| 23 | C | Circles | Equal tangent lengths give P = (5,0); P, midpoint M of AB, and the center are collinear, and AM^2 = PM * MO yields the radius. | |
| 24 | C | Sequences & Series | Set y_n = x_n - 4: then y_{n+1} = y_n (y_n + 9)/(y_n + 10), a shrink factor between 9/10 and 10/11, so (9/10)^n <= y_n <= (10/11)^n. | |
| 25 | C | Recursive Counting | Past the leading 0, the string is k blocks of 10 or 110; the length equation 1 + 2k + j = 19 makes the count a sum of binomials. |
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.