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2019 AMC 10B — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2019 AMC 10B — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | D | Ratios, Percents & Averages — The same water fills 5/6 of the first container and 3/4 of the second. Pick an amount of water that makes both containers whole numbers. |
| 2 | E | Logic Puzzles — To break an if-then statement, the if part must be true and the then part false. So n and n minus 2 must both be not prime. |
| 3 | B | Ratios, Percents & Averages — Two of the facts are about students who do not play. Count the non-players among seniors and non-seniors, and set the total equal to 46.8% of 500. |
| 4 | A | Coordinate Geometry — If every such line passes through one point, any two of them will find it. Use the easy progressions 1, 1, 1 and 1, 2, 3. |
| 5 | E | Coordinate Geometry — Reflecting over y = x swaps the two coordinates of each point. So a line of slope 2 becomes a line of slope 1/2, and those are not perpendicular. |
| 6 | C | Algebraic Manipulation — Both factorials on the left contain n! as a factor. Divide it out to get (n+1)(n+3) = 440, which is 20 times 22. |
| 7 | B | GCD & LCM — Casper's money can be split evenly into 12, 14 or 15 equal prices, so it is a common multiple of those numbers. The smallest is 420 cents. |
| 8 | B | Quadrilaterals & Polygon Areas — Each triangle stands on a side and its tip reaches the center, so its height is half the square's side. So the square has side 2 times root 3. |
| 9 | A | Functions — Try a positive number, a negative whole number, and a negative number with a decimal part. Only the last kind gives something other than 0, and it always gives -1. |
| 10 | A | Geometric Optimization — Area 100 on a base of 10 needs point C at height 20. But then AC and BC are each at least 20, and they cannot both equal 20. |
| 11 | A | Ratios, Percents & Averages — Both jars hold the same number of marbles. Green is 1/10 of Jar 1 and 1/9 of Jar 2, so a jar holding 90 marbles is the natural unit. |
| 12 | C | Bases & Digits — Numbers below 2019 have at most four base-7 digits, each at most 6. Check whether the smallest number with each high digit sum still fits under 2019. |
| 13 | A | Statistics & Data — The median is 6, x, or 8, depending on where x falls. Solve median = mean in each case, then keep only answers that really fall in that case. |
| 14 | C | Divisibility & Factors — 19! ends in three zeros, so H is 0. Then 19! divides evenly by 9 and by 11, and those two digit tests pin down T and M. |
| 15 | A | Triangles: Area & Pythagorean — The shared sides cannot play the same roles in both triangles, so the small triangle's hypotenuse must be a leg of the big one. That doubles a leg. |
| 16 | A | Triangles: Area & Pythagorean — The isosceles triangles copy the angles at A and B over to D, so angle CDE is a right angle. Then CDE is a 3-4-5 triangle and BC = 8. |
| 17 | C | Basic Probability — Red higher and green higher are equally likely. The only other result is a tie, which has probability 1/4 + 1/16 + 1/64 + ... = 1/3. |
| 18 | C | Sequences & Series — In the long run Henry bounces between two fixed turning points. Walking toward the gym from A must end at B, and walking home from B must end at A. |
| 19 | C | Divisibility & Factors — Every divisor is 2 to some power times 5 to some power, both powers 0 to 5. Multiplying two of them adds the powers, giving an 11 by 11 grid. |
| 20 | E | Circles — Cut the big circle along lines EG and AD. Above EG is a half-disk, below AD is a circle slice, and in between only two small gaps are shaded. |
| 21 | B | Conditional Probability & States — The first four flips are forced to be T, H, T, H. After that, list the ways to finish with two heads and add their probabilities. |
| 22 | B | Conditional Probability & States — The money is always split 1, 1, 1 or 2, 1, 0 in some order. From either split, the next ring leaves everyone with 1 dollar with probability 1/4. |
| 23 | C | Circles — Equal tangent lengths put the meeting point at P = (5, 0). The center lies on line PM, with M the midpoint of AB, and Pythagoras gives the radius. |
| 24 | C | Sequences & Series — Track how far each term is above 4. That distance shrinks by a factor between 9/10 and 10/11 each step, so powers of those fractions bound m. |
| 25 | C | Recursive Counting — After the first 0, the string breaks into pieces 10 and 110. Count how many of each fit into 18 more digits, then choose where the long pieces go. |
2019 AMC 10B — 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.