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.
The same key as a PDF 129 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 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. 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.