2015 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.
The same key as a PDF 126 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | C | Exponents, Logarithms & Radicals | Any nonzero number to the power 0 is 1, so the parentheses hold 25. A power of negative one means flip the number over, giving 1/25, then times 5. | |
| 2 | D | Linear Equations & Word Problems | Pretend all 25 tiles are triangles: that is 75 edges. Each triangle swapped for a square adds one edge, so the 9 extra edges mean 9 squares. | |
| 3 | D | Basic Counting | Count the flat toothpicks row by row, then use the fact that there are just as many standing ones. The 3-step staircase has 18, the 5-step one has 40. | |
| 4 | B | Ratios, Percents & Averages | Only the ratio matters, so give Mia 1 egg, Sofia 2 and Pablo 6. Everyone should end with 3, so Sofia gets 1 of Pablo's 6 eggs. | |
| 5 | E | Ratios, Percents & Averages | Turn each average into a total: 14 tests averaging 80 add to 1120, and 15 tests averaging 81 add to 1215. The difference, 95, is Payton's score. | |
| 6 | B | Linear Equations & Word Problems | The sum equals 5 times the difference. Gathering terms gives 4 times the larger equals 6 times the smaller, so the ratio is 3 to 2. | |
| 7 | B | Sequences & Series | From 13 to 73 is a climb of 60, which is 20 jumps of 3. Twenty jumps connect 21 numbers, because you also count the starting number. | |
| 8 | B | Linear Equations & Word Problems | Name the ages from two years ago so Pete is 3 times Claire. The four-years-ago fact then gives Claire 6, so today they are 20 and 8. | |
| 9 | D | Ratios, Percents & Averages | Volume uses the radius squared. A radius 1.1 times as big gives 1.21 times the base area, so the first cylinder must be 1.21 times as tall. | |
| 10 | C | Arrangements with Restrictions | The letter b may only sit next to d, and c only next to a. With one allowed neighbor each, both must sit at the ends. | |
| 11 | C | Triangles: Area & Pythagorean | Try the 3-by-4 rectangle: its diagonal is 5 and its area is 12. Area divided by diagonal squared is 12/25, and that stays the same at every size. | |
| 12 | C | Algebraic Manipulation | The equation rearranges to (y minus x squared) squared equals 1. So y is x squared plus 1 or minus 1, and the two heights always differ by 2. | |
| 13 | C | Basic Counting | Count in nickels, with a dime worth 2. With at least one nickel, every amount up to the total can be made, so the total is 17 nickels. | |
| 14 | C | Circles | The disk turns from rolling and also from traveling around the clock: 3 turns per lap. So the arrow is up again after one third of a lap. | |
| 15 | B | Diophantine Equations | Cross-multiply and rearrange to get y times (10 minus x) equals 11x. So x is at most 9; checking x from 1 to 9 leaves only 5/11 in lowest terms. | |
| 16 | B | Systems of Equations | Subtract the equations: the difference of squares lets you cancel x minus y, leaving x + y = 3. Adding the equations then gives 15. | |
| 17 | D | Coordinate Geometry | The given slope means a 30-degree tilt. The third line tilts 30 degrees downward, so the triangle's vertical side runs between the two lines' heights at x = 1. | |
| 18 | E | Bases & Digits | In base 16, 1000 is 3E8. Every three-digit string of digits 0 to 9 starting with 0 to 3 is below it: 400 strings, minus zero. | |
| 19 | D | Triangles: Area & Pythagorean | Cut along the mirror line, the height from C. Each half is a right triangle with a 15-degree angle, and tan 15 degrees is 2 minus root 3. | |
| 20 | B | Algebraic Manipulation | Area plus perimeter plus 4 always factors as (width + 2) times (length + 2), both at least 3. But 106 is only 2 times 53. | |
| 21 | C | Solid Geometry | Faces ABC and ABD are matching 3-4-5 right triangles on side AB. Their heights to AB meet at one point, and CD shows they are perpendicular. | |
| 22 | A | Recursive Counting | Rows with no two neighbors standing follow 2, 3, 5, 8, 13, 21, 34. Open the circle at one person: 34 ways if they sit, 13 if they stand. | |
| 23 | C | Quadratics | The zeros multiply to twice their sum. That rearranges to (r minus 2)(s minus 2) = 4; use every factor pair of 4, negatives included. | |
| 24 | B | Diophantine Equations | A right triangle inside shows BC squared is 4 times (AD minus 1). So BC = 2k, AD = k squared plus 1, with k from 1 to 31. | |
| 25 | A | Geometric Probability | Split by where the two points land: same side (chance 1/4), next-door sides (1/2), or opposite sides (1/4). Find the success chance in each case and weight it. |
2015 AMC 10A. 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.