2011 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 132 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | D | Linear Equations & Word Problems | Split the bill into three parts. Only the extra half hour costs money, and half an hour is 30 minutes at 10 cents each. | |
| 2 | E | Number Properties | Fourteen small bottles hold only 490 milliliters, which is not quite 500, so she needs one more bottle. | |
| 3 | D | Functions | Work from the inside out. The inner brackets are 2/3 and 1/2; then average those two numbers together with 0. | |
| 4 | A | Sequences & Series | The two sums share every even number from 12 to 100. Those cancel when you subtract, leaving just 102 minus 10. | |
| 5 | C | Ratios, Percents & Averages | The grade sizes are in the ratio 4 : 2 : 1, so pretend there are 4, 2 and 1 students. Add up all their minutes and divide by 7. | |
| 6 | C | Sets, Estimation & Miscellaneous | The union always contains all 20 elements of A. It has exactly 20 when every element of B is already in A. | |
| 7 | B | Absolute Value & Inequalities | An absolute value is never negative, so |−3x| + 5 is always at least 5 and can never equal 0. Each other equation has an actual solution. | |
| 8 | C | Ratios, Percents & Averages | Pretend there are 100 birds. Leaving out the 25 swans leaves 75 birds, and 30 of them are geese, so the answer is 30 out of 75, which is 40%. | |
| 9 | A | Coordinate Geometry | The lines x = −c and x = d sit on opposite sides of the y-axis, so the width is c + d. Likewise the height is a + b. | |
| 10 | B | Divisibility & Factors | Factor 1771 cents as 7 × 11 × 23. A majority of 30 is 16 to 30 students, so 23 bought; each bought 7 pencils at 11 cents. | |
| 11 | B | Triangles: Area & Pythagorean | Make the side 8, so AE = 7 and EB = 1. Each corner triangle has legs 7 and 1; the inner square has area 49 + 1 = 50. | |
| 12 | A | Linear Equations & Word Problems | Call the number of two-point shots y. They give 2y points, three-pointers also give 2y points, and free throws give y + 1 points, so 5y + 1 = 61. | |
| 13 | A | Basic Counting | The first digit must be 2 or 5 and the last digit must be 2 or 8. If the first digit is 2, the last digit can only be 8. | |
| 14 | B | Basic Probability | With diameter d, the area is less than the circumference exactly when d is less than 4. So the dice must sum to 2 or 3: 3 of 36 rolls. | |
| 15 | C | Linear Equations & Word Problems | Miles per gallon for the whole trip is total miles divided by total gallons. Only the miles after the first 40 use any gas. | |
| 16 | B | Exponents, Logarithms & Radicals | Square the sum. The two roots multiply to the square root of 81 − 72 = 9, which is 3, so the square is 18 + 6 = 24. | |
| 17 | C | Sequences & Series | Two overlapping groups of three share two terms, so each term equals the term three places later. So H equals B, and A + B = 30 − 5. | |
| 18 | C | Circles | Circle C passes through the point where A and B touch, so it overlaps each of them in a lens of area π/2 − 1. What is left is 2. | |
| 19 | E | Number Properties | Call 1991 a² and 2001 b² + 9. Then b² − a² = 141, so a = 70 or a = 22. Only 22 also makes 2011 a perfect square. | |
| 20 | D | Geometric Probability | A chord as long as the radius covers 60 degrees. Fix the first point; the chords cross only if the second point lands within 60 degrees of it, either side. | |
| 21 | D | Conditional Probability & States | The two pairs weigh the same only when both are all real coins or each holds exactly one fake. Count those two kinds of outcomes and compare. | |
| 22 | C | Arrangements with Restrictions | The five diagonals join up into one closed loop A, C, E, B, D, back to A. So count colorings where neighbours on this loop differ. | |
| 23 | C | Games & Processes | Each student gets an evenly spaced list. The numbers she skips start one step later and are spaced three times as far apart: 2, 5, 14, 41, 122, 365. | |
| 24 | D | Solid Geometry | Each tetrahedron has volume 1/3. The other tetrahedron's faces cut off its four corners, each a half-size copy of volume 1/24, leaving 1/3 − 4/24 = 1/6. | |
| 25 | C | Triangles: Area & Pythagorean | The rays must pass through all four corners. That forces both coordinates of the point to be multiples of 2/n, so the question becomes counting grid points. |
2011 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.