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2013 AMC 10B — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2013 AMC 10B — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | C | Fractions & Decimals — Work out the small sums first. The problem becomes 12/9 minus 9/12, which is 4/3 minus 3/4. |
| 2 | A | Sets, Estimation & Miscellaneous — Change both sides from steps to feet first, giving 30 feet by 40 feet. Find the area in square feet, then take half of it. |
| 3 | C | Linear Equations & Word Problems — If two numbers average 3, they add up to 6. The high is 16 more than the low, so the low is 8 below the average. |
| 4 | D | Number Properties — Counting down from 201 to 53 counts both end numbers, so the count is 201 minus 53, plus 1. |
| 5 | B | Algebraic Manipulation — Write 2a - ab as a(2 - b). Use b = 5 to make (2 - b) as negative as possible, then the largest a, which is 5. |
| 6 | C | Statistics & Data — Turn each average back into a total: the kids' ages add to 363 and the parents' ages to 1815. Then divide the grand total by all 88 people. |
| 7 | B | Triangles: Area & Pythagorean — The triangle with three different sides uses two neighbors and the point opposite one of them. Its long side is a diameter, so it has a right angle. |
| 8 | B | Ratios, Percents & Averages — Pick an easy distance, like 40 miles each. Ray uses 1 gallon and Tom uses 4, so 80 miles on 5 gallons is 16 miles per gallon. |
| 9 | D | Divisibility & Factors — Relatively prime numbers share no prime, so all the 2s go in one number, all the 3s in another, all the 5s in the third: 8, 27, 125. |
| 10 | C | Linear Equations & Word Problems — Let x be the three-point tries; then there are 1.5x two-point tries. Points are 1.5x + 1.2x = 2.7x, and 2.7x = 54 gives x = 20. |
| 11 | B | Quadratics — Complete the square: the equation becomes (x-5)^2 + (y+3)^2 = 0. Squares are never negative, so both must be 0. |
| 12 | B | Basic Probability — A regular pentagon has 5 sides of one length and 5 diagonals of another length. Whatever segment you pick first, 4 of the other 9 match it. |
| 13 | E | Sequences & Series — Turn k says k numbers, so after 9 turns 45 numbers have been said. The 53rd number is the 8th number of turn 10, which is 8. |
| 14 | E | Algebraic Manipulation — The equation becomes 2xy(x - y) = 0, which is true when x = 0, when y = 0, or when x = y: three lines. |
| 15 | B | Quadrilaterals & Polygon Areas — Write each area using its side length: the triangle's side is a/3 and the hexagon's side is b/6. The hexagon is six small equilateral triangles. |
| 16 | B | Triangle Centers & Cevians — 1.5, 2, 2.5 is half of a 3-4-5 triangle, so the medians cross at a right angle. Doubling PD and PE gives the other pieces of each median. |
| 17 | E | Games & Processes — Every trade gives one silver token, so count trades. Trading must stop at 1 red and 2 blue; two equations then give 59 + 44 = 103 trades. |
| 18 | D | Bases & Digits — For numbers 1bcd, the last digit is 1 + b + c, so b + c is at most 8: that gives 45 numbers. Add 2002. |
| 19 | D | Quadratics — Exactly one root means b^2 = 4ac, and b is halfway between a and c. Take a = 1, solve for c, and the root is -b/(2a). |
| 20 | B | Primes — 2013 has the prime factor 61, so the top needs 61!. That brings in the prime 59, so the bottom needs 59! to cancel it. |
| 21 | C | Sequences & Series — The seventh term is 5a + 8b. Two different starts give the same seventh term only if one start is 8 more in a and 5 less in b. |
| 22 | C | Arrangements with Restrictions — J is on every line, so the lines have equal sums exactly when the four opposite pairs have equal sums. The eight outer numbers add to 45 - J. |
| 23 | B | Circles — Right angles at D and F both face AB, so A, F, D, B lie on one circle. Then angle AFE equals angle B, giving EF = 4. |
| 24 | A | Divisibility & Factors — A number with exactly four divisors is p^3 or p times q for primes. The divisor sums are (1+p)(1+p^2) or (1+p)(1+q); only 2016 works. |
| 25 | E | Bases & Digits — The last digits in base 5 and base 6 must match, so N is 30m plus a remainder from 0 to 4. The tens digits then depend only on m. |
2013 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.