2013 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.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | C | Fractions & Decimals | The two sums are just 12 and 9, so the expression is 4/3 minus 3/4, which is 7/12 over the common denominator 12. | |
| 2 | A | Sets, Estimation & Miscellaneous | Convert each side from steps to feet before finding the area; then the yield is half a pound per square foot. | |
| 3 | C | Linear Equations & Word Problems | An average of 3 means the high and low add to 6, and the high is the low plus 16, so twice the low is 6 minus 16. | |
| 4 | D | Number Properties | Counting backwards from 201, the number 53 is the (201 - 53 + 1)th, because inclusive counts need the extra 1. | |
| 5 | B | Algebraic Manipulation | Factor 2a - ab = a(2 - b); make the factor 2 - b as negative as possible (b = 5) and a as large as possible (a = 5). | |
| 6 | C | Statistics & Data | A combined average is total age over total people: (33*11 + 55*33)/(33 + 55), not the average of 11 and 33. | |
| 7 | B | Triangles: Area & Pythagorean | The only scalene choice uses two adjacent points and the point opposite one of them, giving a 30-60-90 triangle on a diameter of length 2. | |
| 8 | B | Ratios, Percents & Averages | Combined mileage is total miles over total gallons; with equal distances, choose a convenient distance like 40 miles and count gallons. | |
| 9 | D | Divisibility & Factors | Pairwise relatively prime factors cannot share a prime, so each prime power of 27000 = 2^3 3^3 5^3 goes whole to one number: 8, 27, 125. | |
| 10 | C | Linear Equations & Word Problems | Let x be three-point attempts; two-point attempts are 1.5x, and points are 2(0.5)(1.5x) + 3(0.4)x = 2.7x = 54. | |
| 11 | B | Quadratics | Move everything to one side and complete the square in x and y: (x-5)^2 + (y+3)^2 = 0 forces x = 5, y = -3. | |
| 12 | B | Basic Probability | A regular pentagon has 5 equal sides and 5 equal diagonals, so a matching pair is two sides or two diagonals: 2*C(5,2) out of C(10,2). | |
| 13 | E | Sequences & Series | The k-th turn says k numbers, so after k turns 1+2+...+k numbers are spoken; 45 < 53 <= 55 puts the 53rd number in the 10th turn, at position 8. | |
| 14 | E | Algebraic Manipulation | Both sides factor as xy(x-y) and -xy(x-y), so 2xy(x-y) = 0: the three lines x = 0, y = 0, x = y. | |
| 15 | B | Quadrilaterals & Polygon Areas | Write both areas in terms of the side lengths a/3 and b/6; the hexagon is six unit triangles, so the ratio of a^2 to b^2 is 3/2. | |
| 16 | B | Triangle Centers & Cevians | 1.5, 2, 2.5 is a right triangle, so the medians are perpendicular; the centroid's 2:1 ratio gives AD = 6, CE = 4.5, and AEDC has area half their product. | |
| 17 | E | Games & Processes | Exchanges stop only at 1 red and 2 blue; solving 75-2x+y = 1, 75+x-3y = 2 gives x+y = 103 exchanges. | |
| 18 | D | Bases & Digits | For 1bcd the condition is d = 1 + b + c with b + c <= 8, giving 45 pairs; the only extra number in the 2000s is 2002. | |
| 19 | D | Quadratics | One root means b^2 = 4ac with b = (a+c)/2; dividing by a^2 gives a quadratic in t = c/a, and the root is -b/(2a) = -(1+t)/4. | |
| 20 | B | Primes | 2013 = 3*11*61 forces a_1 >= 61; then 61! contains the prime 59, which must be cancelled, so b_1 >= 59, and 61 and 59 work. | |
| 21 | C | Sequences & Series | Seventh term is 5a+8b; two starts with equal N differ by (+8,-5), so b >= a+13 and (0,13), (8,8) give 104. | |
| 22 | C | Arrangements with Restrictions | The center is on every line, so opposite pairs need equal sums; 45-J divisible by 4 forces J in {1,5,9}, then 4!*2^4 each. | |
| 23 | B | Circles | Right angles at D and F put A, B, D, F on the circle with diameter AB, so angle AFE = angle B and EF = 4. | |
| 24 | A | Divisibility & Factors | Four divisors means m = p^3 or m = pq, with divisor sums (1+p)(1+p^2) or (p+1)(q+1); only 2016 = 4 * 504 splits as (3+1)(503+1). | |
| 25 | E | Bases & Digits | Units digits force N mod 5 = N mod 6, so N = 30m+r, r <= 4; tens digits then fix the base-6 tens digit by parity of floor(N/25). |
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