2010 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 | Algebraic Manipulation | Distribute the first product: 100(100-3) = 100*100 - 300, so the two 100*100 terms cancel and only -300 + 3 remains. | |
| 2 | C | Ratios, Percents & Averages | Convert everything to minutes: 45 + 90 = 135 minutes of meetings out of 540, which is exactly one quarter. | |
| 3 | C | Sets, Estimation & Miscellaneous | Four colors means four socks can all be different, but a fifth sock must repeat one of the four colors. | |
| 4 | C | Functions | The heart of x is (x + x^2)/2, so the three values are 1, 3 and 6, the first three triangular numbers. | |
| 5 | B | Clocks, Calendars & Time | 31 days is four weeks plus 3, so only the first three weekdays occur five times; Monday and Wednesday must both be in or both out of that block. | |
| 6 | B | Circles | Angle CAB is an inscribed angle on arc CB, while angle COB is the central angle on the same arc, so it is half of 50 degrees. | |
| 7 | D | Triangles: Area & Pythagorean | The altitude of the 10-10-12 isosceles triangle splits it into two 6-8-10 right triangles, so its area is 48. | |
| 8 | E | Divisibility & Factors | The price must divide both 48 and 64, so it is a divisor of gcd(48, 64) = 16, which has five divisors. | |
| 9 | D | Algebraic Manipulation | Simplifying gives a-b+c-d-e, while Larry computed a-b-c-d+e; the two agree exactly when e = c. | |
| 10 | C | Linear Equations & Word Problems | Work in miles per minute: 1/2 mile per minute in sun and 1/3 in rain; with t rain minutes, (40 - t)/2 + t/3 = 16. | |
| 11 | A | Linear Equations & Word Problems | Express each coupon's savings in terms of the price p; A beats B when p >= 200 and beats C when p <= 250. | |
| 12 | D | Sets, Estimation & Miscellaneous | The net shift is 20% from No to Yes; switchers come in that 20% plus matched pairs who swap in opposite directions, and at most 30% can swap Yes-to-No. | |
| 13 | C | Absolute Value & Inequalities | Split at x = 30 to remove the inner absolute value, then each case is |linear| = x, which yields two candidate roots to verify. | |
| 14 | B | Linear Equations & Word Problems | There are 100 numbers whose sum is 4950 + x, so the average condition reads 4950 + x = 10000x, giving x = 4950/9999. | |
| 15 | C | Linear Equations & Word Problems | With c correct and w wrong, 4c-w = 99 and c+w <= 50; substituting w = 4c-99 gives c <= 29. | |
| 16 | B | Circles | The circle crosses each side at a 60-degree central angle (cos of half-angle is (1/2)/(sqrt3/3) = sqrt3/2), so the four bulges are four 60-degree circular segments. | |
| 17 | B | Statistics & Data | With 3n students Andrea's median rank is (3n+1)/2, so n is odd; she is above 37th, so n < 25, and 64 students exist, so n >= 22. | |
| 18 | E | Modular Arithmetic | Factor as a(b(c+1)+1): divisible by 3 when a is, or else when b(c+1) is 2 mod 3; residues are uniform since 3 divides 2010. | |
| 19 | B | Circles | O, A and the midpoint M of BC are collinear; O outside the triangle means OM = OA + AM, then Pythagoras in triangle OMB. | |
| 20 | D | Circles | Lines BC and FA meet at P in a 60-degree angle; both circles are inscribed in it, so each center is 2r from P. | |
| 21 | E | Bases & Digits | A four-digit palindrome abba equals 1001a + 110b, and 1001 = 7*11*13, so divisibility by 7 depends only on b: b must be 0 or 7. | |
| 22 | C | Inclusion-Exclusion | Every candy independently picks one of 3 bags (3^7 total); subtract the assignments that leave red empty or blue empty, adding back the one that leaves both empty. | |
| 23 | D | Arrangements with Restrictions | 1 and 9 are fixed at the corners; the center must be 4, 5, or 6, and each case leaves two independent small ordering problems. | |
| 24 | E | Sequences & Series | The 100-point cap leaves only ratios 2, 3, 4, 3/2 with tiny first terms; requiring 4a+6d one less singles out a=5, r=2. | |
| 25 | B | Polynomials | Factor P(x)-a = (x-1)(x-3)(x-5)(x-7)Q(x); evaluating at 2, 4, 6, 8 forces 15, 9, 15, 105 to divide 2a, so 315 divides a. |
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