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2010 AMC 10B — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2010 AMC 10B — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | C | Algebraic Manipulation — Multiply out 100(100 - 3) to get 10000 - 300. The 10000 shows up in both parts and cancels, leaving -300 + 3. |
| 2 | C | Ratios, Percents & Averages — Put everything in minutes. The meetings take 45 + 90 = 135 minutes, and the work day is 9 times 60 = 540 minutes, exactly four times as long. |
| 3 | C | Sets, Estimation & Miscellaneous — Four colors means you can pull four socks that are all different. The fifth sock has to match one of them. |
| 4 | C | Functions — The heart of a number is the number plus its square, divided by 2. For 1, 2 and 3 that gives 1, 3 and 6. |
| 5 | B | Clocks, Calendars & Time — 31 days is 4 weeks plus 3 days, so the month's first three weekdays appear 5 times. Monday and Wednesday must both be among those three, or both not. |
| 6 | B | Circles — Angles CAB and COB open onto the same arc CB. The one with its corner on the circle is half the one at the center: 25 degrees. |
| 7 | D | Triangles: Area & Pythagorean — Cut the 10-10-12 triangle down the middle. Each half is a right triangle with sides 6, 8, 10, so the height is 8 and the area is 48. |
| 8 | E | Divisibility & Factors — The ticket price must divide both 48 and 64 evenly. The prices that do are 1, 2, 4, 8 and 16, the numbers that divide 16. |
| 9 | D | Algebraic Manipulation — Plug in the known numbers first. The correct value works out to -2 - e and Larry's value is e - 8; setting them equal gives e = 3. |
| 10 | C | Linear Equations & Word Problems — Change the speeds to miles per minute: 1/2 in the sun and 1/3 in the rain. Then one equation with the rain minutes as the unknown finishes it. |
| 11 | A | Linear Equations & Word Problems — For price p, Coupon A saves 0.15p, B saves 30, C saves 0.25(p - 100). A ties B at p = 200 and ties C at p = 250. |
| 12 | D | Sets, Estimation & Miscellaneous — In a class of 100, Yes grows by 20. At least those 20 switch. At most, all 50 No students switch to Yes and 30 Yes students switch to No. |
| 13 | C | Absolute Value & Inequalities — The inside absolute value changes behavior at x = 30. Split into x up to 30 and x above 30; each case becomes simpler. Check every answer. |
| 14 | B | Linear Equations & Word Problems — There are 100 numbers, not 99. Their sum is 4950 + x, and that sum must equal 100 times 100x, so 9999x = 4950. |
| 15 | C | Linear Equations & Word Problems — With c correct answers, Jesse needs 4c - 99 wrong ones to get 99 points. Correct plus wrong is at most 50, so c is at most 29. |
| 16 | B | Circles — The circle bulges out past each side of the square in four equal caps. Each cap is a 60-degree slice of the circle minus an equilateral triangle. |
| 17 | B | Statistics & Data — Andrea is the one middle student, so the number of students is odd. She beat 37th place, and someone finished 64th; only 23 schools fit all three facts. |
| 18 | E | Modular Arithmetic — Factor the expression as a times (bc + b + 1). It is a multiple of 3 when a is, or when bc + b + 1 is. |
| 19 | B | Circles — The center O, the corner A and the midpoint M of BC lie on one line. Since O is outside the triangle, OM = OA + AM; then use Pythagoras. |
| 20 | D | Circles — Extend lines BC and FA to meet at P. Both circles sit in the same 60-degree corner at P, so each center is twice its radius from P. |
| 21 | E | Bases & Digits — A palindrome abba equals 1001a + 110b. 1001 is a multiple of 7, so only the middle digit b matters, and b must be 0 or 7. |
| 22 | C | Inclusion-Exclusion — Each candy picks one of 3 bags: 3^7 ways. Subtract the ways with red empty and with blue empty, then add back the one way with both empty. |
| 23 | D | Arrangements with Restrictions — 1 and 9 are stuck in opposite corners. The center must be 4, 5 or 6; in each case the leftover cells split into small pieces you can count. |
| 24 | E | Sequences & Series — The 100-point limit leaves the Raiders few options, such as 1, 2, 4, 8 or 8, 12, 18, 27. Only 5, 10, 20, 40 fits a Wildcats total one lower. |
| 25 | B | Polynomials — P(x) - a is zero at 1, 3, 5, 7, so it has the factor (x-1)(x-3)(x-5)(x-7). Plugging in 2, 4, 6, 8 forces a to be a multiple of 315. |
2010 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.