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.
The same key as a PDF 124 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 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. 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.