2009 AMC 10A
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 126 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | E | Linear Equations & Word Problems | Ten cans hold only 120 ounces, which is 8 short, so one more can is needed even though it will not be full. | |
| 2 | A | Number Properties | Every choice is a multiple of 5, so the pennies must total a multiple of 5, which means none; four coins of at least 5 cents make at least 20. | |
| 3 | C | Fractions & Decimals | Work from the innermost part outward: 1 + 1 = 2, then 1 + 1/2 = 3/2, flip it to get 2/3, then 1 + 2/3 = 5/3. | |
| 4 | A | Linear Equations & Word Problems | Find how long the swim and the run take, subtract from 2 hours, and use the time that is left for the 15-mile bike ride. | |
| 5 | E | Bases & Digits | Squaring 11, 111 and 1111 gives 121, 12321 and 1234321; the digits climb to the number of ones and come back down, with no carrying up to nine ones. | |
| 6 | A | Circles | The small circle reaches from the middle of the flat edge up to the top of the arc, so the semicircle's radius is the small circle's diameter, 4. | |
| 7 | C | Ratios, Percents & Averages | Taking 40% away from the whole-milk amount leaves 60% of it, so 60% of the whole-milk amount is 2, and the whole-milk amount is 2 divided by 0.6. | |
| 8 | B | Ratios, Percents & Averages | A senior ticket is 75% of the full price, so the full price is 6 divided by 0.75, which is 8 dollars; a child ticket is half of that. | |
| 9 | B | Sequences & Series | The ratio squared must divide 2009 = 7 x 7 x 41, so the ratio is 7 and a = 2009 / 49 = 41. | |
| 10 | B | Similar & Congruent Triangles | The height BD is a leg of two small right triangles. Using the Pythagorean theorem in all three triangles gives BD squared = 3 x 4 = 12. | |
| 11 | D | Algebraic Manipulation | The new box has volume s(s+1)(s-1), which is s cubed minus s, so it is always s less than the cube. A drop of 5 means s = 5. | |
| 12 | C | Triangles: Area & Pythagorean | In triangle BCD, BD + 5 must be more than 17, so BD > 12. In triangle ABD, BD < 5 + 9 = 14. So BD = 13. | |
| 13 | E | Exponents, Logarithms & Radicals | Split 12 into 2 x 2 x 3, so 12^(mn) = 2^(2mn) times 3^(mn). Then 2^(2mn) = P^(2n) and 3^(mn) = Q^m. | |
| 14 | A | Quadrilaterals & Polygon Areas | The outer square's side is a + b and the inner square's is a - b. Four times the area means twice the side. | |
| 15 | E | Sequences & Series | Each new ring with n diamonds per side holds 4(n - 1) diamonds, so F_20 = 1 + 4(1 + 2 + ... + 19) = 761. | |
| 16 | D | Absolute Value & Inequalities | Put a at 0 and step 2, then 3, then 4, each to the left or right. The possible distances from the start are 1, 3, 5 and 9. | |
| 17 | C | Similar & Congruent Triangles | DB is the height of right triangle DEF to its long side EF; the triangles DBE and DBF are enlarged copies of the 3-4-5 triangles DAB and DCB. | |
| 18 | D | Ratios, Percents & Averages | Pretend there are 100 children: 60 play soccer, 24 of those swim, so 36 soccer players do not swim, out of 70 children who do not swim. | |
| 19 | B | Divisibility & Factors | Rolling without slipping, circle B has turned 100/r times when it gets back to the start; the same point is touching again only if r divides 100. | |
| 20 | D | Linear Equations & Word Problems | The gap closes 1 km per minute, split 3 to 1, so Lauren rides 1/4 km per minute. After 5 minutes, she alone rides the last 15 km. | |
| 21 | C | Circles | The four small centers form a square of side 2r. The big center is half a diagonal, r times root 2, from each, so R = r(1 + root 2). | |
| 22 | D | Basic Probability | Random stickers make the first top number any of the 12, and the second any of the other 11. Exactly 2 of those 11 make 7. | |
| 23 | E | Similar & Congruent Triangles | Adding triangle ABE to both equal areas shows ABD and ABC have equal area, so AB is parallel to CD and AE : EC = 9 : 12. | |
| 24 | C | Basic Probability | Count the bad choices: the plane misses the inside only when all three corners lie on one face. That is 6 faces times 4 choices, 24 out of 56. | |
| 25 | B | Divisibility & Factors | I_k is 10 to the power k+2, plus 64. Pull out the smaller power of 2 and an odd number remains, except at the tie k = 4. |
2009 AMC 10A. 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.