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2004 AMC 10A — answer key
The answer to each problem, with the topic it tests and the idea it turns on.
2004 AMC 10A — answer key
amcstep.com| # | Answer | Topic and the idea it turns on |
|---|---|---|
| 1 | A | Linear Equations & Word Problems — Count the people first. You plus five friends is six people, so each person raises 1500 divided by 6. |
| 2 | B | Functions — Work from the inside out. The three inner values are -1, 1 and -3; then use the same rule once more on those three numbers. |
| 3 | E | Ratios, Percents & Averages — The answer must be in cents, so change 20 dollars into 2000 cents first. Then 1.45 percent of 2000 is 29. |
| 4 | D | Absolute Value & Inequalities — The distance from x to 1 equals the distance from x to 2, so x is exactly halfway between 1 and 2, which is 3/2. |
| 5 | C | Basic Probability — There are 84 ways to pick 3 of the 9 points, and only 8 of them lie on a line: the 3 rows, the 3 columns and the 2 diagonals. |
| 6 | E | Linear Equations & Word Problems — The 24 granddaughters come in groups of 6, so only 4 daughters are mothers. Everyone else, 30 - 4 = 26 women, has no daughters. |
| 7 | C | Sequences & Series — Each orange sits in a gap between four below, so each layer shrinks by one both ways: 5 by 8, 4 by 7, down to 1 by 4. |
| 8 | B | Games & Processes — Every three rounds, each player ends up with exactly one fewer token. So after 36 rounds they hold 3, 2, 1, and round 37 leaves A with nothing. |
| 9 | B | Triangles: Area & Pythagorean — Both big right triangles contain the same piece, triangle ABD. Take it away from each and the difference stays the same: 16 - 12 = 4. |
| 10 | D | Basic Probability — Go through each possible number of heads, 0 to 3. Multiply the ways coin A and coin B can each get that many, then add: 35 out of 128. |
| 11 | C | Ratios, Percents & Averages — Making the jar 5/4 as wide makes the base 25/16 as big, so the height must shrink to 16/25 of what it was. That is a 36 percent drop. |
| 12 | C | Basic Counting — For each of the 8 condiments you just say yes or no, giving 2 to the 8th = 256 choices. Multiply by 3 patty choices. |
| 13 | D | Basic Counting — Count dance pairs two ways. The men make 12 times 3 = 36 pairs; each woman is in 2 pairs, so there are 18 women. |
| 14 | A | Ratios, Percents & Averages — Adding one 25-cent coin raises the average by exactly 1 cent only if there were 4 coins, worth 80 cents in all. That forces three quarters and a nickel. |
| 15 | D | Absolute Value & Inequalities — Split the fraction into 1 + y/x. Since y/x is always negative, make it as close to 0 as possible: y = 2 and x = -4 give 1/2. |
| 16 | D | Basic Counting — Count by size. The numbers of squares that cover the center are 1, 4, 9, 4, 1 for sizes 1 through 5, which adds to 19. |
| 17 | C | Linear Equations & Word Problems — Together they run half a lap before the first meeting and a full lap between meetings, so Brenda runs 200 more; 200 + 150 is one lap. |
| 18 | A | Sequences & Series — The new numbers are 9, 11 + d and 29 + 2d. Making them geometric gives d = 10 or d = -14; the second gives third term 1. |
| 19 | C | Solid Geometry — Unroll the silo into a flat rectangle. The stripe becomes slanted strips 3 feet wide climbing 80 feet in total, so its area is 3 times 80. |
| 20 | D | Triangles: Area & Pythagorean — With side 1 and AE = CF = t, setting sides BE and EF equal gives (1 - t) squared = 2t, which makes the ratio exactly 2. |
| 21 | B | Circles — The shaded area works out to 3 pi + 3 theta, and it is 8 parts out of 8 + 13 = 21 parts of the whole 9 pi. |
| 22 | D | Circles — Sides BC and AD also touch the semicircle, so equal tangent lengths give CE = 2 + x and DE = 2 - x; the Pythagorean theorem finds x. |
| 23 | D | Circles — Circle D has radius 2. Put B's center at (x, r); touching D and touching A give two distance equations, leading to r = 8/9. |
| 24 | D | Functions — Climb the powers of 2 using the rule. Reaching 2 to the 100th multiplies factors 2 to the 0, 1, up to 99, so the exponent is 4950. |
| 25 | B | Solid Geometry — Use the centers. The big center is 3 from each small center, above the middle of an equilateral triangle of side 2, 2 over root 3 from each corner. |
2004 AMC 10A — 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.