2004 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.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | A | Linear Equations & Word Problems | You plus five friends is six people, so each share is 1500 divided by 6. | |
| 2 | B | Functions | Evaluate the three inner operations first to get -1, 1, -3, then apply the rule once more to those three numbers. | |
| 3 | E | Ratios, Percents & Averages | Convert 20 dollars to 2000 cents first; 1.45 percent of 2000 is 29. | |
| 4 | D | Absolute Value & Inequalities | Equal absolute values mean x is the same distance from 1 and from 2, so x is their midpoint 3/2. | |
| 5 | C | Basic Probability | Only 8 of the C(9,3) = 84 triples are collinear: three rows, three columns, and two diagonals. | |
| 6 | E | Linear Equations & Word Problems | There are 24 granddaughters, so exactly 4 daughters had children; the childless women are the other 2 daughters plus all 24 granddaughters. | |
| 7 | C | Sequences & Series | Each layer loses one orange in each direction, so the layers are 5x8, 4x7, 3x6, 2x5, 1x4 and the stack stops at the single row of 4. | |
| 8 | B | Games & Processes | Every three rounds each player loses exactly one token, so after 36 rounds the holdings are 3, 2, 1 and round 37 empties A. | |
| 9 | B | Triangles: Area & Pythagorean | Both triangles ABE and ABC contain triangle ABD, so the difference of the two small pieces equals the difference of the two big right triangles: 16 - 12 = 4. | |
| 10 | D | Basic Probability | Match the head counts 0, 1, 2, 3: the favorable outcomes number C(3,k)C(4,k) for each k, totaling 35 out of 2^7 = 128. | |
| 11 | C | Ratios, Percents & Averages | Scaling the diameter by 5/4 scales the base area by 25/16, so the height must scale by 16/25, a drop of 9/25 = 36 percent. | |
| 12 | C | Basic Counting | Each of the 8 condiments is independently in or out, giving 2^8 subsets, times 3 patty choices. | |
| 13 | D | Basic Counting | Count man-woman dance pairs two ways: 12 men times 3 equals (number of women) times 2, so there are 18 women. | |
| 14 | A | Ratios, Percents & Averages | Adding a 25-cent coin raises a 20-cent average to 21 only if there were 4 coins; 80 cents from 4 coins forces three quarters and a nickel. | |
| 15 | D | Absolute Value & Inequalities | Rewrite as 1 + y/x; since y/x is negative, maximize by making |y/x| as small as possible: y = 2, x = -4 gives 1 - 1/2. | |
| 16 | D | Basic Counting | Count by size: k by k squares containing the center number 1, 4, 9, 4, 1 for k = 1 to 5, totaling 19. | |
| 17 | C | Linear Equations & Word Problems | Runners jointly cover half a lap before the first meeting and a full lap between meetings, so Brenda runs 200 there and 200 + 150 is one lap. | |
| 18 | A | Sequences & Series | Terms are 9, 11 + d, 29 + 2d; the geometric condition gives d = 10 or d = -14, and d = -14 yields third term 1. | |
| 19 | C | Solid Geometry | Unroll the silo's surface into a rectangle; the stripe becomes parallelograms of horizontal width 3 whose heights total 80, so the area is 3 times 80. | |
| 20 | D | Triangles: Area & Pythagorean | Symmetry gives AE = CF = t; equating BE^2 = 1 + t^2 with EF^2 = 2(1 - t)^2 yields (1 - t)^2 = 2t, so the ratio is 2. | |
| 21 | B | Circles | Shaded area is 3 pi + 3 theta, and shaded is 8/21 of the total 9 pi, so 3 theta = 3 pi / 7. | |
| 22 | D | Circles | Sides BC and AD are tangents too, so CE = 2 + x and DE = 2 - x; the Pythagorean theorem in triangle CDE gives x = 1/2. | |
| 23 | D | Circles | Circle D has radius 2; placing B's center at (x, r), the two tangency equations give x = 3r - 2 and 9r^2 = 8r. | |
| 24 | D | Functions | Applying the rule to 2^100 = 2 * 2^99 repeatedly multiplies by 2^99, 2^98, ..., 2^1, so the exponent is 1 + 2 + ... + 99 = 4950. | |
| 25 | B | Solid Geometry | Connect the centers: the big center sits 3 units from each small center, directly above the centroid of an equilateral triangle of side 2, whose circumradius is 2/sqrt(3). |
Problems © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. The topic tags, difficulty ratings and key insights on this page are original to this site. No problem statements are reproduced here — each links to its own page.