2005 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 | D | Ratios, Percents & Averages | A $2 tip at 10% means a $20 bill and at 20% means a $10 bill; the smaller percentage belongs to the larger bill. | |
| 2 | C | Functions | Evaluate the inner operation first: 1 star 2 equals -3, and then -3 star 3 has numerator zero. | |
| 3 | B | Linear Equations & Word Problems | Solve the equation without b to get x = -2, then substitute that x into the second equation and solve for b. | |
| 4 | B | Triangles: Area & Pythagorean | With width w and length 2w, the diagonal satisfies x^2 = 5w^2, and the area 2w^2 is two fifths of x^2. | |
| 5 | A | Linear Equations & Word Problems | Count free windows: 7 and 8 alone each earn one free window, but 15 together earns three, so pooling gains exactly one free window. | |
| 6 | B | Statistics & Data | Convert each average to a total: both groups sum to 600, so all 50 numbers sum to 1200 and average 24. | |
| 7 | B | Linear Equations & Word Problems | Distance is rate times time, so Josh rode 2 times 4/5 = 8/5 of Mike's distance; the two distances total 13 miles. | |
| 8 | C | Triangles: Area & Pythagorean | Each corner right triangle has legs 1 and s+1 (s the inner side) and hypotenuse root 50, so (s+1)^2 = 49. | |
| 9 | B | Basic Probability | Only the positions of the two O's matter: C(5,2) = 10 equally likely patterns, and exactly one of them is XOXOX. | |
| 10 | A | Quadratics | Combine the x-terms into (a+8)x; one solution means discriminant zero, so a + 8 = plus or minus 12. | |
| 11 | B | Solid Geometry | Red faces are the original surface, 6n^2 unit squares, out of 6n^3 unit-cube faces in all, so the fraction is exactly 1/n. | |
| 12 | B | Circles | The trefoil is four 60-degree sectors of radius 1 rearranged: each sector is one small equilateral triangle plus one circular segment. | |
| 13 | E | Exponents, Logarithms & Radicals | Take the 50th root of every part: the chain becomes 130n > n^2 > 16, that is, 4 < n < 130. | |
| 14 | E | Bases & Digits | The middle digit is determined by the outer two, which only need the same parity: 9 first digits times 5 matching last digits. | |
| 15 | E | Divisibility & Factors | Factor into primes, 2^8 3^4 5^2 7; a cube divisor uses exponents that are multiples of 3, giving 3 choices for 2 and 2 for 3. | |
| 16 | D | Bases & Digits | Subtracting the digit sum from 10a + b leaves 9a, independent of b; the only multiple of 9 up to 81 ending in 6 is 36, so a = 4. | |
| 17 | D | Sequences & Series | Every vertex lies on exactly two segments, so the five sums total 2(3+5+6+7+9) = 60, and the middle term of five in arithmetic progression is their average, 12. | |
| 18 | A | Conditional Probability & States | List every series where B wins game 2 and A still wins, weighting each by (1/2)^(games played); BBAAA is 1/32 out of a total 5/32. | |
| 19 | D | Triangles: Area & Pythagorean | The tilted square stops when its lower edges hit the neighbors' top corners, 1 inch apart; a 45-45-90 triangle puts its bottom vertex 1/2 below those corners. | |
| 20 | A | Quadrilaterals & Polygon Areas | All angles are 135 degrees, so extending the four unit sides makes a 2-by-2 square from which four isosceles right triangles with hypotenuse root 2 over 2 are clipped. | |
| 21 | B | Divisibility & Factors | Since 1 + ... + n = n(n+1)/2, the quotient 6n divided by it is 12/(n+1), so n + 1 must be a divisor of 12 greater than 1. | |
| 22 | D | GCD & LCM | Common elements are multiples of 12, and the binding constraint is the smaller set: S tops out at 8020, so count multiples of 12 up to 8020. | |
| 23 | C | Circles | O is the midpoint of DE, so [DCE] = 2[DCO], and triangles DCO and ABD share the height DC with bases CO and AB in ratio 1:6. | |
| 24 | B | Primes | P(m) = sqrt(m) forces m to be the square of a prime, so q^2 - p^2 = 48 with p, q prime; factor as (q-p)(q+p) with both factors even. | |
| 25 | D | Triangles: Area & Pythagorean | Triangles sharing angle A have area ratio (AD/AB)(AE/AC) = 19/75, so ADE is 19 parts of 75 and the quadrilateral is the other 56. |
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.