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.
The same key as a PDF 125 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | D | Ratios, Percents & Averages | The same 2 dollar tip is a smaller percent of a bigger bill. Work backward from each tip to each whole bill, then subtract. | |
| 2 | C | Functions | Do the inner operation first: 1 star 2 is -3. Then -3 star 3 has a top of 0 and a bottom of -6, so the answer is 0. | |
| 3 | B | Linear Equations & Word Problems | The first equation has no b, so solve it to get x = -2. Put that x into the second equation and solve for b. | |
| 4 | B | Triangles: Area & Pythagorean | With width w and length 2w, the Pythagorean theorem gives x squared = 5 w squared. The area 2 w squared is two fifths of that. | |
| 5 | A | Linear Equations & Word Problems | Each group of 5 windows gets one free. Alone, 7 and 8 each earn one free window; together, 15 earns three. Buying together gains one extra free window. | |
| 6 | B | Statistics & Data | Turn each average into a total: 20 times 30 is 600 and 30 times 20 is 600. All 50 numbers add to 1200, so the average is 24. | |
| 7 | B | Linear Equations & Word Problems | Josh rides twice as long at four fifths the speed, so he covers 8 fifths of Mike's distance. Mike's 5 parts plus Josh's 8 parts make 13 miles. | |
| 8 | C | Triangles: Area & Pythagorean | The corner triangle on side AB has legs 1 and (1 + inner side). The Pythagorean theorem with 50 gives the inner side 6. | |
| 9 | B | Basic Probability | Only the spots of the two O tiles matter. There are 10 ways to pick 2 spots out of 5, and XOXOX is exactly one of them. | |
| 10 | A | Quadratics | Combine ax and 8x into (a + 8)x. One solution means the discriminant is zero, so (a + 8) squared is 144 and a + 8 is 12 or -12. | |
| 11 | B | Solid Geometry | Red faces are the outside of the big cube: 6 n squared. All faces: 6 n cubed. The fraction red is 1 over n, so n = 4. | |
| 12 | B | Circles | Each bulge is a sector minus a small triangle, and the four small triangles fill the big one. So the trefoil is just four 60 degree sectors. | |
| 13 | E | Exponents, Logarithms & Radicals | Every exponent is a multiple of 50. Take the 50th root of each part to get 130n > n squared > 16, which means n is from 5 to 129. | |
| 14 | E | Bases & Digits | The middle digit is decided by the other two, and it is a whole number only when the first and last digits are both odd or both even. | |
| 15 | E | Divisibility & Factors | Write the product as 2^8 times 3^4 times 5^2 times 7. A cube divisor uses each prime a multiple of 3 times, giving 3 times 2 choices. | |
| 16 | D | Bases & Digits | A two-digit number minus its digit sum is 9 times the tens digit. Only 36 ends in 6, so the tens digit is 4 and the units digit is free. | |
| 17 | D | Sequences & Series | Each letter is on exactly two segments, so the five sums add to twice 30, which is 60. The middle of five evenly spaced numbers is their average, 12. | |
| 18 | A | Conditional Probability & States | List every way the series can go when B wins game 2 and A wins the series. Weight shorter series more, since each game is a coin flip. | |
| 19 | D | Triangles: Area & Pythagorean | The tilted square rests on its neighbors' inner top corners, at height 1. Its bottom tip is half an inch lower, and B is one diagonal above the tip. | |
| 20 | A | Quadrilaterals & Polygon Areas | Every angle is 135 degrees, so the octagon is a 2 by 2 square with four small corners cut off. Each corner is a right triangle with legs one half. | |
| 21 | B | Divisibility & Factors | The sum is n(n+1)/2. Dividing 6n by it leaves 12 over (n+1), so n+1 must be a divisor of 12 bigger than 1. | |
| 22 | D | GCD & LCM | Common numbers are multiples of 12. S stops at 4 times 2005 = 8020, which is lower than where T stops, so count the multiples of 12 up to 8020. | |
| 23 | C | Circles | O is the midpoint of DE, so triangle DCE is twice triangle DCO. Triangles DCO and ABD share the height DC, so compare their bases CO and AB. | |
| 24 | B | Primes | The largest prime factor equals the square root only for a prime squared. So look for two prime squares 48 apart: 121 and 169. | |
| 25 | D | Triangles: Area & Pythagorean | Triangles ADE and ABC share the angle at A, so their areas compare as 19 times 14 against 25 times 42, which is 19 to 75. The rest is 56. |
2005 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.