2001 AMC 10
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 | E | Statistics & Data | The list is already sorted, so the median is the fifth entry n+6; the mean is n plus the average of the offsets, 63/9 = 7. | |
| 2 | C | Linear Equations & Word Problems | The reciprocal 1/x times the additive inverse -x is always -1, so the number is simply -1 + 2 = 1. | |
| 3 | E | Linear Equations & Word Problems | Adding 3 to each of two numbers adds 6 to their sum, and doubling each number doubles the sum, giving 2(S + 6) = 2S + 12. | |
| 4 | E | Circles | A line meets a circle at most twice, so three sides give at most six points, and a triangle slightly larger than the circle achieves all six. | |
| 5 | D | Transformations & Symmetry | Check each pentomino for a mirror axis, including diagonal ones: I, T, U, V, W and X have one; F, L, N, P, Y and Z do not. | |
| 6 | E | Bases & Digits | Write N = 10a + b; the condition becomes 10a + b = ab + a + b, and cancelling b and dividing by a leaves b = 9. | |
| 7 | C | Fractions & Decimals | Moving the decimal point four places right multiplies by 10^4, so 10^4 x = 4/x gives x^2 = 4 * 10^(-4) and x = 2 * 10^(-2). | |
| 8 | B | GCD & LCM | They coincide again after a number of days divisible by 3, 4, 6 and 7, so the answer is lcm(3,4,6,7) = 84. | |
| 9 | B | Ratios, Percents & Averages | Split the tax as p% of all income plus an extra 2% on the amount over 28000; then the p% terms cancel and 2%(I - 28000) = 0.25% I. | |
| 10 | D | Systems of Equations | Dividing two of the equations cancels a variable: xz/xy = 2 gives z = 2y, and then yz = 2y^2 = 72 pins down y = 6. | |
| 11 | C | Sequences & Series | The nth ring is a (2n+1)-square minus a (2n-1)-square, and (2n+1)^2 - (2n-1)^2 = 8n, so the 100th ring has 800 squares. | |
| 12 | D | Divisibility & Factors | Three consecutive integers always supply a factor of 2 and a factor of 3, but not necessarily 4; test one example like 5*6*7. | |
| 13 | E | Logic Puzzles | All ten digits are used, so ABC is whatever the odd run GHIJ and even run DEF leave behind; only 9753 and 642 leave digits summing to 9. | |
| 14 | A | Diophantine Equations | Doubling the revenue equation gives p(f + 140) = 4002 = 2*3*23*29, and the only divisor between 140 and 280 is 174, so f = 34 and p = 23. | |
| 15 | C | Quadrilaterals & Polygon Areas | The crosswalk is a parallelogram: its area is 15 times 40 using the curbs as bases and 50 times d using the stripes, so d = 12. | |
| 16 | D | Statistics & Data | Write the least and greatest as m - 10 and m + 15 in terms of the mean m; with median 5 the sum 2m + 10 must equal 3m. | |
| 17 | C | Solid Geometry | The sector's radius becomes the slant height and its arc becomes the base circumference: (252/360)(20 pi) = 14 pi, so the base radius is 7. | |
| 18 | D | Quadrilaterals & Polygon Areas | One repeating 3 by 3 block has area 9, of which the four corner unit squares take 4, so the four pentagons cover 5/9, about 55.6 percent. | |
| 19 | D | Distributions & Stars and Bars | Choosing 4 donuts of 3 types is the number of nonnegative solutions to g + c + p = 4, which stars and bars counts as C(6,2). | |
| 20 | B | Angles & Polygons | One side of the square holds two triangle legs plus one octagon side s; each leg is s/sqrt(2), so s(1 + sqrt 2) = 2000. | |
| 21 | B | Solid Geometry | Slice through the axis: the cone becomes a triangle, the cylinder a 2r-by-2r square, and the small triangle above it is similar to the whole. | |
| 22 | D | Systems of Equations | Compare lines that share an unknown: left column versus main diagonal gives x = 22, top row versus left column gives w = 19 and the magic sum 66. | |
| 23 | D | Basic Probability | Pretend all five chips are drawn; the last chip actually drawn is white exactly when the very last chip of the full sequence is red, which has probability 3/5. | |
| 24 | B | Triangles: Area & Pythagorean | Drop the perpendicular from B to CD: a right triangle with legs 7 and CD - AB and hypotenuse AB + CD, so 4(AB)(CD) = 49. | |
| 25 | B | Inclusion-Exclusion | Count multiples of 3 or 4 by inclusion-exclusion, then remove only those that are also multiples of 5 by a second inclusion-exclusion with 15, 20 and 60. |
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.