AMC 10 Step by Step

2002 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.

#AnswerTopicDifficultyKey insight
1DExponents, Logarithms & RadicalsPull the smallest power of 10 out of the top and bottom; the huge exponents cancel and only 101/20 remains.
2CFunctionsSubstitute a=2, b=12, c=9 into the definition and add the three fractions over the common denominator 6.
3BExponents, Logarithms & RadicalsThere are only five ways to parenthesize four 2s; since 2^(2^2) = (2^2)^2 = 16, every grouping evaluates to either 2^16 or 256.
4EDiophantine EquationsTaking n = 1 turns the inequality into m <= m + 1, which is true for every positive integer m.
5CCirclesA diameter of the big circle passes through three unit circles in a row, so its radius is 3; subtract seven unit-circle areas from 9 pi.
6ALinear Equations & Word ProblemsUndo Cindy's steps in reverse (multiply 43 by 3, add 9) to recover the original number 138, then apply the correct steps.
7ACirclesEqual arc lengths mean 45 r_A = 30 r_B, so the radii are in ratio 2:3 and the areas in ratio 4:9.
8AQuadrilaterals & Polygon AreasCut the flag into a 4-by-4 grid of equal squares; the 12 border squares are each exactly half white and half blue, so B = W.
9BSystems of EquationsAdd the two equations: the A terms combine to 1001A, so 1001(A+B+C) = 9009 and the average follows without finding A, B, C.
10AQuadraticsBoth terms share the factor (2x+3); factor it out instead of expanding, and the two roots appear immediately.
11BSets, Estimation & MiscellaneousOnly 0.7+0.7, 0.8+0.4, 0.7+0.4 and 0.4+0.4+0.4 fit on a disk; the 15 large files force at least 9 disks with little room for the 15 small ones.
12BLinear Equations & Word ProblemsThe two trips differ by 6 minutes, so d/40 - d/60 = 1/10 hour gives d = 12 miles and the on-time trip takes 15 minutes.
13BTriangles: Area & Pythagorean15-20-25 is a right triangle, so the area is 150, and the shortest altitude is the one to the longest side: 2*150/25 = 12.
14BPrimesThe roots add to the odd number 63, so one prime root must be the only even prime, 2, forcing the other to be 61.
15EPrimesTwo-digit primes cannot end in 2, 4, 5 or 6, so those are the tens digits and 1, 3, 7, 9 the units digits; the sum is forced.
16BSystems of EquationsName the common value x; then a, b, c, d are x-1, x-2, x-3, x-4, and their sum 4x-10 must also equal x-5.
17DRatios, Percents & AveragesTrack ounces of coffee and cream separately: after the pour-back, cup 1 holds 3 oz coffee and 2 oz cream, so cream is 2/5 of 5 oz.
18DSolid GeometryClassify dice by visible faces: 8 corners show 3 mutually adjacent faces (min 1+2+3), 12 edges show 2 (min 1+2), 6 face-centers show 1; total 8*6+12*3+6*1 = 90.
19ECirclesThe reachable region is a 240-degree sector of radius 2 plus, after the rope wraps around each neighboring vertex, two 60-degree sectors of radius 1.
20DSimilar & Congruent TrianglesBoth HC and JE are parallel to AG, so triangles DCH and FEJ are scaled copies of DAG and FAG with ratios 1/3 and 1/5; divide.
21DStatistics & DataMean fixes the sum at 64, range ties the minimum to max minus 8, and unique mode forces enough 8s to outnumber the repeated low values.
22CNumber PropertiesFrom n^2 tiles, two operations remove n and then n-1 tiles, landing exactly on (n-1)^2; so going from 10^2 down to 1^2 takes 2*9 = 18 steps.
23DTriangles: Area & PythagoreanThe altitude from E to BC has length 8, so AE = sqrt((x+6)^2 + 64), and the perimeter condition AE = 26 - x squares to a linear equation.
24ABasic ProbabilityTina has only 10 possible pairs; tabulate each pair's sum s, and Sergio wins in exactly 10 - s ways, so the answer is a short sum divided by 100.
25CQuadrilaterals & Polygon AreasSlide the two legs together: the legs 5 and 12 with base 52 - 39 = 13 form a 5-12-13 right triangle, whose altitude 60/13 is the trapezoid's height.

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.