AMC 10 Step by Step

2008 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
1DClocks, Calendars & TimeOne third of the job took 2 hours 40 minutes, so the whole job takes three times that, 8 hours, starting from 8:30 AM.
2ARatios, Percents & AveragesSet the square's side to 1; the rectangle is then 2 by 4, so the square covers 1/8 of it.
3ADivisibility & FactorsThe proper divisors of 6 are 1, 2, 3, which add to 6, so applying the operation any number of times returns 6.
4CRatios, Percents & AveragesTwenty thirds of a banana equal 8 oranges, so one banana is worth 6/5 of an orange; five halves of a banana is then 3 oranges.
5BAlgebraic ManipulationEach numerator is the next fraction's denominator, so everything cancels except the last numerator over the first denominator: 2008/4.
6DRatios, Percents & AveragesAverage speed is total distance over total time; with each leg 60 km the times are 20, 3 and 6 hours, giving 180/29, about 6.2.
7EExponents, Logarithms & RadicalsNumerator and denominator have the same shape; the numerator is the denominator with every exponent raised by 2, i.e. multiplied by 3^2 = 9.
8ARatios, Percents & AveragesWrite both prices in terms of the sticker price p: store A charges 0.85p minus 90, store B charges 0.75p, and A is 15 dollars cheaper.
9BDivisibility & FactorsCombine the fractions: 2x/3 - x/6 = x/2, so x is an even integer and nothing more; x = 2 rules out any multiple-of-3 claim.
10EQuadrilaterals & Polygon AreasJoining the midpoints of a square cuts off four corner triangles that together make half the area, so each step halves the area: 16, 8, 4.
11DLinear Equations & Word ProblemsThe trip takes 15 minutes, so the net inflow may be at most 30/15 = 2 gallons per minute, meaning LeRoy bails at least 10 - 2 = 8.
12CRatios, Percents & AveragesRed is 1.25 times blue, so blue is r/1.25 = 0.8r; green is 1.6r; the total is (0.8 + 1 + 1.6)r = 3.4r.
13DLinear Equations & Word ProblemsTheir combined rate is 1/5 + 1/7 rooms per hour, and they paint for only t - 1 of the t hours because lunch takes one hour.
14DTriangles: Area & PythagoreanA 4:3 screen with diagonal 27 is a scaled 3-4-5 triangle, 21.6 by 16.2; a 2:1 movie at full width is 10.8 tall, leaving 5.4 to split.
15DLinear Equations & Word ProblemsExpanding Han's distance gives v + 5t = 65 for Ian's speed v and time t; Jan's excess 2v + 10t + 20 is exactly twice that plus 20.
16BCirclesThe small center lies on the 30-degree bisector, so its distance from O is twice its radius r; internal tangency gives 2r + r = R, hence r = R/3.
17BCirclesThe band around the triangle is three 6-by-3 rectangles plus three 120-degree sectors of radius 3 at the corners, and the sectors together form one full circle.
18BTriangles: Area & PythagoreanUse (a+b)^2 = a^2 + b^2 + 2ab = c^2 + 4(area); with a + b = 32 - c this becomes a linear equation in c.
19CTransformations & SymmetryEach rotation moves P along a quarter circle whose radius is P's distance from the center: first the diagonal 2 sqrt 10, then the side PS = 6.
20DSimilar & Congruent TrianglesTriangles AKB and CKD are similar with ratio 3:4, so the diagonals are split 3:4; the four triangles at K then have areas 18, 24, 24, 32 by shared altitudes.
21ASolid GeometryABCD is a rhombus: all four sides are half a face diagonal's worth, sqrt(5)/2, and its diagonals are the space diagonal sqrt(3) and the midpoint-to-midpoint segment sqrt(2).
22DBasic ProbabilityOnly three flips separate the first and fourth terms, so trace all eight equally likely paths, tracking actual values; five of the eight end at an integer.
23BBasic CountingEach element is in both subsets, only the first, or only the second: choose the 2 shared, 2^3 for the rest, then halve for unordered pairs.
24DModular Arithmetick ends in 4 + 6 = 0, so k^2 ends in 0; 2^k depends on k mod 4, and both summands of k are multiples of 4, giving 6.
25CCirclesThe six inner corners form a regular hexagon with side x, so the inner edge is x sqrt(3)/2 from the center; add 1 and apply Pythagoras.

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.