AMC 10 Step by Step

2011 AMC 10B

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
1CFractions & DecimalsThe two sums are 12 and 9, so the expression is 4/3 - 3/4, a single subtraction over the common denominator 12.
2ERatios, Percents & AveragesThe current average is 77, so six tests must total at least 6 times 80 = 480; the current total is 385, leaving 95.
3AAbsolute Value & InequalitiesBoth dimensions are positive, so the smallest area comes from the smallest possible length and width together: 1.5 times 2.5.
4CLinear Equations & Word ProblemsEach person's fair share is half the total, (A+B)/2, so LeRoy must hand over the gap between that share and what he already paid.
5EDivisibility & Factors161 = 7 x 23 has only one two-digit factor, 23, so the reversed number was 23, the true a is 32, and b = 7.
6ALinear Equations & Word ProblemsUndo the story from the end: 8 plus 4 is 12, two thirds of 18; 18 plus 2 is 20, two thirds of 30.
7BAngles & PolygonsThe two angles total 108 degrees, so they are 39 and 69, and the third angle 180 - 108 = 72 is the largest.
8BLogic PuzzlesTake the contrapositive: not crowded means the hypothesis failed, and the negation of 'hot AND sunny' is 'not hot OR not sunny'.
9DSimilar & Congruent TrianglesTriangle EBD shares angle B with ABC and is right-angled, so it is similar; area ratio 1/3 means side ratio 1/sqrt(3), and BD corresponds to BC = 4.
10BSequences & SeriesThe other ten elements sum to 1111111111, just over 10^10 / 9, so the ratio is a hair under 9.
11DSets, Estimation & MiscellaneousTwelve months can hold at most 4 people each with 48 people total, so 52 people force some month to have at least 5, and 5 is sharp.
12ACirclesThe straight sides are the same length on both edges; only the two semicircles differ, and their radii differ by 6, adding exactly 12 pi meters.
13DGeometric ProbabilityA positive product needs both numbers negative or both positive; the chances of negative and positive are 2/3 and 1/3, so add (2/3)^2 and (1/3)^2.
14CAlgebraic ManipulationDiagonal gives l^2 + w^2 = 625, area gives lw = 168, so (l + w)^2 = 625 + 336 = 961 and l + w = 31.
15EFunctionsExpand each side by the definition: I fails because its right side carries an extra x/2, while II and III both simplify to matching expressions.
16AGeometric ProbabilityWith octagon side s, the center square also has side s, and the octagon is a square of side s(1 + sqrt 2) minus four corner triangles.
17CCirclesAngle EAB is 90 (inscribed in a semicircle), so angle ABE is 50; parallels carry it to angle BED, and cyclic quadrilateral EDCB gives angle BCD = 130.
18ETriangles: Area & PythagoreanSince AB is parallel to DC, triangle MDC is isosceles with CM = CD = 6; with CB = 3, angle CMB = 30, leaving 150 to split evenly.
19AAbsolute Value & InequalitiesSquare both sides and set u = |x|: u^2 - 5u - 24 = 0 forces |x| = 8, so the roots 8 and -8 multiply to -64.
20CQuadrilaterals & Polygon AreasDiagonal BD splits the rhombus into two equilateral triangles of side 2; in each, the perpendicular bisectors from B carve out exactly one third of the triangle around B.
21BNumber PropertiesThe largest difference 9 is w - z, so the three consecutive gaps are three listed differences summing to 9; only {1, 3, 5} works, in two orders.
22ASolid GeometryPyramid height is sqrt2/2, so the cross-section at height s is a square of side 1 - s*sqrt2; equate it to the cube's top face: s = sqrt2 - 1.
23DModular ArithmeticOnly the last three digits matter, and 2011 is 11 = 10 + 1 mod 1000; binomial expansion of (10 + 1)^2011 leaves just three terms mod 1000.
24BCoordinate GeometryA lattice hit at integer x means m = k/x; so a is the smallest fraction above 1/2 with denominator at most 100, which is 50/99.
25DCirclesTangent lengths are s - a, so sides x-1, x, x+1 become x/2-1, x/2, x/2+1; the middle side halves until it drops below 2.

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.