2009 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.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | B | Linear Equations & Word Problems | Five muffins cost $2.50, and each swap to a bagel adds 25 cents, so the total is a whole number of dollars only after two swaps. | |
| 2 | C | Fractions & Decimals | Both differences of unit fractions collapse to single unit fractions, 1/12 over 1/6, whose quotient is 1/2. | |
| 3 | C | Ratios, Percents & Averages | Losing three cans cost her five rooms, so each can paints 5/3 rooms and 25 rooms need 25 divided by 5/3 = 15 cans. | |
| 4 | C | Quadrilaterals & Polygon Areas | The two triangle bases fill the 25 - 15 = 10 meters left over on the long side, so each leg is 5, which is also the yard's width. | |
| 5 | D | Ratios, Percents & Averages | Twenty percent less than 60 is 48, and 'one-third more than n' means 4n/3, so n = 36. | |
| 6 | D | Divisibility & Factors | The product is k times t squared where t is the twins' age and k < t; since 128 = 2^7, only t = 8 with k = 2 works. | |
| 7 | C | Basic Counting | Only the single + can be regrouped, so the distinct results come from which of the two products absorb the addition: 26, 46, 50, 70. | |
| 8 | B | Ratios, Percents & Averages | Percent changes multiply: 1.2 times 0.8 times 1.25 = 1.2, so April must multiply by 5/6, a drop of one-sixth, about 17 percent. | |
| 9 | A | Angles & Polygons | Triangle ABC is isosceles with apex B, so its angles are 5x/2, 5x/2, x giving x = 30; vertical angles then pass 75 degrees to isosceles triangle CDE. | |
| 10 | E | Triangles: Area & Pythagorean | The broken piece is the hypotenuse 5 - x of a right triangle with legs x and 1, and the x squared terms cancel to leave a linear equation. | |
| 11 | A | Basic Counting | The middle digit must be the odd-count 5, and the first three digits are one 2, one 3, one 5 in some order: 3! palindromes. | |
| 12 | A | Triangles: Area & Pythagorean | Every triangle has its base on one line and apex on the other, so its height is the fixed gap and only the base length, 1, 2 or 3, matters. | |
| 13 | C | Modular Arithmetic | Rolling lays the sides along the axis in order AB, BC, CD, DE, EA with period 23; 2009 = 23*87 + 8 falls in CD's span 7 to 13. | |
| 14 | D | Sequences & Series | Non-millet seeds are always exactly 3/4 quart after a refill, while millet obeys m -> 3m/4 + 1/4, so the day it first exceeds 3/4 quart is day 5. | |
| 15 | E | Linear Equations & Word Problems | Subtracting the two weighings isolates one-sixth of the water, so full water is 6(a - b), and the empty bucket is b minus half of that. | |
| 16 | B | Circles | BO bisects the 60-degree angle at B and OA is perpendicular to the tangent, so triangle OAB is 30-60-90: BO = 2r and BD = r. | |
| 17 | C | Coordinate Geometry | The shaded piece is the big right triangle with vertices (c,0), (3,0), (3,3) minus the missing unit square in its corner, so 3(3 - c)/2 - 1 = 5/2. | |
| 18 | D | Similar & Congruent Triangles | Right triangle AME shares angle A with right triangle ABC, so it is a 3-4-5 triangle scaled to have leg AM = 5, giving ME = 15/4. | |
| 19 | A | Basic Counting | The hour and minute are independent, so multiply the fraction of hours without a 1 (8 of 12) by the fraction of minutes without a 1 (45 of 60). | |
| 20 | B | Triangle Centers & Cevians | The angle bisector theorem gives BD : DC = AB : AC = 1 : sqrt5, so BD = 2/(1 + sqrt5), which rationalizes to (sqrt5 - 1)/2. | |
| 21 | D | Modular Arithmetic | Powers of 3 alternate 1, 3, 1, 3 mod 8, so consecutive pairs sum to 4 and the 1005 pairs contribute 1005 times 4, which is 4 mod 8. | |
| 22 | B | Solid Geometry | Vertical cuts make the piece a prism, so everything reduces to triangle B's area 4/5; volume is twice that and icing is that plus the full 2-by-2 side face. | |
| 23 | C | Geometric Probability | Each runner is visible in a time window centered on passing the start: Rachel 618.75 to 641.25 s, Robert 630 to 650 s; overlap 11.25 of 60. | |
| 24 | A | Angles & Polygons | Extending all legs to the arch's center splits 180 degrees into nine 20-degree wedges, so each trapezoid's outer angles are 80 and inner angles 100. | |
| 25 | B | Basic Probability | An encircling stripe is a belt of four faces around one of three axes, each forced; the other two faces are free, and belts cannot coexist: 12/64. |
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.