2007 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 | E | Solid Geometry | Each room's four walls form a band of perimeter 44 feet and height 8 feet; subtract the 60 square feet of openings, then triple. | |
| 2 | E | Functions | Both products share the factor a + b = 8, so the difference is 8 times (5 - 3) = 16. | |
| 3 | B | Ratios, Percents & Averages | Average mileage is total miles over total gallons: 240 miles on 4 + 6 = 10 gallons gives 24, not the naive average 25. | |
| 4 | D | Circles | The three central angles sum to 360, so angle AOC = 100, and the inscribed angle ABC on the same arc is half of it. | |
| 5 | D | Logic Puzzles | The facts chain into nested sets Crups inside Dramps inside Arogs inside Brafs, so anything about Crups being Arogs or Brafs is forced. | |
| 6 | D | Linear Equations & Word Problems | The three blanks bank 4.5 points, so she needs 6c >= 95.5, and the smallest whole number of correct answers is 16. | |
| 7 | E | Angles & Polygons | Drawing CE splits the pentagon into square ABCE and equilateral triangle CDE, so angle E is 90 + 60. | |
| 8 | D | Basic Counting | b is determined by a and c, and b is an integer exactly when a and c have the same parity, so count same-parity pairs a < c. | |
| 9 | D | Modular Arithmetic | The final s is the 12th s in the message, so it shifts by 1 + 2 + ... + 12 = 78, a multiple of 26, and stays s. | |
| 10 | A | Triangles: Area & Pythagorean | Fixed base BC and fixed area force a fixed height, and the points at a fixed distance from line BC form two parallel lines. | |
| 11 | C | Circles | The circumcenter lies on the altitude to the base; equating its distances to the apex and to a base vertex gives R = 9 sqrt(2)/8. | |
| 12 | D | Linear Equations & Word Problems | N years ago Tom was T - N and the three children together were T - 3N; setting T - N = 2(T - 3N) gives T = 5N. | |
| 13 | D | Circles | The circles meet at (0,0) and (2,2); that chord subtends a right angle at each center, so the lens is two quarter-circle segments. | |
| 14 | C | Ratios, Percents & Averages | The group size never changes, so the drop from 40% to 30% of the same total equals the two girls who left: 10% of the group is 2. | |
| 15 | D | Angles & Polygons | Write every angle as a fraction of angle A; the angle sum gives A(1 + 1/2 + 1/3 + 1/4) = 360, so A = 172.8. | |
| 16 | C | Ratios, Percents & Averages | The class average is a weighted average: 0.1 j + 0.9(83) = 84, so the juniors' common score is j = 93. | |
| 17 | D | Triangles: Area & Pythagorean | Joining P to the vertices splits the triangle into three pieces whose areas sum to the whole, forcing the altitude to equal 1 + 2 + 3 = 6. | |
| 18 | B | Circles | The four outer centers form a square of side 2r whose half-diagonal, r sqrt(2), is the center-to-center distance 1 + r. | |
| 19 | C | Basic Probability | Whatever column is chosen, the shaded rows in it are either {1,3} or {2,4}, and each of those row-sets has probability exactly 1/2. | |
| 20 | C | Basic Counting | Pick the 3 rows and 3 columns (10 ways each), then match rows to columns in 3! ways: 10 x 10 x 6 = 600. | |
| 21 | B | Similar & Congruent Triangles | The small triangle above the square is similar to ABC with altitude h - s in place of h = 12/5, so s/5 = (h - s)/h. | |
| 22 | B | Expected Value | Each die shows the chosen number with probability 1/4, so the outcomes have probabilities 9/16, 6/16, 1/16 and the payoff sum is (-9 + 6 + 2)/16. | |
| 23 | E | Solid Geometry | The top pyramid is a scaled copy; half the surface area means scale factor 1/sqrt 2, so its height is H/sqrt 2 and H - H/sqrt 2 = 2. | |
| 24 | C | Bases & Digits | Divisibility by 4 forces the ending 44, and digit sum divisible by 9 forces the number of 4's to be a multiple of 9, so n = 4444444944. | |
| 25 | A | Diophantine Equations | Clearing denominators to 9ab | 9a^2 + 14b^2 and using gcd(a,b) = 1 forces a | 14 and b | 9, and the factor 9 forces 3 | b. |
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.