2025 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.
The same key as a PDF 135 KB — a separate file from the paper, so one can be handed over without the other.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | B | Ratios, Percents & Averages | Each mug needs a full 20 grams. Seventeen mugs use 340 grams, and the 10 grams left over cannot make another proper mug. | |
| 2 | D | Modular Arithmetic | The last digit of a square depends only on the last digit of the number, so the list repeats every ten terms, and each block of ten adds to 45. | |
| 3 | D | Sequences & Series | From the second row on, each row's total is exactly double the one before, so the totals go 11, 22, 44, … and the 11th row totals 5632. | |
| 4 | A | Bases & Digits | Write both numbers in ordinary place value: 7a + b = 9b + a. This simplifies to 3a = 4b, and a digit in base seven is at most 6. | |
| 5 | C | Circles | Angle B is 130 degrees, so the far arc AC is 260 and the near arc is 100. Triangle AOC is isosceles with a 100-degree angle at O. | |
| 6 | C | Coordinate Geometry | The line crosses both side edges, so the lower region is a trapezoid of area 8/3. The piece left of x = a must have area 4/3. | |
| 7 | C | Coordinate Geometry | Measure from the locked gate. The west route is 17 + (x + 8), using the 8-15-17 right triangle; the east route is √(15² + (x+9)²) + 9. | |
| 8 | C | Divisibility & Factors | The total in cents, ABBBA, is a multiple of 36. The test for 4 makes A even; the test for 9 makes A a multiple of 3. | |
| 9 | C | Basic Counting | Add the inequalities in pairs to trap x, y and z between 0 and 2. Then list the few triples, sorted by their sum. | |
| 10 | A | Polynomials | Subtracting gives f − g = (n−4)(n+1), a factor of both. The fraction becomes (n−2)/(n−3), but n = 4 is out because g(4) = 0. | |
| 11 | B | Basic Probability | Only Tuesday matters: compare it with Monday. Choose the 2 students who repeat, then the other 4 must all get a new tutor, which happens in 9 ways. | |
| 12 | C | Circles | Cut the hexagon into six equilateral triangles and the rhombus into two: each disk is the circle inside one triangle, so H = R. | |
| 13 | A | Triangle Centers & Cevians | Put the right angle at the origin. The crossing point and the altitude's foot lie on one line from there; compare how far across each is. | |
| 14 | C | Basic Probability | The three tallest are the captains exactly when they land in three different groups. Only their three wristbands matter: 6/8 × 3/7. | |
| 15 | D | Sequences & Series | The bottom factors as k(k+2)(k+4). Each term is one quarter of the difference 1/(k(k+2)) − 1/((k+2)(k+4)), so the sum cancels down to its first two pieces. | |
| 16 | D | Arrangements with Restrictions | Each color covers two non-neighboring sectors. Split the sectors into three such pairs (4 ways), then give the pairs the three colors (6 ways). | |
| 17 | B | Sequences & Series | The average of the first k terms is 2028 − k, so their total is k(2028 − k). Subtracting totals shows later terms drop by 2. | |
| 18 | D | Sequences & Series | The value n appears 2n + 1 times, from n² to just before (n+1)². So add n(2n + 1) for n = 1 to 44, plus 45. | |
| 19 | D | Solid Geometry | Extend the slanted sides to a point: the container is a pyramid with its tip cut off, and volume below a square of side s grows like s³. | |
| 20 | A | Circles | The four semicircle centers form a square of side 2r around the center, so each is r√2 from the center, and the small circle's radius is r√2 − r. | |
| 21 | C | Arrangements with Restrictions | Renaming colors red to blue, blue to yellow, yellow to red keeps every rule, so count only colorings with a red center and multiply by 3. | |
| 22 | A | Conditional Probability & States | A digit sum of 61 is 2 short of seven 9s, leaving only 28 numbers. The test for 11 passes only when the whole shortfall is in even places. | |
| 23 | C | Diophantine Equations | Horace writes 91(r−1) + c and Vera writes 141(c−1) + r. Setting them equal gives 9r = 14c − 5, solved by c = 1, 10, 19, … | |
| 24 | E | Basic Probability | Let p1, p2, p3 be the chances of reaching 4 from 1, 2, 3. Each is a quarter of the chances one step down and one step up. | |
| 25 | B | Transformations & Symmetry | Flip the square instead of bouncing the path. The path becomes one straight line through a grid of squares, ending at the first grid corner it meets. |
2025 AMC 10B. Problems © Mathematical Association of America, American Mathematics Competitions (maa.org/amc), licensed CC BY-NC-SA 4.0. Typeset and annotated by AMC 10 Step by Step (amcstep.com) and shared under the same licence. Not endorsed by the MAA.