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.
| # | Answer | Topic | Difficulty | Key insight |
|---|---|---|---|---|
| 1 | B | Ratios, Percents & Averages | Dividing 350 by 20 gives 17.5, and half a mug is not a properly brewed mug, so the count is the whole part, 17. | |
| 2 | D | Modular Arithmetic | The ones digit of n^2 depends only on the ones digit of n, so the list repeats with period 10 and each block of ten sums to 45. | |
| 3 | D | Sequences & Series | Building a row copies the two ends and uses every other entry twice, so each row sum doubles: from the second row on, the sums are 11 * 2^(n-2). | |
| 4 | A | Bases & Digits | Place value turns the condition into 7a + b = 9b + a, so 3a = 4b, and a must be a base-seven digit. | |
| 5 | C | Circles | Angle B is obtuse, so it subtends the major arc: the central angle AOC is 360 - 260 = 100 degrees, and OA = OC gives 40. | |
| 6 | C | Coordinate Geometry | The line meets both vertical edges, so the lower region is a trapezoid of area 8/3, and the piece left of x = a has area a + a^2/6. | |
| 7 | C | Coordinate Geometry | Put the locked gate at the origin: the western route is 17 + (x + 8) by the 8-15-17 triple, and the eastern route is sqrt(15^2 + (x+9)^2) + 9. | |
| 8 | C | Divisibility & Factors | The total in cents is ABBBA and must be divisible by 36 = 4 * 9; divisibility by 4 forces A even and by 9 forces 3 | A. | |
| 9 | C | Basic Counting | The last three inequalities force the sum S to satisfy S <= 2 + 2min(x, y, z), which with S >= 2 traps every coordinate in {0, 1, 2}. | |
| 10 | A | Polynomials | f and g differ by (n-4)(n+1), which is a factor of both, so the quotient collapses to (n-2)/(n-3) = 1 + 1/(n-3). | |
| 11 | B | Basic Probability | Relabelling the tutors by Monday's assignment turns 'met the same tutor twice' into 'fixed point of a uniformly random permutation of the six students'. | |
| 12 | C | Circles | The hexagon cuts into six equilateral triangles and the 60-degree rhombus into two, and in each figure the disks are exactly the incircles of those triangles, so H = R. | |
| 13 | A | Triangle Centers & Cevians | The altitude foot and the intersection point both lie on one ray from the right-angle vertex, so the 4:3 split is read straight off their first coordinates. | |
| 14 | C | Basic Probability | The captains are the three tallest exactly when the three tallest land in three different groups, so only the bands of the top three matter. | |
| 15 | D | Sequences & Series | k^3+6k^2+8k = k(k+2)(k+4), and 1/(k(k+2)(k+4)) = (1/4)[1/(k(k+2)) - 1/((k+2)(k+4))], a telescope of step 2. | |
| 16 | D | Arrangements with Restrictions | A colouring is a pairing of the six sectors into three non-neighbouring pairs together with a naming of the colours, and there are exactly 4 such pairings. | |
| 17 | B | Sequences & Series | The averages form the arithmetic sequence A_k = 2028 - k, so S_k = k(2028 - k) and every term from the fourth on is x_k = 2029 - 2k. | |
| 18 | D | Sequences & Series | The value n is taken by exactly 2n+1 consecutive terms, so the sum is sum n(2n+1) for n up to 44 plus the single leftover term 45. | |
| 19 | D | Solid Geometry | Extend the four trapezoids to their common apex: the container is part of a square pyramid, where volume below a cross-section of side s scales as s^3. | |
| 20 | A | Circles | The four semicircle centres form a square of side 2r about the centre, so the centre lies r*sqrt(2) from each and the small radius is (sqrt(2)-1)r. | |
| 21 | C | Arrangements with Restrictions | Writing the colours as 0, 1, 2 turns all three rules into 'every square has a neighbour one greater mod 3', which forces each corner's value. | |
| 22 | A | Conditional Probability & States | A digit sum of 61 is only 2 below the maximum 63, so the number is seven 9s short by 2, and the alternating sum is 7, 9 or 11. | |
| 23 | C | Diophantine Equations | Both fillings are linear in the row and column indices, so setting 91(r-1)+c equal to 141(c-1)+r reduces to 9r = 14c - 5 and hence c = 1 mod 9. | |
| 24 | E | Basic Probability | Let p_n be the chance of reaching 4 from n; the first hop gives p_n = (p_{n+1} + p_{n-1})/4, and writing all in terms of p_1 collapses it. | |
| 25 | B | Transformations & Symmetry | Reflect the square rather than the ray: the path unfolds to a straight line of slope 18/25, dying at the first image point, whose parities name the corner. |
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