2025 AMC 10B Problem 19

Attempt Problem 19 of the 2025 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2025 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

A container has a 1×11 \times 1 square bottom, a 3×33 \times 3 open square top, and four congruent trapezoidal sides, as shown. Starting when the container is empty, a hose that runs water at a constant rate takes 3535 minutes to fill the container up to the midline of the trapezoids.
How many more minutes will it take to fill the remainder of the container?
7070
8585
9090
9595
105105
Answer: D
Concepts:power scaling of length, area, and volumevolumesimilarity
Small hint:
The container is a piece of a pyramid; the volume from the apex up to side length ww scales as w3w^3
Big hint:
Volumes split as 23−13=72^3 - 1^3 = 7 (up to the midline, side 22) and 33−23=193^3 - 2^3 = 19 (remainder); scale the 3535 minutes
Solution:
The container is a square frustum: a horizontal slice at height fraction tt has side 1+2t.1 + 2t. Extend the sides up to their apex, and the volume out to where the side length is ww scales as w3.w^3. So the whole container is 33−13=263^3 - 1^3 = 26 parts, the piece up to the midline (side 22) is 23−13=72^3 - 1^3 = 7 parts, and the rest is 33−23=193^3 - 2^3 = 19 parts. Those 77 parts take 3535 minutes, so each part is 55 minutes. The remaining 1919 parts take 9595 minutes. Thus, D is the correct answer.
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