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2025 AMC 10B Problem 19

Problem 19 of 25HarderGeometry

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?

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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 3313=263^3 - 1^3 = 26 parts, the piece up to the midline (side 22) is 2313=72^3 - 1^3 = 7 parts, and the rest is 3323=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.

More practice

Concepts: power scaling of length, area, and volume · volume · similarity

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.