2025 AMC 12B Problem 15

Problem 15 of 25IntermediateAlgebra Geometry

A container has a

1 \times 1

square bottom, a

3 \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

35

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

At height fraction

t

the square cross-section has side

1 + 2t,

so the volume filled up to height

t

\int_0^t (1 + 2u)^2\,du.

Up to the midline

\left(t = \tfrac{1}{2}\right)

this is

\tfrac{7}{6},

and the full volume is

\tfrac{13}{3}.

The remaining volume is

\tfrac{13}{3} - \tfrac{7}{6} = \tfrac{19}{6},

which is

\tfrac{19}{7}

times the first part. So the remainder takes

35 \cdot \tfrac{19}{7} = 95

more minutes. Thus, the correct answer is D.

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