Question

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?

$70$

$85$

$90$

$95$

$105$

Solution:

The container is a square frustum: a horizontal slice at height fraction $t$ has side $1 + 2t.$ Extend the sides up to their apex, and the volume out to where the side length is $w$ scales as $w^3.$ So the whole container is $3^3 - 1^3 = 26$ parts, the piece up to the midline (side $2$ ) is $2^3 - 1^3 = 7$ parts, and the rest is $3^3 - 2^3 = 19$ parts. Those $7$ parts take $35$ minutes, so each part is $5$ minutes. The remaining $19$ parts take $95$ minutes. Thus, D is the correct answer.

2025 AMC 10B Problem 19

Solution:

Problem 19 in Other Years