2025 AMC 12B Problem 15

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

Answer: D

Concepts:volume similarity ratio and proportion

Difficulty rating: 1800

Small hint:

At height fraction

t

the cross-section is a square of side

1 + 2t,

so its area is

(1 + 2t)^2

Big hint:

Compare the volume up to the midline (

t = \tfrac{1}{2}

) with the total volume; fill time is proportional to volume

Solution:

At height fraction

t

the square cross-section has side

1 + 2t,

so the volume filled up to height

t

is

\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.