2025 AMC 10B Problem 19

The four sides are congruent trapezoids joining a $1 \times 1$ square to a concentric $3 \times 3$ square, so the horizontal cross-section at height $t$ is a square whose side grows linearly from $1$ at the bottom to $3$ at the top. Extending the four faces downwards, the side length shrinks to $0$ at a single point: the container is the part of a square pyramid lying between the cross-sections of side $1$ and side $3$.

For that pyramid, the portion below the cross-section of side $s$ is similar to the whole pyramid with ratio $s$, so its volume is
$$V(s) = c\,s^3$$
for some constant $c$.

Three levels matter. The bottom is $s = 1$. The midline of a trapezoid is the segment joining the midpoints of its legs, so filling "up to the midline" means filling to half the height of the container, where the side is the average $\tfrac{1+3}{2} = 2$. The top is $s = 3$. Hence

• water poured so far: $V(2) - V(1) = c(8 - 1) = 7c$,
• water still to come: $V(3) - V(2) = c(27 - 8) = 19c$.

The hose runs at a constant rate, so time is proportional to volume: $7c$ takes $35$ minutes, i.e. $5$ minutes per unit of $c$, and the remaining $19c$ takes
$$19 \cdot 5 = 95 \text{ minutes}.$$

The answer is $\boxed{\textbf{(D)}\ 95}$.

Every frustum question becomes easy once the missing tip is restored, because a pyramid or cone has the one-line scaling law $V \propto s^3$ and differences of cubes then replace any volume formula. Nothing about the actual height, or the value of $c$, is ever needed — only the three side lengths $1, 2, 3$. The same idea answers "what fraction of a conical cup is filled to half its depth", and it is the reason a frustum's two halves are so unequal.

Use the frustum formula $V = \tfrac{H}{3}\left(A_1 + \sqrt{A_1A_2} + A_2\right)$ directly, with $H$ the height of each half. The lower half joins areas $1$ and $4$:
$$V_{\text{low}} = \frac{H}{3}\left(1 + 2 + 4\right) = \frac{7H}{3},$$
and the upper half joins areas $4$ and $9$:
$$V_{\text{high}} = \frac{H}{3}\left(4 + 6 + 9\right) = \frac{19H}{3}.$$
The ratio is again $7 : 19$, giving $35 \cdot \tfrac{19}{7} = 95$.

• Assuming that filling to half the height fills half the container, so the rest takes about the same 35 minutes again and a guess is pushed toward the 70 in (A).
• Taking the cross-section at the midline to have area $\tfrac{1+9}{2} = 5$ (the average of the end areas) instead of $2^2 = 4$ (the square of the average side).
• Scaling volumes by $s^2$ rather than $s^3$, which gives a ratio of $3 : 5$ and no listed answer.