A container in the shape of a cube has edge length $20 \mathrm{~cm}$ and contains some water. A solid gold cube, with edge length $15 \mathrm{~cm}$, sinks to the bottom of this container, causing the water level to rise just to the top of the solid cube. Which of the following is closest to the original depth of the water?
(A) $6.56 \mathrm{~cm}$
(B) $8.25 \mathrm{~cm}$
(D) $5.31 \mathrm{~cm}$
(E) $7.50 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-119.jpg?height=364&width=384&top_left_y=114&top_left_x=1443)
Let $d$ be the original depth of the water.

Then the total volume of water in the container initially is $20 \times 20 \times d=400 d$.

The volume of the gold cube is $15 \times 15 \times 15=3375$.

After the cube has been added to the water, the total volume inside the container that is filled is $20 \times 20 \times 15=6000$, since the container with base $20 \mathrm{~cm}$ by $20 \mathrm{~cm}$ has been filled to a depth of $15 \mathrm{~cm}$. Therefore, $400 d+3375=6000$ or $400 d=2625$ or $d=6.5625$. So $d$ is closest to $6.56 \mathrm{~cm}$.

Answer: (A)