A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top $\frac{1}{8}$ of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet? $\textbf{(A)}\ 4000 \qquad$ $ \textbf{(B)}\ 2000(4-\sqrt{2}) \qquad$ $ \textbf{(C)}\ 6000 \qquad$ $ \textbf{(D)}\ 6400 \qquad$ $ \textbf{(E)}\ 7000$
Solution
The part above water is a smaller cone similar to the whole, with volume ratio $\left(\frac{h_{\text{top}}}{H}\right)^3=\frac18$, so $h_{\text{top}}=\frac H2=4000$.
Ocean depth $=H-h_{\text{top}}=8000-4000=4000$. The answer is (\textbf{A}).
A cone-shaped mountain has its base on the ocean floor and height $9000$ feet. The top $\frac{1}{27}$ of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain, in feet?
$\textbf{(A)}\ 4000 \qquad$ $ \textbf{(B)}\ 2000(4-\sqrt{2}) \qquad$ $ \textbf{(C)}\ 6000 \qquad$ $ \textbf{(D)}\ 6400 \qquad$ $ \textbf{(E)}\ 7000$ 
