2008 AMC 12B Problems/Problem 11

Problem 11

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

In a cone, radius and height each vary inversely with increasing height (i.e. the radius of the cone formed by cutting off the mountain at $4,000$ feet is half that of the original mountain). Therefore, volume varies as the inverse cube of increasing height (expressed as a percentage of the total height of cone): $V_I\times \text{Height}^3 = V_N$

Plugging in our given condition, $\frac{1}{8} = \text{Height}^3 \Rightarrow \text{Height} = \frac{1}{2}$.

$8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}$.

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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All AMC 12 Problems and Solutions

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