# 2008 AMC 12B Problems/Problem 11

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

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, $\frac{1}{8} = \text{Height}^3 \Rightarrow \text{Height} = \frac{1}{2}$.

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

## Faster Solution

The volume of the cone above water is $\frac 1 8$ that of the entire cone (mountain). These cones are obviously similar so the radius and height of the small cone must be $\sqrt[3]{\frac1 8}= \frac 1 2$ that of the large one. Because the height of the large cone is $8000$ the height of the small cone is $4000$. Thus the depth of the water is $8000-4000 = 4000 \Rightarrow\boxed{\text{A}}$ - AOPqghj

## See Also

 2008 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Invalid username
Login to AoPS