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Difference between revisions of "2008 AMC 12B Problems/Problem 11"

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<math>V_I*Height^3 = V_N</math>
 
<math>V_I*Height^3 = V_N</math>
  
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Plugging in our given condition, <math>1/8 = Height^3 /rightarrow Height = 1/2</math>
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Plugging in our given condition, <math>1/8 = Height^3 -> Height = 1/2</math>
  
 
<math>8000*1/2=4,000</math>, answer choice A.
 
<math>8000*1/2=4,000</math>, answer choice A.

Revision as of 22:13, 1 March 2008

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 (the radius at $4,000$ feet is half that at $0$ feet). Therefore, volume varies as the inverse cube of increasing height (expressed as a percentage of the total height of cone): $V_I*Height^3 = V_N$

Plugging in our given condition, $1/8 = Height^3 -> Height = 1/2$

$8000*1/2=4,000$, answer choice A.

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