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

(Solution)
(Solution)
Line 8: Line 8:
 
<math>V_I*Height^3 = V_N</math>
 
<math>V_I*Height^3 = V_N</math>
  
Plugging in our given condition, <math>1/8 = Height^3 -> Height = 1/2</math>
+
Plugging in our given condition, <math>1/8 = Height^3 \Rightarrow 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 14:25, 2 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 (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): $V_I*Height^3 = V_N$

Plugging in our given condition, $1/8 = Height^3 \Rightarrow Height = 1/2$

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

See Also

2008 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 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