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Difference between revisions of "2008 AMC 12B Problems/Problem 11"
m (→‎Solution)
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==Solution==
 
==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 <math>4,000</math> 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):
 
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 <math>4,000</math> 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):
−
<math>V_I*Height^3 = V_N</math>
+
<math>V_I\times \text{Height}^3 = V_N</math>
  
−
Plugging in our given condition, <math>1/8 = Height^3 \Rightarrow Height = 1/2</math>
+
Plugging in our given condition, <math>\frac{1}{8} = \text{Height}^3 \Rightarrow \text{Height} = \frac{1}{2}</math>. 
  
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<math>8000*1/2=4,000 \Rightarrow \boxed{A}</math>
+
<math>8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}</math>.
  
 
==See Also==
 
==See Also==

Revision as of 10:36, 2 February 2015

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 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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All AMC 12 Problems and Solutions

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