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Difference between revisions of "2008 AMC 12B Problems/Problem 11"
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==Problem 11==
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==Problem==
 
A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top <math>\frac{1}{8}</math> of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?
 
A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top <math>\frac{1}{8}</math> of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?
  
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==Solution==
 
==Solution==
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In a cone, radius and height each vary inversely. with increasing height (the radius at <math>4,000</math> feet is half that at <math>0</math> feet). Therefore, volume varies as the inverse cube of increasing height (expressed as a percentage of the total height of cone):
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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):
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<math>V_I*Height^3 = V_N</math>
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<math>V_I\times \text{Height}^3 = V_N</math>
  
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Plugging in our given condition, <math>1/8 = Height^3 -> Height = 1/2</math>
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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</math>, answer choice A.
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<math>8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}</math>.
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==Faster Solution==
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The volume of the cone above water is <math>\frac 1 8</math> that of the entire cone (mountain). These cones are obviously similar so the radius and height of the small cone must be <math>\sqrt[3]{\frac1 8}= \frac 1 2</math> that of the large one. Because the height of the large cone is <math>8000</math> the height of the small cone is <math>4000</math>. Thus the depth of the water is <math>8000-4000 = 4000 \Rightarrow\boxed{\text{A}}</math> - AOPqghj
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2008|ab=B|num-b=10|num-a=12}}
 
{{AMC12 box|year=2008|ab=B|num-b=10|num-a=12}}
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[[Category:Introductory Geometry Problems]]
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[[Category:3D Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 21:12, 8 May 2021

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 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

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