Difference between revisions of "2008 AMC 12B Problems/Problem 11"
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<math>8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}</math>. | <math>8000\cdot\frac{1}{2}=4000 \Rightarrow \boxed{\textbf{A}}</math>. | ||
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+ | ==Faster Solution== | ||
+ | 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== |
Latest revision as of 21:12, 8 May 2021
Contents
[hide]Problem
A cone-shaped mountain has its base on the ocean floor and has a height of 8000 feet. The top of the volume of the mountain is above water. What is the depth of the ocean at the base of the mountain in feet?
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 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):
Plugging in our given condition, .
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Faster Solution
The volume of the cone above water is that of the entire cone (mountain). These cones are obviously similar so the radius and height of the small cone must be that of the large one. Because the height of the large cone is the height of the small cone is . Thus the depth of the water is - 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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