Difference between revisions of "2017 AMC 12B Problems/Problem 14"
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==Solution== | ==Solution== | ||
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| + | We add up the volumes separately. | ||
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| + | Top cone = <math> \dfrac{1}{3} \pi (2)^2 \times 4 </math>. | ||
| + | |||
| + | Frustum = large inverted cone - small inverted cone = | ||
| + | <math> \dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4</math>. | ||
| + | |||
| + | Adding we get <math>\dfrac{1}{3} \pi (16+32-4) = \dfrac{44}{3} \pi </math>. | ||
{{AMC12 box|year=2017|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2017|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:29, 17 February 2017
Problem
An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?
Solution
We add up the volumes separately.
Top cone = 
.
Frustum = large inverted cone - small inverted cone =
.
Adding we get 
.
| 2017 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
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