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>.
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Frustum  = large inverted cone - small inverted cone =
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<math> \dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4</math>.
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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}}
 
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{{MAA Notice}}

Revision as of 09:29, 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?

$\textbf{(A)}\ 8\pi \qquad \textbf{(B)}\ \frac{28\pi}{3} \qquad \textbf{(C)}\ 12\pi \qquad \textbf{(D)}\ 14\pi \qquad \textbf{(E)}\ \frac{44\pi}{3}$

Solution

We add up the volumes separately.

Top cone = $\dfrac{1}{3} \pi (2)^2 \times 4$.

Frustum = large inverted cone - small inverted cone = $\dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4$.

Adding we get $\dfrac{1}{3} \pi (16+32-4) = \dfrac{44}{3} \pi$.

2017 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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All AMC 12 Problems and Solutions

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