Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 12B Problems/Problem 14"

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
We add up the volumes separately.
+
We add up the volumes separately. The top cone has radius 2 and height 4 so we have
  
−
Top cone = <math> \dfrac{1}{3} \pi (2)^2 \times 4 </math>.  
+
top cone = <math> \dfrac{1}{3} \pi (2)^2 \times 4 </math>.  
  
−
Frustum = large inverted cone - small inverted cone =  
+
The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so
 +
 
 +
frustum = large cone - small cone =  
 
<math> \dfrac{1}{3} \pi (2)^2 \times 8 - \dfrac{1}{3} \pi (1)^2 \times 4</math>.
 
<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>.
+
Adding, we get <math>\dfrac{1}{3} \pi (16+32-4) = \dfrac{44\pi}{3}</math>. Thus, the answer is E.  
  
 
{{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 10:34, 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. The top cone has radius 2 and height 4 so we have

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

The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so

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

Adding, we get $\dfrac{1}{3} \pi (16+32-4) = \dfrac{44\pi}{3}$. Thus, the answer is E.

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12B_Problems/Problem_14&oldid=84022"