Difference between revisions of "1952 AHSME Problems/Problem 41"

Latest revision as of 00:19, 13 September 2022

Problem

Increasing the radius of a cylinder by $6$ units increased the volume by $y$ cubic units. Increasing the height of the cylinder by $6$ units also increases the volume by $y$ cubic units. If the original height is $2$ , then the original radius is:

$\text{(A) } 2 \qquad \text{(B) } 4 \qquad \text{(C) } 6 \qquad \text{(D) } 6\pi \qquad \text{(E) } 8$

Solution 1

We know that the volume of a cylinder is equal to $\pi r^2h$ , where $r$ and $h$ are the radius and height, respectively. So we know that $2\pi (r+6)^2-2\pi r^2=y=\pi r^2(2+6)-2\pi r^2$ . Expanding and rearranging, we get that $2\pi (12r+36)=6\pi r^2$ . Divide both sides by $6\pi$ to get that $4r+12=r^2$ , and rearrange to see that $r^2-4r-12=0$ . This factors to become $(r-6)(r+2)=0$ , so $r=6$ or $r=-2$ . Obviously, the radius cannot be negative, so our answer is $\fbox{(C) 6}$

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 40	Followed by Problem 42
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All AHSME Problems and Solutions

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

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 {{AHSME 50p box|year=1952|num-b=40|num-a=42}}
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+[[Category: Introductory Geometry Problems]]
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Difference between revisions of "1952 AHSME Problems/Problem 41"

Latest revision as of 00:19, 13 September 2022

Problem

Solution 1

See also