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

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Increasing the radius of a cylinder by <math>6</math> units increased the volume by <math>y</math> cubic units. Increasing the altitude of the cylinder by <math>6</math> units also increases the volume by <math>y</math> cubic units. If the original altitude is <math>2</math>, then the original radius is:  
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Increasing the radius of a cylinder by <math>6</math> units increased the volume by <math>y</math> cubic units. Increasing the height of the cylinder by <math>6</math> units also increases the volume by <math>y</math> cubic units. If the original height is <math>2</math>, then the original radius is:  
  
 
<math>\text{(A) } 2 \qquad
 
<math>\text{(A) } 2 \qquad

Revision as of 00:58, 6 August 2016

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 $\fbox{(C) 6}$

See also

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

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