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

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\text{(E) } 8 </math>
 
\text{(E) } 8 </math>
  
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== Solution ==
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== Solution 1==
 
We know that the volume of a cylinder is equal to <math>\pi r^2h</math>, where <math>r</math> and <math>h</math> are the radius and height, respectively. So we know that <math>2\pi (r+6)^2-2\pi r^2=y=\pi r^2(2+6)-2\pi r^2</math>. Expanding and rearranging, we get that <math>2\pi (12r+36)=6\pi r^2</math>. Divide both sides by <math>6\pi</math> to get that <math>4r+12=r^2</math>, and rearrange to see that <math>r^2-4r-12=0</math>. This factors to become <math>(r-6)(r+2)=0</math>, so <math>r=6</math> or <math>r=-2</math>. Obviously, the radius cannot be negative, so our answer is <math>\fbox{(C) 6}</math>
 
We know that the volume of a cylinder is equal to <math>\pi r^2h</math>, where <math>r</math> and <math>h</math> are the radius and height, respectively. So we know that <math>2\pi (r+6)^2-2\pi r^2=y=\pi r^2(2+6)-2\pi r^2</math>. Expanding and rearranging, we get that <math>2\pi (12r+36)=6\pi r^2</math>. Divide both sides by <math>6\pi</math> to get that <math>4r+12=r^2</math>, and rearrange to see that <math>r^2-4r-12=0</math>. This factors to become <math>(r-6)(r+2)=0</math>, so <math>r=6</math> or <math>r=-2</math>. Obviously, the radius cannot be negative, so our answer is <math>\fbox{(C) 6}</math>
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== Solution 2==
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We know that the testmakers want the contestant to be confused between <math>6</math> and <math>6\pi</math>. Since the radius does not have a <math>\pi</math> in it, the answer is <math>\fbox{(C) 6}</math>
  
 
== See also ==
 
== See also ==

Revision as of 02:30, 4 February 2016

Problem

Increasing the radius of a cylinder by $6$ units increased the volume by $y$ cubic units. Increasing the altitude of the cylinder by $6$ units also increases the volume by $y$ cubic units. If the original altitude 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 $\fbox{(C) 6}$

Solution 2

We know that the testmakers want the contestant to be confused between $6$ and $6\pi$. Since the radius does not have a $\pi$ in it, the answer is $\fbox{(C) 6}$

See also

1952 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 40 		Followed by
Problem 42
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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