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 | + | 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 | ||
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== Solution 1== | == 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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== See also == | == See also == | ||
{{AHSME 50p box|year=1952|num-b=40|num-a=42}} | {{AHSME 50p box|year=1952|num-b=40|num-a=42}} | ||
− | [[Category: | + | [[Category: Introductory Geometry Problems]] |
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 01:19, 13 September 2022
Problem
Increasing the radius of a cylinder by units increased the volume by cubic units. Increasing the height of the cylinder by units also increases the volume by cubic units. If the original height is , then the original radius is:
Solution 1
We know that the volume of a cylinder is equal to , where and are the radius and height, respectively. So we know that . Expanding and rearranging, we get that . Divide both sides by to get that , and rearrange to see that . This factors to become , so or . Obviously, the radius cannot be negative, so our answer is
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 40 |
Followed by Problem 42 | |
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