Difference between revisions of "1950 AHSME Problems/Problem 33"
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A circular pipe with diameter 1 inch and height h has a volume of <math>\pi \left(\frac{1}{2}\right)^2h=\frac{\pi h}{4}</math>. A pipe with diameter 6 inches and height h has volume <math>\pi \left(\frac{6}{2}\right)^2h=9\pi h</math>. To find how many 1-pipes fit in a 6-pipe, we divide: <math>\frac{9\pi h}{\frac{\pi h}{4}}=\frac{9*4\pi h}{\pi h}=\frac{36\pi h}{\pi h}=36 \textbf{(D)}</math> | A circular pipe with diameter 1 inch and height h has a volume of <math>\pi \left(\frac{1}{2}\right)^2h=\frac{\pi h}{4}</math>. A pipe with diameter 6 inches and height h has volume <math>\pi \left(\frac{6}{2}\right)^2h=9\pi h</math>. To find how many 1-pipes fit in a 6-pipe, we divide: <math>\frac{9\pi h}{\frac{\pi h}{4}}=\frac{9*4\pi h}{\pi h}=\frac{36\pi h}{\pi h}=36 \textbf{(D)}</math> | ||
| − | If the ratio of similar length of similar shapes is x, then the ratio between area is <math>x^2</math>. Therefore, since the ratio between diameters is <math>1/6</math>, the ratio between area is <math>1/36</math>, so <math>36</math> pipes of diameter <math>1</math> are required. | + | If the ratio of similar length of similar shapes is x, then the ratio between area is <math>x^2</math>. Therefore, since the ratio between diameters is <math>1/6</math>, the ratio between area is <math>1/36 textbf{(D)}</math>, so <math>36</math> pipes of diameter <math>1</math> are required. |
==See Also== | ==See Also== | ||
Revision as of 15:37, 30 July 2017
Problem
The number of circular pipes with an inside diameter of 
inch which will carry the same amount of water as a pipe with an inside diameter of 
inches is:
Solution
It must be assumed that the pipes have an equal height.
A circular pipe with diameter 1 inch and height h has a volume of 
. A pipe with diameter 6 inches and height h has volume 
. To find how many 1-pipes fit in a 6-pipe, we divide: 
If the ratio of similar length of similar shapes is x, then the ratio between area is 
. Therefore, since the ratio between diameters is 
, the ratio between area is 
, so 
pipes of diameter are required.
See Also
| 1950 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 32 |
Followed by Problem 34 | |
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| All AHSME Problems and Solutions | ||
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