1950 AHSME Problems/Problem 33: Difference between revisions
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\textbf{(E)}\ 36\pi</math> | \textbf{(E)}\ 36\pi</math> | ||
==Solution== | ==Solution== | ||
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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 <math>\pi (\frac{1}{2})^2h=\frac{\pi h}{4}</math>. A pipe with diameter 6 inches and height h has volume <math>\pi (\frac{6}{2})^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 \text{(D)}</math> | |||
==See Also== | ==See Also== | ||
Revision as of 14:06, 5 July 2012
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: 
See Also
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