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Difference between revisions of "1950 AHSME Problems/Problem 33"
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It must be assumed that the pipes have an equal height.
 
It must be assumed that the pipes have an equal height.
  
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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>
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We can represent the amount of water carried per unit time by cross sectional area.
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Cross sectional of Pipe with diameter <math>6 in</math>,
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<cmath>\pi r^2 = \pi \cdot 3^2 = 9\pi</cmath>
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Cross sectional area of pipe with diameter <math>1 in</math>
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<cmath>\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}</cmath>  
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So number of 1 in pipes required is the number obtained by dividing their cross sectional areas
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<cmath>\frac{9\pi}{\frac{\pi}{4}} = 36</cmath>
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So the answer is <math>\boxed{\textbf{(D)}\ 36}</math>.
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==Solution 2 (Solution 1 but easier)==
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Knowing that <math>d^2 \alpha A</math>, where <math>d</math> is the pipe diameter and <math>A</math> is the cross-sectional area we simply get <math>6^2 = 36 = \boxed{\textbf{(D)}\ 36}</math>.
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This works because the diameter of one of the other pipes is <math>1</math>, which is not affected by powers.
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~PeterDoesPhysics
  
 
==See Also==
 
==See Also==
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[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Geometry Problems]]
 
[[Category:Introductory Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 23:56, 25 August 2024

Problem

The number of circular pipes with an inside diameter of $1$ inch which will carry the same amount of water as a pipe with an inside diameter of $6$ inches is:

$\textbf{(A)}\ 6\pi \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 36 \qquad \textbf{(E)}\ 36\pi$

Solution

It must be assumed that the pipes have an equal height.

We can represent the amount of water carried per unit time by cross sectional area. Cross sectional of Pipe with diameter $6 in$, \[\pi r^2 = \pi \cdot 3^2 = 9\pi\]

Cross sectional area of pipe with diameter $1 in$

\[\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}\]

So number of 1 in pipes required is the number obtained by dividing their cross sectional areas

\[\frac{9\pi}{\frac{\pi}{4}} = 36\]

So the answer is $\boxed{\textbf{(D)}\ 36}$.

Solution 2 (Solution 1 but easier)

Knowing that $d^2 \alpha A$, where $d$ is the pipe diameter and $A$ is the cross-sectional area we simply get $6^2 = 36 = \boxed{\textbf{(D)}\ 36}$.

This works because the diameter of one of the other pipes is $1$, which is not affected by powers.

~PeterDoesPhysics

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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