1950 AHSME Problems/Problem 33: Difference between revisions

Revision as of 21:19, 24 December 2023

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 $\boxed{\textbf{(D)}\ 36}$ .

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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All AHSME Problems and Solutions

@@ Line 14: / Line 14: @@
 We can represent the amount of water carried per unit time by cross sectional area.
 Cross sectional of Pipe with diameter <math>6 in</math>,
-<cmath>\pir^2 = \pi \cdot 3^2 = 9\pi</cmath>
+<cmath>\pi r^2 = \pi \cdot 3^2 = 9\pi</cmath>
 Cross sectional area of pipe with diameter <math>1 in</math>
-<cmath>\pir^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}</cmath>
+<cmath>\pi r^2 = \pi \cdot 0.5 \cdot 0.5 = \frac{\pi}{4}</cmath>
 So number of 1 in pipes required is the number obtained by dividing their cross sectional areas

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1950 AHSME Problems/Problem 33: Difference between revisions

Revision as of 21:19, 24 December 2023

Problem

Solution

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