1950 AHSME Problems/Problem 33

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 $d^2 ∝ A$ (Error compiling LaTeX. Unknown error_msg), 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

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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1950 AHSME Problems/Problem 33

Contents

Problem

Solution

Solution 2 (Solution 1 but easier)

See Also