1953 AHSME Problems/Problem 10

Problem

The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:

$\textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi\qquad\textbf{(E)}\ \text{none of these}$

Solution

We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\frac{5280}{6\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\boxed{\textbf{(C) } \frac{880}{\pi}}$ .

1953 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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1953 AHSME Problems/Problem 10

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