1956 AHSME Problems/Problem 45

A wheel with a rubber tire has an outside diameter of $25$ in. When the radius has been decreased a quarter of an inch, the number of revolutions in one mile will:

$\text{(A)}\ \text{be increased about }2\% \qquad \\ \text{(B)}\ \text{be increased about }1\% \\ \text{(C)}\ \text{be increased about }20\%\qquad \\ \text{(D)}\ \text{be increased about }\frac{1}{2}\%\qquad \\ \text{(E)}\ \text{remain the same}$

Solution

The circumference of the normal tire is $50\pi$ inches, while the shaved tire has a circumference of $49.5\pi$ inches. Thus, the ratio of the number of rotations the two tires take to go a mile can be expressed as $\frac{5280*12}{50\pi}:\frac{5280*12}{49.5\pi}$ Simplifying the ratio, we get $\frac{1}{100}:\frac{1}{99} \rightarrow 99:100$

The increase in rotations can be expressed as $\frac{100-99}{99} \rightarrow \frac{1}{99}$ . This leads to an increase of $\boxed{\textbf{(B) }\text{about } 1\%}$

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1956 AHSME Problems/Problem 45

Solution

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