1966 AHSME Problems/Problem 34

Problem

Let $r$ be the speed in miles per hour at which a wheel, $11$ feet in circumference, travels. If the time for a complete rotation of the wheel is shortened by $\frac{1}{4}$ of a second, the speed $r$ is increased by $5$ miles per hour. Then $r$ is:

$\text{(A) } 9 \quad \text{(B) } 10 \quad \text{(C) } 10\frac{1}{2} \quad \text{(D) } 11 \quad \text{(E) } 12$

Solution

The circumference of the wheel is $\frac{11}{5280}$ miles. Let the time for the rotation in seconds be $t$ . So $rt=\frac{11}{5280}*3600$ . We also know reducing the time by $\frac{1}{4}$ of a second makes $r$ increase by $5$ . So $(r+5)(t-\frac{1}{4})=\frac{11}{5280}*3600$ . Solving for $r$ we get $r=10$ . So our answer is $(B)$ $10$ .

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 33	Followed by Problem 35
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1966 AHSME Problems/Problem 34

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