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)$$ (Error compiling LaTeX. Unknown error_msg)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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