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