## Precalculus (6th Edition)

$35$ min.
Distance traveled by either runner = (average speed)(time of travel). If the speed is given in mph, the time is expressed in hours. 10 minutes =$\displaystyle \frac{10}{60}=\frac{1}{6}$ hours. At a certain distance from the start, Mary and Janet will meet. Janet will have been running for $t$ hours, covering $5t$ miles. Mary will have been running $(t-\displaystyle \frac{1}{6})$ hours, covering $7(t-\displaystyle \frac{1}{6})$ miles. The distances covered are equal: $5t= 7(t-\displaystyle \frac{1}{6})$ $5t=7t-\displaystyle \frac{7}{6}$ $-2t=-\displaystyle \frac{7}{6}$ $t=\displaystyle \frac{7}{12}$ hours In minutes, $\displaystyle \frac{7}{12}(60)=35$ min.