# 2010 AMC 8 Problems/Problem 8

## Problem

As Emily is riding her bicycle on a long straight road, she spots Emerson skating in the same direction $1/2$ mile in front of her. After she passes him, she can see him in her rear mirror until he is $1/2$ mile behind her. Emily rides at a constant rate of $12$ miles per hour, and Emerson skates at a constant rate of $8$ miles per hour. For how many minutes can Emily see Emerson? $\textbf{(A)}\ 6 \qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 15\qquad\textbf{(E)}\ 16$

## Solution

Because they are both moving in the same direction, Emily is riding relative to Emerson $12-8=4$ mph. Now we can look at it as if Emerson is not moving at all[on his skateboard] and Emily is riding at $4$ mph. It takes her $$\frac12 \ \text{mile} \cdot \frac{1\ \text{hour}}{4\ \text{miles}} = \frac18\ \text{hour}$$

to ride the $1/2$ mile to reach him, and then the same amount of time to be $1/2$ mile ahead of him. This totals to $$2 \cdot \frac18 \ \text{hour} \cdot \frac{60\ \text{minutes}}{1\ \text{hour}} = \boxed{\textbf{(D)}\ 15}\ \text{minutes}$$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 