# 2009 AMC 10B Problems/Problem 23

The following problem is from both the 2009 AMC 10B #23 and 2009 AMC 12B #18, so both problems redirect to this page.

## Problem

Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture? $\mathrm{(A)}\frac {1}{16}\qquad \mathrm{(B)}\frac 18\qquad \mathrm{(C)}\frac {3}{16}\qquad \mathrm{(D)}\frac 14\qquad \mathrm{(E)}\frac {5}{16}$

## Solution

After $10$ minutes $(600$ seconds $),$ Rachel will have completed $6$ laps and be $30$ seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in $22.5$ seconds, she will be in the picture between $18.75$ seconds and $41.25$ seconds of the tenth minute. After 10 minutes, Robert will have completed $7$ laps and be $40$ seconds from completing his eighth lap. Because Robert runs one-fourth of a lap in $20$ seconds, he will be in the picture between $30$ seconds and $50$ seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between $30$ seconds and $41.25$ seconds of the tenth minute. So the probability that both runners are in the picture is $\frac {41.25 - 30} {60} = \boxed{\frac {3}{16}}$. The answer is $\mathrm{(C)}$.

## Solution 2 (Video solution)

Video: https://youtu.be/eZjJ5MQV47o ~DaBobWhoLikeMath

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