# 1960 AHSME Problems/Problem 34

## Problem 34

Two swimmers, at opposite ends of a $90$-foot pool, start to swim the length of the pool, one at the rate of $3$ feet per second, the other at $2$ feet per second. They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.

$\textbf{(A)}\ 24\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 18$

## Solution

First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side. Also, note that after $180$ seconds (or $3$ minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.

$[asy] draw((0,0)--(0,105),EndArrow); draw((0,0)--(105,0),EndArrow); for (int i=0; i<6;++i) { dot((0,18i)); } for (int j=0;j<7;++j) { dot((15j,0)); } label("0",(0,0),SW); label("90",(0,90),W); label("180",(90,0),S); draw((0,0)--(15,90)--(30,0)--(45,90)--(60,0)--(75,90)--(90,0),red); draw((0,90)--(22.5,0)--(45,90)--(67.5,0)--(90,90),blue); [/asy]$

At this point, find the number of meeting points in the first $3$ minutes, then multiply by four to get the answer. From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, so the two swimmers will pass each other $20$ times, which is answer choice $\boxed{\textbf{(C)}}$.