# 1950 AHSME Problems/Problem 28

## Problem

Two boys $A$ and $B$ start at the same time to ride from Port Jervis to Poughkeepsie, $60$ miles away. $A$ travels $4$ miles an hour slower than $B$. $B$ reaches Poughkeepsie and at once turns back meeting $A$ $12$ miles from Poughkeepsie. The rate of $A$ was:

$\textbf{(A)}\ 4\text{ mph}\qquad \textbf{(B)}\ 8\text{ mph} \qquad \textbf{(C)}\ 12\text{ mph} \qquad \textbf{(D)}\ 16\text{ mph} \qquad \textbf{(E)}\ 20\text{ mph}$

## Solution

Let the speed of boy $A$ be $a$, and the speed of boy $B$ be $b$. Notice that $A$ travels $4$ miles per hour slower than boy $B$, so we can replace $b$ with $a+4$.

Now let us see the distances that the boys each travel. Boy $A$ travels $60-12=48$ miles, and boy $B$ travels $60+12=72$ miles. Now, we can use $d=rt$ to make an equation, where we set the time to be equal: $$\frac{48}{a}=\frac{72}{a+4}$$ Cross-multiplying gives $48a+192=72a$. Isolating the variable $a$, we get the equation $24a=192$, so $a=\boxed{\textbf{(B) }8 \text{ mph}}$.

## Alternate Solution

Note that $A$ travels $60-12=48$ miles in the time it takes $B$ to travel $60+12=72$ miles. Thus, $B$ travels $72-48=24$ more miles in the given time, meaning $\frac{24\text{miles}}{4\text{miles}/\text{hour}} = 6 \text{hours}$ have passed, as $B$ goes $4$ miles per hour faster. Thus, $A$ travels $48$ miles per $6$ hours, or $8$ miles per hour.