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 $a=\boxed{\textbf{(B) }8 \text{ mph}}$ .

1950 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 27	Followed by Problem 29
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

Page

Toolbox

Search

1950 AHSME Problems/Problem 28

Problem

Solution

See Also