Track / Stage 3 / 45 of 260 #45 of 1964

Problem 45

AMC 10/12, early questions

Algebra Difficulty 3.1 Multiple choice

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:

Pick one

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

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.