1962 AHSME Problems/Problem 10

Problem

A man drives $150$ miles to the seashore in $3$ hours and $20$ minutes. He returns from the shore to the starting point in $4$ hours and $10$ minutes. Let $r$ be the average rate for the entire trip. Then the average rate for the trip going exceeds $r$ in miles per hour, by:

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 4\frac{1}{2}\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 1$

Solution

Since the equation for rate is $r=\frac{d}{t}$, you only need to find $d$ and $t$. The distance from the seashore to the starting point is $150$ miles, and since he makes a round trip, $d=300$. Also, you know the times it took him to go both directions, so when you add them up ($3$ hours and $20$ minutes $+$ $4$ hours and $10$ minutes), you get $7\frac{1}{2}$ hours. Since $r=\frac{d}{t}$, after plugging in the values of $d$ and $t$, you get $r=\frac{300}{7\frac{1}{2}}$, or $r=40$. But what the question asks is for the positive difference between the speed going to the seashore and the average speed. The equation is $r=\frac{d}{t}$, so when you plug in $d$ and $t$, you get $r=\frac{150}{3\frac{1}{3}}$. Simplifying, you get $r=45$. $45-40=5 \rightarrow \boxed{\text{A}}$

See Also

1962 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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