1966 AHSME Problems/Problem 27

Problem

At his usual rate a man rows 15 miles downstream in five hours less time than it takes him to return. If he doubles his usual rate, the time downstream is only one hour less than the time upstream. In miles per hour, the rate of the stream's current is:

$\text{(A) } 2 \quad \text{(B) } \frac{5}{2} \quad \text{(C) } 3 \quad \text{(D) } \frac{7}{2} \quad \text{(E) } 4$

Solution

$\fbox{A}$

See also

1966 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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