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

Let the speed of rowing in still water and water speed be $r, w$ .

Then:

$\frac{15}{r-w}-\frac{15}{r+w}=5$ .

$\frac{15}{2r-w}-\frac{15}{2r+w=1$ (Error compiling LaTeX. Unknown error_msg).

Solving (remember: both variables must be positive!), we have $r=4, w=2$ . Select $\boxed{A}$ .

~hastapasta

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1966 AHSME Problems/Problem 27

Problem

Solution

See also