1966 AMC 12 Problem 27

Question

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:

2

\dfrac52

3

\dfrac72

4

Answer: A

Concepts:rational equation relative speed system of equations

Difficulty rating: 1850

Small hint:

Let

v

be the rower’s still-water speed and

c

the current speed

Big hint:

Translate the two time differences using speeds

v\pm c

and

2v\pm c

Solution:

The two time differences give

\begin{aligned} \frac{15}{v-c}-\frac{15}{v+c}&=5,\\ \frac{15}{2v-c}-\frac{15}{2v+c}&=1. \end{aligned}

These simplify to

v^2-c^2=6c

and

4v^2-c^2=30c.

Substituting

v^2=c^2+6c

into the second gives

3c^2-6c=0.

Since

c\gt0,

we have

c=2.

Therefore, the correct answer is A.