A man rows in still water at a certain speed V (km/hour). He takes as much…

Concept

For a boat moving on a flowing river, its own speed in still water is V and the stream contributes a speed s. Rowing against the current the effective speed is V - s, and rowing with the current the effective speed is V + s.

Time equals distance divided by speed, so two journeys take the same time exactly when their distance-divided-by-effective-speed quotients are equal. Setting those two quotients equal turns the wording of such a problem into a single linear equation in V.

Application

1. The stream flows at 2 km/hour, so the upstream speed is (V - 2) km/hour and the downstream speed is (V + 2) km/hour.

2. The time to row 2 km upstream is 2 / (V - 2) hours, and the time to row 3 km downstream is 3 / (V + 2) hours.

3. The problem states that these two times are the same, so 2 / (V - 2) = 3 / (V + 2).

4. Cross-multiplying the equation gives 2 × (V + 2) = 3 × (V - 2).

5. Expanding both sides gives 2V + 4 = 3V - 6.

6. Subtracting 2V from both sides and adding 6 to both sides gives V = 10, so the rowing speed in still water is 10 km/hour.

Cross-check

Substituting V = 10 back into the journey times: the upstream speed is 8 km/hour, so 2 km upstream takes 2 / 8 = 0.25 hour, and the downstream speed is 12 km/hour, so 3 km downstream takes 3 / 12 = 0.25 hour. The two times agree, and V = 10 is larger than the stream speed of 2 km/hour, so upstream travel is physically possible.

A second route to the same result: when two journeys take equal time, the distances are in the same ratio as the speeds, so (V - 2) : (V + 2) = 2 : 3, which again gives 3V - 6 = 2V + 4 and V = 10.