A man rows a boat at a speed of 5 km/hr in still water. He takes a total of 1…

Concept

For a boat with speed b in still water crossing a stream of speed s, the effective speed is (b + s) when moving with the current (downstream) and (b − s) when moving against it (upstream). Time equals Distance divided by Speed, so for a round trip covering the same distance d each way, the total time is d/(b+s) + d/(b−s).

Application

1. Let the speed of the river current be y km/hr. The boat speed in still water is 5 km/hr and the one-way distance is 2.4 km.

2. Downstream speed = (5 + y) km/hr, and upstream speed = (5 − y) km/hr.

3. Total time equation: 2.4/(5 + y) + 2.4/(5 − y) = 1.

4. Combine the fractions over a common denominator: 2.4 × [(5 − y) + (5 + y)] divided by (5 + y)(5 − y) = 1, which simplifies to 2.4 × 10 divided by (25 − y2) = 1.

5. So 24 = 25 − y2, which gives y2 = 1.

6. Since speed must be positive, y = 1 km/hr.

Cross-check

With y = 1 km/hr: downstream speed = 6 km/hr, taking 2.4/6 = 0.4 hr; upstream speed = 4 km/hr, taking 2.4/4 = 0.6 hr. The total is 0.4 + 0.6 = 1 hr, matching the time given in the question.