A sailor goes 12 km downstream in 1 hour and returns in 1 hour 12 minutes.…

Concept: a boat moving in a stream carries two component speeds — its own speed in still water, b, and the speed of the current, s. Travelling with the current the two add and travelling against it they subtract, so downstream speed = b + s and upstream speed = b - s. Adding those two relations and halving isolates the still-water speed, b = (downstream speed + upstream speed) / 2, while subtracting and halving isolates the current, s = (downstream speed - upstream speed) / 2. Each leg supplies one of those speeds through that leg’s own distance divided by that leg’s own time.

Application: both legs cover the same 12 km, so each leg yields one component speed and the two relations above then give the answer.

1. Downstream leg: 12 km is covered in 1 hour, so the downstream speed is 12 ÷ 1 = 12 km/hr.

2. Convert the return time into hours: 1 hour 12 minutes = 1 + 12/60 hour = 1.2 hours, that is 6/5 hour.

3. Upstream leg: the same 12 km is covered in 1.2 hours, so the upstream speed is 12 ÷ 1.2 = 10 km/hr.

4. Apply the still-water relation: b = (12 + 10) ÷ 2 = 22 ÷ 2 = 11 km/hr.

Cross-check: recover the current from the same pair of leg speeds and feed both numbers back into the original journey.

• Current: s = (12 - 10) ÷ 2 = 1 km/hr.

• Onward journey: b + s = 11 + 1 = 12 km/hr, and 12 km ÷ 12 km/hr = 1 hour, which is the time stated for the downstream trip.

• Return journey: b - s = 11 - 1 = 10 km/hr, and 12 km ÷ 10 km/hr = 1.2 hours = 1 hour 12 minutes, which is the time stated for the return trip.

The sailor’s speed in still water is therefore 11 km/hr.