A motor boat can travel at 18 km/h in still water. It goes from point P to point Q, which is 84 km downstream, and then returns from Q to P upstream in the river. On the downstream journey, the boat remains stationary for 30 minutes due to fog (engine switched off), and on the upstream journey it remains stationary for 60 minutes for an engine check. If the total time taken for the entire round trip (including stoppages) is 12 hours, what is the speed of the stream?

Solution

Let speed of stream = s km/h Then: Downstream speed = 18 + s Upstream speed = 18 − s Distance each way = 84 km. Let actual moving time (excluding stoppages) be T hours. Given stoppages = 30 min + 60 min = 1.5 hours Total time = moving time + stoppages = 12 ⇒ Moving time = 12 − 1.5 = 10.5 hours Time to go downstream (moving time only) = 84 / (18 + s) Time to come upstream (moving time only) = 84 / (18 − s) So: 84/(18 + s) + 84/(18 − s) = 10.5 Divide both sides by 84: 1/(18 + s) + 1/(18 − s) = 10.5 / 84 = 105/840 = 1/8 Now compute LHS: 1/(18 + s) + 1/(18 − s) = [(18 − s) + (18 + s)] / [(18 + s)(18 − s)] = 36 / (324 − s²) So: 36 / (324 − s²) = 1/8 Cross-multiply: 36 × 8 = 324 − s² 288 = 324 − s² s² = 324 − 288 = 36 s = 6 km/h (positive value) So, speed of stream = 6 km/h.