A motorboat, whose speed is 15 km/hr in still water goes 30 km downstream and…

Concept: For a boat travelling between two points along a stream, if b is the boat's speed in still water and v is the stream's speed, the boat's effective speed with the current (downstream) is (b + v) and against the current (upstream) is (b − v). For any leg of the journey, time taken = distance ÷ speed.

1. Here the boat's speed in still water b = 15 km/hr. Let the stream's speed be v km/hr.

2. Downstream speed = (15 + v) km/hr and upstream speed = (15 − v) km/hr.

3. Time for the 30 km downstream leg = 30/(15 + v) hours; time for the 30 km upstream leg = 30/(15 − v) hours.

4. The total time is 4 hours 30 minutes = 9/2 hours, so 30/(15 + v) + 30/(15 − v) = 9/2.

5. Dividing throughout by 30: 1/(15 + v) + 1/(15 − v) = 9/60.

6. Combining the left side over a common denominator: [(15 − v) + (15 + v)] / [(15 + v)(15 − v)] = 30/(225 − v2) = 9/60.

7. Cross-multiplying: 30 × 60 = 9 × (225 − v2), so 1800 = 2025 − 9v2, so 9v2 = 225, so v2 = 25.

8. Since a speed must be positive, v = 5.

Cross-check: with v = 5, downstream speed = 20 km/hr and upstream speed = 10 km/hr, so total time = 30/20 + 30/10 = 1.5 + 3 = 4.5 hours, exactly the 4 hours 30 minutes given — confirming the value.

Hence, the speed of the stream is 5 km/hr.