A boat goes 28 km downstream and, while returning, covers only 75% of the…

Concept

In boats-and-streams problems, if the boat's speed in still water is b and the stream's speed is s, then downstream speed = (b + s) and upstream speed = (b - s). For any leg, time = distance / speed. The governing condition here is that the upstream time is greater than the downstream time by the stated gap.

Application

1. Convert the stream speed to km/hr. To change m/sec into km/hr, multiply by 18/5, so (5/9) x (18/5) = 2 km/hr. Hence s = 2 km/hr.

2. Find the two distances. The downstream distance is 28 km. The return (upstream) leg covers 75% of it, i.e. 0.75 x 28 = 21 km.

3. Write the time equation. With b the still-water speed, downstream time = 28/(b + 2) and upstream time = 21/(b - 2). Since the upstream leg takes 3 hr longer: 21/(b - 2) - 28/(b + 2) = 3.

4. Clear the denominators. Multiply through by (b - 2)(b + 2): 21(b + 2) - 28(b - 2) = 3(b2 - 4), which simplifies the left side to 21b + 42 - 28b + 56 = -7b + 98.

5. Form the quadratic. So -7b + 98 = 3b2 - 12, giving 3b2 + 7b - 110 = 0.

6. Solve it. Factor: 3b2 + 7b - 110 = (b - 5)(3b + 22) = 0, so b = 5 or b = -22/3. A speed cannot be negative, so b = 5 km/hr.

Cross-check

With b = 5 and s = 2: downstream time = 28/(5 + 2) = 4 hr and upstream time = 21/(5 - 2) = 7 hr. Their difference 7 - 4 = 3 hr matches the condition exactly, confirming the still-water speed is 5 km/hr.