A motorboat whose speed is 20 km/h in still water takes

Solution

According to the question, the motorboat takes 30 minutes more to go 24 km upstream than to cover the same distance downstream. Let, the speed of the water = x km/h So, 24/(20 - x) = 24/(20 + x) + (1/2) [∵ 30 minutes = 1/2 hour] ⇒ 24/(20 - x) - 24/(20 + x) = (1/2) => {24(20+x) - 24(20-x)}/(400 - x²) = 1/2 => 24(20+x-20+x)/400 - x² = 1/2 => (24 x 2x)/400-x² = 1/2 ⇒ 400 - x² = 96x ⇒ x² + 96x - 400 = 0 ⇒ x² + 100x - 4x - 400 = 0 ⇒ x (x + 100) - 4 (x + 100) = 0 ⇒ (x + 100) (x - 4) = 0 ⇒ x + 100 = 0 ⇒ x = -100 ["-" is neglacted] ⇒ x - 4 = 0 ⇒ x = 4 ∴ The speed of the water = 4 km/h The speed of the motorboat in still water increased 2 km/h = 20 + 2 = 22 km/h The time for 39 km downstream and 30 km upstream = 39/(22 + 4) + 30/(22 - 4) hours = (39/26) + (30/18) hours = 3/2 + 5/3 hours = 19/6 hours = (19/6) × 60 minutes = 190 minutes = 3 hours 10 minutes