The ratio of the time taken by a boat to cover D km downstream to the time taken to cover (D/2 + 28) km upstream is 1 : 6. The ratio of the downstream speed of the boat to its speed in still water is 5 : 3. If the difference between the time taken to cover (D − 15) km upstream and the same distance downstream is 5 hours, find the distance covered by the boat in still water in 6 hours.

ATQ, Let the downstream speed of the boat be 5a. Let the speed of the boat in still water be 3a. Speed of the current = 5a − 3a = 2a Upstream speed = 3a − 2a = a Time taken to cover D km downstream = D/5a Time taken to cover (D/2 + 28) km upstream = (D/2 + 28)/a Given ratio of these times = 1 : 6 So, D/5a : (D/2 + 28)/a = 1 : 6 D/5a ÷ (D/2 + 28)/a = 1/6 D/[5(D/2 + 28)] = 1/6 6D = 5(D/2 + 28) 6D = 5D/2 + 140 Multiply both sides by 2. 12D = 5D + 280 12D − 5D = 280 7D = 280 D = 40 Now, D − 15 = 40 − 15 = 25 km Time taken to cover 25 km upstream = 25/a Time taken to cover 25 km downstream = 25/5a Given difference = 5 hours 25/a − 25/5a = 5 25/a − 5/a = 5 20/a = 5 a = 20/5 = 4 Speed in still water = 3a = 3 × 4 = 12 km/hr Distance covered in still water in 6 hours = 12 × 6 = 72 km