Two motorboats A and B, start simultaneously from the two ends P and Q of a river in downstream and upstream respectively. The speed of the motorboat A in still water is 200% more than that of motorboat B. If the distance between P and Q is 120 Km and they meet each other after 5 hours then how long the motorboat A will take to travel 90 km in still water?

Let motorboat B's speed in still water be v km/h. Then motorboat A's speed = v + 200% of v = 3v km/h. Let river current speed be c km/h. Downstream speed of A = 3v + c. Upstream speed of B = v - c. When they meet after 5 hours: (3v + c) × 5 + (v - c) × 5 = 120. Solving: 15v + 5c + 5v - 5c = 120, so 20v = 120, v = 6 km/h. Motorboat A's speed in still water = 3v = 18 km/h. Time for 90 km = 90/18 = 5 hours.