Suppose A, B and C are three taps fixed to the bottom of a tank with draining capacity 1 : 2 : 3. When all three of them are on, it takes 1 hour to drain out the full tank. If A and C are on but B is off, then how much time, in minutes, will it take to empty out a full tank of water? (a) 75 (b) 90 (c) 105 (d) 120

The problem involves calculating the time to empty a tank based on the relative flow rates of three taps. The draining capacities of taps A, B, and C are given in the ratio 1: 2: 3. Let the rates be 1k, 2k, and 3k respectively. When all three taps are open, their combined rate is 1k + 2k + 3k = 6k. Since they empty the tank in 1 hour (60 minutes), the total capacity of the tank can be represented as 6k * 60 = 360k. If only taps A and C are open, their combined draining rate is 1k + 3k = 4k. To find the time required for A and C to empty the full tank, we divide the total capacity by their combined rate: 360k / 4k = 90 minutes. Thus, it will take 90 minutes to empty the tank.