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 ?
Let the draining capacities (rates) of taps A, B, and C be R_A, R_B, and R_C respectively.
The ratio of capacities is 1:2:3, so we can write R_A = k, R_B = 2k, and R_C = 3k for some constant k.
When all three taps are on, the total draining rate is R_A + R_B + R_C = k + 2k + 3k = 6k.
It takes 1 hour (60 minutes) to drain the full tank.
So, V = (Total Rate) * Time = (6k) * 60 = 360k.
When taps A and C are on, and B is off, the total draining rate is R_A + R_C = k + 3k = 4k.
Let T be the time in minutes it takes to empty the full tank with A and C on.
The volume drained is V.
V = (Rate A+C) * T = (4k) * T.
We know V = 360k.
So, 360k = 4k * T.
Divide both sides by 4k (assuming k > 0):
T = 360k / 4k = 360 / 4 = 90 minutes.