Three taps I J and K can fill a tank in 20 30and 40 minutes respectively All the taps are opened sim...........

Quant (SSC) Pipes&cistern

Three taps I, J and K can fill a tank in 20,30and 40 minutes respectively. All the taps are opened simultaneously and after 5 minutes tap A was closed and then after 6 minutes tab B was closed .At the moment a leak developed which can empty the full tank in 70 minutes. What is the total time taken for the completely full ?

24.315 minutes

26.166 minutes

22.154 minutes

24 minutes

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Explanation:

Upto first 5 minutes I, J and K will fill => 5[(1/20)+(1/30)+(1/40)] = 65/120
For next 6 minutes, J and K will fill => 6[(1/30)+(1/40)] = 42/120
So tank filled upto first 11 minutes = (65/120) + (42/120) = 107/120
So remaining tank = 13/120
Now at the moment filling with C and leakage @ 1/60 per minute= (1/40) - (1/70) = 3/280.
So time taken to fill remaining 13/120 tank =(13/120) /(3/280) = 91/6 minutes
Hence total time taken to completely fill the tank = 5 + 6 + 91/6 = 26.16 minutes.