Given taps X and Y can fill the tank in 30 and 40 minutes respectively. Therefore,

part filled by tap X in 1 minute = 1/30

part filled by tap Y in 1 minute = 1/40

Tap Z can empty the tank in 60 minutes. Therefore,

part emptied by tap Z in 1 minute = 1/60

Net part filled by Pipes X,Y,Z together in 1 minute = [1/30 +1/40 - 1/60] = 5/120 = 1/24

i.e., the tank can be filled in 24 minutes.

Let filling capacity of the pump be 'F' m 3 / min

Then, the filling capacity of the pump will be (F + 10) m 3 / min

From the given data,

$$\begin{gathered} 2400 F - 2400 F + 10 = 8 \\ ? 2400 F + 10 - F F ( F + 10 ) = 8 \\ ? F 2 + 10 F - 3000 = 0 \\ ? F = 50 m 3 / min \end{gathered}$$

Let the pipe B be closed after 'K' minutes.

30/16 - K/24 = 1 => K/24 = 30/16 - 1 = 14/16

=> K = 14/16 x 24 = 21 min.

Let the efficiencies of filling pipes is 4p and 5p respectively.

Efficiency of pipe which empty the tank = 2/3 x 9p/2 = 3p

Total work = 3p x 36 = 108p

Time to fill the tank by both the pipes = 108p/9p = 12 min.

Time taken by one tap to fill half of the tank = 3 hrs.

Part filled by the taps in 1 hour = 4 x 1/6 = 2/3

Remaining part = 1 - 1/2 = 1/2

2/3 : 1/2 :: 1 : p

p = 1/2 x 1 x 3/2 = 3/4 hrs. i.e., 45 min

So, total time taken = 3 hrs 45 min.

Let pipe A takes p min to fill

Then,

pipe B takes 3p min to fill

=> 3p - p = 32

=> p = 16 min => 3p = 48 min

Required, both pipes to fill = (48 x 16)/(48 + 16) min = 12 min.

The time taken by the leak to empty the tank =  1 8 - 1 10 = 5 - 4 40 = 1 40

Therefore, the leak empties the tank in 40 hours.

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.

1/k - 1/20 = -1/24

k = 120

120 x 4 = 480

Therefore, the capacity of the cistern is 480 liters.

Given that the waste tap can empty the filled tank in 32 min.

Now, the rate at which the waste tap can empty the tank = (40 + 60)8/32 = 100/4 = 25 lit/min.