**a)** Pipe C empties 1 tank in 20 h

⇒ 2/5 th tank in 2/5 x 20 = 8 hours

**b)** Part of tank filled in 1 hour = \(\frac 1{15} + \frac 1{12} - \frac1{20} = \frac 1{10}th\)

⇒ time taken to fill tank completely = 10 hours

**c) **At 5 am,

Let the tank be completely filled in ‘t’ hours

⇒ pipe A is opened for ‘t’ hours

pipe B is opened for ‘t−3’ hours

And, pipe C is opened for ‘t−4’ hours

⇒ In one hour,

part of tank filled by pipe A = t/15 th

part of tank filled by pipe B = t−3/15 th

and, part of tank emptied by pipe C = t−4/15 th

Therefore \(\frac t{15} + \frac{t-3}{12} - \frac{t-4}{20} =1\)

⇒ t = 10.5

Total time to fill the tank = 10 hours 30 minutes

**OR**

6 am, pipe C is opened to empty 1/2 filled tank

Time to empty = 10 hours

Time for cleaning = 1 hour

Part of tank filled by pipes A and B in 1 hour = \(\frac1{15} + \frac 1{12} = \frac 3{20}\)th tank

⇒ time taken to fill the tank completely = 20/3 hours

Total time taken in the process = 10 + 1 + 20/3 = 17 hour 40 minutes