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