Two pipes A and can fill an empty cistern in 18 and 27 hours, respectively. Pipe C can drain the entire cistern in 45 hours when no other pipe is in operation. Initially, when the cistern was empty Pipe A and Pipe C were turned on. After a few hours Pipe A was turned off and Pipe B was turned on instantly. In all, it took 55 hours to fill the cistern. For how many hours was Pipe B turned on?

Pipe a can fill in 1 hr = 1/18

Pipe b can fill in 1 hr = 1/27

Pipe c can drain in 1 hr = 1/45

Let say for x hrs Pipe a & c were turned on

x(1/18 - 1/45)

=(x/90)(5-2)

= x/30

Now pipe a turned off & pipe b turned on

remaining hrs = 55-x

(55-x) ( 1/27 - 1/45)

= (55-x) (1/135)(5 -3)

= (55-x)2/135

x/30  + (55-x)2/135 = 1

=>(1/270) ( 9x  + 220 - 4x) =  1

=> 5x + 220 = 270

=> 5x = 50

=> x = 10

pipe b turned for 55 - 10 = 45 hrs