Two pipes A and B can fill an empty cistern in 32 and 48 hours, respectively. Pipe C can
drain the entire cistern in 64 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 112 hours to fill the cistern. For how many hours was Pipe B turned on?

A can fill the tank in 32 hours , 

$$\frac{1}{A} = \frac{1}{32}$$

B can fill the tank in 48 hours 

$$\frac{1}{B} = \frac{1}{48}$$

C can drain the tank in 64 hours

$$\frac{1}{C} = \frac{1}{64}$$

According to question ,

$$ x  \times (\frac{1}{32} - \frac{1}{64}) + (112 - x) \times (\frac{1}{48} - \frac{1}{64}) = \frac{1}{112}$$

$$ x \times \frac{1}{64} + (112 - x) \times \frac{1}{192} = \frac{1}{112}$$

$$\frac{1}{x} = \frac{1}{40}$$

Pipe B was turned on for 72 hours.

So, the answer would be Option a)72 hours.