Two pipes fill a tank when working individually in 25 and 40 hours, respectively while a third pipe can drain the filled tank in 16 hours. If all the three pipes are turned on at the same time when the tank is empty, how long will it take to fill the tank completely?

Let Pipes are A, B, C

Pipes A and B can fill the tank in 25 and 40 hours respectively. Therefore,
part filled by pipe A in 1 hour 
part filled by pipe B in 1 hour $$= \frac{1}{40} $$

Pipe C can empty the tank in 16 hours. Therefore,
part emptied by pipe C in 1 hour 

Net part filled by Pipes A,B,C together in 1 hour

$$= \frac{1}{25} +\frac{1}{40} - \frac{1}{16} $$

$$= \frac{16+10-25}{400} $$

$$= \frac{1}{400} $$

Required time $=400hour$s

As 1 day is = 24 hours.

So, we will get $$\frac {400}{24} = 16 days 16 hours $$