Two pipes can fill a tank in 12 hours and 16 hours respectively. A third pipe can empty the tank in 30 hours. If all three pipes are opened and functions simultaneously, how much time will the tank take to be full?( in hours )

Solution (By Examveda Team)

First pipe fill the tank in 1 hour = $$\frac{1}{{12}}$$ part of tank
Second pipe fill the tank in 1 hour = $$\frac{1}{{16}}$$ part of tank
Third pipe empty the tank in 1 hour = $$\frac{1}{{30}}$$ part of tank
When all three pipes are opened simultaneously, part of the tank filled in 1 hour
$$ = \frac{1}{{12}} + \frac{1}{{16}} - \frac{1}{{30}}$$
LCM of 12, 16 and 30 = 240
$$\eqalign{ & {\text{ = }}\frac{{20 + 15 - 8}}{{240}} \cr & = \frac{{27}}{{240}} \cr} $$
∴ Required time taken by all the three pipes
$${\text{ = }}\frac{{240}}{{27}} = \frac{{80}}{9} = 8\frac{8}{9}\,{\text{Hours}}$$