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 )

Answer: Option C

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
$$\begin{array}{rl}& \text{ = }\frac{20+15-8}{240}\\ & =\frac{27}{240}\end{array}$$
∴ Required time taken by all the three pipes
$$\text{ = }\frac{240}{27}=\frac{80}{9}=8\frac{8}{9}\text{Hours}$$