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 )

110⁴/₉
29¹/₂
38⁸/₉
47²/₉

Answer:- 1

Explanation:-

Solution:

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

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