A cistern has two inlets A and B which can fill it in 12 hours and 15 hours respectively. An outlet can empty the full cistern in 10 hours. If all the three pipes are opened together in the empty cistern, how much time will they take to fill the cistern completely?
Time taken by tap A to fill the cistern $=12$ hours
Time taken by tap B to fill the cistern $=15$ hours
Let $C$ be the outlet that can empty the cistern in 10 hours.
Time taken by tap $\mathrm{C}$ to empty the cistern $=10$ hours
Now,
Tap A fills $\frac{1}{12}$ th part of the cistern in 1 hour.
Tap B fills $\frac{1}{15}$ th part of the cistern in 1 hour.
Tap C empties out $\frac{1}{10}$ th part of the cistern in 1 hour.
Thus, in one hour, $\left(\frac{1}{12}+\frac{1}{15}-\frac{1}{10}\right)$ th part of the cistern is filled.
We have:
$\frac{1}{12}+\frac{1}{15}-\frac{1}{10}=\frac{10+8-12}{120}=\frac{6}{120}=\frac{1}{20}$
Thus, in 1 hour, $\frac{1}{20}$ th part of the cistern is filled.
Hence, the cistern will be filled completely in 20 hours if all the three taps are opened together.