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 C to empty the cistern = 10 hours
Now,
Tap A fills \(\frac{1}{12}\text{th}\) part of the cistern in 1 hour.
Tap B fills \(\frac{1}{15}\text{th}\) part of the cistern in 1 hour.
Tap C empties out \(\frac{1}{10}\text{th}\) part of the cistern in 1 hour.
Thus, in one hour, \(\Big(\frac{1}{12}+\frac{1}{15}-\frac{1}{10}\Big)\text{th}\) 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, \(\frac{1}{20}\text{th}\) part of the cistern is filled.
Hence, the cistern will be filled completely in 20 hours if all the three taps are opened together.
