Question

two taps can fill a cistern in 10 hours and 8 hours respectively. A third tap can empty it in 15 hours. How long will it take to fill the empty cistern, if all of them are opened together?

Answer 1

It is given that

The first tap can fill a cistern in 10 hours.

The second tap can fill a cistern in 8 hours.

The third tap can empty the cistern in 15 hours.

First tap's one hour work = 1/10

second's tap's one hour work = 1/8

Third tap's one hour work = 1/15

Here if all the tap are opened together then their one hour work $=1/10+1/8-1/15$

Taking LCM $=\frac{(12+5-8)}{120}$

So we get $=\frac{(27-8)}{120}=\frac{19}{120}$

All the taps can fill the cistern in $=120/19$ hours $=6\frac{6}{19}$ hours