An overhead tank, which supplies water to a settlement, is filled by three bore wells. First two bore wells operating together fill the tank in the same time as taken by third bore well to fill it. The second bore well fills the tank 10 hours faster then the first one and 8 hours slower than the third one. The time required by the third bore well to fill the tank alone is:

Let the time taken by second bore well be t hours, then the time taken by first bore well will be t + 10 and the time taken by the third bore well will be t - 8. Now since first two bore wells operating together take same time as taken by the third bore well we can say that work done in unit time will be same in both cases :

$$\frac{1}{t+10}+\frac{1}{t}=\frac{1}{t-8}$$

$$\frac{(t)+(t+10)}{(t)(t+10)}=\frac{1}{t-8}$$

$$(2t+10)(t-8)=(t)(t+10)$$

$$t^{2}-16t-80=0$$

$$(t-20)(t+4)=0$$

Since t $$\neq$$ 4, hence t = 20

So the time taken by the third bore well will be t - 8 = 20 - 8 =12 hours