Two pipes X and Y can fill a cistern in 6 hours and 10 hours respectively. Pipe Z can empty the tank in 4 hours. If all the three pipes are open then in how many hours the cistern will be full?

Let total capacity of cistern is L.C.M. (6,10,4) = 60 units

X alone can fill in 6 hours, => X's efficiency = $$\frac{60}{6}=10$$ units/hr

Similarly Y's efficiency = $$\frac{60}{10}=6$$ units/hr

and Z's efficiency = $$\frac{60}{-4}=-15$$ units/hr

Now, (X+Y+Z)'s 1 hour's work = $$10+6-15=1$$ unit/hr

$$\therefore$$ Time required by X,Y and Z together to fill the cistern = $$\frac{60}{1}=60$$ hours

=> Ans - (D)