Two pipes A and B can fill a tank in 6 hours and 4 hours respectively. If they are opened on alternate hours and if pipe A is opened first, then the tank shall be fill in

Let total capacity of tank is L.C.M. (6,4) = 12 units

A alone can fill the tank in 6 hours, => A's efficiency = $$\frac{12}{6}=2$$ units/hr

Similarly B's efficiency = $$\frac{12}{4}=3$$ units/hr

Now, (A+B)'s 2 hour's work (A in first hour and B in second) = $$2+3=5$$ units

=> In 4 hours, tank filled = $$5\times2=10$$ units

=> Remaining part = $$12-10=2$$ units, which will be filled by A in the next complete hour.

$$\therefore$$ Total time = $$4+1=5$$ hours

=> Ans - (B)