Pipe A can fill a tank in 6 hours and pipe B can fill it in 4 hours. If they are opened on alternate hours and if pipe A is opened first, in how many hours, the tank shall be full ?

A's work in 1 hour = $$\frac{1}{6}$$

B's work in 1 hour = $$\frac{1}{4}$$

(A+B)'s 2 hour's work when opened alternatively = $$\frac{1}{6}$$+$$\frac{1}{4}$$ = $$\frac{5}{12}$$

Now, (A+B)'s 4 hour's work when opened alternatively = $$\frac{5}{12}$$*2 = $$\frac{5}{6}$$

=> Remaining part = (1 - $$\frac{5}{6}$$) = $$\frac{1}{6}$$

Now, it is A's turn and it fills $$\frac{1}{6}$$ part in 1 hour.

$$\therefore$$ Total time taken to fill the tank = (4+1) = 5 hours