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

Answer: Option B

A → 4 hours
B → 6 hours

According to question,
⇒ For the first hour tap A is opened and B for second hour
⇒ Work done by both in 2 hours
$$\begin{array}{rl}& \to \left(3\text{lit/h}+2\text{lit/h}\right)\times 2=10\text{units}\\ & \underset{\text{4 hours}}{\overset{\text{2 hours}}{|\times 2}}\underset{\text{10 litres}}{\overset{\text{5 liters}}{|\times 2}}\end{array}$$
⇒ Remaining part
= 12 - 10 = 2 liters
⇒ Again 5th hour A will be opened Tap A will fill the 2 liters water with its efficiency = $$\frac{2}{3}$$ hours
⇒ Therefore tank will be filled in
= $$\left(4+\frac{2}{3}\right)$$  hours
= $$\text{4}\frac{2}{3}$$ hours