Pipe $A$ can fill a tank in $3$ hours and pipe $B$ can fill it in $4$ hours. If they are opened at alternate hours and if pipe $A$ is opened first, in how many hours will the tank be filled?

Let the total capacity of the tank be the LCM of $3$ and $4$, which is $12$ units.
* Rate of pipe $A=\frac{12}{3}=4$ units/hour
* Rate of pipe $B=\frac{12}{4}=3$ units/hour

The pipes operate alternately starting with $A$:

* Hour 1 (Pipe $A$): $4$ units filled
* Hour 2 (Pipe $B$): $3$ units filled

Total work completed in one $2$ -hour cycle:

$$\text{Work in }2\text{ hours}=4+3=7\text{ units}$$

Remaining work after the first $2$ -hour cycle:

$$\text{Remaining Work}=12-7=5\text{ units}$$

In the next hour (Hour 3), Pipe $A$ turns on and works at its full capacity:

* Hour 3 (Pipe $A$): $4$ units filled

$$\text{Remaining Work after hour 3}=5-4=1\text{ unit}$$

In the fourth hour, Pipe $B$ turns on. It fills $3$ units per hour, so the time needed to complete the final $1$ unit is:

$$\text{Time}=\frac{\text{Remaining Work}}{\text{Rate of B}}=\frac{1}{3}\text{ hour}$$

Total time taken to completely fill the tank:

$$\text{Total Time}=3+\frac{1}{3}=3\frac{1}{3}\text{ hours}$$