An empty tank is connected to two inlet pipes. The first fills it in $12\text{ hours}$ and the second in $20\text{ hours}$. If the second pipe is opened $4\text{ hours}$ after the first, how long will it take to fill the tank completely from the very beginning?

Let the total capacity of the tank be the LCM of $12$ and $20=60\text{ units}$.
* Efficiency of the first pipe ($E_1$) = $\frac{60}{12}=5\text{ units/hour}$
* Efficiency of the second pipe ($E_2$) = $\frac{60}{20}=3\text{ units/hour}$

For the first $4\text{ hours}$, only the first pipe is open:

$$\text{Work done in first 4 hours}=5\text{ units/hour}\times 4\text{ hours}=20\text{ units}$$

$$\text{Remaining Work}=60-20=40\text{ units}$$

Now, the second pipe is also opened. Combined efficiency = $E_1+E_2=5+3=8\text{ units/hour}$.

$$\text{Time taken to fill the remaining part}=\frac{40}{8}=5\text{ hours}$$

$$\text{Total time taken from the very beginning}=4\text{ hours (initial)}+5\text{ hours (together)}=9\text{ hours}$$

Therefore, it takes $9\text{ hours}$ to fill the tank completely.