C can do a piece of work in $24\text{ days}$ and D in $16\text{ days}$. They began the work together, but D left the work $4\text{ days}$ before its completion. In how many days was the work completed?

Let the total work be the least common multiple (LCM) of $24$ and $16$.
$$\text{Total Work}=\text{LCM}(24,16)=48\text{ units}$$

Now, find the individual efficiencies per day:

$$\text{Efficiency of C}=\frac{48}{24}=2\text{ units/day}$$

$$\text{Efficiency of D}=\frac{48}{16}=3\text{ units/day}$$

Let the total time taken to complete the work be $x\text{ days}$.
Since D left $4\text{ days}$ before completion, D worked for $(x-4)\text{ days}$, while C worked for the entire $x\text{ days}$.

According to the question:

$$(\text{Efficiency of C}\times x)+(\text{Efficiency of D}\times (x-4))=\text{Total Work}$$

$$2x+3(x-4)=48$$

$$2x+3x-12=48$$

$$5x=60$$

$$x=12\text{ days}$$

Thus, the work was completed in $12\text{ days}$.