A work is done by $3\text{ males}$ in $16\text{ days}$ and the same work is done by $6\text{ females}$ in $16\text{ days}$. If $12\text{ males}$ and $8\text{ females}$ work together, then they will complete the work in how many days?

Let the efficiency of $1\text{ male}$ be $M$ and $1\text{ female}$ be $F$.
According to the problem, the total work is equal in both cases:

$$\text{Total Work}=3M\times 16=6F\times 16$$

$$3M=6F⟹1M=2F$$

This means the work done by $1\text{ male}$ is equal to the work done by $2\text{ females}$.
Let's express the total work in terms of female days:

$$\text{Total Work}=6\times 16=96\text{ female-days}$$

Now, let's find the combined efficiency of $12\text{ males}$ and $8\text{ females}$ in terms of females:

$$12M+8F=12(2F)+8F=24F+8F=32F$$

Let the number of days required by them together be $D$.

$$32F\times D=96F$$

$$D=\frac{96}{32}=3\text{ days}$$