Eight years ago, the ratio of the ages of $A$ and $B$ was $9:10$. The ratio of their ages $4\text{ years}$ from now will be $12:13$. If the age of $C$ is $4\text{ years}$ less than the present age of $A$, then what is the present age (in years) of $C$?

Let the ages of $A$ and $B$ eight years ago be $9k$ and $10k$, respectively.Their present ages would be $9k+8$ and $10k+8$.Their ages $4\text{ years}$ from now will be:
$$\text{Age of }A=9k+8+4=9k+12$$
$$\text{Age of }B=10k+8+4=10k+12$$
The ratio of their ages $4\text{ years}$ from now is given as $12:13$:
$$\frac{9k+12}{10k+12}=\frac{12}{13}$$
$$13(9k+12)=12(10k+12)$$
$$117k+156=120k+144$$
$$156-144=120k-117k$$
$$12=3k⟹k=4$$
Now, find the present age of $A$:
$$\text{Present age of }A=9(4)+8=36+8=44\text{ years}$$
The age of $C$ is $4\text{ years}$ less than the present age of $A$:
$$\text{Present age of }C=44-4=40\text{ years}$$