[Algebra]
Problem: Qing is twice as old as Rayna. Qing is 4 years younger than Paolo. The average age of Paolo, Qing and Rayna is 13. Determine their ages.

Solution: Suppose that Rayna's age is $x$ years.

Since Qing is twice as old as Rayna, Qing's age is $2 x$ years.

Since Qing is 4 years younger than Paolo, Paolo's age is $2 x+4$ years.

Since the average of their ages is 13 years, we obtain

$$
\frac{x+(2 x)+(2 x+4)}{3}=13
$$

This gives $5 x+4=39$ and so $5 x=35$ or $x=7$.

Therefore, Rayna is 7 years old, Qing is 14 years old, and Paolo is 18 years old.

(Checking, the average of 7,14 and 18 is $\frac{7+14+18}{3}=\frac{39}{3}=13$.)

Final answer: 7,14,18