13.016. The numerators of three fractions are proportional to the numbers 1, 2, and 3, while the reciprocals of the corresponding denominators are proportional to the numbers $1, 1 / 3$, and 0.2. Find these fractions, given that their arithmetic mean is $136 / 315$.
Solution.

Let's compile the following table:

| Fraction | Numerator | Denominator |
| :--- | :---: | :---: |
| First | $x$ | $1 / y$ |
| Second | $2 x$ | $3 / y$ |
| Third | $3 x$ | $1 / 0.2 y$ |

First fraction $-x y$; second $-\frac{2}{3} x y$; third $-0.6 x y$. The arithmetic mean of the fractions $\frac{x y+\frac{2}{3} x y+0.6 x y}{3}=\frac{136}{315}$, from which $x y=\frac{136}{238}=\frac{4}{7}$. Therefore, the first fraction is $-\frac{4}{7}$; the second is $-\frac{8}{21}$; the third is $-\frac{12}{35}$.

Answer: $\frac{4}{7} ; \frac{8}{21} ; \frac{12}{35}$.