13.294. Two workers were assigned to manufacture a batch of identical parts; after the first worked for $a$ hours, and the second for $0.6 a$ hours, it turned out that they had completed $5 / n$ of the entire work. After working together for another $0.6 a$ hours, they found that they still had to manufacture $1 / n$ of the entire batch of parts. How many hours will each of them, working separately, need to complete the entire work? The number $n$ is a natural number; find it.
## Solution.

Let the entire volume of work be 1. Let $x, y$ be the labor productivity (the number of parts produced per hour) of the first and second workers, respectively. According to the conditions, we have the system

$$
\left\{\begin{array}{l}
a x+0.6 a y=\frac{5}{n} \\
a x+0.6 a y+0.6 a(x+y)=1-\frac{1}{n}
\end{array} \text {, from which } x=\frac{11-n}{0.4 a n}, y=\frac{n-9}{0.24 a n} .\right.
$$

Then the first worker will manufacture all the parts in $\frac{1}{x}=\frac{0.4 a n}{11-n}$ (hours), and the second worker will do so in $\frac{1}{y}=\frac{0.24 a n}{n-9}$ (hours). The value of $n$ must satisfy the system of inequalities

$$
\left\{\begin{array}{l}
11-n>0, \\
n-9>0,
\end{array} \text { i.e., } n=10\right.
$$

Answer: $\frac{0.4 a n}{11-n} ; \frac{0.24 a n}{n-9} ; n=10$.