13.291. If the execution of an order for a set of several books is entrusted to one of three typesetters, then the first will complete the work 10 hours faster, and the third will complete it 6 hours faster than the second. If one of the ordered books is set by the first typesetter, and another book is set simultaneously by the second, then in 9 hours they will set as many pages as the second and third, working together, will set in 10 hours under the same conditions. How much time will each typesetter need to set all the ordered books when working separately?
## Solution.

Let's assume the entire volume of work is 1. Let $\alpha, \beta, \gamma$ be the work rates (number of pages typed per hour) of the first, second, and third typists, respectively.

According to the problem, we have the system

$$
\left\{\begin{array}{l}
\frac{1}{\beta}-\frac{1}{\alpha}=10, \\
\frac{1}{\beta}-\frac{1}{\gamma}=6, \\
9(\alpha+\beta)=10(\beta+\gamma)
\end{array} \quad \text { from which } \alpha=\frac{1}{20}, \beta=\frac{1}{30}, \gamma=\frac{1}{24}\right.
$$

Then the first typist will complete the entire work in $\frac{1}{\alpha}=20$ (hours), the second $-3a \frac{1}{R}=30$ (hours), and the third $-3a \frac{1}{\gamma}=24$ (hours).

Answer: 20, 30, and 24 hours.