13.330. Two trucks were supposed to transport a certain cargo in 6 hours. The second truck was delayed in the garage, and when it arrived at the loading site, the first truck had already transported $3 / 5$ of the total cargo; the remaining part of the cargo was transported by the second truck, and the entire cargo was transported in this way in 12 hours. How much time would each truck need separately to transport the cargo?
Solution.

Let's take the entire volume of work as 1. Let $x, y$ be the weight of cargo transported by each machine in one trip. According to the conditions, we have the system

$$
\left\{\begin{array}{l}
\frac{1}{x+y}=6, \\
\frac{3}{5 x}+\frac{2}{5 y}=12,
\end{array} \text { from which } x=\frac{1}{10}, y=\frac{1}{15} \text { or } x=y=\frac{1}{12} .\right. \text { Then the first ma- }
$$

chine will transport the entire cargo in 10 hours, and the second in 15 hours, or each in 12 hours.

Answer: 10 and 15 hours or 12 hours each.