13.317. One tourist set out at 6 a.m., and the second - towards him at 7 a.m. They met at 8 a.m. and, without stopping, continued their journey. How much time did each of them spend on the entire journey, if the first arrived at the place where the second started 28 minutes later than the second arrived at the place where the first started? It is assumed that each walked without stopping at a constant speed.
## Solution.

Let $V_{1}, V_{2}$ be the speeds of the first and second tourists, respectively, and $x$ be the time it takes for the second tourist to cover the distance that the first tourist covers in 2 hours. According to the problem, we have the system $\left\{\begin{array}{l}x V_{2}=120 V_{1}, \\ V_{2} x+60 V_{2}=120 V_{1}+V_{1}(x+28) .\end{array}\right.$ From the first equation, we express $\frac{V_{1}}{V_{2}}$ and substitute it into the second equation. From the resulting equation, we find $x=72$ (min). Then the first tourist spent a total of $120+72+28=220$ (min) on the entire journey, and the second tourist spent $-60+72=132$ (min).

Answer: 3 hours 40 minutes and 2 hours 12 minutes.