10.1. Two athletes with constant speeds run on an oval track of a sports ground, the first of them runs the track completely 5 seconds faster than the second. If they run on the track from the same starting point in the same direction, they will meet again for the first time after 30 seconds. How many seconds will it take for them to meet again for the first time if they run on the track from the same starting point in opposite directions
Answer. In 6 seconds.

Solution. Let the length of the track be $S$ meters, and the speeds of the first and second runners be $x$ and $y$ meters per second, respectively. From the first condition: $\frac{S}{x} + 5 = \frac{S}{y}$, and from the second condition $\frac{S}{x-y} = 30$, since in this case the first runner catches up with the second runner who is initially $S$ meters behind. From the second equation, we express $S = 30(x - y)$, substitute it into the first equation, make the substitution $t = \frac{x}{y} > 1$, and obtain the equation $t^2 - \frac{13}{6} t + 1 = 0$, from which $t = \frac{3}{2} = \frac{x}{y}$. Express $y = \frac{2}{3} x$, substitute it into the second equation, we have $\frac{S}{x} = 10$, that is, $S = 10 x$.

If the runners run in opposite directions, their speeds will add up, and they will meet after covering the track in $\frac{S}{x+y} = \frac{S}{\frac{5}{3} x} = \frac{3}{5} \frac{S}{x} = 6$ seconds.

Grading criteria. Correct formulation of the system of equations: 3 points.