1. Two athletes are running around a track, each at a constant speed. If they run in opposite directions, they meet every 10 minutes; if they run in the same direction, they meet every 40 minutes. How long does it take the faster athlete to run the circuit?
1. Let the speeds of the runners be $v_{1}$ and $v_{2}$ such that $v_{1}>v_{2}$ (speeds are given in laps per minute). Suppose the athletes start from the same point but in opposite directions. At the moment of their next meeting after 10 minutes, the sum of the lengths of the two segments they have run will exactly correspond to the length of one lap, i.e., $10 v_{1}+10 v_{2}=1$.

If the athletes start from the same point in the same direction, the next meeting will occur when the faster athlete has run one lap more than the slower one. Therefore, $40 v_{1}-40 v_{2}=1$.

We have obtained a system of two linear equations with unknowns $v_{1}, v_{2}$:

$$
\begin{aligned}
& 10 v_{1}+10 v_{2}=1 \\
& 40 v_{1}-40 v_{2}=1
\end{aligned}
$$

We can solve this system, for example, by adding the second equation to four times the first equation, which gives $80 v_{1}=5$ or $v_{1}=\frac{1}{16}$. We are interested in how long it takes the faster runner to complete one lap, i.e., the value of the ratio $1 / v_{1}$. By substituting the calculated value of $v_{1}$, we get the answer: 16 minutes.

Note. The problem can also be solved by reasoning: in 40 minutes, the athletes together run 4 laps (which follows from the first condition), while the faster one runs 1 lap more than the slower one (which follows from the second condition). This means that the first one runs 2.5 laps and the second one 1.5 laps in the given time, so the faster one completes one lap in 40/2.5, or 16 minutes.

For a complete solution, award 6 points, with 2 points for each equation and 2 points for calculating the required value.