1. The infantry column stretched out to 1 km. Sergeant Kim, riding out on a gyro-scooter from the end of the column, reached its beginning and returned to the end. The infantrymen walked $4 / 3$ km during this time. How far did the sergeant travel during this time?
Answer: $8 / 3$ km

Solution. Let the speed of the column be $x$ km/h, and the corporal rides $k$ times faster, i.e., at a speed of $k x$ km/h. Kim rode to the end of the column for $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction for $t_{2}=\frac{1}{k x+x}$ hours (meeting head-on). During this time, the column covered $4 / 3$ km, i.e., $x\left(t_{1}+t_{2}\right)=4 / 3$. Substituting the expressions for $t_{1}$ and $t_{2}$, we get

$$
\frac{x}{k x-x}+\frac{x}{k x+x}=4 / 3 ; \quad \frac{1}{k-1}+\frac{1}{k+1}=4 / 3 ; \quad 2 k=\frac{4}{3}\left(k^{2}-1\right)
$$

The resulting quadratic equation has a single positive root $k=2$. The corporal rides twice as fast as the column. Therefore, the distance he covers will also be twice as much.

Evaluation. Full solution: 12 points.