1. The infantry column stretched out over 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 2 km 400 m during this time. How far did the sergeant travel during this time?
Answer: 3 km $600 \mathrm{~m}$

Solution. Let the speed of the column be $x$ km/h, and the sergeant travels $k$ times faster, i.e., at a speed of $k x$ km/h. To reach the end of the column, Kim traveled $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction, $t_{2}=\frac{1}{k x+x}$ hours (meeting head-on). During this time, the column covered 2.4 km, i.e., $x\left(t_{1}+t_{2}\right)=2.4$. Substituting the expressions for $t_{1}$ and $t_{2}$, we get

$$
\frac{x}{k x-x}+\frac{x}{k x+x}=2.4 ; \quad \frac{1}{k-1}+\frac{1}{k+1}=2.4 ; \quad 2 k=2.4\left(k^{2}-1\right) .
$$

The obtained quadratic equation has a single positive root $k=\frac{3}{2}$. The sergeant travels 1.5 times faster than the column. Therefore, the distance he covers will also be 1.5 times greater.

Evaluation. Full solution: 12 points.