468. Two schoolchildren left home for school at the same time and with the same speed. After 3 minutes, one of them remembered that he had forgotten his notebook at home, and ran back at a speed greater than the initial one by 60 m/min. After picking up the notebook, he ran back at the same speed and caught up with his friend, who was walking at the initial speed, at the school door. The distance from home to school is $400 \mathrm{m}$. Find the initial speed of the schoolchildren.
$\triangle$ Let's denote the initial speed of the schoolchildren as $v$ m/min. Then in 3 minutes, they covered a distance of $3 v \, \text{m}$. Therefore, the schoolchild who ran for the notebook traveled a distance of $(400 + 3 v)$ m at a speed of $(v + 60)$ m/min. The second schoolchild, in the same time, covered a distance of $(400 - 3 v)$ m at a speed of $v$ m/min. Expressing this time in terms of $v$ in two ways, we get the equation:

$$
\frac{400 + 3 v}{v + 60} = \frac{400 - 3 v}{v}
$$

Solving this, we get: $v = 50 \, \text{m/min} = 3 \, \text{km/h}$.

Answer: 3 km/h.