3.21 A motorcyclist set off from point $A$ to point $B$, which is 120 km away from $A$. On the way back, he set off at the same speed, but had to stop for 10 minutes one hour after departure. After this stop, he continued his journey to $A$, increasing his speed by 6 km/h. What was the initial speed of the motorcyclist, given that he spent as much time on the return trip as he did on the trip from $A$ to $B$?
3.21 According to the condition, $A C=C D$ (Fig. 3.6).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-100.jpg?height=482&width=792&top_left_y=1151&top_left_x=196)

Fig. 3.6

We have $A C=\frac{120}{x}$, where $x$ is the initial speed;

$C D=1+\frac{1}{6}+\frac{120-x}{x+6}$. Therefore,

$$
\frac{120}{x}=\frac{7}{6}+\frac{120-x}{x+6}, \text { from which } x=48 \text { (km/h). }
$$

Answer: $48 \mathrm{~km} / \mathrm{h}$.