## Task B-1.6.

Luka and Pero decided to visit a friend and set off from place $A$ to place $B$, which is $170 \mathrm{~km}$ away. Luka started walking, while Pero was driven by a motorcyclist. After some time, the motorcyclist returned to pick up Luka, and Pero continued on foot, so that both arrived at place $B$ at the same time. Both pedestrians walk at the same speed of $5 \mathrm{~km} / \mathrm{h}$, and the motorcyclist drives at a speed of $85 \mathrm{~km} / \mathrm{h}$. At what distance from place $B$ and after how much time did the motorcyclist stop and return to pick up Luka? How long did it take for Luka and Pero to travel from place $A$ to place $B$?
## Solution.

Since both arrived at the same time traveling at the same speed, it means they walked the same part of the path on foot and the same part on the motor.

![](https://cdn.mathpix.com/cropped/2024_05_30_8cadc17e7912dc7108beg-04.jpg?height=69&width=846&top_left_y=948&top_left_x=565)

Let's introduce some notations: $s=$ distance $|A B|$ (from place $A$ to place $B)$.

$|A D|=|C B|=x$ is the part of the path that Luka, respectively Pero, walked on foot.

Then $|A C|=|B D|=s-x,|C D|=s-2 x$.

While Luka was walking the distance $|A D|$, the motorcyclist was traveling the distance

$$
|A C|+|C D|=s-x+s-2 x=2 s-3 x
$$

From this, we have

$$
\frac{x}{5}=\frac{2 s-3 x}{85} \Rightarrow x=\frac{s}{10} \Rightarrow x=\frac{170}{10}=17 \mathrm{~km}
$$

From $|\overline{A D}|=17 \mathrm{~km}$ it follows that $|\overline{A C}|=s-x=153 \mathrm{~km}$.

The time the motorcyclist drove Pero is $t_{A C}=\frac{s-x}{85}=\frac{153}{85}=1 \mathrm{~h}$ and $48 \mathrm{~min}$.

The time Luka or Pero walked is $t_{C B}=t_{A D}=\frac{x}{5}=\frac{17}{5}=3 \mathrm{~h}$ and $24 \mathrm{~min}$. 1 point

The time of arrival from place $A$ to place $B$ is $t_{A C}+t_{C B}=\frac{153}{85}+\frac{17}{5}=5 \mathrm{~h}$ and $12 \mathrm{~min}$.