## Task A-1.1.

The path connecting place $A$ with place $B$ is flat in the first part, and the rest is downhill. The cyclist arrived from place $A$ to place $B$ in 1 hour and 15 minutes. On the return trip, it took him half an hour longer. On the flat part of the road, he rode at a speed 4 km/h faster than on the uphill part. When going downhill, he was twice as fast as when going uphill and 50% faster than on the flat part of the road. What is the distance between places $A$ and $B$?
## First solution.

Let $v$ be the speed on the uphill. Then the speed on the downhill is $2v$, and on the flat section $v+4$. \quad 1 point It also holds that $2v=1.5(v+4)$.

From the above equation, it follows that $v=12$. This means the speed on the uphill is $12 \mathrm{~km} / \mathrm{h}$, the speed on the downhill is $24 \mathrm{~km} / \mathrm{h}$, and on the flat section of the road $16 \mathrm{~km} / \mathrm{h}$.

If we denote the length of the flat section of the road and the length of the downhill section from place $A$ to place $B$ by $s_{1}$ and $s_{2}$, respectively, we have

$$
\begin{aligned}
& \frac{s_{1}}{16}+\frac{s_{2}}{24}=1.25 \\
& \frac{s_{1}}{16}+\frac{s_{2}}{12}=1.75
\end{aligned}
$$

By adding the above equations, we get

$$
\begin{aligned}
\frac{s_{1}}{16}+\frac{s_{1}}{16}+\frac{s_{2}}{12}+\frac{s_{2}}{24} & =3, & & 1 \text { point } \\
s_{1}\left(\frac{1}{16}+\frac{1}{16}\right)+s_{2}\left(\frac{1}{12}+\frac{1}{24}\right) & =3, & & \\
\frac{1}{8}\left(s_{1}+s_{2}\right) & =3, & & \\
s_{1}+s_{2} & =24 . & & 1 \text { point }
\end{aligned}
$$

The distance between place $A$ and place $B$ is $24 \mathrm{~km}$.