The road from city $A$ to city $B$ has an uphill, a flat, and a downhill section. A pedestrian walked from $A$ to $B$ in 2.3 hours and returned in 2.6 hours. On the uphill, they walked at 3 km/h, on the flat section at 4 km/h, and on the downhill at 6 km/h. What can we determine about the length of the $A B$ road section? What can we determine if the pedestrian only walked at 5 km/h on the downhill?

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I. Let the horizontal section of the road be $x \mathrm{~km}$, the ascending section towards $B$ be $y$, and the descending section be $z \mathrm{~km}$, so the ascending section towards $A$ is $z \mathrm{~km}$ and the descending section is $y \mathrm{~km}$. The time taken to travel each section can be expressed using the relationship time $=\frac{\text { distance }}{\text { speed }}$, resulting in two equations for the three unknowns:

$$
\frac{x}{4}+\frac{y}{3}+\frac{z}{6}=2.3, \quad \frac{x}{4}+\frac{y}{6}+\frac{z}{3}=2.6
$$

However, we can still determine the total length of the route $s=x+y+z$ precisely because when we add the two equations, the coefficients of $x, y$, and $z$ become equal:

$$
\frac{1}{2}(x+y+z)=\frac{s}{2}=4.9, \quad s=9.8 \mathrm{~km}
$$

We cannot determine the lengths of the individual sections of the route, but we can calculate the difference $z-y$ by subtracting the first equation from the second:

$$
\frac{1}{6}(z-y)=0.3, \quad z-y=1.8 \mathrm{~km}
$$

Thus, the descending section towards $B$ is $1.8 \mathrm{~km}$ longer.

II. In the second case, we use a comma to distinguish the distances marked with the above symbol. We have two equations, and their sum is as follows:

$$
\begin{gathered}
\frac{x^{\prime}}{4}+\frac{y^{\prime}}{3}+\frac{z^{\prime}}{5}=2.3 \\
\frac{x^{\prime}}{4}+\frac{y^{\prime}}{5}+\frac{z^{\prime}}{3}=2.6 \\
\frac{x^{\prime}}{2}+\frac{8}{15}\left(y^{\prime}+z^{\prime}\right)=4.9
\end{gathered}
$$

Here, we cannot determine $s^{\prime}$ uniquely, but we can provide bounds for its value.

To find a lower bound, we increase the coefficient of $x^{\prime}$ as much as the common coefficient of $y^{\prime}$ and $z^{\prime}$, thereby increasing the left side:

$$
\frac{8}{15}\left(x^{\prime}+y^{\prime}+z^{\prime}\right)=\frac{8 s^{\prime}}{15}>4.9, \quad \quad s^{\prime}>9 \frac{3}{16} \mathrm{~km}
$$

The value of $s^{\prime}$ could only be equal to the lower bound if the modification did not increase the left side, i.e., if $x^{\prime}=0$, meaning there would be no horizontal section on the route; however, the problem excludes this possibility.

To find an upper bound for $s^{\prime}$, we determine the difference $z^{\prime}-y^{\prime}$. Subtracting (1) from (2):

$$
\frac{2}{15}\left(z^{\prime}-y^{\prime}\right)=0.3, \quad z^{\prime}=y^{\prime}+2.25 \mathrm{~km}
$$

With this, (3) becomes:

$$
\begin{gathered}
\frac{x^{\prime}}{2}+\frac{8}{15} \cdot 2 y^{\prime}+\frac{8 \cdot 2.25}{15}=4.9 \\
\frac{x^{\prime}}{2}+\frac{8}{15} \cdot 2 y^{\prime}=3.7
\end{gathered}
$$

and the length of the route $s^{\prime}=x^{\prime}+2 y^{\prime}+2.25$. In (4), we decrease the coefficient of $2 y^{\prime}$ as much as the coefficient of $x^{\prime}$, thereby decreasing the left side:

$$
\frac{1}{2}\left(x^{\prime}+2 y^{\prime}\right)<3.7, \quad x^{\prime}+2 y^{\prime}<7.4
$$

and thus

$$
s^{\prime}<9.65 \mathrm{~km}
$$

The value of $s^{\prime}$ could only be equal to the upper bound if $y^{\prime}=0$, meaning there would be no ascending section towards $B$; however, the problem excludes this possibility. Therefore,

$$
9.1875 \mathrm{~km}<s^{\prime}<9.65 \mathrm{~km}
$$

Remark. In the first case, the speed $v_{0}$ exerted on the horizontal section is the harmonic mean of the speeds $v_{e}$ and $v_{i}$ exerted on the ascending and descending sections, respectively:

$$
\frac{1}{v_{0}}+\frac{1}{v_{0}}=\frac{1}{v_{e}}+\frac{1}{v_{i}}, \quad v_{0}=\frac{2 v_{e} v_{i}}{v_{e}+v_{i}}
$$

This explains why we obtained a precise value for the length of the route. Essentially, the lower and upper bounds according to the second calculation would be equal in the first case.