A tourist group organized a three-day cycling trip. On the first day, they wanted to cover $\frac{1}{3}$ of the entire route, but unfortunately, they fell short by $4 \mathrm{~km}$. On the second day, they aimed to cover half of the remaining distance, but it ended up being $2 \mathrm{~km}$ less. On the third day, however, they caught up, covering $\frac{10}{11}$ of the remaining distance plus an additional $4 \mathrm{~km}$. How long was the route, and how much did they cover on the first, second, and third day? ( $M$. Volfová)
We proceed by reasoning "backwards".

$\frac{1}{11}$ of the remaining distance after the second day is equal to $4 \text{ km}$, so the entire remainder was $11 \cdot 4 = 44 (\text{ km})$. If on the second day they had traveled (as planned) $2 \text{ km}$ more, the remainder after the second day would have been $2 \text{ km}$ less, i.e., only $42 \text{ km}$, and would have been half of what remained after the first day. After the first day, $84 \text{ km}$ remained to the destination.

If on the first day they had traveled the entire $\frac{1}{3}$ of the route, i.e., $4 \text{ km}$ more, the remainder would have been $4 \text{ km}$ less, i.e., only $80 \text{ km}$, and would have represented $\frac{2}{3}$ of the entire route. $\frac{1}{3}$ of the route was thus $40 \text{ km}$ and the entire route was $120 \text{ km}$.

We summarize, calculating the distances traveled each day. On the first day, they planned to travel $40 \text{ km}$, but only traveled $36 \text{ km}$; $84 \text{ km}$ remained to the destination. On the second day, they planned to travel half of the remainder, i.e., $42 \text{ km}$, but only traveled $40 \text{ km}$; $44 \text{ km}$ remained. On the third day, they traveled $\frac{10}{11}$ of the remainder, i.e., $40 \text{ km}$, and another $4 \text{ km}$; they were thus at the destination.

In the individual days, they traveled $36 \text{ km}$, $40 \text{ km}$, and $44 \text{ km}$, a total of $120 \text{ km}$.

Alternative solution. We proceed using algebra "from the front"; let the length of the entire route in km be $x$.

On the first day, they traveled $\frac{1}{3} x - 4$, leaving them $\frac{2}{3} x + 4$.

On the second day, they traveled $\frac{1}{2} \cdot \left(\frac{2}{3} x + 4\right) - 2 = \frac{1}{3} x$, leaving them $\frac{1}{3} x + 4$.

On the third day, they traveled $\frac{10}{11} \cdot \left(\frac{1}{3} x + 4\right) + 4 = \frac{10}{33} x + \frac{40}{11} + 4$.

The length of the entire route was thus

$$
x = \frac{1}{3} x - 4 + \frac{1}{3} x + \frac{10}{33} x + \frac{40}{11} + 4.
$$

After simplification, we get

$$
\begin{aligned}
\frac{1}{33} x & = \frac{40}{11} \\
x & = 120.
\end{aligned}
$$

Substituting into the above expressions, we find that in the individual days they traveled $36 \text{ km}$, $40 \text{ km}$, and $44 \text{ km}$, a total of $120 \text{ km}$.

Evaluation. 1 point for the length of the route on the first, second, and third day, and the total length of the route; the remaining 2 points according to the completeness of the commentary.