A pedestrian walked for 3.5 hours. In any 1-hour period, he covered $5 \mathrm{~km}$. Could his average speed be greater than $5 \mathrm{~km} /$ hour?
Let's divide the 3.5 hours into half-hour segments. If the pedestrian walks at a speed of $(5+v) \frac{\mathrm{km}}{\text{ h }}$ (where $v>0$) during the first, third, fifth, and seventh half-hours, and at a speed of $(5-v) \frac{\mathrm{km}}{\text{ h }}$ during the second, fourth, and sixth half-hours, then it is true that in any 1-hour period, they cover $5 \mathrm{~km}$.

The average speed is:

$$
\frac{4 \cdot 0.5 \cdot (5+v) + 3 \cdot 0.5 \cdot (5-v)}{3.5} = \frac{17.5 + 0.5v}{3.5} = 5 + \frac{0.5v}{3.5} > 5.
$$

Thus, the average speed can be greater than $5 \frac{\mathrm{km}}{\text{ h }}$.

Biczó Mihály (Paks, Vak Bottyán Gymn., 11th grade)

Note: Many of the solvers stated that this is "naturally" not possible.