47 (1289). Two brothers walk together from school to home at the same speed. One day, 15 minutes after leaving school, the first brother ran back to school and, upon reaching it, immediately set off to catch up with the second. Left alone, the second continued walking home at half his usual speed. When the first brother caught up with the second, they resumed their initial speed and arrived home 6 minutes later than usual. How many times faster is the first brother's running speed compared to their usual walking speed?
Solution. Let the initial speed of the brothers be denoted by $v$ (m/min), and the segment of the path that the second brother traveled at a speed of $\frac{v}{2}$ be denoted by $s$. For this segment, he spent $\frac{2 s}{v}$ minutes $\left(s: \frac{v}{2}\right)$, which is 6 minutes more than the usual time spent (at speed $v$). We have the equation:

$$
\frac{2 s}{v}-\frac{s}{v}=6 \text {, i.e., } \frac{s}{v}=6, s=6 v \text {. }
$$

The first brother, in the same time $\left(\frac{2 s}{v}=\frac{2 \cdot 6 v}{v}=12\right.$ minutes), ran a distance of $15 v+15 v+s=30 v+6 v=36 v$. His running speed is $36 v: 12=3 v$, i.e., three times the usual walking speed of the brothers.

Answer: three times.

■ 48(1134)'. In a school mathematics olympiad, 9 sixth-grade students participated. For each solved problem, a student received 2 points, and for each unsolved problem, 1 point was deducted. A total of 10 problems were offered. Prove that among the participants from the sixth grade, there were at least 2 students who scored the same number of points. (It is assumed that a student who received more penalty points than earned points scored zero points.)

Solution. To prove this, we will create a table showing the number of points scored depending on the number of solved problems.

| Number of solved problems | 10 | 9 | 8 | 7 | 6 | 5 | 4 | 3 | 2 | 1 |
| :--- | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Number of points scored | 20 | 17 | 14 | 11 | 8 | 5 | 2 | 0 | 0 | 0 |

From the table, it is clear that there are only 8 different possibilities for scoring points. Since there were 9 students, at least two of them must have scored the same number of points.

49(1135). Two tourists, with one bicycle, need to travel a distance of 12 km in 1.5 hours. On the bicycle, each can reach a speed of 20 km/h, and on foot, 5 km/h. Can the tourists complete the entire journey without being late, if they cannot ride the bicycle together? (The bicycle can be left unattended.)

Solution. Let the first tourist ride the bicycle for $x$ km and walk the remaining $12-x$ km. Then the second tourist will ride the bicycle for $12-x$ km and walk $x$ km. Since the tourists will gain the most time if they arrive at the final destination simultaneously, we have the equation:

$$
\frac{x}{20}+\frac{12-x}{5}=\frac{x}{5}+\frac{12-x}{20}, \text { from which } x=6
$$

Thus, the tourists can complete the entire journey without being late if each of them rides the bicycle for half the distance (6 km) and walks the other half. They can switch modes of transportation, for example, every 100 meters.