3. Dima was supposed to arrive at the station at 18:00. By this time, his father was supposed to pick him up in a car. However, Dima managed to catch an earlier train and arrived at the station at 17:05. He didn't wait for his father and started walking towards him. On the way, they met, Dima got into the car, and they arrived home 10 minutes earlier than the planned time. At what speed was Dima walking before the meeting with his father, if the car's speed was $60 \mathrm{km} / \mathrm{h}$?
Answer: 6 km/h

Solution. Dima arrived home 10 minutes earlier, during which time the car would have traveled the distance Dima walked twice. Therefore, on the way to the station, the father saved 5 minutes and met Dima at 17:55. This means Dima walked the distance from the station to the meeting point in 50 minutes, so he walked 10 times slower than the car, and his speed was 6 km/h.

## Grading Criteria.

- Complete correct solution - 7 points.
- Generally correct solution with insufficient justifications (in particular, a diagram of the movement is drawn with incomplete justifications) - 5 points.
- Correct approach but incorrect answer due to arithmetic error - 4 points.
- It is found that the meeting time is 17:55, but there is no further progress - 2 points.
- It is found that the father saved 5 minutes on the way to the station, but it is incorrectly assumed that Dima walked for 55 minutes before the meeting - 2 points.
- Only the correct answer is provided - 1 point.