## Task 4 - 190724

A driver drove his car from A to B. After 20 minutes of driving, he had a breakdown that took 30 minutes to fix. After another 12 minutes of driving, he had to wait 4 minutes at a closed railway crossing. By then, he had covered $40 \mathrm{~km}$. The journey from the railway crossing to B started at 11:06 AM and proceeded without any stops. Upon arrival in B, the driver noticed that he had taken exactly half the time from the departure at the railway crossing to the arrival in B compared to the total time from the departure from A to the arrival in B.

It is assumed that the driver maintained the same average speed on each segment of the journey.
a) At what time did the driver arrive in B?

b) What was the average speed, expressed in $\mathrm{km} \mathrm{h}$?

c) How many kilometers did he cover in total from A to B?
a) The truck driver needed $20+30+12+4=66$ minutes until the departure from the railway crossing. Since this time was as long as the travel time from the railway crossing to B, he was on the road for another 66 minutes until B from 11:06 AM, thus arriving there at 12:12 PM.

b) For the first $40 \mathrm{~km}$, the pure driving time was $66-30-4=32$ minutes, which is $\frac{8}{15}$ hours. Because $40: \frac{8}{15}$, his average speed was therefore $75 \frac{\mathrm{km}}{\mathrm{h}}$.

c) Since he covered the rest of the distance at the same average speed and it took him 66 minutes, which is $\frac{11}{10}$ hours, he covered $75 \cdot \frac{11}{10}=82.5$ more kilometers. Thus, he had traveled a total of $40 \mathrm{~km}+82.5 \mathrm{~km}=122.5 \mathrm{~km}$ from A to B.

Solutions of the 2nd Round 1979 adopted from [5]

### 4.21.3 3rd Round 1979, Class 7