At 12 15, Oscar left Town B for Town C, driving at a speed of 66 km/h. At 14 15, Paul also left Town B for Town C driving at a certain speed. Both of them did not change their speed throughout the journey. At 17 15, both of them passed a cafe that was 198 km away from Town C. How many minutes earlier did Paul reach Town C than Oscar?

Distance that Oscar travelled
= 5 x 66
= 330 km

Distance between Town B and Town C
= 330 + 198
= 528 km

Since Paul passed the cafe at the same time as Oscar, it means that Paul travelled 330 km in 3 hours.

Paul's speed during the journey
= 330 ÷ 3
= 110 km/h

Duration that Paul took to reach Town C after passing the cafe
= 198 ÷ 110
= 1.8 h

Duration that Oscar took to reach Town C after passing the cafe
= 198 ÷ 66
= 3 h

Difference between the time they took to reach Town C
= 3 - 1.8
= 1.2 h

1 h = 60 min
1.2 h = 1.2 x 60 = 72 min

Paul reached Town C 72 minutes earlier than Oscar.

Answer(s): 72 min