City X and City Y is 296 km apart. Paul left City X for City Y at 11.00 a.m. travelling at an average speed of 74 km/h. Vincent left City X later than Paul and caught up with him at 1.00 p.m. Vincent was travelling at a speed of 111 km/h.

1. At what time did Vincent leave City X?
2. How much later did Paul arrive in City Y than Vincent? Express your answer in h as a fraction or mixed number.

(a)
From 11.00 a.m. to 1.00 p.m. = 2 h

Distance that Paul covered when Vincent caught up with Paul
= 2 x 74
= 148 km

Time that Vincent took to travel 148 km
= 148 ÷ 111
= 1¹₃ h

1 h = 60 min
¹₃ h = ¹₃ x 60 = 20 min
1¹₃ h = 1 h 20 min

Time that Vincent left City X:
1 h 20 min before 1.00 p.m. = 11.40 a.m.

(b)
Remaining distance that they had to cover to reach City Y
= 296 - 148
= 148 km

Duration that Vincent had to travel before reaching City Y
= 148 ÷ 111
= 1¹₃ h

Duration that Paul had to travel before reaching City Y
= 148 ÷ 74
= 2 h

Duration that Paul took to arrive later than Vincent in City Y
= 2 - 1¹₃
= ²₃ h

Answer(s): (a) 11.40 a.m.; (b) ²₃ h