City D and City E is 264 km apart. Tom left City D for City E at 8.00 a.m. travelling at an average speed of 66 km/h. Paul left City D later than Tom and caught up with him at 10.00 a.m. Paul was travelling at a speed of 99 km/h.

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

(a)
From 8.00 a.m. to 10.00 a.m. = 2 h

Distance that Tom covered when Paul caught up with Tom
= 2 x 66
= 132 km

Time that Paul took to travel 132 km
= 132 ÷ 99
= 1¹₃ h

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

Time that Paul left City D:
1 h 20 min before 10.00 a.m. = 8.40 a.m.

(b)
Remaining distance that they had to cover to reach City E
= 264 - 132
= 132 km

Duration that Paul had to travel before reaching City E
= 132 ÷ 99
= 1¹₃ h

Duration that Tom had to travel before reaching City E
= 132 ÷ 66
= 2 h

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

Answer(s): (a) 8.40 a.m.; (b) ²₃ h