City D and City E is 432 km apart. Fred left City D for City E at 10.00 a.m. travelling at an average speed of 72 km/h. Paul left City D later than Fred and caught up with him at 12.00 p.m. Paul was travelling at a speed of 96 km/h.

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

(a)
From 10.00 a.m. to 12.00 p.m. = 2 h

Distance that Fred covered when Paul caught up with Fred
= 2 x 72
= 144 km

Time that Paul took to travel 144 km
= 144 ÷ 96
= 1¹₂ h

1 h = 60 min
¹₂ h = ¹₂ x 60 = 30 min
1¹₂ h = 1 h 30 min

Time that Paul left City D:
1 h 30 min before 12.00 p.m. = 10.30 a.m.

(b)
Remaining distance that they had to cover to reach City E
= 432 - 144
= 288 km

Duration that Paul had to travel before reaching City E
= 288 ÷ 96
= 3 h

Duration that Fred had to travel before reaching City E
= 288 ÷ 72
= 4 h

Duration that Fred took to arrive later than Paul in City E
= 4 - 3
= 1 h

Answer(s): (a) 10.30 a.m.; (b) 1 h