City Z and City A is 288 km apart. Tom left City Z for City A at 10.00 a.m. travelling at an average speed of 72 km/h. Sean left City Z later than Tom and caught up with him at 12.00 p.m. Sean was travelling at a speed of 108 km/h.

1. At what time did Sean leave City Z?
2. How much later did Tom arrive in City A than Sean? 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 Tom covered when Sean caught up with Tom
= 2 x 72
= 144 km

Time that Sean took to travel 144 km
= 144 ÷ 108
= 1¹₃ h

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

Time that Sean left City Z:
1 h 20 min before 12.00 p.m. = 10.40 a.m.

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

Duration that Sean had to travel before reaching City A
= 144 ÷ 108
= 1¹₃ h

Duration that Tom had to travel before reaching City A
= 144 ÷ 72
= 2 h

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

Answer(s): (a) 10.40 a.m.; (b) ²₃ h