City S and City T is 480 km apart. Ahmad left City S for City T at 10.00 a.m. travelling at an average speed of 60 km/h. Caden left City S later than Ahmad and caught up with him at 1.00 p.m. Caden was travelling at a speed of 108 km/h.

1. At what time did Caden leave City S?
2. How much later did Ahmad arrive in City T than Caden? Express your answer in h as a fraction or mixed number.

(a)
From 10.00 a.m. to 1.00 p.m. = 3 h

Distance that Ahmad covered when Caden caught up with Ahmad
= 3 x 60
= 180 km

Time that Caden took to travel 180 km
= 180 ÷ 108
= 1²₃ h

1 h = 60 min
²₃ h = ²₃ x 60 = 40 min
1²₃ h = 1 h 40 min

Time that Caden left City S:
1 h 40 min before 1.00 p.m. = 11.20 a.m.

(b)
Remaining distance that they had to cover to reach City T
= 480 - 180
= 300 km

Duration that Caden had to travel before reaching City T
= 300 ÷ 108
= 2⁷₉ h

Duration that Ahmad had to travel before reaching City T
= 300 ÷ 60
= 5 h

Duration that Ahmad took to arrive later than Caden in City T
= 5 - 2⁷₉
= 2²₉ h

Answer(s): (a) 11.20 a.m.; (b) 2²₉ h