City A and City B is 528 km apart. Riordan left City A for City B at 10.00 a.m. travelling at an average speed of 66 km/h. Wesley left City A later than Riordan and caught up with him at 2.00 p.m. Wesley was travelling at a speed of 99 km/h.

1. At what time did Wesley leave City A?
2. How much later did Riordan arrive in City B than Wesley? Express your answer in h as a fraction or mixed number.

(a)
From 10.00 a.m. to 2.00 p.m. = 4 h

Distance that Riordan covered when Wesley caught up with Riordan
= 4 x 66
= 264 km

Time that Wesley took to travel 264 km
= 264 ÷ 99
= 2²₃ h

1 h = 60 min
²₃ h = ²₃ x 60 = 40 min
2²₃ h = 2 h 40 min

Time that Wesley left City A:
2 h 40 min before 2.00 p.m. = 11.20 a.m.

(b)
Remaining distance that they had to cover to reach City B
= 528 - 264
= 264 km

Duration that Wesley had to travel before reaching City B
= 264 ÷ 99
= 2²₃ h

Duration that Riordan had to travel before reaching City B
= 264 ÷ 66
= 4 h

Duration that Riordan took to arrive later than Wesley in City B
= 4 - 2²₃
= 1¹₃ h

Answer(s): (a) 11.20 a.m.; (b) 1¹₃ h