City C and City D is 594 km apart. Archie left City C for City D at 8.00 a.m. travelling at an average speed of 66 km/h. Brandon left City C later than Archie and caught up with him at 12.00 p.m. Brandon was travelling at a speed of 99 km/h.

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

(a)
From 8.00 a.m. to 12.00 p.m. = 4 h

Distance that Archie covered when Brandon caught up with Archie
= 4 x 66
= 264 km

Time that Brandon 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 Brandon left City C:
2 h 40 min before 12.00 p.m. = 9.20 a.m.

(b)
Remaining distance that they had to cover to reach City D
= 594 - 264
= 330 km

Duration that Brandon had to travel before reaching City D
= 330 ÷ 99
= 3¹₃ h

Duration that Archie had to travel before reaching City D
= 330 ÷ 66
= 5 h

Duration that Archie took to arrive later than Brandon in City D
= 5 - 3¹₃
= 1²₃ h

Answer(s): (a) 9.20 a.m.; (b) 1²₃ h