City H and City J is 480 km apart. Vaidev left City H for City J at 10.00 a.m. travelling at an average speed of 60 km/h. Brandon left City H later than Vaidev and caught up with him at 1.00 p.m. Brandon was travelling at a speed of 108 km/h.

1. At what time did Brandon leave City H?
2. How much later did Vaidev arrive in City J than Brandon? 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 Vaidev covered when Brandon caught up with Vaidev
= 3 x 60
= 180 km

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

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

Duration that Brandon had to travel before reaching City J
= 300 ÷ 108
= 2⁷₉ h

Duration that Vaidev had to travel before reaching City J
= 300 ÷ 60
= 5 h

Duration that Vaidev took to arrive later than Brandon in City J
= 5 - 2⁷₉
= 2²₉ h

Answer(s): (a) 11.20 a.m.; (b) 2²₉ h