City S and City T is 608 km apart. Ken left City S for City T at 8.00 a.m. travelling at an average speed of 76 km/h. Fabian left City S later than Ken and caught up with him at 12.00 p.m. Fabian was travelling at a speed of 114 km/h.

1. At what time did Fabian leave City S?
2. How much later did Ken arrive in City T than Fabian? 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 Ken covered when Fabian caught up with Ken
= 4 x 76
= 304 km

Time that Fabian took to travel 304 km
= 304 ÷ 114
= 2²₃ h

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

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

(b)
Remaining distance that they had to cover to reach City T
= 608 - 304
= 304 km

Duration that Fabian had to travel before reaching City T
= 304 ÷ 114
= 2²₃ h

Duration that Ken had to travel before reaching City T
= 304 ÷ 76
= 4 h

Duration that Ken took to arrive later than Fabian in City T
= 4 - 2²₃
= 1¹₃ h

Answer(s): (a) 9.20 a.m.; (b) 1¹₃ h