City Y and City Z is 455 km apart. Seth left City Y for City Z at 11.00 a.m. travelling at an average speed of 65 km/h. Ken left City Y later than Seth and caught up with him at 2.00 p.m. Ken was travelling at a speed of 117 km/h.

1. At what time did Ken leave City Y?
2. How much later did Seth arrive in City Z than Ken? Express your answer in h as a fraction or mixed number.

(a)
From 11.00 a.m. to 2.00 p.m. = 3 h

Distance that Seth covered when Ken caught up with Seth
= 3 x 65
= 195 km

Time that Ken took to travel 195 km
= 195 ÷ 117
= 1²₃ h

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

Time that Ken left City Y:
1 h 40 min before 2.00 p.m. = 12.20 p.m.

(b)
Remaining distance that they had to cover to reach City Z
= 455 - 195
= 260 km

Duration that Ken had to travel before reaching City Z
= 260 ÷ 117
= 2²₉ h

Duration that Seth had to travel before reaching City Z
= 260 ÷ 65
= 4 h

Duration that Seth took to arrive later than Ken in City Z
= 4 - 2²₉
= 1⁷₉ h

Answer(s): (a) 12.20 p.m.; (b) 1⁷₉ h