City J and City K is 552 km apart. Cody left City J for City K at 11.00 a.m. travelling at an average speed of 69 km/h. Oscar left City J later than Cody and caught up with him at 2.00 p.m. Oscar was travelling at a speed of 108 km/h.

1. At what time did Oscar leave City J?
2. How much later did Cody arrive in City K than Oscar? 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 Cody covered when Oscar caught up with Cody
= 3 x 69
= 207 km

Time that Oscar took to travel 207 km
= 207 ÷ 108
= 1¹¹₁₂ h

1 h = 60 min
¹¹₁₂ h = ¹¹₁₂ x 60 = 55 min
1¹¹₁₂ h = 1 h 55 min

Time that Oscar left City J:
1 h 55 min before 2.00 p.m. = 12.05 p.m.

(b)
Remaining distance that they had to cover to reach City K
= 552 - 207
= 345 km

Duration that Oscar had to travel before reaching City K
= 345 ÷ 108
= 3⁷₃₆ h

Duration that Cody had to travel before reaching City K
= 345 ÷ 69
= 5 h

Duration that Cody took to arrive later than Oscar in City K
= 5 - 3⁷₃₆
= 1²⁹₃₆ h

Answer(s): (a) 12.05 p.m.; (b) 1²⁹₃₆ h