City X and City Y is 483 km apart. Cody left City X for City Y at 10.00 a.m. travelling at an average speed of 69 km/h. Michael left City X later than Cody and caught up with him at 1.00 p.m. Michael was travelling at a speed of 108 km/h.

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

Time that Michael 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 Michael left City X:
1 h 55 min before 1.00 p.m. = 11.05 a.m.

(b)
Remaining distance that they had to cover to reach City Y
= 483 - 207
= 276 km

Duration that Michael had to travel before reaching City Y
= 276 ÷ 108
= 2⁵₉ h

Duration that Cody had to travel before reaching City Y
= 276 ÷ 69
= 4 h

Duration that Cody took to arrive later than Michael in City Y
= 4 - 2⁵₉
= 1⁴₉ h

Answer(s): (a) 11.05 a.m.; (b) 1⁴₉ h