City X and City Y is 414 km apart. Reggie left City X for City Y at 9.00 a.m. travelling at an average speed of 69 km/h. Tim left City X later than Reggie and caught up with him at 12.00 p.m. Tim was travelling at a speed of 108 km/h.

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

(a)
From 9.00 a.m. to 12.00 p.m. = 3 h

Distance that Reggie covered when Tim caught up with Reggie
= 3 x 69
= 207 km

Time that Tim 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 Tim left City X:
1 h 55 min before 12.00 p.m. = 10.05 a.m.

(b)
Remaining distance that they had to cover to reach City Y
= 414 - 207
= 207 km

Duration that Tim had to travel before reaching City Y
= 207 ÷ 108
= 1¹¹₁₂ h

Duration that Reggie had to travel before reaching City Y
= 207 ÷ 69
= 3 h

Duration that Reggie took to arrive later than Tim in City Y
= 3 - 1¹¹₁₂
= 1¹₁₂ h

Answer(s): (a) 10.05 a.m.; (b) 1¹₁₂ h