City T and City U is 396 km apart. Reggie left City T for City U at 11.00 a.m. travelling at an average speed of 66 km/h. Michael left City T later than Reggie and caught up with him at 3.00 p.m. Michael was travelling at a speed of 99 km/h.

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

(a)
From 11.00 a.m. to 3.00 p.m. = 4 h

Distance that Reggie covered when Michael caught up with Reggie
= 4 x 66
= 264 km

Time that Michael took to travel 264 km
= 264 ÷ 99
= 2²₃ h

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

Time that Michael left City T:
2 h 40 min before 3.00 p.m. = 12.20 p.m.

(b)
Remaining distance that they had to cover to reach City U
= 396 - 264
= 132 km

Duration that Michael had to travel before reaching City U
= 132 ÷ 99
= 1¹₃ h

Duration that Reggie had to travel before reaching City U
= 132 ÷ 66
= 2 h

Duration that Reggie took to arrive later than Michael in City U
= 2 - 1¹₃
= ²₃ h

Answer(s): (a) 12.20 p.m.; (b) ²₃ h