City T and City U is 462 km apart. Luis left City T for City U at 11.00 a.m. travelling at an average speed of 66 km/h. Reggie left City T later than Luis and caught up with him at 1.00 p.m. Reggie was travelling at a speed of 99 km/h.

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

(a)
From 11.00 a.m. to 1.00 p.m. = 2 h

Distance that Luis covered when Reggie caught up with Luis
= 2 x 66
= 132 km

Time that Reggie took to travel 132 km
= 132 ÷ 99
= 1¹₃ h

1 h = 60 min
¹₃ h = ¹₃ x 60 = 20 min
1¹₃ h = 1 h 20 min

Time that Reggie left City T:
1 h 20 min before 1.00 p.m. = 11.40 a.m.

(b)
Remaining distance that they had to cover to reach City U
= 462 - 132
= 330 km

Duration that Reggie had to travel before reaching City U
= 330 ÷ 99
= 3¹₃ h

Duration that Luis had to travel before reaching City U
= 330 ÷ 66
= 5 h

Duration that Luis took to arrive later than Reggie in City U
= 5 - 3¹₃
= 1²₃ h

Answer(s): (a) 11.40 a.m.; (b) 1²₃ h