City R and City S is 450 km apart. David left City R for City S at 8.00 a.m. travelling at an average speed of 75 km/h. Reggie left City R later than David and caught up with him at 10.00 a.m. Reggie was travelling at a speed of 100 km/h.

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

(a)
From 8.00 a.m. to 10.00 a.m. = 2 h

Distance that David covered when Reggie caught up with David
= 2 x 75
= 150 km

Time that Reggie took to travel 150 km
= 150 ÷ 100
= 1¹₂ h

1 h = 60 min
¹₂ h = ¹₂ x 60 = 30 min
1¹₂ h = 1 h 30 min

Time that Reggie left City R:
1 h 30 min before 10.00 a.m. = 8.30 a.m.

(b)
Remaining distance that they had to cover to reach City S
= 450 - 150
= 300 km

Duration that Reggie had to travel before reaching City S
= 300 ÷ 100
= 3 h

Duration that David had to travel before reaching City S
= 300 ÷ 75
= 4 h

Duration that David took to arrive later than Reggie in City S
= 4 - 3
= 1 h

Answer(s): (a) 8.30 a.m.; (b) 1 h