City K and City L is 600 km apart. Lee left City K for City L at 7.00 a.m. travelling at an average speed of 75 km/h. Oscar left City K later than Lee and caught up with him at 10.00 a.m. Oscar was travelling at a speed of 100 km/h.

1. At what time did Oscar leave City K?
2. How much later did Lee arrive in City L than Oscar? Express your answer in h as a fraction or mixed number.

(a)
From 7.00 a.m. to 10.00 a.m. = 3 h

Distance that Lee covered when Oscar caught up with Lee
= 3 x 75
= 225 km

Time that Oscar took to travel 225 km
= 225 ÷ 100
= 2¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
2¹₄ h = 2 h 15 min

Time that Oscar left City K:
2 h 15 min before 10.00 a.m. = 7.45 a.m.

(b)
Remaining distance that they had to cover to reach City L
= 600 - 225
= 375 km

Duration that Oscar had to travel before reaching City L
= 375 ÷ 100
= 3³₄ h

Duration that Lee had to travel before reaching City L
= 375 ÷ 75
= 5 h

Duration that Lee took to arrive later than Oscar in City L
= 5 - 3³₄
= 1¹₄ h

Answer(s): (a) 7.45 a.m.; (b) 1¹₄ h