City W and City X is 360 km apart. Fred left City W for City X at 7.00 a.m. travelling at an average speed of 60 km/h. Luis left City W later than Fred and caught up with him at 10.00 a.m. Luis was travelling at a speed of 80 km/h.

1. At what time did Luis leave City W?
2. How much later did Fred arrive in City X than Luis? 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 Fred covered when Luis caught up with Fred
= 3 x 60
= 180 km

Time that Luis took to travel 180 km
= 180 ÷ 80
= 2¹₄ h

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

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

(b)
Remaining distance that they had to cover to reach City X
= 360 - 180
= 180 km

Duration that Luis had to travel before reaching City X
= 180 ÷ 80
= 2¹₄ h

Duration that Fred had to travel before reaching City X
= 180 ÷ 60
= 3 h

Duration that Fred took to arrive later than Luis in City X
= 3 - 2¹₄
= ³₄ h

Answer(s): (a) 7.45 a.m.; (b) ³₄ h