City C and City D is 240 km apart. Luis left City C for City D at 8.00 a.m. travelling at an average speed of 60 km/h. John left City C later than Luis and caught up with him at 10.00 a.m. John was travelling at a speed of 96 km/h.

1. At what time did John leave City C?
2. How much later did Luis arrive in City D than John? 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 Luis covered when John caught up with Luis
= 2 x 60
= 120 km

Time that John took to travel 120 km
= 120 ÷ 96
= 1¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
1¹₄ h = 1 h 15 min

Time that John left City C:
1 h 15 min before 10.00 a.m. = 8.45 a.m.

(b)
Remaining distance that they had to cover to reach City D
= 240 - 120
= 120 km

Duration that John had to travel before reaching City D
= 120 ÷ 96
= 1¹₄ h

Duration that Luis had to travel before reaching City D
= 120 ÷ 60
= 2 h

Duration that Luis took to arrive later than John in City D
= 2 - 1¹₄
= ³₄ h

Answer(s): (a) 8.45 a.m.; (b) ³₄ h