City S and City T is 340 km apart. Albert left City S for City T at 9.00 a.m. travelling at an average speed of 68 km/h. Wesley left City S later than Albert and caught up with him at 11.00 a.m. Wesley was travelling at a speed of 102 km/h.

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

(a)
From 9.00 a.m. to 11.00 a.m. = 2 h

Distance that Albert covered when Wesley caught up with Albert
= 2 x 68
= 136 km

Time that Wesley took to travel 136 km
= 136 ÷ 102
= 1¹₃ h

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

Time that Wesley left City S:
1 h 20 min before 11.00 a.m. = 9.40 a.m.

(b)
Remaining distance that they had to cover to reach City T
= 340 - 136
= 204 km

Duration that Wesley had to travel before reaching City T
= 204 ÷ 102
= 2 h

Duration that Albert had to travel before reaching City T
= 204 ÷ 68
= 3 h

Duration that Albert took to arrive later than Wesley in City T
= 3 - 2
= 1 h

Answer(s): (a) 9.40 a.m.; (b) 1 h