City S and City T is 396 km apart. Charlie left City S for City T at 10.00 a.m. travelling at an average speed of 66 km/h. Lee left City S later than Charlie and caught up with him at 12.00 p.m. Lee was travelling at a speed of 99 km/h.

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

(a)
From 10.00 a.m. to 12.00 p.m. = 2 h

Distance that Charlie covered when Lee caught up with Charlie
= 2 x 66
= 132 km

Time that Lee took to travel 132 km
= 132 ÷ 99
= 1¹₃ h

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

Time that Lee left City S:
1 h 20 min before 12.00 p.m. = 10.40 a.m.

(b)
Remaining distance that they had to cover to reach City T
= 396 - 132
= 264 km

Duration that Lee had to travel before reaching City T
= 264 ÷ 99
= 2²₃ h

Duration that Charlie had to travel before reaching City T
= 264 ÷ 66
= 4 h

Duration that Charlie took to arrive later than Lee in City T
= 4 - 2²₃
= 1¹₃ h

Answer(s): (a) 10.40 a.m.; (b) 1¹₃ h