City R and City S is 264 km apart. Brandon left City R for City S at 9.00 a.m. travelling at an average speed of 66 km/h. Lee left City R later than Brandon and caught up with him at 11.00 a.m. Lee was travelling at a speed of 99 km/h.

1. At what time did Lee leave City R?
2. How much later did Brandon arrive in City S than Lee? 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 Brandon covered when Lee caught up with Brandon
= 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 R:
1 h 20 min before 11.00 a.m. = 9.40 a.m.

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

Duration that Lee had to travel before reaching City S
= 132 ÷ 99
= 1¹₃ h

Duration that Brandon had to travel before reaching City S
= 132 ÷ 66
= 2 h

Duration that Brandon took to arrive later than Lee in City S
= 2 - 1¹₃
= ²₃ h

Answer(s): (a) 9.40 a.m.; (b) ²₃ h