City T and City U is 396 km apart. Brandon left City T for City U at 7.00 a.m. travelling at an average speed of 66 km/h. Paul left City T later than Brandon and caught up with him at 9.00 a.m. Paul was travelling at a speed of 99 km/h.

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

(a)
From 7.00 a.m. to 9.00 a.m. = 2 h

Distance that Brandon covered when Paul caught up with Brandon
= 2 x 66
= 132 km

Time that Paul 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 Paul left City T:
1 h 20 min before 9.00 a.m. = 7.40 a.m.

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

Duration that Paul had to travel before reaching City U
= 264 ÷ 99
= 2²₃ h

Duration that Brandon had to travel before reaching City U
= 264 ÷ 66
= 4 h

Duration that Brandon took to arrive later than Paul in City U
= 4 - 2²₃
= 1¹₃ h

Answer(s): (a) 7.40 a.m.; (b) 1¹₃ h