City X and City Y is 272 km apart. Paul left City X for City Y at 8.00 a.m. travelling at an average speed of 68 km/h. Tommy left City X later than Paul and caught up with him at 10.00 a.m. Tommy was travelling at a speed of 96 km/h.

1. At what time did Tommy leave City X?
2. How much later did Paul arrive in City Y than Tommy? 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 Paul covered when Tommy caught up with Paul
= 2 x 68
= 136 km

Time that Tommy took to travel 136 km
= 136 ÷ 96
= 1⁵₁₂ h

1 h = 60 min
⁵₁₂ h = ⁵₁₂ x 60 = 25 min
1⁵₁₂ h = 1 h 25 min

Time that Tommy left City X:
1 h 25 min before 10.00 a.m. = 8.35 a.m.

(b)
Remaining distance that they had to cover to reach City Y
= 272 - 136
= 136 km

Duration that Tommy had to travel before reaching City Y
= 136 ÷ 96
= 1⁵₁₂ h

Duration that Paul had to travel before reaching City Y
= 136 ÷ 68
= 2 h

Duration that Paul took to arrive later than Tommy in City Y
= 2 - 1⁵₁₂
= ⁷₁₂ h

Answer(s): (a) 8.35 a.m.; (b) ⁷₁₂ h