City X and City Y is 434 km apart. Sam left City X for City Y at 11.00 a.m. travelling at an average speed of 62 km/h. Luis left City X later than Sam and caught up with him at 2.00 p.m. Luis was travelling at a speed of 72 km/h.

1. At what time did Luis leave City X?
2. How much later did Sam arrive in City Y than Luis? Express your answer in h as a fraction or mixed number.

(a)
From 11.00 a.m. to 2.00 p.m. = 3 h

Distance that Sam covered when Luis caught up with Sam
= 3 x 62
= 186 km

Time that Luis took to travel 186 km
= 186 ÷ 72
= 2⁷₁₂ h

1 h = 60 min
⁷₁₂ h = ⁷₁₂ x 60 = 35 min
2⁷₁₂ h = 2 h 35 min

Time that Luis left City X:
2 h 35 min before 2.00 p.m. = 11.25 a.m.

(b)
Remaining distance that they had to cover to reach City Y
= 434 - 186
= 248 km

Duration that Luis had to travel before reaching City Y
= 248 ÷ 72
= 3⁴₉ h

Duration that Sam had to travel before reaching City Y
= 248 ÷ 62
= 4 h

Duration that Sam took to arrive later than Luis in City Y
= 4 - 3⁴₉
= ⁵₉ h

Answer(s): (a) 11.25 a.m.; (b) ⁵₉ h