City Z and City A is 504 km apart. Peter left City Z for City A at 10.00 a.m. travelling at an average speed of 72 km/h. Luis left City Z later than Peter and caught up with him at 1.00 p.m. Luis was travelling at a speed of 96 km/h.

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

(a)
From 10.00 a.m. to 1.00 p.m. = 3 h

Distance that Peter covered when Luis caught up with Peter
= 3 x 72
= 216 km

Time that Luis took to travel 216 km
= 216 ÷ 96
= 2¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
2¹₄ h = 2 h 15 min

Time that Luis left City Z:
2 h 15 min before 1.00 p.m. = 10.45 a.m.

(b)
Remaining distance that they had to cover to reach City A
= 504 - 216
= 288 km

Duration that Luis had to travel before reaching City A
= 288 ÷ 96
= 3 h

Duration that Peter had to travel before reaching City A
= 288 ÷ 72
= 4 h

Duration that Peter took to arrive later than Luis in City A
= 4 - 3
= 1 h

Answer(s): (a) 10.45 a.m.; (b) 1 h