City Z and City A is 441 km apart. Elijah left City Z for City A at 8.00 a.m. travelling at an average speed of 63 km/h. Ethan left City Z later than Elijah and caught up with him at 10.00 a.m. Ethan was travelling at a speed of 72 km/h.

1. At what time did Ethan leave City Z?
2. How much later did Elijah arrive in City A than Ethan? 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 Elijah covered when Ethan caught up with Elijah
= 2 x 63
= 126 km

Time that Ethan took to travel 126 km
= 126 ÷ 72
= 1³₄ h

1 h = 60 min
³₄ h = ³₄ x 60 = 45 min
1³₄ h = 1 h 45 min

Time that Ethan left City Z:
1 h 45 min before 10.00 a.m. = 8.15 a.m.

(b)
Remaining distance that they had to cover to reach City A
= 441 - 126
= 315 km

Duration that Ethan had to travel before reaching City A
= 315 ÷ 72
= 4³₈ h

Duration that Elijah had to travel before reaching City A
= 315 ÷ 63
= 5 h

Duration that Elijah took to arrive later than Ethan in City A
= 5 - 4³₈
= ⁵₈ h

Answer(s): (a) 8.15 a.m.; (b) ⁵₈ h