City D and City E is 462 km apart. Elijah left City D for City E at 9.00 a.m. travelling at an average speed of 66 km/h. Seth left City D later than Elijah and caught up with him at 11.00 a.m. Seth was travelling at a speed of 72 km/h.

1. At what time did Seth leave City D?
2. How much later did Elijah arrive in City E than Seth? Express your answer in h as a fraction or mixed number.

(a)
From 9.00 a.m. to 11.00 a.m. = 2 h

Distance that Elijah covered when Seth caught up with Elijah
= 2 x 66
= 132 km

Time that Seth took to travel 132 km
= 132 ÷ 72
= 1⁵₆ h

1 h = 60 min
⁵₆ h = ⁵₆ x 60 = 50 min
1⁵₆ h = 1 h 50 min

Time that Seth left City D:
1 h 50 min before 11.00 a.m. = 9.10 a.m.

(b)
Remaining distance that they had to cover to reach City E
= 462 - 132
= 330 km

Duration that Seth had to travel before reaching City E
= 330 ÷ 72
= 4⁷₁₂ h

Duration that Elijah had to travel before reaching City E
= 330 ÷ 66
= 5 h

Duration that Elijah took to arrive later than Seth in City E
= 5 - 4⁷₁₂
= ⁵₁₂ h

Answer(s): (a) 9.10 a.m.; (b) ⁵₁₂ h