City C and City D is 462 km apart. Jack left City C for City D at 8.00 a.m. travelling at an average speed of 77 km/h. Eric left City C later than Jack and caught up with him at 11.00 a.m. Eric was travelling at a speed of 99 km/h.

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

(a)
From 8.00 a.m. to 11.00 a.m. = 3 h

Distance that Jack covered when Eric caught up with Jack
= 3 x 77
= 231 km

Time that Eric took to travel 231 km
= 231 ÷ 99
= 2¹₃ h

1 h = 60 min
¹₃ h = ¹₃ x 60 = 20 min
2¹₃ h = 2 h 20 min

Time that Eric left City C:
2 h 20 min before 11.00 a.m. = 8.40 a.m.

(b)
Remaining distance that they had to cover to reach City D
= 462 - 231
= 231 km

Duration that Eric had to travel before reaching City D
= 231 ÷ 99
= 2¹₃ h

Duration that Jack had to travel before reaching City D
= 231 ÷ 77
= 3 h

Duration that Jack took to arrive later than Eric in City D
= 3 - 2¹₃
= ²₃ h

Answer(s): (a) 8.40 a.m.; (b) ²₃ h