City D and City E is 434 km apart. Neave left City D for City E at 8.00 a.m. travelling at an average speed of 62 km/h. Charlie left City D later than Neave and caught up with him at 11.00 a.m. Charlie was travelling at a speed of 72 km/h.

1. At what time did Charlie leave City D?
2. How much later did Neave arrive in City E than Charlie? 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 Neave covered when Charlie caught up with Neave
= 3 x 62
= 186 km

Time that Charlie 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 Charlie left City D:
2 h 35 min before 11.00 a.m. = 8.25 a.m.

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

Duration that Charlie had to travel before reaching City E
= 248 ÷ 72
= 3⁴₉ h

Duration that Neave had to travel before reaching City E
= 248 ÷ 62
= 4 h

Duration that Neave took to arrive later than Charlie in City E
= 4 - 3⁴₉
= ⁵₉ h

Answer(s): (a) 8.25 a.m.; (b) ⁵₉ h