City T and City U is 462 km apart. Tim left City T for City U at 7.00 a.m. travelling at an average speed of 66 km/h. Seth left City T later than Tim and caught up with him at 10.00 a.m. Seth was travelling at a speed of 108 km/h.

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

(a)
From 7.00 a.m. to 10.00 a.m. = 3 h

Distance that Tim covered when Seth caught up with Tim
= 3 x 66
= 198 km

Time that Seth took to travel 198 km
= 198 ÷ 108
= 1⁵₆ h

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

Time that Seth left City T:
1 h 50 min before 10.00 a.m. = 8.10 a.m.

(b)
Remaining distance that they had to cover to reach City U
= 462 - 198
= 264 km

Duration that Seth had to travel before reaching City U
= 264 ÷ 108
= 2⁴₉ h

Duration that Tim had to travel before reaching City U
= 264 ÷ 66
= 4 h

Duration that Tim took to arrive later than Seth in City U
= 4 - 2⁴₉
= 1⁵₉ h

Answer(s): (a) 8.10 a.m.; (b) 1⁵₉ h