City X and City Y is 476 km apart. Cole left City X for City Y at 7.00 a.m. travelling at an average speed of 68 km/h. Seth left City X later than Cole and caught up with him at 9.00 a.m. Seth was travelling at a speed of 96 km/h.

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

(a)
From 7.00 a.m. to 9.00 a.m. = 2 h

Distance that Cole covered when Seth caught up with Cole
= 2 x 68
= 136 km

Time that Seth took to travel 136 km
= 136 ÷ 96
= 1⁵₁₂ h

1 h = 60 min
⁵₁₂ h = ⁵₁₂ x 60 = 25 min
1⁵₁₂ h = 1 h 25 min

Time that Seth left City X:
1 h 25 min before 9.00 a.m. = 7.35 a.m.

(b)
Remaining distance that they had to cover to reach City Y
= 476 - 136
= 340 km

Duration that Seth had to travel before reaching City Y
= 340 ÷ 96
= 3¹³₂₄ h

Duration that Cole had to travel before reaching City Y
= 340 ÷ 68
= 5 h

Duration that Cole took to arrive later than Seth in City Y
= 5 - 3¹³₂₄
= 1¹¹₂₄ h

Answer(s): (a) 7.35 a.m.; (b) 1¹¹₂₄ h