City D and City E is 380 km apart. Oscar left City D for City E at 11.00 a.m. travelling at an average speed of 76 km/h. Sam left City D later than Oscar and caught up with him at 1.00 p.m. Sam was travelling at a speed of 96 km/h.

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

(a)
From 11.00 a.m. to 1.00 p.m. = 2 h

Distance that Oscar covered when Sam caught up with Oscar
= 2 x 76
= 152 km

Time that Sam took to travel 152 km
= 152 ÷ 96
= 1⁷₁₂ h

1 h = 60 min
⁷₁₂ h = ⁷₁₂ x 60 = 35 min
1⁷₁₂ h = 1 h 35 min

Time that Sam left City D:
1 h 35 min before 1.00 p.m. = 11.25 a.m.

(b)
Remaining distance that they had to cover to reach City E
= 380 - 152
= 228 km

Duration that Sam had to travel before reaching City E
= 228 ÷ 96
= 2³₈ h

Duration that Oscar had to travel before reaching City E
= 228 ÷ 76
= 3 h

Duration that Oscar took to arrive later than Sam in City E
= 3 - 2³₈
= ⁵₈ h

Answer(s): (a) 11.25 a.m.; (b) ⁵₈ h