City C and City D is 272 km apart. Tommy left City C for City D at 10.00 a.m. travelling at an average speed of 68 km/h. Fred left City C later than Tommy and caught up with him at 12.00 p.m. Fred was travelling at a speed of 96 km/h.

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

(a)
From 10.00 a.m. to 12.00 p.m. = 2 h

Distance that Tommy covered when Fred caught up with Tommy
= 2 x 68
= 136 km

Time that Fred 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 Fred left City C:
1 h 25 min before 12.00 p.m. = 10.35 a.m.

(b)
Remaining distance that they had to cover to reach City D
= 272 - 136
= 136 km

Duration that Fred had to travel before reaching City D
= 136 ÷ 96
= 1⁵₁₂ h

Duration that Tommy had to travel before reaching City D
= 136 ÷ 68
= 2 h

Duration that Tommy took to arrive later than Fred in City D
= 2 - 1⁵₁₂
= ⁷₁₂ h

Answer(s): (a) 10.35 a.m.; (b) ⁷₁₂ h