City B and City C is 483 km apart. Ian left City B for City C at 11.00 a.m. travelling at an average speed of 69 km/h. Luke left City B later than Ian and caught up with him at 1.00 p.m. Luke was travelling at a speed of 72 km/h.

1. At what time did Luke leave City B?
2. How much later did Ian arrive in City C than Luke? 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 Ian covered when Luke caught up with Ian
= 2 x 69
= 138 km

Time that Luke took to travel 138 km
= 138 ÷ 72
= 1¹¹₁₂ h

1 h = 60 min
¹¹₁₂ h = ¹¹₁₂ x 60 = 55 min
1¹¹₁₂ h = 1 h 55 min

Time that Luke left City B:
1 h 55 min before 1.00 p.m. = 11.05 a.m.

(b)
Remaining distance that they had to cover to reach City C
= 483 - 138
= 345 km

Duration that Luke had to travel before reaching City C
= 345 ÷ 72
= 4¹⁹₂₄ h

Duration that Ian had to travel before reaching City C
= 345 ÷ 69
= 5 h

Duration that Ian took to arrive later than Luke in City C
= 5 - 4¹⁹₂₄
= ⁵₂₄ h

Answer(s): (a) 11.05 a.m.; (b) ⁵₂₄ h