City B and City C is 624 km apart. Brandon left City B for City C at 8.00 a.m. travelling at an average speed of 78 km/h. Sam left City B later than Brandon and caught up with him at 11.00 a.m. Sam was travelling at a speed of 104 km/h.

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

(a)
From 8.00 a.m. to 11.00 a.m. = 3 h

Distance that Brandon covered when Sam caught up with Brandon
= 3 x 78
= 234 km

Time that Sam took to travel 234 km
= 234 ÷ 104
= 2¹₄ h

1 h = 60 min
¹₄ h = ¹₄ x 60 = 15 min
2¹₄ h = 2 h 15 min

Time that Sam left City B:
2 h 15 min before 11.00 a.m. = 8.45 a.m.

(b)
Remaining distance that they had to cover to reach City C
= 624 - 234
= 390 km

Duration that Sam had to travel before reaching City C
= 390 ÷ 104
= 3³₄ h

Duration that Brandon had to travel before reaching City C
= 390 ÷ 78
= 5 h

Duration that Brandon took to arrive later than Sam in City C
= 5 - 3³₄
= 1¹₄ h

Answer(s): (a) 8.45 a.m.; (b) 1¹₄ h