City S and City T is 546 km apart. Brandon left City S for City T at 10.00 a.m. travelling at an average speed of 78 km/h. Tom left City S later than Brandon and caught up with him at 2.00 p.m. Tom was travelling at a speed of 117 km/h.

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

(a)
From 10.00 a.m. to 2.00 p.m. = 4 h

Distance that Brandon covered when Tom caught up with Brandon
= 4 x 78
= 312 km

Time that Tom took to travel 312 km
= 312 ÷ 117
= 2²₃ h

1 h = 60 min
²₃ h = ²₃ x 60 = 40 min
2²₃ h = 2 h 40 min

Time that Tom left City S:
2 h 40 min before 2.00 p.m. = 11.20 a.m.

(b)
Remaining distance that they had to cover to reach City T
= 546 - 312
= 234 km

Duration that Tom had to travel before reaching City T
= 234 ÷ 117
= 2 h

Duration that Brandon had to travel before reaching City T
= 234 ÷ 78
= 3 h

Duration that Brandon took to arrive later than Tom in City T
= 3 - 2
= 1 h

Answer(s): (a) 11.20 a.m.; (b) 1 h