City Z and City A is 300 km apart. Paul left City Z for City A at 10.00 a.m. travelling at an average speed of 60 km/h. Jack left City Z later than Paul and caught up with him at 12.00 p.m. Jack was travelling at a speed of 96 km/h.
- At what time did Jack leave City Z?
- How much later did Paul arrive in City A than Jack? 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 Paul covered when Jack caught up with Paul
= 2 x 60
= 120 km
Time that Jack took to travel 120 km
= 120 ÷ 96
= 1
14 h
1 h = 60 min
14 h =
14 x 60 = 15 min
1
14 h = 1 h 15 min
Time that Jack left City Z:
1 h 15 min before 12.00 p.m. = 10.45 a.m.
(b)
Remaining distance that they had to cover to reach City A
= 300 - 120
= 180 km
Duration that Jack had to travel before reaching City A
= 180 ÷ 96
= 1
78 h
Duration that Paul had to travel before reaching City A
= 180 ÷ 60
= 3 h
Duration that Paul took to arrive later than Jack in City A
= 3 - 1
78= 1
18 h
Answer(s): (a) 10.45 a.m.; (b) 1
18 h