City E and City F is 300 km apart. Jack left City E for City F at 8.00 a.m. travelling at an average speed of 75 km/h. Paul left City E later than Jack and caught up with him at 10.00 a.m. Paul was travelling at a speed of 100 km/h.
- At what time did Paul leave City E?
- How much later did Jack arrive in City F than Paul? Express your answer in h as a fraction or mixed number.
(a)
From 8.00 a.m. to 10.00 a.m. = 2 h
Distance that Jack covered when Paul caught up with Jack
= 2 x 75
= 150 km
Time that Paul took to travel 150 km
= 150 ÷ 100
= 1
12 h
1 h = 60 min
12 h =
12 x 60 = 30 min
1
12 h = 1 h 30 min
Time that Paul left City E:
1 h 30 min before 10.00 a.m. = 8.30 a.m.
(b)
Remaining distance that they had to cover to reach City F
= 300 - 150
= 150 km
Duration that Paul had to travel before reaching City F
= 150 ÷ 100
= 1
12 h
Duration that Jack had to travel before reaching City F
= 150 ÷ 75
= 2 h
Duration that Jack took to arrive later than Paul in City F
= 2 - 1
12=
12 h
Answer(s): (a) 8.30 a.m.; (b)
12 h
