City M and City N is 468 km apart. Xavier left City M for City N at 11.00 a.m. travelling at an average speed of 78 km/h. Paul left City M later than Xavier and caught up with him at 2.00 p.m. Paul was travelling at a speed of 104 km/h.
- At what time did Paul leave City M?
- How much later did Xavier arrive in City N than Paul? Express your answer in h as a fraction or mixed number.
(a)
From 11.00 a.m. to 2.00 p.m. = 3 h
Distance that Xavier covered when Paul caught up with Xavier
= 3 x 78
= 234 km
Time that Paul took to travel 234 km
= 234 ÷ 104
= 2
14 h
1 h = 60 min
14 h =
14 x 60 = 15 min
2
14 h = 2 h 15 min
Time that Paul left City M:
2 h 15 min before 2.00 p.m. = 11.45 a.m.
(b)
Remaining distance that they had to cover to reach City N
= 468 - 234
= 234 km
Duration that Paul had to travel before reaching City N
= 234 ÷ 104
= 2
14 h
Duration that Xavier had to travel before reaching City N
= 234 ÷ 78
= 3 h
Duration that Xavier took to arrive later than Paul in City N
= 3 - 2
14=
34 h
Answer(s): (a) 11.45 a.m.; (b)
34 h