City P and City Q is 483 km apart. Jenson left City P for City Q at 11.00 a.m. travelling at an average speed of 69 km/h. Paul left City P later than Jenson and caught up with him at 1.00 p.m. Paul was travelling at a speed of 92 km/h.
- At what time did Paul leave City P?
- How much later did Jenson arrive in City Q than Paul? Express your answer in h as a fraction or mixed number.
(a)
From 11.00 a.m. to 1.00 p.m. = 2 h
Distance that Jenson covered when Paul caught up with Jenson
= 2 x 69
= 138 km
Time that Paul took to travel 138 km
= 138 ÷ 92
= 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 P:
1 h 30 min before 1.00 p.m. = 11.30 a.m.
(b)
Remaining distance that they had to cover to reach City Q
= 483 - 138
= 345 km
Duration that Paul had to travel before reaching City Q
= 345 ÷ 92
= 3
34 h
Duration that Jenson had to travel before reaching City Q
= 345 ÷ 69
= 5 h
Duration that Jenson took to arrive later than Paul in City Q
= 5 - 3
34= 1
14 h
Answer(s): (a) 11.30 a.m.; (b) 1
14 h
