City D and City E is 315 km apart. Luis left City D for City E at 11.00 a.m. travelling at an average speed of 63 km/h. Wesley left City D later than Luis and caught up with him at 2.00 p.m. Wesley was travelling at a speed of 84 km/h.
- At what time did Wesley leave City D?
- How much later did Luis arrive in City E than Wesley? 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 Luis covered when Wesley caught up with Luis
= 3 x 63
= 189 km
Time that Wesley took to travel 189 km
= 189 ÷ 84
= 2
14 h
1 h = 60 min
14 h =
14 x 60 = 15 min
2
14 h = 2 h 15 min
Time that Wesley left City D:
2 h 15 min before 2.00 p.m. = 11.45 a.m.
(b)
Remaining distance that they had to cover to reach City E
= 315 - 189
= 126 km
Duration that Wesley had to travel before reaching City E
= 126 ÷ 84
= 1
12 h
Duration that Luis had to travel before reaching City E
= 126 ÷ 63
= 2 h
Duration that Luis took to arrive later than Wesley in City E
= 2 - 1
12=
12 h
Answer(s): (a) 11.45 a.m.; (b)
12 h
