City L and City M is 504 km apart. Wesley left City L for City M at 11.00 a.m. travelling at an average speed of 63 km/h. David left City L later than Wesley and caught up with him at 2.00 p.m. David was travelling at a speed of 108 km/h.
- At what time did David leave City L?
- How much later did Wesley arrive in City M than David? 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 Wesley covered when David caught up with Wesley
= 3 x 63
= 189 km
Time that David took to travel 189 km
= 189 ÷ 108
= 1
34 h
1 h = 60 min
34 h =
34 x 60 = 45 min
1
34 h = 1 h 45 min
Time that David left City L:
1 h 45 min before 2.00 p.m. = 12.15 p.m.
(b)
Remaining distance that they had to cover to reach City M
= 504 - 189
= 315 km
Duration that David had to travel before reaching City M
= 315 ÷ 108
= 2
1112 h
Duration that Wesley had to travel before reaching City M
= 315 ÷ 63
= 5 h
Duration that Wesley took to arrive later than David in City M
= 5 - 2
1112= 2
112 h
Answer(s): (a) 12.15 p.m.; (b) 2
112 h
