City U and City V is 540 km apart. Howard left City U for City V at 9.00 a.m. travelling at an average speed of 60 km/h. Luis left City U later than Howard and caught up with him at 1.00 p.m. Luis was travelling at a speed of 90 km/h.
- At what time did Luis leave City U?
- How much later did Howard arrive in City V than Luis? Express your answer in h as a fraction or mixed number.
(a)
From 9.00 a.m. to 1.00 p.m. = 4 h
Distance that Howard covered when Luis caught up with Howard
= 4 x 60
= 240 km
Time that Luis took to travel 240 km
= 240 ÷ 90
= 2
23 h
1 h = 60 min
23 h =
23 x 60 = 40 min
2
23 h = 2 h 40 min
Time that Luis left City U:
2 h 40 min before 1.00 p.m. = 10.20 a.m.
(b)
Remaining distance that they had to cover to reach City V
= 540 - 240
= 300 km
Duration that Luis had to travel before reaching City V
= 300 ÷ 90
= 3
13 h
Duration that Howard had to travel before reaching City V
= 300 ÷ 60
= 5 h
Duration that Howard took to arrive later than Luis in City V
= 5 - 3
13= 1
23 h
Answer(s): (a) 10.20 a.m.; (b) 1
23 h