City C and City D is 455 km apart. Will left City C for City D at 8.00 a.m. travelling at an average speed of 65 km/h. Julian left City C later than Will and caught up with him at 11.00 a.m. Julian was travelling at a speed of 90 km/h.
- At what time did Julian leave City C?
- How much later did Will arrive in City D than Julian? Express your answer in h as a fraction or mixed number.
(a)
From 8.00 a.m. to 11.00 a.m. = 3 h
Distance that Will covered when Julian caught up with Will
= 3 x 65
= 195 km
Time that Julian took to travel 195 km
= 195 ÷ 90
= 2
16 h
1 h = 60 min
16 h =
16 x 60 = 10 min
2
16 h = 2 h 10 min
Time that Julian left City C:
2 h 10 min before 11.00 a.m. = 8.50 a.m.
(b)
Remaining distance that they had to cover to reach City D
= 455 - 195
= 260 km
Duration that Julian had to travel before reaching City D
= 260 ÷ 90
= 2
89 h
Duration that Will had to travel before reaching City D
= 260 ÷ 65
= 4 h
Duration that Will took to arrive later than Julian in City D
= 4 - 2
89= 1
19 h
Answer(s): (a) 8.50 a.m.; (b) 1
19 h
