City C and City D is 300 km apart. Caden left City C for City D at 7.00 a.m. travelling at an average speed of 75 km/h. Fred left City C later than Caden and caught up with him at 9.00 a.m. Fred was travelling at a speed of 90 km/h.
- At what time did Fred leave City C?
- How much later did Caden arrive in City D than Fred? Express your answer in h as a fraction or mixed number.
(a)
From 7.00 a.m. to 9.00 a.m. = 2 h
Distance that Caden covered when Fred caught up with Caden
= 2 x 75
= 150 km
Time that Fred took to travel 150 km
= 150 ÷ 90
= 1
23 h
1 h = 60 min
23 h =
23 x 60 = 40 min
1
23 h = 1 h 40 min
Time that Fred left City C:
1 h 40 min before 9.00 a.m. = 7.20 a.m.
(b)
Remaining distance that they had to cover to reach City D
= 300 - 150
= 150 km
Duration that Fred had to travel before reaching City D
= 150 ÷ 90
= 1
23 h
Duration that Caden had to travel before reaching City D
= 150 ÷ 75
= 2 h
Duration that Caden took to arrive later than Fred in City D
= 2 - 1
23=
13 h
Answer(s): (a) 7.20 a.m.; (b)
13 h