City A and City B is 260 km apart. Caden left City A for City B at 11.00 a.m. travelling at an average speed of 65 km/h. Dylan left City A later than Caden and caught up with him at 1.00 p.m. Dylan was travelling at a speed of 104 km/h.
- At what time did Dylan leave City A?
- How much later did Caden arrive in City B than Dylan? Express your answer in h as a fraction or mixed number.
(a)
From 11.00 a.m. to 1.00 p.m. = 2 h
Distance that Caden covered when Dylan caught up with Caden
= 2 x 65
= 130 km
Time that Dylan took to travel 130 km
= 130 ÷ 104
= 1
14 h
1 h = 60 min
14 h =
14 x 60 = 15 min
1
14 h = 1 h 15 min
Time that Dylan left City A:
1 h 15 min before 1.00 p.m. = 11.45 a.m.
(b)
Remaining distance that they had to cover to reach City B
= 260 - 130
= 130 km
Duration that Dylan had to travel before reaching City B
= 130 ÷ 104
= 1
14 h
Duration that Caden had to travel before reaching City B
= 130 ÷ 65
= 2 h
Duration that Caden took to arrive later than Dylan in City B
= 2 - 1
14=
34 h
Answer(s): (a) 11.45 a.m.; (b)
34 h