City H and City J is 455 km apart. Cody left City H for City J at 10.00 a.m. travelling at an average speed of 65 km/h. Seth left City H later than Cody and caught up with him at 12.00 p.m. Seth was travelling at a speed of 78 km/h.
- At what time did Seth leave City H?
- How much later did Cody arrive in City J than Seth? Express your answer in h as a fraction or mixed number.
(a)
From 10.00 a.m. to 12.00 p.m. = 2 h
Distance that Cody covered when Seth caught up with Cody
= 2 x 65
= 130 km
Time that Seth took to travel 130 km
= 130 ÷ 78
= 1
23 h
1 h = 60 min
23 h =
23 x 60 = 40 min
1
23 h = 1 h 40 min
Time that Seth left City H:
1 h 40 min before 12.00 p.m. = 10.20 a.m.
(b)
Remaining distance that they had to cover to reach City J
= 455 - 130
= 325 km
Duration that Seth had to travel before reaching City J
= 325 ÷ 78
= 4
16 h
Duration that Cody had to travel before reaching City J
= 325 ÷ 65
= 5 h
Duration that Cody took to arrive later than Seth in City J
= 5 - 4
16=
56 h
Answer(s): (a) 10.20 a.m.; (b)
56 h