City B and City C is 576 km apart. Justin left City B for City C at 10.00 a.m. travelling at an average speed of 64 km/h. John left City B later than Justin and caught up with him at 2.00 p.m. John was travelling at a speed of 96 km/h.
- At what time did John leave City B?
- How much later did Justin arrive in City C than John? Express your answer in h as a fraction or mixed number.
(a)
From 10.00 a.m. to 2.00 p.m. = 4 h
Distance that Justin covered when John caught up with Justin
= 4 x 64
= 256 km
Time that John took to travel 256 km
= 256 ÷ 96
= 2
23 h
1 h = 60 min
23 h =
23 x 60 = 40 min
2
23 h = 2 h 40 min
Time that John left City B:
2 h 40 min before 2.00 p.m. = 11.20 a.m.
(b)
Remaining distance that they had to cover to reach City C
= 576 - 256
= 320 km
Duration that John had to travel before reaching City C
= 320 ÷ 96
= 3
13 h
Duration that Justin had to travel before reaching City C
= 320 ÷ 64
= 5 h
Duration that Justin took to arrive later than John in City C
= 5 - 3
13= 1
23 h
Answer(s): (a) 11.20 a.m.; (b) 1
23 h
