City Z and City A is 512 km apart. Tim left City Z for City A at 11.00 a.m. travelling at an average speed of 64 km/h. Pierre left City Z later than Tim and caught up with him at 3.00 p.m. Pierre was travelling at a speed of 96 km/h.
- At what time did Pierre leave City Z?
- How much later did Tim arrive in City A than Pierre? Express your answer in h as a fraction or mixed number.
(a)
From 11.00 a.m. to 3.00 p.m. = 4 h
Distance that Tim covered when Pierre caught up with Tim
= 4 x 64
= 256 km
Time that Pierre 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 Pierre left City Z:
2 h 40 min before 3.00 p.m. = 12.20 p.m.
(b)
Remaining distance that they had to cover to reach City A
= 512 - 256
= 256 km
Duration that Pierre had to travel before reaching City A
= 256 ÷ 96
= 2
23 h
Duration that Tim had to travel before reaching City A
= 256 ÷ 64
= 4 h
Duration that Tim took to arrive later than Pierre in City A
= 4 - 2
23= 1
13 h
Answer(s): (a) 12.20 p.m.; (b) 1
13 h
