City H and City J is 372 km apart. Wesley left City H for City J at 11.00 a.m. travelling at an average speed of 62 km/h. Jack left City H later than Wesley and caught up with him at 1.00 p.m. Jack was travelling at a speed of 93 km/h.
- At what time did Jack leave City H?
- How much later did Wesley arrive in City J than Jack? 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 Wesley covered when Jack caught up with Wesley
= 2 x 62
= 124 km
Time that Jack took to travel 124 km
= 124 ÷ 93
= 1
13 h
1 h = 60 min
13 h =
13 x 60 = 20 min
1
13 h = 1 h 20 min
Time that Jack left City H:
1 h 20 min before 1.00 p.m. = 11.40 a.m.
(b)
Remaining distance that they had to cover to reach City J
= 372 - 124
= 248 km
Duration that Jack had to travel before reaching City J
= 248 ÷ 93
= 2
23 h
Duration that Wesley had to travel before reaching City J
= 248 ÷ 62
= 4 h
Duration that Wesley took to arrive later than Jack in City J
= 4 - 2
23= 1
13 h
Answer(s): (a) 11.40 a.m.; (b) 1
13 h