City S and City T is 345 km apart. Luis left City S for City T at 11.00 a.m. travelling at an average speed of 69 km/h. Henry left City S later than Luis and caught up with him at 1.00 p.m. Henry was travelling at a speed of 92 km/h.
- At what time did Henry leave City S?
- How much later did Luis arrive in City T than Henry? 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 Luis covered when Henry caught up with Luis
= 2 x 69
= 138 km
Time that Henry took to travel 138 km
= 138 ÷ 92
= 1
12 h
1 h = 60 min
12 h =
12 x 60 = 30 min
1
12 h = 1 h 30 min
Time that Henry left City S:
1 h 30 min before 1.00 p.m. = 11.30 a.m.
(b)
Remaining distance that they had to cover to reach City T
= 345 - 138
= 207 km
Duration that Henry had to travel before reaching City T
= 207 ÷ 92
= 2
14 h
Duration that Luis had to travel before reaching City T
= 207 ÷ 69
= 3 h
Duration that Luis took to arrive later than Henry in City T
= 3 - 2
14=
34 h
Answer(s): (a) 11.30 a.m.; (b)
34 h