City R and City S is 476 km apart. Asher left City R for City S at 9.00 a.m. travelling at an average speed of 68 km/h. Henry left City R later than Asher and caught up with him at 11.00 a.m. Henry was travelling at a speed of 96 km/h.
- At what time did Henry leave City R?
- How much later did Asher arrive in City S than Henry? Express your answer in h as a fraction or mixed number.
(a)
From 9.00 a.m. to 11.00 a.m. = 2 h
Distance that Asher covered when Henry caught up with Asher
= 2 x 68
= 136 km
Time that Henry took to travel 136 km
= 136 ÷ 96
= 1
512 h
1 h = 60 min
512 h =
512 x 60 = 25 min
1
512 h = 1 h 25 min
Time that Henry left City R:
1 h 25 min before 11.00 a.m. = 9.35 a.m.
(b)
Remaining distance that they had to cover to reach City S
= 476 - 136
= 340 km
Duration that Henry had to travel before reaching City S
= 340 ÷ 96
= 3
1324 h
Duration that Asher had to travel before reaching City S
= 340 ÷ 68
= 5 h
Duration that Asher took to arrive later than Henry in City S
= 5 - 3
1324= 1
1124 h
Answer(s): (a) 9.35 a.m.; (b) 1
1124 h