City B and City C is 272 km apart. Xavier left City B for City C at 7.00 a.m. travelling at an average speed of 68 km/h. Fred left City B later than Xavier and caught up with him at 9.00 a.m. Fred was travelling at a speed of 96 km/h.
- At what time did Fred leave City B?
- How much later did Xavier arrive in City C than Fred? Express your answer in h as a fraction or mixed number.
(a)
From 7.00 a.m. to 9.00 a.m. = 2 h
Distance that Xavier covered when Fred caught up with Xavier
= 2 x 68
= 136 km
Time that Fred 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 Fred left City B:
1 h 25 min before 9.00 a.m. = 7.35 a.m.
(b)
Remaining distance that they had to cover to reach City C
= 272 - 136
= 136 km
Duration that Fred had to travel before reaching City C
= 136 ÷ 96
= 1
512 h
Duration that Xavier had to travel before reaching City C
= 136 ÷ 68
= 2 h
Duration that Xavier took to arrive later than Fred in City C
= 2 - 1
512=
712 h
Answer(s): (a) 7.35 a.m.; (b)
712 h