Fred and Albert started on a 48-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 38 km, Fred cycled faster than Albert. He arrived at the finishing point 40 minutes before Albert who was 10 km behind him. Albert did not change his speed throughout and completed it at 07 30.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Fred's average speed for the remaining 38 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Albert's average speed
= 10 ÷
23 = 15 km/h
Time that Albert took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 07 30 is 04 18.
(b)
To find the time that Fred took for the remaining 38 km of the trip, we need to use the time that Albert took subtract that of the time taken for first 10 km and the time that Fred was faster than Albert.
Time that Fred took for the remaining 38 km of the trip
= 3
15 -
23 -
23 = 1
1315 h
Fred's average speed
= 38 ÷ 1
1315 = 20
514 km/h
Answer(s): (a) 04 18; (b) 20
514 km/h