Harry and Michael started on a 55-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 45 km, Harry cycled faster than Michael. He arrived at the finishing point 24 minutes before Michael who was 10 km behind him. Michael 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 Harry's average speed for the remaining 45 km of the trip in km/h?
(a)
60 min = 1 h
24 min =
2460 h =
25 Michael's average speed
= 10 ÷
25 = 25 km/h
Time that Michael took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 07 30 is 05 18.
(b)
To find the time that Harry took for the remaining 45 km of the trip, we need to use the time that Michael took subtract that of the time taken for first 10 km and the time that Harry was faster than Michael.
Time that Harry took for the remaining 45 km of the trip
= 2
15 -
25 -
25 = 1
25 h
Harry's average speed
= 45 ÷ 1
25 = 32
17 km/h
Answer(s): (a) 05 18; (b) 32
17 km/h