Owen and Justin started on a 55-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 50 km, Owen cycled faster than Justin. He arrived at the finishing point 12 minutes before Justin who was 5 km behind him. Justin did not change his speed throughout and completed it at 09 25.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Owen's average speed for the remaining 50 km of the trip in km/h?
(a)
60 min = 1 h
12 min =
1260 h =
15 Justin's average speed
= 5 ÷
15 = 25 km/h
Time that Justin took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 09 25 is 07 13.
(b)
To find the time that Owen took for the remaining 50 km of the trip, we need to use the time that Justin took subtract that of the time taken for first 5 km and the time that Owen was faster than Justin.
Time that Owen took for the remaining 50 km of the trip
= 2
15 -
15 -
15 = 1
45 h
Owen's average speed
= 50 ÷ 1
45 = 27
79 km/h
Answer(s): (a) 07 13; (b) 27
79 km/h