Justin and George 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, Justin cycled faster than George. He arrived at the finishing point 12 minutes before George who was 5 km behind him. George 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 Justin's average speed for the remaining 50 km of the trip in km/h?
(a)
60 min = 1 h
12 min =
1260 h =
15 George's average speed
= 5 ÷
15 = 25 km/h
Time that George 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 Justin took for the remaining 50 km of the trip, we need to use the time that George took subtract that of the time taken for first 5 km and the time that Justin was faster than George.
Time that Justin took for the remaining 50 km of the trip
= 2
15 -
15 -
15 = 1
45 h
Justin's average speed
= 50 ÷ 1
45 = 27
79 km/h
Answer(s): (a) 05 18; (b) 27
79 km/h