Justin and Lee 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 Lee. He arrived at the finishing point 12 minutes before Lee who was 5 km behind him. Lee did not change his speed throughout and completed it at 09 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 Lee's average speed
= 5 ÷
15 = 25 km/h
Time that Lee took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 09 30 is 07 18.
(b)
To find the time that Justin took for the remaining 50 km of the trip, we need to use the time that Lee took subtract that of the time taken for first 5 km and the time that Justin was faster than Lee.
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) 07 18; (b) 27
79 km/h