Luke and Cody started on a 50-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 45 km, Luke cycled faster than Cody. He arrived at the finishing point 20 minutes before Cody who was 5 km behind him. Cody 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 Luke's average speed for the remaining 45 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Cody's average speed
= 5 ÷
13 = 15 km/h
Time that Cody took for the trip
= 50 ÷ 15
= 3
515 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 09 30 is 06 10.
(b)
To find the time that Luke took for the remaining 45 km of the trip, we need to use the time that Cody took subtract that of the time taken for first 5 km and the time that Luke was faster than Cody.
Time that Luke took for the remaining 45 km of the trip
= 3
13 -
13 -
13 = 2
23 h
Luke's average speed
= 45 ÷ 2
23 = 16
78 km/h
Answer(s): (a) 06 10; (b) 16
78 km/h