Ken and Caden started on a 48-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 43 km, Ken cycled faster than Caden. He arrived at the finishing point 20 minutes before Caden who was 5 km behind him. Caden 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 Ken's average speed for the remaining 43 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Caden's average speed
= 5 ÷
13 = 15 km/h
Time that Caden took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 09 25 is 06 13.
(b)
To find the time that Ken took for the remaining 43 km of the trip, we need to use the time that Caden took subtract that of the time taken for first 5 km and the time that Ken was faster than Caden.
Time that Ken took for the remaining 43 km of the trip
= 3
15 -
13 -
13 = 2
815 h
Ken's average speed
= 43 ÷ 2
815 = 16
3738 km/h
Answer(s): (a) 06 13; (b) 16
3738 km/h