Charlie and Sean started on a 50-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 40 km, Charlie cycled faster than Sean. He arrived at the finishing point 40 minutes before Sean who was 10 km behind him. Sean 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 Charlie's average speed for the remaining 40 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Sean's average speed
= 10 ÷
23 = 15 km/h
Time that Sean took for the trip
= 50 ÷ 15
= 3
515 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 07 30 is 04 10.
(b)
To find the time that Charlie took for the remaining 40 km of the trip, we need to use the time that Sean took subtract that of the time taken for first 10 km and the time that Charlie was faster than Sean.
Time that Charlie took for the remaining 40 km of the trip
= 3
13 -
23 -
23 = 2 h
Charlie's average speed
= 40 ÷ 2
= 20 km/h
Answer(s): (a) 04 10; (b) 20 km/h