Brandon and Paul 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, Brandon cycled faster than Paul. He arrived at the finishing point 40 minutes before Paul who was 10 km behind him. Paul 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 Brandon's average speed for the remaining 40 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Paul's average speed
= 10 ÷
23 = 15 km/h
Time that Paul 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 Brandon took for the remaining 40 km of the trip, we need to use the time that Paul took subtract that of the time taken for first 10 km and the time that Brandon was faster than Paul.
Time that Brandon took for the remaining 40 km of the trip
= 3
13 -
23 -
23 = 2 h
Brandon's average speed
= 40 ÷ 2
= 20 km/h
Answer(s): (a) 06 10; (b) 20 km/h