Perry and Paul 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, Perry cycled faster than Paul. He arrived at the finishing point 20 minutes before Paul who was 5 km behind him. Paul did not change his speed throughout and completed it at 08 30.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Perry's average speed for the remaining 43 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Paul's average speed
= 5 ÷
13 = 15 km/h
Time that Paul took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 08 30 is 05 18.
(b)
To find the time that Perry took for the remaining 43 km of the trip, we need to use the time that Paul took subtract that of the time taken for first 5 km and the time that Perry was faster than Paul.
Time that Perry took for the remaining 43 km of the trip
= 3
15 -
13 -
13 = 2
815 h
Perry's average speed
= 43 ÷ 2
815 = 16
3738 km/h
Answer(s): (a) 05 18; (b) 16
3738 km/h