Perry and George started on a 48-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 38 km, Perry cycled faster than George. He arrived at the finishing point 40 minutes before George who was 10 km behind him. George did not change his speed throughout and completed it at 08 25.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Perry's average speed for the remaining 38 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 George's average speed
= 10 ÷
23 = 15 km/h
Time that George took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 08 25 is 05 13.
(b)
To find the time that Perry took for the remaining 38 km of the trip, we need to use the time that George took subtract that of the time taken for first 10 km and the time that Perry was faster than George.
Time that Perry took for the remaining 38 km of the trip
= 3
15 -
23 -
23 = 1
1315 h
Perry's average speed
= 38 ÷ 1
1315 = 20
514 km/h
Answer(s): (a) 05 13; (b) 20
514 km/h