Perry and Oscar 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 Oscar. He arrived at the finishing point 20 minutes before Oscar who was 5 km behind him. Oscar 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 43 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Oscar's average speed
= 5 ÷
13 = 15 km/h
Time that Oscar 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 43 km of the trip, we need to use the time that Oscar took subtract that of the time taken for first 5 km and the time that Perry was faster than Oscar.
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 13; (b) 16
3738 km/h