Peter and Henry started on a 55-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 50 km, Peter cycled faster than Henry. He arrived at the finishing point 12 minutes before Henry who was 5 km behind him. Henry did not change his speed throughout and completed it at 07 25.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Peter's average speed for the remaining 50 km of the trip in km/h?
(a)
60 min = 1 h
12 min =
1260 h =
15 Henry's average speed
= 5 ÷
15 = 25 km/h
Time that Henry took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 07 25 is 05 13.
(b)
To find the time that Peter took for the remaining 50 km of the trip, we need to use the time that Henry took subtract that of the time taken for first 5 km and the time that Peter was faster than Henry.
Time that Peter took for the remaining 50 km of the trip
= 2
15 -
15 -
15 = 1
45 h
Peter's average speed
= 50 ÷ 1
45 = 27
79 km/h
Answer(s): (a) 05 13; (b) 27
79 km/h