Seth and Peter 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, Seth cycled faster than Peter. He arrived at the finishing point 40 minutes before Peter who was 10 km behind him. Peter did not change his speed throughout and completed it at 07 30.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Seth's average speed for the remaining 38 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Peter's average speed
= 10 ÷
23 = 15 km/h
Time that Peter took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 07 30 is 04 18.
(b)
To find the time that Seth took for the remaining 38 km of the trip, we need to use the time that Peter took subtract that of the time taken for first 10 km and the time that Seth was faster than Peter.
Time that Seth took for the remaining 38 km of the trip
= 3
15 -
23 -
23 = 1
1315 h
Seth's average speed
= 38 ÷ 1
1315 = 20
514 km/h
Answer(s): (a) 04 18; (b) 20
514 km/h