Seth and Jenson started on a 50-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 40 km, Seth cycled faster than Jenson. He arrived at the finishing point 40 minutes before Jenson who was 10 km behind him. Jenson 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 Seth's average speed for the remaining 40 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Jenson's average speed
= 10 ÷
23 = 15 km/h
Time that Jenson took for the trip
= 50 ÷ 15
= 3
515 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 08 30 is 05 10.
(b)
To find the time that Seth took for the remaining 40 km of the trip, we need to use the time that Jenson took subtract that of the time taken for first 10 km and the time that Seth was faster than Jenson.
Time that Seth took for the remaining 40 km of the trip
= 3
13 -
23 -
23 = 2 h
Seth's average speed
= 40 ÷ 2
= 20 km/h
Answer(s): (a) 05 10; (b) 20 km/h