Asher and Justin 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, Asher cycled faster than Justin. He arrived at the finishing point 20 minutes before Justin who was 5 km behind him. Justin 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 Asher's average speed for the remaining 43 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Justin's average speed
= 5 ÷
13 = 15 km/h
Time that Justin took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 07 25 is 04 13.
(b)
To find the time that Asher took for the remaining 43 km of the trip, we need to use the time that Justin took subtract that of the time taken for first 5 km and the time that Asher was faster than Justin.
Time that Asher took for the remaining 43 km of the trip
= 3
15 -
13 -
13 = 2
815 h
Asher's average speed
= 43 ÷ 2
815 = 16
3738 km/h
Answer(s): (a) 04 13; (b) 16
3738 km/h