Elijah and Dylan started on a 50-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 45 km, Elijah cycled faster than Dylan. He arrived at the finishing point 20 minutes before Dylan who was 5 km behind him. Dylan 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 Elijah's average speed for the remaining 45 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Dylan's average speed
= 5 ÷
13 = 15 km/h
Time that Dylan took for the trip
= 50 ÷ 15
= 3
515 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 07 30 is 04 10.
(b)
To find the time that Elijah took for the remaining 45 km of the trip, we need to use the time that Dylan took subtract that of the time taken for first 5 km and the time that Elijah was faster than Dylan.
Time that Elijah took for the remaining 45 km of the trip
= 3
13 -
13 -
13 = 2
23 h
Elijah's average speed
= 45 ÷ 2
23 = 16
78 km/h
Answer(s): (a) 04 10; (b) 16
78 km/h