Elijah and Henry started on a 55-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 45 km, Elijah cycled faster than Henry. He arrived at the finishing point 24 minutes before Henry who was 10 km behind him. Henry did not change his speed throughout and completed it at 09 25.
- 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
24 min =
2460 h =
25 Henry's average speed
= 10 ÷
25 = 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 09 25 is 07 13.
(b)
To find the time that Elijah took for the remaining 45 km of the trip, we need to use the time that Henry took subtract that of the time taken for first 10 km and the time that Elijah was faster than Henry.
Time that Elijah took for the remaining 45 km of the trip
= 2
15 -
25 -
25 = 1
25 h
Elijah's average speed
= 45 ÷ 1
25 = 32
17 km/h
Answer(s): (a) 07 13; (b) 32
17 km/h