Elijah and Sam started on a 55-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 50 km, Elijah cycled faster than Sam. He arrived at the finishing point 12 minutes before Sam who was 5 km behind him. Sam 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 50 km of the trip in km/h?
(a)
60 min = 1 h
12 min =
1260 h =
15 Sam's average speed
= 5 ÷
15 = 25 km/h
Time that Sam took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 07 30 is 05 18.
(b)
To find the time that Elijah took for the remaining 50 km of the trip, we need to use the time that Sam took subtract that of the time taken for first 5 km and the time that Elijah was faster than Sam.
Time that Elijah took for the remaining 50 km of the trip
= 2
15 -
15 -
15 = 1
45 h
Elijah's average speed
= 50 ÷ 1
45 = 27
79 km/h
Answer(s): (a) 05 18; (b) 27
79 km/h