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