Gabriel and Pierre 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 Pierre. He arrived at the finishing point 20 minutes before Pierre who was 5 km behind him. Pierre did not change his speed throughout and completed it at 08 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 Pierre's average speed
= 5 ÷
13 = 15 km/h
Time that Pierre took for the trip
= 50 ÷ 15
= 3
515 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 08 30 is 05 10.
(b)
To find the time that Gabriel took for the remaining 45 km of the trip, we need to use the time that Pierre took subtract that of the time taken for first 5 km and the time that Gabriel was faster than Pierre.
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) 05 10; (b) 16
78 km/h