Pierre and Tim started on a 60-km cycling trip at the same time. They cycled at the same speed for first 12 km. For the remaining 48 km, Pierre cycled faster than Tim. He arrived at the finishing point 40 minutes before Tim who was 12 km behind him. Tim 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 Pierre's average speed for the remaining 48 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Tim's average speed
= 12 ÷
23 = 18 km/h
Time that Tim took for the trip
= 60 ÷ 18
= 3
618 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 08 30 is 05 10.
(b)
To find the time that Pierre took for the remaining 48 km of the trip, we need to use the time that Tim took subtract that of the time taken for first 12 km and the time that Pierre was faster than Tim.
Time that Pierre took for the remaining 48 km of the trip
= 3
13 -
23 -
23 = 2 h
Pierre's average speed
= 48 ÷ 2
= 24 km/h
Answer(s): (a) 05 10; (b) 24 km/h