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