Liam and Paul 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, Liam cycled faster than Paul. He arrived at the finishing point 20 minutes before Paul who was 5 km behind him. Paul did not change his speed throughout and completed it at 09 30.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Liam's average speed for the remaining 45 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Paul's average speed
= 5 ÷
13 = 15 km/h
Time that Paul took for the trip
= 50 ÷ 15
= 3
515 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 09 30 is 06 10.
(b)
To find the time that Liam took for the remaining 45 km of the trip, we need to use the time that Paul took subtract that of the time taken for first 5 km and the time that Liam was faster than Paul.
Time that Liam took for the remaining 45 km of the trip
= 3
13 -
13 -
13 = 2
23 h
Liam's average speed
= 45 ÷ 2
23 = 16
78 km/h
Answer(s): (a) 06 10; (b) 16
78 km/h