Luis and Daniel 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, Luis cycled faster than Daniel. He arrived at the finishing point 40 minutes before Daniel who was 12 km behind him. Daniel 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 Luis's average speed for the remaining 48 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Daniel's average speed
= 12 ÷
23 = 18 km/h
Time that Daniel took for the trip
= 60 ÷ 18
= 3
618 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 09 30 is 06 10.
(b)
To find the time that Luis took for the remaining 48 km of the trip, we need to use the time that Daniel took subtract that of the time taken for first 12 km and the time that Luis was faster than Daniel.
Time that Luis took for the remaining 48 km of the trip
= 3
13 -
23 -
23 = 2 h
Luis's average speed
= 48 ÷ 2
= 24 km/h
Answer(s): (a) 06 10; (b) 24 km/h