Billy and Pierre started on a 48-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 43 km, Billy 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 Billy's average speed for the remaining 43 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
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 08 30 is 05 18.
(b)
To find the time that Billy took for the remaining 43 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 Billy was faster than Pierre.
Time that Billy took for the remaining 43 km of the trip
= 3
15 -
13 -
13 = 2
815 h
Billy's average speed
= 43 ÷ 2
815 = 16
3738 km/h
Answer(s): (a) 05 18; (b) 16
3738 km/h