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