Bobby and Gabriel started on a 55-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 50 km, Bobby cycled faster than Gabriel. He arrived at the finishing point 12 minutes before Gabriel who was 5 km behind him. Gabriel 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 Bobby's average speed for the remaining 50 km of the trip in km/h?
(a)
60 min = 1 h
12 min =
1260 h =
15 Gabriel's average speed
= 5 ÷
15 = 25 km/h
Time that Gabriel took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 08 30 is 06 18.
(b)
To find the time that Bobby took for the remaining 50 km of the trip, we need to use the time that Gabriel took subtract that of the time taken for first 5 km and the time that Bobby was faster than Gabriel.
Time that Bobby took for the remaining 50 km of the trip
= 2
15 -
15 -
15 = 1
45 h
Bobby's average speed
= 50 ÷ 1
45 = 27
79 km/h
Answer(s): (a) 06 18; (b) 27
79 km/h