Brandon and Tim 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, Brandon cycled faster than Tim. He arrived at the finishing point 20 minutes before Tim who was 5 km behind him. Tim did not change his speed throughout and completed it at 07 30.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Brandon's average speed for the remaining 45 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Tim's average speed
= 5 ÷
13 = 15 km/h
Time that Tim took for the trip
= 50 ÷ 15
= 3
515 h
= 3
13 h
= 3 h 20 min
3 h 20 min before 07 30 is 04 10.
(b)
To find the time that Brandon took for the remaining 45 km of the trip, we need to use the time that Tim took subtract that of the time taken for first 5 km and the time that Brandon was faster than Tim.
Time that Brandon took for the remaining 45 km of the trip
= 3
13 -
13 -
13 = 2
23 h
Brandon's average speed
= 45 ÷ 2
23 = 16
78 km/h
Answer(s): (a) 04 10; (b) 16
78 km/h