Tommy and Wesley 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, Tommy cycled faster than Wesley. He arrived at the finishing point 20 minutes before Wesley who was 5 km behind him. Wesley 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 Tommy's average speed for the remaining 43 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Wesley's average speed
= 5 ÷
13 = 15 km/h
Time that Wesley took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 09 30 is 06 18.
(b)
To find the time that Tommy took for the remaining 43 km of the trip, we need to use the time that Wesley took subtract that of the time taken for first 5 km and the time that Tommy was faster than Wesley.
Time that Tommy took for the remaining 43 km of the trip
= 3
15 -
13 -
13 = 2
815 h
Tommy's average speed
= 43 ÷ 2
815 = 16
3738 km/h
Answer(s): (a) 06 18; (b) 16
3738 km/h