Tommy and Luis 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 Luis. He arrived at the finishing point 20 minutes before Luis who was 5 km behind him. Luis did not change his speed throughout and completed it at 07 25.
- 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 Luis's average speed
= 5 ÷
13 = 15 km/h
Time that Luis took for the trip
= 48 ÷ 15
= 3
315 h
= 3
15 h
= 3 h 12 min
3 h 12 min before 07 25 is 04 13.
(b)
To find the time that Tommy took for the remaining 43 km of the trip, we need to use the time that Luis took subtract that of the time taken for first 5 km and the time that Tommy was faster than Luis.
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) 04 13; (b) 16
3738 km/h