Tim and Will started on a 55-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 45 km, Tim cycled faster than Will. He arrived at the finishing point 24 minutes before Will who was 10 km behind him. Will 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 Tim's average speed for the remaining 45 km of the trip in km/h?
(a)
60 min = 1 h
24 min =
2460 h =
25 Will's average speed
= 10 ÷
25 = 25 km/h
Time that Will took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 07 25 is 05 13.
(b)
To find the time that Tim took for the remaining 45 km of the trip, we need to use the time that Will took subtract that of the time taken for first 10 km and the time that Tim was faster than Will.
Time that Tim took for the remaining 45 km of the trip
= 2
15 -
25 -
25 = 1
25 h
Tim's average speed
= 45 ÷ 1
25 = 32
17 km/h
Answer(s): (a) 05 13; (b) 32
17 km/h