Henry and Fred started on a 48-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 38 km, Henry cycled faster than Fred. He arrived at the finishing point 40 minutes before Fred who was 10 km behind him. Fred 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 Henry's average speed for the remaining 38 km of the trip in km/h?
(a)
60 min = 1 h
40 min =
4060 h =
23 Fred's average speed
= 10 ÷
23 = 15 km/h
Time that Fred 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 Henry took for the remaining 38 km of the trip, we need to use the time that Fred took subtract that of the time taken for first 10 km and the time that Henry was faster than Fred.
Time that Henry took for the remaining 38 km of the trip
= 3
15 -
23 -
23 = 1
1315 h
Henry's average speed
= 38 ÷ 1
1315 = 20
514 km/h
Answer(s): (a) 04 13; (b) 20
514 km/h