Sean and Tommy started on a 55-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 50 km, Sean cycled faster than Tommy. He arrived at the finishing point 12 minutes before Tommy who was 5 km behind him. Tommy did not change his speed throughout and completed it at 08 30.
- At what time did the trip begin? Give the answer in 12-hour format.
- What was Sean's average speed for the remaining 50 km of the trip in km/h?
(a)
60 min = 1 h
12 min =
1260 h =
15 Tommy's average speed
= 5 ÷
15 = 25 km/h
Time that Tommy took for the trip
= 55 ÷ 25
= 2
525 h
= 2
15 h
= 2 h 12 min
2 h 12 min before 08 30 is 06 18.
(b)
To find the time that Sean took for the remaining 50 km of the trip, we need to use the time that Tommy took subtract that of the time taken for first 5 km and the time that Sean was faster than Tommy.
Time that Sean took for the remaining 50 km of the trip
= 2
15 -
15 -
15 = 1
45 h
Sean's average speed
= 50 ÷ 1
45 = 27
79 km/h
Answer(s): (a) 06 18; (b) 27
79 km/h