Seth and Brandon started on a 35-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 30 km, Seth cycled faster than Brandon. He arrived at the finishing point 20 minutes before Brandon who was 5 km behind him. Brandon 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 Seth's average speed for the remaining 30 km of the trip in km/h?
(a)
60 min = 1 h
20 min =
2060 h =
13 Brandon's average speed
= 5 ÷
13 = 15 km/h
Time that Brandon took for the trip
= 35 ÷ 15
= 2
515 h
= 2
13 h
= 2 h 20 min
2 h 20 min before 08 30 is 06 10.
(b)
To find the time that Seth took for the remaining 30 km of the trip, we need to use the time that Brandon took subtract that of the time taken for first 5 km and the time that Seth was faster than Brandon.
Time that Seth took for the remaining 30 km of the trip
= 2
13 -
13 -
13 = 1
23 h
Seth's average speed
= 30 ÷ 1
23 = 18 km/h
Answer(s): (a) 06 10; (b) 18 km/h