Dylan and Will started on a 50-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 40 km, Dylan cycled faster than Will. He arrived at the finishing point 40 minutes before Will who was 10 km behind him. Will did not change his speed throughout and completed it at 08 30.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Dylan's average speed for the remaining 40 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

Will's average speed
= 10 ÷ ²₃
= 15 km/h

Time that Will took for the trip
= 50 ÷ 15
= 3⁵₁₅ h
= 3¹₃ h
= 3 h 20 min

3 h 20 min before 08 30 is 05 10.

(b)
To find the time that Dylan took for the remaining 40 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 Dylan was faster than Will.

Time that Dylan took for the remaining 40 km of the trip
= 3¹₃ - ²₃ - ²₃
= 2 h

Dylan's average speed
= 40 ÷ 2
= 20 km/h

Answer(s): (a) 05 10; (b) 20 km/h