Asher and Ryan 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, Asher cycled faster than Ryan. He arrived at the finishing point 40 minutes before Ryan who was 10 km behind him. Ryan did not change his speed throughout and completed it at 09 30.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 20 min before 09 30 is 06 10.

(b)
To find the time that Asher took for the remaining 40 km of the trip, we need to use the time that Ryan took subtract that of the time taken for first 10 km and the time that Asher was faster than Ryan.

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

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

Answer(s): (a) 06 10; (b) 20 km/h