Riordan and Zane 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, Riordan cycled faster than Zane. He arrived at the finishing point 40 minutes before Zane who was 10 km behind him. Zane did not change his speed throughout and completed it at 07 30.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 20 min before 07 30 is 04 10.

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

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

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

Answer(s): (a) 04 10; (b) 20 km/h