Riordan and Michael started on a 48-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 38 km, Riordan cycled faster than Michael. He arrived at the finishing point 40 minutes before Michael who was 10 km behind him. Michael 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 Riordan's average speed for the remaining 38 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

Time that Michael took for the trip
= 48 ÷ 15
= 3³₁₅ h
= 3¹₅ h
= 3 h 12 min

3 h 12 min before 09 30 is 06 18.

(b)
To find the time that Riordan took for the remaining 38 km of the trip, we need to use the time that Michael took subtract that of the time taken for first 10 km and the time that Riordan was faster than Michael.

Time that Riordan took for the remaining 38 km of the trip
= 3¹₅ - ²₃ - ²₃
= 1¹³₁₅ h

Riordan's average speed
= 38 ÷ 1¹³₁₅
= 20⁵₁₄ km/h

Answer(s): (a) 06 18; (b) 20⁵₁₄ km/h