Rael and Ken started on a 48-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 43 km, Rael cycled faster than Ken. He arrived at the finishing point 20 minutes before Ken who was 5 km behind him. Ken did not change his speed throughout and completed it at 08 25.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Rael's average speed for the remaining 43 km of the trip in km/h?

(a)
60 min = 1 h
20 min =²⁰₆₀ h = ¹₃ 

Ken's average speed
= 5 ÷ ¹₃
= 15 km/h

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

3 h 12 min before 08 25 is 05 13.

(b)
To find the time that Rael took for the remaining 43 km of the trip, we need to use the time that Ken took subtract that of the time taken for first 5 km and the time that Rael was faster than Ken.

Time that Rael took for the remaining 43 km of the trip
= 3¹₅ - ¹₃ - ¹₃
= 2⁸₁₅ h

Rael's average speed
= 43 ÷ 2⁸₁₅
= 16³⁷₃₈ km/h

Answer(s): (a) 05 13; (b) 16³⁷₃₈ km/h