Julian and Michael 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, Julian cycled faster than Michael. He arrived at the finishing point 20 minutes before Michael who was 5 km behind him. Michael 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 Julian's average speed for the remaining 43 km of the trip in km/h?

(a)
60 min = 1 h
20 min =²⁰₆₀ h = ¹₃ 

Michael's average speed
= 5 ÷ ¹₃
= 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 07 30 is 04 18.

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

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

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

Answer(s): (a) 04 18; (b) 16³⁷₃₈ km/h