Bryan and Michael started on a 55-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 50 km, Bryan cycled faster than Michael. He arrived at the finishing point 12 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 Bryan's average speed for the remaining 50 km of the trip in km/h?

(a)
60 min = 1 h
12 min =¹²₆₀ h = ¹₅ 

Michael's average speed
= 5 ÷ ¹₅
= 25 km/h

Time that Michael took for the trip
= 55 ÷ 25
= 2⁵₂₅ h
= 2¹₅ h
= 2 h 12 min

2 h 12 min before 07 30 is 05 18.

(b)
To find the time that Bryan took for the remaining 50 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 Bryan was faster than Michael.

Time that Bryan took for the remaining 50 km of the trip
= 2¹₅ - ¹₅ - ¹₅
= 1⁴₅ h

Bryan's average speed
= 50 ÷ 1⁴₅
= 27⁷₉ km/h

Answer(s): (a) 05 18; (b) 27⁷₉ km/h