Andy and Albert 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, Andy cycled faster than Albert. He arrived at the finishing point 40 minutes before Albert who was 10 km behind him. Albert did not change his speed throughout and completed it at 08 30.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 20 min before 08 30 is 05 10.

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

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

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

Answer(s): (a) 05 10; (b) 20 km/h