Charlie and Xavier 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, Charlie cycled faster than Xavier. He arrived at the finishing point 40 minutes before Xavier who was 10 km behind him. Xavier 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 Charlie's average speed for the remaining 40 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

Time that Xavier 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 Charlie took for the remaining 40 km of the trip, we need to use the time that Xavier took subtract that of the time taken for first 10 km and the time that Charlie was faster than Xavier.

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

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

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