Fred and Tim 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, Fred cycled faster than Tim. He arrived at the finishing point 40 minutes before Tim who was 10 km behind him. Tim did not change his speed throughout and completed it at 09 30.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 20 min before 09 30 is 06 10.

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

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

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

Answer(s): (a) 06 10; (b) 20 km/h