Lee and Wesley started on a 60-km cycling trip at the same time. They cycled at the same speed for first 12 km. For the remaining 48 km, Lee cycled faster than Wesley. He arrived at the finishing point 40 minutes before Wesley who was 12 km behind him. Wesley 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 Lee's average speed for the remaining 48 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

Wesley's average speed
= 12 ÷ ²₃
= 18 km/h

Time that Wesley took for the trip
= 60 ÷ 18
= 3⁶₁₈ h
= 3¹₃ h
= 3 h 20 min

3 h 20 min before 07 30 is 04 10.

(b)
To find the time that Lee took for the remaining 48 km of the trip, we need to use the time that Wesley took subtract that of the time taken for first 12 km and the time that Lee was faster than Wesley.

Time that Lee took for the remaining 48 km of the trip
= 3¹₃ - ²₃ - ²₃
= 2 h

Lee's average speed
= 48 ÷ 2
= 24 km/h

Answer(s): (a) 04 10; (b) 24 km/h