Lee and Cody 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 Cody. He arrived at the finishing point 40 minutes before Cody who was 12 km behind him. Cody 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 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 

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

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

3 h 20 min before 08 30 is 05 10.

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

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) 05 10; (b) 24 km/h