Sam and Wesley started on a 48-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 38 km, Sam cycled faster than Wesley. He arrived at the finishing point 40 minutes before Wesley who was 10 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 Sam's average speed for the remaining 38 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

Time that Wesley took for the trip
= 48 ÷ 15
= 3³₁₅ h
= 3¹₅ h
= 3 h 12 min

3 h 12 min before 07 30 is 04 18.

(b)
To find the time that Sam took for the remaining 38 km of the trip, we need to use the time that Wesley took subtract that of the time taken for first 10 km and the time that Sam was faster than Wesley.

Time that Sam took for the remaining 38 km of the trip
= 3¹₅ - ²₃ - ²₃
= 1¹³₁₅ h

Sam's average speed
= 38 ÷ 1¹³₁₅
= 20⁵₁₄ km/h

Answer(s): (a) 04 18; (b) 20⁵₁₄ km/h