Wesley and Elijah 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, Wesley cycled faster than Elijah. He arrived at the finishing point 40 minutes before Elijah who was 10 km behind him. Elijah 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 Wesley's average speed for the remaining 38 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 12 min before 09 30 is 06 18.

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

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

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

Answer(s): (a) 06 18; (b) 20⁵₁₄ km/h