Elijah and Peter 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, Elijah cycled faster than Peter. He arrived at the finishing point 40 minutes before Peter who was 10 km behind him. Peter did not change his speed throughout and completed it at 08 25.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Elijah's average speed for the remaining 38 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 12 min before 08 25 is 05 13.

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

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

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

Answer(s): (a) 05 13; (b) 20⁵₁₄ km/h