Billy and Asher 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, Billy cycled faster than Asher. He arrived at the finishing point 40 minutes before Asher who was 10 km behind him. Asher did not change his speed throughout and completed it at 07 25.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 12 min before 07 25 is 04 13.

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

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

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

Answer(s): (a) 04 13; (b) 20⁵₁₄ km/h