Asher and Gabriel 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, Asher cycled faster than Gabriel. He arrived at the finishing point 40 minutes before Gabriel who was 10 km behind him. Gabriel did not change his speed throughout and completed it at 09 25.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 12 min before 09 25 is 06 13.

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

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

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

Answer(s): (a) 06 13; (b) 20⁵₁₄ km/h