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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

Time that Reggie 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 Peter took for the remaining 38 km of the trip, we need to use the time that Reggie took subtract that of the time taken for first 10 km and the time that Peter was faster than Reggie.

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

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

Answer(s): (a) 06 13; (b) 20⁵₁₄ km/h