Reggie and Ahmad started on a 50-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 40 km, Reggie cycled faster than Ahmad. He arrived at the finishing point 40 minutes before Ahmad who was 10 km behind him. Ahmad did not change his speed throughout and completed it at 08 30.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Reggie's average speed for the remaining 40 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

Time that Ahmad took for the trip
= 50 ÷ 15
= 3⁵₁₅ h
= 3¹₃ h
= 3 h 20 min

3 h 20 min before 08 30 is 05 10.

(b)
To find the time that Reggie took for the remaining 40 km of the trip, we need to use the time that Ahmad took subtract that of the time taken for first 10 km and the time that Reggie was faster than Ahmad.

Time that Reggie took for the remaining 40 km of the trip
= 3¹₃ - ²₃ - ²₃
= 2 h

Reggie's average speed
= 40 ÷ 2
= 20 km/h

Answer(s): (a) 05 10; (b) 20 km/h