Reggie and Andy started on a 48-km cycling trip at the same time. They cycled at the same speed for first 5 km. For the remaining 43 km, Reggie cycled faster than Andy. He arrived at the finishing point 20 minutes before Andy who was 5 km behind him. Andy 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 43 km of the trip in km/h?

(a)
60 min = 1 h
20 min =²⁰₆₀ h = ¹₃ 

Andy's average speed
= 5 ÷ ¹₃
= 15 km/h

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

3 h 12 min before 08 30 is 05 18.

(b)
To find the time that Reggie took for the remaining 43 km of the trip, we need to use the time that Andy took subtract that of the time taken for first 5 km and the time that Reggie was faster than Andy.

Time that Reggie took for the remaining 43 km of the trip
= 3¹₅ - ¹₃ - ¹₃
= 2⁸₁₅ h

Reggie's average speed
= 43 ÷ 2⁸₁₅
= 16³⁷₃₈ km/h

Answer(s): (a) 05 18; (b) 16³⁷₃₈ km/h