Nick and Justin 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, Nick cycled faster than Justin. He arrived at the finishing point 40 minutes before Justin who was 10 km behind him. Justin did not change his speed throughout and completed it at 08 25.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 12 min before 08 25 is 05 13.

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

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

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

Answer(s): (a) 05 13; (b) 20⁵₁₄ km/h