Luis and Justin started on a 45-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 35 km, Luis cycled faster than Justin. He arrived at the finishing point 30 minutes before Justin who was 10 km behind him. Justin did not change his speed throughout and completed it at 07 30.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Luis's average speed for the remaining 35 km of the trip in km/h?

(a)
60 min = 1 h
30 min =³⁰₆₀ h = ¹₂ 

Justin's average speed
= 10 ÷ ¹₂
= 20 km/h

Time that Justin took for the trip
= 45 ÷ 20
= 2⁵₂₀ h
= 2¹₄ h
= 2 h 15 min

2 h 15 min before 07 30 is 05 15.

(b)
To find the time that Luis took for the remaining 35 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 Luis was faster than Justin.

Time that Luis took for the remaining 35 km of the trip
= 2¹₄ - ¹₂ - ¹₂
= 1¹₄ h

Luis's average speed
= 35 ÷ 1¹₄
= 28 km/h

Answer(s): (a) 05 15; (b) 28 km/h