Perry 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, Perry 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 09 30.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Perry'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 09 30 is 07 15.

(b)
To find the time that Perry 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 Perry was faster than Justin.

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

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

Answer(s): (a) 07 15; (b) 28 km/h