Justin and Jenson started on a 55-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 45 km, Justin cycled faster than Jenson. He arrived at the finishing point 24 minutes before Jenson who was 10 km behind him. Jenson 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 Justin's average speed for the remaining 45 km of the trip in km/h?

(a)
60 min = 1 h
24 min =²⁴₆₀ h = ²₅ 

Jenson's average speed
= 10 ÷ ²₅
= 25 km/h

Time that Jenson took for the trip
= 55 ÷ 25
= 2⁵₂₅ h
= 2¹₅ h
= 2 h 12 min

2 h 12 min before 09 30 is 07 18.

(b)
To find the time that Justin took for the remaining 45 km of the trip, we need to use the time that Jenson took subtract that of the time taken for first 10 km and the time that Justin was faster than Jenson.

Time that Justin took for the remaining 45 km of the trip
= 2¹₅ - ²₅ - ²₅
= 1²₅ h

Justin's average speed
= 45 ÷ 1²₅
= 32¹₇ km/h

Answer(s): (a) 07 18; (b) 32¹₇ km/h