Owen and Oscar 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, Owen cycled faster than Oscar. He arrived at the finishing point 24 minutes before Oscar who was 10 km behind him. Oscar 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 Owen's average speed for the remaining 45 km of the trip in km/h?

(a)
60 min = 1 h
24 min =²⁴₆₀ h = ²₅ 

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

Time that Oscar 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 Owen took for the remaining 45 km of the trip, we need to use the time that Oscar took subtract that of the time taken for first 10 km and the time that Owen was faster than Oscar.

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

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

Answer(s): (a) 07 18; (b) 32¹₇ km/h