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

(a)
60 min = 1 h
24 min =²⁴₆₀ h = ²₅ 

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

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

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

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

Answer(s): (a) 07 18; (b) 32¹₇ km/h