Jack and Howard 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, Jack cycled faster than Howard. He arrived at the finishing point 24 minutes before Howard who was 10 km behind him. Howard did not change his speed throughout and completed it at 07 25.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Jack's average speed for the remaining 45 km of the trip in km/h?

(a)
60 min = 1 h
24 min =²⁴₆₀ h = ²₅ 

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

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

2 h 12 min before 07 25 is 05 13.

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

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

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

Answer(s): (a) 05 13; (b) 32¹₇ km/h