Tommy and Owen started on a 48-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 38 km, Tommy cycled faster than Owen. He arrived at the finishing point 40 minutes before Owen who was 10 km behind him. Owen did not change his speed throughout and completed it at 08 30.

1. At what time did the trip begin? Give the answer in 12-hour format.
2. What was Tommy's average speed for the remaining 38 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

Owen's average speed
= 10 ÷ ²₃
= 15 km/h

Time that Owen took for the trip
= 48 ÷ 15
= 3³₁₅ h
= 3¹₅ h
= 3 h 12 min

3 h 12 min before 08 30 is 05 18.

(b)
To find the time that Tommy took for the remaining 38 km of the trip, we need to use the time that Owen took subtract that of the time taken for first 10 km and the time that Tommy was faster than Owen.

Time that Tommy took for the remaining 38 km of the trip
= 3¹₅ - ²₃ - ²₃
= 1¹³₁₅ h

Tommy's average speed
= 38 ÷ 1¹³₁₅
= 20⁵₁₄ km/h

Answer(s): (a) 05 18; (b) 20⁵₁₄ km/h