Eric and Luke 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, Eric cycled faster than Luke. He arrived at the finishing point 40 minutes before Luke who was 10 km behind him. Luke did not change his speed throughout and completed it at 07 30.

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

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

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

3 h 12 min before 07 30 is 04 18.

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

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

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

Answer(s): (a) 04 18; (b) 20⁵₁₄ km/h