Seth and Nick started on a 50-km cycling trip at the same time. They cycled at the same speed for first 10 km. For the remaining 40 km, Seth cycled faster than Nick. He arrived at the finishing point 40 minutes before Nick who was 10 km behind him. Nick 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 Seth's average speed for the remaining 40 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

Time that Nick took for the trip
= 50 ÷ 15
= 3⁵₁₅ h
= 3¹₃ h
= 3 h 20 min

3 h 20 min before 09 30 is 06 10.

(b)
To find the time that Seth took for the remaining 40 km of the trip, we need to use the time that Nick took subtract that of the time taken for first 10 km and the time that Seth was faster than Nick.

Time that Seth took for the remaining 40 km of the trip
= 3¹₃ - ²₃ - ²₃
= 2 h

Seth's average speed
= 40 ÷ 2
= 20 km/h

Answer(s): (a) 06 10; (b) 20 km/h