Howard and Bobby 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, Howard cycled faster than Bobby. He arrived at the finishing point 40 minutes before Bobby who was 10 km behind him. Bobby 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 Howard's average speed for the remaining 38 km of the trip in km/h?

(a)
60 min = 1 h
40 min =⁴⁰₆₀ h = ²₃ 

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

Time that Bobby 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 Howard took for the remaining 38 km of the trip, we need to use the time that Bobby took subtract that of the time taken for first 10 km and the time that Howard was faster than Bobby.

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

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

Answer(s): (a) 05 18; (b) 20⁵₁₄ km/h