A man had to travel 49 km, which he divided into six stretches which were exact multiples of either 4 km or 5 km. The first four stretches were 4 km 5 km, 8 km and 10 km. Multiples of 4 km and 5 km were travelled at the rates of 40 and 60 km/h, respectively, and between two consecutive stretches, he rested for six minutes. What was his average speed (in km/h) for the whole journey?

Remaining distance = 49 - (4 + 5 + 8 + 10) ⇒ 22 km Only 12 km and 10 km is possible according to the question, Total time taken to complete the journey = (4/40 + 5/60 + 8/40 + 10/60 + 12/40 + 10/60)h + 6 × 5 m [Total 5 times halt of 6 minutes each] ⇒ (1/10 + 1/12 + 1/5 + 1/6 + 3/10 + 1/6)h + 30 m ⇒ (6 + 5 + 12 + 10 + 18 + 10)/60 + 1/2 ⇒ 61/60 + 1/2 ⇒ 91/60 hours Average speed = 49 × 60/91 ⇒ 420/13 ⇒ 32(4/13) ∴ His average speed (in km/h) for the whole journey was 32(4/13).