X travels 6 km on a bicycle with average speeds of 5 km per hour, 10 km per hour and 4 km per hour during the first 1 km, the next 2 km and the remaining 3 km, respectively. Y travels the same distances with average speeds of 4 km per hour, 10 km per hour and 5 km per hour, respectively. How many minutes early will Y complete the journey if both X and Y start at the same time?

Time taken by X = 1/5 + 2/10 + 3/4 = 0.2 + 0.2 + 0.75 = 1.15 hours. Time taken by Y = 1/4 + 2/10 + 3/5 = 0.25 + 0.2 + 0.6 = 1.05 hours. Difference = 1.15 - 1.05 = 0.10 hours = 6 minutes. Wait — let me recheck: X takes 0.2 + 0.2 + 0.75 = 1.15 hr; Y takes 0.25 + 0.2 + 0.6 = 1.05 hr. Difference = 0.10 hr = 6 minutes. So Y finishes 6 minutes earlier. Answer: (d) 6. The marked answer in the key may differ, but the computation gives 6 minutes.