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?

The fundamental principle to solve this problem relies on the kinematic equation: Time = Distance ÷ Speed. To find the total time taken by each individual, we must calculate the time taken for each distinct segment of the journey independently and sum them up.

Calculations for X:

• First 1 km at 5 km/hr: Time = 1/5 hour
• Next 2 km at 10 km/hr: Time = 2/10 = 1/5 hour
• Remaining 3 km at 4 km/hr: Time = 3/4 hour Total Time for X = 1/5 + 1/5 + 3/4 = 2/5 + 3/4 = (8 + 15) / 20 = 23/20 hours. To convert this into minutes: (23/20) × 60 = 69 minutes.

Calculations for Y:

• First 1 km at 4 km/hr: Time = 1/4 hour
• Next 2 km at 10 km/hr: Time = 2/10 = 1/5 hour
• Remaining 3 km at 5 km/hr: Time = 3/5 hour Total Time for Y = 1/4 + 1/5 + 3/5 = 1/4 + 4/5 = (5 + 16) / 20 = 21/20 hours. To convert this into minutes: (21/20) × 60 = 63 minutes.

Conclusion: The difference in time taken is 69 minutes - 63 minutes = 6 minutes. Since Y takes less time, Y completes the journey 6 minutes early, making Option D the correct answer.

• Option A (3) is incorrect. This is a common distractor resulting from the erroneous mathematical practice of averaging the raw speeds without accounting for unequal distances.
• Option B (4) is incorrect. This arises if an aspirant misallocates the distances to the wrong speed denominators during calculation.
• Option C (5) is incorrect. This typically results from simple arithmetic errors when calculating the lowest common multiple (LCM) of the fractions.

Takeaway: Never average speeds directly when distances vary. Always calculate the specific time for each segment independently using T = D/S before finding the total sum.