A worker reaches his factory 3 minutes late if his speed from his house to the factory is 5 km/hr. If he walks at a speed of 6 km/hr, then he reaches the factory 7 minutes early. The distance of the factory from his house is

To solve this elegantly, I first determine the effective time difference. One scenario is 3 minutes late (+3) and the other is 7 minutes early (-7). The gap between them is 3 + 7 = 10 minutes.\n\nSince speed is in km/hr, I convert 10 minutes to hours: 10/60 = 1/6 hours.\n\nNow, applying the 'Product of Speeds' shortcut which is efficient for this specific exam pattern:\nDistance = (Time Difference) * (Speed1 * Speed2) / (Speed2 - Speed1)\n\nCalculation:\nDistance = (1/6) * (5 * 6) / (6 - 5)\nDistance = (1/6) * (30) / 1\nDistance = 5 km.\n\nAlternatively, looking at the options: I need a distance that, when divided by 5 and 6, gives a time difference of 10 minutes. \nOption (C) 5 km is the most intuitive:\nAt 5 km/hr, time = 1 hour (60 mins).\nAt 6 km/hr, time = 5/6 hour (50 mins).\nThe difference is exactly 10 minutes. This confirms the answer immediately.