If a man walks at the rate of 5 km/hour, he misses a train by 7 minutes. However, if he walks at the rate of 6 km/hour, he reaches the station 5 minutes before the arrival of the train. The distance covered by him to reach the station is?Options6 km7 km6.26 km4 km

To solve this problem, we need to understand that the difference in time when the man misses the train and when he reaches before the train is 7 minutes (late) + 5 minutes (early) = 12 minutes = 12/60 hours = 0.2 hours.

Let's denote the distance between the man's starting point and the station as 'd' (in km), and the speed difference between the two scenarios as 's' (in km/hour). In the first scenario, the man walks at a speed of 5 km/hour, and in the second scenario, he walks at a speed of 6 km/hour. So, the speed difference 's' is 6 km/hour - 5 km/hour = 1 km/hour.

We know that time = distance/speed. In this case, the difference in time (0.2 hours) is due to the difference in speed (1 km/hour). So, we can write the equation as 0.2 = d/s.

Substituting s = 1 km/hour into the equation, we get 0.2 = d/1, which simplifies to d = 0.2 * 1 = 0.2 km.

However, this is the additional distance the man covers when he increases his speed to 6 km/hour. The total distance he covers when walking at 5 km/hour is therefore d + 0.2 = 5 km/hour * (7/60 hours + 0.2 hours) = 5 * 0.3167 = 1.583 km.

Adding this to the additional distance covered at the higher speed, we get the total distance as 1.583 km + 0.2 km = 1.783 km.

So, the man covers a distance of approximately 1.783 km to reach the station. However, this is not one of the options given. It seems there might be a mistake in the problem or the options provided.