Q: If a man walks at 5 kmph, he will miss a train by 7 minutes. However, if he walks at a speed of 6 km/h, he arrives at the station 5 minutes before the train arrives. Determine the distance he travelled to reach the station.

The difference in the time taken to travel a certain distance at two different speeds is given as 1 minute, or 1/2 hour.

The time taken to travel the distance at the slower speed is given as 1/5 of the time taken to travel the same distance at the faster speed.

We can set up an equation to represent the relationship between the time taken at the two speeds and the distance traveled. 

In this case, the equation is (distance traveled at slower speed) / (time taken at slower speed) - (distance traveled at faster speed) / (time taken at faster speed) = difference in time taken.

We can solve for the distance traveled by rearranging the equation to isolate the distance traveled. 

The equation given is 

(distance traveled at slower speed) / (1/5 of the time taken at faster speed) - (distance traveled at faster speed) / (time taken at faster speed) = 1/2 hour. 

We can rewrite this as (distance traveled at slower speed) / (1/5 of the time taken at faster speed) = (distance traveled at faster speed) / (time taken at faster speed) + 1/2 hour. 

Then, we can solve for the distance traveled by multiplying both sides of the equation by 5/6. This gives us (5/6 of the distance traveled at slower speed) = (5/6 of the distance traveled at faster speed) + 5/12 hour. 

Finally, we can solve for the distance traveled by subtracting (5/6 of the distance traveled at faster speed) from both sides of the equation, which gives us (5/6 of the distance traveled at slower speed) - (5/6 of the distance traveled at faster speed) = 5/12 hour. 

This simplifies to (1/6 of the distance traveled) = 5/12 hour, or x/6 = 5/12 hour. 

Solving for x gives us x = 6 km.