Rates

Doesn't the thief have a 500 meter head start?

Yes, I think "m" stands for "meters" in this problem so no conversion to miles necessary

Here's how I would work through it:

The unknown is time = x
The rate of the cop is 1/6th of a km per minute
The rate of the thief is 1/10th of a km per minute
But the thief has 1/2 of a km head start (500m)

So:

1/6*x - (1/10*x + 1/2) = 0
x/6 - x/10 - 1/2 = 0
5x/30 - 3x/30 = 1/2
2x/30 = 1/2
4x = 30
x = 7.5 minutes = 7 and 1/2 minutes

Is that the answer?

Shucks! You don't have to beat yourself up for something like this.

One thing you may want to do is to convert everything to a base where things are easy to calculate.

You have distance and speed ... you need time.

First i convert the speeds to a common unit ... kmph...

If the policeman runs 1 km in 6 mins. He will run 10 km in 1 hr. So his speed is 10kmph.

Similarly the thief will run 6 km in that same hr.

The difference is speed is 4kmph

The extra distance the policeman has to cover is 500m = 0.5 km

So to catch the thief, the time he will take is 0.5/4 = 1/8 of an hr or 1/8 * 60 min = 7.5 mins