A policeman chases a thief who is running at a speed of $7\text{km/h}$. The speed of the policeman is $12\text{km/h}$. If the thief is $280\text{meters}$ ahead of the policeman, then how much time will the policeman take to catch the thief?

The initial distance between the policeman and the thief is $d=280\text{meters}$.
Converting the distance into kilometers:

$$d=\frac{280}{1000}\text{km}=\frac{14}{50}\text{km}=\frac{7}{25}\text{km}$$

Since both are running in the same direction, their relative speed ($v_{\text{rel}}$) is:

$$v_{\text{rel}}=\text{Speed of policeman}-\text{Speed of thief}$$

$$v_{\text{rel}}=12\text{km/h}-7\text{km/h}=5\text{km/h}$$

The time ($t$) taken by the policeman to catch the thief is:

$$t=\frac{\text{Distance}}{\text{Relative Speed}}=\frac{\frac{7}{25}}{5}\text{hours}=\frac{7}{120}\text{hours}$$

Converting hours into minutes (by multiplying by $60$):

$$t=\frac{7}{120}\times 60\text{minutes}=\frac{7}{2}\text{minutes}=3.5\text{minutes}$$

Let us verify the given mixed fraction formats with a denominator of $25$:

$$\frac{7}{2}\text{minutes}=\frac{7\times 12.5}{2\times 12.5}=\frac{87.5}{25}\text{minutes}$$

Alternatively, converting option D ($3\frac{9}{25}$):

$$3\frac{9}{25}=\frac{3\times 25+9}{25}=\frac{84}{25}\text{minutes}$$

Let's re-verify the fractional conversion from standard time metrics to match option markings:

$$\text{Time in minutes}=\frac{\text{Distance in meters}}{\text{Relative speed in m/s}}$$

$$v_{\text{rel}}=5\text{km/h}=5\times \frac{5}{18}\text{m/s}=\frac{25}{18}\text{m/s}$$

$$t=\frac{280}{\frac{25}{18}}=\frac{280\times 18}{25}\text{seconds}$$

$$t=\frac{5040}{25}\text{seconds}$$

Converting seconds to minutes by dividing by $60$:

$$t=\frac{5040}{25\times 60}\text{minutes}=\frac{84}{25}\text{minutes}=3\frac{9}{25}\text{minutes}$$