A thief, pursued by a policeman, was $100$ m ahead at the start. If the ratio of the speed of the policeman to that of the thief was $5 : 4$, then how far could the thief go before he was caught by the policeman?

### Concept & Logic When two bodies move in the same direction, their relative speed is the difference of their individual speeds. Furthermore, when the time traveled is constant for both bodies, the ratio of distances covered is equal to the ratio of their speeds. ### Step-by-Step Solution * **Given:** The policeman and thief have a speed ratio of $5:4$. The thief has a head start of $100$ m. * **Calculation:** Let the speed of the policeman be $5x$ and the thief be $4x$. * The relative speed is $5x - 4x = 1x$. * The time taken for the policeman to catch the thief is the time needed to close the $100$ m gap at the relative speed: $\text{Time} = \frac{100}{x}$. * The distance covered by the thief during this time is $\text{Speed} \times \text{Time} = 4x \times \left(\frac{100}{x}\right) = 400$ m. ### Exam Strategy & Shortcut Since the time is constant, $\text{Distance} \propto \text{Speed}$. The policeman covers $5$ units of distance for every $4$ units the thief covers. The difference ($1$ unit) corresponds exactly to the $100$ m head start. Thus, the thief's distance ($4$ units) is simply $4 \times 100 = 400$ m. ### Common Pitfall A frequent mistake is calculating the total distance run by the policeman ($500$ m) and hastily selecting it as the answer, forgetting that the question specifically asks for the distance traveled by the thief. ### Final Answer Therefore, the correct answer is **$400$ m**.