Two men start together to walk to a certain destination, one at 3 kmph and another at 3.75 kmph. The latter arrives half an hour before the former. The distance is

### Concept & Time Difference for Constant Distance When the distance ($d$) is constant, the difference in time taken at two different speeds ($S_1$ and $S_2$) can be expressed algebraically. $$\frac{d}{S_1} - \frac{d}{S_2} = \Delta T$$ ### Step-by-Step Solution 1. Let the total distance to the destination be $d$ km. 2. The speed of the first man is $S_1 = 3$ kmph. Time taken by him is $T_1 = \frac{d}{3}$ hours. 3. The speed of the second man is $S_2 = 3.75$ kmph (which is $\frac{15}{4}$ kmph). Time taken by him is $T_2 = \frac{d}{3.75} = \frac{4d}{15}$ hours. 4. We are given that the second man arrives half an hour ($\frac{1}{2}$ hour) before the first. $$T_1 - T_2 = \frac{1}{2}$$ $$\frac{d}{3} - \frac{4d}{15} = \frac{1}{2}$$ 5. Find a common denominator (15) to solve for $d$: $$\frac{5d - 4d}{15} = \frac{1}{2}$$ $$\frac{d}{15} = \frac{1}{2}$$ $$d = \frac{15}{2} = 7.5 \text{ km}$$ ### Exam Strategy & Shortcut Use the specific distance formula based on speeds and time difference: $$Distance = \frac{S_1 \times S_2}{S_1 \sim S_2} \times \Delta T$$ $$Distance = \frac{3 \times 3.75}{3.75 - 3} \times \frac{1}{2} = \frac{11.25}{0.75} \times \frac{1}{2} = 15 \times \frac{1}{2} = 7.5 \text{ km}$$ This yields the answer directly without algebraic setup. ### Common Pitfall A common error is keeping the time difference as $30$ minutes instead of converting it to $\frac{1}{2}$ an hour. Always ensure your units (km/h and hours) are consistent. ### Final Answer Therefore, the correct answer is **7.5 km**.