A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is

### Concept & Speed-Time Variations When speed changes cause corresponding changes in time for a constant distance, we can use the formula relating distance, speed, and time difference: $$Distance = \frac{v_1 \times v_2}{|v_1 - v_2|} \times \Delta t$$ ### Step-by-Step Solution * **Establish Variables:** Let actual speed be $v$ kmph and distance be $d$ km. Convert 40 minutes to hours: $\frac{40}{60} = \frac{2}{3}$ hours. * **Set up Equations:** * Case 1 (Faster): $d = \frac{v(v + 3)}{3} \times \frac{2}{3}$ * Case 2 (Slower): $d = \frac{v(v - 2)}{2} \times \frac{2}{3}$ * **Equate and Solve for $v$:** * $\frac{v(v + 3)}{3} = \frac{v(v - 2)}{2}$ * Divide both sides by $v$ (since $v \neq 0$): * $\frac{v + 3}{3} = \frac{v - 2}{2}$ * $2(v + 3) = 3(v - 2)$ * $2v + 6 = 3v - 6$ * $v = 12$ kmph. * **Calculate Distance:** * Substitute $v = 12$ into the first equation: * $d = \frac{12(12 + 3)}{3} \times \frac{2}{3}$ * $d = \frac{12 \times 15}{3} \times \frac{2}{3} = 4 \times 15 \times \frac{2}{3} = 40$ km. ### Exam Strategy & Shortcut Since both cases have the same time difference (40 mins), the ratio of the speed changes directly maps to the equations. You can quickly balance $\frac{v+3}{3} = \frac{v-2}{2}$ to find original speed mentally ($v=12$). Then plug it into the distance formula. ### Common Pitfall Forgetting to convert minutes to hours before multiplying with speeds in kmph, which throws off the final distance value entirely. ### Final Answer Therefore, the correct answer is **40**.