A train covered a certain distance at a uniform speed. If the train had been 6 km/hr faster, then it would have taken 4 hours less than the scheduled time. And, if the train were slower by 6 km/hr, then the train would have taken 6 hours more than the scheduled time. The length of the journey is (M.A.T., 2006)

### Concept & Speed-Time Variations Just like the previous problem, use the formula for distance when variations in speed and time are given: $$Distance = \frac{v \times (v \pm \Delta v)}{\Delta v} \times \Delta t$$ ### Step-by-Step Solution * **Establish Variables:** Let original speed be $v$ km/hr and distance be $d$ km. * **Set up Equations:** * Case 1 (Faster): $d = \frac{v(v + 6)}{6} \times 4$ * Case 2 (Slower): $d = \frac{v(v - 6)}{6} \times 6$ * **Equate and Solve for $v$:** * $\frac{v(v + 6)}{6} \times 4 = \frac{v(v - 6)}{6} \times 6$ * $4(v + 6) = 6(v - 6)$ * $4v + 24 = 6v - 36$ * $2v = 60 \Rightarrow v = 30$ km/hr. * **Calculate Distance:** * Substitute $v = 30$ into the Case 1 equation: * $d = \frac{30(30 + 6)}{6} \times 4$ * $d = 5 \times 36 \times 4 = 180 \times 4 = 720$ km. ### Exam Strategy & Shortcut Equate the two conditions directly by focusing on the remaining terms: $\frac{\Delta t_1}{\Delta v_1}(v + \Delta v_1) = \frac{\Delta t_2}{\Delta v_2}(v - \Delta v_2)$. Here, $\frac{4}{6}(v+6) = \frac{6}{6}(v-6)$, simplifying to $\frac{2}{3}(v+6) = v-6$, which quickly yields $v=30$. ### Common Pitfall Misinterpreting the variables and trying to solve for total time instead of substituting the speed directly back into the distance formula. ### Final Answer Therefore, the correct answer is **720 km**.