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 then the scheduled time. And, if the train were slower by 6 km/hr, then the train would have take 6 hours more than the scheduled time. The length of the journey is :

Solution(By Examveda Team)

Let distance = x km and usual speed = y kmph
$$\eqalign{ & \Rightarrow \frac{x}{y} - \frac{x}{{y + 6}} = 4 \cr & \Rightarrow 6x = 4y\left( {y + 6} \right).....(i) \cr} $$
And,
$$\eqalign{ & \Rightarrow \frac{x}{{y - 6}} - \frac{x}{y} = 6 \cr & \Rightarrow 6x = 6y\left( {y - 6} \right).....(ii) \cr} $$
From (i) and (ii), we get :
$$\eqalign{ & \Rightarrow 4y\left( {y + 6} \right) = 6y\left( {y - 6} \right) \cr & \Rightarrow 2\left( {y + 6} \right) = 3\left( {y - 6} \right) \cr & \Rightarrow y = 30 \cr} $$
∴ Length of the journey :
$$\eqalign{ & = x{\text{ km}} \cr & = \left( {\frac{{4y\left( {y + 6} \right)}}{6}} \right){\text{ km}} \cr & = \left( {\frac{{4 \times 30 \times 36}}{6}} \right){\text{ km}} \cr & = 720{\text{ km}} \cr} $$