A train travelled a certain distance at a uniform speed. Had the speed been 6 km per hour more, it would have needed 4 hours less. Had the speed been 6 km per hour less, it would have needed 6 hours more. The distance, in km, travelled by the train is

Let us assume that the distance is D, speed of the train is S and time taken by the train is t. 

t is nothing but $$\frac{D}{S}$$

Statement 1: Had the speed been 6 km per hour more, it would have needed 4 hours less
$$\frac{D}{S+6}=t-4$$

$$\frac{D}{S+6}=\frac{D}{S}-4$$

$$4=\frac{D}{S}-\frac{D}{S+6}$$

$$\frac{S+6-S}{S\left(S+6\right)}=\frac{4}{D}$$

$$\frac{6}{S\left(S+6\right)}=\frac{4}{D}$$

$$D=\frac{2S\left(S+6\right)}{3}$$

Statement 2:  Had the speed been 6 km per hour less, it would have needed 6 hours more
$$\frac{D}{S-6}=t+6$$

$$D\left[\frac{1}{S-6}-\frac{1}{S}\right]=6$$

$$\frac{S-S+6}{S\left(S-6\right)}=6$$

$$D=S\left(S-6\right)$$

Equating the two equations for distance, 
$$S\left(S-6\right)=\frac{2S\left(S+6\right)}{3}$$

$$3S-18=2S+12$$

$$S=30$$

Hence the speed is 30 kmph

We know that the distance $$D=S\left(S-6\right)$$ = 30*24  =720 km