Journey on a train!

Let the required distance be $d$ km. and the original speed of the train be $u$ km/hr. Then

$d\left (\dfrac{1}{u}-\dfrac{1}{u+10}\right ) =1\implies d=\dfrac{u^2+10u}{10}$

$d\left (\dfrac{1}{u-6}-\dfrac{1}{u}\right ) =1\implies d=\dfrac{u^2-6u}{6}$ .

So, $\dfrac{u^2+10u}{10}=\dfrac{u^2-6u}{6}$

$\implies 3u^2+30u=5u^2-30u\implies u=30$ km/hr. (Since $u\neq 0$ ).

Hence, $d=\dfrac{900+300}{10}=\boxed {120}$ km.

$d=vt$ $[1]$

$d=(v+10)(t-1)=vt-v+10t-10$ $[2]$

$d=(v-6)(t+1)=vt+v-6t-6$ $[3]$

Subtracting $[3]$ from $[2]$ , we get

$2v+4=16t$ $[4]$

Subtracting $[1]$ from $[2]$ , we get

$v=10t-10$ $[5]$

Substituting $[5]$ into $[4]$ , we get

$t=4$

It follows that

$v=10t-10=10(4) - 10=30$

Finally,

$d=vt=(30)(4)=\text{\boxed{120 km}}$