A fast train takes one hour less than a slow train for a journey of 200 km.

Solution:

Let the speed of the fast train be $x \mathrm{~km} / \mathrm{hr}$ then

the speed of the slow train be $=(x-10) \mathrm{km} / \mathrm{hr}$

Time taken by the fast train to cover $200 \mathrm{~km}=\frac{200}{x} \mathrm{hr}$

Time taken by the slow train to cover $200 \mathrm{~km}=\frac{200}{(x-10)} \mathrm{hr}$

Therefore,

$\frac{200}{(x-10)}-\frac{200}{x}=1$

$\Rightarrow \frac{200 x-200(x-10)}{x(x-10)}=1$

$\Rightarrow \frac{200 x-220 x+2000}{x^{2}-10 x}=1$

$\Rightarrow x^{2}-10 x=2000$

 

$\Rightarrow x^{2}-10 x-2000=0$

$\Rightarrow x^{2}-50 x+40 x-2000=0$

$\Rightarrow x(x-50)+40(x-50)=0$

 

$\Rightarrow(x-50)(x+40)=0$

So, either

$(x-50)=0$

$x=50$

Or

$(x+40)=0$

$x=-40$

But, the speed of the train can never be negative.

Thus, when $x=50$ then

$=(x-10)$

$=(50-10)$

 

$=40$

Hence, the speed of the fast train is $x=50 \mathrm{~km} / \mathrm{hr}$

and the speed of the slow train is $x=40 \mathrm{~km} / \mathrm{hr}$ respectively.