A fast train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.
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.