A faster train takes one hour less than a slower train for a journey of 200 km.

Solution:

Let the speed of faster train be x km/h.
Then, the speed of slower train is (x − 10) km/h.

Given:
A faster train takes one hour less than a slower train for a journey of 200 km.

$\frac{\text { Distance }}{\text { Speed }}=$ Time

Time taken by faster train to cover $200 \mathrm{~km}=\frac{200}{x} \mathrm{~h}$

Time taken by slower train to cover $200 \mathrm{~km}=\frac{200}{x-10} \mathrm{~h}$

According to the question,

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

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

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

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

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

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

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

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

$\Rightarrow x=50,-40$

But $(x)$ is the speed of the train, which is always positive.

Thus, $x=50$

and $x-10=40$

Hence, the speed of fast train is 50 km/h and the speed of slow train is 40 km/h.