A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed.

Let the usual speed be $x \mathrm{~km} / \mathrm{hr}$.

According to the question:

$\frac{300}{x}-\frac{300}{(x+5)}=2$

$\Rightarrow \frac{300(x+5)-300 x}{x(x+5)}=2$

$\Rightarrow \frac{300 x+1500-300 x}{x^{2}+5 x}=2$

$\Rightarrow 1500=2\left(x^{2}+5 x\right)$

$\Rightarrow 1500=2 x^{2}+10 x$

$\Rightarrow x^{2}+5 x-750=0$

$\Rightarrow x^{2}+(30-25) x-750=0$

$\Rightarrow x^{2}+30 x-25 x-750=0$

$\Rightarrow x(x+30)-25(x+30)=0$

$\Rightarrow(x+30)(x-25)=0$

$\Rightarrow x=-30$ or $x=25$

The usual speed cannot be negative; therefore, the speed is $25 \mathrm{~km} / \mathrm{hr}$.