A passenger train takes one hour less for a journey of 150 km

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

Increased speed of the train $=(x+5) \mathrm{km} / \mathrm{hr}$

Time taken by the train under usual speed to cover $150 \mathrm{~km}=\frac{150}{x} \mathrm{hr}$

Time taken by the train under increased speed to cover $150 \mathrm{~km}=\frac{150}{(x+5)} \mathrm{hr}$

Therefore,

$\frac{150}{x}-\frac{150}{(x+5)}=1$

$\frac{\{150(x+5)-150 x\}}{x(x+5)}=1$

$\frac{150 x+750-150 x}{x^{2}+5 x}=1$

$750=x^{2}+5 x$

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

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

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

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

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

So, either

$(x-25)=0$

$x=25$

Or

$(x+30)=0$

$x=-30$

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

Hence, the usual speed of train is  $x=25 \mathrm{~km} / \mathrm{hr}$