An aeroplane take 1 hour less for a journey of 1200 km

Solution:

Let the usual speed of aero plane be $x \mathrm{~km} / \mathrm{hr}$. Then,

Increased speed of the aero plane $=(x+100) \mathrm{km} / \mathrm{hr}$

Time taken by the aero plane under usual speed to cover $1200 \mathrm{~km}=\frac{1200}{x} \mathrm{hr}$

Time taken by the aero plane under increased speed to cover $1200 \mathrm{~km}=\frac{1200}{(x+100)} \mathrm{hr}$

Therefore,

$\frac{1200}{x}-\frac{1200}{(x+100)}=1$

$\frac{\{1200(x+100)-1200 x\}}{x(x+100)}=1$

$\frac{1200 x+120000-1200 x}{x^{2}+100 x}=1$

$120000=x^{2}+100 x$

$x^{2}+100 x-120000=0$

$x^{2}+100 x-120000=0$

$x^{2}-300 x+400 x-120000=0$

$x(x-300)+400(x-300)=0$

 

$(x-300)(x+400)=0$

So, either 

$(x-300)=0$

$x=300$

Or

$(x+400)=0$

$x=-400$

But, the speed of the aero plane can never be negative.

 

Hence, the usual speed of train is $x=300 \mathrm{~km} / \mathrm{hr}$