An aeroplane take 1 hour less for a journey of 1200 km if its speed is increased by 100 km/hr from its usual speed. Find its usual speed.
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 $x=300 \mathrm{~km} / \mathrm{hr}$
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