A plane left 40 minutes late due to bad weather and in order to reach its destination,

Solution:

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

Increased speed of the plane $=(x+400) \mathrm{km} / \mathrm{hr}$

Time taken by the plane under usual speed to cover $1600 \mathrm{~km}=\frac{1600}{x} \mathrm{hr}$

Time taken by the plane under increased speed to cover $1600 \mathrm{~km}=\frac{1600}{(x+400)} \mathrm{hr}$

Therefore,

$\frac{1600}{x}-\frac{1600}{(x+400)}=\frac{40}{60}$

$\frac{\{1600(x+400)-1600 x\}}{x(x+400)}=\frac{2}{3}$

$\frac{1600 x+640000-1600 x}{x^{2}+400 x}=\frac{2}{3}$

$1920000=2 x^{2}+800 x$

$2 x^{2}+800 x-1920000=0$

 

$2\left(x^{2}+400 x-960000\right)=0$

$x^{2}+400 x-960000=0$

$x^{2}-800 x+1200 x-960000=0$

$x(x-800)+1200(x-800)=0$

 

$(x-800)(x+1200)=0$

So, either 

$(x-800)=0$

$x=800$

Or

$(x+1200)=0$

$x=-1200$

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

Hence, the usual speed of train is  $x=800 \mathrm{~km} / \mathrm{hr}$