Question

An aeroplane left $50$ minutes later than its scheduled time, and in order to reach the destination, $1250km$ away, in time, it had to increase its speed by $250km/hr$ from its usual speed. Its usual speed is -___.$km/hr$

Answer 1

$\Rightarrow$ Let the usual speed of airoplane be $xkm/hr.$

$\Rightarrow$ Then, increased speed of the airoplane $=(x+250)km/hr$

$\Rightarrow$ Time taken by the airoplane under usual speed to cover $1250km=\frac{1250}{x}hr$

$\Rightarrow$ Time taken by the airoplane under increased speed to cover $1250km=\frac{1250}{x+250}hr$

$\Rightarrow$ $\frac{1250}{x}-\frac{1250}{x+250}=\frac{50}{60}$

$\Rightarrow$ $\frac{1250(x+250)-1250x}{x(x+250)}=\frac{5}{6}$

$\Rightarrow$ $\frac{1250x+312500-1250x}{x^2+250x}=\frac{5}{6}$

$\Rightarrow$ $\frac{312500}{x^2+250x}=\frac{5}{6}$

$\Rightarrow$ $312500\left(6\right)=5(x^2+250x)$

$\Rightarrow$ $1875000=5x^2+1250x$

$\Rightarrow$ $5x^2+1250x-1875000=0$

$\Rightarrow$ $x^2+250x-375000=0$

$\Rightarrow$ $x^2-500x+750x-375000=0$

$\Rightarrow$ $x(x-500)+750(x-500)=0$

$\Rightarrow$ $(x-500)(x+750)=0$

$\therefore$ $x=500$ and $x=-750$

Speed can not be negative.

$\therefore$ The usual speed of airoplane is $500km/hr$.