A train covers a distance of $480 \mathrm{~km}$ at a uniform speed. If the speed had been $8 \mathrm{~km} / \mathrm{hr}$ less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.

Home » A train covers a distance of at a uniform speed. If the speed had been less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.

A train covers a distance of $480 \mathrm{~km}$ at a uniform speed. If the speed had been $8 \mathrm{~km} / \mathrm{hr}$ less then it would have taken 3 hours more to cover the same distance. Find the usual speed of the train.

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

$\therefore$ Reduced speed of the train $=(x-8) \mathrm{km} / \mathrm{h}$

Total distance to be covered $=480 \mathrm{~km}$

Time taken by the train to cover the distance at usual speed $=\frac{480}{x} h \quad\left(\right.$ Time $\left.=\frac{\text { Distance}}{\text { Speed }}\right)$

Time taken by the train to cover the distance at reduced speed $=\frac{480}{x-8} h$

According to the given condition,

Time taken by the train to cover the distance at reduced speed = Time taken by the train to cover the distance at usual speed $+3 \mathrm{~h}$

$\therefore \frac{480}{x-8}=\frac{480}{x}+3$
$\Rightarrow \frac{480}{x-8}-\frac{480}{x}=3$
$\Rightarrow \frac{480 x-480 x+3840}{x(x-8)}=3$
$\Rightarrow \frac{3840}{x^{2}-8 x}=3$
$\Rightarrow x^{2}-8 x=1280$
$\Rightarrow x^{2}-8 x-1280=0$
$\Rightarrow x(x-40)+32(x-40)=0$
$\Rightarrow(x-40)(x+32)=0$
$\Rightarrow x-40=0$ or $x+32=0$
$\Rightarrow x=40$ or $x=-32$
$\therefore x=40 \quad$ (Speed cannot be negative)