Question

A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hour more, it would have taken 30 minutes less for the journey. The original speed of the train is ----- km/hr.

Answer 1

Let original speed $=xkm/hr$

$\Rightarrow$ Time taken to travel $=\frac{90}{x}hr$

$\Rightarrow$ New speed $=(x+15)km/hr$

$\Rightarrow$ Time taken to travel $=\frac{90}{x+15}hr$

$\therefore$ $\frac{90}{x}=\frac{90}{x+15}+\frac{1}{2}$

$\Rightarrow$ $\frac{90}{x}-\frac{90}{x+15}=\frac{1}{2}$

$\Rightarrow$ $90(\frac{1}{x}-\frac{1}{x+15})=\frac{1}{2}$

$\Rightarrow$ $\frac{x+15-x}{x(x+15)}=\frac{1}{180}$

$\Rightarrow$ $15\times 180=x(x+15)$

$\Rightarrow$ $2700=x^2+15x$

$\Rightarrow$ $x^2+15x-2700=0$

$\Rightarrow$ $x^2+60x-45x-2700=0$

$\Rightarrow$ $x(x+60)-45(x+60)=0$

$\Rightarrow$ $(x+60)(x-45)=0$

Speed can not be negative.

$\therefore$ The original speed of the train is $45km/hr$