Question

A train covers a distance of $90$ km at a uniform speed. Had the speed been $15$ km/hr more, it would have taken $30$ minutes less for the journey. Find the original speed of the train(in km/hr).

Answer 1

$\Rightarrow$ Let the original speed of train be $xkm/hr$

$\Rightarrow$ Then, increased speed will be $x+15km/hr$

$\Rightarrow$ The original time $=\frac{90}{x}hours$

$\Rightarrow$ New time $=\frac{90}{x+15}hours$

According to the question,

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

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

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

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

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

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

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

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

$\Rightarrow$ $x=-60or45$

$\Rightarrow$ Since speed can't be negative, we have $x=45$

$\Rightarrow$ Hence original speed of train $=45km/hr.$