Question

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

Answer 1

Let the original speed of the train be $x$kmphr

$\therefore$ time take to cover a distance of $90$km$=\frac{90}{x}$hrs.

The new speed of the train$=x+15$kmph

$\therefore$ new time taken to cover $90$km$=\frac{90}{x+15}$hrs.

Since the train takes $30$ minutes

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

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

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

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

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

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

$\therefore x=-60$ or $x=45$ but $x$ cannot be negative.

$\therefore x=45$

Hence the original speed of the train $=45kmph$