A train covers a distance of $90 \; km$ at an uniform speed. Had the speed been $15 \; kmph$ more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.
Let the original speed of the train be $x \; kmph$
Original time taken to cover $90 \; km$ $= t_1 = \dfrac{\text{distance}}{\text{speed}} = \dfrac{90}{x}$ hour
New speed of train $= x + 15 \; kmph$
New time taken to cover $90 \; km$ $= t_2 = \dfrac{90}{x + 15}$ hour
As per question, $\;$ $t_2 = t_1 - 30 \; \text{minutes} = t_1 - \dfrac{1}{2} \; \text{hour}$
i.e. $\;$ $\dfrac{90}{x + 15} = \dfrac{90}{x} - \dfrac{1}{2}$
i.e. $\;$ $\dfrac{90}{x + 15} = \dfrac{180 - x}{2x}$
i.e. $\;$ $180x = 180x - x^2 + 2700 - 15x$
i.e. $\;$ $x^2 + 15x - 2700 = 0$
i.e. $\;$ $x^2 + 60x - 45x - 2700 = 0$
i.e. $\;$ $x \left(x + 60\right) - 45 \left(x + 60\right) = 0$
i.e. $\;$ $\left(x + 60\right) \left(x - 45\right) = 0$
i.e. $\;$ $x + 60 = 0$ $\;$ or $\;$ $x - 45 = 0$
i.e. $\;$ $x = -60$ $\;$ or $\;$ $x = 45$
Since the speed of the train cannot be negative,
$\therefore \;$ original speed of train $= 45 \; kmph$