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A train travels a distance of $90$ km at a constant speed. Had the speed been $15$ km/h more, it would have taken $30$ minutes less for the journey. Find the original speed of the train.
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Let the original speed be $x$ km/h
New speed $= (x + 15)$ km/h
A.T.Q.
$$\begin{aligned}& \frac{90}{x} - \frac{90}{x+15} = \frac{1}{2} \\ & \Rightarrow x^2 + 15x - 2700 = 0 \\ & \Rightarrow(x + 60) (x - 45) = 0 \\ & x \neq -60 \text{ , } x = 45 \\ & \text{The original speed of the train } = 45\text{km/h}\end{aligned}$$
New speed $= (x + 15)$ km/h
A.T.Q.
$$\begin{aligned}& \frac{90}{x} - \frac{90}{x+15} = \frac{1}{2} \\ & \Rightarrow x^2 + 15x - 2700 = 0 \\ & \Rightarrow(x + 60) (x - 45) = 0 \\ & x \neq -60 \text{ , } x = 45 \\ & \text{The original speed of the train } = 45\text{km/h}\end{aligned}$$