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.

Answer 1

$$\begin{gathered} \begin{array}{l}\\ \text{Let the original speed of the train be}x\text{km/hr}\text{.}\end{array} \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{question}: \\ \frac{90}{x}-\frac{90}{(x+15)}=\frac{1}{2} \\ \Rightarrow \frac{90(x+15)-90x}{x(x+15)}=\frac{1}{2} \\ \Rightarrow \frac{90x+1350-90x}{x^2+15x}=\frac{1}{2} \\ \Rightarrow \frac{1350}{x^2+15x}=\frac{1}{2} \\ \Rightarrow 2700=x^2+15x \\ \Rightarrow x^2+(60-45)x-2700=0 \\ \Rightarrow x^2+60x-45x-2700=0 \\ \Rightarrow x(x+60)-45(x+60)=0 \\ \Rightarrow (x+60)(x-45)=0 \\ \Rightarrow x=-60\text{or}x=45 \\ x\text{cannot be negative; therefore, the original speed of train is 45 km/hr.} \end{gathered}$$