Question

A train and covers a distance of 600 kilometre at x kilometre per hour. had the speed been (x + 20) kilometre per hour the time taken to cover the distance would have been reduced by 5 hours. write down an equation in x and solve if to evaluate x.

Answer 1

$$\begin{gathered} \mathrm{Total}\mathrm{distance}\mathrm{to}\mathrm{be}\mathrm{covered}=600\mathrm{km} \\ \mathrm{Usual}\mathrm{spped}\mathrm{of}\mathrm{the}\mathrm{train}=x\mathrm{km}/\mathrm{h} \\ \mathrm{Increased}\mathrm{speed}\mathrm{of}\mathrm{train}=\left(x+20\right)\mathrm{km}/\mathrm{h} \\ \mathrm{Now},\mathrm{time}\mathrm{taken}\mathrm{to}\mathrm{cover}600\mathrm{km}\mathrm{with}\mathrm{usual}\mathrm{speed}=\left(\frac{600}{x}\right)\mathrm{hours} \\ \mathrm{Time}\mathrm{taken}\mathrm{to}\mathrm{cover}600\mathrm{km}\mathrm{with}\mathrm{increased}\mathrm{speed}=\left(\frac{600}{x+20}\right)\mathrm{hours} \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{question}, \\ \frac{600}{x}-\left(\frac{600}{x+20}\right)=5 \\ \Rightarrow \frac{1}{x}-\frac{1}{x+20}=\frac{1}{120} \\ \Rightarrow \frac{x+20-x}{x\left(x+20\right)}=\frac{1}{120} \\ \Rightarrow \frac{20}{x^2+20x}=\frac{1}{120} \\ \Rightarrow x^2+20x-2400=0 \\ \Rightarrow x^2+60x-40x-2400=0 \\ \Rightarrow x\left(x+60\right)-40\left(x+60\right)=0 \\ \Rightarrow \left(x-40\right)\left(x+60\right)=0 \\ \Rightarrow x-40=0\mathrm{or}x+60=0 \\ \Rightarrow x=40\mathrm{or}x=-60\left[\mathrm{rejected}\right] \\ \mathrm{So},\mathrm{usual}\mathrm{speed}\mathrm{of}\mathrm{train}=40\mathrm{km}/\mathrm{hr} \end{gathered}$$