Question

If a fast takes 3 hours less than a slow train for a journey of 900km and the speed of slow train is 15km per hour less than that of the fast train, then the speed of two trains is

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{fast}\mathrm{train}=x\mathrm{km}/\mathrm{hr} \\ \mathrm{then}\mathrm{the}\mathrm{speed}\mathrm{of}\mathrm{slow}\mathrm{train}=\left(x-15\right)\mathrm{km}/\mathrm{hr} \\ \mathrm{then}\mathrm{time}\mathrm{taken}\mathrm{by}\mathrm{fast}\mathrm{train}\mathrm{to}\mathrm{travel}900\mathrm{km}=\left(\frac{900}{x}\right)\mathrm{hours} \\ \mathrm{and}\mathrm{time}\mathrm{taken}\mathrm{by}\mathrm{slow}\mathrm{train}\mathrm{to}\mathrm{travel}900\mathrm{km}=\frac{900}{\left(x-15\right)}\mathrm{hours} \\ \mathrm{But}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}\mathrm{fast}\mathrm{train}\mathrm{takes}3\mathrm{hours}\mathrm{less}\mathrm{than}\mathrm{a}\mathrm{slow}\mathrm{train}\mathrm{for} \\ \mathrm{a}\mathrm{journey}\mathrm{of}900\mathrm{km} \\ \mathrm{Hence},\frac{900}{x-15}-\frac{900}{x}=3 \\ \Rightarrow 900\left[\frac{1}{x-15}-\frac{1}{x}\right]=3 \\ \Rightarrow \frac{1}{x-15}-\frac{1}{x}=\frac{3}{900} \\ \Rightarrow \frac{x-\left(x-15\right)}{x\left(x-15\right)}=\frac{1}{300} \\ \Rightarrow \frac{x-x+15}{x^2-15x}=\frac{1}{300} \\ \Rightarrow \frac{15}{x^2-15x}=\frac{1}{300} \\ \Rightarrow x^2-15x=4500 \\ \Rightarrow x^2-15x-4500=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow x^2-75x+60x-4500=0 \\ \Rightarrow x\left(x-75\right)+60\left(x-75\right)=0 \\ \Rightarrow \left(x-75\right)\left(x+60\right)=0 \\ \Rightarrow x-75=0\mathrm{or}x+60=0 \\ \Rightarrow x=75\mathrm{or}x=-60 \\ \mathrm{But}x=-60\mathrm{is}\mathrm{rejected} \\ \mathrm{then},x=75 \\ \mathrm{Hence},\mathrm{speed}\mathrm{of}\mathrm{the}\mathrm{fast}\mathrm{train}=75\mathrm{km}/\mathrm{hr} \\ \mathrm{and}\mathrm{speed}\mathrm{of}\mathrm{slow}\mathrm{train}=75-15=60\mathrm{km}/\mathrm{hr} \end{gathered}$$