Question

The distance of a road between two towns A and B is 206km.A car travel at a speed of x km/hr .A train travels at a speed which is 16km/hr faster than the car calculate
1—the time taken by car to reach town
2—the time taken by train to reach the town B from A in terms of x
3—if the train takes 2 hours less than the car to reach town B, Obtain an equation in terms of x and solve it
4— Find speed of train.

Answer 1

$$\begin{gathered} \mathrm{Distance}=206\mathrm{km}. \\ \mathrm{Time}\mathrm{taken}=\frac{\mathrm{Distance}}{\mathrm{Speed}} \\ 1).\mathrm{Time}\mathrm{taken}\mathrm{by}\mathrm{car}=\frac{206}{\mathrm{x}}\mathrm{hours}. \\ 2).\mathrm{Speed}\mathrm{of}\mathrm{train}=\left(\mathrm{x}+16\right)\mathrm{km}/\mathrm{hr} \\ \mathrm{Time}\mathrm{taken}\mathrm{by}\mathrm{train}=\frac{206}{\mathrm{x}+16} \\ 3).\mathrm{Train}\mathrm{takes}2\mathrm{hours}\mathrm{less}\mathrm{than}\mathrm{car} \\ \mathrm{So}, \\ \frac{206}{\mathrm{x}}-\frac{206}{\mathrm{x}+16}=2 \\ \frac{103}{\mathrm{x}}-\frac{103}{\mathrm{x}+16}=1 \\ \frac{103\mathrm{x}+1648-103\mathrm{x}}{\mathrm{x}\left(\mathrm{x}+16\right)}=1 \\ 1648={\mathrm{x}}^2+16\mathrm{x} \\ {\mathrm{x}}^2+16\mathrm{x}-1648=0 \\ \mathrm{x}=\frac{-16\pm \sqrt{{\left(16\right)}^2-4\left(1\right)\left(-1648\right)}}{2}=\frac{-16\pm \sqrt{256+6592}}{2} \\ =\frac{-16\pm \sqrt{6848}}{2}=\frac{-16\pm 8\sqrt{107}}{2} \\ =-8\pm 4\sqrt{107} \\ \mathrm{Speed}\mathrm{can}\text{'}\mathrm{t}\mathrm{be}\mathrm{negative} \\ \mathrm{So},\mathrm{x}=-8+4\sqrt{107}. \\ 4).\mathrm{Speed}\mathrm{of}\mathrm{train}=\mathrm{x}+16=-8+4\sqrt{107}+16=\left(8+4\sqrt{107}\right)\mathrm{km}/\mathrm{hour}. \end{gathered}$$