A convex lens of focal length 20 cm produced images of the same magnification 2 when an object is kept at two distances $x_1\text{ and }{\mathrm{x}}_2\left({\mathrm{x}}_1>{\mathrm{x}}_2\right)$ from the lens. The ratio of x₁ and x₂ is 

Since both images have same magnification. one is real and the other one is virtual images 
$\text{ for }\mathrm{m}=-2=\frac{\mathrm{v}}{\mathrm{u}}\Rightarrow \mathrm{v}=-2\mathrm{x}\left[\because \mathrm{u}={\mathrm{x}}_1\right]\mathrm{Using}$
lens formula, 
$\begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{f}=\frac{1}{-2x_1}-\frac{1}{x_1}\\ \Rightarrow x_1=\frac{-3f}{2}=-30\mathrm{cm}\end{array}$
$\text{ for }\mathrm{m}=2\Rightarrow \mathrm{v}=2{\mathrm{x}}_2$
$\begin{array}{l}\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\Rightarrow \frac{1}{f}=\frac{1}{2x_2}-\frac{1}{x_2}\\ \Rightarrow x_2=\frac{-f}{2}=-10\mathrm{cm}\therefore \frac{x_1}{x_2}=\frac{30}{10}=\frac{3}{1}\end{array}$