An object is kept at a distance of 16 cm from a thin lens and the image formed is real. If the object is kept at a distance of 6 cm from the same lens the image formed is virtual. If the size of the images formed are equal, the focal length of the lens will be

Only convex lens can form a real well as virtual image. So, the given lens is a convex lens. Let f is the focal length of the lens and n the magnitude of magnification.
In the first case u,hen the image is real
u = -16 cm
v = +16 n
So, applying 

$\begin{array}{l}\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\\ \frac{1}{16\mathrm{n}}+\frac{1}{16}=\frac{1}{\mathrm{f}}\\ \mathrm{or}1+\frac{1}{\mathrm{n}}=\frac{16}{\mathrm{f}}....\left(1\right)\\ \mathrm{in}\mathrm{the}\mathrm{second}\mathrm{case}\mathrm{when}\mathrm{image}\mathrm{is}\mathrm{virtual}\\ \mathrm{u}=-6\mathrm{cm}\\ \mathrm{v}=-6\mathrm{n}\\ \mathrm{f}=+\mathrm{f}\\ \therefore \frac{1}{-6\mathrm{n}}+\frac{1}{6}=\frac{1}{\mathrm{f}}\\ \therefore 1-\frac{1}{\mathrm{n}}=\frac{6}{\mathrm{f}}....\left(2\right)\\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right)\mathrm{we}\mathrm{get}\\ 2=\frac{22}{\mathrm{f}}\mathrm{or}\mathrm{f}=11\mathrm{cm}\end{array}$