A concave lens forms the image of an object such that the distance between the object and image is 10 cm and the magnification produced is 1/4. The focal length of the lens will be

Concave lens forms the virtual image of a real object.
So let u = 4x and v =  - x , then 3x = 10 cm

$\begin{array}{l}\mathrm{or}\mathrm{x}=\frac{10}{3}\mathrm{cm}\\ \therefore \mathrm{u}=-\frac{40}{3}\mathrm{cm}\mathrm{and}\mathrm{v}=-\frac{10}{3}\mathrm{cm}\end{array}$

$\begin{array}{l}\mathrm{substituting}\mathrm{in}\frac{1}{\mathrm{f}}=\frac{-3}{10}+\frac{3}{40}\\ \mathrm{or}\mathrm{f}=-\frac{40}{9} \mathrm{or} \mathrm{f}=-4.4\mathrm{cm}\end{array}$