Answer :
According to the question, the object given is virtual and the image formed is real.
Object distance is u= +12 cm
(i) The focal length of the convex lens is f =20 cm
Image distance is denoted by v
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$ \frac{1}{v}-\frac{1}{12}=\frac{1}{20}$
$ \frac{1}{v}=\frac{1}{20}+\frac{1}{12}=\frac{3+5}{60}=\frac{8}{60}$
$ Therefore\;v=\frac{60}{8}=7.5cm$
Therefore, the image formed will be 7.5cm away from the lens, to its right.
(ii) Focal length of the concave lens is f =-16 cm
Image distance is denoted by v
$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$
$ \frac{1}{v}=-\frac{1}{16}+\frac{1}{12}$
$ =\frac{-3+4}{48}=\frac{1}{48}$
∴ v=48cm.
Therefore, the image so formed is 48cm away from the lens towards its right.