Explanation

(b) As a convex lens alone can form a real images as well as a virtual images, therefore, the lens in the present question is a convex lens. Let $f$ be the focal length of the lens and $m$ be the magnification produced
In the first case, when image is real,
$u = - 16\ $cm,$\ v = (m \times 16)$cm
As $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\therefore\frac{1}{16m} + \frac{1}{16} = \frac{1}{f}\ \ $or$\ 1 + \frac{1}{m} = \frac{16}{f}\ \ $...(i)
In the second case, when image is virtual
$u = - 6\ cm,v = ( - 6\ $m)cm
From $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{- 6m} + \frac{1}{6} = \frac{1}{f}\ \ $or$\ \ 1 - \frac{1}{m} = \frac{6}{f}$, …(ii)
Add Eq.(i) and Eq.(ii) we have
$2 = \frac{22}{f}\ \ $or$\ \ f = \frac{22}{2} = 11\ $cm