(d) Let focal length of convex lens is +$f$ then focal length of concave lens would be $- \frac{3}{2}f$.
From the given condition,
$\frac{1}{30} = \frac{1}{f} - \frac{2}{3f} = \frac{1}{3f}$
$\therefore\ f = 10$ cm
Therefore, focal length of convex lens = + 10 cm and that of concave lens = -15 cm.