Here, focal length of concave lens, `f = -15 cm`.

object distance, `u = ?`, image distance, `v = -10 cm`, magnification of lens, `m = ?`

As `(1)/(f) = (1)/(v) - 1/u, 1/u = (1)/(v) -(1)/(f) = 1/(-10) + 1/15 = (-1)/30 or u = -30 cm`

Thus the object should be placed at a distance of 30 cm on the left side of the concave lens.

Linear magnification, `m = v/u = (-10)/(-30) = 1/3`.

The positive sign of m shows that the image is virtual and erect, and its size is `(1//3)` of the size of the object.