A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also find the magnification of the lens.

For a concave lens:

• Focal length ($f$) is negative. So, $f = -15 \text{ cm}$.
• The image formed by a concave lens is always virtual and on the same side as the object. So, the image distance ($v$) is negative. $v = -10 \text{ cm}$.

The lens formula is:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

Rearrange the formula to solve for $\frac{1}{u}$:

$$\frac{1}{u} = \frac{1}{f} - \frac{1}{v}$$

Substitute the values:

$$\frac{1}{u} = \frac{1}{-15 \text{ cm}} - \frac{1}{-10 \text{ cm}}$$ $$\frac{1}{u} = -\frac{1}{15 \text{ cm}} + \frac{1}{10 \text{ cm}}$$

Find a common denominator, which is 30:

$$\frac{1}{u} = -\frac{2}{30 \text{ cm}} + \frac{3}{30 \text{ cm}}$$ $$\frac{1}{u} = \frac{1}{30 \text{ cm}}$$ $$u = 30 \text{ cm}$$

The magnification formula is:

$$M = -\frac{v}{u}$$

Substitute the values:

$$M = -\frac{-10 \text{ cm}}{30 \text{ cm}}$$ $$M = \frac{10}{30}$$ $$M = \frac{1}{3}$$ $$M \approx 0.33$$

The object should be placed at a distance of $\boxed{30 \text{ cm}}$ from the lens.
The magnification of the lens is $\boxed{0.33}$.