Situated at a distance of 15 cm from a convex lens of focal length...

For the first lens:

$u_1 = -15 \text{ cm}$, $f_1 = 10 \text{ cm}$

Using the lens formula, $\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}$

$\dfrac{1}{v_1} - \dfrac{1}{-15} = \dfrac{1}{10}$

$\dfrac{1}{v_1} = \dfrac{1}{10} - \dfrac{1}{15} = \dfrac{1}{30}$

$v_1 = 30 \text{ cm}$

The image formed by the first lens acts as an object for the second lens. The distance between the lenses is $40 \text{ cm}$.

Object distance for the second lens, $u_2 = -(40 - 30) = -10 \text{ cm}$

Focal length of the second lens, $f_2 = 5 \text{ cm}$

Using the lens formula again, $\dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f_2}$

$\dfrac{1}{v_2} - \dfrac{1}{-10} = \dfrac{1}{5}$

$\dfrac{1}{v_2} = \dfrac{1}{5} - \dfrac{1}{10} = \dfrac{1}{10}$

$v_2 = 10 \text{ cm}$

The positive sign indicates that the final image is formed $10 \text{ cm}$ from the second lens on the side opposite to the object, which is further away from the first lens.