An object AB is placed 15 cm on the left of a convex lens P of...

For lens $P$:

$\dfrac{1}{v_1} - \dfrac{1}{u_1} = \dfrac{1}{f_1}$

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

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

Magnification $m_1 = \dfrac{v_1}{u_1} = \dfrac{30}{-15} = -2$

The image formed by lens $P$ acts as a virtual object for lens $Q$. Since the distance between the lenses is $15\text{ cm}$, the object distance for lens $Q$ is:

$u_2 = 30 - 15 = +15\text{ cm}$

For lens $Q$:

$\dfrac{1}{v_2} - \dfrac{1}{u_2} = \dfrac{1}{f_2}$

$\dfrac{1}{v_2} - \dfrac{1}{15} = \dfrac{1}{15}$

$\dfrac{1}{v_2} = \dfrac{2}{15} \Rightarrow v_2 = 7.5\text{ cm}$

The positive sign indicates that the final image is real and formed $7.5\text{ cm}$ to the right of lens $Q$.

Magnification $m_2 = \dfrac{v_2}{u_2} = \dfrac{7.5}{15} = \dfrac{1}{2}$

Total magnification $m = m_1 \times m_2 = -2 \times \dfrac{1}{2} = -1$

Since $|m| = 1$, the size of the final image is the same as that of the object $AB$.

Answer: real, formed at $7.5\text{ cm}$ right of lens $Q$, with a size same as that of $AB$