A plane mirror is placed $22.5cm$ in front of a concave mirror of focal length $10cm$. An object can be placed between the two mirrors so that the first image in both the mirror coincides. The object distance from the spherical mirror is

It can seen from the figure that, if the object is placed at a distance $x$ from the concave mirror, its distance from the plane mirror will be $(22.5-x)$. So plane mirror equal and erect image of object at a distance $(22.5-x)$ behind the mirror. Now according to the given problem, the image coincided with the image formed by the plane mirror, therefore for concave mirror.
$v=-[22.5+22.5-x]=-(45-x)$ 

and object distance
$u=-x$

given focal length is

$f=-10$

So using mirror formula,

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

$\frac{1}{-(45-x)}+\frac{1}{-x}=\frac{1}{-10}⟹\frac{45}{45x-x^2}=\frac{1}{10}$

$x^2-45x+450=0$

$(x-30)(x-15)=0$

Hence $x=30cm$ or $x=15cm$

.
But as the distance between the two mirror is $22.5$, $x=30$ is not admissible . So the object must be at a distance of $15$ cm from the concave mirror