Teacher I was having a doubt regarding this question. So can u please explain me in detailed . As my exam's are going to start so can u explain me in a easy and fast way

Solution,

The question is of geometric optics and can be solved using the standard mirror formula for concave mirrors, which can be stated as follows:
$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$
Given,
the object distance, u = -25 cm ( since, the object is on the left side of the mirror)
Focal length of the mirror, f = -15 cm 
Thus, its image distance (v) would be:
$$\begin{gathered} \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\ \frac{1}{-25}+\frac{1}{v}=\frac{1}{-15} \\ \frac{1}{v}=\frac{1}{-15}+\frac{1}{25}=\frac{-5+3}{75}=-\frac{2}{75} \\ Thus, \\ v=-37.5 cm \\ Thus, the image is on the same side as that of the object. \\ It is on the left side of the mirror. \end{gathered}$$
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Magnification is defined as:
$$\begin{gathered} M=\frac{image height}{object height}=-\frac{v}{u}=-\frac{-37.5}{-25} \\ Object height = 7 cm \\ Hence, \\ image height=-\frac{-37.5}{-25}\times 7 cm=-10.5 cm \\ Negative sign implies that the image is inverted which is the case when the image is real. \\ Thus, \\ The image is 10.5 cm in height and real. \end{gathered}$$