Why is mass totally neglected when center of gravity is shifted? A square sheet of metal has a square of one quarter of the original area cut from 
one corner as shown in the figure. Which of the following statements is true about 
the position of the centre of gravity of the remaining portion of the sheet?

a) center of gravity lies at a distance of 5/12 of the side of the original square from each uncut side.
b) center of gravity lies at a distance 7/12 of the side of the original square from each uncut side.
c) center of gravity lies at a distance of 63/4 of the side of the original square from each uncut side.
d) None of these.
Ans: 

$\overline X = \large \frac{A_{1}\overline{x_{1}} - A_{2}\overline{x_{2}}}{A_{1}-A_{2}} = 
\frac{(4x^{2}).(x) - (x^{2})(1.5x)}{4x^{2}-x^{2}}$
$\overline X = \large\frac{5x}{6} = \frac{5}{12}\normalsize(2x)$
$\overline Y = \large\frac{A_{1}\overline{y_{1}} - A_{1}\overline{y_{2}}}{A_{1}-A_{2}} 
= \frac{5x}{6} = \frac{5}{12}\normalsize(2x)$
original side of the square = $2x$
Actual center of gravity forumula = $\large\frac{m_{1}d_{1}+m_{2}.d_{2} + m_{3}.d_{3}.....}{m}$
But above problem is about center of gravity shifting, where mass is totally neglected, why so?
 A: That formula applies to discrete point masses. You don't have an m1, d1, m2, d2, etc., you have a continuous distribution of mass over a continuous area. Since the density of the sheet is constant, finding the "mean X" and "mean Y" positions is equivalent to finding the position of the center of mass.
A: The mass seems to be neglected, but that is because of an implicit assumption in the question. The assumption is that the density, $\rho$, of the object is uniform. That means that for any given small volume of a piece of the object, $\Delta V$, the mass of each volume is the same, $$\Delta m = \rho\Delta V.$$
If you were to split the object into thousands of pieces, then perform the sums indicated by the center of gravity, you would end up with a $\rho$ factor in both the numerator and denominator, and they would divide away to 1. That would leave you with a set of $x\Delta x,~ y\Delta y,~z\Delta z$ factors.
So it's not that the mass has been neglected, but that uniform density has been an assumption. If the density is not uniform, then one would have to account for that.
A: In fact, here we have $m_i = \text{areal density}_i \times \text{area}_i$
But for any piece $i$ of your sheet, the areal density is the same (That is to say that you suppose your sheet to be of uniform thickness and density. It isn't said in your question but it has to be supposed). 
Therefore you have 
$\frac{\text{areal density} \times \text{area}_1 + \text{areal density} \times \text{area}_2 + \ldots}{\text{areal density} \times \text{total area}}$ $=$ $\frac{\text{area}_1 + \text{area}_2 + \ldots}{\text{total area}}$
So the center of gravity becomes the center of surface. And finding the center of surface is nothing but finding the mean $X$ and the mean $Y$
