To find the centre of mass of a rigid body,why is it not possible to know $\sum m_i \vec{r}_i$ for all the particles?

Four particles of masses $m_1 = 2m$,$m_2 = 4m$,$m_3 = m$,and $m_4$ are placed at the four corners of a square. What should be the value of $m_4$ so that the centre of mass of all the four particles is exactly at the centre of the square?

An isolated particle of mass $m$ is moving in a horizontal $x-y$ plane along the $x-$axis at a certain height above the ground. It suddenly explodes into two fragments of masses $m/4$ and $3m/4$. An instant later,the smaller fragment is at $y = 15 \ cm$. The larger fragment at this instant is at $y = \dots \ cm$.

Look at the drawing given in the figure which has been drawn with ink of uniform line-thickness. The mass of ink used to draw each of the two inner circles,and each of the two line segments is $m$. The mass of the ink used to draw the outer circle is $6m$. The coordinates of the centres of the different parts are: outer circle $(0,0)$,left inner circle $(-a, a)$,right inner circle $(a, a)$,vertical line $(0,0)$ and horizontal line $(0,-a)$. The $y$-coordinate of the centre of mass of the ink in this drawing is

Two particles of masses $200 \ g$ and $500 \ g$ are moving with velocities $10 \ \hat{i} \ m/s$ and $3 \ \hat{i} + 5 \ \hat{j} \ m/s$ respectively. The velocity of the center of mass of the system is:

Four particles of masses $1\,kg, 2 \,kg, 3 \,kg$ and $4\, kg$ are placed at the four vertices $A, B, C$ and $D$ of a square of side $1\, m$ as shown in the figure. The coordinates of the centre of mass of the particles are: