Imbalance Puzzle - Part 2

The left arm: $\LARGE{\circ}$ $+\triangle > \square + \triangle \quad \Rightarrow \LARGE{\circ}$ $>\square$

The right arm: $\triangle + \LARGE{\circ}$ $> \LARGE \circ$ $+ \LARGE \circ$ $\quad \Rightarrow \triangle > \LARGE \circ$

Therefore, $\triangle > \LARGE \circ$ $>\square$

This is basically simple algebra represented in a real-life example. Replace the shapes with variables. Let "s" represent the square, "c" for circle and "t" for triangle.

In the first equation (left arm) we have the following, s+t<c+t we subtract 't' from both sides and get s<c

In the second equation (right arm) we have following, c+c<t+c same thing again and this time we get c<t

we combine the two equations and get: s<c<t (square<circle<triangle)

{ (Rectangle = x) (Circle = y) (Triangle = z) }

SIMPLE: Rectangle + Triangle < Circle + Triangle
SO: Rectangle < Circle + Triangle - Triangle
THEN: Rectangle < Circle + [ z - z ( value of 0 ) ]
RESULT: Rectangle < Circle

AND

SIMPLE: Circle + Circle < Triangle + Circle
SO: Circle < Triangle + ( Circle - Circle )
THEN: Circle < Triangle + [ y - y ( value of 0 ) ]
RESULT: Circle < Triangle

CONCLUSION:

Rectangle < Circle AND Circle < Triangle
SO: Rectangle < Circle < Triangle
OR: Triangle > Circle > Rectangle