Let R, B, G, X be the weights of the red, blue, green and gray
objects respectively. From the figures, we have the following equations:
(1) 2*(B+R) = B+G
(2) 2*X = R + X + G
(3) 4*R = B + X
Combining equations (2) and (3) yields
(4) 3*R = B+G
Combining equations (1) and (4) yields
2*B = R
Since the length of level on left side is twice that
on the right side, only one red weight is needed to counter balance the blue
weight on the left side.
