Cool puzzle, this is very similar to the raven progressive matrix items, I think the solution is D.
Explanation:
RED + RED = RED (IN THE SAME POSITION ON THE SQUARES, THIS WORK FOR THE LAST COLUMN AND THE LAST ROW)
Otherwise is BLACK (IN THE SAME POSITION ON THE SQUARES, THIS WORK FOR THE LAST COLUMN AND THE LAST ROW)
Choice (d) is correct. The cells in the third column are formed by “adding” the black squares from the first two columns, and the cells in the third row are formed by “adding” the first two rows as well.
I don’t see a pattern that will emerge if the shapes are rearranged, so I’m going with uniformity. There are 3 boxes that have equal amounts of red and black, 3 boxes that have an abundance of red over black at 3:1, and only 2 boxes that have more black than red at 3:1. I choose (C) to level the playing field and bring the total area of the final 3×3 grid into solidarity with equal amounts of red and black.
C
..simple,
3 squares with 3 reds
3 squares with 2 reds
2 squares with 1 red (1 is mising)
-therefore square with 1 red is the best choice to complete the equilibrium of system.
D
BOX “D”…
above box + middle box = bottom box
also left box + middle box = right box
D.
The third box in each row equals the sum of the first two boxes’ black squares.
(d)
d
Cool puzzle, this is very similar to the raven progressive matrix items, I think the solution is D.
Explanation:
RED + RED = RED (IN THE SAME POSITION ON THE SQUARES, THIS WORK FOR THE LAST COLUMN AND THE LAST ROW)
Otherwise is BLACK (IN THE SAME POSITION ON THE SQUARES, THIS WORK FOR THE LAST COLUMN AND THE LAST ROW)
Choice (d) is correct. The cells in the third column are formed by “adding” the black squares from the first two columns, and the cells in the third row are formed by “adding” the first two rows as well.
C
I don’t see a pattern that will emerge if the shapes are rearranged, so I’m going with uniformity. There are 3 boxes that have equal amounts of red and black, 3 boxes that have an abundance of red over black at 3:1, and only 2 boxes that have more black than red at 3:1. I choose (C) to level the playing field and bring the total area of the final 3×3 grid into solidarity with equal amounts of red and black.
D
Looking at columns
The black square(s) in the top row of each column combines with the middle row to form the figures in the third row.
d. Black adds vertically.
Answer1:
3rd row or column = add up the black areas of 1st and 2nd
Hence the answer is (d)
Answer2:
There are 3 squares with one black area, 3 squares with 2 black areas, and hence 3 squares with black area as well.
Answer is (c)
d
Oh, I see it!! It’s D.
Bottom row is a superimposition of the top two rows, with black dominating over red.
The answer is D.
In each row, you add the black squares from the first two squares and get the third square.
The expected answer was ‘D’, as most of you said, but Hex and Shawn show how ‘C’ is also reasonable
i had C for the same reason as Hex
Box “a”
simply D… :S
C
..simple,
3 squares with 3 reds
3 squares with 2 reds
2 squares with 1 red (1 is mising)
-therefore square with 1 red is the best choice to complete the equilibrium of system.
its d box
d
D
D