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Here's the puzzle

Anyone happen to know how this works? We can't decide whether to go top down or left to right.

Source of puzzle: https://test.mensa.no/

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  • $\begingroup$ This puzzle might be harder for people reading from right to left...that would mean different operations in the right-to-left direction than in the top-to-bottom direction. $\endgroup$
    – Klaws
    Apr 18, 2020 at 8:19

4 Answers 4

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I believe the answer is

C

because

Treating squares to the left of the line as negative and squares to the right as positive, and counting black and white squares separately, the number of squares in the right column is the sum of the squares in the cells left of it, and the number of squares in the bottom row is the sum of the squares in the cells above it.

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    $\begingroup$ Sweet that's what we figured as well. Thanks! edit It does appear to work the same way left -> right as well, I believe $\endgroup$
    – Fool Tbs
    Apr 16, 2020 at 21:56
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    $\begingroup$ Yeah, the point of "We can't decide whether to go top down or left to right" is that you don't need to decide because a correct solution works both ways. $\endgroup$
    – Peteris
    Apr 17, 2020 at 17:43
  • $\begingroup$ You can also view the figures as representation of "fraction". "Black" and "white" squares are coprime numbers, when on one side of the bar they're part of the product. Third(right) column is the result (reduced) of multiplication of "fractions" from the first two columns. And third(bottom) row is the result(reduced) or multiplication of "fractions" from first two rows. Just remember how to multiply and reduce fractions and that would be easy to solve. $\endgroup$
    – user28434
    Apr 17, 2020 at 18:27
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If we treat each bar divised group as a number, with left side as positive and right side as negative value, then:
First cell "plus" second cell equals third cell -> black squares on left are added while two white squares on left and right nullify each other.
This is also true for first row "plus" second row equals third row.
The last cell is thus
Horizontally: 1B 1W | 0 "plus" 0 | 3W = 1B | 2W
Black white "plus" negative three white. Black remains, one white gets nullified, two whites remaining on right side, giving "Black | 2 white"
Vertically: 2B | 0 "plus" 0 | 1B 2W = 1B | 2W which is the same as horizontally.
Caveat: In all but one cell the Black square is always on top of white ones. Yet the middle cell has Black square underneath white. It could mean the order is also important here, but then I have no theory how it would work.

Answer C - 1B | 2W, but unsure.

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Very late reply but might help someone in the future!

Treat the puzzle as columns. Overlay the top image on the middle image and remove any squares that appear on both sides of the line. The bottom image is the result.

So: top left image overlaid on middle left gives 1 black and 2 whites left of the line and 1 white right of the line. so 1 white on left and 1 white on right cancel each other out leaving 1 black and 1 white on left of line.

Top middle overlaid on middle middle gives 1 black left of the line, 1 black and 3 white right of the line. The black squares cancel out to give 3 whites right of the line.

For the 3rd column, we overlay the images and get 2 black left of line and 1 black 2 white right of the line. the duplicate black squares cancel out leaving 1 black left of line and 2 white right of line. So answer is 'C'

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Some contributors might have used top to bottom approach i.e solving the 1st two images in the column.

My approach is from left to right with respect to the images in the first row.

Here's my interpretation of the solution's approach to give a little clarity as to how the first 2 rows' images were possible.

1st Row: First image, 1 black and 2 whites on the left of the bar. Second image, 1 black on the left and 2 whites on the right of the bar.

Adding the blacks as they are on the same side i.e left. And since the whites are on the opposite sides of the bar and are two in number they cancel each other out.

Therefore final image is of the two blacks on the left.

Applying the same logic, the 2nd rows' final image is as follows:

1 white on right from 1st image + 1 white and 1 black on right from the 2nd image.

Since all are on the same side (right side) they aren't cancelled out and you get 1 black and 2 whites with black on top in the final image.

If you've understood my approach to the solution, hopefully you can complete the figures given in the 3rd row all by yourself.

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