Mensa question - groups of three white, shaded, black squares [closed]

Question

I've tried to solve this question (Exercise 31) in many ways but I couldn't figure out the logic behind this.

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Answer 1

My solution:

D

My reasoning:

Treat the triples as vectors over a set with elements B (black), W (white) and S (striped). The vectors in the first two rows are "added" (with some noncommutative binary operation), the third row is the result. From the first elements in the first column, we get S + S = W. This leaves the solutions B, D and E. B and E can be eliminated because W + S can't be two values at the same time.

Answer 2

I think the answer is

'A'. There are 9 white squares - 3 in the right position, 3 in the middle, 3 in the left spot. There are 10 striped - 4 left, 3 middle, 3 right. There are 5 blacks - 1 left, 2 middle, 2 right.
Option A would give you:
6 blacks, 2 at each position.
9 whites, 3 at each position.
12 striped, 4 at each position.

Answer 3

I need help with finding a pattern among the black squares, what I have found so far:

'B' is the correct answer.

Why?

I took the test 3 times in a row giving the exact same answers for problem 1 to 15 (14 correct + 1 wrong) and then gave the answer A, B and D on problem 31 for in each of the tests. A and D gives 88 iq, B gives 92 iq -> B is correct. (double and triple checked this, its B)

A, B or D is the correct answer, why?

There is a clear pattern on the diagonals leaving B or D (A/B/D on new version) as the only possible correct answer. NOTICE: THIS SPECIFIC TEST IS THE OLD VERSION AND ON THE NEW ONE A is replaced with another 3 black tiled picture. So in the picture over we can deduce that B or D is the correct answer.

On the diagonals top left to bottom right we have an inversion of stripes + white on the main diagonal (diagonal from top right to bottom left including picture 3, 5 and 7) -> inversion: white becomes striped, striped becomes white. this is if we look past the black tiles.

So our answer gives white middle square because of picture 1 having a white middle square and striped outer squares because of picture 5 having white outer squares (inverse of the diagonal -> change to striped).

Therefore we know that its (Striped, White, Striped) but in not knowing the black pattern: there is possible black squares which could cover, leaving B or D as the only possible correct answer (A/B/D in new version).

The only possible solution: the rows give a (3 black, 2 black, 1 black pattern) and columns give (2 black, 3 black, 1 black pattern) -> answer must have one black -> B is the correct answer. {THIS IS VERY ARBITRARY)

I feel like this theory behind the black squares is very arbitrary and might be incorrect. I would love some other explanation behind the black squares.

Last notes:

• there is also a pattern on the opposite diagonal also yielding B or D (A/B/D on new version).
• if the legit answer of the old version was D, then the puzzle might be just a random solution created to waste time with the logic that someone smart would see there is no pattern and waste little time and someone less smart would spent a lot of time and therefore having almost no time for the last problems.

Answer 4

For this task, I would argue that it is important to ignore the black tiles. If you look at the blank space, the black will not provide pointers towards the logical conclusion (as there are no black tiles leading to the solution). There is consistency in everything except how the black tiles relate to other tiles. My reasoning is the following: How does white relate to grey/striped, how does white relate to white, and how does grey relate to grey. When examining the tiles, the logical conclusion is D. White relates to white by giving a white tile, grey relates to white by providing a grey tile, and grey relates to grey by giving a white tile. Thus, both vertically and horizontally, the conclusion is grey, white, grey.

Answer 5

Other possible answers [not mentioned as an option]

Answer 1:

My reasoning:

Treat them as base 3 numbers. Like: striped = 1, white = 2, black = 0. Now, from left to right and top to bottom, we have, $$120, 010, 212$$ $$101, 202, 111$$ $$222, 121, ?$$ Translating them to decimal, we have: $$15, 3, 23$$$$10, 20, 13$$$$26,16,?$$ Now sum of the two smaller number + 1 = largest number in any column. i.e. $$15+10+1 = 26$$$$16+3+1=20$$ So the next number must be either $37$ or $10$ but $37$ will have $4$ blocks which is not the answer. So, the answer is $10$ whose base $3$ representation is $101$ and hence [striped, black, striped]

Answer 2:

Consider the decimal numbers of decoding base 3 numbers. Finding their digit sum, we get $$6,3,5$$$$1,2,4$$$$8,7,?$$. Sum of each row is a multiple of $7$. Hence, the possible answer is 6 (others are not possible since it would take a lot of digits shooting above the 3 block mark). So, possibility 1: $$6 = 0+6 = 020_3$$ Thus, black, white, black. Possibility 2: $$6 = 1+5 = 110$$[striped, striped, black]. Possibility 3: $$6=2+4 = 220$$ i.e. white, white, black.

Answer 6

I would say possibly A because then the number of whites are decreasing by -1 if you add the number of white squares per subsequent column from left to right. Also there is another contingent pattern it solves with the sum of blacks per subsequent row from up to down decreasing by -1. No other answer satisfies the first logic anyway. So it is sufficient enough to stand on its own as the sole reason.

Answer 7

My solution:

D

My reasoning:

The patters works both in rows and columns: The first and third figures in the third row and third columns are white & grey. Therefore, whatever the results is, the first and third figure must be the same, which leaves us with options D and F. From the first row, we know that white & grey is grey, which leaves us D.

Answer 8

This is not a full solution. It is brainstorming that others might be able to use.. if I come up with a solution that meets all requirements I will delete this comment and append the solution to the bottom.

Let's first look at the solution posted by Alexander Fasching.

- Overlay the left column with the center column.
- Let G be striped (grey), B be black, and W be white.
- Wherever G and B overlap the result is W
- Wherever W and G overlap the result is G
- W and W overlapping is undefined.
* - But if you take the first horizontal row and overlay it with the middle horizontal row you get B+W=W on column 1 and B+W=G on column 2.
* - Another break is to work on the columns left to right, where it breaks because on Row 1 B+B=W but on Row 2 B+B=G

Therefore I am thinking there must be another solution..

That being said, what if we look at Deepthinker101's solution.

- First off, as they pointed out choosing A allows for all small boxes to be grouped into 3 same color sets evenly.
- 3 sets of whites, 2 sets of black, and 4 sets of greys.
- The only issue I have is that it seems arbitrary to break all of the groups apart to meet this goal.. but perhaps there's a pattern that explains it.

So, looking even further into it..

- I found a definite pattern.. - I assigned G as -1, W as 0 and B as 1. I then subtracted C1R1 from C4R2 and arrived at -1 (C9R3). Then I subtracted C1R2 from C4R3 and got C9R1. Then I subtracted C1R3 from C4R1 and got C9R2. You can repeat this system for the entire grid and it works out.. Except it requires a solution of Grey, Grey, Black.
- As this is not an option it can't be the area, but perhaps the pattern can still be used.

Answer 9

D because horizontally, white + grey -> grey

Answer 10

My answer:

B

My reasoning:

Looking horizontally, we can see that the first two pictures of each row have a tile or two that the third picture does not have. Knowing this, we can conclude that the first two pictures in each row are "same" in that sense and the third picture is "different". The third picture in the third row(which is represented with the question mark) must contain a black tile to be "different" from the first two pictures as the first two pictures do not contain any black tiles, leaving us with options A, B, E and F. Looking horizontally at the first two rows, we can see that either one or all of the striped tiles in the first two pictures goes to the third picture. From this, we can conclude that there must at least one striped tile in the third picture of the third row(which is represented with the question mark). Knowing this, we can rule out options E and F as they both do not contain any striped tiles, leaving us with options A and B. Looking horizontally at the second row, we can see that in the third picture, there is a striped tile in the middle that does not exist in the first two pictures. The reason why there is such a thing is that there are black tiles in the middle of the first two pictures of the second row and from what we know, black tile + black tile -> white/striped tile. However, there are no black tiles in the middle of the first two pictures of the third row. There is also no evidence that tells us that white tile + white tile -> striped tile, and hence there should not be a striped tile in the middle of the picture represented with the question mark. Knowing this, we can rule out A, as it has a striped tile in the middle. This leaves us with option B.

Answer 11

Answer:

Let us number the squares 1, 2, 3, where 1 is striped, 2 is white, and 3 is black. The missing combination in the big picture where the question mark is are 3, 2, 3 and 3, 3, 3. By replacing the question mark with one of these, we answer the question.

Reasoning:

Here we have the case of combinations by repetition. Specifically, we have three objects taken three at a time, in which case we have ten combinations in total.

Answer 12

1. There are 9 cells in the matrix.
2. A cell contains a set of three squares b,w,s.
3. There are no duplicate cells in the matrix.
4. Each row in the matrix contains all 3 types of squares b,w,s.
5. Each row of the matrix contains only 2 cells with the same center square.

A is my answer.