What's the most reasonable solution for this pattern puzzle?
Source: https://www.polizeitest.de/einstellungstest-polizei-matrizentest/
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4 Answers
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Could be d because the rule is that the amount of empty tiles contained in the previous two columns leads to the amount in the last and on top of that every tile has two shapes in the top left corner and every row would have all 4 shapes and all shapes have all 3 colours
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$\begingroup$ Hi, one remark on your last argument. You wrote: "...and on top of that every tile has two shapes in the top left corner and every row would have all 4 shapes and all shapes have all 3 colours..." The last sentence I not understand. What do you mean by that in a row all 4 shapes with all 3 colours occure? Do you probably mean in the sense that separately in every upper row all figures occure and all colours? The formulation suggests that in every row every possible symbol with each posible color for each symbol occure, and this is of course wrong $\endgroup$ Commented Jul 20, 2020 at 1:27
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$\begingroup$ No I mean in terms of the total grid if you look at all the tiles $\endgroup$– PDTCommented Jul 20, 2020 at 3:18
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It could be option d), but also thinking it could be option b) as every row from top-to-bottom has 3 filled in shapes (the top one has 3 circles and the bottom has 3 triangles), makes me think the 2nd row is the inconsistent one, which could may mean it needs a filled in square for that last box. So option
b.
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$\begingroup$ I not understand why this leads you to (b). Your suggested logic is that every row (exept the middle one) should contain a box where 3 same elements occure, right? $\endgroup$ Commented Jul 20, 2020 at 1:34
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$\begingroup$ that is correct, where there are 3 shaded elements (in grey) maybe i forgot to mention that :) ... combined on each row, there's at least 3, the first row has 3 shaded in grey for circles, then the 3 shaded triangles in grey in last row but the middle row has 1 shaded grey circle, 1 shaded grey triangle which led me to think last one should be shaded grey square for uniqueness... $\endgroup$ Commented Jul 20, 2020 at 2:24
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I took the test and looked at the results. According to the source, the answer is
D. The explanation being that the top left corner always has two elements while other positions have arbitrary number of shapes. Which makes sense considering this is only the fifth (our of 15) question.
My original guess was
e for a reason similar to that of deepthinker's. Thought the blank space for the third figure is the addition of the first two (excluding overlapping), and then rotated 90 degrees clockwise.
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Answer is
D.
It's a multi layered pattern comprised of two rules:
1. The top left box always has two shapes in it.
2. The third box from left to right has the sum of first and seconds' blank boxes.
Turning the blank squares clockwise is not a part of the pattern as the middle box of all rows breaks such supposed pattern and leaves you with an answer breaking the more consistent pattern of two shapes in all the top left boxes.
