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enter image description here

Can anyone help me solve this puzzle?

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closed as off-topic by Rubio Mar 3 '18 at 2:26

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  • $\begingroup$ Hi, and welcome to the Puzzling Stack Exchange! Please remember to always state the source of any puzzles that aren't your own creation! You might also want to take the tour to see how things work here. Cheers, have fun! $\endgroup$ – Bass Mar 2 '18 at 10:39
  • $\begingroup$ I think "tough" is subjective. I usually cannot solve puzzles on this site at all - not very good at it, but this one, was so obvious to me that I knew the answer within a minute. I'm genuinely surprised this one called tough. $\endgroup$ – Andrew Savinykh Mar 2 '18 at 21:02
  • $\begingroup$ I'm putting this question on hold until proper attribution of its original source is provided. This looks like you're asking us to solve a puzzle you found elsewhere. For content that you did not create yourself, please provide attribution - at minimum you need to let us know where this came from, and any additional context you can provide is usually a big help to solvers. Posts which use someone else's content without disclosing where it came from are generally deleted. $\endgroup$ – Rubio Mar 3 '18 at 2:26
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    $\begingroup$ Ok, well, I'm trying to help. This puzzle can be found on the Preview section (second image) of this web: jobtestprep.co.uk/shl-logical-inductive-reasoning-test-practice $\endgroup$ – athin Mar 3 '18 at 3:15
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The answer is:

C

Because:

If you rotate the hexagon which contains $6$ middle figures counter-clockwise for $60$ degrees, you can see that each of the figures in the middle will be the half-right of its outer figure. The half-left is its mirror.

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    $\begingroup$ Shouldn't that be 90 degrees? $\endgroup$ – ABcDexter Mar 2 '18 at 7:50
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    $\begingroup$ Actually what I want to say is the hexagon which contains the 6 middle figures. $\endgroup$ – athin Mar 2 '18 at 7:51
  • $\begingroup$ Yes, that makes sense :) $\endgroup$ – ABcDexter Mar 2 '18 at 7:52
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    $\begingroup$ Same solution, but I'd explain it this way: Rotate every figure in the hexagon 90 degrees to the left and you get the right half of the outer figure on its right. $\endgroup$ – dessert Mar 2 '18 at 11:08
  • $\begingroup$ @dessert I thought the same thing, but athin's answer is more precise than our impression. Consider if the B option had been flipped horizontally - athin's answer rules that out, whereas the "rotate and cut" pattern allows for asymmetric figures. $\endgroup$ – Brilliand Mar 2 '18 at 23:52
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Meta-argumentation incoming...

C is the right answer. It is a common error in creating "wrong" answers to derive wrong answers by changing just a bit of the correct answer for most of the answers. This seems to make it more difficult, but in fact, it makes it more easier to guess the correct solution.

To explain this in detail:

"A" is the only one depicting a simple form. All others are more complex. So "A" is improbable. Apart from "A", "E" is the only one not consisting of six little squares, but only four. So we ignore "E" as well. "B" is the only one that is not symmetric. Drop this as well. "D" is the only one upside down. This leaves "C" as the only one that has no unique difference to the rest, which makes it the correct answer with high probability.

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  • $\begingroup$ There are some good arguments. Still, I'm not sure which one from C or D is upside-down. $\endgroup$ – Eric Duminil Mar 2 '18 at 16:27
  • $\begingroup$ @EricDuminil D is the only one with downward protrusions, hence it being the upside down one. $\endgroup$ – Carl Kevinson Mar 2 '18 at 16:36
  • $\begingroup$ @CarlKevinson: And to me, C is an upside-down bridge ;) $\endgroup$ – Eric Duminil Mar 2 '18 at 16:44
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    $\begingroup$ @EricDuminil D is upside down with respect to the other possible answers. $\endgroup$ – Dan Henderson Mar 2 '18 at 22:27
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The answer is:

C

Reasoning: Step 1

The outer shapes relate to the inner shapes following this pattern: pattern

Step 2

Most image in their original position have a vertical symmetry axis in their middle. So you rotate them either left or right 90 degrees and mirror them where their top side used to be. At this point it could be either C or D depending on the rotation direction. symmetry

Step 3

One of the provided images is also asymmetric and only works with left rotation. enter image description here

Step 4

Applying the rules to the missing piece, we get C.
enter image description here

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This is an hexagram: a compound of 12 triangles, 6 outward-oriented and 6 inward-oriented (forming an hexagon).

For each outward-oriented triangle, there can be an uniform transformation that matches the image of an inward-oriented triangle. First, take the right vertex of the base of the top outward-oriented triangle, rotate counter-clockwise 120 degrees around this point; divide the figure inside (for instance, the ovale) in 2 symmetrical parts through its horizontal axis, keep the bottom part and mirror it on its vertical axis (no change here for the ovale). Repeat the transformation for each outward triangle, in clockwise direction: next is the rectangle, then the irregular hexagon, your missing piece, trapezoid, lozenge (knowing that the rotation axis will always be the next vertex on the hexagon). So now we are looking at the L shape figure in the hexagon: it’s the one related to our missing piece. Conversely, mirror this L shape on its vertical axis, mirror it again horizontally and keep both parts; rotate clockwise 120 degrees: the missing piece is the U shape (third image from the top among the possible solutions).

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I'll add this as my thought pattern was different as far as I can tell from the other answers. You can simply:

Take each shape in a star point, rotate it clockwise 90 degrees, then chop the top half off.
The result of that operation is the inner symbol one triangle clockwise.

I figured this out starting with the trapezium as that's the only non-symmetric shape, so governs which part of the shape is taken.

With this operation, there is only one option that fits the bill.

Therefore the middle one, C, is the answer

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  • $\begingroup$ Hmm, I think on closer inspection, this is very similar to @PaulDaPigeon's answer. $\endgroup$ – Tom Carpenter Mar 2 '18 at 22:28
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    $\begingroup$ It does seem to be the same as both athin and paul's except you are starting from the outside and translating to the inner whereas they started at the inner and came to the outside. $\endgroup$ – Chris Mar 2 '18 at 22:44

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