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I have a question about this Mensa IQ question:

enter image description here

I know that the correct answer is F, but I don't completely understand why.

The solution guide says the logic is to rotate around 180 and duplicate bars. Problem: in which direction are the bars duplicated? To say clockwise or counterclockwise fails in both cases.

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Try "folding" them in half, and duplicating there. They're just exactly opposite and duplicated.

I.E. The bottom left of b is 12 because the 4+2(6) duplicates are overlapping at that point in the circle. Top of b is 18 as the duplicates of both 3 and 6 (9).

As I'm trying to edit this and clean it up I realize I think I misunderstood your question... But why does it have to be clockwise/counterclockwise? If you duplicate outward from the midpoint of each set, maybe that helps clarify?

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  • $\begingroup$ this idea with duplicate outward from the midpoint of each set went also through my head. Problems: how we duplicate if we have a group with odd number of bars. Futhermore for example the two bars group on the left seem to be duplicated only in one direction. Up to now the most promising idea was to draw a vertical line with splits the circle perfectly and deside to which of the both intersetions of the vertical line with the circle is closer to a group of bars, the upper or the lower. $\endgroup$ Feb 21 '20 at 1:57
  • $\begingroup$ And then - after mirroring horizontally - doplicate in the direction with shorter distance between intersection and the group of bars. That works but seems to circumstaneous for me so I was curious if it can be done in more simple way. $\endgroup$ Feb 21 '20 at 1:57
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I think the procedure is like so:

  1. Visit every group of bars
  2. For each group, replicate it and add it clockwise from the original group
  3. Rotate the whole circle by 180 degrees

You could probably swap steps 2 and 3, but I find it harder to understand that way.

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  • $\begingroup$ hehe, maybe I need magnifying glass but I think there is a problem: let's look at the two group of bars in A at the bottom: so the 3 and the 6 group. We duplicate each of them and add them clockwise (2.) and rotate (3.). The new group on the top in B has no gap, on the other hand 6 bars seem not to be enough to fill the gap between considered 3 and 6 groups in A below. Indeed, by your logic the 3 new bars are added not between 3 and 6 groups, but on the other side, since clockwise. Or did I misunderstood your procedure? $\endgroup$ Feb 21 '20 at 4:47
  • $\begingroup$ You understood correctly. I also feel like there should be a gap. Perhaps the drawings are not supposed to be to scale. Otherwise I don't see how it can work. Well at least the counts make sense, each group always doubles in size. $\endgroup$ Feb 21 '20 at 5:18
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I think you want more from this "riddle" that it actually is. First of all the goal of this puzzle is to evaluate your ability to see patterns. To do that it's enough for one answer to fit better that all the rest. And this is what F does.

It doesn't have to be ideal both:
1. because it's a test-like question, you have discrete amount of options
2. because all patterns in the real life are approximate and it may test your ability to find patterns despite their distortions.

If you wish one can formulate the relation A->B and C->F like this:

  1. Rotatate around 180 degree.
  2. Duplucate amount of bars in each group.
  3. Shift each group arbitrary by any angle in between -30 and 30 degree
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