I have two exercises with solutions (red) from a German assessment centre logic test which I not understand:

enter image description here

Does anybody get the logic behind them?

  • $\begingroup$ Can you please link to the original source? $\endgroup$
    – bobble
    Jun 4, 2020 at 20:37
  • $\begingroup$ mputraining.de/mpu_tests/neunkaestchentest $\endgroup$
    – user267839
    Jun 4, 2020 at 20:42
  • $\begingroup$ ...but they occure (pseudo) randomly $\endgroup$
    – user267839
    Jun 4, 2020 at 20:43
  • $\begingroup$ Are you sure that the answer for the first one is the first image, @TimGrosskreutz ? $\endgroup$ Jun 4, 2020 at 21:39
  • $\begingroup$ @JohnBrookfields: yes, start the linked test a couple times. The missing part is varying and if you face the first one with another part missing you will see that the middle part is the X $\endgroup$
    – user267839
    Jun 4, 2020 at 21:46

2 Answers 2


For the second puzzle, Positions The position of the regions are given in the image above. The logic is that each shaded region makes a copy of itself and moves the copy to its right. For example, consider the image $(1,1)$. The shaded region $4$ makes a copy and moves the copy to its right (region $1$). Now, add this new image with the second given image. It produces an image with shaded regions in $4,1,2$ which is exactly the third image. Note : All the shaded region in the first region must be replicated and all these replicas must be moved to the right (refering to the image $(3,1)$). I mean in the image $(3,1)$, copy of the shaded region $4$ moves to $1$ and the copy of the shaded region moves to $2$. [Only the copies move not the original one. This is the point stressed from the beginning.] Now, the shaded regions in this new imaged formed from $(3,1)$ are : $4,1,2$. Now, adding this with the second gives the third. First puzzle For the first puzzle (as shown above), the rules are :

    1. Big square with a cross destroys the smaller white square and only the bigger one remains.
    2. The smaller square adds to a cross to produce a blue small square.
    3. Big square adds to small blue square to get a big square containing the blue square with a cross.
Confimation of the rules:
    1. In images $(1,1), (1,2)$, big white square with a cross destroys the smaller white square insde another big white square i.e. cross and small white square are annihilated and big squares are merged. This is equivalent to the image $(1,3)$
    2. In images $(1,1), (2,1)$, the above happens along with rule #3 and the duplicate cross (due to the cross in small blue square) is merged with the one produced due to addition of bigger white square and the smaller blue square.
    3. Since in images $(1,2),(3,2)$, the difference is only the change of color of the smaller square. Thus, there must be cross and again this confirms rule #2 for images $(2,1),(2,3)$.
    4. And finally, images $(1,3),(2,3)$ combine according to rule #3 to form the image $(3,3)$.

  • 3
    $\begingroup$ I think I'd call both of these solutions "overfitting", that is, the logic is convoluted enough that any answer could be justified with an explanation that's no more complex than the one given. That said, I don't have any simpler answers to give, so in no way am I going to blame the answerer: +1. $\endgroup$
    – Bass
    Jun 7, 2020 at 13:54

I have a different idea of the second puzzle:

In row 1 and 3, the "inclusive sum" (that is, superimposing one over another without cancelling out overlapping parts) of the 1st and 2nd image will yield a shape that is exactly one triangle short of the final image in the 3rd column. The idea is that on row 1, there is no overlap for the first two shape; but for row 2 and 3, the number of overlapping triangles increases from 1 to 2.


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