# Mensa Norway last 3 questions

I had no problems with solving first 32 questions from http://test.mensa.no but the last 3 questions are totally bugging me out even knowing the answers. Could anyone help me getting the logic behind them?   E A D

For the last one:

All the diagonals and broken diagonals have a shape in common. For those going NE we have:   And for those going SE we have:   For the second one:

The black balls follow the rule of diagonals going NE described above. The white balls follow a separate rule: There are two white balls in every cell and they move in a counter-clockwise direction in the column they are in, as shown by the yellow balls above. If a white ball is at the same location as a black ball, only the black ball is shown.

The link provided by @Duck in the comments gives the answer to the first problem, so all three problems should now be answered.

• I saw this answer in comments but tbh I don't get the explanation for the first riddle. Could you explain it better for me? – Michał Pilarek Sep 19 '19 at 8:32
• Look at the $3$ NE diagonals and assume the triangles can flip across the line they are connected to. Only one flip is allowed per row change and you start in the top row, moving down. Assume also that a striped triangle flipped onto a white triangle gives a striped triangle. Hope that helps! – Jens Sep 20 '19 at 1:55
• Ok, now I get it – Michał Pilarek Sep 20 '19 at 11:26
• The answer could also be E for the last one since because each row has in total then (of the rectangle) two tops, one bottom and one left and one right. This is why iq tests often annoy me because you can have a logically consistent answer and still be wrong. Oh well they are for mainlynfor people who don't know that they are smart (which probably means they are not if they take it). – Deepthinker101 Sep 20 '19 at 23:13

3 is:

D. The matrix can be interpreted as a determinant, with the 'factor' of a cell being formed from $$(TL \land BR) \lor (TR \land BL)$$, where $$\land$$ returns the common components of the two images, and $$\lor$$ returns all components of the images.

The correct answer for the first one is

E. Look at the 2nd block and see the figure which is shaded out and now see the other two figures, you will see that the other two figures are just opposite/reverse of each other. Now come to the 3rd block and see the figure which is shaded out and see the figure which is not shaded out. So the correct answer should be the figure which is the opposite/reverse of the non-shaded figure in the 3rd block, which is option E.

For the first problem a.k.a problem #33 I looked it up in downward direction. For each column we can see that there is a triangle that never “moves” (red circle) and that's our reference point and it follows the clockwise pattern . So we can expect that in the third column there will be a plain triangle at the upper left corner.

The rest is finding the alternating triangle pattern start from the first column and I guess you can just follow it with few adjustments for the rest of the columns. 