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?
Answers are:
E A D
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asked Sep 18, 2019 at 15:32
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2$\begingroup$ Your first answer is here (with explanation): puzzling.stackexchange.com/questions/87706/mensa-iq-test-puzzle $\endgroup$– DuckCommented Sep 18, 2019 at 23:28
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12 Answers
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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.
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$\begingroup$ 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? $\endgroup$ Commented Sep 19, 2019 at 8:32
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$\begingroup$ 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! $\endgroup$– JensCommented Sep 20, 2019 at 1:55
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1$\begingroup$ 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). $\endgroup$– PDTCommented Sep 20, 2019 at 23:13
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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.
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$\begingroup$ That is an awesome observation, for those wondering: the given Matrix "M" is its own adjacency matrix : "M = adj(M)", i.e. $M_{ij}$ is obtained by calculating the determinant of the sub matrix of $M$ obtained by removing the i-th row and j-th column of $M$. $\endgroup$ Commented Mar 23, 2023 at 4:40
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Made 3 videos explaining the puzzles, thought it would be easier than reading it over text:
Puzzle 33:
E is the correct answer: https://www.youtube.com/watch?v=cZUr0JYe7zg
Puzzle 34:
A is the correct answer: https://www.youtube.com/watch?v=ks7Js7mJHz4
Puzzle 35:
D is the correct answer: https://www.youtube.com/watch?v=sW-uSRaIcRs
Hopefully anyone who were confused will understand them now :)
answered May 3, 2020 at 16:32
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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.
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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.
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I got A and here is my reasoning (of course, no guarantee it's correct or even CLOSE to the most logical solution):
Before we delve into the main explanation, let us establish these two main variables:
- The number of triangles ABOVE and BELOW the line
- The number of triangles BLACK and WHITE
For the following equations, we subtract the LARGER figure by the SMALLER one.
Focusing on the 1st ROW, we can see there are 5 triangles on the bottom half and 4 on the top. 5 - 4 = 1 2nd ROW: 4 - 3 = 1 3rd ROW (Replace the question mark with A) 4 - 2 = 2
Now let's trying adding the answers of these three equations together: 1 + 1 +2 = 4
Keep the answer 4 in mind as we check the same variable for the columns
Observing the 1st COLUMN, there are 4 on the BOTTOM and 2 at the TOP, so I would answer the equation 4 - 2 = 2; The LARGER figure goes FIRST in these subtraction equations. 2nd COLUMN: 5-3 = 2 3rd COLUMN (once again replacing the missing tile with the one from A): 4 - 4 = 0
Time to add these 3 answers up again! 2 + 2 + 0 = 4
As you can see, the sum for these TWO groups of THREE subtraction equations are EQUAL!!!!
Also notice how there is a PAIR of equal answers to the subtraction equations in both the ROWS and COLUMNS (1 for the ROWS, 2 for the COLUMNS). I'll attempt to postulate why I don't think this is a coincidence later, but for now, let's move on the 2nd VARIABLE!
Let's recap the second variable: "The no. of triangles that are BLACK and WHITE"
Why don't we take a shot at doing the same style of subtraction equations for the ROWS and COLUMNS once again, except this time for the no. of BLACK and WHITE triangles?
Taking a look at the 1st ROW, I can see 6 BLACK and 3 WHITE. May someone double check for me? 🤣 Let's perform subtraction yet again, remembering to put the larger figure first as we'd done for all our earlier equations. 6 - 3 = 3
2nd ROW: 5 - 2 = 3
3rd ROW (remember once again to replace tile A into the unknown space): 3 - 1 = 2
Adding the answers to these 3 equations gives us 3 + 3 + 2 = 8 (EIGHT)! Keep this number in mind as we move onto the COLUMNS
1st COLUMN: 4 - 2 = 2 2nd COLUMN: 5 - 3 = 2 3rd COLUMN (Testing tile A): 6-2 = 4
Let's sum it up! 2 + 2 + 4 = 8. THE ANSWER IS EIGHT AGAIN!!!!
Also notice how there's a pair of common answers in BOTH the ROWS and COLUMNS, just as with our 1st variable?! TAKE A LOOK! For the ROWS, there are a PAIR of 3s and in the COLUMNS, there are a PAIR of 2s! Just in case you forgot, here's the first variable: "The number of triangles ABOVE and BELOW the line".
It appears to me adding the answers of the subtraction equations from The ROWs and the COLUMNs separately gives us the same ANSWER!!!
Remember how I postulated at the beginning that I don't think the pairs of equal answers for each triplet of subtraction equations are a coincidence?
My reasoning is a PAIR is composed of TWO entities. There are ALSO TWO pairs of equal numbers for each variable. Summing up the answers to the subtraction equations in the ROWs and COLUMNS separately for the 2nd VARIABLE gave us 8 for BOTH, which is DOUBLE 4 (the answer we obtained summing up the answers of the subtraction equations in either the ROWS or COLUMNS individually for the 1st VARIABLE!
My final answer is A because none of the other tiles allow these rules to apply for 100% of cases.
Hope my response helped! 🤗👍🏼
If you (reading this right now) manage to poke holes through this theory, feel free to adduce a more logical explanation!
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Puzzle 35 explained with a bigger picture:
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1$\begingroup$ Welcome to Puzzling.SE! I'm afraid I don't see how the image you provided is actually supposed to solve the puzzle in question. Could you elaborate? $\endgroup$– F1KrazyCommented May 14, 2020 at 12:46
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1$\begingroup$ I see looking diagonally \ , there is validity to this idea. The side pieces do move in a regular manner. However, a written explanation is probably needed for others, including myself, to fully grasp this. Great idea, though. Thinking outside the box. -- Note: If that is even what you meant. Otherwise, your idea got me thinking. :) $\endgroup$ Commented May 14, 2020 at 14:51
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$\begingroup$ I wrote a too long explanation so I will try and make it easier. You have these symbols recurring in diagonals, they skip 2 rows. ^ V = | | ¯ So the diagonal from top left to bottom right is V and then it's = and then ^, then it repeats itself. But if we start from top left but going down left it's V, | | and ¯. So it creates the same square over and over again in every direction. $\endgroup$ Commented Jun 6, 2020 at 11:17
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Answer for the first Puzzle:
The amount of triangles that are directed up & down should be balanced. 7 stripped triangles UP - 7 stripped triangles DOWN To make the balance for the blank one you need to choose an answer D.
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We have to put some rules:
1: In rows there is always one triangle white and black and they are moving in a clock wise direction
2: If the white triangle reached this site only (_/) it turns into a stripped one and when leave it returns to be white.
The original triangles I spotted them with orange dots, the arrows show how they move irrespective of the other triangles that appears(which i could an explanation to them if someone wants), finally the answer should be E but upside-down. I hope you get it.
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Another presentation for the solution of 3 is: rearrange the diagonals and solve this easier one:
which is just a cartesian product, as explained by @puzzlesandsolutionsYT
The same rearrangement can help, or not, for the other two puzzles:
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Answer to the third puzzle: D
The simplest solution is to treat each line independently. Ignore everything else. Find the pattern in columns and rows. Is the line there? Is it not there?
Give it a try! :)
Solution rule and steps below
Rule: In all rows and all columns, each line will appear exactly once or twice. Never zero, and never three times.
How to Find Clues: In our solution column and row, we need lines that appear twice, or lines that are missing twice. If a line appears twice, it will not appear in our solution. If a line is missing twice, it will appear in our solution. Ignore everything else.
Solve:
Step 1 - Narrow our answers
Look down column 3 for a common line or common missing line.
There are two top line segments, so we cannot have an answer with a top line. That eliminates A, B, E and F, leaving only C and D as possible solutions.
Step 2 - Find our solution
Look across row 3 for a common line or common missing line.
The left vertical line is missing twice, so our answer must have a left vertical line. C has no left vertical line. Therefore the answer is D.
Step 3 - Verify our solution
We can then "work backwards" and verify the solution both horizontally and vertically for all other individual lines
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For Puzzle 33, I got B. Here is my reasoning. In every pattern, we should first find the constant. The constant is something that doesn't vary in all squares. The top left corner in all squares is shaded, otherwise absent. So our Rule 1 is
R1/ If there is a top left corner, it has to be shaded.
Now we need to confirm if there is a top left corner.
We use S for shaded and N for not-shaded, and 0 for absent.
S___0___S
0___S___S
S___0___?
The patter in this is clear. The answer is S. In the top left corner places, there are only Ss and 0s. And in each row, we have two Ss and one 0. We use correlational logic and conclude to have an S here too.
Only B and C have a top left shaded corner.
Now we need to check the top right corner. Here is the grid for top right corner.
0___S___S
0___0___N
0___N___?
From left to right, in this grid the pattern is that any 2 variables consecutively occur. In row 1, S occurs twice consecutively. In row 2, zero occurs twice consecutively. In row 3, only N can occur twice consecutively.
So it is B. Top-left shaded. Top-right not-shaded.
The only doubt that remains is - why take patterns horizontally, not vertically or diagnolly? Because there is no decipherable pattern in vertical. Diagnol supports the S top-left corner theory.
Check
S___0___S
0___S___S
S___0___?
Diagnolly and horizontally it's S. Vertically no such pattern can be surely confirmed.
For top-right corner
0___S___S
0___0___N
0___N___?
Diagnolly here, both S and N are equally valid. Symmetrically S would apply and asymmetrically N would apply. Vertically too, both S and N are equally valid. No S in column 1, one S in column 2, so two S in column 3. Same for N. No Ns in column 1, one N in column two, two Ns in column 3.
So it's settled. It's B.
