What are the solutions to these 2 puzzles. My brain hurts from trying to figure these out.
The choices are below the box.
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8
I think the first one is
second row, third entry
on the following grounds which I very much suspect are equivalent to something clearer and neater if one thinks about it right:
Think of the white squares as mere background. Each box has a thing of length 3, a thing of length 2, and a thing of length 1, either horizontally or vertically, with some permutation of {black, dark grey, light grey}.
In each row and in each column, the colour of the length-3 thing takes all three possible values once each. So in the bottom right it must be black. (This rules out three of the 8 possibilities given.)
The colour of the length-2, colour of the length-1, and horizontal-or-vertical all appear to fit a different pattern: in each row or column two of the entries are the same and the third is different, and the odd one out is placed differently in all three rows and in all three columns.
So the colour of the length-2 thing has to be not-black (looking at rows) and has to be dark grey (looking at columns). This leaves only row 1 column 2 and row 2 column 3 as possible answers.
And the orientation has to be horizontal (same conclusion looking either at rows or at columns). That leaves only row 2 column 3 as a possible answer.
Again, I suspect that there's a better way to describe what's going on that leads to a nicer-sounding answer.
For the second one, I wonder whether the intended answer is
again row 2, column 3
but again I am not much satisfied with my reasoning:
Count a dot outside the circle as +1 and one inside as -1. Then on each row the third column is the sum of the first two. Therefore, on the third row the sum is zero, which we presumably represent with a circle and no dots.
I find this unsatisfactory for at least two reasons.
First, this doesn't explain anything about where the dots are placed; that turns out not to matter for the bottom-right box which has no dots, but it seems a bit ugly if the positioning is just arbitrary and serves only to confuse. Second, the amount of evidence I'm going off -- basically just the first two rows -- seems insufficient: it could perfectly well just be coincidence that they work out.
I confess that
I think I've seen another IQ-test-like question similar to this second one, with exactly this "addition" rule. This is one reason why I think it likely to be the right answer. (I don't remember that question clearly enough to know whether it's actually the same question.)
answered Jun 5 '16 at 13:32
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1$\begingroup$ Will I be able to answer the first one without any predefined options? Because I didn't think of eliminating the options, my thought process went straight to finding the complete pattern, which I failed to do. $\endgroup$ – William Wino Jun 6 '16 at 17:51
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$\begingroup$ Quite possibly not -- which is another reason why I find my answer unsatisfactory. $\endgroup$ – Gareth McCaughan♦ Jun 6 '16 at 22:27
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$\begingroup$ Allright, thanks for your answer, at least it gives me a little peace of mind. :) $\endgroup$ – William Wino Jun 7 '16 at 14:53
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$\begingroup$ I wonder if this specific kind(9 shapes in a grid, one missing, a set of alternatives) of test have a name? Googled 'spatial IQ test' but there has to be something more specific than that $\endgroup$ – Adam Jun 20 '17 at 14:09
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$\begingroup$ Not that I know of, but that's no reason why there shouldn't be one. $\endgroup$ – Gareth McCaughan♦ Jun 20 '17 at 14:14
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4
For the puzzle on the left
Call the choices from top left to bottom right $(a,b,c,d,e,f,g,h)$
Each entry is made from three bars, one of each of $\{\text{black, dark, light}\}$ (leaving three $\text{white}$ cells). The same is true of all choices.
The bars in each entry are either horizontal or vertical. The same is true for all choices.
1. Each complete row and column contains $1$ vertical and $2$ horizontal entries. Choices maintaining this pattern are $\{d,e,f,g,h\}$.
2. Of the $2$ horizontal entries in each complete row or column $1$ has the three bars aligned on the left and $1$ has them aligned on the right. Choices maintaining this pattern are $\{d,f,g,h\}$.
3. The horizontal entries in each complete row have $\text{black}$ swapped with $\text{light}$, those in each column have $\text{dark}$ swapped with $\text{light}$. Choices maintaining this pattern are $\{e,g\}$
The only choice maintaining all three of these patterns is $g$:
answered Jun 5 '16 at 18:00
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$\begingroup$ Could you elaborate on the third rule please? $\endgroup$ – William Wino Jun 6 '16 at 2:09
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$\begingroup$ @William ignoring bar length the two horizontal entries in a row have the dark bar in the same position and the others swapped. Ignoring bar length the two horizontal entries in a column have the black bar in the same position and the others swapped. $\endgroup$ – Jonathan Allan Jun 6 '16 at 2:26
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$\begingroup$ I see, btw will I be able to answer the first one without any predefined options? Because I didn't think of eliminating the options, my thought process went straight to finding the complete pattern, which I failed to do. $\endgroup$ – William Wino Jun 6 '16 at 17:52
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$\begingroup$ I'm new to solving puzzles like this, and I enjoyed reading your solution, but upon reading your answer and the answer above, I realized it seems like both of you have found viable criteria to eliminate certain options, and then eliminated them, and it somewhat feels like coincidence that both of your criteria resulted in both of you eliminating exactly the same options, resulting in the same answer. Perhaps someone else can construct another set of criteria to eliminate a different set of options, yielding an alternate solution. $\endgroup$ – porridgemathematics Jun 7 '16 at 5:42
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This is for First Puzzle.
Bottom And Second From right
There is no concrete linear pattern though.
But if You consider diagonals as above you can see each diagonal has 1,2 and 3 Blocks of different length but same colour Also orientation of all bars in digonal remains same.
Summarizing.
Blue diagonal : Orientation Vertical
Black Diagonal : Orientation Horizontal
Red Diagonal : Orientation Horizontal
As our Matrix is in Black Diagonal orientation of all bars would be Horizontal and we can see that 3 Black Blocks,2 Dark Gray Block, 1 Light Gray Blocks are missing.
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I have simpler solution to the left side puzzle. There are total 15 blacks, 16 grays and 17 lights. To make them equal, per pattern we need to add 3 blacks, 2 grays and 1 light. That leaves only two options b (row=1, column=2) and g (row=2, column=3). Now, 3- same-colored pattern are once appeared in each column and twice appeared in each row. That means all black needs to be horizontal....answer = g
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1
The right one:
The solution is the empty circle. In each row, there is summation to the right - the last one is a sum of the previous two (and subtraction to the left - the first one is the third minus the second). Dots inside are positive, dots outside negative. A similar summation/subtraction applies for columns.
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$\begingroup$ Hi Vojta, welcome to PSE! Please note that when you answer questions, please avoid spoilers using ">!" to hide your answers. Also note, this question has already been answered (it's almost 3 years old). You only answered half of the question and didn't provide a new answer (Gareth answered the second part the same). $\endgroup$ – eye_am_groot Feb 1 '19 at 12:46
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I think the answer on the first question will be the fifth(bottom row first one) and I will quess for the second on the 3th answer. Not sure!
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It is the
Second one
Because
Always turn the highest placed dot to the right (in the first row) and left (in the second row). Overlapping dots are erased meaning that third row are the dots that remain.
