Can't figure this one out.. What is the missing box?
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I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
This was taken from a Korn Ferry Leadership Assessment practice trial.
logical-deduction pattern visual geometry progressive-matrix
New contributor
$endgroup$
add a comment |
$begingroup$
I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
This was taken from a Korn Ferry Leadership Assessment practice trial.
logical-deduction pattern visual geometry progressive-matrix
New contributor
$endgroup$
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
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– Steve
Apr 14 at 21:43
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I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
Apr 14 at 21:43
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Rrz0, please edit in where this puzzle came from. See the close notice for more info.
$endgroup$
– Rubio♦
Apr 15 at 5:49
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@Rubio, apologies for this, will update. Do let me know if further action should be taken from my end.
$endgroup$
– Rrz0
Apr 15 at 6:55
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Thanks Rrz0. Reopened.
$endgroup$
– Rubio♦
Apr 15 at 7:47
add a comment |
$begingroup$
I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
This was taken from a Korn Ferry Leadership Assessment practice trial.
logical-deduction pattern visual geometry progressive-matrix
New contributor
$endgroup$
I've been stuck on this for ages, and can't figure this out.
What is the missing box, and the logic behind the answer?
This was taken from a Korn Ferry Leadership Assessment practice trial.
logical-deduction pattern visual geometry progressive-matrix
logical-deduction pattern visual geometry progressive-matrix
New contributor
New contributor
edited Apr 15 at 6:56
Rrz0
New contributor
asked Apr 14 at 15:38
Rrz0Rrz0
1587
1587
New contributor
New contributor
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
Rrz0, please edit in where this puzzle came from. See the close notice for more info.
$endgroup$
– Rubio♦
Apr 15 at 5:49
$begingroup$
@Rubio, apologies for this, will update. Do let me know if further action should be taken from my end.
$endgroup$
– Rrz0
Apr 15 at 6:55
$begingroup$
Thanks Rrz0. Reopened.
$endgroup$
– Rubio♦
Apr 15 at 7:47
add a comment |
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
Rrz0, please edit in where this puzzle came from. See the close notice for more info.
$endgroup$
– Rubio♦
Apr 15 at 5:49
$begingroup$
@Rubio, apologies for this, will update. Do let me know if further action should be taken from my end.
$endgroup$
– Rrz0
Apr 15 at 6:55
$begingroup$
Thanks Rrz0. Reopened.
$endgroup$
– Rubio♦
Apr 15 at 7:47
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
Rrz0, please edit in where this puzzle came from. See the close notice for more info.
$endgroup$
– Rubio♦
Apr 15 at 5:49
$begingroup$
Rrz0, please edit in where this puzzle came from. See the close notice for more info.
$endgroup$
– Rubio♦
Apr 15 at 5:49
$begingroup$
@Rubio, apologies for this, will update. Do let me know if further action should be taken from my end.
$endgroup$
– Rrz0
Apr 15 at 6:55
$begingroup$
@Rubio, apologies for this, will update. Do let me know if further action should be taken from my end.
$endgroup$
– Rrz0
Apr 15 at 6:55
$begingroup$
Thanks Rrz0. Reopened.
$endgroup$
– Rubio♦
Apr 15 at 7:47
$begingroup$
Thanks Rrz0. Reopened.
$endgroup$
– Rubio♦
Apr 15 at 7:47
add a comment |
3 Answers
3
active
oldest
votes
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I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
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1
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Yay! Thanks @Rrz0 for the $color{green}{checkmark}$! It was a nice puzzle :D
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– user477343
Apr 14 at 17:33
1
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Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
Apr 14 at 21:03
1
$begingroup$
Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks!
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– Rubio♦
Apr 15 at 5:47
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@Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed.
$endgroup$
– user477343
Apr 15 at 5:53
1
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Nah, you can leave it; just a note for next time. Thanks :)
$endgroup$
– Rubio♦
Apr 15 at 5:58
add a comment |
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Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Box 1 to 4) show an increase in shaded number of shapes and then from Boxes 5 to 8 there is a decrease in the number of shaded shapes,in a clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of sides (+1) that the newly shaded shapes have.
New contributor
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1
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I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$
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– user477343
Apr 15 at 3:50
1
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@user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices.
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– Rubio♦
Apr 15 at 7:50
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@Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!!
$endgroup$
– user477343
Apr 15 at 8:01
add a comment |
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I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
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1
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Thanks for your contribution. I think this directly relates to my rule number 2.
$endgroup$
– Rrz0
Apr 15 at 6:58
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
$endgroup$
1
$begingroup$
Yay! Thanks @Rrz0 for the $color{green}{checkmark}$! It was a nice puzzle :D
$endgroup$
– user477343
Apr 14 at 17:33
1
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
Apr 14 at 21:03
1
$begingroup$
Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks!
$endgroup$
– Rubio♦
Apr 15 at 5:47
$begingroup$
@Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed.
$endgroup$
– user477343
Apr 15 at 5:53
1
$begingroup$
Nah, you can leave it; just a note for next time. Thanks :)
$endgroup$
– Rubio♦
Apr 15 at 5:58
add a comment |
$begingroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
$endgroup$
1
$begingroup$
Yay! Thanks @Rrz0 for the $color{green}{checkmark}$! It was a nice puzzle :D
$endgroup$
– user477343
Apr 14 at 17:33
1
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
Apr 14 at 21:03
1
$begingroup$
Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks!
$endgroup$
– Rubio♦
Apr 15 at 5:47
$begingroup$
@Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed.
$endgroup$
– user477343
Apr 15 at 5:53
1
$begingroup$
Nah, you can leave it; just a note for next time. Thanks :)
$endgroup$
– Rubio♦
Apr 15 at 5:58
add a comment |
$begingroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
$endgroup$
I think the answer is
A
Reasoning:
If you look at each diagonal going from top left to bottom right, there appears to be a pattern of opposing ratios in colour, and corresponding ratios in edges.
Let $rm B : W$ be the ratio of Black shapes to White shapes, then the first diagonal (bottom left corner) is $1 : 3$.
Next diagonal (left middle box and bottom middle box) is $4 : 0$ and $0 : 4$ respectively.
Next diagonal (the middle diagonal with the missing box) have the ratios $1 : 3$ and $3 : 1$, so to me it makes sense that the missing box has a ratio of $1 : 3$.
That leaves either
A or F
Due to how the shapes in these boxes are positioned, I am leaning towards
A
But also, in order to explain the colour scheme...
...it has to do with the edges! Here is how we will count them for each shape:
White circle's edges: 0
Black circle's edges: 1
Donut's edges: 2 (because the outer edge and inner edge are accounted for since the donut is coloured black, because the circle coloured black has its outer edge accounted for as well).
All other shapes have the number of edges you see they have.
And then one last rule:
Every shape corresponds to another shape with 4+ the edges (with alternating colour).
Let me explain:
The top left box has a black circle in the top right and the rest are white circles. That leaves a $1 : 0$ edge ratio (in the same ratio form as $rm B : W$ but not in terms of colour anymore). Therefore, the next box in this diagonal (the middle box) must share this same edge ratio.
All the white circles correspond to black diamonds (white circles have 0 edges, 0 + 4 = 4 so the next shape has 4 edges, and the opposite of white is black, so we have black diamonds). So, in the top left box, we replace the white circles with black diamonds, and then change the black circle to a white pentagon (black circles have 1 edge, 1 + 4 = 5 so the next shape has 5 edges, and the opposite of black is white, so this makes the next shape a white pentagon). Now we can maintain our edge ratio, and violá! Lo and behold the box in the middle!
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
white circles (0 edges) $to$ black diamonds (0 + 4 edges);
black circles (1 edge) $to$ white pentagons (1 + 4 edges).
Same concept applies for the first parallel diagonals above and below the middle diagonal, if you want to try this pattern with the other boxes!
Get it, now?
Repeating this process in the middle diagonal from our middle box, black diamonds go to white octagons (black 4 $to$ white $8$) and white pentagons go to black nonagons (white 5 $to$ black $9$).
$LARGEstackrel{implies}{vphantom{prod_{n=1}^Uparrow}}$
Recognise the pattern? :)
This is true
for every diagonal. You can count the bottom left corner and top right corner as part of the same diagonal, and the pattern still works.
edited Apr 14 at 16:40
answered Apr 14 at 15:56
user477343user477343
3,08411064
3,08411064
1
$begingroup$
Yay! Thanks @Rrz0 for the $color{green}{checkmark}$! It was a nice puzzle :D
$endgroup$
– user477343
Apr 14 at 17:33
1
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
Apr 14 at 21:03
1
$begingroup$
Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks!
$endgroup$
– Rubio♦
Apr 15 at 5:47
$begingroup$
@Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed.
$endgroup$
– user477343
Apr 15 at 5:53
1
$begingroup$
Nah, you can leave it; just a note for next time. Thanks :)
$endgroup$
– Rubio♦
Apr 15 at 5:58
add a comment |
1
$begingroup$
Yay! Thanks @Rrz0 for the $color{green}{checkmark}$! It was a nice puzzle :D
$endgroup$
– user477343
Apr 14 at 17:33
1
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
Apr 14 at 21:03
1
$begingroup$
Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks!
$endgroup$
– Rubio♦
Apr 15 at 5:47
$begingroup$
@Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed.
$endgroup$
– user477343
Apr 15 at 5:53
1
$begingroup$
Nah, you can leave it; just a note for next time. Thanks :)
$endgroup$
– Rubio♦
Apr 15 at 5:58
1
1
$begingroup$
Yay! Thanks @Rrz0 for the $color{green}{checkmark}$! It was a nice puzzle :D
$endgroup$
– user477343
Apr 14 at 17:33
$begingroup$
Yay! Thanks @Rrz0 for the $color{green}{checkmark}$! It was a nice puzzle :D
$endgroup$
– user477343
Apr 14 at 17:33
1
1
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
Apr 14 at 21:03
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
Apr 14 at 21:03
1
1
$begingroup$
Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks!
$endgroup$
– Rubio♦
Apr 15 at 5:47
$begingroup$
Please remember, we require attribution for puzzles that are not the creation of the person posting them here. There is a close reason for that; it's there for a reason; please use it. If you don't have close privilege, you can flag it for the same reason. We don't want to encourage people posting unattributed content by providing answers, so .... close or flag, don't answer. Thanks!
$endgroup$
– Rubio♦
Apr 15 at 5:47
$begingroup$
@Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed.
$endgroup$
– user477343
Apr 15 at 5:53
$begingroup$
@Rubio do you wish for me to delete my answer? I can do that. I flagged the question, albeit it has already been closed.
$endgroup$
– user477343
Apr 15 at 5:53
1
1
$begingroup$
Nah, you can leave it; just a note for next time. Thanks :)
$endgroup$
– Rubio♦
Apr 15 at 5:58
$begingroup$
Nah, you can leave it; just a note for next time. Thanks :)
$endgroup$
– Rubio♦
Apr 15 at 5:58
add a comment |
$begingroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Box 1 to 4) show an increase in shaded number of shapes and then from Boxes 5 to 8 there is a decrease in the number of shaded shapes,in a clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of sides (+1) that the newly shaded shapes have.
New contributor
$endgroup$
1
$begingroup$
I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$
$endgroup$
– user477343
Apr 15 at 3:50
1
$begingroup$
@user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices.
$endgroup$
– Rubio♦
Apr 15 at 7:50
$begingroup$
@Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!!
$endgroup$
– user477343
Apr 15 at 8:01
add a comment |
$begingroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Box 1 to 4) show an increase in shaded number of shapes and then from Boxes 5 to 8 there is a decrease in the number of shaded shapes,in a clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of sides (+1) that the newly shaded shapes have.
New contributor
$endgroup$
1
$begingroup$
I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$
$endgroup$
– user477343
Apr 15 at 3:50
1
$begingroup$
@user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices.
$endgroup$
– Rubio♦
Apr 15 at 7:50
$begingroup$
@Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!!
$endgroup$
– user477343
Apr 15 at 8:01
add a comment |
$begingroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Box 1 to 4) show an increase in shaded number of shapes and then from Boxes 5 to 8 there is a decrease in the number of shaded shapes,in a clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of sides (+1) that the newly shaded shapes have.
New contributor
$endgroup$
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Rule 1:
Each box increases in the number of shaded shapes in a clock-wise fashion. When all four shapes are shaded, each following box increases in the number of un-shaded shapes in a clock-wise manner.
Let us take the shaded circle in box 1 is the starting point.
(Box 1 to 4) show an increase in shaded number of shapes and then from Boxes 5 to 8 there is a decrease in the number of shaded shapes,in a clock-wise manner. Therefore, by simply following this rule, the answer would be A
since it has one shaded shape, which is also in the correct starting position.
Rule 2:
With each increase in shaded shape, there is an also an increase in the number of sides (+1) that the newly shaded shapes have.
New contributor
edited Apr 15 at 7:00
New contributor
answered Apr 14 at 21:10
Rrz0Rrz0
1587
1587
New contributor
New contributor
1
$begingroup$
I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$
$endgroup$
– user477343
Apr 15 at 3:50
1
$begingroup$
@user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices.
$endgroup$
– Rubio♦
Apr 15 at 7:50
$begingroup$
@Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!!
$endgroup$
– user477343
Apr 15 at 8:01
add a comment |
1
$begingroup$
I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$
$endgroup$
– user477343
Apr 15 at 3:50
1
$begingroup$
@user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices.
$endgroup$
– Rubio♦
Apr 15 at 7:50
$begingroup$
@Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!!
$endgroup$
– user477343
Apr 15 at 8:01
1
1
$begingroup$
I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$
$endgroup$
– user477343
Apr 15 at 3:50
$begingroup$
I think my answer is a consequence of your answer; i.e. yours is actually the intended one. Ha! I suppose I was merely overthinking. $(+1)$
$endgroup$
– user477343
Apr 15 at 3:50
1
1
$begingroup$
@user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices.
$endgroup$
– Rubio♦
Apr 15 at 7:50
$begingroup$
@user477343 .... please stop the mathjax +1s, and +1s in general. Your vote suffices.
$endgroup$
– Rubio♦
Apr 15 at 7:50
$begingroup$
@Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!!
$endgroup$
– user477343
Apr 15 at 8:01
$begingroup$
@Rubio oh s***! Sorry, sorry. I am just so used to it, especially after writing such comments on the Math Stack Exchange. No more!!!
$endgroup$
– user477343
Apr 15 at 8:01
add a comment |
$begingroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
$endgroup$
1
$begingroup$
Thanks for your contribution. I think this directly relates to my rule number 2.
$endgroup$
– Rrz0
Apr 15 at 6:58
add a comment |
$begingroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
$endgroup$
1
$begingroup$
Thanks for your contribution. I think this directly relates to my rule number 2.
$endgroup$
– Rrz0
Apr 15 at 6:58
add a comment |
$begingroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
$endgroup$
I don't think this is really an answer. By only looking at the question I arrived at the right answer. It could be luck, but I think the puzzle might be improved slightly...
There is a simple pattern, if the boxes are "read" left-to-right, top-to-bottom. The first one contains shapes containing maximum one side. Block 3 has triangles, 4 has squares, etc. I lazily justified that the second box was "weird" because lines aren't used, but circles in circles ... two sides?
The choices to select an answer from only contain a single selection with a 9-sided polygon.
I'd recommend adding a nonagon to at least one other of the choices :)
New contributor
edited Apr 14 at 22:11
New contributor
answered Apr 14 at 21:51
SteveSteve
1212
1212
New contributor
New contributor
1
$begingroup$
Thanks for your contribution. I think this directly relates to my rule number 2.
$endgroup$
– Rrz0
Apr 15 at 6:58
add a comment |
1
$begingroup$
Thanks for your contribution. I think this directly relates to my rule number 2.
$endgroup$
– Rrz0
Apr 15 at 6:58
1
1
$begingroup$
Thanks for your contribution. I think this directly relates to my rule number 2.
$endgroup$
– Rrz0
Apr 15 at 6:58
$begingroup$
Thanks for your contribution. I think this directly relates to my rule number 2.
$endgroup$
– Rrz0
Apr 15 at 6:58
add a comment |
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
Rrz0 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I was able to come up with the correct answer, and with a much, much simpler reason than the other answers. I'd love to share (and get feedback), it's not worth its own answer and I don't think there's a way to hide spoilers in comments... :(
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
Apr 14 at 21:43
$begingroup$
Rrz0, please edit in where this puzzle came from. See the close notice for more info.
$endgroup$
– Rubio♦
Apr 15 at 5:49
$begingroup$
@Rubio, apologies for this, will update. Do let me know if further action should be taken from my end.
$endgroup$
– Rrz0
Apr 15 at 6:55
$begingroup$
Thanks Rrz0. Reopened.
$endgroup$
– Rubio♦
Apr 15 at 7:47