Can't figure this one out.. What is the missing box? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Difficult IQ test question: What is the box suggesting?Mensa IQ test questionMissing number in figureSeemingly difficult culture fair question5008 out of the boxFigure out the missing cardFind the missing boxFind the missing brickFind the missing brick, againShapes and letters. What is the missing box?
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Can't figure this one out.. What is the missing box?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Difficult IQ test question: What is the box suggesting?Mensa IQ test questionMissing number in figureSeemingly difficult culture fair question5008 out of the boxFigure out the missing cardFind the missing boxFind the missing brickFind the missing brick, againShapes and letters. What is the missing box?
$begingroup$
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?
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?
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... :(
$endgroup$
– Steve
10 mins ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
9 mins ago
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?
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?
logical-deduction pattern visual geometry progressive-matrix
logical-deduction pattern visual geometry progressive-matrix
New contributor
New contributor
edited 4 hours ago
user477343
3,05511061
3,05511061
New contributor
asked 6 hours ago
Rrz0Rrz0
1285
1285
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
10 mins ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
9 mins ago
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
10 mins ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
9 mins ago
$begingroup$
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
10 mins ago
$begingroup$
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
10 mins ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
9 mins ago
$begingroup$
I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
9 mins ago
add a comment |
2 Answers
2
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!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
4 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
50 mins ago
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.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) 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
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
4 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
50 mins ago
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!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
4 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
50 mins ago
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!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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!
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 $9$).
$LARGEstackrelimpliesvphantomprod_n=1^Uparrow$
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 5 hours ago
answered 5 hours ago
user477343user477343
3,05511061
3,05511061
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
4 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
50 mins ago
add a comment |
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
4 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
50 mins ago
1
1
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
4 hours ago
$begingroup$
Yay! Thanks @Rrz0 for the $colorgreencheckmark$! It was a nice puzzle :D
$endgroup$
– user477343
4 hours ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
50 mins ago
$begingroup$
Thank you @user477343 for the time you took to solve the puzzle.
$endgroup$
– Rrz0
50 mins ago
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.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) 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
$endgroup$
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.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) 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
$endgroup$
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.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) 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.
(Square 1 to 4) and then decreases the number of shaded shapes (square 5 to 8) 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
New contributor
answered 43 mins ago
Rrz0Rrz0
1285
1285
New contributor
New contributor
add a comment |
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... :(
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– Steve
10 mins ago
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I'll post it as an answer anyway, I can always delete it.
$endgroup$
– Steve
9 mins ago