The IQ test was pretty jokes until the last question
This one got me stumped.
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4 Answers
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The answer might (there is often multiple solution in these kind of questions) be
D
Explanation
with the rules:
1: moves to the right one square every step
2: moves left one square every two steps
3: moves clockwise as the knight in chess every step
4: moves counter-clockwise only on corners every step
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$\begingroup$ I had a hunch the answer was d because of the movement of the bottom two squares $\endgroup$– PDTCommented Sep 3, 2019 at 15:03
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$\begingroup$ Just could not see the logic of the top two $\endgroup$– PDTCommented Sep 3, 2019 at 15:04
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$\begingroup$ I was trying the same logic identifying the moves of each cube. but that knight move of chess :p making me illogical and also unable to figure out that more than one block can be moved to the same block holder. $\endgroup$ Commented Sep 3, 2019 at 18:04
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I think you all overthink it... how about just
4-3-4-4-3-4
the answer is H
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How about this:
G
Rules:
1: there is an integer sequence capturing the number of total coloured squares: 4-3-4-4-3-3
2: there is an integer sequence capturing the maximum number of connected coloured squares: 2-3-3-4-2-3
3: there is an integer sequence that captures the number of connected coloured squares: 3-1-2-1-2-1 it is not clear how it progresses from there onwards but it is clear that there is an alternation between some integer k and 1.
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I think the logic is quite vague. It seems each cube piece has its own local rule all from 1-9. The choices being, of course, b+b, w+w, b+w.
- b+w=w, w+w=b b+b is not given
- b+w=w others n/a
- b+w=b, but we require w+w which is not provided for this. Another corner piece (corner 8, picture 2,3) which behaves similar to b+w=b has w+b=b, so from picture 3 and 4 we get that b+b=w for this corner and corner 3 also.
- w+w=b all we need
- here is a issue w+w=w from pic1 and pic2 this behaviour is dissimilar from all corners. But we can assume that w+w=b is the anti of this so the b+w property we need is whatever the anti would have for corner 5 , it would be the anti of that. We use corner 4 w+w=b and b+w=w so it has to be for corner 5 b+w = b.
- w+b=w all we require
- w+b = w. Corner 7 is weird. It shows b+w = b and w in different pictures, however we only require w+b=w.
- similar to corner 4 so b+w = w.
- again same as corner 4 so w+b = w.
Then we can piece together the last cube. Succinctly that would be
- Corner 1: b+w=w, w+w=b
- Corner 2: b+w=w
- Corner 3: b+w=b, b+b=w, w+w=b
- Corner 4: w+w=b, b+w=w
- Corner 5: w+w=w, Anti-rule (assumed of Corner 4): b+w=b
- Corner 6: w+b=w
- Corner 7: w+b=w, b+w=b or w(ambiguous{this was used}, but w+b=w is consistent)
- Corner 8: b+w=w, b+b=w
- Corner 9: w+b=w, b+w=w
From that our given cubes semi-final and final are 1-b+w, 2-b+w, 3-w+w, 4-w+w, 5-b+w, 6-w+b, 7-w+b, 8-b+w, 9-w+b. If we apply the rule we get 1-w, 2-w, 3-b, 4-b, 5-w, 6-b, 7-w, 8-w, 9-w. So, wwb, bwb, www. Okay some nuance. 6-b was changed due to the ambiguity in corner 5 that can be used to generate this answer which is B. In short I treated this as a 9 systems of equation and did it one by one, by imposing 1 picture onto another.
