This is from https://www.publicserviceprep.com/SampleTests/canarmy/TestPage_1.aspx?TESTID=496 question 6. I know the answer (C) but don't know why or the reasoning behind it. Can someone please explain to me why it is the answer?
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2 Answers
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This is my best guess at the rules which govern this pattern.
- The number of black balls must always increase by one.
- If the number of black balls is even, subtract six white balls.
- If the number of black balls is odd, add three white balls.
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6
The answer is
It alternates between removing the first column and adding a column but removing one ball.
Using that principle, you will see that the colors are not important and are just a distraction. (Which I have seen often in such puzzles)
C is the only one that satisfies that condition.
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$\begingroup$ This can't be right, even ignoring the colours. If you were correct, then C would have 10 balls, not 9; two columns of five. $\endgroup$ Jul 25 at 11:19
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$\begingroup$ I think you're close though. If we do ignore colour, then the pattern could be expressed more accurately as "Alternate between removing 5 balls and adding 4 balls", which is nice and simple. $\endgroup$ Jul 25 at 11:24
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1$\begingroup$ Yup, I know. Check the pattern. If you're right, then the pattern goes: A -> [Remove Last Column] -> B -> [Add Column, Remove Ball] -> C -> [Remove Last Column] -> ???. To remove the last column from pattern C would leave 10 balls, not 9 as required by answer C. Changing to [Remove 5] and [Add 4] resolves this problem. $\endgroup$ Jul 25 at 13:29
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1$\begingroup$ Then, provided that the colours are indeed a red herring, this would work. $\endgroup$ Jul 25 at 13:47