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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 2

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This is my best guess at the rules which govern this pattern.

  1. The number of black balls must always increase by one.
  2. If the number of black balls is even, subtract six white balls.
  3. If the number of black balls is odd, add three white balls.
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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
  • $\begingroup$ 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
  • $\begingroup$ "adding a column but removing one ball" $\endgroup$
    – WOWOW
    Jul 25 at 13:16
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    $\begingroup$ 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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    $\begingroup$ Then, provided that the colours are indeed a red herring, this would work. $\endgroup$ Jul 25 at 13:47

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