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If x and y are consecutive negative integers, is x greater

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If x and y are consecutive negative integers, is x greater [#permalink]

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If x and y are consecutive negative integers, is x greater than y?

(1) x + 1 and y – 1 are consecutive negative integers.
(2) x is an even integer.
[Reveal] Spoiler: OA

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Last edited by aiming4mba on 04 Aug 2010, 23:18, edited 1 time in total.
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Re: DS question [#permalink]

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aiming4mba wrote:
If x and y are consecutive negative integers, is x greater than y?
a. x + 1 and y – 1 are consecutive negative integers.
b. x is an even integer.


\(x\) and \(y\) are consecutive integers means that either:
\(y=x+1\) (for example \(x=-5\) and \(y=-4\)), in this case \(x<y\);
OR:
\(x=y+1\) (for example \(x=-5\) and \(y=-6\)), in this case \(x>y\).

So the we should determine which case we have.

(1) \(x+1\) and \(y-1\) are consecutive negative integers --> again either \(x+1=(y-1)+1\) --> \(x+1=y\) (first case so \(x<y\)) or \(y-1=(x+1)+1\) --> \(y=x+3\), which is not possible as \(x\) and \(y\) are consecutive integers (difference between two consecutive integers can not equal to 3). So only first case is possible: \(x<y\). Sufficient.

(2) \(x\) is an even integer. Clearly insufficient.

Answer: A.
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Re: If x and y are consecutive negative integers, is x greater [#permalink]

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New post 19 May 2016, 23:36
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Re: If x and y are consecutive negative integers, is x greater   [#permalink] 19 May 2016, 23:36
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