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

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Joined: 20 Nov 2009
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If x and y are consecutive negative integers, is x greater [#permalink]  04 Aug 2010, 21:58
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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, 22:18, edited 1 time in total.
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Kudos [?]: 42645 [1] , given: 6052

Re: DS question [#permalink]  04 Aug 2010, 22:51
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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.

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Re: DS question   [#permalink] 04 Aug 2010, 22:51
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