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# GMAT Diagnostic Test Question 43

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Re: GMAT Diagnostic Test Question 43 [#permalink]

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25 Feb 2012, 12:56
T740qc wrote:
Quote:
Does anyone know how to easily derive from statement (2) the following possibilities: $$x = 1$$ and $$y = 0$$, and $$x = 2$$ and $$y = 2$$ ? I don't think it's that obvious. So it's easy to miss one possibility and answer the question differently.

i didnt see an answer to this q. is there an easier way to get to 0,1 or 2? thanks.

This question is also discussed here: if-x-and-y-are-integers-is-x-y-1-x-y-93713.html#p761563

But I wouldn't worry about it since it needs a revision and will be updated for the next edition.
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Re: GMAT Diagnostic Test Question 43 [#permalink]

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17 Oct 2012, 09:30
Answer is E,for the first one is not suffice to solve the question cos give if /x/=/y+1/ implies /y+1/ greater than /y/ which is wrong if y is -2.the second one is not suffice to do it so cos if the second statement is right which implies absolute y = x^y-x! which is that /x/ must be greater than x^y-x! from here we know x must be greater or equal to 0,so if x=2,y=5 then the statement is wrong.so if you put them together would still yield the same result,the solution is indeterminant!
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Re: GMAT Diagnostic Test Question 43 [#permalink]

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17 Oct 2012, 09:33
Answer is E,for the first one is not suffice to solve the question cos give if /x/=/y+1/ implies /y+1/ greater than /y/ which is wrong if y is -2.the second one is not suffice to do it so cos if the second statement is right which implies absolute y = x^y-x! which is that /x/ must be greater than x^y-x! from here we know x must be greater or equal to 0,so if x=2,y=5 then the statement is wrong.so if you put them together would still yield the same result,the solution is indeterminant!
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Re: GMAT Diagnostic Test Question 43 [#permalink]

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31 Dec 2012, 12:30
GMAT TIGER wrote:
Explanation
 Rating:

Statement (1):
Given that $$|x| = |y+1|$$
If $$x$$ and $$(y+1)$$ both have same signs (either positive or negative), then $$x = y+1$$ and $$x>y$$, which makes $$x$$ and $$y$$ consecutive integers.
If $$x$$ and $$(y+1)$$ both do not have the same signs (one is negative and the other is positive), then $$x = -(y+1)$$ or $$(x + y + 1) = 0$$. In that case, either one could be greater in absolute value. S1 is not sufficient by itself.

Statement (2):
If $$x^y = x! + |y|$$, $$x$$ and $$y$$ both have to be positive because $$x! + |y|$$ is always a positive integer. If $$y$$ is negative, $$x^y$$ can never be an integer. So for $$x^y$$ to be an integer, $$y$$ has to be positive. Let's see what values satisfy the equation:
If $$x = 1$$, and $$y = 0$$, $$x^y = x! + |y| = 1$$.
If $$x = 2$$, and $$y = 2$$, $$x^y = x! + |y| = 4$$.
So $$x$$ and $$y$$ have at least two possibilities. S2 is not sufficient by itself either.

Statement (1)+Statement (2):
Because of Statement (2), $$x$$ and $$y$$ have to be positive, in which case, by Statement (1) $$x > y$$. Sufficient. Therefore, the answer is C.

I don't get it..Statement 2 gives you 2 solutions 1,0 and 2,2 so x and y are non-negative .. even combining with statement 2 you get 2 solutions.. so answer is E..
am I missing something ?
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Re: GMAT Diagnostic Test Question 43 [#permalink]

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07 Mar 2013, 18:07
There's no way this is a 700-level only question, is there? I think I'm a high scorer and the analysis of the correct answer by GMAT Tiger is very sophisticated.
Re: GMAT Diagnostic Test Question 43   [#permalink] 07 Mar 2013, 18:07

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# GMAT Diagnostic Test Question 43

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