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GMAT Diagnostic Test Question 43 [#permalink]
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07 Jun 2009, 01:10
GMAT Diagnostic Test Question 43Field: algebra Difficulty 700
If \(x\) and \(y\) are integers, is \(x>y\)? 1. \(x = y+1\) 2. \(x^y = x! + y\) A. Statement (1) ALONE is sufficient, but Statement (2) ALONE is not sufficient B. Statement (2) ALONE is sufficient, but Statement (1) ALONE is not sufficient C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient D. EACH statement ALONE is sufficient E. Statements (1) and (2) TOGETHER are NOT sufficient
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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14 Jun 2009, 09:41
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Explanation
Official Answer: СStatement (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.
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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21 Jul 2009, 21:37
i think the answer should be D. We do not need any of the statements. both x and y are consecutive integers, So , lets x = n and y = n+1;
put that in the given equation :
5n = n+1 + 7
this gives n = 2
therefore, x=2 and y = 3
so, xy = 1..
both statements are redundant here..hence, the answer should be D..



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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23 Jul 2009, 09:13
vishalgupta wrote: i think the answer should be D. We do not need any of the statements. both x and y are consecutive integers, So , lets x = n and y = n+1;
put that in the given equation :
5n = n+1 + 7
this gives n = 2
therefore, x=2 and y = 3
so, xy = 1..
both statements are redundant here..hence, the answer should be D.. The difference of consecutive integers is always 1  agree. Need to change this question.
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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23 Jul 2009, 21:01
I think the question is still not correct, it should be given that x and y are positive integers (not consecutive). otherwise, we can not ignore the value of y = 10.



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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23 Jul 2009, 21:14
Using first option we get x = 2, 3/5 and y = 3 or 10
but using second option alone we get x = 2 and y = 3
So 2 alone is sufficient but not 1 and the answer should be B



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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29 Jul 2009, 23:31
This question had some revisions and now it should be OK. I don't see any problem with the OE. B is the answer and GMAT TIGER explained it all. Is anything still unclear? Please read the OE carefully. sher676 wrote: I also dont get d as the answer. The question does not say that x and y are positive or integers. So A can't be sufficient.
Please explain GMATTIGER
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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30 Jul 2009, 00:15
Final OA/OE after a couple of changes and updates. Let me know if there is still an issue. GMAT TIGER wrote: Explanation
Official Answer: BStatement 1: \(xy = 6\) is sufficient to answer the question because solving the equation gives \(y = 3\) when \(x\) is 2 and \(y\) is 10 when \(x\) is \(\frac{3}{5}\). \(5x = y + 7\) \(x = \frac{y + 7}{5}\) ................i \(xy = 6\) ........................ ii Replacing the value of \(x\) on eq ii: \(\frac{y(y + 7)}{5} = 6\) \(y^2 + 7y = 30\) \(y^2 + 10y  3y  30= 0\) \(y (y + 10)  3 (y + 10) = 0\) \((y + 10) (y  3) = 0\) \(y =  10\) or \(3\) If \(y =  10\), \(x = 3/5\). In this case, \(x  y = 3/5  (10) = 47/5\). Yes. If \(y = 3\), \(x = 2\). In this case, \(x  y = 2  3 = 1\). No. In each case (xy) > 0 and < 0. Hence statement 1 is not sufficient. Statement 2: If \(x\) and \(y\) are consecutive integers, \(x = 2\) and \(y = 3\). When x and y are consecutive integers, either x = y+1 or y = x+1 is possible. (i) If x = y+1 \(5x = y + 7\) \(5(y+1) = y + 7\) \(4y = 2\) \(y = 1/2\). Then \(x = 3/2\). However this is not possible becase x and y are even not integers. So this option is not valid. (ii) If y = x+1 \(5x = y + 7\) \(5x = x + 1 + 7\) \(4x = 8\) \(x = 2\) Then, \(y = 3\). This is valid because x and y are consecutive integers. So (xy) = 23 = 1. Hence statement 2 is sufficient. Therefore B is correct.
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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30 Jul 2009, 00:45



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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10 Aug 2009, 17:20
in your solution for xy=6, yo found y=10 and x =3/5. But this is not possible because X is not an integer. So answer should be D (I guess).



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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11 Aug 2009, 00:05
I think you're mixing the data from S1 with the data given in S2. Neither the question stem nor the S1 states that \(x\) is an integer. So the OE is right. Do you agree? pk wrote: in your solution for xy=6, yo found y=10 and x =3/5. But this is not possible because X is not an integer. So answer should be D (I guess).
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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12 Oct 2009, 08:19
Thanks a lot! +1. Corrected. Talinhuu wrote: in the online test, you put the answer A instead of B, please check it.
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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30 Apr 2010, 13:39
The question and OA/OE have been completely changed and now all the posts have nothing to do with the current question. Maybe it's time to create a new thread for question 43. Also, I think the current OE could be made clearer: GMAT TIGER wrote: Explanation Official Answer: С
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 different values. S2 is not sufficient by itself either.
Statement (1)+Statement (2): \(x\) and \(y\) are not equal and \(x > y\). Sufficient. Therefore its C. It would be better to write: Quote: 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.
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.



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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13 May 2010, 01:14
Do you think it's closer to 700? bibha wrote: Is this a 600 level question?????
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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22 May 2010, 10:47
bibha wrote: Is this a 600 level question????? No, it is of 700 level.
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Re: GMAT Diagnostic Test Question 43 [#permalink]
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14 Dec 2010, 12:06
Can someone tell me if x ! is defined if x is negative? I thought it was. If it is defined, then I don't see how in ii) both x and y need to be positive. I can see why y needs to positive in order to make the left side of the equation an integer, but not x. otherwise, for ii): Let x= 1, y=2 x^y = x ! + y (1)^2 = (1) ! + 2 1 = 1 + 2 1 = 1 and x > y is not true because 1 < 2 If this is accurate, then it changes the parameters of the two statements together. I had chosen E, not C. Any help?



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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14 Dec 2010, 12:18



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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19 Dec 2010, 10:50
Phew. Surely 44 is easier than this and maybe you consider changing the order. I think this is a 700 type problem.



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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06 Jun 2011, 06:41
Based on the example presented and the logic I think it should be 'y' is nonnegative rather than y must be positive. Ie y=0 can leave x^y=1 as a positive integer as shown in the S2 example.? (doesn't change the answer). is this correct?



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Re: GMAT Diagnostic Test Question 43 [#permalink]
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25 Feb 2012, 12:30
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.




Re: GMAT Diagnostic Test Question 43
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