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If x*y+ 9 > 0, and x and y are integers, is x < 6?
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Updated on: 27 Feb 2012, 02:01
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If x*y+ 9 > 0, and x and y are integers, is x < 6? (1) y is negative (2) y <= 1 I solved the question as below: If x.y+ 9 > 0 x.y > 9 As x and y are integers then x.y > 9 * 1 or x.y > 3 * 3 Case 1 y is negative then x can be 9, 1 and 3. Insufficient Case 2 y < 1 implies 1 < y < 1 Clearly not sufficient (1) & (2) not sufficient together either. Is my approach correct?
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Originally posted by rohitgoel15 on 01 Feb 2012, 04:11.
Last edited by Bunuel on 27 Feb 2012, 02:01, edited 1 time in total.
Edited the question




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If x*y+ 9 > 0, and x and y are integers, is x < 6?
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01 Feb 2012, 04:24
rohitgoel15 wrote: If x.y+ 9 > 0, and x and y are integers, is x < 6? (1) y is negative (2) y < 1
I solved the question as below: If x.y+ 9 > 0 x.y > 9 As x and y are integers then x.y > 9 * 1 or x.y > 3 * 3
Case 1 y is negative then x can be 9, 1 and 3. Insufficient
Case 2 y < 1 implies 1 < y < 1 Clearly not sufficient
(1) & (2) not sufficient together either. Is my approach correct? Statement (2) should read \(y\leq{1}\) and not \(y<{1}\). Because: (1) \(y<0\), (2) \(y<{1}\) (1<y<1) and \(y=integer\) (from the stem) cannot be true simultaneously (there is no integer between 1 and 0, not inclusive). If x*y+ 9 > 0, and x and y are integers, is x < 6?(1) \(y<0\). If \(y=1\), then \(x+9>0\) > \(x<9\), so \(x\) can be 8, 8, 7, 7, and so on. So \(x\) may or may not be less than 6. Not sufficient. (2) \(y\leq{1}\) > \(1\leq{y}\leq{1}\) > as \(y\) is an integer, then \(y\) is 1, 0, or 1. As we've seen above if \(y=1\), then \(x\) may or may not be less than 6. Not sufficient. (1)+(2) Again if \(y=1\) (which satisfies both stem and statements) then \(x\) may or may not be less than 6. Not sufficient. Answer: E. P.S. rohitgoel15 please tag the questions properly: it's not "Statistics and Sets Problems", it's "Absolute Values/Modules" and "Inequalities" question. Tags changed.
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Re: If x*y+ 9 > 0, and x and y are integers, is x < 6?
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25 Aug 2013, 14:00
I am wondering if there is way to algebraicaly do this problem? Here is one appraoch:
Does this algebraic approach make sense on this problem?
Given x (y) > 9 Or (x)*(y) > 9 ==> xy < 9 Also, xy > 9
Thus we have 9 < xy < 9
Now, with statement 1, y < 0 that means:  9 < x ( some int) < 9 or each separately,  9 <  x (some int)*9 > x AND  x < 9 or x > 9 (some int)
Thus, x can be < 6 or > 6. NS.
Similarly, statement 2, n.s.
E.



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Re: If x*y+ 9 > 0, and x and y are integers, is x < 6?
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29 Jul 2014, 19:24
minkathebest wrote: I am wondering if there is way to algebraicaly do this problem? Here is one appraoch:
Does this algebraic approach make sense on this problem?
Given x (y) > 9 Or (x)*(y) > 9 ==> xy < 9 Also, xy > 9
Thus we have 9 < xy < 9
Now, with statement 1, y < 0 that means:  9 < x ( some int) < 9 or each separately,  9 <  x (some int)*9 > x AND  x < 9 or x > 9 (some int)
Thus, x can be < 6 or > 6. NS.
Similarly, statement 2, n.s.
E. +1 kudos for your solution
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Re: If x*y+ 9 > 0, and x and y are integers, is x < 6?
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19 May 2016, 19:06
x*y>9 x is always positive y can be negative or positive.
1. y is negative y=1 x=7 x*y>9 true x=5 x*y>9 true
2 outcomes. not sufficient.
2. y<=1 since y is an integer, y can be 1, 0, or 1.
we can have similar cases as in A. since we can't give a definite answer, B is out.
1+2 y=1 in 1 we already demonstrated that it's not sufficient.
E



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If x*y+ 9 > 0, and x and y are integers, is x < 6?
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06 Jun 2017, 22:26
Hi, I believe statement 1 should be enough. Can someone guide me as to where I am wrong:
x * y > 9
statement 1 = y is negative
so x< 9 / y
Now since LHS has to be positive, and has to be an integer the only values that y satisfies are 9 and 3. The equation becomes x < 1 or x < 3.
In both the cases x<6, hence statement 1 is sufficient.
The reason why I did not consider y = 1 is because we are explicitly given in the question stem that X * y + 9 > 0 (or X * y > 9) and hence considered the values y = 3, 9, to satisfy the given equation



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Re: If x*y+ 9 > 0, and x and y are integers, is x < 6?
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27 Oct 2018, 11:06
mvictor wrote: x*y>9 x is always positive y can be negative or positive.
1. y is negative y=1 x=7 x*y>9 true x=5 x*y>9 true
2 outcomes. not sufficient.
2. y<=1 since y is an integer, y can be 1, 0, or 1.
we can have similar cases as in A. since we can't give a definite answer, B is out.
1+2 y=1 in 1 we already demonstrated that it's not sufficient.
E Very simply explained. Great ! Sent from my SMG935F using GMAT Club Forum mobile app




Re: If x*y+ 9 > 0, and x and y are integers, is x < 6?
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27 Oct 2018, 11:06






