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Hello. I did not see this question under the search, although I'm not certain I searched correctly. Please explain: If \(x\) is an integer, is \(x \gt 1\) ? 1. \((1  2x)(1 + x) \lt 0\) 2. \((1  x)(1 + 2x) \lt 0\) Source: GMAT Club Tests  hardest GMAT questions Answer is C because combined we come conclude that neither 0,1 and 1 can be X and therefore [x]>1 My question: What is the most effiecient way of answering this question? It it best to just plug in numbers and by doing so, conclude that between S1 and S2 X is neither 0, 1, or 1? My route was: 1) Simplifying S1 to.... 2X>1 so X>1/2 OR X<1 2) Simplifying S2 to....X>1 OR 2X<1 so X>1/2 Is that incorrect? And if so, how would you combine S1 and S2 and exclude numbers? Thank you.



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12 Nov 2008, 00:00
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I will try possible options.
From stmt1: either (12x) > 0 and (1+x) < 0 or, (12x) < 0 and (1+x) > 0 Hence, x > 1/2 or x < 1...insufficient.
Similarly, from stmt2: x > 1 or x < 1/2....insufficient.
Combining stmt1 and stmt2: x > 1 or, x < 1....sufficient.



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12 Nov 2008, 11:12
jallenmorris... I was using the wrong keys on my computer, but yes, I meant the absolute value of X. scthakur... thank you for the clarification. You can see that I messed up by decreasing S2 to X>1/2....You only change signs when you divide by a negative but not when you divide into a negative. Thank you. Now, it makes sense.



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25 Jan 2010, 06:48
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lets rephrase the question x>1 is x> 1 or x<1 1.(12x)(1+x) <0 so x<1 or x>1/2 statement 1 statisfies x<1 but it says x>1/2 not x>1 so insufficient 2.(1x)(1+2x)<0 so x>1 or x<1/2 statement 2 statisfies x>1 but it says x<11/2 not x<1 so insufficient
but combining both we will know that x<1 and x>1
so ans is C



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This post might help to get the ranges for (1) and (2)  "How to solve quadratic inequalities  Graphic approach": x24x94661.html#p731476If x is an integer, is x > 1?First of all: is \(x > 1\) means is \(x<1\) (2, 3, 4, ...) or \(x>1\) (2, 3, 4, ...), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1) (1  2x)(1 + x) < 0 > rewrite as \((2x1)(x+1)>0\) (so that the coefficient of x^2 to be positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\) > "\(>\)" sign means that the given inequality holds true for: \(x<1\) and \(x>\frac{1}{2}\). \(x\) could still equal to 1, so not sufficient. (2) (1  x)(1 + 2x) < 0 > rewrite as \((x1)(2x+1)>0\): roots are \(x=\frac{1}{2}\) and \(x=1\) > "\(>\)" sign means that the given inequality holds true for: \(x<\frac{1}{2}\) and \(x>1\). \(x\) could still equal to 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x<1\) and \(x>1\). Sufficient. Answer: C.
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28 Jan 2011, 10:46
Bunuel wrote: This post might help to get the ranges for (1) and (2)  "How to solve quadratic inequalities  Graphic approach": x24x94661.html#p731476If x is an integer, is x > 1?First of all: is \(x > 1\) means is \(x<1\) (2, 3, 4, ...) or \(x>1\) (2, 3, 4, ...), so for YES answer \(x\) can be any integer but 1, 0, and 1. (1) (1  2x)(1 + x) < 0 > rewrite as \((2x1)(x+1)>0\) (so that the coefficient of x^2 to be positive after expanding): roots are \(x=1\) and \(x=\frac{1}{2}\) > "\(>\)" sign means that the given inequality holds true for: \(x<1\) and \(x>\frac{1}{2}\). \(x\) could still equal to 1, so not sufficient. (2) (1  x)(1 + 2x) < 0 > rewrite as \((x1)(2x+1)>0\): roots are \(x=\frac{1}{2}\) and \(x=1\) > "\(>\)" sign means that the given inequality holds true for: \(x<\frac{1}{2}\) and \(x>1\). \(x\) could still equal to 1, so not sufficient. (1)+(2) Intersection of the ranges from (1) and (2) is \(x<1\) and \(x>1\). Sufficient. Answer: C. nice explanation bunuel. i followed the same procedure.
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29 Jan 2011, 08:17
I suppose it could be (D). We are told that \(x\) is an integer. So, from (1) we have \(x<1\) and \(x>1/2\), but if x is an int, so second part turns to be \(x>1\). The same logic for (2) \(x<1/2\) and \(x>1\) turns into \(x<1\) and \(x>1\)



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29 Jan 2011, 08:31
ulm wrote: I suppose it could be (D). We are told that \(x\) is an integer. So, from (1) we have \(x<1\) and \(x>1/2\), but if x is an int, so second part turns to be \(x>1\). The same logic for (2) \(x<1/2\) and \(x>1\) turns into \(x<1\) and \(x>1\) OA for this question is C, not D. x>1/2 and x=integer means that x can be any integer more than 1/2: 1, 2, 3, ... so x could still equal to 1 for statement (1), so this statement is not sufficient. Similarly for statement (2) x could still equal to 1, so it's also not sufficient. Hope it's clear.
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23 May 2011, 15:08
sonnco wrote: silasaaa2 wrote: lets rephrase the question x>1 is x> 1 or x<1 1.(12x)(1+x) <0 so x<1 or x>1/2 statement 1 statisfies x<1 but it says x>1/2 not x>1 so insufficient 2.(1x)(1+2x)<0 so x>1 or x<1/2 statement 2 statisfies x>1 but it says x<11/2 not x<1 so insufficient
but combining both we will know that x<1 and x>1
so ans is C That was a very good setup. Thank you! Thanks a lot. However, for statement one you have to consider that either (12x) is < 0 and (1+x) > 0 OR (12x) is > 0 and (1 + x) < 0. Considering all four possible values will make you arrive at X < 1 or X > 1/2. You have to do the same for statement 2. However, the problem with this approach is that its too time consuming. Will take around 3 minutes I believe. Is there any short cut?



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01 Feb 2012, 06:44
Why isn't E instead of C ? when you combine the two statements, don't you still have the possibility that x could be = +/ 1 ?



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01 Feb 2012, 07:34
thanks, could you please simplify more?



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01 Feb 2012, 11:41
imadkho wrote: thanks, could you please simplify more? Please refer to my solution above. The question asks: if x is an integer, is x > 1? So, basically the question asks whether: x is an integer more than 1: 2, 3, 4, 5, ... or an integer less than 1: 2, 3, 4, 5, ... For (1)+(2) we get that \(x>1\) or \(x<1\), which is exactly what we wanted to know. Hope it's clear.
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01 Feb 2012, 15:20
In statement 1) x <1 or x>1/2 ; In statement 2) x <1/2 or x>1 ; so my question is simply the following: why would we, upon combining the two statements, consider only x<1 and x>1 while not considering x>1/2 and x<1/2, and thus choose E instead of C (the same logic used with each of the statements alone)? thanks for your help.



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01 Feb 2012, 18:55
I was wondering how you go about combing the statements? Do you just know that we must satisfy the greater of statements? E.g. X>1 must be used because everything can be satisfied in this where x>1/2 cannot. And same as other. Is this right? Posted from my mobile device
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01 Feb 2012, 20:07
C or 3. So because x>1 it cannot be x>1/2 and x<1 it cannot be x<1/2.
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27 Feb 2012, 20:07
Hey Bunuel or anyone,
In statement (1) I found the domain of x by finding the roots and plugging different values within the sections to find if it satified the inequility. But how do we find the domain of x through this other approach???
either {(12x) > 0 and (1+x) < 0 } OR { (12x) < 0 and (1+x) > 0 }
Hence, x > 1/2 or x < 1 ???
What I arrived at using this approach was:
{ 1/2 > x and x < 1 } OR { 1/2 < x and x>1} >>> { 1/2 > x and x < 1 } OR { 1/2 < x }
but how do we combine this OR statement to arrive >>>> x > 1/2 or x < 1 ????
We have two contradictory expressions 1/2 > x or 1/2 < x
Please Help!!!



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28 Feb 2012, 00:39
alphabeta1234 wrote: Hey Bunuel or anyone,
In statement (1) I found the domain of x by finding the roots and plugging different values within the sections to find if it satified the inequility. But how do we find the domain of x through this other approach???
either {(12x) > 0 and (1+x) < 0 } OR { (12x) < 0 and (1+x) > 0 }
Hence, x > 1/2 or x < 1 ???
What I arrived at using this approach was:
{ 1/2 > x and x < 1 } OR { 1/2 < x and x>1} >>> { 1/2 > x and x < 1 } OR { 1/2 < x }
but how do we combine this OR statement to arrive >>>> x > 1/2 or x < 1 ????
We have two contradictory expressions 1/2 > x or 1/2 < x
Please Help!!! \((12x)(1+x)<0\) > two multiples have the opposite signs: \((12x)>0\) and \((1 + x)<0\) > \(x<\frac{1}{2}\) and \(x<1\) > \(x<1\) (you take the intersection of the ranges since both must be true simultaneously); \((12x)<0\) and \((1 + x)>0\) > \(x>\frac{1}{2}\) and \(x>1\) > \(x>\frac{1}{2}\) (you take the intersection of the ranges since both must be true simultaneously); So the the ranges for which \((12x)(1+x)<0\) holds true are: \(x<1\) and \(x>\frac{1}{2}\). Hope it's clear.
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