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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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lets rephrase the question |x|>1 is x> 1 or x<-1
1.(1-2x)(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.(1-x)(1+2x)<0
so x>1 or x<-1/2
statement 2 statisfies x>1 but it says x<1-1/2 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": x2-4x-94661.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 (2x-1)(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 (x-1)(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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I will try possible options.
From stmt1:
either (1-2x) > 0 and (1+x) < 0
or, (1-2x) < 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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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 {(1-2x) > 0 and (1+x) < 0 } OR { (1-2x) < 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!!! (1-2x)(1+x)<0 --> two multiples have the opposite signs: (1-2x)>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); (1-2x)<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 (1-2x)(1+x)<0 holds true are: x<-1 and x>\frac{1}{2}. Hope it's clear.
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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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Bunuel wrote: This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.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 (2x-1)(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 (x-1)(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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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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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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sonnco wrote: silasaaa2 wrote: lets rephrase the question |x|>1 is x> 1 or x<-1
1.(1-2x)(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.(1-x)(1+2x)<0
so x>1 or x<-1/2
statement 2 statisfies x>1 but it says x<1-1/2 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 (1-2x) is < 0 and (1+x) > 0 OR (1-2x) 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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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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thanks, could you please simplify more?
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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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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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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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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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