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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.

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.

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

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If 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.

This post might help to get the ranges for (1) and (2) - "How to solve quadratic inequalities - Graphic approach": x2-4x-94661.html#p731476

If 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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Attitude determine everything. all the best and God bless you.

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

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.

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?

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.

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.

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.

When statements are taken together we should consider intersection of the ranges, so the ranges which are true for both statements. _________________

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?

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?

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Consider this: If x is an integer, what is the value of x?

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

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}.