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Will really appreciate if answer is supported by explanation.

\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\) Two cases: A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values: 1 for \(x>0\) --> so we would have: \(1<x\); Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again: \(x<0\)--> \(-x>x\)--> \(x<0\). \(x>0\) --> \(x>x\): never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Re: If x is not equal to 0, is |x| less than 1? [#permalink]

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28 Aug 2010, 03:36

1

This post was BOOKMARKED

Bunuel wrote:

Hussain15 wrote:

If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\) Two cases: A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

OR: \(\frac{x}{|x|}< x\) multiply both sides of inequality by \(|x|\) (side note: we can safely do this as absolute value is non-negative and in this case we know it's not zero too) --> \(x<x|x|\) --> \(x(|x|-1)>0\): Either \(x>0\) and \(|x|-1>0\), so \(x>1\) or \(x<-1\) --> \(x>1\); Or \(x<0\) and \(|x|-1<0\), so \(-1<x<1\) --> \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again: \(x<0\)--> \(-x>x\)--> \(x<0\). \(x>0\) --> \(x>x\): never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.

Hi Bunuel, Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

Hi Bunuel, Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Re: If x is not equal to 0, is |x| less than 1? [#permalink]

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13 Sep 2010, 20:33

1

This post received KUDOS

Nice explaination. I had used numbers and came to wrong conclusion. I am inclined to say that for all mod questions I should consider positive and negative like Bunuel suggested _________________

If you like my post, consider giving me some KUDOS !!!!! Like you I need them

It seemed to me that (1) was enough. Since x is not 0, |x| must be greater than 0, and it should be safe to reorder the inequality like this by multiplying both sides by |x|:

x < |x| * x

And my thinking went: if |x| * x is greater than x, then x must be positive (if it weren't, then |x| * x would be a more negative number than x). And also x must be >= 1, for if it were between 0 and 1 then |x| * x would be a smaller number than x.

So, (1) should be sufficient -- (A). But according to the answer explanation, the answer is (C). Have I made a mistake in my logic here?

Hi, and welcome to Gmat Club.

You are right inequality \(x<|x|*x\) holds true when \(x>1\), but it's not the only range, it also holds true when \(-1<x<0\). So from (1) we can not be sure whether \(|x|<1\).

You should have spotted that your range for (1) is not correct when dealing with statement (2): \(|x| > x\), which basically implies that \(x\) is negative, \(x<0\). So you would have from (1) \(x>1\) and from (2) \(x<0\): statements clearly contradict each other, but on the GMAT, two data sufficiency statements always provide TRUE information and these statements never contradict each other.

You are right inequality \(x<|x|*x\) holds true when \(x>1\), but it's not the only range, it also holds true when \(-1<x<0\). So from (1) we can not be sure whether \(|x|<1\).

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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07 Jan 2012, 21:50

Hi Bunuel - Firstly thanks for the wonderful collection and the explanation you provided in the word format.

1. In this kind of inequalities involving ‘mod’ values is it ever advisable to first square both side and then proceed ? I saw somewhere this approach was quite fast to solve the problem. 2. In this specific example as it involves negative & decimal , squaring and multiplying makes things erroneous .

Please suggest if there has any general rule/comment on this?

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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09 Jan 2012, 12:21

Rephrase of stem: Is x a positive or negative fraction.

Following AD/BCE strategy, 2 looks easy so lets start with that.

2. This means that x is negative. But, x could be integer. insuff 1. This one is a bit complicated. If x = positive integer, this ineq holds and it does not for a positive fraction. So, for positive numbers, x=integer. For x = negative integer, this ineq does not hold. It does hold for negative fractions. Insuff.

Combining 1 and 2, x = negative and x = integer. Satisfies the stem. C. _________________

I am the master of my fate. I am the captain of my soul. Please consider giving +1 Kudos if deserved!

DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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09 Jan 2012, 18:31

+1 C.

question stem rephrase: is x a fraction?

(1) try some numbers x = 5 , x = -5, x = 1/5, x = -1/5 x = 5 5/5 => 1 < 5 . works. answer to question stem NO. x = -5 -5/5 => -1 < -5. does not work x = 1/5 => 1 < 1/5 does not work x = -1/5 => -1 < -1/5 . works. answer to question stem YES. We get YES and NO so INSUFF.

(2) This means x is negative so it may be integer or fraction. Again YES and NO. INSUFF. (3) x is negative and the only negative value that works is negative fraction. SUFF. _________________

We will either find a way, or make one.

Please consider giving kudos if you find my post hepful

Re: If x is not equal to 0, is |x| less than 1? [#permalink]

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19 Jan 2012, 09:29

Bunuel wrote:

udaymathapati wrote:

Hi Bunuel, Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.

Hi Bunuel...I still didnot got it..red area says -1<x<0 but we need that -1<x<1 Please explain

Hi Bunuel - Firstly thanks for the wonderful collection and the explanation you provided in the word format.

1. In this kind of inequalities involving ‘mod’ values is it ever advisable to first square both side and then proceed ? I saw somewhere this approach was quite fast to solve the problem. 2. In this specific example as it involves negative & decimal , squaring and multiplying makes things erroneous .

Please suggest if there has any general rule/comment on this?

Thanks, VCG

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality). For example: \(2<4\) --> we can square both sides and write: \(2^2<4^2\); \(0\leq{x}<{y}\) --> we can square both sides and write: \(x^2<y^2\);

But if either of side is negative then raising to even power doesn't always work. For example: \(1>-2\) if we square we'll get \(1>4\) which is not right. So if given that \(x>y\) then we can not square both sides and write \(x^2>y^2\) if we are not certain that both \(x\) and \(y\) are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality). For example: \(-2<-1\) --> we can raise both sides to third power and write: \(-2^3=-8<-1=-1^3\) or \(-5<1\) --> \(-5^2=-125<1=1^3\); \(x<y\) --> we can raise both sides to third power and write: \(x^3<y^3\).

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hi Bunuel, Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.

Hi Bunuel...I still didnot got it..red area says -1<x<0 but we need that -1<x<1 Please explain

The question asks "is \(-1<x<1\)?" We got that \(x\) is in the range \(-1<x<0\) (red area). Now, as ANY \(x\) from this range (from red area) is indeed in \(-1<x<1\), then we can answer YES to our original question.

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