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Love your answers. But I want to suggest another method that is giving me some trouble when combining statesment (1) & (2)

1) \(\frac{x}{|x|}< x\) 2) \(x<|x|\)

Since \(0<|x|\) (x cannot equal 0), then we can rewrite statement 2, \(x<|x|\) as \(\frac{x}{|x|}< 1\).

We then subtract statment (1) and (2) as

1) \(\frac{x}{|x|}-\frac{x}{|x|}< x-1\) to get \(0< x-1\) or \(1<x\) showing that x is outside the boundary of \(-1<x<1\) and making the combined statements suffient. But \(1<x\), the derived statement of (1) and (2), contradicts statement (2), which claims that \(0<x\). What am I doing wrong?

Thank you!

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So, you cannot subtract \(\frac{x}{|x|}< 1\) from \(\frac{x}{|x|}< x\) since their signs are NOT in the opposite direction.

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

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14 Jun 2012, 05:21

1

This post received KUDOS

Hello Bunuel, I tried to solve like this: Basically the question asks whether -1<x<1? Condition 1: x/|x|<x case 1: if x>0; x/x<x ==> 1<x ==> x>1 case 2: if x<0; x/-x<x; since x<0, -x >0 therefore, x<-x2; x+x2 <0 ==> x(1+x)<0, therefore, -1<x<0 {Please confirm if I have derived this right} Given that we do not know the sign we cannot determine, hence condition 1 is insufficient

Condition 2: |x|>x ==> which means x cannot be > then 0. thus, x<0, independently condition 2 is not sufficient;

Combining C1 and C2: we arrive at case 2; but that proves that x does not lie between -1 and 1 but between -1 and 0.

Hence C. Please let me know in case I have done anything conceptually wrong!

Hello Bunuel, I tried to solve like this: Basically the question asks whether -1<x<1? Condition 1: x/|x|<x case 1: if x>0; x/x<x ==> 1<x ==> x>1 case 2: if x<0; x/-x<x; since x<0, -x >0 therefore, x<-x2; x+x2 <0 ==> x(1+x)<0, therefore, -1<x<0 {Please confirm if I have derived this right} Given that we do not know the sign we cannot determine, hence condition 1 is insufficient

Condition 2: |x|>x ==> which means x cannot be > then 0. thus, x<0, independently condition 2 is not sufficient;

Combining C1 and C2: we arrive at case 2; but that proves that x does not lie between -1 and 1 but between -1 and 0.

Hence C. Please let me know in case I have done anything conceptually wrong!

You've done everything right.

Though for case 2 you could do as follows: \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-\frac{x}{x}<x\) --> x is simply reduced: \(-1<x\). Since we consider the range \(x<0\) then \(-1<x<0\).

Also for (1)+(2) we have that \(-1<x<0\). So we can answer yes to the question whether \(-1<x<1\).

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

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24 Apr 2013, 18:25

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 \(3\))

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.

Answer: C.

Buenel terrific explanation. This system, that you use for solving inequations, is quite fundamental and hard to be wrong. But is it really always the quickest way?? _________________

When you feel like giving up, remember why you held on for so long in the first place.

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

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25 Apr 2013, 09:17

Bunuel wrote:

When \(x<0\), then \(|x|=-x\), thus \(\frac{x}{|x|}<x\) becomes \(\frac{x}{-x}<x\) --> \(-1<x\) but since \(x<0\), then \(-1<x<0\).

Hope it's clear.

Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?

When \(x<0\), then \(|x|=-x\), thus \(\frac{x}{|x|}<x\) becomes \(\frac{x}{-x}<x\) --> \(-1<x\) but since \(x<0\), then \(-1<x<0\).

Hope it's clear.

Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?

First of all \(|-x|=|x|\). Next, if \(x<0\), then \(|x|=|negative|=-x=-negative=positive\).

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

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23 Jun 2013, 11:37

For number 1 why do we have two values for x<0? In other words why do we have x< 0 and -1<x<0?

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 \(3\))

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? (1) x/|x|< x [#permalink]

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11 Jul 2013, 14:43

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

|x|<1 Is -1<x<1

(1) x/|x|< x

x/|x|< x Two cases: x>0, x<0

x>0 x/|x|< x x/x<x 1<x

If x>0 and x>1 then x>1 If x>1 then |x| is NOT less than 1

x<0 x/|x|< x -x/|-x|<x -1<x

-1<x<0 if -1<x<0 then |x| IS less than 1 INSUFFICIENT

(2) |x| > x

If the absolute value of x is greater than x than x must be negative. However, |-x| could be less than 1 or greater than 1 depending on the value of x. INSUFFICIENT

1+2 x must be negative and x from #1 x is either >1 or between -1 and 0. Therefore, we know -1<x<0 which means that |x| is always less than 1. SUFFICIENT

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

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06 Aug 2013, 08:42

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 \(3\))

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.

Answer: C.

Hi bunel,

hw is that we can have two ranges x<0 & x>0. How we got this ranges?

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

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29 Sep 2014, 07:08

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.

Hope it's clear.[/quote]

Exactly, based on your explanation above, Though the answer is YES it is sufficient to answer but the answer to the question is

That for x=0.5 it is in the range -1<x<1 but it is NOT part of -1<x<0 which we found out. So it is not possible to say that all possible values of -1<x<0 come under -1<x<1 unless ofcourse it is specified that x is an integer.

Exactly, based on your explanation above, Though the answer is YES it is sufficient to answer but the answer to the question is the range -1<x<1 for x is NO as -1<x<0 is the range and so x could be 0.5 and not be part of the -1<x<0 .

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