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

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24 Apr 2013, 17:25

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

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25 Apr 2013, 08:17

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

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11 Jul 2013, 13:43

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

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06 Aug 2013, 07:42

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?

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

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29 Sep 2014, 06:08

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 .

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

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29 Sep 2014, 08:52

Bunuel wrote:

earnit wrote:

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 .

When we combine the statements we get that -1 < x < 0, then HOW can x be 0.5???

I'm sorry if it comes across like that but as per my understanding, the Question is if x lies between -1 and 1 and we find out that x lies between -1 and 0 so we are not entirely going by the asked range -1 < x < 1 is what i think.

But yes, the range of -1 < x < 0 comes under the broad -1 < x < 1 and so that is why it is correct.

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

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21 Oct 2014, 14:46

Thoughtosphere 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.

1 - x / |x| < x This would mean that -1 < x < 0 & 0 < x <infinite Thus 1 is insufficient

2 - |x| > x - This would mean that negative infinite values < x < 0 This is also not sufficient.

Combining 1 & 2 - the overlapping region is -1 < x < 0

We can conclude that |x| < 1

So C

Solving the Question is not much of a problem as deciding whether our solved range: \(-1 < x < 0\) is SUFFICIENT to answer the asked range: \(-1 < x < 1\)

I need to understand that: Is our answer sufficient to conclude that YES x lies between (-1,1) even though our solution was that x lies between (-1,0) ?

Will really appreciate if answer is supported by explanation.

1 - x / |x| < x This would mean that -1 < x < 0 & 0 < x <infinite Thus 1 is insufficient

2 - |x| > x - This would mean that negative infinite values < x < 0 This is also not sufficient.

Combining 1 & 2 - the overlapping region is -1 < x < 0

We can conclude that |x| < 1

So C

Solving the Question is not much of a problem as deciding whether our solved range: \(-1 < x < 0\) is SUFFICIENT to answer the asked range: \(-1 < x < 1\)

I need to understand that: Is our answer sufficient to conclude that YES x lies between (-1,1) even though our solution was that x lies between (-1,0) ?

The question asks whether -1 < x< 1. We got that -1 < x < 0. So, the answer is YES: x is definitely from the range (-1, 1).
_________________

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

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24 Nov 2014, 10:26

The question is asking if -1 < x < 1

Statement 1) x/|x| < x Now if x + ve, x/|x| = 1 and if x is -ve x / |x| = -1 So essentially, the equation becomes +1 < x or -1 < -x or 1 > x (multiplying both sides by -ve) In two cases, one case says x > 1 and the other one says x < 1, hence Not sufficient.

Statement 2. |x| > x , this will only happen when x is -ve id x is -ve |x| can be > 1 or |x| can be < 1 or equal to 1. - Not sufficient.

Combining the two statements: x is -ve 1 > x hence sufficient. C)
_________________

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

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26 Nov 2014, 13:05

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.

Bunuel I'm very unclear about this. if x<0, then -x/|x| <-x (since x is negative in our scenario 1)...so where do I go from here? if i cut both -x's, I get 1/|x| > 1....so x>1....I can't understand how we got x>-1... If you could help, that'll be great. I tried plugging in for this question and got in right, but I've been trying to teach myself your conceptual way as I believe it is much better than plugging in.