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# If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x

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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x  [#permalink]

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27 Dec 2017, 02:15
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Similar topic: https://gmatclub.com/forum/if-x-is-diff ... 26715.html
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x  [#permalink]

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29 Mar 2018, 05:33
I did it this way.

Is |x| < 1?

Statement 1

x/|x| < x

For, x>1
Divide by x on both sides, without changing the sign.
1/|x| < 1
|x| > 1

For x<1
Divide by x on both sides, and change the sign.
1/|x| > 1
|x| < 1

Insufficient, as we have both |x| > 1 (x is positive) and |x| < 1 (x is negative)

Statement 2

|x| > x
This is possible only when x < 0. (x is negative)

This is still insufficient to prove the given question.

Taking them both,

From (2), we get (x is negative) and from (1), the corresponding answer is |x| < 1.

Hence, C.

Hussain15 wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

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Joined: 02 Sep 2009
Posts: 62380
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x  [#permalink]

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02 May 2018, 08:24
cruiseav wrote:
Bunuel wrote:
If x is not equal to 0, is |x| less than 1?

Is $$|x|<1$$?
Is $$-1<x<1$$? ($$x\neq{0}$$)
So, the question asks whether x is in the range shown below:

(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 consider 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 $$-1<x<1$$. So, we have a definite YES answer to the question. Sufficient.

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Hi Bunuel

Surely i am missing something here..my basic doubt is in scenario when we take X<0

If we assume, x<0, then shouldn't the statement x/|x|<x = -x/-x <-x (my thinking here is if we considering x<0, then x should be negative throughout )
So, the equation becomes 1<-x = -1>x, we can keep this option as we have initially assumed X<0
Then if we take any values of x less then -1, |x| will always be greater than 1

if we assume X>0, then x/|x|<x = x/x<x.....which is 1<x, we can keep this option as initially we have assumed X>0
Then we take any values of x greater than 1 then |x| will always be greater than 1.

So statement 1 in either case is sufficient..

Negative x does not mean that you should replace x with -x. x just represents a negative number. You should replace |x| with -x, though because if x < 0, then |x| = -x.
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x  [#permalink]

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03 May 2018, 07:37
Bunuel wrote:
cruiseav wrote:
Bunuel wrote:
If x is not equal to 0, is |x| less than 1?

Is $$|x|<1$$?
Is $$-1<x<1$$? ($$x\neq{0}$$)
So, the question asks whether x is in the range shown below:

(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 consider 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 $$-1<x<1$$. So, we have a definite YES answer to the question. Sufficient.

Attachment:
MSP340910303aagbh2c14gf0000420i437if9d12bhg.gif

Attachment:
MSP2380151735322d2i2d790000280i6giebh17fdf3.gif

Attachment:
MSP8506173e0f30207c4a2700005eg6g59c48d51i5b.gif

Attachment:
MSP111401047g43cgaehag920000640367fgf148e668.gif

Hi Bunuel

Surely i am missing something here..my basic doubt is in scenario when we take X<0

If we assume, x<0, then shouldn't the statement x/|x|<x = -x/-x <-x (my thinking here is if we considering x<0, then x should be negative throughout )
So, the equation becomes 1<-x = -1>x, we can keep this option as we have initially assumed X<0
Then if we take any values of x less then -1, |x| will always be greater than 1

if we assume X>0, then x/|x|<x = x/x<x.....which is 1<x, we can keep this option as initially we have assumed X>0
Then we take any values of x greater than 1 then |x| will always be greater than 1.

So statement 1 in either case is sufficient..

Negative x does not mean that you should replace x with -x. x just represents a negative number. You should replace |x| with -x, though because if x < 0, then |x| = -x.

Hi Bunuel

Thanks for your reply. But i am still not able to get why we shouldn't consider negative x's throughout the equation.
If we assuming |x|= -x, we are assuming x<0, which means we are assuming the variable "X" as negative . So, we consider it negative only for Mode but keep it positive in other parts of the equation it seems inconsistent

Thanks again..
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Joined: 02 Sep 2009
Posts: 62380
Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x  [#permalink]

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03 May 2018, 08:53
cruiseav wrote:
Hi Bunuel

Thanks for your reply. But i am still not able to get why we shouldn't consider negative x's throughout the equation.
If we assuming |x|= -x, we are assuming x<0, which means we are assuming the variable "X" as negative . So, we consider it negative only for Mode but keep it positive in other parts of the equation it seems inconsistent

Thanks again..

Consider simple example: x = -1. So, we know that x is negative. Do you replace x there by -x and write -x = -1?
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x  [#permalink]

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13 Mar 2020, 00:52
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x   [#permalink] 13 Mar 2020, 00:52

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