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

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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, 12:39
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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 $$-1<x<0$$
.

if x<0, doesn't x/|x|<x become -x/x<-x which simplifies to -1<-x or x<1 ?

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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.

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??
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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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25 Apr 2013, 03:14
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Expert's post
oyabu wrote:
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 $$-1<x<0$$
.

if x<0, doesn't x/|x|<x become -x/x<-x which simplifies to -1<-x or x<1 ?

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.
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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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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?

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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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25 Apr 2013, 23:56
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Expert's post
oyabu wrote:
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?

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

It seems that you need to go through basics: math-absolute-value-modulus-86462.html
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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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23 Jun 2013, 00:36
One question:

if you pug in numbers, for the first statement is the statement not only true for number between -1 and 0? I just do not get it completely!

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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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23 Jun 2013, 01:02
BankerRUS wrote:
One question:

if you pug in numbers, for the first statement is the statement not only true for number between -1 and 0? I just do not get it completely!

It's also true if x>1. For example, if x=2, then x/|x|=1 < 2=x.
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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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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

(C)

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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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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.

Hi bunel,

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

regards,
rrsnathan.

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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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11 Mar 2014, 03:27
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| < 1
if we plot this then we get the range of x as (-1,1)

1) $$\frac{x}{|x|} < x$$
the range of x here is (1, infinity) U (-1,0)
hence, it is not sufficient.

2) |x| > x
very clearly this would be true for all x < 0

if we combine 1 and 2, the common range we get is (-1,0)
which satisfies |x| < 1
hence, C.
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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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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.

Bunuel,

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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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29 Sep 2014, 06:15
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 .

Bunuel,

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

Bunuel,

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.

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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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20 Oct 2014, 20:51
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
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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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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) ?

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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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22 Oct 2014, 00:15
Expert's post
1
This post was
BOOKMARKED
earnit wrote:
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) ?

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).
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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 Nov 2014, 10:26
-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)
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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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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.

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.

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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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14 Dec 2014, 11:04

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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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15 Dec 2014, 07:04
manojpandey80 wrote:

Check this: if-4x-12-x-9-which-of-the-following-must-be-true-101732.html
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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x   [#permalink] 15 Dec 2014, 07:04

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