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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? [#permalink]

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09 Jun 2012, 09:10
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alphabeta1234 wrote:
Hey Bunuel,

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.

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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09 Jun 2012, 12:56
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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.

FOR THIS QUESTION: is |x|< 1

From (1) - Two cases.
Case 1 x>0,
x/|x|< x => x/x < |x| => |x|> 1 (Ans to Q stem No)

Case 2 x<0,
x/|x|< x => x/x > |x| => |x|< 1 (Ans to Q stem Yes)

Hence Insuff.

From (2) - We simply get that x < 0, Insuff.

Combined (1) and (2), we see that only x is <0
Hence case 2. Sufficient

Ans 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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14 Jun 2012, 04:21
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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!
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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 Jun 2012, 04:31
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pavanpuneet wrote:
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 $$-1<x<1$$.

Hope it helps.
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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 Jul 2012, 03:36
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Here is a video explanation that takes a slightly different approach using graphs of basic functions:

http://www.gmatquantum.com/shared-posts ... -xx-x.html

Cheers,
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Re: If x is not equal to 0, is |x| less than 1? [#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, 14:22
This to me seems like a 700 difficulty level. I almost put C but in the end decided on B.

I honestly hate these types of questions but know they will come.

unrelated: wat does CAT mean?
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Re: If x is not equal to 0, is |x| less than 1? [#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? [#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? [#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? [#permalink]

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25 Apr 2013, 23:56
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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? [#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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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? [#permalink]

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23 Jun 2013, 10:37
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.

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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? [#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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07 Aug 2013, 03:56
metallicafan wrote:
+1 C

Statement 1: x./|x| < x
We could have:
x > 1
or
x > -1
INSUFF.

Statement 2: |x| > x
So, x < 0
INSUFF.

Combining (1) and (2)
-1 < x < 0
SUFF.

Most complete explanation by Bunnel. Thanks to you. pretty good
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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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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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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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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x   [#permalink] 29 Sep 2014, 06:15

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