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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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New post 09 Jun 2012, 13:56
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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 (2) |x| > x [#permalink]

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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 (2) |x| > x [#permalink]

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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 (2) |x| > x [#permalink]

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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

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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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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 (2) |x| > x [#permalink]

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New post 25 Apr 2013, 04:14
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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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If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x [#permalink]

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New post 01 Jan 2015, 22:46
Hi,

There are a handful of Number Property rules in this DS question that you can find by either TESTing VALUES or taking lots of notes.

We're told that X cannot = 0. We're asked if |X| < 1. This is a YES/NO question.

Fact 1: X/|X| < X

This inequality requires a thorough examination of the possibilities.

X CANNOT = 1 since 1/|1| is NOT < 1

X can be ANY number > 1 though

If X = 2....
2/|2| < 2
And the answer to the question is NO.

X CANNOT be a positive fraction...

If X = 1/2
(1/2)/|1/2| = 1 which is NOT < 1/2

X could be ANY negative fraction though

If X = -1/2
(-1/2)/|-1/2| = -1 which IS < -1/2
And the answer to the question is YES.

X CANNOT be -1 since -1/|-1| is NOT < -1

X CANNOT be < -1...

X = -2
(-2)/|-2| = -1 which is NOT < -2

According to all of this data, the possibilities are X > 1 or -1 < X < 0
Fact 1 is INSUFFICIENT

Fact 2: |X|> X

Here, we know that X can be ANY negative, but CANNOT be positive.

If X = -1/2 then the answer to the question is YES (we did the work already in Fact 1)
If X = -2 then the answer to the question is NO
Fact 2 is INSUFFICIENT

Combining Facts, we have...
X > 1 or -1 < X < 0
AND
X MUST be negative

The ONLY possibilities that fit BOTH Facts ARE negative fractions. Thus the answer to the question will be ALWAYS YES.
Combined, SUFFICIENT

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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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New post 13 May 2015, 07:19
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.



Thanks for your nice explanations. I am not clear on one issue though. As we can multiply an inequality by a variable only if we know its sign, how can we multiply both side of an inequality by an absolute value? Would you please explain..
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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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ranaazad 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 \(-1<x<0\)

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.



Thanks for your nice explanations. I am not clear on one issue though. As we can multiply an inequality by a variable only if we know its sign, how can we multiply both side of an inequality by an absolute value? Would you please explain..


An absolute value of a number cannot be negative (it's 0 or positive), and since we are given that x is not 0, then |x| is positive only.
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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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New post 05 Jul 2015, 16:45
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.



I did mine a little differently.

If x ≠ 0, is |x| < 1?


Means: Is -1 < x < 1?

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

x < |x|*x
If x is positive, then: \(\frac{x}{x} < |x|\) which is the same as \(1 < |x|\)
This means that \(-1 > x > 1\)

If x is negative, then: \(\frac{x}{x} > |x|\) which is the same as \(1 > |x|\)
This means that \(-1 < x < 1\) (we switch the direction of < to > because we divided by -1 to put |x| by itself).

These two answers are inconsistent: x is both less than 1 and greater than 1. So, insufficient.

(2) \(|x| > x\)
This means that x is negative since the negative value of something is always less than the absolute value of something; whereas, a positive or 0 value of something is equal to its absolute value.

If x is equal to a negative value less than -1, then it's within the range -1 < x < 1. But, if x is equal to a large negative value, |x| can be out of range, for example, x = -1,000. Therefore, (2) is insufficient.

(1) + (2)
|x| > x and |x| > 1
x is negative, and we know that x < -1. So, pick an arbitrary value less than -1, say, -5.
Plug it into equation (1): \(\frac{-5}{|-5|} < -5\). Is \(-1< -5\)? False. x does not refer to values < -1.

|x| > x and |x| < 1
x is negative, and we know that x > -1. Therefore, \(-1 < x < 0\).
Pick an arbitrary value between -1 and 0, non-inclusive, say, -0.5.
Plug it into equation (1): \(\frac{-0.5}{|-0.5|} < -0.5\). Is \(-1 < -0.5\)? Yes, this works. Sufficient.

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

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New post 17 Nov 2015, 10:24
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) x/|x| < x
(2) |x| > x

If the range of the condition falls into that of the condition in terms of inequalities, the condition is sufficient.

There is 1 variable in the original condition, and there are 2 equations provided by the 2 conditions, so there is high chance (D) will be our answer.
For condition 1, if x>0, x/|x|<x --> x/x<x --> 1<x, then 1<x
if x<0, x/|x|<x --> x/-x<x --> -1<x, then -1<x<0. Therefore this condition is insufficient because this range does not fall into that of the question.
For condition 2, |x|>x --> x<0. This is insufficient for the same reason.
Looking at them together, however, -1<x<0 falls into the range of the question, so this is sufficient, and the answer becomes (C).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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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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New post 08 Mar 2016, 17:30
Hi Bunuel,

For the A, how did you assume that remember that x<0

Thanks
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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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New post 08 Mar 2016, 23:48
marvas8581 wrote:
Hi Bunuel,

For the A, how did you assume that remember that x<0

Thanks


There we consider two cases: (A) when x<0 and (B) when x>0. So, A we discard the range which is not less than 0 and for B discard the range which is not more than 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 (2) |x| > x [#permalink]

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New post 09 Mar 2016, 01:08
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If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x



Given : x is NOT equal to zero

Question : Is |x| < 1?

Statement 1: x/|x|< x

Case 1: If x <0, x/|x| = -1 i.e. x/|x|< x can be rewritten as -1< x i.e. -1< x < 0 Giving answer to the question as YES
Case 2: If x >0, x/|x| = +1 i.e. x/|x|< x can be rewritten as 1< x i.e. 1< x Giving answer to the question as NO
NOT SUFFICIENT

Statement 2: |x| > x
|x| can be greater than x only if x is Negative because in all other cases both will be equal
i.e. x < 0 but x may be -0.5 Giving answer to the question as YES and x may be -1.5 Giving answer to the question as NO. Hence,
NOT SUFFICIENT

Combining the two statements
Combining -1< x < 0 and 1< x and x < 0

we get, only -1< x < 0 Giving answer to the question as YES. Hence,
SUFFICIENT

Answer: option C
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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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New post 15 Oct 2017, 14:02
For (1), wouldn't x<0 give you:

-x/x < -x
-1 < -x
1 > x?

When I plug in numbers for that, it doesn't work, but I don't see how we get x/-x < x when plugging in 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.

Answer: C.
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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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New post 15 Oct 2017, 21:17
cgarmestani wrote:
For (1), wouldn't x<0 give you:

-x/x < -x
-1 < -x
1 > x?

When I plug in numbers for that, it doesn't work, but I don't see how we get x/-x < x when plugging in 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.

Answer: C.


If x < 0, then |x| = -x.

Substitute |x| by -x in x/|x|< x to get x/(-x) < x and then to get -1 < x. Since we consider the range when x < 0, then -1 < x < 0.
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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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New post 29 Nov 2017, 07:09
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?


Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.


the red zone indicates that the range is -1<x<0 , how do we arrive at -1<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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New post 29 Nov 2017, 07:22
yousufkhan wrote:
Bunuel wrote:
udaymathapati wrote:
Hi Bunuel,
Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?


Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.


the red zone indicates that the range is -1<x<0 , how do we arrive at -1<x<1?


The question asks whether \(-1<x<1\) is true. We got that \(-1<x<0\). Any, x from \(-1<x<0\) IS in the range from -1 to 1, so we have an YES answer to the question.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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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New post 27 Dec 2017, 03:15
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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New post 29 Mar 2018, 06:33
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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Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x (2) |x| > x   [#permalink] 29 Mar 2018, 06:33

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