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ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT

1) this only holds true if x is a negative fraction or greater than 1. The question however is asking if |x| is less than 1. The abs val of a number less than -1 will be greater than 1, whereas the abs val of a negative fraction will be less than 1.

x/|x| < x

try -2, -2/|-2| = -1, -1 is greater than -2, so does not hold try -1/2, (-1/2)/|(-1/2)| = -1, -1 is less than -1/2 so this holds. try 1/2, (1/2)/|(1/2)| = 1, 1 is greater than 1/2, so this does not hold try 2, 2/|2| = 1, 1 is less than 2, so this holds

so, -1 < x < 0 and x > 1

the question asks: is |x| < 1?

-1 < x < 0 --> |x| will be a positive fraction, i.e. less than 1 x > 1 --> |x| will be greater than 1

hence insufficient.

2) the abs val of a positive number equals that number so this only holds true for negative numbers (including fractions). The question is asking if |x| is less than 1. Same as in (1), |x| > x

x < 0

try a negative integer, -2, | -2 | = 2, --> |x| will be greater 1

try a negative fraction, -(1/2), | -(1/2) | = 1/2 --> |x| will be less than 1

hence insufficient.

putting both together,

x < 0 and -1< x < 0 and x > 1 this limits x to -1< x < 0, thus |x| will be a positive fraction and will always be less than 1.

therefore the answer is C _________________

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ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT

1) this only holds true if x is a negative fraction or greater than 1. The question however is asking if |x| is less than 1. The abs val of a number less than -1 will be greater than 1, whereas the abs val of a negative fraction will be less than 1.

x/|x| < x

try -2, -2/|-2| = -1, -1 is greater than -2, so does not hold try -1/2, (-1/2)/|(-1/2)| = -1, -1 is less than -1/2 so this holds. try 1/2, (1/2)/|(1/2)| = 1, 1 is greater than 1/2, so this does not hold try 2, 2/|2| = 1, 1 is less than 2, so this holds

so, -1 < x < 0 and x > 1

the question asks: is |x| < 1?

-1 < x < 0 --> |x| will be a positive fraction, i.e. less than 1 x > 1 --> |x| will be greater than 1

hence insufficient.

2) the abs val of a positive number equals that number so this only holds true for negative numbers (including fractions). The question is asking if |x| is less than 1. Same as in (1), |x| > x

x < 0

try a negative integer, -2, | -2 | = 2, --> |x| will be greater 1

try a negative fraction, -(1/2), | -(1/2) | = 1/2 --> |x| will be less than 1

hence insufficient.

putting both together,

x < 0 and -1< x < 0 and x > 1 this limits x to -1< x < 0, thus |x| will be a positive fraction and will always be less than 1.

therefore the answer is C

I second that... my slip.... i missed the greater part in ST 1...... ! Yes.. it should be C.... _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

1. For Condition 1, x can either be positive integer or -ve fraction. Its true for both i.e if x = 2,3, etc or x = -1/2, -1/3, etc. Hence not sufficient.

2. For condition 2, X has to be -ve fraction or integer. Its true only when x = -1/2, -1/3, etc. or x=-1, -2, etc. Hence B not sufficient.

3.Combine 1 & 2. True only when x is -ve fraction i.e. x=-1/2, -1/3, etc. Hence |x| < 1. Hence C should be the answer.

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

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06 Feb 2014, 03:32

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

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03 Oct 2015, 11:35

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

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13 Nov 2015, 09:21

Bunuel wrote:

jeeteshsingh wrote:

ugimba wrote:

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

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

IMO A...

ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT

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

Q: is \(-1<x<1\) true?

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

\(x<0\) --> \(\frac{x}{-x} < x\) --> \(x>-1\), but as \(x<0\), then --> \(-1<x<0\);

\(x>0\) --> \(\frac{x}{x} < x\) --> \(x>1\).

So x can be as in the range {-1,1} as well as out of this range. Not sufficient.

(2) \(|x| > x\) --> \(x<0\). Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is: \(-1<x<0\). Every \(x\) from this range is in the range {-1,1}. Sufficient.

Hi Bunuel, Thanks for this explanation. Bunuel, I am still having a hard time trying to understand the concepts behind the process followed to solve statement 1 in this question.

My reasoning is the following: for example if |x-3|=2, then x-3=2 and x-3=-2. If I follow this same reasoning then for statement one I get the following:

Hi Bunuel, Thanks for this explanation. Bunuel, I am still having a hard time trying to understand the concepts behind the process followed to solve statement 1 in this question.

My reasoning is the following: for example if |x-3|=2, then x-3=2 and x-3=-2. If I follow this same reasoning then for statement one I get the following:

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

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

What is the flaw in my reasoning?

Thanks so much for your help!

Let me try to answer your question.

Firstly, |x|<1 ---> -1<x<1 . So the question asks whether x is between -1 and 1 with x\(\neq\)0

Your interpretation of |x-3|=2 is correct by taking the 2 cases. But the way you are approaching statement 1 is not correct. Look below.

Statement 1, x/|x|<x .

Take 2 cases.

Case 1: when x > 0 ---> |x| = x ---> x/|x|<x ---> x>1 . Now you had assumed that x>0 and it gave you x>1. Thus the range for x satisfying this case will thus become x>1. This will give a "no" to "is |x|<1 or is -1<x<1?".

case 2: when x < 0 ---> |x| = -x ---> x/|x|<x ---> x>-1 . Now you had assumed that x<0 and it gave you x>-1. Thus the range for x satisfying this case will thus become -1<x<0. This will give a "yes" to "is |x|<1 or is -1<x<1?".

Thus you get 2 different cases for the same statement, making statement 1 not sufficient.

The issue with your solution is that you are not taking the intersection of the possible values as I have shown in 2 cases above.

|x-3| = 2 --> x-3=2 ONLY when x-3 \(\geq\)0 .

Similarly, |x-3| = 2 --> x-3=-2 ONLY when x-3 < 0.

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

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15 Nov 2015, 20:14

Engr2012 wrote:

angierch wrote:

Hi Bunuel, Thanks for this explanation. Bunuel, I am still having a hard time trying to understand the concepts behind the process followed to solve statement 1 in this question.

My reasoning is the following: for example if |x-3|=2, then x-3=2 and x-3=-2. If I follow this same reasoning then for statement one I get the following:

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

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

What is the flaw in my reasoning?

Thanks so much for your help!

Let me try to answer your question.

Firstly, |x|<1 ---> -1<x<1 . So the question asks whether x is between -1 and 1 with x\(\neq\)0

Your interpretation of |x-3|=2 is correct by taking the 2 cases. But the way you are approaching statement 1 is not correct. Look below.

Statement 1, x/|x|<x .

Take 2 cases.

Case 1: when x > 0 ---> |x| = x ---> x/|x|<x ---> x>1 . Now you had assumed that x>0 and it gave you x>1. Thus the range for x satisfying this case will thus become x>1. This will give a "no" to "is |x|<1 or is -1<x<1?".

case 2: when x < 0 ---> |x| = -x ---> x/|x|<x ---> x>-1 . Now you had assumed that x<0 and it gave you x>-1. Thus the range for x satisfying this case will thus become -1<x<0. This will give a "yes" to "is |x|<1 or is -1<x<1?".

Thus you get 2 different cases for the same statement, making statement 1 not sufficient.

The issue with your solution is that you are not taking the intersection of the possible values as I have shown in 2 cases above.

|x-3| = 2 --> x-3=2 ONLY when x-3 \(\geq\)0 .

Similarly, |x-3| = 2 --> x-3=-2 ONLY when x-3 < 0.

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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