|x| less than 1

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by GMATGuruNY » Sun Mar 27, 2016 2:35 am
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x
Question rephrased: Is x between -1 and 1?

Statement 1: x/|x| < x
x < x|x|

0 < x|x| - x

0 < x (|x| - 1)

The CRITICAL POINTS are -1, 0 and 1.
These are the only values where x(|x|-1) = 0.
To determine the ranges where x(|x|-1) > 0, test one value to the left and right of each critical point.

Case 1: x<-1
Plug x = -2 into x/|x| < x:
-2/ |-2| < -2
-1 < -2.
Doesn't work.
Thus, x < -1 is not a valid range.

Case 2: -1<x<0
Plug x = -1/2 into x/|x| < x:
-1/2/ |-1/2| < -1/2
-1 < -1/2.
This works.
Thus, -1<x<0 is a valid range.

Case 3: 0<x<1
Plug x = 1/2 into x/|x| < x:
(1/2)/ |1/2| < 1/2
1 < 1/2
Doesn't work.
Thus, 0<x<1 is not a valid range.

Case 4: x>1
Plug x = 2 into x/|x| < x:
2/ |2| < 2
1 < 2.
This works.
Thus, x > 1 is a valid range.

Thus, -1<x<0 or x>1.
INSUFFICIENT.

Statement 2: |x| > x
Any negative value will satisfy this inequality.
If x=-1/2, then x is between -1 and 1.
If x=-2, then x is NOT between -1 and 1.
INSUFFICIENT.

Statements combined:
The only range that satisfies both statements is -1<x<0.
Thus, x is between -1 and 1.
SUFFICIENT.

The correct answer is C.
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by Mo2men » Sun Mar 27, 2016 8:23 am
GMATGuruNY wrote:
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x
Question rephrased: Is x between -1 and 1?

Statement 1: x/|x| < x
x < x|x|

0 < x|x| - x

0 < x (|x| - 1)

The CRITICAL POINTS are -1, 0 and 1.
These are the only values where x(|x|-1) = 0.
To determine the ranges where x(|x|-1) > 0, test one value to the left and right of each critical point.

Case 1: x<-1
Plug x = -2 into x/|x| < x:
-2/ |-2| < -2
-1 < -2.
Doesn't work.
Thus, x < -1 is not a valid range.

Case 2: -1<x<0
Plug x = -1/2 into x/|x| < x:
-1/2/ |-1/2| < -1/2
-1 < -1/2.
This works.
Thus, -1<x<0 is a valid range.

Case 3: 0<x<1
Plug x = 1/2 into x/|x| < x:
(1/2)/ |1/2| < 1/2
1 < 1/2
Doesn't work.
Thus, 0<x<1 is not a valid range.

Case 4: x>1
Plug x = 2 into x/|x| < x:
2/ |2| < 2
1 < 2.
This works.
Thus, x > 1 is a valid range.

Thus, -1<x<0 or x>1.
INSUFFICIENT.

Statement 2: |x| > x
Any negative value will satisfy this inequality.
If x=-1/2, then x is between -1 and 1.
If x=-2, then x is NOT between -1 and 1.
INSUFFICIENT.

Statements combined:
The only range that satisfies both statements is -1<x<0.
Thus, x is between -1 and 1.
SUFFICIENT.

The correct answer is C.
Hi Mitch,

In statement 1, how did you multiply by |x|. X is variable and we do not know its sign. So we should not multiply inequality by variable.

Thanks

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by [email protected] » Sun Mar 27, 2016 11:10 am
Hi jain2016,

This DS question is built around some interesting Number Properties and patterns. If you can spot those patterns, then solving this problem should take considerably less time. This also looks like a question that can be beaten by TESTing VALUES.

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

Fact 1: X/|X| < X

Before TESTing VALUES, I want to note a pattern in this inequality:

X/|X| will either equal 1 (if X is positive) OR -1 (if X is negative). This will save us some time when it comes to TESTing VALUES, since there are many values of X that will NOT fit this information.

If X = 2, then the answer to the question is NO.

X cannot be 1, any positive fraction, 0, or any negative integer.....

So what's left to TEST....?

If X = -1/2, then the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: |X| > X

This tells us that X CANNOT be positive or 0.

If X = -1, then the answer to the question is NO.
If X = -1/2, then the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we have deal with the "overlapping restrictions" that we noted in the two Facts:
X cannot be....anything positive, 0, or any negative integer.
X can ONLY BE negative fractions between 0 and -1.
ALL of those answers (e.g. -1/2, -.4, etc.) lead to a YES answer.
Combined SUFFICIENT.

Final Answer: C

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by GMATGuruNY » Sun Mar 27, 2016 3:07 pm
Mo2men wrote:In statement 1, how did you multiply by |x|. X is variable and we do not know its sign. So we should not multiply inequality by variable.
|x| > 0 for all nonzero values of x.
The prompt indicates that x≠0, implying that |x| > 0.
Since |x| > 0, we can safely multiply each side of x/|x| < x by |x|.
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