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

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

by gmattesttaker2 » Tue Jan 28, 2014 8:08 pm
Hello,

Can you please tell me where I am going wrong here:

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

(1) x / |x| < x

(2) |x| > x

OA: C


I tried to solve this as follows:

1) x/|x| < x

Since |x| > 0
=> x < x |x|

If x = 2 => 2 < 2 |2| ? - Yes. Hence, |x| is not less than 1.

If x = 1 => 1 < 1 |1| ? - No

If x = -1 => -1 < -1 |1| ? - No

If x = -2 => -2 < -2 |-2| ? - No

Hence, |x| is not less than 1. - Sufficient


2) |x| > x

If x = 2 => |2| > 2 ? - No

If x = -2 => |-2| > -2 - Yes - Hence, |x| is not less than 1.

If x = 1 => |1| > 1 ? - No

Hence, |x| is not less than 1. - Sufficient

Hence, D


Thanks,
Sri
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Source: — Data Sufficiency |
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by GMATGuruNY » Wed Jan 29, 2014 3:50 am
You neglected to consider a NEGATIVE FRACTION such as -1/2.
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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