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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 wholelottalove » Sat Jun 29, 2013 1:45 pm
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

I'm not entirely sure how to solve this. One thing that threw me off were the ranges obtained for (1) The explanation states:

A. x<0: x/-x<x ===> -1<x. But remember that x<0, so -1<x<0

B. x>0: x/x<x ===> 1<x.

Generally, when I test for range I do so to make sure the inequality is valid. For example, in (1) I tested for where x is negative and got x>-1. The problem here is that because x>-1 then it could be positive thus making the solution invalid. Also, I've never encountered a problem where you include the range tested (in this case x<0) in the solution (i.e. -1<x<0)

Could someone tell me what I am doing wrong?

Thanks!
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by GMATGuruNY » Sat Jun 29, 2013 5:21 pm
wholelottalove 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.
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by fcabanski » Sat Jun 29, 2013 6:02 pm
This problem isn't about solving equations. It's about determining if a statement or statements are sufficient for answering the question.

"Generally, when I test for range I do so to make sure the inequality is valid. For example, in (1) I tested for where x is negative and got x>-1. The problem here is that because x>-1 then it could be positive thus making the solution invalid. " It can't be positive, because x>-1 only when x <0 (yes, |x| <1). Part B. of the explanation shows you that x could also be > 1 when x is positive (no, |x| is not < 1. Statement 1 is therefore not sufficient to answer the question "is |x|<1" . Statement 1 leads to both yes and no.
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by wholelottalove » Sun Jun 30, 2013 7:22 am
Ahh! That makes a lot more sense now. Thanks!

GMATGuruNY wrote:
wholelottalove 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.
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by gmaster328 » Mon Jul 01, 2013 8:46 pm
That makes sense! Thank you!
GMATGuruNY wrote:
wholelottalove 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.
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by pims » Wed Jul 16, 2014 1:20 am
Rich, could you provide a solution as well?
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by Matt@VeritasPrep » Wed Jul 16, 2014 10:35 am
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x
Here's a pretty quick approach.

Let's start with S1. Since |x| is never negative, we can multiply both sides by |x| without flipping the inequality. This gives us x < |x|*x. Now we have TWO cases.

Case 1: x is negative. In this case, dividing both sides by x gives us 1 > |x|. (The sign has to FLIP, since x is negative!)

Case 2: x is positive. In this case, dividing both sides by x gives us 1 < |x|.

Since we have two different answers (sometimes |x| > 1, sometimes 1 > |x|), S1 is INSUFFICIENT.

Now let's move to S2.

By the definition of absolute value, if x > 0, |x| = x. But we have |x| > x, so x CANNOT be positive. Hence x is negative ... but we still can't answer the question! (x could be -1/2 or -1,000,000, or any other negative number.)

Taking the two statements together, S2 tells us x < 0, and S1 tells us that if x is negative, 1 > |x|. SUFFICIENT!
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by [email protected] » Wed Jul 16, 2014 6:32 pm
HI pims,

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 Blackboxx » Fri Nov 07, 2014 8:31 am
[email protected] wrote:... we have deal with the "overlapping restrictions"
Rich
Rich - May I know what you meant by "overlapping restrictions"? For e.g., in the cases above,

Case 1 x/|x| < x holds true for x = all positives > 1 and x = all negative fractions
Case 2 |x| > x holds true for x = all negatives < -1 and x = all negative fractions

Now, when you say overlapping restrictions, should I look for those conditions that yield a YES in both the cases,eh? Like x= all negative fractions is the only option that satisfies both the cases. Is that what it is?
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by Matt@VeritasPrep » Mon Nov 10, 2014 11:43 am
Blackboxx wrote:
[email protected] wrote:... we have deal with the "overlapping restrictions"
Rich
Rich - May I know what you meant by "overlapping restrictions"? For e.g., in the cases above,

Case 1 x/|x| < x holds true for x = all positives > 1 and x = all negative fractions
Case 2 |x| > x holds true for x = all negatives < -1 and x = all negative fractions

Now, when you say overlapping restrictions, should I look for those conditions that yield a YES in both the cases,eh? Like x= all negative fractions is the only option that satisfies both the cases. Is that what it is?
That seems to be the implication, yeah.

Put more simply, suppose we have these two statements, both of which are true:

S1: I am a man or a woman.
S2: I am a man or an elephant.

I can't be a woman, since S2 doesn't give that possibility. I can't be an elephant, since S1 doesn't give that possibility. So I must be the only thing the statements agree on: a man.

This is important in DS. If x, y, and z are mutually exclusive -- that is, at most ONE of these three things is true -- and we have the statements "This is either x or y" and "this is either y or z", then we can conclude "this is y", since it's the only thing on which the two (true) statements agree.
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by [email protected] » Mon Nov 10, 2014 3:37 pm
Hi Blackboxx,

In a DS question, if you have proof that BOTH Facts are INSUFFICIENT, then you have to combine the Facts. We're only allowed to use values that satisfy BOTH conditions. In this DS question, each Fact restricted what X could possibly be, so combining them required us to deal with both restrictions.

You'll end up combining Facts on about 40% of your DS questions, so strong organization (and proper note-taking) is a must.

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by Mathsbuddy » Tue Nov 11, 2014 4:11 am
Statement (1) rearranged could give 1 < x, so |x| is not less than 1
However, if x = -0.5 (for example), then -0.5/0.5 = -1 < -0/5, so x is less than 1
Therefore, by contradiction, Statement (1) is INSUFFICIENT

Statement (2): dividing both sides by |x| gives 1 > +- 1
Due to the +- (i.e. plus or minus), this is INSUFFICIENT

Combining statements, a positive value of x does not comply with both statements.
However a negative value (between -1 and 0, not inclusive) does work.
SUFFICIENT
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by GMATinsight » Thu Nov 27, 2014 8:15 pm
If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x
Question: is |x| less than 1?
Question: is -1<x<1 ?

Statement 1: x/|x|< x

LHS will always be +1 or -1
@ x= 2 LHS<RHS and answer to Question is NO
@ x= -0.5 LHS<RHS and answer to Question is YES
NOT SUFFICIENT

Statement 2: |x| > x
This is true for all negative values of x therefore not sufficient as x may be -0.5 or -2

Combining two statements
x must be negative and still greater than -1 i.e. -1<x<0
SUfficient

Answer: Option C
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