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If x is not equal to 0 or -1, what is the value of |x-2|?

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If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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If x is not equal to 0 or -1, what is the value of |x-2|?

(1) \(\frac{(|x+1|)}{(-|x|x^3)}>0\)

(2) \(x^4 = 16\)

This is Question 1 for the e-GMAT Question Series on Absolute Value.

Provide your solution below. Kudos for participation. Happy Solving! :-D

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Last edited by EgmatQuantExpert on 13 Dec 2016, 23:56, edited 3 times in total.

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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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EgmatQuantExpert wrote:
If x is not equal to 0 or -1, what is the value of |x-2|?

(1) \(\frac{(|x+1|)}{(-|x|x^3)}>0\)

(2) \(x^4 = 16\)

This is Question 1 for the e-GMAT Question Series on Absolute Value.

Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 20th May.

Till then, Happy Solving! :-D

Best Regards
The e-GMAT Team


Statement 1
Does't give you a final value of |x-2|. I tried to rearange the inequality but you can not multiply by \(-|x|x^3\) because the term can be positive or negative (you don't know). Therefore the whole Statement 1 just tells you that some Division with X is > 0 and is therefore insufficient because it doesn't provide a VALUE for |x-2|.

Statement 2
x can be 2 or -2. Which is insufficient.

If you combine Statement 1 and 2 you know that in order to be true, Statement 1 needs a positive denominator which is just possible if X equals -2 (x^3 would then be -8 and the denominator would be -8*-2 = 16.

Answer Choice C
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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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Statement 1 says x is negative.
Statement 2 gives 2 and -2 as values.
Combining the two we get x as -2

So Answer must be C
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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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EgmatQuantExpert wrote:
If x is not equal to 0 or -1, what is the value of |x-2|?

(1) \(\frac{(|x+1|)}{(-|x|x^3)}>0\)

(2) \(x^4 = 16\)

This is Question 1 for the e-GMAT Question Series on Absolute Value.

Provide your solution below. Kudos for participation. The Official Answer and Explanation will be posted on 20th May.

Till then, Happy Solving! :-D

Best Regards
The e-GMAT Team


hi,
(1) \(\frac{(|x+1|)}{(-|x|x^3)}>0\)...
we know that for fraction to be greater than 0 both deno and numerator should be of same sign....
numeratos has to be +ive as it is mod.. so -|x|x^3 should be positive so x is a negative integer.. no other info ... insuff

(2) \(x^4 = 16\)
x can be 2 or -2... insuff

combined we get x as -2 .. suff ans C
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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Stmt 1
fraction is > 0 then Numerator & Denominator need to be same sign.

Numerator |x-2| > 0

Denominator \(-|x|x^3 > 0\) for this ( \(x^3 < 0\)) , implies x < 0 insufficient.

Stmt 2
x can be 2 or -2. insufficient.

combine 1 and 2 we get the answer, from 1 we know x < 0, and from 2 we know (x = 2 or -2)

Ans : C
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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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EgmatQuantExpert wrote:
If x is not equal to 0 or -1, what is the value of |x-2|?

(1) \(\frac{(|x+1|)}{(-|x|x^3)}>0\)

(2) \(x^4 = 16\)

This is Question 1 for the e-GMAT Question Series on Absolute Value.



Official Explanation

In order to find the value of |x-2|, we first need to know the value of x. From question statement, we know that x is not equal to 0 or -1. Let's see if St. 1 and/or 2 can get us a unique value of x.

Analyzing St. 1 Independently

\(\frac{(|x+1|)}{(-|x|x^3 )}>0\)


Multiplying both sides by a positive number doesn’t impact the sign of inequality. So, multiplying both sides by \(\frac{(|x|)}{(|x+1|)}\), we get:

\(\frac{(-1)}{x^3} >0\)

(Note that we could multiply both sides by this number because we were told that x ≠ -1. Therefore, x + 1 ≠ 0 (division by 0 is not defined))

Now, multiplying both sides by -1 will reverse the sign of inequality. We get:

\(\frac{1}{x^3}<0\)

\(x^3\) will have the same +/- sign as x

This means, \(1/x\) is negative.

This means, x is negative.

This information is not sufficient to find the value of x.

Analyzing St. 2 independently
\(x^4 = 16\)


That is, \(x^4 - 2^4 = 0\)
\((x^2 + 2^2)(x+2)(x-2) = 0\)

This gives us, x = 2 or -2

Since we don’t get a unique value of x, not sufficient.

Combining St. 1 and St. 2

From St. 1, x is negative
From St. 2, x = 2 or -2

Combining both, x = -2

Sufficient.
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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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New post 11 Jul 2016, 08:05
from the 1 statement we come to know that x is negative as we have denominator as negative.So the expression to be grater than 0 x^3 should be negative.then only it will br=e greater than zero.We do not get any definite value except x is negative.So NS

From2 statement we get that x could be 2 or -2 and both give different answers when put inside the expression.So NS

Combined we get x is negative which is only -2.And we get definite value.So answer is C.

Let me know if i am wrong.

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Re: If x is not equal to 0 or -1, what is the value of |x-2|? [#permalink]

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New post 17 Aug 2017, 18:28
I know this is unnecessary, but can anyone confirm that the below is correct if it was necessary to do so. Trying to make sure I understand the 3 step approach to absolute modulus.

For statement 1:

Since l x+1 l is always positive, then l x+ 1 l >0

critical point is -1,

a) so if x>-1, then x+1 is positive, so x+1>0, and x>-1,
b) if x<-1, then X+1 is negative, so -(x+1)>0, and x<-1,

so x>-1 or x<-1 right?

Denominator needs to have the same sign as the numerator, so x must be negative for denominator to be positive. so x>-1, but doesn't solve what lx-2l is.

2) x^4=16, x= 2 or -2, insuff.

c: x>-1, so x=2. Suff.

Did I handle the splitting up the modulus correctly in statement 1? Thank you! Kudos to whoever answers.

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Re: If x is not equal to 0 or -1, what is the value of |x-2|?   [#permalink] 17 Aug 2017, 18:28
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