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Re: Inequality - Interesting one (Pay attention) [#permalink]

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24 Feb 2010, 07:56

gurpreetsingh wrote:

B- I only

this equality will hold true only when x =0 as if you take x>0 then x cannot be - |x| and if you take x<0 then x cannot be |x|

Yes, but with statement 1 if x=0, the absolute value of x will be 0 .. and that can't have a negative or positive value. I agree statements II and III are impossible but am weary of statment I.

I would vote for A none of the above. Am I overthinking this problem?

Re: Inequality - Interesting one (Pay attention) [#permalink]

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24 Feb 2010, 13:48

For me the answer is 0, I agree with all of you.

This question is from Nova guide, I will post explanation as given in the nova tomorrow. The OA is 0 and x < 0 though but I do not agree with it.That's why wanted to know the opinion of others.

Re: Inequality - Interesting one (Pay attention) [#permalink]

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24 Feb 2010, 14:11

GMATMadeeasy wrote:

If x = ± |x| , then which one of the following statements could be true? I.

x = 0 II.

x < 0 III.

x > 0 (A) None (B) I only (C) III only (D) I and II (E) II and III

Not too sure... but if the question was y = ± |x| - x, I would have gone with D as Answer.

If you plot y = ± |x| - x , we get y <= 0.... _________________

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|

Re: Inequality - Interesting one (Pay attention) [#permalink]

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24 Feb 2010, 16:29

One way of solving modulus questions, square on both sides.

x = +- |x|, implies x^2 = x^2, you can't cancel both sides because you can't divide with 0. So, x^2 - x^2 = 0, this is of the form a^2 - b^2. so, (x - x)(x + x) = 0. 2x(x - x)=0.

X can be 0 or x can be any number in the whole wide universe. There's no option specifying all three statements together, so go with the safer bet, since we have definitely proved one possibility that x can be 0. So B should be the answer. _________________

Re: Inequality - Interesting one (Pay attention) [#permalink]

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25 Feb 2010, 03:44

Answer copied from NOVA guide

Statement I could be true because ± 0 = −(+0) = −(0) = 0 . Statement II could be true because the right side of the equation is always negative [ ± x = –(a positive number) = a negative number]. Now, if one side of an equation is always negative, then the other side must always be negative, otherwise the opposite sides of the equation would not be equal. Since Statement III is the opposite of Statement II, it must be false. But let’s show this explicitly: Suppose x were positive. Then x = x, and the equation x = ± x becomes x = –x. Dividing both sides of this equation by x yields 1 = –1. This is contradiction.

Re: If x=-|x|, then which one of the following statements [#permalink]

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16 Dec 2013, 11:36

I understand clearly that x = - !x! would mean x is negative since an absolute value always is positive.. But why is 0 even a valid answer? x = -0? What does -0 even mean? Zero is zero regardless, there is no "negative" zero. Now, in this case I picked the right answer since I couldn't pick II alone. But HAD I been able to do so, I cannot say I would have picked "I and II".

So if someone has a good, intuitive explanation, feel free to explain (preferbly Bunuel)

I understand clearly that x = - !x! would mean x is negative since an absolute value always is positive.. But why is 0 even a valid answer? x = -0? What does -0 even mean? Zero is zero regardless, there is no "negative" zero. Now, in this case I picked the right answer since I couldn't pick II alone. But HAD I been able to do so, I cannot say I would have picked "I and II".

So if someone has a good, intuitive explanation, feel free to explain (preferbly Bunuel)

0=-0.

\(x=-|x|\) means that \(x\leq{0}\). If \(x=0\), then we'd have \(0=-|0|\) --> \(0=-0\). _________________

Re: If x=-|x|, then which one of the following statements [#permalink]

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22 Aug 2015, 06:56

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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