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My point was to tell you that statement 1 is insufficient. I just tried to highlight different values of x and their relations. Nothing else.
_________________

Hit kudos if my post helps you. You may send me a PM if you have any doubts about my solution or GMAT problems in general.

Searched for this multiple times before posting but couldnt find it...

What is the value of |x| ? (1) x = –|x| (2) \(x^2\) = 4

It is not mandatory to solve it algebraically. I have seen modulus question can be solved using PIN quite well.

Here, Say; x=0; |x|=0; -|x|=0; x=-|x|; So, x can be 0. x=1; |x|=1; -|x|=-1; x<>-|x|; So, x can't be 1. x=2; |x|=2; -|x|=-2; x<>-|x|; So, x can't be 2. We see that the modulus of a +ve number is always +ve. When we flip the sign of it we get a -ve. Thus, x can't be +ve because a +ve will not be equal to -ve.

x=-2; |x|=|-2|=2; -|x|=-2; So, x=-|x|; x can be 2. x=-3; |x|=|-3|=3; -|x|=-3; So, x=-|x|; x can be 3. In fact x can be any -ve number.

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Searched for this multiple times before posting but couldnt find it...

What is the value of |x| ? (1) x = –|x| (2) \(x^2\) = 4

It is not mandatory to solve it algebraically. I have seen modulus question can be solved using PIN quite well.

Here, Say; x=0; |x|=0; -|x|=0; x=-|x|; So, x can be 0. x=1; |x|=1; -|x|=-1; x<>-|x|; So, x can't be 1. x=2; |x|=2; -|x|=-2; x<>-|x|; So, x can't be 2. We see that the modulus of a +ve number is always +ve. When we flip the sign of it we get a -ve. Thus, x can't be +ve because a +ve will not be equal to -ve.

x=-2; |x|=|-2|=2; -|x|=-2; So, x=-|x|; x can be 2. x=-3; |x|=|-3|=3; -|x|=-3; So, x=-|x|; x can be 3. In fact x can be any -ve number.

x <= 0 AND |x| >= 0 Not Sufficient.

2) x^2=4 x=+-2 So, |x|=|-2|=2 Or |x|=|+2|=2

Either way, |x|=2 Sufficient.

Ans: "B"

Sorry but i am having a hard time understanding whats going on here and have got more confused...

Searched for this multiple times before posting but couldnt find it...

What is the value of |x| ? (1) x = –|x| (2) \(x^2\) = 4

It is not mandatory to solve it algebraically. I have seen modulus question can be solved using PIN quite well.

Here, Say; x=0; |x|=0; -|x|=0; x=-|x|; So, x can be 0. x=1; |x|=1; -|x|=-1; x<>-|x|; So, x can't be 1. x=2; |x|=2; -|x|=-2; x<>-|x|; So, x can't be 2. We see that the modulus of a +ve number is always +ve. When we flip the sign of it we get a -ve. Thus, x can't be +ve because a +ve will not be equal to -ve.

x=-2; |x|=|-2|=2; -|x|=-2; So, x=-|x|; x can be 2. x=-3; |x|=|-3|=3; -|x|=-3; So, x=-|x|; x can be 3. In fact x can be any -ve number.

x <= 0 AND |x| >= 0 Not Sufficient.

2) x^2=4 x=+-2 So, |x|=|-2|=2 Or |x|=|+2|=2

Either way, |x|=2 Sufficient.

Ans: "B"

Sorry but i am having a hard time understanding whats going on here and have got more confused...

x=1; |x|=1; -|x|=-1; x<>-|x|; So, x can't be 1. x=-2; |x|=|-2|=2; -|x|=-2; So, x=-|x|; x can be 2.

These 2 above statements confused me a lot...

if x=-2 ; whats the reason for doing this step |x| = |-2|? and then -|x| =-2 ??

What is the value of |x|? (1) x = –|x| (2) \(x^2\) = 4

Sorry but i am having a hard time understanding whats going on here and have got more confused...

x=1; |x|=1; -|x|=-1; x<>-|x|; So, x can't be 1. x=-2; |x|=|-2|=2; -|x|=-2; So, x=-|x|; x can be 2.

These 2 above statements confused me a lot...

if x=-2 ; whats the reason for doing this step |x| = |-2|? and then -|x| =-2 ??

What is modulus OR ||?

If a variable is -ve, and we wrap it around with ||, it becomes +ve. If a variable is +ve, and we wrap it around with ||, it remains +ve. If a variable is 0, and we wrap it around with ||, it remains 0.

So, -2; Wrap it around; |-2|=2 +2; Wrap it around; |+2|=+2=2 0; Wrap it around; |0|=0

Statement 1: (1) x = –|x|

Let's say x=1; L.H.S.=x=1 R.H.S.=-|x|=-|1|=-1 We know; 1 <> -1, so the expression x=-|x| doesn't hold good for x=1; And it won't hold good for any +ve number.

Now, let's say x=-1; L.H.S.=x=-1 R.H.S.=-|x|=-|-1|=-1 We know; -1 = -1, so the expression x=-|x| does indeed hold good for x=-1; And it will hold good for any -ve number.

For x=-0.464654 OR x=-100000; this expression will hold good. Thus, we won't be able to find a conclusive value for x. Not Sufficient.
_________________

What is the value of |x|? (1) x = –|x| (2) \(x^2\) = 4

Sorry but i am having a hard time understanding whats going on here and have got more confused...

x=1; |x|=1; -|x|=-1; x<>-|x|; So, x can't be 1. x=-2; |x|=|-2|=2; -|x|=-2; So, x=-|x|; x can be 2.

These 2 above statements confused me a lot...

if x=-2 ; whats the reason for doing this step |x| = |-2|? and then -|x| =-2 ??

What is modulus OR ||?

If a variable is -ve, and we wrap it around with ||, it becomes +ve. If a variable is +ve, and we wrap it around with ||, it remains +ve. If a variable is 0, and we wrap it around with ||, it remains 0.

So, -2; Wrap it around; |-2|=2 +2; Wrap it around; |+2|=+2=2 0; Wrap it around; |0|=0

Statement 1: (1) x = –|x|

Let's say x=1; L.H.S.=x=1 R.H.S.=-|x|=-|1|=-1 We know; 1 <> -1, so the expression x=-|x| doesn't hold good for x=1; And it won't hold good for any +ve number.

Now, let's say x=-1; L.H.S.=x=-1 R.H.S.=-|x|=-|-1|=-1 We know; -1 = -1, so the expression x=-|x| does indeed hold good for x=-1; And it will hold good for any -ve number.

For x=-0.464654 OR x=-100000; this expression will hold good. Thus, we won't be able to find a conclusive value for x. Not Sufficient.

Thats a great explnation!!!... I knew what modulus is but confused on the cases you had mentioned earlier... Do you know how can we do this algebrically too? Just curious to know

Preparation for final battel: GMAT PREP-1 750 Q50 V41 - Oct 16 2011 GMAT PREP-2 710 Q50 V36 - Oct 22 2011 ==> Scored 50 in Quant second time in a row MGMAT---- -1 560 Q28 V39 - Oct 29 2011 ==> Left Quant half done and continued with Verbal. Happy to see Q39

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