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# What is x? (1) |x| < 2 (2) |x| = 3x – 2

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What is x? (1) |x| < 2 (2) |x| = 3x – 2 [#permalink]  31 Jan 2012, 18:48
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What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?
[Reveal] Spoiler: OA
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Re: What is x? [#permalink]  31 Jan 2012, 18:56
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rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> -2<x<2. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

Hope it's clear.
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Re: What is x? (1) |x| < 2 (2) |x| = 3x – 2 [#permalink]  31 Jan 2012, 20:38
Vow! that is a very good solution It opened my eyes on how to read modulus based problems...
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Re: What is x? [#permalink]  01 Feb 2012, 03:42
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> -2<x<2. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me.
Generally in modulus questions we do as below:
(1) |x| < 2
I agree with u. -2<x<2
(2) |x| = 3x – 2
In this case i did
a. x is positive so x = 3x - 2 , upon solving x = 1
a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2.
I am confused !
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Re: What is x? [#permalink]  01 Feb 2012, 04:05
rohitgoel15 wrote:
Bunuel wrote:
rohitgoel15 wrote:
What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Can anyone justify the OA?

What is x?

(1) |x| < 2 --> -2<x<2. Not sufficient.

(2) |x| = 3x – 2 --> LHS is an absolute value, which is always non negative, hence RHS must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x --> x=3x-2 --> x=1. Sufficient

Hope it's clear.

Hi Bunuel,

Thanks for reply .. But this is not clear to me.
Generally in modulus questions we do as below:
(1) |x| < 2
I agree with u. -2<x<2
(2) |x| = 3x – 2
In this case i did
a. x is positive so x = 3x - 2 , upon solving x = 1
a. x is negative so -x = 3x - 2 , upon solving x = 1/2

Why have you considered the bold part in this particular question. for eg. if |x| = 2 we always consider x = +2/-2.
I am confused !

Your way is more common, orthodox. Though you should discard x=1/2 as you get this value when considering x<0 and 1/2 is not less than 0. Also, if x=1/2 then 3x-2<0 and you get that |x|=negative, which is never true.

Next, the way I'm ding is based on the fact that absolute value is always non negative, thus 3x-2 which equals to some absolute value must also be non-negative: 3x-2\geq{0} --> x\geq{\frac{2}{3}} --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.
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what is the value of x? [#permalink]  27 Aug 2012, 10:38
What is the value of x?

(1) |x| < 2

(2) |x| = 3x – 2

I marked E but i was wrong, my reasoning was as follows:

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0
x=3x-2
x=1

when x<0
-x=3x-2 ( |x|=-x if x is -ve)
x=1/2

again 2 values not sufficient.
Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong.
Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.
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Last edited by conty911 on 27 Aug 2012, 11:20, edited 2 times in total.
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Re: what is x? [#permalink]  27 Aug 2012, 10:42
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conty911 wrote:
Sorry, if this question has already been posted earlier, i tried searching the forum but it never came up.

What is x?

(1) |x| < 2

(2) |x| = 3x – 2

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0
x=3x-2
x=1

when x<0
-x=3x-2 ( |x|=-x if x is -ve)
x=1/2

again 2 values not sufficient.
Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong.
Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.

Merging similar topics. Please refer to the solution above.
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Re: what is x? [#permalink]  27 Aug 2012, 11:24
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Bunuel wrote:
conty911 wrote:
Sorry, if this question has already been posted earlier, i tried searching the forum but it never came up.

What is x?

(1) |x| < 2

(2) |x| = 3x – 2

statement one has a range (-2<x<2) of values so it is insufficient.

Statement 2: i solved for two cases

when x>0
x=3x-2
x=1

when x<0
-x=3x-2 ( |x|=-x if x is -ve)
x=1/2

again 2 values not sufficient.
Using 1 and 2 it also does not help either. SO I marked E. But somewhere i went wrong.
Can some one point out my mistake and also let me know how to best deal with combo of inequalities ad absolute values such as "|x| = 3x – 2" in DS problems.

Merging similar topics. Please refer to the solution above.

Thanks for merging the topic .
By the way the post is tagged in the wrong section , it should be in DS rather than PS.section?
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Whatever one does in life is a repetition of what one has done several times in one's life!
If my post was worth it, then i deserve kudos

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Joined: 02 Sep 2009
Posts: 12116
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Kudos [?]: 10132 [0], given: 965

Re: what is x? [#permalink]  27 Aug 2012, 11:48
conty911 wrote:
Thanks for merging the topic .
By the way the post is tagged in the wrong section , it should be in DS rather than PS.section?

Topic moved to the DS forum. Thank you.
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Re: what is x?   [#permalink] 27 Aug 2012, 11:48
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