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# Is |x|<1 ?

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Is |x|<1 ? [#permalink]

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31 Dec 2009, 08:24
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Is |x|<1 ?

(1)|x + 1|= 2|x - 1|
(2) |x - 3| ≠ 0

Can anybody suggest a general approach to solve this kind of problems ? I am facing problems to solve these type of questions where equations are containing absolute values together with inequalities... please help.
[Reveal] Spoiler: OA

Last edited by msand on 31 Dec 2009, 10:30, edited 1 time in total.
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Re: DS question [#permalink]

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31 Dec 2009, 08:51
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Is $$|x| < 1$$?

Is $$|x| < 1$$, means is $$x$$ in the range (-1,1) or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
Two key points: $$x=-1$$ and $$x=1$$ (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

So we got TWO values of $$x$$ (two solutions): $$\frac{1}{3}$$ and $$3$$, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Check Walker's post: Absolute value and also to practice in absolute value questions check the link of Inequalities in my signature.

Hope it helps.
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Re: Tough Absolute value problem [#permalink]

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07 Jun 2010, 14:31
|X|<1 means -1<x<1

a. |x+2| = 2 |X-1|

Case 1. X+2 = 2(X-1) => X=3
Case 2. X+2 = -2(X-1) => X=1/3

Not Sufficient

b. | X-3| not equals 0 means X not equals 3 => Not sufficient

From both a & B,

X = 3, 1/3, and X not equals 3 => Hence X can have only one value X=1/3 which lies between -1 and 1

So, Ans. C
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Re: DS question [#permalink]

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08 Jun 2010, 06:16
Thank you Bunuel for your great explanation!

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Re: DS question [#permalink]

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26 Jun 2010, 03:14
Bunuel wrote:
Is $$|x| < 1$$?

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

I am not clear about how you derived this.
$$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$

Same with other two ranges (green and red)

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Re: DS question [#permalink]

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26 Jun 2010, 04:13
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JoyLibs wrote:
Bunuel wrote:
Is $$|x| < 1$$?

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

I am not clear about how you derived this.
$$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$

Same with other two ranges (green and red)

We have: $$|x + 1| = 2|x - 1|$$.

Absolute value properties:
If $$x\geq{0}$$, then $$|x|=x$$ and if $$x\leq{0}$$, then $$|x|=-x$$.

For the range $$x<-1$$ (blue range) --> $$x+1<0$$ so $$|x+1|=-(x+1)$$ and $$(x - 1)<0$$ so $$|x-1|=-(x-1)$$ --> $$|x + 1| = 2|x - 1|$$ becomes: $$-(x+1)=2(-x+1)$$

The same for other ranges.

Check this for more: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: DS question [#permalink]

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26 Jun 2010, 04:18
Thanks a lot

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Re: absolute value [#permalink]

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23 Oct 2010, 09:09
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tatane90 wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

(1) |x+1| = 2|x-1|
Lets solve this thing
+/-(x+1) = +/-2 (x-1)

Case 1 : x+1 = 2(x-1) .. x=3
Case 2 : x+1 = -2(x-1) .. 3x=1 .. x=1/3
Case 3 : -x-1 = 2(x-1) .. 3x=1 .. x=1/3
Case 4 : -x-1 = -2x+2 .. x=3

So x is either 3 or 1/3. Not enough to answer question

(2) This only implies x cannot be 3. Clearly insufficient

(1+2) x can only be 1/3. Sufficient to answer YES

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Re: how the answer C why not B [#permalink]

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28 Feb 2011, 09:35
Merging similar topics.
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Re: Is |x|<1 ? [#permalink]

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13 Sep 2013, 23:36
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Re: Is |x|<1 ? [#permalink]

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27 Nov 2013, 06:40
msand wrote:
Is |x|<1 ?

(1)|x + 1|= 2|x - 1|
(2) |x - 3| ≠ 0

Can anybody suggest a general approach to solve this kind of problems ? I am facing problems to solve these type of questions where equations are containing absolute values together with inequalities... please help.

Question --> Is |x|<1 ?
Is -1<x<1?

Statement 1
Square both sides
You end up with quadratic 3x^2-10x+30
Factorize (3x-1)(x-3)
x= 3 or x=1/3.
Two answers, therefore Insuff

Statement 2
x cannot be 3

Statement 1 and 2 together
Then it has to be x 1/3 and it is within the range so answer is YES

Suff

Hence C the correct answer
Cheers!
J
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Re: Is |x|<1 ? [#permalink]

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16 Dec 2014, 02:51
Hello from the GMAT Club BumpBot!

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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Re: Is |x|<1 ?   [#permalink] 16 Dec 2014, 02:51
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