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Is |x|<1 ?

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Is |x|<1 ? [#permalink] New post 31 Dec 2009, 07:24
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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, 09:30, edited 1 time in total.
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Re: DS question [#permalink] New post 31 Dec 2009, 07:51
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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.

Answer: C.

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] New post 07 Jun 2010, 13:31
|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] New post 08 Jun 2010, 05:16
Thank you Bunuel for your great explanation!

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Re: DS question [#permalink] New post 26 Jun 2010, 02:14
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)

Appreciate your time.
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Re: DS question [#permalink] New post 26 Jun 2010, 03:13
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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)

Appreciate your time.


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] New post 26 Jun 2010, 03:18
Thanks a lot

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Re: absolute value [#permalink] New post 23 Oct 2010, 08:09
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tatane90 wrote:
Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

thanks in advance


(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

Answer : (c)
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Re: how the answer C why not B [#permalink] New post 28 Feb 2011, 08:35
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Re: Is |x|<1 ? [#permalink] New post 13 Sep 2013, 22:36
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Re: Is |x|<1 ? [#permalink] New post 27 Nov 2013, 05:40
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 :)
Re: Is |x|<1 ?   [#permalink] 27 Nov 2013, 05:40
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