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# Is |x|<1?

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Senior Manager
Joined: 08 Nov 2010
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Re: Inequalities, Is |X| < 1 ?  [#permalink]

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14 Sep 2011, 07:08
mustu wrote:
scbguy wrote:
I see the answer as A, obviously I'm wrong but I don't see how x is 1/3 in statement 1

Posted from GMAT ToolKit

(1) |x + 1| = 2|x – 1|

This has 2 cases.. X>0 and X<0
If X>0 , then X+1 = 2(x-1)
If X<0 , then X+1 = -2(x-1)

Solving these equations we get X= 3 or X= 1/3. Since we have YES and NO situation => Not sufficient

(2) |x – 3| > 0

Solving this equation , we get x>3 or X<3, in either cases, X<> 3. So not sufficient.

(1) + (2) ==> X= 1/3 . Since X<> 3.

Regards,
Mustu

Mustu - if im not wrong - u should check not for X><0 but X<>1,-1
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Re: Inequalities, Is |X| < 1 ?  [#permalink]

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15 Sep 2011, 21:58
5
mustu wrote:
Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

As I have said before, most mod questions are best tackled using a number line. You don't need to do many calculations then.
|x| means the distance from 0.
|x-3| means the distance from 3.
etc. For details of this approach, check out:

Let's go on to this question now.
Is |x| < 1 i.e. Is the distance of point x from 0 less than 1?

Statement 1: |x + 1| = 2|x – 1|
This means 'distance of x from -1 is twice the distance of x from 1'. Draw the number line now. There will be 2 points where the distance from -1 will be twice the distance from 1.
Attachment:

Ques6.jpg [ 5.12 KiB | Viewed 3250 times ]

For one of these points, distance from 0 is less than 1, for the other it is more than 1. So not sufficient.

Statement 2: |x – 3| > 0
This statement tells us that distance of x from 3 is more than 0 i.e. x does not lie at 3. It can lie anywhere else.
You can look at it in another way: Mods are always more than or equal to 0. All this statement tells us is that this mod is not equal to zero i.e. x is not equal to 3.
For some of these points, distance from 0 will be less than 1, for the others it will be more than 1. So not sufficient.

Using both statements together, statement 1 says that x is either 3 or a point between 0 and 1 (which I don't really need to calculate). Statement 2 tells us that x is not 3. So together, x must be a point between 0 and 1 and its distance from 0 must be less than 1. Sufficient.
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Re: Inequalities, Is |X| < 1 ?  [#permalink]

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15 Sep 2011, 22:23
mustu wrote:
Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

(1) There are 2 cases:
1/ x+1= 2(x-1) ---> x=3
2/ x+1= -2(x-1) ---> x=1/3
(2) X is not 3

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Re: Inequalities, Is |X| < 1 ?  [#permalink]

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23 Sep 2011, 08:10
Karishma,

etc. For details of this approach, check out:
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Re: Inequalities, Is |X| < 1 ?  [#permalink]

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23 Sep 2011, 20:24
shikari wrote:
Karishma,

etc. For details of this approach, check out:

I apologize. Here you go:

http://www.veritasprep.com/blog/2011/01 ... edore-did/
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Re: Inequalities, Is |X| < 1 ?  [#permalink]

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24 Sep 2011, 17:40
Am I doing this right:

Statement 1:

(1) |x + 1| = 2|x – 1|

(x+1)=2(x-1)
x+1=2x-2
3=x (NO)
&
(x+1)=-2(x-1)
x+1=-2x+2
3x=1
x=1/3 (YES)

Insufficient
B,C,orE

Statement 2

(2) |x – 3| > 0

(x-3)>0
x>3 (NO)

-(x-3)>0
-x+3>0
-x>-3
x<3 (MAYBE)

Not sufficient

C or E

(

Combined:

Statement 1: x=1/3,3
Statement 2: x <> 3

Since x CANNOT equal 3, x = 1/3

Since |1/3| < 1, both statements are sufficient to answer the prompt.

C

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Re: Inequalities, Is |X| < 1 ?  [#permalink]

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25 Sep 2011, 22:13
This is fine as long as you know why you are doing this:

4LEX wrote:
Am I doing this right:

Statement 1:

(1) |x + 1| = 2|x – 1|

(x+1)=2(x-1) (When both the terms are positive, x > 1)
x+1=2x-2
3=x (NO) (Valid value for x since 3 >1)
&
(x+1)=-2(x-1) (When -1 < x < 1, (x+1) is positive but (x-1) is negative so you are put a negative sign here)
x+1=-2x+2
3x=1
x=1/3 (YES) (Valid value since -1 < 1/3 < 1)

There would be another case x < -1. In that case both the terms will be negative.
-(x+1)=-2(x-1)
giving x = 3 (Not a valid value since 3 is not less than -1)
I am assuming that you saw the two negatives will get canceled out and give x = 3 which will not be valid so you skipped this step. In some questions, you could get a valid value here.
So you have only 2 values for x (3 and 1/3).

Insufficient
B,C,orE

Statement 2

(2) |x – 3| > 0

(x-3)>0
x>3 (NO)

-(x-3)>0
-x+3>0
-x>-3
x<3 (MAYBE)

Not sufficient

C or E

Combined:

Statement 1: x=1/3,3
Statement 2: x <> 3

Since x CANNOT equal 3, x = 1/3

Since |1/3| < 1, both statements are sufficient to answer the prompt.

C

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Re: Is |x| < 1 ?  [#permalink]

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31 Dec 2012, 19:57
monir6000 wrote:
Is |x| < 1 ?
(1) |x + 1| = 2|x – 1|
(2) |x – 3| > 0

From statement 1, we are able to get two values of x; they are $$x=3$$ and $$x=1/3$$. Two values of x, hence insufficient.
From statement 2, all we know is that the distance from x is more than 0 or it indirectly implies that x is not 0. Not enough information. Hence insufficient.

On combining these two statements, we come to know that x cannot be 3 and x=1/3. Since $$1/3$$ < 1, hence $$|x|<1$$.
+1C.

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07 Feb 2013, 12:22
This is my approach:
Is |x|<1?
1st start from statement 2, cause it is easier,
|x – 3| > 0 just tell us x is note equal to 3, so it is insufficient to solve the target question
2nd for statement 2: |x + 1| = 2|x – 1|
we have to separate the condition to x<-1, -1<x<1, x>1 that is , |x|<1 and |x|> 1 to do further thinking
1) when |x|<1, we could know we will get the solution in this range after solving equation, thus get the answer "YES" for question |x|<1
2) when |x|>1, we could know we will get the same answer in x<-1 and x>1 condition, and we could assure the answer is "NO" for target question
so based on above, statement 2 is insufficient to solve the target question

we only left option C and E now.
To test whether statements together will help to solve target question, we could use the denied solution x=3 in statement 2 to statement 1 to see whether it is one of the two solutions of equation.

If it is one of the solutions, then statement 2 will help to reduce the two solutions to one, thus, support the target question. We could feel free to choose option C
If it is not one of the solution, then statement 2 will not help to reduce the number of solutions, thus, we could feel free to choose option E.

Let us test now.
LS 3+1|=4 RS:2*|3-1|=4,
we could know x=3 is one of the two answers.
Thus we could choose C
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Joined: 28 Dec 2011
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Re: Inequality - Data Sufficiency Problem 3  [#permalink]

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26 Feb 2014, 14:02
faceharshit wrote:
Is |x| < 1 ?
1. |x+1| = 2|x-1| 2. |x-3| > 0

How to approach and solve this kind of problem ..

Dear faceharshit,
I'm happy to respond. I dare say, this problem is a little bit harder than what the GMAT will ask of you.

Statement #1: |x+1| = 2|x-1|
If we are given |P| = |Q|, this means: P = Q OR P = -Q. Notice that the word "or" is not a piece of garnish there: rather, it is an essential piece of mathematical equipment.

|x + 1| = 2|x - 1|

Case I
(x + 1) = 2(x - 1)
x + 1 = 2x - 2
x = 3

Case II
(x + 1) = -2(x - 1)
x + 1 = -2x + 2
3x = 1
x = 1/3

This, from statement #1, we have x = 3 or x = 1/3. With this, we do not have sufficient information to answer the prompt question. This statement, by itself, is insufficient.

Statement #2: |x-3| > 0
Forget about everything we did in statement #1. Here, x could equal 10, in which case |x| is not less than 1, or x could equal 0, in which cases |x| is less than 1. We can pick different values that satisfy |x-3| > 0, x = 10 and x = 0, that give two different answers to the prompt question. Therefore, we do not have sufficient information to answer the prompt question. This statement, by itself, is insufficient.

Combined:
#1 gives us x = 3 or x = 1/3
The value x = 3 does not satisfy the second statement, so we reject that value.
The value x = 1/3 is only value that satisfies both statements, and with this, |x| < 1.

Combined, the statements are sufficient.

Does all this make sense?
Mike
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Re: Inequality - Data Sufficiency Problem 3  [#permalink]

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26 Feb 2014, 20:50
faceharshit wrote:
Is |x| < 1 ?
1. |x+1| = 2|x-1| 2. |x-3| > 0

How to approach and solve this kind of problem ..

Use the number line to solve it quickly. Check: http://www.veritasprep.com/blog/2011/01 ... edore-did/

'Is |x| < 1' implies 'Is distance of x from 0 less than 1?' i.e. does x lie within -1 and 1 (excluding the points -1 and 1)?

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

This tells you that distance of x from -1 is twice the distance of x from 1. There are two values of x for which this is possible:
Attachment:

Ques3.jpg [ 8.77 KiB | Viewed 1224 times ]

The red line is twice the length of the blue line in both the cases. For the first case, x lies somewhere between 0 and 1 but for the second case, x lies at 3. Hence we can't answer whether x will lie between -1 and 1 from this statement alone.

2. |x-3| > 0
This tells us that x is a point whose distance from 3 is more than 0. That means it is not at 3 but on its left or right. This statement alone doesn't tell us whether x lies between -1 and 1.

Both statements together: Stmnt 1 tells us that x lies between -1 and 1 or at 3. Stmnt 2 tells us that x doesn't lie at 3. Then there is only one option left: x must lie between -1 and 1.

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Re: Is |x|<1?   [#permalink] 06 Sep 2018, 09:43

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