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

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(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.

So the answer is (c).

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] New post 15 Sep 2011, 21:58
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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
Ques6.jpg [ 5.12 KiB | Viewed 2075 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.
Answer C.
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Re: Inequalities, Is |X| < 1 ? [#permalink] New post 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

==> The answer is C
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Re: Inequalities, Is |X| < 1 ? [#permalink] New post 23 Sep 2011, 08:10
Karishma,

In this question, you were going to add a link to read up more about the approach. Seems like you forgot to add the link. Can you please send me the link. Thank you.

etc. For details of this approach, check out:
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Re: Inequalities, Is |X| < 1 ? [#permalink] New post 23 Sep 2011, 20:24
Expert's post
shikari wrote:
Karishma,

In this question, you were going to add a link to read up more about the approach. Seems like you forgot to add the link. Can you please send me the link. Thank you.

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] New post 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] New post 25 Sep 2011, 22:13
Expert's post
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] New post 31 Dec 2012, 19:57
Expert's post
monir6000 wrote:
Is |x| < 1 ?
(1) |x + 1| = 2|x – 1|
(2) |x – 3| > 0


Answer is C.

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.

Please do add the OA while posting the questions.
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Re: Is |x|<1? [#permalink] New post 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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Inequality - Data Sufficiency Problem 3 [#permalink] New post 26 Feb 2014, 08:58
Is |x| < 1 ?
1. |x+1| = 2|x-1| 2. |x-3| > 0



How to approach and solve this kind of problem ..
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Re: Inequality - Data Sufficiency Problem 3 [#permalink] New post 26 Feb 2014, 14:02
Expert's post
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.
Answer = (C)

Does all this make sense?
Mike :-)
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Re: Inequality - Data Sufficiency Problem 3 [#permalink] New post 26 Feb 2014, 20:50
Expert's post
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
Ques3.jpg [ 8.77 KiB | Viewed 177 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.

Answer (C)
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Re: Inequality - Data Sufficiency Problem 3 [#permalink] New post 27 Feb 2014, 05:13
Expert's post
faceharshit wrote:
Is |x| < 1 ?
1. |x+1| = 2|x-1| 2. |x-3| > 0



How to approach and solve this kind of problem ..


Merging similar topics.

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Re: Inequality - Data Sufficiency Problem 3   [#permalink] 27 Feb 2014, 05:13
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