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# What is x?

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31 Jan 2012, 18:48
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What is x?

(1) |x| < 2
(2) |x| = 3x – 2
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Jul 2013, 00:52, edited 1 time in total.
Edited the question.
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31 Jan 2012, 18:56
12
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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]

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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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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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01 Feb 2012, 04:05
5
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1
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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 doing 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]

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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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27 Aug 2012, 10:42
2
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Expert's post
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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27 Aug 2012, 11:24
1
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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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27 Aug 2012, 11:48
Expert's post
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? (1) |x| < 2 (2) |x| = 3x – 2 [#permalink]

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11 Jul 2013, 00:51
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Think of statement 2 this way:

If we consider x to be negative, then -x=3x-2 which translates to x=1/2. Now when we test this situation 1/2=3*1/2-2 the outcome is =3/2-2= -1/2. The RHS is not equal to the LHS. Hence x has to be 1, because when we test x=3x-2 we get x=1.
and the RHS and LHS are equal to each other
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11 Jul 2013, 11:22
2
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What is x?

(1) |x| < 2

x<2 x>-2
Invalid, x could be any number greater than 2 or less than 2.
INSUFFICIENT

(2) |x| = 3x – 2
x>=0
|x| = 3x – 2
x = 3x - 2
-2x = -2
x = 1 Valid
OR
x<0
-x = 3x - 2
-4x = -2
x = (1/2) Invalid as 1/2 is not less than zero.
SUFFICIENT

(B)
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21 Jul 2013, 06:29
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

[Reveal] Spoiler: Variable inside absolute value
I solved equation 2 and got two values. Hence I thought 2 was insufficient. However, mgmat says "When dealing with absolute values that contain variables, you should always check that the solution is valid".They solve the equation, and then test the value once again in the equation and one the solutions is now invalid. Why is such a test required? Usually when you solve a equation, what you get is the SOLUTION right? Can someone please clarify why one solution will be 'invalid' in case of 'variables with absolute values'? I am unable to visualize that.
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21 Jul 2013, 06:32
Expert's post
chk86 wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

[Reveal] Spoiler: Variable inside absolute value
I solved equation 2 and got two values. Hence I thought 2 was insufficient. However, mgmat says "When dealing with absolute values that contain variables, you should always check that the solution is valid".They solve the equation, and then test the value once again in the equation and one the solutions is now invalid. Why is such a test required? Usually when you solve a equation, what you get is the SOLUTION right? Can someone please clarify why one solution will be 'invalid' in case of 'variables with absolute values'? I am unable to visualize that.

Merging similar topics. Please refer to the solution above.

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28 Nov 2013, 18:35
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,
I have a question.
(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

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

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28 Nov 2013, 19:53
sunita123 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,
I have a question.
(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

If it is true as above then why we are taking as -2<X<2 for |x| ,i mean why we are considering negative value (<0) for |X|?

We never considered negative value for |x|.... we considered negative value for just x...... By definition, mod never has a negative value....
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25 Mar 2014, 21:38
rohitgoel15 wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

When absolute values are given I guess it's better to go for number substitution.

1) |x| < 2 implies x can be either 1 or -1. 0 won't come here as it is neither positive nor negative and there is no need to take it here.
2) |x| = 3x - 2 is possible only when x = 1.

So, the final answer will be B
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15 Jul 2015, 10:53
Hi Bunuel

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

Can anyone justify the OA?

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>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify
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15 Jul 2015, 11:16
Expert's post
Hi Bunuel

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

Can anyone justify the OA?

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>=0 --> x>=2/3 --> x positive, so |x|=x and no need to consider |x|=-x --> x=3x-2 --> x=1.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

I dont think so. You need to follow the following 2 step procedure for ALL absolute value questions:

Treat |x| as following:

1. Let x $$\geq$$0 ----> |x| = x ---> x=3x-2 ---> x=1 and as x=1 satisfies x $$\geq$$0, this is a valid answer.

2. Let |x| < 0 ----> |x| = -x ----> -x=3x+2 ---> x = 1/2 and as x= 1/2 does not satisfy x<0, this solution is rejected. Thus we only 1 value, x = 1 and thus combining the 2 statements is sufficient. Thus C is the correct answer.

ok my concern is if by cosidering |x| as positive and solving equation , if we get x as -negative value then , will we put |x| = -x and solve equation as we did in case of +x. please clarify

If you consider |x| as positive (or >0) and you get x <0 , then this solution is not possible as you have 2 conflicting scenarios.

Hope this helps.
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15 Jul 2015, 22:48
HI Enge2012
I completely agree what you have mentioned, and that is way we normally solve mod equation, but if you see the Bunuel solution, he consider |x| as positive first and put equation >= 0 through which he gets +ve X and gain he solve the equation. Now my concern is over above equation what if we get -ve value of X example
instead of 3x-2>=0 if we have equation as 3x +2 >=0 now we get x as negative. hope i am cleared with doubt.
Re: What is x?   [#permalink] 15 Jul 2015, 22:48

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