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Is x > 0?

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Manager
Joined: 04 Nov 2012
Posts: 52
Schools: NTU '16 (A)
Is x > 0?  [#permalink]

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Updated on: 01 Aug 2013, 03:43
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Difficulty:

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Question Stats:

64% (01:08) correct 36% (01:17) wrong based on 281 sessions

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Is $$x > 0$$?

(1) $$|x + 3| = 4x - 3$$

(2) $$|x + 1| = 2x - 1$$

Now, I approached the question in the following manner:

Statement 1:
|x+3|>=0 always
So,

4x-3>=0
so X>=3/4

Sufficient

Statement 2

similarly 2X-1>=0
x>1/2
Sufficient

So: D each statement is enough.

Can you tell me whether I am missing something conceptually?
Or a case in which this approach might fail?

Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??

Thanks..

Originally posted by 12bhang on 01 Aug 2013, 03:31.
Last edited by Zarrolou on 01 Aug 2013, 03:43, edited 2 times in total.
Edited the question.
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Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: Is x > 0?  [#permalink]

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06 May 2014, 07:52
3
3
jlgdr wrote:
Shortcut LHS, RHS for this one anyone?

Cheers!
J

Is $$x > 0$$?

(1) $$|x + 3| = 4x - 3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient.

(2) $$|x + 1| = 2x - 1$$. The smae here: LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$2x-1\geq{0}$$ --> $$x\geq{\frac{1}{2}}$$, hence $$x>0$$. Sufficient.

Answer: D. (You can see that you don't even need to find exact value(s) of x to answer the question.)

Similar question to practice: is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html
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Senior Manager
Joined: 24 Aug 2009
Posts: 469
Schools: Harvard, Columbia, Stern, Booth, LSB,
Re: Is x > 0?  [#permalink]

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01 Aug 2013, 03:41
4
12bhang wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x + 1| = 2x – 1

Statement 1
If |x + 3| >= 0 -----> x+3 = 4x-3 ----->3x=6----->x=2
If |x + 3| < 0 -----> -x-3 = 4x-3 ----->5x=0----->x=0
x=0 is not possible because we assumed that x+3<0 or x<-3. Thus the only possible value of x is 2.
Sufficient

Statement 2
If |x + 1| >= 0 -----> x+1 = 2x-1 ----->x=2----->x=2
If |x + 1| < 0 -----> -x-1 = 2x-1 ----->3x=0----->x=0
x=0 is not possible because we assumed that x+1<0 or x<-1. Thus the only possible value of x is 2.
Sufficient

Answer D
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Joined: 04 Nov 2012
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Re: Is x > 0?  [#permalink]

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01 Aug 2013, 06:04
2
fameatop wrote:
12bhang wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x + 1| = 2x – 1

Statement 1
If |x + 3| >= 0 -----> x+3 = 4x-3 ----->3x=6----->x=2
If |x + 3| < 0 -----> -x-3 = 4x-3 ----->5x=0----->x=0
x=0 is not possible because we assumed that x+3<0 or x<-3. Thus the only possible value of x is 2.
Sufficient

Statement 2
If |x + 1| >= 0 -----> x+1 = 2x-1 ----->x=2----->x=2
If |x + 1| < 0 -----> -x-1 = 2x-1 ----->3x=0----->x=0
x=0 is not possible because we assumed that x+1<0 or x<-1. Thus the only possible value of x is 2.
Sufficient

Answer D

Ok. that is the correct conventional way. Here's what i thought.

|x+3| will always be either > or = to 0.
So , 4x-3>=0
So, x>3/4, which is greater than 0.
So sufficient.

statement 2:
|x+1|>=0
So , 2x-1>=0
or x>=1/2
which is greater than 0
so sufficient

D

What I want to know is whether there is any loophole in my approach. I.e am i missing out on some hidden value of the modulus or something and whether it is better to solve for X and then check whether it falls in the range.

Please help.

Thanks
Senior Manager
Joined: 10 Jul 2013
Posts: 315
Re: Is x > 0?  [#permalink]

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03 Aug 2013, 04:32
12bhang wrote:
Is $$x > 0$$?

(1) $$|x + 3| = 4x - 3$$

(2) $$|x + 1| = 2x - 1$$

Now, I approached the question in the following manner:

Statement 1:
|x+3|>=0 always
So,

4x-3>=0
so X>=3/4

Sufficient

Statement 2

similarly 2X-1>=0
x>1/2
Sufficient

So: D each statement is enough.

Can you tell me whether I am missing something conceptually?
Or a case in which this approach might fail?

Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??

Thanks..

this is a wrong process. 3/4 or 1/2 never satisfies the two equations of statements.... use x = 3/4 0r 1/2 in those equations you will see these aren't the solutions.......
some good process already members publishes here, use them.........
thanks
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Asif vai.....

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Joined: 04 Nov 2012
Posts: 52
Schools: NTU '16 (A)
Re: Is x > 0?  [#permalink]

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04 Aug 2013, 02:53
Asifpirlo wrote:
12bhang wrote:
Is $$x > 0$$?

(1) $$|x + 3| = 4x - 3$$

(2) $$|x + 1| = 2x - 1$$

Now, I approached the question in the following manner:

Statement 1:
|x+3|>=0 always
So,

4x-3>=0
so X>=3/4

Sufficient

Statement 2

similarly 2X-1>=0
x>1/2
Sufficient

So: D each statement is enough.

Can you tell me whether I am missing something conceptually?
Or a case in which this approach might fail?

Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??

Thanks..

this is a wrong process. 3/4 or 1/2 never satisfies the two equations of statements.... use x = 3/4 0r 1/2 in those equations you will see these aren't the solutions.......
some good process already members publishes here, use them.........
thanks

Hi, Thanks for your insight.

I agree that the equation does not hold true at the given values. However, the question does not need us to find the actual values.

merely proving that the value of x>0 is enough. This will be proved, as the LHS in both statements are in the modulus and will return values that are greater than or equal to zero.
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Joined: 06 Sep 2013
Posts: 1720
Concentration: Finance
Re: Is x > 0?  [#permalink]

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26 Apr 2014, 11:56
Shortcut LHS, RHS for this one anyone?

Cheers!
J
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Joined: 06 Sep 2013
Posts: 1720
Concentration: Finance
Re: Is x > 0?  [#permalink]

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19 May 2014, 11:32
Bunuel wrote:
jlgdr wrote:
Shortcut LHS, RHS for this one anyone?

Cheers!
J

Is $$x > 0$$?

(1) $$|x + 3| = 4x - 3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient.

(2) $$|x + 1| = 2x - 1$$. The smae here: LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$2x-1\geq{0}$$ --> $$x\geq{\frac{1}{2}}$$, hence $$x>0$$. Sufficient.

Answer: D. (You can see that you don't even need to find exact value(s) of x to answer the question.)

Similar question to practice: is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html

Is it because you have 2x - 1 and therefore 2x>=1, x>=1/2?

What if you had 2x + 1? Will it still work? Don't thimk so right? We would need to analyze both scenarios

Please advice
Cheers!
J
Math Expert
Joined: 02 Sep 2009
Posts: 51218
Re: Is x > 0?  [#permalink]

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20 May 2014, 00:34
1
jlgdr wrote:
Bunuel wrote:
jlgdr wrote:
Shortcut LHS, RHS for this one anyone?

Cheers!
J

Is $$x > 0$$?

(1) $$|x + 3| = 4x - 3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient.

(2) $$|x + 1| = 2x - 1$$. The smae here: LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$2x-1\geq{0}$$ --> $$x\geq{\frac{1}{2}}$$, hence $$x>0$$. Sufficient.

Answer: D. (You can see that you don't even need to find exact value(s) of x to answer the question.)

Similar question to practice: is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html

Is it because you have 2x - 1 and therefore 2x>=1, x>=1/2?

What if you had 2x + 1? Will it still work? Don't thimk so right? We would need to analyze both scenarios

Please advice
Cheers!
J

We could still solve $$|x + 1| = 2x + 1$$ with this approach to get x. LHS is an absolute value, which is always non-negative, so RHS must also be non-negative --> $$2x+1\geq{0}$$ --> $$x\geq{-\frac{1}{2}}$$. For, $$x\geq{-\frac{1}{2}}$$, $$x+1>0$$, hence $$|x + 1| = x + 1$$ --> $$x + 1 = 2x + 1$$ --> $$x=0$$.

But if it were $$|x - 1| = 2x + 1$$, then getting $$x\geq{-\frac{1}{2}}$$ from LHS, wouldn't be useful to determine whether $$|x - 1| = x - 1$$ or $$|x - 1| = -(x - 1)$$ because $$x-1$$ could be positive as well as negative for different x'es which are more than or equal to -1/2. Therefore we should use usual method of solving for $$|x - 1| = 2x + 1$$, not RHS/LHS.

So, this method is not universal and can be applied only when non-modulus part defines x in such way that helps to get the sign of absolute value expression on the other side.

Does this make sense?
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Re: Is x > 0?  [#permalink]

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26 Sep 2018, 08:11
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