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# Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1

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ashikaverma13 wrote:
Is x > 0?

1) |3 - x| < |x+5|
2) |3-2x| < x - 1

Statement 1: $$|3-x|<|x+5|$$, square both sides to get

$$9+x^2-6x<x^2+25+10x$$

$$=>16x>-16 =>x>-1$$. Hence $$x>0$$ or $$x<0$$. Insufficient

Statement 2: implies $$-(x-1)<3-2x<x-1$$

$$=>3-2x<x-1 =>3x>4 =>x>\frac{4}{3}$$

and $$3-2x>1-x =>x<2$$

So range $$\frac{4}{3}<x<2$$. Clearly $$x>0$$. Sufficient

Option B
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink]
pushpitkc wrote:
Is x > 0?

(1) |3 - x| < |x + 5|
(2) |3 - 2x| < x - 1

Source: Experts Global

Another approach

(1) |3 - x| < |x + 5|

I spotted quickly the following:

Let x = 0 ......|3| < |5|..Answer is No

Let x =1 .......|2| < |6|..Answer is Yes

Insufficient

|3 - 2x| < x - 1

|3 - 2x| is equal or greater than zero.......x-1 is also equal or greater than zero

x -1 > 0...x >1>0

Sufficient

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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink]
For statement 1, how do we know that we need to square both side ? I was trying to solve it and could not get the answer.
any tips on identifying this.
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink]
2
Kudos
For statement 1, how do we know that we need to square both side ? I was trying to solve it and could not get the answer.
any tips on identifying this.

by squaring a mod function you can get rid of the mod and then it becomes a simple equation which is easier to work with.
when you remove the mod function, then you need to be careful that you test both the positive and negative values. but by squaring, you can turn the function to a positive value as square is always positive or 0
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink]
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink]
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