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

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

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New post Updated on: 16 Oct 2019, 07:05
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Is x > 0 ?

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

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

Can some one please explain the best way to tackle inequalities with modulus. A solution that would not require substitution...

Originally posted by sachin1114GMCKS on 16 Oct 2019, 06:30.
Last edited by Bunuel on 16 Oct 2019, 07:05, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is x > 0 ? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

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New post 17 Oct 2019, 06:53
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Is x > 0 ?

(1) |3 - x| < |x + 5|
We have MODs on both sides, so both sides are Non-negative, and we can square the two sides..
\(|3 - x|^2 < |x + 5|^2=x^2-6x+9<x^2+10x+25.......16x>-14....x>\frac{-14}{16}\)
Thus, x can be -1/2. -1/16, 0, or 10 etc
Insuff

(2) |3 - 2x| < x - 1
Just looking at the inequality should tell you that x>1..
The RHS has to be greater than 0 as |3-2x| will ne Non-negative..So, x-1>0..x>1
Let us do by opening MOD..
\(3-2x<x-1.....3x>4.....x>\frac{4}{3}>0\)..
\(2x-3<x-1......x<2\)..
So, \(\frac{4}{3}<x<2\)......Yes, x>0
Suff

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

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New post 16 Oct 2019, 11:18
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1
Is x > 0 ?

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

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

From (1)
Case I: Opening mod for same sign
3-x < x+5
=> -2 < 2x
=> -1< x

Case II: Opening mod for different signs
3-x < -x-5
Not possible

So statement (1) is Insufficient

From Statement (2)
3-2x < x-1
=> 4/3 < x

& -3+2x < x-1
=> x < 2

So 4/3 < x < 2
Definitely x>0 Sufficient

Answer(B)

Hope this is helpful.
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Re: Is x > 0 ? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

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New post 16 Oct 2019, 12:30
Eternal wrote:
Is x > 0 ?

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

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

From (1)
Case I: Opening mod for same sign
3-x < x+5
=> -2 < 2x
=> -1< x

Case II: Opening mod for different signs
3-x < -x-5
Not possible

So statement (1) is Insufficient

From Statement (2)
3-2x < x-1
=> 4/3 < x

& -3+2x < x-1
=> x < 2

So 4/3 < x < 2
Definitely x>0 Sufficient

Answer(B)

Hope this is helpful.



Thanks for writing.

The Answer is B as you correctly pointed out. But in Scenario one, we have two mods and that is what confuses me. I think we need to have 4 equations for two mods. But doing that is very time confusing.

Hope someone sheds some light on that. :)
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Is x > 0 ? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

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New post 16 Oct 2019, 13:37
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Consider this equation for example.

|x-2| = |2x-3|
What are the possible values of x?

Overall four cases

I) +VE, +VE
(x-2)=(2x-3)

II) +VE, -VE
(x-2)= -(2x-3)

III) -VE, +VE
-(x-2)=(2x-3)

IV) -VE, -VE
-(x-2)= -(2x-3)

If you look carefully or try to solve,

Equation (I) and (IV) are same

Also (II) and (III) are same

So we can take only two cases

Case (A): Same signs
Case (B): Different signs

Hope this helps.
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Re: Is x > 0 ? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

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New post 17 Oct 2019, 05:12
Hello nice explanation but note that your case 11 should be
-(3-x) < x+5 for different signs

Since +ve can never be less than -ve as you have indicated
3-x < -(x+5)

Thanks :)
Eternal wrote:
Is x > 0 ?

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

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

From (1)
Case I: Opening mod for same sign
3-x < x+5
=> -2 < 2x
=> -1< x

Case II: Opening mod for different signs
3-x < -x-5
Not possible

So statement (1) is Insufficient

From Statement (2)
3-2x < x-1
=> 4/3 < x

& -3+2x < x-1
=> x < 2

So 4/3 < x < 2
Definitely x>0 Sufficient

Answer(B)

Hope this is helpful.


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Re: Is x > 0 ? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1   [#permalink] 17 Oct 2019, 05:12
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