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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 27 Dec 2017, 11:16
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A
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Difficulty:

  75% (hard)

Question Stats:

54% (02:29) correct 46% (02:19) wrong based on 112 sessions

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Is x > 0?

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

Source: Experts Global

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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 27 Dec 2017, 19:02
pushpitkc wrote:
Is x > 0?

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


Source: Experts Global


1....|3-x|<|x+5|
Square both sides
\(9+x^2-6x<x^2+25+10x....16x>-16....x>-1\)
x can be -0.5 or 0......Ans NO
x can be 1,2,3.... Ans YES
Insufficient

2..|3-2x|<x-1
Now |3-2x| will always be positive or 0, so x-1>0...x>1
Suff

B
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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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 27 Dec 2017, 19:10
St1: |3 - x| < |x + 5|
Square on both sides: 9 + x^2 - 6x < x^2 + 25 + 10x --> x > -1
Not Sufficient

St2: |3 - 2x| < x - 1
3 - 2x < x - 1 --> x > 4/3
3 - 2x > 1 - x --> x < 2
4/3 < x < 2
Hence x > 0
Sufficient

Answer: B
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Is x > 0?  [#permalink]

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New post 31 Dec 2017, 08:11
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]

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New post 01 Jan 2018, 03:00
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

Answer: B
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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 07 Jan 2018, 10:17
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]

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New post 07 Jan 2018, 10:48
2
adityanahan wrote:
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.


Hi adityanahan

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 &nbs [#permalink] 07 Jan 2018, 10:48
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