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pushpitkc
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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 27 Dec 2017, 11:16
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

53% (02:20) correct 47% (02:20) wrong based on 135 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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B
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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 27 Dec 2017, 19:02
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pushpitkc
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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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 27 Dec 2017, 19:10
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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? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 31 Dec 2017, 08:11
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ashikaverma13
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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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 01 Jan 2018, 03:00
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pushpitkc
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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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 07 Jan 2018, 10:17
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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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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 07 Jan 2018, 10:48
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adityanahan
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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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 03 Dec 2019, 20:15
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