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

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pushpitkc
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Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink] New post  27 Dec 2017, 11:16
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Is x > 0?

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

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niks18
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink] New post  07 Jan 2018, 10:48
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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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chetan2u
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink] New post  27 Dec 2017, 19:02
Expert Reply
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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Kurtosis
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink] 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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niks18
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Is x > 0? [#permalink] 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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Mo2men
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink] 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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utopian25
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Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1 [#permalink] 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] New post  03 Dec 2019, 20:15
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