GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 22 Oct 2019, 13:04

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

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

Author Message
TAGS:

### Hide Tags

Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3333
Location: India
GPA: 3.12
Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

### Show Tags

27 Dec 2017, 11:16
1
4
00:00

Difficulty:

75% (hard)

Question Stats:

53% (02:30) correct 47% (02:21) wrong based on 116 sessions

### HideShow timer Statistics

Is x > 0?

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

Source: Experts Global

_________________
You've got what it takes, but it will take everything you've got
Math Expert
Joined: 02 Aug 2009
Posts: 8006
Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

### Show Tags

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
_________________
Marshall & McDonough Moderator
Joined: 13 Apr 2015
Posts: 1683
Location: India
Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

### Show Tags

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

Retired Moderator
Joined: 25 Feb 2013
Posts: 1177
Location: India
GPA: 3.82

### Show Tags

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
SVP
Joined: 26 Mar 2013
Posts: 2343
Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

### Show Tags

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

Intern
Joined: 03 Nov 2017
Posts: 12
GMAT 1: 590 Q48 V23
Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

### Show Tags

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.
Retired Moderator
Joined: 25 Feb 2013
Posts: 1177
Location: India
GPA: 3.82
Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1  [#permalink]

### Show Tags

07 Jan 2018, 10:48
2
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
Re: Is x > 0? (1) |3 - x| < |x + 5| (2) |3 - 2x| < x - 1   [#permalink] 07 Jan 2018, 10:48
Display posts from previous: Sort by