It is currently 14 Dec 2017, 22:37

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# Is x > 0?

Author Message
TAGS:

### Hide Tags

Retired Moderator
Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [1], given: 49

### Show Tags

15 Aug 2009, 13:48
1
KUDOS
2
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

58% (01:08) correct 42% (01:17) wrong based on 243 sessions

### HideShow timer Statistics

Is x > 0?

(1) |x + 3| = 4x – 3

(2) |x + 1| = 2x – 1

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-0-1-x-3-4x-3-2-x-1-2x-1-can-100357.html
[Reveal] Spoiler: OA

Kudos [?]: 447 [1], given: 49

Intern
Joined: 01 Aug 2009
Posts: 29

Kudos [?]: 18 [0], given: 3

Location: Australia
Re: Is x > 0? [#permalink]

### Show Tags

16 Aug 2009, 08:08
IMO D.

Statement 1: x = 2
For (x+3) >= 0
x >= -3 [Not a solution as it doesn't satisfies the condition]
x+3=4x-3
x = 2

For (x+3) < 0
x < -3 [Not a solution as it doesn't satisfies the condition]
-(x+3) = 4x-3
x = 0 [Not a solution as it doesn't satisfies the condition]

Statement 2: x = 2
For (x+1) >= 0
x >= -1 [Not a solution as it doesn't satisfies the condition]
x+1 = 2x-1
x = 2

For (x+1) < 0
x < -1 [Not a solution as it doesn't satisfies the condition]
-(x+1) = 2x - 1
x = 0 [Not a solution as it doesn't satisfies the condition]

To save time, we should ignore all -ve cases of x as LHS is +ve and equal to RHS. So, for RHS to be positive x must be +ve.
_________________

The three most significant times in your life are:
1. When you fall in love
2. The birth of your first child
3. When you prepare for your GMAT

Kudos [?]: 18 [0], given: 3

Senior Manager
Joined: 17 Mar 2009
Posts: 299

Kudos [?]: 611 [0], given: 22

Re: Is x > 0? [#permalink]

### Show Tags

16 Aug 2009, 09:45
stmt1

|x+3| = 4x - 3
when |x+3|>0
x+3= 4x -3
3x-6=0
x=2

when |x+3|<0
-(x+3) = 4x-3
5x=0,x=0

not sufficient

stmt2
|x+1| = 2x - 1
when |x+1|>0
x+1=2x-1
x=2

when |x+1|<0
-(x+1)=2x-1
3x=0, x=0
not sufficient

Kudos [?]: 611 [0], given: 22

CEO
Joined: 17 Nov 2007
Posts: 3583

Kudos [?]: 4718 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Re: Is x > 0? [#permalink]

### Show Tags

16 Aug 2009, 10:09
crejoc wrote:
stmt1
...
when |x+3|<0
-(x+3) = 4x-3
5x=0,x=0
....

Here is a typical mistake with moduli. You forgot to check your answer.
It is possible to use -(x+3) only if |x+3|<0 ! But at x=0, |0+3| > 0.
_________________

HOT! GMAT TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

Kudos [?]: 4718 [0], given: 360

Senior Manager
Joined: 17 Mar 2009
Posts: 299

Kudos [?]: 611 [0], given: 22

Re: Is x > 0? [#permalink]

### Show Tags

16 Aug 2009, 10:46
walker wrote:
crejoc wrote:
stmt1
...
when |x+3|<0
-(x+3) = 4x-3
5x=0,x=0
....

Here is a typical mistake with moduli. You forgot to check your answer.
It is possible to use -(x+3) only if |x+3|<0 ! But at x=0, |0+3| > 0.

thanks walker for pointing out the error,.. it was a stupid mistake..

Kudos [?]: 611 [0], given: 22

Director
Joined: 25 Oct 2008
Posts: 593

Kudos [?]: 1206 [0], given: 100

Location: Kolkata,India
Re: Is x > 0? [#permalink]

### Show Tags

23 Oct 2009, 17:28
The question stem asks is x>0??

Stmt 1:|x + 3| = 4x – 3
Two cases:
a. x+3=4x-3 ,x=0
b. -(x+3)=4x-3 ,x=0.

Both cases gives us x=0 hence SUFF.

Stmt 2:|x + 1| = 2x – 1
Again two cases:
a.x+1=2x-1
x-2x=-1-1
x=2

b. -(x+1)=2x-1
x=0
Both case gives us diff. values hence insuff.
IMO: A

_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Kudos [?]: 1206 [0], given: 100

Math Expert
Joined: 02 Sep 2009
Posts: 42606

Kudos [?]: 135685 [4], given: 12706

### Show Tags

23 Oct 2009, 18:38
4
KUDOS
Expert's post
4
This post was
BOOKMARKED
tejal777 wrote:
The question stem asks is x>0??

Stmt 1:|x + 3| = 4x – 3
Two cases:
a. x+3=4x-3 ,x=0
b. -(x+3)=4x-3 ,x=0.

Both cases gives us x=0 hence SUFF.

Stmt 2:|x + 1| = 2x – 1
Again two cases:
a.x+1=2x-1
x-2x=-1-1
x=2

b. -(x+1)=2x-1
x=0
Both case gives us diff. values hence insuff.
IMO: A

First you made a mistake in calculation: see the red part, solution there is x=2.
Second, which is more important the way of solving, from my point of view, is incorrect.

Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient.

(2) $$|x+1|=2x-1$$ --> the same here --> $$2x-1\geq{0}$$ --> $$x\geq{\frac{1}{2}}$$, hence $$x>0$$. Sufficient.

So you see that you don't even need to find exact value(s) of $$x$$ to answer the question.

_________________

Kudos [?]: 135685 [4], given: 12706

Director
Joined: 25 Oct 2008
Posts: 593

Kudos [?]: 1206 [0], given: 100

Location: Kolkata,India
Re: Is x > 0? [#permalink]

### Show Tags

23 Oct 2009, 18:42
Nice!Thank you so much!Kudos!:)
_________________

http://gmatclub.com/forum/countdown-beginshas-ended-85483-40.html#p649902

Kudos [?]: 1206 [0], given: 100

Senior Manager
Affiliations: PMP
Joined: 13 Oct 2009
Posts: 296

Kudos [?]: 177 [0], given: 37

Re: Is x > 0? [#permalink]

### Show Tags

23 Oct 2009, 19:03
I used substituting values as below and got D as answer:

S1)

|x+3| = 4x-3
when x=
-2 eq becomes 1 = -7 -->absurd
-1 eq becomes 2 = -7 -->absurd
0 eq becomes 3 = -3 -->absurd
1 eq becomes 5 = 5
2 eq becomes 6 = 9 -->absurd

so we can tell only x=2 satisifies the eq and hence S1 is sufficient.

similarly for S2)

|x+1| = 2x -1

when x =
-2 eq becomes 1 = -5 -->absurd
-1 eq becomes 0 = -3 -->absurd
0 eq becomes 1 = -1 -->absurd
1 eq becomes 2= 0 -->absurd
2 eq becomes 3 = 3
3 eq becomes 4 = 5 -->absurd

again we can tell x = 2 alone satisifes eq, hence S2 is sufficient.
_________________

Thanks, Sri
-------------------------------
keep uppp...ing the tempo...

Press +1 Kudos, if you think my post gave u a tiny tip

Kudos [?]: 177 [0], given: 37

Intern
Joined: 11 Dec 2012
Posts: 36

Kudos [?]: [0], given: 59

Re: Is x > 0? [#permalink]

### Show Tags

09 Jan 2013, 10:03
One more easier exp.

Stmt (1)
First we squire both sides of the equality:
|x + 3| = 4x – 3
x^2+6x+9=16x^2-24x+9
15x^2-30x=0 divide both sides to 15
and we have:
x^2-2x=0
x(x-2)=0
x=0
but x=0 is not a root of the equation since |0 + 3| is not equal to 4*0 – 3= -3 because absolute value cannot equal to digit below zero
x-2=0
x=2 is a root of the equality, since |2 + 3| = 4*2 – 3=> 5 both sides
hence (1) is sufficient

stmt (2)
|x + 1| = 2x – 1
squire both sides again and we have:
x^2+2x+1=4x^2-4x+1
3x^2-6x=0, divide both sides to 3 and have:
X^2-2x=0
x(x-2)=0
x=0
but x=0 is not a root of the equation since |0 + 1| is not equal to 2*0 – 1= -1 because absolute value cannot equal to digit below zero
x-2=0
x=2 is a root of the equality, since |2 + 1| = 2*2 – 1=> 3 both sides

hence stmt (2) is sufficient

D

Kudos [?]: [0], given: 59

Non-Human User
Joined: 09 Sep 2013
Posts: 14832

Kudos [?]: 287 [0], given: 0

Re: Is x > 0? [#permalink]

### Show Tags

17 Oct 2014, 23:54
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 287 [0], given: 0

Intern
Joined: 09 Aug 2014
Posts: 22

Kudos [?]: 1 [0], given: 59

Re: Is x > 0? [#permalink]

### Show Tags

22 Nov 2014, 19:32
crejoc wrote:
stmt1

|x+3| = 4x - 3
when |x+3|>0
x+3= 4x -3
3x-6=0
x=2

when |x+3|<0
-(x+3) = 4x-3
5x=0,x=0

not sufficient

stmt2
|x+1| = 2x - 1
when |x+1|>0
x+1=2x-1
x=2

when |x+1|<0
-(x+1)=2x-1
3x=0, x=0
not sufficient

seem you go wrong with it
Statement 1

|x+3| = 4x - 3
when x+3>=0 <->x>=-3
thus
x+3= 4x -3
3x-6=0
x=2 ( satisfy with condition x>=-3)

when x+3<0 <-> x<-3
-(x+3) = 4x-3
5x=0 -> x=0 ( dissatisfy with condition x<-3)

=> Statement 1 sufficient

Statement 2
do the same
=> statement 2 sufficient

Kudos [?]: 1 [0], given: 59

Math Expert
Joined: 02 Sep 2009
Posts: 42606

Kudos [?]: 135685 [1], given: 12706

Re: Is x > 0? [#permalink]

### Show Tags

23 Nov 2014, 04:48
1
KUDOS
Expert's post
Is x > 0?

(1) $$|x+3|=4x-3$$ --> LHS is an absolute value, which is always non-negative ($$|some \ expression|\geq{0}$$), so RHS must also be non-negative --> $$4x-3\geq{0}$$ --> $$x\geq{\frac{3}{4}}$$, hence $$x>0$$. Sufficient.

(2) $$|x+1|=2x-1$$ --> the same here --> $$2x-1\geq{0}$$ --> $$x\geq{\frac{1}{2}}$$, hence $$x>0$$. Sufficient.

So you see that you don't even need to find exact value(s) of $$x$$ to answer the question.

OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-0-1-x-3-4x-3-2-x-1-2x-1-can-100357.html
_________________

Kudos [?]: 135685 [1], given: 12706

Re: Is x > 0?   [#permalink] 23 Nov 2014, 04:48
Display posts from previous: Sort by