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# Is x > 0?

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SVP
Joined: 05 Jul 2006
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Is x > 0? [#permalink]  15 Aug 2009, 13:48
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Difficulty:

55% (hard)

Question Stats:

58% (02:16) correct 42% (01:20) wrong based on 202 sessions
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
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Intern
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Re: Is x > 0? [#permalink]  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.
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Senior Manager
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Re: Is x > 0? [#permalink]  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

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Kudos [?]: 2506 [0], given: 359

Re: Is x > 0? [#permalink]  16 Aug 2009, 10:09
Expert's post
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.
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Senior Manager
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Posts: 309
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Kudos [?]: 339 [0], given: 22

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

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Is x > 0? [#permalink]  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.

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Re: Is x > 0? [#permalink]  23 Oct 2009, 18:42
Nice!Thank you so much!Kudos!:)
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Re: Is x > 0? [#permalink]  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.
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Re: Is x > 0? [#permalink]  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
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Re: Is x > 0? [#permalink]  17 Oct 2014, 23:54
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Re: Is x > 0? [#permalink]  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
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Re: Is x > 0? [#permalink]  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
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Re: Is x > 0?   [#permalink] 23 Nov 2014, 04:48
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