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# What is x? (1) |x| < 2 (2) |x| = 3x 2

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Director
Joined: 11 Jun 2007
Posts: 595
What is x? (1) |x| < 2 (2) |x| = 3x 2 [#permalink]

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20 Oct 2007, 21:50
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

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Director
Joined: 11 Jun 2007
Posts: 891
Re: MGMAT Math Modulus [#permalink]

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20 Oct 2007, 22:09
eyunni wrote:
What is x?

(1) |x| < 2

(2) |x| = 3x – 2

i get E

(1)
for x > 0 => x < 2
for x < 0 => x > -2
so range of x = -2 < x < 2
not sufficient

(2)
for x > 0 => x = 3x - 2
2x = 2 : x = 1
for x < 0 => -x = 3x - 2
4x = 2 : x = 1/2
x could be 1 or 1/2, not sufficient

putting (1) and (2) together, both 1 and 1/2 fits in -2 < x < 2
so both not sufficient
Director
Joined: 03 May 2007
Posts: 832
Schools: University of Chicago, Wharton School
Re: MGMAT Math Modulus [#permalink]

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Updated on: 20 Oct 2007, 22:14
eyunni wrote:
What is x?

(1) |x| < 2
(2) |x| = 3x – 2

E.

From 1, x could be anything smaller than 2 but grater than -2. nsf
From 2, x = 1/2 and 1. nsf

from 1 and 2 also nsf.

Originally posted by Fistail on 20 Oct 2007, 22:11.
Last edited by Fistail on 20 Oct 2007, 22:14, edited 3 times in total.
VP
Joined: 28 Mar 2006
Posts: 1329

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20 Oct 2007, 22:14
Fistail wrote:
trivikram wrote:
yes it should be B

how???

Mistake....I messed a calculation here..

squaring both sides in (B) and applyingA would lead us to E
SVP
Joined: 01 May 2006
Posts: 1786

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21 Oct 2007, 01:27
Perfect explanation from beckee529 .... Nothing to add
Manager
Joined: 07 Mar 2007
Posts: 189

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21 Oct 2007, 09:38
Statement 1 is an absolute so we have 2 possibilities. When X is positive and when X is negative.

So, |x| < 2
1. x<2 or
-x<2>-2

So, when x > 0 => x < 2
And if x <0> x > -2
Manager
Joined: 27 Jul 2007
Posts: 103

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21 Oct 2007, 10:05
Statement 1 is an absolute so we have 2 possibilities. When X is positive and when X is negative.

So, |x| < 2
1. x<2 or
-x<2>-2

So, when x > 0 => x < 2
And if x <0> x > -2

But from stmt 2
for x > 0 => x = 3x - 2
2x = 2 : x = 1
for x <0> -x = 3x - 2
4x = 2 : x = 1/2

here with x<0 we r gettin x=1/2 ????
Manager
Joined: 27 Jul 2007
Posts: 103

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21 Oct 2007, 10:07
Statement 1 is an absolute so we have 2 possibilities. When X is positive and when X is negative.

So, |x| < 2
1. x<2 or
-x<2>-2

So, when x > 0 => x < 2
And if x <0> x > -2

But from stmt 2
for x > 0 => x = 3x - 2
2x = 2 : x = 1
for x <0> -x = 3x - 2
4x = 2 : x = 1/2

here with x<0 we r gettin x=1/2 ????
Manager
Joined: 07 Mar 2007
Posts: 189

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21 Oct 2007, 10:24
Yes you're right.
Because when you take the absolute into consideration, the equations is:
-x=3x-2 --> -4x=-2--> x=1/2
Manager
Joined: 07 Mar 2007
Posts: 189

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22 Oct 2007, 10:27
But How can B alone be sufficient? We get two values - x=1 or x=1/2

Eyunni- could you please share the source of this question?
Manager
Joined: 18 Jun 2007
Posts: 55

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22 Oct 2007, 11:04
The question probably assumed X was an integer.
Manager
Joined: 07 Mar 2007
Posts: 189

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22 Oct 2007, 11:23
But in DS how can we ASSUME x is an integer? I thought we had to be told if this was the case. Am i wrong?
SVP
Joined: 01 May 2006
Posts: 1786

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22 Oct 2007, 12:10
(B) it is ... I didnt look in all details of the explanations given precedently...

What is x?
(1) |x| < 2
(2) |x| = 3x – 2

Stat1
|x| < 2
<=> -2 < x < 2

INSUFF.

Stat1
|x| = 3x – 2

o If x >= 0, then
|x| = 3x – 2
<=> x = 3x - 2
<=> x = 1 >>>> Solution ok as 1 > 0 (x>=0)

o If x < 0, then
|x| = 3x – 2
<=> -x = 3x - 2
<=> x = 1/2 >>>> Solution out as x must be negative 0

So, we have unique solution for x : 1

SUFF.
SVP
Joined: 01 May 2006
Posts: 1786

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22 Oct 2007, 12:12
But in DS how can we ASSUME x is an integer? I thought we had to be told if this was the case. Am i wrong?

Actually u are right. If not stated, we can never assume that a variable is an integer.

Here, it's just because x = 1/2 is ruled out as x < 0.
Director
Joined: 11 Jun 2007
Posts: 595

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22 Oct 2007, 12:43
But How can B alone be sufficient? We get two values - x=1 or x=1/2

Eyunni- could you please share the source of this question?

That was a touch tricky. I don't remember the source exactly but I guess it is from MGMAT/Kaplan.
Manager
Joined: 28 Apr 2008
Posts: 110
Re: MGMAT Math Modulus [#permalink]

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25 Nov 2008, 12:15
B

1--> -2<x<2 insuff

2-->
x=3x-2, x>0
x=2-3x, x<0

for x<0, 2-3x is allways +ve--> thus x>0

sufficient
VP
Joined: 05 Jul 2008
Posts: 1333

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25 Nov 2008, 21:24
Fig wrote:
(B) it is ... I didnt look in all details of the explanations given precedently...

What is x?
(1) |x| -2 = 0, then[/b]
|x| = 3x – 2
x = 3x - 2
x = 1 >>>> Solution ok as 1 > 0 (x>=0)

[b]o If x -x = 3x - 2
x = 1/2 >>>> Solution out as x must be negative 0

So, we have unique solution for x : 1

SUFF.

Whew! What the Q did is lot of trickery. Lets us start with x 0 and then rule out that case as it leads to a contradiction.

--== Message from GMAT Club Team ==--

This is not a quality discussion. It has been retired.

If you would like to discuss this question please re-post it in the respective forum. Thank you!

To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative | Verbal Please note - we may remove posts that do not follow our posting guidelines. Thank you.
Re:   [#permalink] 25 Nov 2008, 21:24
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# What is x? (1) |x| < 2 (2) |x| = 3x 2

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