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Is |x|<1? 1) |x+1|=2|x - 1| 2) |x-3| not equal to zero

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Is |x|<1? 1) |x+1|=2|x - 1| 2) |x-3| not equal to zero [#permalink] New post 22 Aug 2006, 01:53
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E

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Is |x|<1?

1) |x+1|=2|x-1|

2) |x-3| not equal to zero

[edited by Hong to reflect the correct question.]
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 [#permalink] New post 22 Aug 2006, 02:58
A) is equivalent to (x+1)^2=0 so A is SUFF
B) X is diff from 3 - NOT SUFFICIENT
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Re: Inequality with absolute value DS [#permalink] New post 22 Aug 2006, 03:01
apollo168 wrote:
Is |x|<1?

1) |x+1|=2|x+1|

2) |x-3| not equal to zero


Actually the answer is c

From 1
x+1= 2x+2 or x+1=-2x-2 x=3 or 1/3 insuff

From 2 x not equal to 3 insuff
Combined x=1/3
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 [#permalink] New post 22 Aug 2006, 03:06
You have 4 combs for A . Right side can be pos or neg, and so is left. Better square both modules to eliminate the modulus sign.
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 [#permalink] New post 22 Aug 2006, 03:15
Hehe you'll find this funny. Decided I had enough of gmat shortcuts and dusted off my high school algebra book. Anyway there is a theorem regarding this type of problems

it gave the example |2x-3|=|7-3x| According to the book you can simplify it by simply:
2x-3=7-3x or 2x-3=-(7-3x)

And the OA is C but OE wasnt given downloaded one of the paper base gmat tests from this site only gave the answer key
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 [#permalink] New post 22 Aug 2006, 03:32
in this problem 4 cases exist
a.RHS +ve ,LHS +ve
b.RHS -ve,LHS +ve
c.LHS +ve,RHS -ve
d.RHS -ve,LHS -ve

we can observe a and d are alike and similarly,b and c are alike
therefore,
we get (x+1)=2(x+1) or (x+1)= -2(x+1),
=>x+1=2x+2 or x+1= -2x-2
=>x= -1 or x= -1
hence,we can say A is sufficient
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Re: Inequality with absolute value DS [#permalink] New post 22 Aug 2006, 03:33
apollo168 wrote:
apollo168 wrote:
Is |x|<1?

1) |x+1|=2|x+1|

2) |x-3| not equal to zero


Actually the answer is c

From 1
x+1= 2x+2 or x+1=-2x-2 x=3 or 1/3 insuff

From 2 x not equal to 3 insuff
Combined x=1/3


x+1=2x+2=> x= -1 and x+1= -2x-2=> 3x= -3=>x= -1
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Re: Inequality with absolute value DS [#permalink] New post 22 Aug 2006, 03:40
Guys sorry there is a typographical error. Question should read

Is |x|<1?

1) |x+1|=2|x-1|

2) |x-3| not equal to zero
My fault the author of the test disabled the ability to cut and paste so I have to retype.
So from here C will stand.

And I really researched that particular theorem I was talking about. When face with two absolute values on either side of the equation. You dont need to break it down into its four scenarios. I did that as well and it took me too much time, which prompted me to research the topic. Thanks for the replies

apollo168 wrote:
Is |x|<1?

1) |x+1|=2|x+1|

2) |x-3| not equal to zero
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 [#permalink] New post 22 Aug 2006, 05:27
ok thn i can giv it ..... the answer is C
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 [#permalink] New post 22 Aug 2006, 05:38
Sorry about the typo. Im sure it threw off a lot of people. My apologies
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 [#permalink] New post 22 Aug 2006, 06:28
For the first problem, statement 1 is sufficient. (A)

For the revised problem, we need both satements. (C)
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 [#permalink] New post 22 Aug 2006, 12:52
Can someone please explain me how C is the answer for the modified question ?
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 [#permalink] New post 23 Aug 2006, 07:09
for problem 1 (A) is the answer.

for problem 2,
stmt 1 gives us 2 values of x -> 3, 1/3. thus we can't say if |x|<1
stmt 2 tells us that |x-3| <> 0, for which x can take any value other than 3.
combining these 2 stmts we can say that since x cannot be 3, it has to be 1/3 and |x| < 1 is true.
hence (C).
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  [#permalink] 23 Aug 2006, 07:09
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