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m08, #11

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m08, #11 [#permalink] New post 06 Nov 2008, 21:41
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Is |x - 1| \lt 1 ?

1. (x - 1)^2 \le 1
2. x^2 - 1 \gt 0

[Reveal] Spoiler: OA
E

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Re: m08, #11 [#permalink] New post 06 Nov 2008, 22:23
ritula wrote:
Is modulus(x-1) < 1?

(1) (x - 1)^2 <= 1
(2) x^2 - 1 > 0


(1) if (x - 1)^2 <= 1, x is >= 0 but smaller than or equal to 2.
if x = 0 or 2, then lx-1l = 1. if x = 0.5, lx-1l <1. so nsf..

(2) if (x^2 - 1) > 0 (or x^2 > 1), x is not equal to 1 or 0 or -1. so nsf.

togather also x can be 0 or 2 or a +ve fraction.

So E.
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Re: m08, #11 [#permalink] New post 07 Nov 2008, 00:56
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ritula wrote:
Is modulus(x-1) <1 ?

(1) (x-1)^2<=1
(2) x^2-1>0


From stmt1: (x-1)^2 <= 1
or, -1 <= (x-1) <= 1
or, |x-1| <=1
Insufficient.

From stmt2: (x-1)(x+1) > 0. This means, either (x-1)>0 or, (x-1) < 0
or, |x-1| > 0 and is no additional information (as modulus will always be greater than 0).

Hence, E.
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m08 q11 IMO OA need to be be B) and not E) [#permalink] New post 09 Dec 2008, 20:47
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Is |X-1| < 1?

1) (x-1) ^ 2 <= 1
2) x^2 - 1 > 0

Stmt I

x=0
(0-1)^2 is equal to 1 and not less than 1
x=1/2

(1/2-1) ^ 2 < 1

INSUFF

Stmt II

x^2 - 1>0
x^2 > 1

|x| > 1

x>1 (or) -x>1 i.e x<-1

A defnite NO since its equal to 1 or greater than 1. The question is |x-1| < 1.

DEFINITE NO

My choice B)

OA is given as E)

Please correct me if I am mistaken.
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Re: m08 q11 IMO OA need to be be B) and not E) [#permalink] New post 09 Dec 2008, 21:57
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Hi Cramya, this is one of the most confusing qs in GMAT, the absolute DS ones. Here's what i would have done to solve it.

First lets look at the qs statement and simplify it
Is |X-1| < 1?
|x-1| = -(x-1) or x-1
solving both
-x+1 < 1 , 1-1 < x , 0 < x or x >0
x-1 < 1 , x < 2

so Is 0 < x < 2 ? is the actual qs .

Now lets look at the statements
1) (x-1) ^ 2 <= 1
x-1 <= +- 1
x <= 0 , x<= 2 So not sufficient

2) x^2 - 1 > 0
x^2 > 1 , x > +-1 , -1 < x < 1
Here x is < 2 so that satisfies one side of the qs, but x can be between -1 and 0 , so that fails the x > 0 part. Hence this is not sufficient

Together , they dont do much to satisfy the qs above, hence the answer should be E.

I guess the way you are trying to solve the statements is different, so i hope you got what i am trying to do here.

Correct me someone if they feel i am wrong here. I am still getting used to solving absolutes as well ... It will be good to know if i have reached the correct answer using incorrect methods afterall.

Thanks.
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Re: m08 q11 IMO OA need to be be B) and not E) [#permalink] New post 09 Dec 2008, 23:03
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My bad! Just considered intege values for x which is big red flag in a DS when no info on x is given

x could be a non integer

I agree its E)

Thanks for the clarification!
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Re: m08 q11 IMO OA need to be be B) and not E) [#permalink] New post 27 Dec 2008, 23:22
gamecode,

Looking at your statement:-

2) x^2 - 1 > 0
x^2 > 1 , x > +-1 , -1 < x < 1
Here x is < 2 so that satisfies one side of the qs, but x can be between -1 and 0 , so that fails the x > 0 part. Hence this is not sufficient

How did you get to from x > +-1 to -1 < x < 1?
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Re: m08 q11 IMO OA need to be be B) and not E) [#permalink] New post 04 Jan 2009, 15:59
[Can someone confirm the above equation? x > +-1 to -1 < x < 1 ???]

its obvious that above is not correct...

Now:
result of I. x<=2, x<=0
result of II. x>-1, x>1

both together as well can not confirm if 0<x<2, whether x is integer or non-integer.
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Re: m08 q11 IMO OA need to be be B) and not E) [#permalink] New post 24 Aug 2009, 13:25
gmattarget700 wrote:
[Can someone confirm the above equation? x > +-1 to -1 < x < 1 ???]

its obvious that above is not correct...

Now:
result of I. x<=2, x<=0
result of II. x>-1, x>1

both together as well can not confirm if 0<x<2, whether x is integer or non-integer.


These equations who have are wrong as well
1 says 0<=x<=2
2 say x<-1 or x>1
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Re: m08, #11 [#permalink] New post 06 Oct 2009, 02:37
ritula wrote:
Is modulus(x-1) <1 ?

(1) (x-1)^2<=1
(2) x^2-1>0


I always solve the stem first.In this we solve the two equations:
x-1<1 and -(x-1)<1
Solving we get,
x<2,x>0

QUERIE:Does the above two values represent:
1.Is the value of X<2 OR x>0 ??
2.The actual question is whther 0<x<2 ??

Solving stmt 1:
(x-1) ^ 2 <= 1
x-1 <= +- 1
x <= 0 , x<= 2 ..Hmm..is'nt this SUFF??It's telling us what the ques. stem is asking.
Confused!!
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Re: m08, #11 [#permalink] New post 07 Oct 2009, 06:31
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tejal777 wrote:
ritula wrote:
Is modulus(x-1) <1 ?

(1) (x-1)^2<=1
(2) x^2-1>0


I always solve the stem first.In this we solve the two equations:
x-1<1 and -(x-1)<1
Solving we get,
x<2,x>0

QUERIE:Does the above two values represent:
1.Is the value of X<2 OR x>0 ??
2.The actual question is whther 0<x<2 ??

Solving stmt 1:
(x-1) ^ 2 <= 1
x-1 <= +- 1
x <= 0 , x<= 2 ..Hmm..is'nt this SUFF??It's telling us what the ques. stem is asking.
Confused!!


You are forgetting the equality sign on statement 1.

Question asks if |x-1|<1, which is the same as asking if 0<x<2.

Stat1 gives 0<=x<=2. If x<2 the answer is yes; if x=2 the answer is no => not sufficient.
Stat2 gives x<-1 or x>1
Combined statements give 1<x<=2. Still not sufficient.
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Re: m08, #11 [#permalink] New post 03 Jun 2010, 09:29
Is modulus(x-1) <1 ?

(1) (x-1)^2<=1
(2) x^2-1>0


From Stmt1:
x can be 0,1,2
No Definite Value

From Stmt2:
(x+1)(x-1)>0
=> either x<-1 or x>1
No Definite Value


Thus , OA is E.
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Re: m08, #11 [#permalink] New post 28 Jun 2010, 02:23
powerka wrote:
tejal777 wrote:
ritula wrote:
Is modulus(x-1) <1 ?

(1) (x-1)^2<=1
(2) x^2-1>0


I always solve the stem first.In this we solve the two equations:
x-1<1 and -(x-1)<1
Solving we get,
x<2,x>0

QUERIE:Does the above two values represent:
1.Is the value of X<2 OR x>0 ??
2.The actual question is whther 0<x<2 ??

Solving stmt 1:
(x-1) ^ 2 <= 1
x-1 <= +- 1
x <= 0 , x<= 2 ..Hmm..is'nt this SUFF??It's telling us what the ques. stem is asking.
Confused!!


You are forgetting the equality sign on statement 1.

Question asks if |x-1|<1, which is the same as asking if 0<x<2.

Stat1 gives 0<=x<=2. If x<2 the answer is yes; if x=2 the answer is no => not sufficient.
Stat2 gives x<-1 or x>1
Combined statements give 1<x<=2. Still not sufficient.


Totally agree, with you. The pitfall is here the difference in meaning of signs "less than" and "Less than or equal." I missed this part and thought the answer was A because the first statement repeats the stem, well kind of. So the main trick lies here. Well, what can I say, you have to be really careful.
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Re: m08, #11 [#permalink] New post 06 Jun 2011, 04:20
(1)

x^2 -2x + 1 <= 1

x(x-2) <= 0

so 0 <= x <= 2

Insufficient

as if x = 1, then |x-1| < 1

but if x = 2, then |x-1| = 1

(2)

(x-1)(x+1) > 0

=> x < -1 or x > 1

Insufficient as if x = 2, then |x-1| = 1

x = 1.5, then |x-1| < 1

(1) + (2)

1 < x <=2

Insufficient

Answer - E
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Re: m08, #11 [#permalink] New post 06 Jun 2011, 04:57
ritula wrote:
Is |x - 1| \lt 1 ?

1. (x - 1)^2 \le 1
2. x^2 - 1 \gt 0



1. not sufficient. If (x - 1)^2 \le 1, then -1< x <= 2;
2. sufficient. -1 > x > 1

Ans: B
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Re: m08, #11 [#permalink] New post 06 Jun 2011, 05:22
seku wrote:
ritula wrote:
Is |x - 1| \lt 1 ?

1. (x - 1)^2 \le 1
2. x^2 - 1 \gt 0



1. not sufficient. If (x - 1)^2 \le 1, then -1< x <= 2;
2. sufficient. -1 > x > 1

Ans: B


|z|<k. It means, -k<z<k
|z| \le k. It means, -k \le z \le k
|z|>k. It means, z>k \hspace{3} OR \hspace{3} z<-k
|z| \ge k. It means, z \ge k \hspace{3} OR \hspace{3} z \le -k
\sqrt{z^2}=|z|
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Re: m08, #11 [#permalink] New post 06 Jun 2011, 18:16
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Re: m08, #11 [#permalink] New post 22 Apr 2012, 02:59
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Is |x - 1|<{1} ?

There is no need to find the ranges for each inequality to solve this question.

(1 (x - 1)^2\leq1 --> since both sides of the inequality are non-negative then we can take square root from both parts: |x-1|\leq{1}, so |x-1| can be less than 1 (answer YES), as well as equal to 1, for x=2 or x=0 (answer NO). Not sufficient. Notice that |x-1|\leq{1}, means 0\leq{x}\leq{2}.

(2) x^2 - 1>0 --> x^2>1 --> again, since both sides of the inequality are non-negative then we can take square root from both parts: |x|>1. If x=1.5 then the answer is YES but if x=2 then the answer is NO. Not sufficient.

(1)+(2) x=1.5 and x=2 satisfy both statements and give different answer to the question. Not sufficient.

Answer: E.
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Re: m08, #11 [#permalink] New post 19 Jun 2012, 23:51
Bunuel wrote:
(2) x^2 - 1>0 --> x^2>1 --> again, since both sides of the inequality are non-negative then we can take square root from both parts: |x|>1. If x=1.5 then the answer is YES but if x=2 then the answer is NO. Not sufficient.


Did you mean 0.5 instead of 1.5 or did I miss something?
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Re: m08, #11 [#permalink] New post 20 Jun 2012, 00:31
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solarzj wrote:
Bunuel wrote:
(2) x^2 - 1>0 --> x^2>1 --> again, since both sides of the inequality are non-negative then we can take square root from both parts: |x|>1. If x=1.5 then the answer is YES but if x=2 then the answer is NO. Not sufficient.


Did you mean 0.5 instead of 1.5 or did I miss something?


No, everything is correct there: if x=1.5 then 1.5^2>1 (so the second statement is satisfied) and |1.5-1|=0.5<1 (so the answer to the original question is YES).

Hope it's clear.
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Re: m08, #11   [#permalink] 20 Jun 2012, 00:31
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