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# m08, #11

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06 Nov 2008, 22:41
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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

Source: GMAT Club Tests - hardest GMAT questions

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06 Nov 2008, 23:23
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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07 Nov 2008, 01:56
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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]

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09 Dec 2008, 21:47
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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]

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09 Dec 2008, 22:57
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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]

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10 Dec 2008, 00:03
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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]

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28 Dec 2008, 00:22
gamecode,

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]

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04 Jan 2009, 16:59
[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]

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24 Aug 2009, 14:25
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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06 Oct 2009, 03:37
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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07 Oct 2009, 07:31
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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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03 Jun 2010, 10:29
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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28 Jun 2010, 03:23
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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06 Jun 2011, 05:20
(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

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06 Jun 2011, 05:57
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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06 Jun 2011, 06:22
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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06 Jun 2011, 19:16
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E for me
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22 Apr 2012, 03:59
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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.

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20 Jun 2012, 00:51
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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20 Jun 2012, 01:31
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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, 01:31

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# m08, #11

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