Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 02 Aug 2015, 02:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# m08, #11

Author Message
VP
Joined: 18 May 2008
Posts: 1293
Followers: 14

Kudos [?]: 200 [0], given: 0

m08, #11 [#permalink]  06 Nov 2008, 21:41
3
This post was
BOOKMARKED
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
 Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes GMAT Pill GMAT Discount Codes
SVP
Joined: 29 Aug 2007
Posts: 2493
Followers: 59

Kudos [?]: 584 [0], given: 19

Re: m08, #11 [#permalink]  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.
_________________
SVP
Joined: 17 Jun 2008
Posts: 1570
Followers: 12

Kudos [?]: 207 [1] , given: 0

Re: m08, #11 [#permalink]  07 Nov 2008, 00:56
1
KUDOS
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.
Intern
Joined: 25 Nov 2008
Posts: 7
Followers: 1

Kudos [?]: 10 [1] , given: 0

m08 q11 IMO OA need to be be B) and not E) [#permalink]  09 Dec 2008, 20:47
1
KUDOS
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.
Manager
Joined: 14 Oct 2008
Posts: 160
Followers: 1

Kudos [?]: 33 [5] , given: 0

Re: m08 q11 IMO OA need to be be B) and not E) [#permalink]  09 Dec 2008, 21:57
5
KUDOS
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.
Intern
Joined: 25 Nov 2008
Posts: 7
Followers: 1

Kudos [?]: 10 [1] , given: 0

Re: m08 q11 IMO OA need to be be B) and not E) [#permalink]  09 Dec 2008, 23:03
1
KUDOS
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!
Intern
Joined: 30 Nov 2008
Posts: 8
Followers: 0

Kudos [?]: 7 [0], given: 0

Re: m08 q11 IMO OA need to be be B) and not E) [#permalink]  27 Dec 2008, 23:22
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?
Intern
Joined: 02 Jan 2009
Posts: 5
Followers: 0

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

Re: m08 q11 IMO OA need to be be B) and not E) [#permalink]  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.
Manager
Joined: 08 Jul 2009
Posts: 176
Followers: 3

Kudos [?]: 36 [0], given: 13

Re: m08 q11 IMO OA need to be be B) and not E) [#permalink]  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
Director
Joined: 25 Oct 2008
Posts: 610
Location: Kolkata,India
Followers: 9

Kudos [?]: 337 [0], given: 100

Re: m08, #11 [#permalink]  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!!
_________________
Manager
Joined: 22 Jul 2009
Posts: 193
Followers: 4

Kudos [?]: 217 [1] , given: 18

Re: m08, #11 [#permalink]  07 Oct 2009, 06:31
1
KUDOS
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.
_________________

Please kudos if my post helps.

Intern
Joined: 29 Mar 2010
Posts: 16
Schools: UCLA, USC
WE 1: 3 Yr at leading SAAS company
Followers: 0

Kudos [?]: 5 [0], given: 0

Re: m08, #11 [#permalink]  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.
Intern
Joined: 08 Jun 2010
Posts: 31
GMAT 1: 690 Q48 V36
GPA: 3.59
WE: Accounting (Accounting)
Followers: 0

Kudos [?]: 7 [0], given: 13

Re: m08, #11 [#permalink]  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.
SVP
Joined: 16 Nov 2010
Posts: 1678
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 31

Kudos [?]: 365 [0], given: 36

Re: m08, #11 [#permalink]  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

_________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 24 May 2010
Posts: 46
Followers: 0

Kudos [?]: 5 [0], given: 6

Re: m08, #11 [#permalink]  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
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2028
Followers: 140

Kudos [?]: 1182 [0], given: 376

Re: m08, #11 [#permalink]  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, $$z \ge k \hspace{3} OR \hspace{3} z \le -k$$
$$\sqrt{z^2}=|z|$$
_________________
Intern
Joined: 15 Jul 2010
Posts: 6
Location: Bangkok, Thailand
Followers: 0

Kudos [?]: 13 [4] , given: 3

Re: m08, #11 [#permalink]  06 Jun 2011, 18:16
4
KUDOS
E for me
Attachments

A.png [ 17.06 KiB | Viewed 6297 times ]

_________________

"You can do it if you believe you can!" - Napoleon Hill
"Insanity: doing the same thing over and over again and expecting different results." - Albert Einstein

Math Expert
Joined: 02 Sep 2009
Posts: 28781
Followers: 4594

Kudos [?]: 47491 [1] , given: 7123

Re: m08, #11 [#permalink]  22 Apr 2012, 02:59
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.

_________________
Intern
Joined: 14 Mar 2012
Posts: 28
GMAT 1: 590 Q32 V30
GMAT 2: 700 Q48 V37
GPA: 3.6
Followers: 0

Kudos [?]: 39 [0], given: 28

Re: m08, #11 [#permalink]  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?
_________________

Give kudos if you find my post helpful

Math Expert
Joined: 02 Sep 2009
Posts: 28781
Followers: 4594

Kudos [?]: 47491 [1] , given: 7123

Re: m08, #11 [#permalink]  20 Jun 2012, 00:31
1
KUDOS
Expert's post
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.
_________________
Re: m08, #11   [#permalink] 20 Jun 2012, 00:31

Go to page    1   2    Next  [ 21 posts ]

Similar topics Replies Last post
Similar
Topics:
1 m08 #28 12 15 Dec 2008, 02:07
8 m08 q#6 11 25 Nov 2008, 10:19
19 m08#27 18 02 Nov 2008, 14:48
3 m08#23 16 02 Nov 2008, 14:43
8 m08 Q#7 16 01 Aug 2008, 09:56
Display posts from previous: Sort by

# m08, #11

Moderators: WoundedTiger, Bunuel

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.