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

 It is currently 30 Jun 2015, 00:21

### 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

# m20 # 33

Author Message
Intern
Joined: 10 Sep 2008
Posts: 37
Followers: 1

Kudos [?]: 34 [2] , given: 0

m20 # 33 [#permalink]  25 Feb 2009, 19:02
2
KUDOS
3
This post was
BOOKMARKED
Question:

Is $$|X+1|<2$$?

1. $$(X-1)^2 < 1$$
2. $$X^2-2 < 0$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions
____________________

Would someone please explain to me what the simplest way of solving this problem is?
1) From S1, is there a quick way to come to the conclusion that 0<x<2
2) From S2, is there a quick way to come to the conclusion that -2<X< square root of 2
3) When the question asks if |X+1|< 2, do we conclude that it is asking if X is either <1 OR <-3?

Thank you.
 Kaplan GMAT Prep Discount Codes Knewton GMAT Discount Codes Veritas Prep GMAT Discount Codes
Director
Joined: 04 Jan 2008
Posts: 919
Followers: 58

Kudos [?]: 271 [3] , given: 17

Re: m20 # 33 [#permalink]  28 Feb 2009, 23:02
3
KUDOS
IMO A
as (1) is SUFFICIENT

from (1), we get 1<x<2
from (2), we get 0<x<1.414

dczuchta wrote:
Would someone please explain to me what the simplest way of solving this problem is?
1) From S1, is there a quick way to come to the conclusion that 0<x<2
2) From S2, is there a quick way to come to the conclusion that -2<X< square root of 2
3) When the question asks if |X+1|< 2, do we conclude that it is asking if X is either <1 OR <-3?

Thank you,
____________________
Question:

Is |X+1|<2?

1) (X-1)^2 < 1
2) X^2-2 < 0

_________________
SVP
Joined: 29 Aug 2007
Posts: 2493
Followers: 59

Kudos [?]: 575 [2] , given: 19

Re: m20 # 33 [#permalink]  10 Mar 2009, 20:50
2
KUDOS
dczuchta wrote:
Would someone please explain to me what the simplest way of solving this problem is?
1) From S1, is there a quick way to come to the conclusion that 0<x<2
2) From S2, is there a quick way to come to the conclusion that -2<X< square root of 2
3) When the question asks if |X+1|< 2, do we conclude that it is asking if X is either <1 OR >-3?

Thank you,
____________________
Question:

Is |X+1|<2?

1) (X-1)^2 < 1
2) X^2-2 < 0

1) From S1, is there a quick way to come to the conclusion that 0<x<2
(x-1)^2 < 1
(x-1) < 1
x < 2

(x-1) > -1
x > 0

sufff....

2) From S2, is there a quick way to come to the conclusion that -2<X< square root of 2
x^2 - 2 < 0
x^2 < 2
x < sqrt2 or x > -sqrt2

insuff..........

3) When the question asks if |X+1|< 2, do we conclude that it is asking if X is either <1 OR >-3? yes, |x+1|< 2 means -3< x <1.

This is how it is solved:

If |x+1| is +ve, (x+1) < 2.
Or, x < 2-1
Or, x < 1

If |x+1| is -ve, -(x+1) < 2.
Or, -(x+1) < 2.
Or, -x-1 < 2.
Or, -2-1 < x
Or, -3 < x

Hence -3 < x < 1.
If any typo, thats mine.
_________________
Intern
Joined: 10 Sep 2008
Posts: 37
Followers: 1

Kudos [?]: 34 [0], given: 0

Re: m20 # 33 Kudos? [#permalink]  12 Mar 2009, 10:03
Thank you. Makes sense. Looking at it now, I'm not even sure what I was confused about-maybe a wrong calculation. What happened to the Kudos?; the button seemed to have been deleted.
CIO
Joined: 02 Oct 2007
Posts: 1217
Followers: 93

Kudos [?]: 766 [0], given: 334

Re: m20 # 33 [#permalink]  30 Mar 2009, 10:10
Kudos system is currently unavailable. We're working on getting it back.
_________________

Welcome to GMAT Club!

Want to solve GMAT questions on the go? GMAT Club iPhone app will help.
Result correlation between real GMAT and GMAT Club Tests
Are GMAT Club Test sets ordered in any way?

Take 15 free tests with questions from GMAT Club, Knewton, Manhattan GMAT, and Veritas.

GMAT Club Premium Membership - big benefits and savings

VP
Joined: 07 Apr 2009
Posts: 1155
Concentration: General Management, Strategy
Schools: Duke (Fuqua) - Class of 2012
Followers: 33

Kudos [?]: 362 [1] , given: 19

Re: m20 # 33 [#permalink]  28 Apr 2009, 00:29
1
KUDOS
Expert's post
maybe I'm having a brainfart, but I think the answer is E

(1) 0<x<2
(2) -sqrt(2)<x<sqrt(2)

we need to know if -3<x<1

(1) and (2) tells us nothing that can be definite proof.
Intern
Joined: 06 Jan 2010
Posts: 19
Schools: Wake Forest Evening
WE 1: ~12 years total
Followers: 0

Kudos [?]: 3 [0], given: 33

Re: m20 # 33 [#permalink]  05 Mar 2010, 06:45
Is |X+1|<2?

1. (X-1)^2 < 1
2. X^2-2 < 0

E.

We have to have -3 < x < 1

1. Consider x=1.1. (1.1-1)^2 = .1^2 = 0.01 (less than one) falls outside -3 < x < 1
Also consider 0.1. (0.1-1)^2 = (-0.5)^2 = .25 (less than one) falls inside -3 < x < 1

This is insuff because it can fall in and it can fall out of our -3 < x < 1.

2. Consider x = 1.1. 1.1^2 - 2 = 1.21 - 2 = (less than 0) falls outside -3 < x < 1
Also Consider x = 0.1. .1^2 - 2 = 0.01 - 2 = (less than 0) falls inside -3 < x < 1

This is insuff because it can fall in and it can fall out of our -3 < x < 1.

Also remember you aren't answering "Is |X+1|<2?". You are answering if 1 or 2 can answer that question.
Manager
Joined: 26 Nov 2009
Posts: 178
Followers: 3

Kudos [?]: 53 [1] , given: 5

Re: m20 # 33 [#permalink]  05 Mar 2010, 10:58
1
KUDOS
is |x+1|<2 can be rewritten as

x+1<2 or -x-1<2

x<1 or x>-3

1)(x-1)^2<1
(x-1)<1
x<2 so insufficient

2)x^2-2<0
x+\sqrt{2} x-\sqrt{2} <0
x+\sqrt{2} <0 or x-\sqrt{2}<0
x<-1.414 or x<1.414
so insufficient

combining both 1 and 2 x can still be 1.2 or 0.5 so both are also insufficient

so ans is E
Manager
Joined: 27 Aug 2009
Posts: 146
Followers: 2

Kudos [?]: 14 [0], given: 1

Re: m20 # 33 [#permalink]  05 Mar 2010, 11:35
Ans is E

Q is if -3<X<1

1 gives 0<x<2
2 gives -sqroot(2) < X < sqroot(2)
Manager
Joined: 04 Dec 2009
Posts: 71
Location: INDIA
Followers: 2

Kudos [?]: 9 [0], given: 4

Re: m20 # 33 [#permalink]  05 Mar 2010, 20:51
what is OA

for me ANS is : E
_________________

MBA (Mind , Body and Attitude )

Senior Manager
Joined: 21 Dec 2009
Posts: 268
Location: India
Followers: 10

Kudos [?]: 109 [0], given: 25

Re: m20 # 33 [#permalink]  08 Mar 2010, 11:38
asimov wrote:
maybe I'm having a brainfart, but I think the answer is E

(1) 0<x<2
(2) -sqrt(2)<x<sqrt(2)

we need to know if -3<x<1

(1) and (2) tells us nothing that can be definite proof.

Absolutely right. So E is the answer.
_________________

Cheers,
SD

Senior Manager
Joined: 13 Dec 2009
Posts: 263
Followers: 10

Kudos [?]: 126 [0], given: 13

Re: m20 # 33 [#permalink]  13 Mar 2010, 06:08
Both statements 1 and 2 give us that X can be atleast 1 or 0. If X = 1 it fails and if X = 0 it passes the |X+1|^2 test. Hence data is not sufficient even with both.
_________________

My debrief: done-and-dusted-730-q49-v40

Manager
Joined: 12 Jul 2010
Posts: 67
Followers: 1

Kudos [?]: 2 [0], given: 3

Re: m20 # 33 [#permalink]  07 Sep 2010, 22:54
Its E.

Question: It becomes -3<x<1

1) Solving 1 says 0<x<2 Insufficient
2) Solving 2 says x< sqrt2 Insufficient
SVP
Joined: 16 Nov 2010
Posts: 1680
Location: United States (IN)
Concentration: Strategy, Technology
Followers: 31

Kudos [?]: 350 [0], given: 36

Re: m20 # 33 [#permalink]  10 Mar 2011, 07:05
|x+1| < 2

x + 1 < 2 or -x-1 < 2

x < 1 or x > -3

From (1)

(x - 1)^2 < 1

x^2 -2x + 1 < 1

=> x(x-2) < 0

So 0 < x < 2, so not enough

From (2)

-1.4 < x < 1.4, so not enough

From (1) and (2) also, not enough as x can fall in 1 < x < 1.4, so answer is E.
_________________

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

GMAT Club Premium Membership - big benefits and savings

Intern
Joined: 03 Apr 2009
Posts: 10
Followers: 0

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

Re: m20 # 33 [#permalink]  13 Mar 2012, 16:22
IMO E.

Both the conditions are not sufficient to conclude.
Intern
Joined: 15 Sep 2011
Posts: 17
Followers: 0

Kudos [?]: 3 [1] , given: 6

Re: m20 # 33 [#permalink]  16 Apr 2012, 13:43
1
KUDOS
Most authors agree that the stem asks if -3<x<1, me too.

However, since it seems that most posts struggle with the first statement, here is my take.

(1) states (x-1)2 < 1

If (x-1)2 < 1, then most would agree that (x-1)2 is also either equal or greater than 0, since squares cannot be negative.

0<= (x-1)2 < 1 is what needs to be solved.

(x-1)(x-1) < 1
x2 - 2x + 1 < 1
x2 - 2x < 0
x2 < 2x

At this point we can see that 0 < x <= 1, which makes this statement insufficient as x, per stem cannot be 1.

(2) x < sqrt2 or x > -sqrt2

(C) To analyze them together, we have to look at both.
0 < x <= 1
(-1.41) -sqrt2 < x < sqrt2 (1.41)

Here we can see, that 1 is still part of both, which makes this question E.
Manager
Status: I will not stop until i realise my goal which is my dream too
Joined: 25 Feb 2010
Posts: 235
Schools: Johnson '15
Followers: 2

Kudos [?]: 33 [0], given: 16

Re: m20 # 33 [#permalink]  25 Apr 2012, 06:56
dczuchta wrote:
Question:

Is $$|X+1|<2$$?

1. $$(X-1)^2 < 1$$
2. $$X^2-2 < 0$$

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions
____________________

Would someone please explain to me what the simplest way of solving this problem is?
1) From S1, is there a quick way to come to the conclusion that 0<x<2
2) From S2, is there a quick way to come to the conclusion that -2<X< square root of 2
3) When the question asks if |X+1|< 2, do we conclude that it is asking if X is either <1 OR <-3?

Thank you.

let me know if my method is OKay...though it is learnt now only by going through the above observations and my own calculations

1) (x-1)sq < 1
(x-1) < + or - 1
so x-1 is between - 1 to +1
hence x is between 0 to 2 ( exclusive)
so it can take the value 1 , then the solution gives YES
if it takes 1.99, then the solution gives NO

hence insufficient... A and D are gone

2) xsq - 2 < 0

xsq < 2
x is between -1.41 to + 1.41, so this also gives both YES and NO as answer
so B is gone too

answer should be b/w C and E

taking together

x would now be between o and 1.41
so here too we get YES and NO for the solution, hence E is the answer
_________________

Regards,
Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Satyameva Jayate - Truth alone triumphs

Math Expert
Joined: 02 Sep 2009
Posts: 28183
Followers: 4452

Kudos [?]: 44824 [3] , given: 6618

Re: m20 # 33 [#permalink]  25 Apr 2012, 08:33
3
KUDOS
Expert's post
1
This post was
BOOKMARKED
harshavmrg wrote:
let me know if my method is OKay...though it is learnt now only by going through the above observations and my own calculations

1) (x-1)sq < 1
(x-1) < + or - 1
so x-1 is between - 1 to +1
hence x is between 0 to 2 ( exclusive)
so it can take the value 1 , then the solution gives YES
if it takes 1.99, then the solution gives NO

hence insufficient... A and D are gone

2) xsq - 2 < 0

xsq < 2
x is between -1.41 to + 1.41, so this also gives both YES and NO as answer
so B is gone too

answer should be b/w C and E

taking together

x would now be between o and 1.41
so here too we get YES and NO for the solution, hence E is the answer

Is $$|x+1|<2$$?

Is $$|x+1|<2$$? --> is $$-3<x<1$$?

(1) $$(x-1)^2 < 1$$ --> since both sides of the inequality are non-negative we can take square root from it: $$|x-1|<1$$ --> $$0<x<2$$. Not sufficient.

(2) $$x^2-2 < 0$$ --> $$x^2<2$$ --> again, since both sides of the inequality are non-negative we can take square root from it: $$|x|<1.4$$ (approximately) --> $$-1.4<x<1.4$$. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is $$0<x<1.4$$. So, $$x$$ could be in the range $$-3<x<1$$ (for example if $$x=1$$) as well as out of the range $$-3<x<1$$ (for example if $$x=1.2$$). Not sufficient.

_________________
Intern
Joined: 10 Aug 2012
Posts: 19
Location: India
Concentration: General Management, Technology
GPA: 3.96
Followers: 0

Kudos [?]: 6 [0], given: 15

Re: m20 # 33 [#permalink]  15 Mar 2013, 19:52
Complete solution of the problem,

Given in question stem,

$$|X+1|$$$$<$$ $$2$$

If, $$(X+1)$$$$< 0$$
$$-(X+1) < 2$$
$$X+1 > -2$$
$$X > -3$$

If$$(X+1) > 0$$
$$(X+1) > 2$$
$$X > 1$$

So we have to check that if any equation proves $$-3 < X < 1$$, that will be sufficient condition

I - $$(X -1)^2 < 1$$
$$(X -1 ) < 1$$ or $$(X - 1) > -1$$

On solving, $$0 < X < 2$$

This is not sufficient condition, So A and D are incorrect choices.

II - $$X^2 -2 < 0$$
$$X^2 < 2$$

On solving, $$\sqrt{-2} < X < \sqrt{2}$$

This is not sufficient condition, So B and D are incorrect choices.

I and II together,

$$0 < X < \sqrt{2}$$

This is also not sufficient, so Ans is E
Re: m20 # 33   [#permalink] 15 Mar 2013, 19:52
Similar topics Replies Last post
Similar
Topics:
m20 #13 5 19 Nov 2009, 05:46
14 m20 #12 23 24 Feb 2009, 08:23
11 M20 Q35 21 18 Nov 2008, 03:49
29 M20Q31 29 18 Nov 2008, 03:46
3 M20 #12 8 18 Nov 2008, 03:30
Display posts from previous: Sort by

# m20 # 33

Moderators: Bunuel, WoundedTiger

 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®.