It is currently 11 Dec 2017, 12:13

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

For everyone interested in modulus practices, I have listed,

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

2 KUDOS received
Retired Moderator
User avatar
B
Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [2], given: 49

GMAT ToolKit User Premium Member
For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

New post 03 Dec 2006, 05:22
2
This post received
KUDOS
13
This post was
BOOKMARKED
For everyone interested in modulus practices, I have listed, at the bottom of this topic, several threads of modulus in a graduated level of difficulty.

Keep in mind that it's for trainning purposes and that the most advanced posts are not reprensative of the real GMAT.

Finally, enjoy !..... Fig :)


a) |1/(x-2)| >= 4,
b)|x-3|<2x-4
c)x + 2 < |x2 - 4|

solve for x

I will highly appreciate detailed explanations, thanks

Kudos [?]: 447 [2], given: 49

1 KUDOS received
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [1], given: 0

 [#permalink]

Show Tags

New post 03 Dec 2006, 05:45
1
This post received
KUDOS
A. |1/(x-2)| >= 4

As we are working positive value and as the function f(x) = 1/x decreases when x increases on the domain of x > 0, we can imply that:

|x-2| =< 1/4

That means,

o if x > 2 then,
x-2 =< 1/4
<=> x <= 1/4+2
<=> x <= 9/4

o if x < 2 then,
-x+2 =< 1/4
<=> x >= 2 - 1/4
<=> x >= 7/4

Thus, 7/4 <= x <= 9/4

Last edited by Fig on 29 Dec 2006, 13:57, edited 2 times in total.

Kudos [?]: 171 [1], given: 0

SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [0], given: 0

 [#permalink]

Show Tags

New post 04 Dec 2006, 10:39
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<=> x > 7/3
=> 7/3 < x < 3

o If x >= 3, then the inequation becomes:
x-3<2x-4
<=> x > 1
=> x >= 3

Finally, we have : x > 7/3

Last edited by Fig on 04 Dec 2006, 13:40, edited 1 time in total.

Kudos [?]: 171 [0], given: 0

1 KUDOS received
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [1], given: 0

 [#permalink]

Show Tags

New post 04 Dec 2006, 13:37
1
This post received
KUDOS
C. x + 2 < |x2 - 4|

|x2 - 4| = |(x+2)(x-2)|

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus,
o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2
o Sign((x+2)(x-2)) < 0 when -2 < x < 2

If -2 < x < 2, the inequation becomes :
x + 2 < -(x^2 - 4)
<=> x^2 + x - 2 < 0 (B)

Delta = 1 + 4*2 = 9.

The roots are:
o Root1 = (-1 - 3)/2 = -2
o Root2 = (-1 + 3)/2 = 1

So, (B) is true when x is between the roots of x^2 + x - 2. That implies -2 < x < 1.

If x < -2 or x > 2, the inequation becomes :
x + 2 < x^2 - 4
<=> x^2 - x - 6 > 0 (A)

Delta = 1 + 4*6 = 25.

The roots are:
o Root1 = (1 - 5)/2 = -2
o Root2 = (1 + 5)/2 = 3

(A) : x^2 - x - 6 > 0
<=> (x+2)(x-3) > 0

One more time, a is positive here. Thus (x+2)(x-3) > 0 outside of the roots.
In other words, x > 3 or x < -2


Finally, the domain of solutions of x is : x > 3 or x < -2 or -2 < x < 1.

Last edited by Fig on 29 Dec 2006, 04:26, edited 2 times in total.

Kudos [?]: 171 [1], given: 0

2 KUDOS received
Retired Moderator
User avatar
B
Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [2], given: 49

GMAT ToolKit User Premium Member
 [#permalink]

Show Tags

New post 05 Dec 2006, 02:51
2
This post received
KUDOS
FIG I HAVE A QUESTION HERE ABOUT THE ANSWER

THE RANGE , 7/4 <= x <= 9/4 INCLUDES 2 , AND THE FUNCTION IS NOT DEFINED AT X=2 AM I RIGHT???

Kudos [?]: 447 [2], given: 49

SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [0], given: 0

 [#permalink]

Show Tags

New post 05 Dec 2006, 04:20
yezz wrote:
FIG I HAVE A QUESTION HERE ABOUT THE ANSWER

THE RANGE , 7/4 <= x <= 9/4 INCLUDES 2 , AND THE FUNCTION IS NOT DEFINED AT X=2 AM I RIGHT???


Yes :).... U are right... My conclusion should clarify it :).... We have to remove the x=2 :)

Kudos [?]: 171 [0], given: 0

Retired Moderator
User avatar
B
Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [0], given: 49

GMAT ToolKit User Premium Member
 [#permalink]

Show Tags

New post 10 Dec 2006, 05:22
Hello Fig , hope u r fine and in best health ,

FOR THE THIRD PROBLEM , I BELIEVE THE ANSWER MISS THE RANGE

-2<X<1

Kudos [?]: 447 [0], given: 49

SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [0], given: 0

 [#permalink]

Show Tags

New post 10 Dec 2006, 10:06
yezz wrote:
Hello Fig , hope u r fine and in best health ,

FOR THE THIRD PROBLEM , I BELIEVE THE ANSWER MISS THE RANGE

-2<X<1


I'm ok thanks... it follows the way :)

Great catch :).... Yes, I did a "silly" mistake 2-4 = 6 and ended up with a wrong x^2 + x + 6 :).... I have corrected my post following your catch... very good :D :)

Kudos [?]: 171 [0], given: 0

Retired Moderator
User avatar
B
Joined: 05 Jul 2006
Posts: 1748

Kudos [?]: 447 [0], given: 49

GMAT ToolKit User Premium Member
 [#permalink]

Show Tags

New post 13 Dec 2006, 06:07
Thanks man , this is no catch this the fruits of your help and coaching :wink:

thanks a lot

Kudos [?]: 447 [0], given: 49

3 KUDOS received
SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [3], given: 0

 [#permalink]

Show Tags

New post 28 Dec 2006, 23:54
3
This post received
KUDOS
3
This post was
BOOKMARKED

Kudos [?]: 171 [3], given: 0

Manager
Manager
avatar
Joined: 14 Mar 2007
Posts: 59

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

 [#permalink]

Show Tags

New post 29 Jul 2007, 06:56
Fig wrote:
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<x> 7/3
=> 7/3 < x <3>= 3, then the inequation becomes:[/b]
x-3<2x-4
<x> 1
=> x >= 3

Finally, we have : x > 7/3


Hi Fig,


Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question :? but hope you would answer..

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

SVP
SVP
User avatar
Joined: 01 May 2006
Posts: 1794

Kudos [?]: 171 [0], given: 0

 [#permalink]

Show Tags

New post 29 Jul 2007, 13:48
ajay_gmat wrote:
Fig wrote:
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<x> 7/3
=> 7/3 < x <3>= 3, then the inequation becomes:[/b]
x-3<2x-4
<x> 1
=> x >= 3

Finally, we have : x > 7/3


Hi Fig,


Why have you mentioned that x >=3 and not that x > 3.. I see absolute Questions which sometimes have the >= sign and sometimes the > sign only.. Might be a very mindless question :? but hope you would answer..


Not meaningless but rather precised question :)...

Actually, it's just to avoid adding 1 solution alone from another interval... and so to meet a higher degree of "perfection" by making it a smoother way to the final solution and a straightforward simplicity to understand :)

Note that nothing is wrong if we say > or >= as soon as we do not miss to consider all values of x, especially here the ones that limit 2 intervals :)

Kudos [?]: 171 [0], given: 0

Intern
Intern
avatar
Joined: 30 Jul 2009
Posts: 19

Kudos [?]: 15 [0], given: 18

Re: [#permalink]

Show Tags

New post 09 Sep 2009, 18:01
Fig wrote:
A. |1/(x-2)| >= 4

As we are working positive value and as the function f(x) = 1/x decreases when x increases on the domain of x > 0, we can imply that:

|x-2| =< 1/4



can someone please explain why this happens. Why can't I just use |1/x-2| >=4. However, when x<2, I get x<=7/4. Also I agree with everyone else as the range include 2, doesn't lie outside of our limitations.

Kudos [?]: 15 [0], given: 18

Manager
Manager
avatar
Joined: 24 Aug 2009
Posts: 149

Kudos [?]: 126 [0], given: 46

Re: [#permalink]

Show Tags

New post 15 Dec 2009, 19:31
Fig wrote:
B. |x-3|<2x-4

o If x < 3, then the inequation becomes:
-(x-3)<2x-4
<=> 7 < 3x
<=> x > 7/3
=> 7/3 < x < 3

o If x >= 3, then the inequation becomes:
x-3<2x-4
<=> x > 1
=> x >= 3

Finally, we have : x > 7/3


Hi buddy

could you please explain this,
<=> x > 7/3
=> 7/3 < x < 3
<=> x > 1
=> x >= 3

i solved and got x>1 & x>7/3

Kudos [?]: 126 [0], given: 46

Intern
Intern
avatar
Joined: 27 Sep 2009
Posts: 6

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

Re: Absolute Mpractices [Modules] - List of GMAT Questions [#permalink]

Show Tags

New post 12 Aug 2010, 05:10
in case If -2 < x < 2, shouldn't it also have one more solution to problem.i.e. x<-2 and x>1

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

Intern
Intern
avatar
Joined: 24 Feb 2012
Posts: 31

Kudos [?]: 19 [0], given: 18

Re: [#permalink]

Show Tags

New post 24 Feb 2012, 23:27
Fig wrote:
C. x + 2 < |x2 - 4|

|x2 - 4| = |(x+2)(x-2)|

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus,
o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2
o Sign((x+2)(x-2)) < 0 when -2 < x < 2


Trying hard to grasp this concept, but don't think I'm understanding what you're trying to convey here.
Ideally, I would try to solve this as such:



Given that |x^2 - 4| is same as |(x+2)(x-2)|

Solving for x in |x^2 - 4| > x + 2 in the following two scenarios:
1. When (x+2)(x-2) >= 0
==> x^2 - 4 > x + 2
==> x>-1 and x>2

2. When (x+2)(x-2) < 0
==> -(x^2 - 4) > x + 2
==> 4 - x^2 > x + 2
==> x<1 and x<-2

Does the combination of these solutions for x represent the final answer? Am I missing a step?

Kudos [?]: 19 [0], given: 18

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 42544

Kudos [?]: 135256 [0], given: 12679

Re: Re: [#permalink]

Show Tags

New post 25 Feb 2012, 02:08
Expert's post
1
This post was
BOOKMARKED
fortsill wrote:
Fig wrote:
C. x + 2 < |x2 - 4|

|x2 - 4| = |(x+2)(x-2)|

Since the sign of a*x^2+b*x+c is that of a for any values of x not between the 2 roots, solutions of the equation a*x^2+b*x+c = 0 if they exist.

Thus,
o Sign((x+2)(x-2)) > 0 when x > 2 or x < -2
o Sign((x+2)(x-2)) < 0 when -2 < x < 2


Trying hard to grasp this concept, but don't think I'm understanding what you're trying to convey here.
Ideally, I would try to solve this as such:



Given that |x^2 - 4| is same as |(x+2)(x-2)|

Solving for x in |x^2 - 4| > x + 2 in the following two scenarios:
1. When (x+2)(x-2) >= 0
==> x^2 - 4 > x + 2
==> x>-1 and x>2

2. When (x+2)(x-2) < 0
==> -(x^2 - 4) > x + 2
==> 4 - x^2 > x + 2
==> x<1 and x<-2

Does the combination of these solutions for x represent the final answer? Am I missing a step?


First of all you won't see such equation on the GMAT, but anyway:

\(|x^2-4|>x+2\).

We should consider the cases when \(x^2-4<0\) and \(x^2-4\geq{0}\):

\(x^2-4<0\) for \(-2<x<2\) --> \(-(x^2-4)>x+2\) --> \(x^2+x-2<0\) --> \((x+2)(x-1)<0\)--> \(-2<x<1\). Since we are considering the range \(-2<x<2\) then the solution for this case will be intersections of these two (common part): \(-2<x<1\);

\(x^2-4\geq{0}\) for \(x\leq{-2}\) or \(x\geq{2}\) --> \((x^2-4)>x+2\) --> \(x^2-x-6>0\) --> \((x+2)(x-3)>0\)--> \(x<-2\) or \(x>3\). Since we are considering the range \(x\leq{-2}\) or \(x\geq{2}\) then the solution for this case will be intersections of these four (common parts): \(x<-2\) or \(x>3\);

Finally the ranges of \(x\) for which \(|x^2-4|>x+2\) true are: \(x<-2\), \(-2<x<1\) and \(x>3\).

Solving inequalities:
x2-4x-94661.html#p731476
inequalities-trick-91482.html
data-suff-inequalities-109078.html
range-for-variable-x-in-a-given-inequality-109468.html?hilit=extreme#p873535
everything-is-less-than-zero-108884.html?hilit=extreme#p868863

Hope it helps.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Kudos [?]: 135256 [0], given: 12679

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 14937

Kudos [?]: 287 [0], given: 0

Premium Member
Re: For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

New post 27 Aug 2014, 04:40
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 287 [0], given: 0

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 14937

Kudos [?]: 287 [0], given: 0

Premium Member
Re: For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

New post 02 Feb 2016, 23:21
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 287 [0], given: 0

Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 14937

Kudos [?]: 287 [0], given: 0

Premium Member
Re: For everyone interested in modulus practices, I have listed, [#permalink]

Show Tags

New post 31 Aug 2017, 08:18
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

GMAT Books | GMAT Club Tests | Best Prices on GMAT Courses | GMAT Mobile App | Math Resources | Verbal Resources

Kudos [?]: 287 [0], given: 0

Re: For everyone interested in modulus practices, I have listed,   [#permalink] 31 Aug 2017, 08:18
Display posts from previous: Sort by

For everyone interested in modulus practices, I have listed,

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

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®.