GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Oct 2019, 00:37

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

How to Solve: Inequalities

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

Hide Tags

Find Similar Topics 
GMAT Tutor
User avatar
G
Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 622
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE: Information Technology (Computer Software)
How to Solve: Inequalities  [#permalink]

Show Tags

New post Updated on: 19 Dec 2014, 05:44
15
31
How to Solve: Inequalities

Hi All,

I have learned a lot from gmatclub and am done with my gmat too. So, i have decided to contribute back.
As part of this i have decided to share the knowledge i have regarding various topics related to gmat quant.
Hope it will be useful. This post is about how to solve "Inequalities"

Theory

There are mainly four types of inequality problems which you would need to solve:--

TYPE 1
x*y > 0
When xy > 0 then we know that both x and y can be either positive or both can be negative
i.e. both x and y have the same sign
so, we have
x>0, y>0 or x<0,y<0

Example Problem
(x-1)*(x-2) > 0
So, we have two cases
Case 1
both (x-1) and (x-2) are positive
so, x-1 > 0 => x > 1
and x-2 > 0 => x > 2
Intersection of the two cases is x >2

Case 2
both (x-1) and (x-2) are negative
so, x-1 < 0 => x < 1
and x-2 < 0 => x < 2
Intersection of the two cases is x < 1

So, Solution to the question is x < 1 or x > 2


TYPE 2
x/y > 0
When x/y > 0 then we know that both x and y can be either positive or both can be negative
i.e. both x and y have the same sign
so, we have
x>0, y>0 or x<0,y<0

Example Problem
\(\frac{(x-3)}{(x-4)}\) > 0
So, we have two cases
Case 1
Both (x-3) and (x-4) are positive
=> x-3 > 0 => x>3
And x-4 > 0 => x>4
Intersection of the two cases is x > 4

Case 2
Both (x-3) and (x-4) are negative
=> x-3 < 0 => x < 3
and x-4 < 0 => x < 4
Intersection of the two cases is x < 3

So, solution to the question is x < 3 or x > 4


TYPE 3
x*y < 0
When x*y < 0 then we know that that
(x can be positive and y will be negative) or (x can be negative and y will be positive)
i.e. x and y have opposite signs
so, we have
x>0, y<0 or x<0,y>0

Example Problem
(x+1)(x-1) < 0
So, we will have two cases
Case 1
(x+1) is positive and (x-1) is negative
=> x + 1 > 0 => x > -1
And x - 1 < 0 => x < 1
Intersection of the two cases is
-1 < x < 1

Case 2
(x+1) is negative and (x-1) is positive
=> x+1 < 0 => x < -1
And x-1 > 0 => x > 1
The two cases have no intersection. So, no solution from this case

So, solution of the problem is -1 < x < 1


TYPE 4
x/y < 0
When x/y < 0 then we know that that
(x can be positive and y will be negative) or (x can be negative and y will be positive)
i.e. x and y have opposite signs
so, we have
x>0, y<0 or x<0,y>0

Example Problem
\(\frac{(x-2)}{(x+3)}\) < 0
So, we will have two cases
Case 1
(x-2) is positive and (x+3) is negative
=> x-2 > 0 => x > 2
And x+3 < 0 or x < -3
There is no intersection of the two cases. So, no solution from this case

Case 2
(x-2) is negative and (x+3) is positive
=> x-2 < 0 => x < 2
And x+3 > 0 => x > -3
Intersection of the two cases is -3 < x < 2

So, Solution of the question is -3 < x < 2


SUGGESTION: Try solving inequalities, they are not tough after all! :)

Problems:

1. x(x-1) > 0. Then value of x will be?

A. x > 0 and x > 1
B. x < 0 and x > 1
C. x < 0 and x < 1
D. x > 0 and x < 1

Solution:
x*(x-1) > 0
this is of the form xy>0 i.e. x and y have the same sign
so,
(1) either, x > 0 and x-1 >0
i.e. x >0 or x>1
taking intersection of the two possibilities we have x >1

(2)or x <0, and x-1 < 0
i.e. x <0 or x<1
taking intersection of the two possibilities we have x < 0

So, Answer will be B

link to the problem:
x-x-189656.html


2. Which of the following describes all the values of y for which y < y^2 ?

A. 1 < y
B. −1 < y < 0
C. y < −1
D. 1/y < 1
E. 0 < y < 1

Solution:
The question can be written as
y^2 - y > 0
s=> y*(y-1) > 0
It is of the form xy > 0
So, we will have two cases
Case 1
Both y and y-1 are positive
=> y > 0
And y-1 > 0 => y > 1
Intersection of the two cases is y > 1

Case 2
Both y and y-1 are negative
=>y < 0
And y -1 < 0 => y < 1
Intersection of the two cases is y <0

So, solution to the problem is y < 0 or y > 1

So, Answer will be D
(As option D can be written as
1/y - 1 < 0
or, (1-y)/y < 0
or (y-1)/y > 0
And solution to this will be same as that of y*(y-1) > 0)

Link to the problem:
which-of-the-following-describes-all-the-values-of-y-for-whi-161602.html


3. Which of the following describes all values of x for which 1–x^2 >= 0?

(A) x >= 1
(B) x <= –1
(C) 0 <= x <= 1
(D) x <= –1 or x >= 1
(E) –1 <= x <= 1

Solution:
Question can be written as
x^2 - 1 <=0
=> (x+1)*(x-1) <=0
Case 1
x+1 is positive or 0 and x-1 is negative or 0
=> x+1 >= 0 => x >= -1
And x-1 <= 0 => x <= 1
Intersection is -1 <= x <= 1

Case 2
x+1 is negative or 0 and x-1 is positive or 0
x+1 <=0 => x <= -1
And x-1 >= 0 => x >= 1
No intersection in this case

So, solution to the problem is -1 <= x <= 1
So, Answer will be E

Link to the problem
which-of-the-following-describes-all-values-of-x-for-which-144461.html


4. If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Solution:
(3+5y)/(x−1) < −7
=> (3+5y)/(x−1) + 7 < 0
=> ((3+5y) + 7*(x-1) )/ (x-1) < 0
=> ( 7x + 5y -4 )/ (x-1) < 0
Now, we know that x < 0 so, x- 1 < 0
in (7x + 5y -4 )/ (x-1) < 0
we know that x - 1 < 0
=> (7x + 5y -4 ) > 0
So, Answer will be B

Link to the problem
if-y-0-x-and-3-5y-x-1-7-then-which-of-the-following-155220.html


5. Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2

Solution:
k^2 + k - 2 > 0
=> (k+2)*(k-1) > 0
So, we will have two cases
Case 1
Both k+2 and k -1 positive
k+2 > 0 and k-1 > 0
=> k > -2 and k > 1
Intersection is k > 1

Case 2
Both k+2 and k-1 negative
k+2 < 0 and k -1 < 0
=> k < - 2 and k < 1
intersection is k < -2
So, Solution to the problem is k> 1 or k < -2

So, STAT1 is not SUFFICIENT
STAT2 is SUFFICIENT

So, Answer will be B

Link to the problem
is-k-2-k-147133.html


Looking for a Quant Tutor?


Check out my post for the same
starting-gmat-quant-classes-tutoring-bangalore-online-135537.html

Hope it helps!
Good Luck!
_________________

Originally posted by BrushMyQuant on 07 Dec 2014, 08:22.
Last edited by BrushMyQuant on 19 Dec 2014, 05:44, edited 3 times in total.
Formatting.
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 58395
Re: How to Solve: Inequalities  [#permalink]

Show Tags

New post 08 Dec 2014, 04:12
nktdotgupta wrote:
How to Solve: Inequalities

Hi All,

I have learned a lot from gmatclub and am done with my gmat too. So, i have decided to contribute back.
As part of this i have decided to share the knowledge i have regarding various topics related to gmat quant.
Hope it will be useful. This post is about how to solve "Inequalities"

Theory

There are mainly four types of inequality problems which you would need to solve:--

TYPE 1
x*y > 0
When xy > 0 then we know that both x and y can be either positive or both can be negative
i.e. both x and y have the same sign
so, we have
x>0, y>0 or x<0,y<0

Example Problem
(x-1)*(x-2) > 0
So, we have two cases
Case 1
both (x-1) and (x-2) are positive
so, x-1 > 0 => x > 1
and x-2 > 0 => x > 2
Intersection of the two cases is x >2

Case 2
both (x-1) and (x-2) are negative
so, x-1 < 0 => x < 1
and x-2 < 0 => x < 2
Intersection of the two cases is x < 1

So, Solution to the question is x < 1 or x > 2


TYPE 2
x/y > 0
When x/y > 0 then we know that both x and y can be either positive or both can be negative
i.e. both x and y have the same sign
so, we have
x>0, y>0 or x<0,y<0

Example Problem
\(\frac{(x-3)}{(x-4)}\) > 0
So, we have two cases
Case 1
Both (x-3) and (x-4) are positive
=> x-3 > 0 => x>3
And x-4 > 0 => x>4
Intersection of the two cases is x > 4

Case 2
Both (x-3) and (x-4) are negative
=> x-3 < 0 => x < 3
and x-4 < 0 => x < 4
Intersection of the two cases is x < 3

So, solution to the question is x < 3 or x > 4


TYPE 3
x*y < 0
When x*y < 0 then we know that that
(x can be positive and y will be negative) or (x can be negative and y will be positive)
i.e. x and y have opposite signs
so, we have
x>0, y<0 or x<0,y>0

Example Problem
(x+1)(x-1) < 0
So, we will have two cases
Case 1
(x+1) is positive and (x-1) is negative
=> x + 1 > 0 => x > -1
And x - 1 < 0 => x < 1
Intersection of the two cases is
-1 < x < 1

Case 2
(x+1) is negative and (x-1) is positive
=> x+1 < 0 => x < -1
And x-1 > 0 => x > 1
The two cases have no intersection. So, no solution from this case

So, solution of the problem is -1 < x < 1


TYPE 4
x/y < 0
When x/y < 0 then we know that that
(x can be positive and y will be negative) or (x can be negative and y will be positive)
i.e. x and y have opposite signs
so, we have
x>0, y<0 or x<0,y>0

Example Problem
\(\frac{(x-2)}{(x+3)}\) < 0
So, we will have two cases
Case 1
(x-2) is positive and (x+3) is negative
=> x-2 > 0 => x > 2
And x+3 < 0 or x < -3
There is no intersection of the two cases. So, no solution from this case

Case 2
(x-2) is negative and (x+3) is positive
=> x-2 < 0 => x < 2
And x+3 > 0 => x > -3
Intersection of the two cases is -3 < x < 2

So, Solution of the question is -3 < x < 2


SUGGESTION: Try solving inequalities, they are not tough after all! :)

Problems:

1. x(x-1) > 0. Then value of x will be?

A. x > 0 and x > 1
B. x < 0 and x > 1
C. x < 0 and x < 1
D. x > 0 and x < 1

Solution:
x*(x-1) > 0
this is of the form xy>0 i.e. x and y have the same sign
so,
(1) either, x > 0 and x-1 >0
i.e. x >0 or x>1
taking intersection of the two possibilities we have x >1

(2)or x <0, and x-1 < 0
i.e. x <0 or x<1
taking intersection of the two possibilities we have x < 0

So, Answer will be B

link to the problem:
x-x-189656.html


2. Which of the following describes all the values of y for which y < y^2 ?

A. 1 < y
B. −1 < y < 0
C. y < −1
D. 1/y < 1
E. 0 < y < 1

Solution:
The question can be written as
y^2 - y > 0
s=> y*(y-1) > 0
It is of the form xy > 0
So, we will have two cases
Case 1
Both y and y-1 are positive
=> y > 0
And y-1 > 0 => y > 1
Intersection of the two cases is y > 1

Case 2
Both y and y-1 are negative
=>y < 0
And y -1 < 0 => y < 1
Intersection of the two cases is y <0

So, solution to the problem is y < 0 or y > 1

So, Answer will be D
(As option D can be written as
1/y - 1 < 0
or, (1-y)/y < 0
or (y-1)/y > 0
And solution to this will be same as that of y*(y-1) > 0)

Link to the problem:
which-of-the-following-describes-all-the-values-of-y-for-whi-161602.html


3. Which of the following describes all values of x for which 1–x^2 >= 0?

(A) x >= 1
(B) x <= –1
(C) 0 <= x <= 1
(D) x <= –1 or x >= 1
(E) –1 <= x <= 1

Solution:
Question can be written as
x^2 - 1 <=0
=> (x+1)*(x-1) <=0
Case 1
x+1 is positive or 0 and x-1 is negative or 0
=> x+1 >= 0 => x >= -1
And x-1 <= 0 => x <= 1
Intersection is -1 <= x <= 1

Case 2
x+1 is negative or 0 and x-1 is positive or 0
x+1 <=0 => x <= -1
And x-1 >= 0 => x >= 1
No intersection in this case

So, solution to the problem is -1 <= x <= 1
So, Answer will be E

Link to the problem
which-of-the-following-describes-all-values-of-x-for-which-144461.html


4. If y>0>x, and (3+5y)/(x−1) < −7, then which of the following must be true?

A. 5y−7x+4 < 0
B. 5y+7x−4 > 0
C. 7x−5y−4 < 0
D. 4+5y+7x > 0
E. 7x−5y+4 > 0

Solution:
(3+5y)/(x−1) < −7
=> (3+5y)/(x−1) + 7 < 0
=> ((3+5y) + 7*(x-1) )/ (x-1) < 0
=> ( 7x + 5y -4 )/ (x-1) < 0
Now, we know that x < 0 so, x- 1 < 0
in (7x + 5y -4 )/ (x-1) < 0
we know that x - 1 < 0
=> (7x + 5y -4 ) > 0
So, Answer will be B

Link to the problem
if-y-0-x-and-3-5y-x-1-7-then-which-of-the-following-155220.html


5. Is k^2 + k - 2 > 0 ?

(1) k < 1
(2) k < -2

Solution:
k^2 + k - 2 > 0
=> (k+2)*(k-1) > 0
So, we will have two cases
Case 1
Both k+2 and k -1 positive
k+2 > 0 and k-1 > 0
=> k > -2 and k > 1
Intersection is k > 1

Case 2
Both k+2 and k-1 negative
k+2 < 0 and k -1 < 0
=> k < - 2 and k < 1
intersection is k < -2
So, Solution to the problem is k> 1 or k < -2

So, STAT1 is not SUFFICIENT
STAT2 is SUFFICIENT

So, Answer will be B

Link to the problem
is-k-2-k-147133.html


Hope it helps!
Good Luck!


Good job! Thank you!
_________________
Intern
Intern
avatar
Joined: 04 Mar 2011
Posts: 13
GMAT ToolKit User Reviews Badge
Re: How to Solve: Inequalities  [#permalink]

Show Tags

New post 16 Dec 2014, 14:33
absolutely fantastic!!
Intern
Intern
avatar
Joined: 24 Mar 2013
Posts: 22
Re: How to Solve: Inequalities  [#permalink]

Show Tags

New post 05 May 2015, 20:12
Very helpful synopsis. Thanks!
Intern
Intern
avatar
Joined: 14 Oct 2013
Posts: 43
GMAT ToolKit User
Re: How to Solve: Inequalities  [#permalink]

Show Tags

New post 08 May 2015, 08:27
In number 4 - why did you multiply 7 by (x-1)? I multiplied the whole left hand side of the expression by (x-1) to get rid of (x-1) in the denominator but it seems as if you kept the (x-1) in the denominator when you did so.
EMPOWERgmat Instructor
User avatar
V
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15294
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: How to Solve: Inequalities  [#permalink]

Show Tags

New post 08 May 2015, 10:52
Hi All,

GMAT Quant questions are typically designed in a way that allows the Test Taker more than one way to solve. In this article, the solutions are essentially all based on "doing algebra", but each of these questions can ALSO be solved by TESTing VALUES.

In many cases, TESTing VALUES is an easier and faster way to get to the correct answer, so beyond learning how to "do the math", you should also put in the proper time to learn (and practice) tactics. You'll find that it's easier to score at a higher level when you have more approaches to choose from.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com
Image


The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 13323
Re: How to Solve: Inequalities  [#permalink]

Show Tags

New post 01 Oct 2018, 02:38
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 Club Bot
Re: How to Solve: Inequalities   [#permalink] 01 Oct 2018, 02:38
Display posts from previous: Sort by

How to Solve: Inequalities

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





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