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How to Solve: Inequalities

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Director
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Status: Tutor - BrushMyQuant
Joined: 05 Apr 2011
Posts: 610
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]

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New post Updated on: 19 Dec 2014, 05:44
14
30
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


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Hope it helps!
Good Luck!
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How to Solve :
Statistics || Reflection of a line || Remainder Problems || Inequalities


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
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Joined: 02 Sep 2009
Posts: 50002
Re: How to Solve: Inequalities  [#permalink]

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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!
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Re: How to Solve: Inequalities  [#permalink]

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New post 16 Dec 2014, 14:33
absolutely fantastic!!
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Re: How to Solve: Inequalities  [#permalink]

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New post 05 May 2015, 20:12
Very helpful synopsis. Thanks!
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Re: How to Solve: Inequalities  [#permalink]

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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.
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Re: How to Solve: Inequalities  [#permalink]

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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.

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Re: How to Solve: Inequalities  [#permalink]

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New post 01 Oct 2018, 02:38
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Re: How to Solve: Inequalities &nbs [#permalink] 01 Oct 2018, 02:38
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