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"

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

TYPE 1x*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 2x/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 3x*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 4x/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.html2. 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.html3. 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.html4. 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.html5. 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.htmlHope it helps!

Good Luck!