# Quantitative Aptitude > Inequalities

 Refresher Test - Inequalities Discuss

Practice Tests
 Practice Test - Inequalities Discuss

Free Online Preparation for CAT 2009 with Minglebox e-CAT Prep. Cover basic concepts of Inequalities under Quantitative Aptitude for MBA Entrance Exam Preparation with Study material, solved examples and tests prepared by CAT coaching experts.

A comparison relationship between two algebraic expression or quantities is known as an Inequalities.

For example

etc.

In any inequality problem variable x is assumed to be real number unless otherwise . Any inequality is generally either one of the following

'greater then' (>)

'greater than or equal to '( >)

'less than' (<)

' less than or equal to' ( < )

Solution of an inequalities

By solution of an inequalities we seek to find a set of value for the variable involve in the problem so that inequalities holds true for all the values lying in the solution set.

An Example

Question

3x+7 > x+1

Solution

We find that any value of x > -3 is a solution of the above inequalities. For instance

let us take x = -2, -1, 0, .5 etc. We find

3(-2)+7 > -2+1 1 > -1 hence true

3(-1)+7 > -1+1 4 > 0  hence true

3(0)+7 > 0+1 7 > 0    hence true

Law of inequalities

if x > y then for all c x+c > y+c and x-c > y-c

If x > y and c > 0 then cx > cy and

If x > y then for c < 0 , xc < yc and that is sign of inequalities gets reversed when both sides of

the inequalities are multiplied or divided by the same negative quantity. If x > y >0 then

8 > 2 > 0

also if 0 > x > y then

but x> 0 > y then

Relationship Between Power of a Number

If x > 1 then …. ............. …........

but if 0< 1 then …. .....

In general if x > 1 then if m> n and if 0 < x < 1 then if m < n

Let us study how the sign of a quadratic expression changes as we very the value of x.

let us consider that is

(x-2) (x-1) for x > 2 we find x-2 > 0 and x-1>0 therefore for all x >2 (x-2) (x-1) > 0

for 1 < x < 2 x-2 < 0 but x-1 > 0 (x-2) (x-1) < 0

where for x < 1 then x-2 < 0 and x-1 < 0 (x-2) (x-1) < 0

thus if we have to solve > 0 we proceed as follows.

• factorize the given expression linear factors.

• Equate each linear term equal to zero and find values of x.

• Arrange the value of x is directly order , so if values are and and if then for all expression is > 0 for all then expression is < 0 and for all expression > 0.

While Factorizing the expression in linear factor make sure to make the sign of coefficient of as positive.

Solved Examples

Question

Solve

Solution

Observe for all and for , > 0. Therefore is the positive factor of the

above expression hence we can divided both sides by we get also

we again observe for all and for

Hence we just need to consider (x-1) (x-3) for solving inequalities now let's equal each linear term equal to zero we get x = 1 and x = 3 therefore for all x >3 (x-1) > 0 and (x-3) > 0 hence (x-1) (x-3) > 0

also for x < 1 , (x-1) < 0 and (x-3) < 0 hence again (x-1) (x-3) > 0 therefore solution set is given by

Question

Solve - |x| + 2 < 0

Solution

We know that for all x > 0

=

Hence become factorizing into linear factor we get

Equality each linear factor equal to zero , we get

and

Hence if then

but for x > 0 and

= -x for -x < 0

above solution is 1 < x < 2 or 1 < -x < -2 or -1 > x > -2 hence following we get solution is

Question

Solve

Solution

Adding -(4 – x) to both side we get

now for all therefore are multiply both side by we get

Discriminant of is given by

= 4 – 48 = -44 < 0

for all real value of 2.

dividing both sides by as the solution . Hence is given by

Question

If x > 3, y > -2 then which of the following holds good

(1) xy > -6

(2) xy < -6

(3)

(4)none of these

Solution

We observe that y > -2 y can be positive as well as negative that is if y > -2 y can value living between 0 > y > -2 as well as y > 0

Hence we solve this using method of options let x = 4 and y = -1

then xy = -4 > -6 therefore 'a' is true but 'b' is not also 4 > 3 hence c also holds let us take x = 5 and

y = - ½ we get

but

hence '3' is not trueFinally if we take x = 18 & y = -1/2 we get

hence 'a' also does not s good

Question

Solve If x > 5 and y < -1 , hence which of the following statements is true ?

(1)(x + 4y) > 1,

1. (x > -4y)

2. (3)-4x < 5y

3. none of these

Solution

Let us go by each option and take specified values of x and y

If x = 6 and y = -10

Then x+4y = 6 + 4(-10) = -34 < 1 hence (1) is not true

also if y < 1

-4y > 4

but x > 5 therefore taking x =6 , y = -10 we get 6 < 40 hence (2) is also not true

From (3) we find that

x > 5

-4x < -20

as y < -1

5y c-5

again taking x = 6 and y = - 10

we find that - 24 > -50 hence (3) is also not true

 Refresher Test - Inequalities Discuss

 Practice Test - Inequalities Discuss