# NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

NCERT solutions for class 11 maths chapter 6 Linear Inequalities: In earlier classes, you have studied equations of one variable and two variables and have solved many problems based on this. In this article, you will get NCERT solutions for class 11 maths chapter 6 linear inequalities. Many real life problems can be solved by converting a problem into a mathematical equation but some problems like the height of all the members in your family is less than 180 cm, auditorium can occupy at most 120 tables or chairs or both can't be converted into equations. Statements which involve sign ‘’ '>' (greater than), ‘≤’ (less than or equal) and ≥ (greater than or equal), '<' (less than) are known as inequalities. The concept of inequality is used in formulating the constraints. In solutions of NCERT for class 11 maths chapter 6 linear inequalities you will understand questions based on inequalities in one variable and two variables. This chapter is useful in various field of mathematics, science and solving real life problems like cost estimation subjected to many constraints. There are 50 problems in 3 exercises of this chapter. All these questions are explained in the CBSE NCERT solutions for class 11 maths chapter 6 linear inequalities in a detailed manner. It will help you understand the concepts in a much easier way. Check all NCERT Solutions from class 6 to 12 which will help you to learn science and maths.

Let's understand this chapter with help of an example.

A manufacturing unit makes two models p and q of a product. Each piece of p requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of q requires 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The manufacturing unit makes a profit of Rs 8000 on each piece of p and Rs 12000 on each piece of Model q. Formulate this problem in linear equalities to maximize the profit.

The above problem can be formulated using linear inequalities and can be solved using linear programming which you will study in NCERT solutions for class 11 maths chapter 6 linear inequalities.

The above problem is formulated as follows.

Let x is the number of pieces of Model p and y is the number of pieces of Model q

We have to maximize the profit Z= 8000x+12000y subjected to the following constraints

$\\9x+12y\leq 180 \(fabricating \ constraint)\\x+3y\leq 30\ (finishing\ constraint)$

## The main topics of the NCERT Class 11 Maths Chapter 6 Linear Inequalities are given below:

6.1 Introduction

6.2 Inequalities

6.3 Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation

6.4 Graphical Solution of Linear Inequalities in Two Variables

6.5 Solution of System of Linear Inequalities in Two Variables

## NCERT solutions for class 11 maths chapter 6 linear inequalities-Exercise: 6.1

Given :   $24x < 100$

$\Rightarrow$      $24x < 100$

Divide by 24 from both sides

$\Rightarrow \, \, \, \frac{24}{24}x< \frac{100}{24}$

$\Rightarrow \, \, \, x< \frac{25}{6}$

$\Rightarrow \, \, \, x< 4.167$

$x$ is a natural number which is less than 4.167.

Hence, values of x can be $\left \{ 1,2,3,4 \right \}$

$x$ is an integer.

Given :   $24x < 100$

$\Rightarrow$      $24x < 100$

Divide by 24 from both sides

$\Rightarrow \, \, \, \frac{24}{24}x< \frac{100}{24}$

$\Rightarrow \, \, \, x< \frac{25}{6}$

$\Rightarrow \, \, \, x< 4.167$

$x$ is are integers which are less than 4.167.

Hence, values of x can be $\left \{..........-3,-2,-1,0, 1,2,3,4 \right \}$

Given :   $-12x>30$

$\Rightarrow$      $-12x>30$

Divide by  -12  from both side

$\Rightarrow \, \, \, \frac{-12}{-12}x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< -2.5$

$x$ is a natural number which is less than - 2.5.

Hence, the values of x do not exist for given inequality.

$x$ is an integer.

Given :   $-12x>30$

$\Rightarrow$      $-12x>30$

Divide by  -12  from both side

$\Rightarrow \, \, \, \frac{-12}{-12}x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< -2.5$

$x$ are integers less than - 2.5 .

Hence, values of x can be  $\left \{ .............,-6,-5,-4,-3 \right \}$

$x$ is an integer.

Given :   $5x - 3 < 7$

$\Rightarrow$      $5x - 3 < 7$

$\Rightarrow \, \, \, 5x< 10$

Divide by 5 from both sides

$\Rightarrow \, \, \, \frac{5}{5}x< \frac{10}{5}$

$\Rightarrow \, \, \, x< 2$

$x$ are  integers less than 2

Hence, values of x can be $\left \{.........-3,-2-1,0,1,\right \}$

$x$ is a real number.

Given :   $5x - 3 < 7$

$\Rightarrow$      $5x - 3 < 7$

$\Rightarrow \, \, \, 5x< 10$

Divide by 5 from both sides

$\Rightarrow \, \, \, \frac{5}{5}x< \frac{10}{5}$

$\Rightarrow \, \, \, x< 2$

$x$ are  real numbers less than 2

i.e.$x\in (-\infty ,2)$

Given :   $3x + 8 >2$

$\Rightarrow$      $3x + 8 >2$

$\Rightarrow \, \, \, 3x> -6$

Divide by 3 from both sides

$\Rightarrow \, \, \, \frac{3}{3}x> \frac{-6}{3}$

$\Rightarrow \, \, \, x> - 2$

$x$ are  integers greater  than -2

Hence, the values of x can be $\left \{-1,0,1,2,3,4...............\right \}$ .

Given :   $3x + 8 >2$

$\Rightarrow$      $3x + 8 >2$

$\Rightarrow \, \, \, 3x> -6$

Divide by 3 from both side

$\Rightarrow \, \, \, \frac{3}{3}x> \frac{-6}{3}$

$\Rightarrow \, \, \, x> - 2$

$x$ are  real numbers  greater  than -2

Hence , values of x can be as  $x\in (-2,\infty )$

Given : $4x + 3 < 5x + 7$

$\Rightarrow$      $4x + 3 < 5x + 7$

$\Rightarrow \, \, \, 4x-5x< 7-3$

$\Rightarrow \, \, \, x> -4$

$x$ are  real numbers greater  than -4.

Hence, values of x can be as   $x\in (-4 ,\infty )$

Given :   $3x - 7 > 5x -1$

$\Rightarrow$      $3x - 7 > 5x -1$

$\Rightarrow \, \, \, -2x> 6$

$\Rightarrow \, \, \, x< \frac{6}{-2}$

$\Rightarrow \, \, \, x< -3$

$x$ are  real numbers less than  -3.

Hence, values of x can be  $x\in (-\infty ,-3)$

Given :   $3(x-1) \leq 2(x-3)$

$\Rightarrow$      $3(x-1) \leq 2(x-3)$

$\Rightarrow \, \, \, 3x-3\leq 2x-6$

$\Rightarrow \, \, \, 3x-2x\leq -6+3$

$\Rightarrow \, \, \, x\leq -3$

$x$ are  real numbers less than equal to  -3

Hence , values of x can be as ,  $x\in (-\infty ,-3]$

Given :   $3(2- x) \geq 2(1-x)$

$\Rightarrow$      $3(2- x) \geq 2(1-x)$

$\Rightarrow \, \, \, 6-3x\geq 2-2x$

$\Rightarrow \, \, \, 6-2\geq 3x-2x$

$\Rightarrow \, \, \, 4\geq x$

$x$ are  real numbers less than equal to 4

Hence, values of x can be  as  $x\in (-\infty ,4]$

Given :   $x + \frac{x}{2} + \frac{x}{3} < 11$

$\Rightarrow$      $x + \frac{x}{2} + \frac{x}{3} < 11$

$\Rightarrow \, \, \, x(1+\frac{1}{2}+\frac{1}{3})< 11$

$\Rightarrow \, \, \, x(\frac{11}{6})< 11$

$\Rightarrow \, \, \, 11 x< 11\times 6$

$\Rightarrow \, \, \, x< 6$

$x$ are  real numbers less than 6

Hence, values of x can be  as  $x\in (-\infty ,6)$

Given :   $\frac{x}{3} > \frac{x}{2} + 1$

$\Rightarrow$      $\frac{x}{3} > \frac{x}{2} + 1$

$\Rightarrow \, \, \, \frac{x}{3}-\frac{x}{2}> 1$

$\Rightarrow \, \, \,x (\frac{1}{3}-\frac{1}{2})> 1$

$\Rightarrow \, \, \,x (-\frac{1}{6})> 1$

$\Rightarrow \, \, \, -x > 6$

$\Rightarrow \, \, \, x< -6$

$x$ are  real numbers less than -6

Hence, values of x can be  as  $x\in (-\infty ,-6)$

Given :   $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$

$\Rightarrow$      $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$

$\Rightarrow \, \, \, 9(x-2)\leq 25(2-x)$

$\Rightarrow \, \, \, 9x-18\leq 50-25x$

$\Rightarrow \, \, \, 9x+25x\leq 50+18$

$\Rightarrow \, \, \, 34x\leq 68$

$\Rightarrow \, \, \, x\leq 2$

$x$ are  real numbers less than equal to 2.

Hence, values of x can be  as  $x\in (-\infty ,2]$

Given :   $\frac{1}{2}\left(\frac{3x}{5} + 4 \right ) \geq \frac{1}{3}(x - 6)$

$\Rightarrow$      $\frac{1}{2}\left(\frac{3x}{5} + 4 \right ) \geq \frac{1}{3}(x - 6)$

$\Rightarrow \, \, 3\left(\frac{3x}{5} + 4 \right ) \geq 2(x - 6)$

$\Rightarrow \, \, \frac{9x}{5} + 12 \geq 2x-12$

$\Rightarrow \, \, 12+12 \geq 2x-\frac{9x}{5}$

$\Rightarrow \, \, 24 \geq \frac{x}{5}$

$\Rightarrow \, \, 120 \geq x$

$x$ are  real numbers less than equal to 120.

Hence, values of x can be  as  $x\in (-\infty,120 ]$.

Given :   $2(2x + 3) - 10 < 6(x-2)$

$\Rightarrow$      $2(2x + 3) - 10 < 6(x-2)$

$\Rightarrow \, \, \, 4x+6-10 < 6x-12$

$\Rightarrow \, \, \, 6-10+12 < 6x-4x$

$\Rightarrow \, \, \, 8 < 2x$

$\Rightarrow \, \, \, 4 < x$

$x$ are  real numbers greater than 4

Hence , values of x can be  as  $x\in (4,\infty )$

Given :   $37 - (3x + 5) \geq 9x - 8(x-3)$

$\Rightarrow$      $37 - (3x + 5) \geq 9x - 8(x-3)$

$\Rightarrow \, \, \, 37 - 3x - 5 \geq 9x - 8x+24$

$\Rightarrow \, \, \, 32 - 3x \geq x+24$

$\Rightarrow \, \, \, 32 - 24 \geq x+3x$

$\Rightarrow \, \, \, 8 \geq 4x$

$\Rightarrow \, \, \, 2\geq x$

$x$ are  real numbers less than equal to 2.

Hence , values of x can be  as  $x\in (-\infty ,2]$

Given :   $\frac{x}{4}< \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

$\Rightarrow$      $\frac{x}{4}< \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

$\Rightarrow \, \, \, \, 15x< 20(5x-2)-12(7x-3)$

$\Rightarrow \, \, \, \, 15x< 100x-40-84x+36$

$\Rightarrow \, \, \, \, 15x< 16x-4$

$\Rightarrow \, \, \, \, 4< x$

$x$ are  real numbers greater than 4.

Hence, values of x can be  as  $x\in (4,\infty)$

Given :   $\frac{(2x - 1)}{3} \geq \frac{3x-2}{4} - \frac{(2-x)}{5}$

$\Rightarrow$      $\frac{(2x - 1)}{3} \geq \frac{3x-2}{4} - \frac{(2-x)}{5}$

$\Rightarrow \, \, \, 20(2x - 1) \geq 15(3x-2) - 12(2-x)$

$\Rightarrow \, \, \, 40x - 20 \geq 45x-30 - 24+12x$

$\Rightarrow \, \, \, 30+24 - 20 \geq 45x-40x+12x$

$\Rightarrow \, \, \, 34 \geq 17x$

$\Rightarrow \, \, \, 2 \geq x$

$x$ are  real numbers less than equal 2.

Hence, values of x can be  as  $x\in (-\infty,2 ]$.

Given :   $3x - 2 < 2x + 1$

$\Rightarrow$      $3x - 2 < 2x + 1$

$\Rightarrow \, \, \, 3x - 2x< 2 + 1$

$\Rightarrow \, \, \, x< 3$

$x$ are  real numbers less than 3

Hence, values of x can be  as  $x\in (-\infty ,3)$

The graphical representation of solutions of the given inequality is as :

Given :   $5x - 3 \geq 3x -5$

$\Rightarrow$      $5x - 3 \geq 3x -5$

$\Rightarrow \, \, \, 5x - 3x \geq 3 -5$

$\Rightarrow \, \, \, 2x \geq -2$

$\Rightarrow \, \, \, x \geq -1$

$x$ are real numbers greater than equal to  -1.

Hence, values of x can be  as  $x\in [-1,\infty )$

The graphical representation of solutions of the given inequality is as :

Given :   $3(1-x) < 2 (x +4)$

$\Rightarrow$      $3(1-x) < 2 (x +4)$

$\Rightarrow \, \, \, 3- 3x< 2x + 8$

$\Rightarrow \, \, \, 3- 8< 2x + 3x$

$\Rightarrow \, \, \, -5< 5 x$

$\Rightarrow \, \, \, -1< x$

$x$ are  real numbers greater than -1

Hence, values of x can be  as  $x\in (-1,\infty )$

The graphical representation of solutions of given inequality is as :

Given :   $\frac{x}{2} \geq \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

$\Rightarrow$      $\frac{x}{2} \geq \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

$\Rightarrow \, \, \, 15x \geq 10(5x-2) - 6(7x-3)$

$\Rightarrow \, \, \, 15x \geq 50x-20 - 42x+18$

$\Rightarrow \, \, \, 15x+42x-50x \geq 18-20$

$\Rightarrow \, \, \, 7x \geq -2$

$\Rightarrow \, \, \, x \geq \frac{-2}{7}$

$x$ are  real numbers greater than  equal to $= \frac{-2}{7}$

Hence, values of x can be  as  $x\in (-\frac{2}{7},\infty )$

The graphical representation of solutions of the given inequality is as :

Let x be marks obtained by Ravi in the third test.

The student should have an average of at least 60 marks.

$\therefore \, \, \, \frac{70+75+x}{3}\geq 60$

$\, \, \, 145+x\geq 180$

$x\geq 180-145$

$x\geq 35$

the student should have minimum marks of 35 to have an average of 60

Sunita’s marks in the first four examinations are 87, 92, 94 and 95.

Let x be marks obtained in the fifth examination.

To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations.

$\therefore \, \, \, \frac{87+92+94+95+x}{5}\geq 90$

$\Rightarrow \, \, \, \frac{368+x}{5}\geq 90$

$\Rightarrow \, \, \, 368+x\geq 450$

$\Rightarrow \, \, \, x\geq 450-368$

$\Rightarrow \, \, \, x\geq 82$

Thus, Sunita must obtain 82 in the fifth examination to get grade ‘A’ in the course.

Let x be smaller of two consecutive odd positive integers. Then the other integer is x+2.

Both integers are smaller than 10.

$\therefore \, \, \, x+2< 10$

$\Rightarrow \, \, \, \, x< 10-2$

$\Rightarrow \, \, \, \, x< 8$

Sum of both integers is more than 11.

$\therefore \, \, \, x+(x+2)> 11$

$\Rightarrow \, \, \, (2x+2)> 11$

$\Rightarrow \, \, \, 2x> 11-2$

$\Rightarrow \, \, \, 2x> 9$

$\Rightarrow \, \, \, x> \frac{9}{2}$

$\Rightarrow \, \, \, x> 4.5$

We conclude  $\, \, \, \, x< 8$  and  $\, \, \, x> 4.5$  and x is odd integer number.

x can be 5,7.

The two pairs of consecutive odd positive integers are $(5,7)\, \, \, and\, \, \, (7,9)$.

Let x be smaller of two consecutive even positive integers. Then the other integer is x+2.

Both integers are larger than 5.

$\therefore \, \, \, x> 5$

Sum of both integers is less than 23.

$\therefore \, \, \, x+(x+2)< 23$

$\Rightarrow \, \, \, (2x+2)< 23$

$\Rightarrow \, \, \, 2x< 23-2$

$\Rightarrow \, \, \, 2x< 21$

$\Rightarrow \, \, \, x< \frac{21}{2}$

$\Rightarrow \, \, \, x< 10.5$

We conclude  $\, \, \, \, x< 10.5$  and  $\, \, \, x> 5$  and x is even integer number.

x can be 6,8,10.

The pairs of consecutive even positive integers are $(6,8),(8,10),(10,12)$.

Let the length of the smallest side be x cm.

Then largest side = 3x cm.

Third side = 3x-2  cm.

Given: The perimeter of the triangle is at least 61 cm.

$\therefore\, \, \, x+3x+(3x-2)\geq 61$

$\Rightarrow \, \, \, 7x-2\geq 61$

$\Rightarrow \, \, \, 7x\geq 61+2$

$\Rightarrow \, \, \, 7x\geq 63$

$\Rightarrow \, \, \, x\geq \frac{63}{7}$

$\Rightarrow \, \, \, x\geq 9$

Minimum length of the shortest side is 9 cm.

[Hint: If x is the length of the shortest board, then $x$ , $(x + 3)$and $2x$ are the lengths of the second and third piece, respectively. Thus,$x + (x + 3) + 2x \leq 91$ and $2x \geq (x + 3) + 5$].

Let  x is the length of the shortest board,

then  $(x + 3)$and $2x$ are the lengths of the second and third piece, respectively.

The  man wants to cut three lengths from a single piece of board of length 91cm.

Thus,$x + (x + 3) + 2x \leq 91$

$4x+3\leq 91$

$\Rightarrow \, \, \, \, 4x\leq 91-3$

$\Rightarrow \, \, \, \, 4x\leq 88$

$\Rightarrow \, \, \, \, x\leq \frac{88}{4}$

$\Rightarrow \, \, \, \, x\leq 22$

if the third piece is to be at least 5cm longer than the second, than

$2x \geq (x + 3) + 5$

$\Rightarrow \, \, \, \, 2x\geq x+8$

$\Rightarrow \, \, \, \, 2x-x\geq 8$

$\Rightarrow \, \, \, \, x\geq 8$

We conclude that  $\, \, \, \, x\geq 8$  and   $\, \, \, \, x\leq 22$.

Thus , $8\leq x\leq 22$.

Hence, the length of the shortest board is greater than equal to 8 cm and less than equal to 22 cm.

Solutions of NCERT for class 11 maths chapter 6 linear inequalities-Exercise: 6.2

$x + y < 5$

Graphical representation of  $x+y=5$ is given in the graph below.

The line $x+y=5$ divides plot in two half planes.

Select a point (not on line $x+y=5$) which lie in one of the half planes, to determine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

$1+2< 5$    i.e. $3< 5$  , which is true.

Therefore, half plane (above the line) is not a solution region of given inequality i.e. $x + y < 5$.

Also, the point on the line does not satisfy the inequality.

Thus, the solution to this inequality is half plane below the line $x+y=5$ excluding points on this line represented by the green part.

This can be represented as follows:

$2x + y \geq 6$

Graphical representation of  $2x+y=6$ is given in the graph below.

The line $2x+y=6$ divides plot in two half-planes.

Select a point (not on the line $2x+y=6$) which lie in one of the half-planes, to determine whether the point satisfies the inequality.

Let there be a  point $(3,2)$

We observe

$6+2\geq 6$    i.e. $8\geq 6$  , which is true.

Therefore, half plane II  is not a solution region of given inequality i.e. $2x + y \geq 6$

Also, the point on the line does satisfy the inequality.

Thus, the solution to this inequality is the half plane I, above  the line $2x+y=6$  including points on this line , represented by green colour.

This can be represented as follows:

$3x + 4y \leq 12$

Graphical representation of  $3x + 4y = 12$ is given in the graph below.

The line $3x + 4y = 12$ divides plot into two half-planes.

Select a point (not on the line $3x + 4y = 12$) which lie in one of the half-planes, to determine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

$1+2\leq 12$    i.e. $3\leq 12$  , which is true.

Therefore, the half plane I(above the line) is not a solution region of given inequality i.e. $3x + 4y \leq 12$.

Also, the point on the line does satisfy the inequality.

Thus, the solution to this inequality is half plane II (below the line $3x + 4y = 12$ ) including points on this line, represented by green colour.

This can be represented as follows:

$y + 8 \geq 2x$

Graphical representation of  $y + 8 = 2x$ is given in the graph below.

The line $y + 8 = 2x$  divides plot in two half-planes.

Select a point (not on the line $y + 8 = 2x$) which lie in one of the half-planes, to determine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

$2+8\geq 2\times 1$    i.e. $10\geq 2$  , which is true.

Therefore, half plane II is not solution region of given inequality i.e. $y + 8 \geq 2x$.

Also, the point on the line does satisfy the inequality.

Thus, the solution to this inequality is the half plane I including points on this line, represented by green colour.

This can be represented as follows:

$x - y \leq 2$

Graphical representation of  $x - y =2$ is given in the graph below.

The line $x - y =2$  divides plot in two half planes.

Select a point (not on the line $x - y =2$) which lie in one of the half-planes, to determine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

$1-2\leq 2$    i.e. $-1\leq 2$  , which is true.

Therefore, half plane Ii is not solution region of given inequality i.e. $x - y \leq 2$.

Also, the point on the line does satisfy the inequality.

Thus, the solution to this inequality is the half plane I including points on this line, represented by green colour

This can be represented as follows:

$2x - 3y > 6$

Graphical representation of  $2x - 3y = 6$ is given in the graph below.

The line $2x - 3y = 6$ divides plot in two half planes.

Select a point (not on the line $2x - 3y = 6$)which lie in one of the half-planes, to determine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

$2-6> 6$    i.e. $-4 > 6$  , which is false .

Therefore, half plane I is not solution region of given inequality i.e. $2x - 3y > 6$.

Also point on line does not satisfy the inequality.

Thus, the solution to this inequality is half plane II excluding points on this line, represented by green colour.

This can be represented as follows:

$-3x + 2y \geq -6$

Graphical representation of  $-3x + 2y = -6$ is given in the graph below.

The line $-3x + 2y = -6$ divides plot in two half planes.

Select a point (not on the line $-3x + 2y = -6$) which lie in one of the half planes, to determine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

$-3+4\geq -6$    i.e. $1\geq -6$  , which is true.

Therefore, half plane II  is not solution region of given inequality i.e. $-3x + 2y \geq -6$.

Also, the point on the line does satisfy the inequality.

Thus, the solution to this inequality is the half plane I including points on this line, represented by green colour

This can be represented as follows:

$3y - 5x < 30$

Graphical representation of  $3y - 5x =30$ is given in graph below.

The line $3y - 5x =30$ divides plot in two half planes.

Select a point (not on the line $3y - 5x =30$) which lie in one of the half plane , to detemine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

$6-5< 30$    i.e. $1< 30$  , which is true.

Therefore, half plane II  is not solution region of given inequality i.e.  $3y - 5x < 30$.

Also point on the line does not satisfy the inequality.

Thus, solution to this inequality is half plane I excluding points on this line, represented by green colour.

This can be represented as follows:

$y < -2$

Graphical representation of  $y=-2$  is given in graph below.

The line $y < -2$  divides plot in two half planes.

Select a point (not on the line $y < -2$) which lie in one of the half plane , to detemine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

i.e. $2< -2$  , which is false.

Therefore, the half plane I  is not a solution region of given inequality i.e. $y < -2$.

Also, the point on the line does not satisfy the inequality.

Thus, the solution to this inequality is half plane II  excluding points on this line, represented by green colour.

This can be represented as follows:

$x > - 3$

Graphical representation of  $x=-3$ is given in the graph below.

The line $x=-3$ divides plot into two half-planes.

Select a point (not on the line $x=-3$) which lie in one of the half-planes, to determine whether the point satisfies the inequality.

Let there be a  point $(1,2)$

We observe

i.e. $1> -3$  , which is true.

Therefore, half plane II  is not a solution region of given inequality i.e.  $x > - 3$.

Also, the point on the line does not satisfy the inequality.

Thus, the solution to this inequality is the half plane I excluding points on this line.

This can be represented as follows:

## CBSE NCERT solutions for class 11 maths chapter 6 linear inequalities-Exercise: 6.3

$x \geq 3,\ y\geq 2$

$x \geq 3,\ y\geq 2$

Graphical representation of  $x=3$ and  $y=2$  is given in the graph below.

The line $x=3$ and $y=2$ divides plot in four regions i.e.I,II,III,IV.

For $x \geq 3$ ,

The  solution to this inequality is region  II and  III including points on this line because points on the line also satisfy the inequality.

For $y \geq 2$,

The solution to this inequality is region  IV and  III including points on this line because points on the line also satisfy the inequality.

Hence, solution to  $x \geq 3,\ y\geq 2$ is common region of graph i.e. region III.

Thus, solution of $x \geq 3,\ y\geq 2$ is region III.

This can be represented as follows:

The below green colour represents the solution

$3x +2y \leq 12,\ x \geq 1, \ y\geq 2$

Graphical representation of  $x=1 \, \, ,3x+2y=12$ and  $y=2$  is given in graph below.

For $x \geq 1$ ,

The  solution to this inequality is region on right hand side  of line $(x=1)$ including points on this line because  points on the line also satisfy the inequality.

For $y \geq 2$,

The  solution to this inequality is region above the line $(y=2)$  including points on this line because  points on the line also satisfy the inequality.

For $3x+2y\leq 12$

The  solution to this inequality is region below  the line $(3x+2y= 12)$ including points on this line because  points on the line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$2x +y \geq 6, 3x +4y\leq 12$

Graphical representation of  $2x +y =6\, \, and\, \, 3x +4y=12$  is given in the graph below.

For  $2x +y \geq 6$,

The  solution to this inequality is region above  line $(2x +y =6)$ including points on this line because  points on the line also satisfy the inequality.

For     $3x +4y\leq 12$ ,

The  solution to this inequality is region   below the line $( 3x +4y= 12)$  including points on this line because points on the line also satisfy the inequality.

Hence, the solution to these linear inequalities is the shaded region(ABC) as shown in figure including points on the respective lines.

This can be represented as follows:

$x + y \geq 4, 2x - y <0$

Graphical representation of  $x +y =4\, \, and\, \, 2x -y=0$  is given in the graph below.

For  $x + y \geq 4,$,

The  solution to this inequality is region above  line $(x +y =4)$ including points on this line because points on the line also satisfy the inequality.

For     $2x - y <0$ ,

The solution to this inequality is half plane corresponding to the line  $( 2x -y=0)$ containing point $(1,0)$ excluding points on this line because points on the line does not  satisfy the inequality.

Hence, the solution to these linear inequalities is the shaded region as shown in figure including points on line  $(x +y =4)$  and excluding points on the line$( 2x -y=0)$.

This can be represented as follows:

$2x - y > 1, \ x -2y < -1$

Graphical representation of  $x -2y =-1\, \, and\, \, 2x -y=1$  is given in graph below.

For  $2x - y > 1,$

The  solution to this inequality is region below  line $( 2x -y=1)$ excluding points on this line because  points on line does not  satisfy the inequality.

For     $\ x -2y < -1$ ,

The  solution to this inequality is region above  the line  $(x -2y =-1)$ excluding points on this line because  points on line does not  satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure excluding points on the lines.

This can be represented as follows:

$x + y \leq 6, x + y \geq 4$

Graphical representation of  $x + y = 6,\, \, and\, \, \, x + y = 4$  is given in the graph below.

For  $x + y \leq 6,$

The  solution to this inequality is region below  line $( x+y=6)$ including points on this line because points on the line also satisfy the inequality.

For     $x + y \geq 4$ ,

The  solution to this inequality is region above  the line  $( x+y=4)$ including points on this line because points on the line also satisfy the inequality.

Hence, the solution to these linear inequalities is shaded region as shown in figure including points on the lines.

This can be represented as follows:

$2x + y \geq 8 , x + 2y \geq 10$

Graphical representation of  $2x + y = 8\, \, and\, \, x + 2y =10$  is given in graph below.

For  $2x + y \geq 8 ,$

The  solution to this inequality is region above line $(2x + y = 8)$  including points on this line because  points on line also satisfy the inequality.

For     $x + 2y \geq 10$ ,

The  solution to this inequality is region above  the line  $( x + 2y =10)$  including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure  including points on the lines.

This can be represented as follows:

$x + y \leq 9, y > x, x\geq 0$

Graphical representation of  $x+y=9,x=y$ and  $x=0$  is given in graph below.

For $x + y \leq 9$ ,

The  solution to this inequality is region below  line $(x+y=9)$ including points on this line because  points on line also satisfy the inequality.

For $y > x$,

The  solution to this inequality represents half plane corresponding to the line $(x=y)$  containing point$(0,1)$ excluding points  on this line because  points on line does not  satisfy the inequality.

For $x\geq 0$

The  solution to this inequality is region on right hand side of the line $(x=0)$ including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure.

This can be represented as follows:

$5x+4y\leq20, \ x\geq 1, \ y\geq 2$

Graphical representation of  $\, ,5x+4y=20,\, \, \, x=1\, \, and \, \, y=2$ is given in graph below.

For $5x+4y\leq20,$ ,

The  solution to this inequality is region below the  line $(5x+4y=20)$ including points on this line because  points on line also satisfy the inequality.

For $\ x\geq 1,$,

The  solution to this inequality is region right hand side of the  line $(x=1)$  including points on this line because  points on line also satisfy the inequality.

For $\ y\geq 2,$

The  solution to this inequality is region above  the line $(y=2)$ including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$3x + 4y \leq 60,\ x + 3y \leq 30, \ x \geq 0, \ y\geq 0$

Graphical representation of  $3x+4y=60 \, \, ,x+3y=30\, \, \, ,x=0\, \, and\, \, y=0$  is given in graph below.

For $3x + 4y \leq 60$ ,

The  solution to this inequality is region below the  line $(3x+4y=60)$ including points on this line because  points on line also satisfy the inequality.

For $\ x + 3y \leq 30$,

The  solution to this inequality is region below the line $(x+3y=30)$  including points on this line because  points on line also satisfy the inequality.

For $\ x \geq 0,$

The  solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because  points on line also satisfy the inequality.

For $\ y \geq 0,$

The  solution to this inequality is region above  the line $(y=0)$ including points on this line because  points on line also satisfy the inequality.

Hence, the solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$2x +y \geq 4, \ x + y \leq 3, \ 2x - 3y \leq 6$

Graphical representation of  $2x+y=4 \, \, ,x+y=3$ and  $2x-3y=6$  is given in graph below.

For $2x +y \geq 4,$ ,

The  solution to this inequality is region above the line $(2x+y=4)$ including points on this line because  points on line also satisfy the inequality.

For $\ x + y \leq 3,$,

The  solution to this inequality is region below the line $(x+y=3)$  including points on this line because  points on line also satisfy the inequality.

For $\ 2x - 3y \leq 6,$

The  solution to this inequality is region above  the line $(2x-3y= 6)$ including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$x -2y \leq 3, 3x + 4y \geq 12, x \geq 0, y\geq 1$

Graphical representation of  $x-2y=3 \, \, ,3x+4y=12\, \, \, ,x=0\, \, and\, \, y=1$  is given in graph below.

For $x -2y \leq 3$ ,

The  solution to this inequality is region above the  line $(x-2y=3)$ including points on this line because  points on line also satisfy the inequality.

For $3x + 4y \geq 12$,

The  solution to this inequality is region above the line $(3x+4y=12)$  including points on this line because  points on line also satisfy the inequality.

For $\ x \geq 0,$

The  solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because  points on line also satisfy the inequality.

For $\ y \geq 1,$

The  solution to this inequality is region above  the line $(y=1)$ including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$4x + 3y \leq 60,\ y\geq 2x,\ x\geq 3,\ x,y\geq 0$

Graphical representation of  $4x+3y=60 \, \, ,y=2x\, \, \,,x=3\, \, ,x=0\, \, and\, \, y=0$  is given in graph below.

For $4x + 3y \leq 60,$

The  solution to this inequality is region below the  line $(4x+3y=60)$ including points on this line because points on the line also satisfy the inequality.

For $y\geq 2x$,

The  solution to this inequality is region above the line $(y=2x)$  including points on this line because  points on the line also satisfy the inequality.

For $x\geq 3$,

The  solution to this inequality is region right hand side of  the line $(x=3)$  including points on this line because  points on the line also satisfy the inequality.

For $\ x \geq 0,$

The  solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because  points on  the line also satisfy the inequality.

For $\ y \geq 0,$

The  solution to this inequality is region above  the line $(y=0)$ including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$3x + 2y \leq 150, \ x +4y \leq 80,\ x\leq 15 \ y\geq 0, \ x\geq 0$

Graphical representation of  $3x+2y=150 \, \, ,x+4y=80\, \, \,,x=15\, \, ,x=0\, \, and\, \, y=0$  is given in graph below.

For $3x + 2y \leq 150,$

The  solution to this inequality is region below the  line $(3x+2y=150)$ including points on this line because points on the line also satisfy the inequality.

For $x+4y\leq 80$,

The  solution to this inequality is region below the line $(x+4y=80)$  including points on this line because points on the line also satisfy the inequality.

For $x\leq 15$,

The  solution to this inequality is region left hand side of  the line $(x=15)$  including points on this line because points on the line also satisfy the inequality.

For $\ x \geq 0,$

The  solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because points on the line also satisfy the inequality.

For $\ y \geq 0,$

The  solution to this inequality is region above  the line $(y=0)$ including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

$x+2y \leq 10, \ x +y \geq 1, \ x-y\leq 0, x\geq 0, \ y\geq 0$

Graphical representation of  $x+2y=10 \, \, ,x+y=1\, \, \,,x-y=0\, \, ,x=0\, \, and\, \, y=0$  is given in graph below.

For $x+2y \leq 10,$

The  solution to this inequality is region below the  line $(x+2y=10)$ including points on this line because  points on line also satisfy the inequality.

For $\ x +y \geq 1,$,

The  solution to this inequality is region above the line $(x+y=1)$  including points on this line because  points on line also satisfy the inequality.

For $\ x-y\leq 0,$,

The  solution to this inequality is region above the line $(x-y=0)$  including points on this line because  points on line also satisfy the inequality.

For $\ x \geq 0,$

The  solution to this inequality is region right hand side of the line $(x=0)$ including points on this line because  points on line also satisfy the inequality.

For $\ y \geq 0,$

The  solution to this inequality is region above  the line $(y=0)$ including points on this line because  points on line also satisfy the inequality.

Hence, solution to these linear inequalities is shaded region as shown in figure including points on the respective lines.

This can be represented as follows:

NCERT solutions for class 11 maths chapter 6 linear inequalities-Miscellaneous Exercise

Given :  $2\leq 3x-4\leq5$

$2\leq 3x-4\leq5$

$\Rightarrow\, \, 2+4\leq 3x\leq 5+4$

$\Rightarrow\, \, 6\leq 3x\leq 9$

$\Rightarrow\, \, \frac{6}{3}\leq x\leq \frac{9}{3}$

$\Rightarrow\, \, 2\leq x\leq 3$

Thus, all the real numbers greater than equal to 2 and less than equal to 3 are solutions to this inequality.

Solution set is $\left $2,3 \right$$

Given  $6 \leq -3(2x - 4) < 12$

$6 \leq -3(2x - 4) < 12$

$\Rightarrow\, \ \frac{6}{3}\leq -(2x-4)< \frac{12}{3}$

$\Rightarrow\, \ -2\geq (2x-4)> -4$

$\Rightarrow\, \ -2+4\geq 2x> -4+4$

$\Rightarrow\, \ 2\geq 2x> 0$

$\Rightarrow\, \ 1\geq x> 0$

Solution set is $(01]$

Given   $-3 \leq 4 - \frac{7x}{2}\leq 18$

$\Rightarrow \, \, -3 \leq 4 - \frac{7x}{2}\leq 18$

$\Rightarrow \, \, -3-4 \leq - \frac{7x}{2}\leq 18-4$

$\Rightarrow \, \, -7 \leq - \frac{7x}{2}\leq 14$

$\Rightarrow \, \, 7 \geq \frac{7x}{2} \geq -14$

$\Rightarrow \, \, 7\times 2 \geq 7x\geq -14\times 2$

$\Rightarrow \, \, 14 \geq 7x \geq -28$

$\Rightarrow \, \, \frac{14}{7} \geq x \geq \frac{-28}{7}$

$\Rightarrow \, \, 2 \geq x \geq -4$

Solution set is $[-4,2]$

Given The inequality

$-15 < \frac{3(x-2)}{5} \leq 0$

$-15 < \frac{3(x-2)}{5} \leq 0$

$\Rightarrow\, \ -15\times 5< 3(x-2)\leq 0\times 5$

$\Rightarrow\, \ -75< 3(x-2)\leq 0$

$\Rightarrow\, \ \frac{-75}{3}< (x-2)\leq \frac{0}{3}$

$\Rightarrow\, \ -25< (x-2)\leq 0$

$\Rightarrow\, \ -25+2< x\leq 0+2$

$\Rightarrow\, \ -23< x\leq 2$

The solution set is $(-23,2]$

Given the inequality

$-12<4-\frac{3x}{-5} \leq 2$

$-12<4-\frac{3x}{-5} \leq 2$

$\Rightarrow\, \, -12-4< -\frac{3x}{-5}\leq 2-4$

$\Rightarrow\, \, -16< -\frac{3x}{-5}\leq -2$

$\Rightarrow\, \, -16< \frac{3x}{5}\leq -2$

$\Rightarrow\, \, -16\times 5< 3x\leq -2\times 5$

$\Rightarrow\, \, -80< 3x\leq -10$

$\Rightarrow\, \, \frac{-80}{3}< 3x\leq \frac{-10}{3}$

Solution set is $(\frac{-80}{3}, \frac{-10}{3}]$

Given the linear inequality

$7 \leq \frac{(3x+ 11)}{2}\leq 11$

$7 \leq \frac{(3x+ 11)}{2}\leq 11$

$\Rightarrow \, \, 7\times 2 \leq (3x+ 11)\leq 11\times 2$

$\Rightarrow \, \, 14 \leq (3x+ 11)\leq 22$

$\Rightarrow \, \, 14-11 \leq (3x)\leq 22-11$

$\Rightarrow \, \, 3 \leq 3x\leq 11$

$\Rightarrow \, \, 1 \leq x\leq \frac{11}{3}$

The solution set  of the given inequality is  $[1,\frac{11}{3}]$

Given : $5x + 1 > -24,\ 5x - 1 <24$

$5x + 1 > -24\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x - 1 <24$

$\Rightarrow 5x > -24-1\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x <24+1$

$\Rightarrow 5x > -25\, \, \, \, \, \, \, and\, \, \, \, \, \, \ 5x <25$

$\Rightarrow x > \frac{-25}{5}\, \, \, \, \, \, \, and\, \, \, \, \, \, \ x <\frac{25}{5}$

$\Rightarrow x > -5\, \, \, \, \, \, \, and\, \, \, \, \, \, \ x <5$

$(-5,5)$

The solution graphically on the number line is as shown :

Given : $2(x-1) 2 -x$

$2(x-1) 2 -x$

$\Rightarrow \, \, 2x-2 2 -x$

$\Rightarrow \, \, 2x-x<2+5\, \, \, \, and\, \, \, \, \, \ 3x+x> 2 -6$

$\Rightarrow \, \, x<7\, \, \, \, and\, \, \, \, \, \ 4x> -4$

$\Rightarrow \, \, x<7\, \, \, \, and\, \, \, \, \, \ x> -1$

$(-1,7)$

The solution graphically on the number line is as shown :

Given : $3x - 7 > 2(x-6),\ 6-x > 11 - 2x$

$3x - 7 > 2(x-6)\, \, \, \, and\, \, \, \, \, \ 6-x > 11 - 2x$

$\Rightarrow \, \, 3x - 7 > 2x-12\, \, \, \, and\, \, \, \, \, \ 6-x > 11 - 2x$

$\Rightarrow \, \, 3x - 2x >7-12\, \, \, \, and\, \, \, \, \, \ 2x-x > 11 - 6$

$\Rightarrow \, \, x >-5\, \, \, \, and\, \, \, \, \, \ x > 5$

$x\in (5,\infty )$

The solution graphically on the number line is as shown :

$5(2x-7)-3(2x+3)\leq 0,\quad 2x + 19 \leq 6x +47$

Given : $5(2x-7)-3(2x+3)\leq 0,\quad 2x + 19 \leq 6x +47$

$5(2x-7)-3(2x+3)\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad 2x + 19 \leq 6x +47$

$\Rightarrow \, \, 10x-35-6x-9\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad 2x -6x\leq 47-19$

$\Rightarrow \, \, 4x-44\leq 0\, \, \, \, \, and\, \, \, \, \, \, \, \quad -4x\leq 28$

$\Rightarrow \, \, 4x\leq 44\, \, \, \, \, and\, \, \, \, \, \, \, \quad 4x\geq - 28$

$\Rightarrow \, \, x\leq 11\, \, \, \, \, and\, \, \, \, \, \, \, \quad x\geq - 7$

$x\in [-7,11]$

The solution graphically on the number line is as shown :

Since the solution is to be kept between 68° F and 77° F.

$68< F< 77$

Putting the value of  $F = \frac{9}{5}C + 32$ , we have

$\Rightarrow \, \, \, 68< \frac{9}{5}C + 32< 77$

$\Rightarrow \, \, \, 68-32< \frac{9}{5}C < 77-32$

$\Rightarrow \, \, \, 36< \frac{9}{5}C < 45$

$\Rightarrow \, \, \, 36\times 5< 9C < 45\times 5$

$\Rightarrow \, \, \, 180< 9C < 225$

$\Rightarrow \, \, \, \frac{180}{9}< C < \frac{225}{9}$

$\Rightarrow \, \, \, 20< C < 25$

the range in temperature in degree Celsius (C) is between  20 to 25.

Let x litres of 2% boric acid solution is required to be added.

Total mixture = (x+640) litres

The resulting mixture is to be more than 4% but less than 6% boric acid.

$\therefore \, 2\%x+8\%\, of\, 640> 4\%\, of\, (640+x)$                    and       $2\%x+8\%\, of\, 640< 6\%\, of\, (x+640)$

$\Rightarrow \, 2\%x+8\%\, of\, 640> 4\%\, of\, (640+x)$                  and      $2\%x+8\%\, of\, 640< 6\%\, of\, (x+640)$

$\Rightarrow \, \frac{2}{100}x+(\frac{8}{100}) 640> \frac{4}{100} (640+x)$                                 $\Rightarrow \, \frac{2}{100}x+(\frac{8}{100}) 640< \frac{6}{100} (640+x)$

$\Rightarrow \, 2x+5120> 4x+2560$                                                        $\Rightarrow \, 2x+5120< 6x+3840$

$\Rightarrow \, 5120-2560> 4x-2x$                                                       $\Rightarrow \, 5120-3840< 6x-2x$

$\Rightarrow \, 2560> 2x$                                                                                   $\Rightarrow \, 1280< 4x$

$\Rightarrow \, 1280> x$                                                                                     $\Rightarrow \, 320< x$

Thus, the number of litres  2% of boric acid solution that is to be added will have to be more than 320 and less than 1280 litres.

Let  x  litres of water is required to be added.

Total mixture = (x+1125) litres

It is evident that amount of acid contained in the resulting mixture is 45% of 1125 litres.

The resulting mixture contain  more than 25 % but less than 30%  acid.

$\therefore \, 30\%\, of\, (1125+x) > 45\%\, of\, (1125)$                    and       $25\%\, of\, (1125+x)< 45\%\, of\, 1125$

$\Rightarrow \, 30\%\, of\, (1125+x) > 45\%\, of\, (1125)$                  and       $25\%\, of\, (1125+x)< 45\%\, of\, 1125$

$\Rightarrow \, \frac{30}{100}(1125+x)> \frac{45}{100} (1125)$                                 $\Rightarrow \, (\frac{25}{100}) (1125+x)< \frac{45}{100} (1125)$

$\Rightarrow \, 30\times 1125+30x> 45\times (1125)$                              $\Rightarrow \, 25 (1125+x)< 45(1125)$

$\Rightarrow \, 30x> (45-30)\times (1125)$                                            $\Rightarrow \, 25 x< (45-25)1125$

$\Rightarrow \, 30x> (15)\times (1125)$                                                        $\Rightarrow \, 25 x< (20)1125$

$\Rightarrow \, x> \frac{15\times 1125}{30}$                                                                   $\Rightarrow \, x< \frac{20\times 1125}{25}$

$\Rightarrow \, x> 562.5$                                                                              $\Rightarrow \, x< 900$

Thus, the number of litres water that is to be added will have to be more than 562.5 and less than 900 litres.

Given that group of 12 years old children.

$80\leq IQ\leq140$

For a group of 12 years old children,   CA =12 years

$IQ= \frac{MA}{CA}\times 100$

Putting the value of IQ, we obtain

$80\leq IQ\leq140$

$\Rightarrow \, \, 80\leq \frac{MA}{CA}\times 100\leq140$

$\Rightarrow \, \, 80\leq \frac{MA}{12}\times 100\leq140$

$\Rightarrow \, \, 80\times 12\leq MA\times 100\leq140\times 12$

$\Rightarrow \, \, \frac{80\times 12}{100}\leq MA\leq \frac{140\times 12}{100}$

$\Rightarrow \, \, 9.6\leq MA\leq 16.8$

Thus, the range of mental age of the group  of 12 years old children is $\, \, 9.6\leq MA\leq 16.8$

## NCERT solutions for class 11 mathematics

 chapter-1 NCERT solutions for class 11 maths chapter 1 Sets chapter-2 Solutions of NCERT for class 11 chapter 2 Relations and Functions chapter-3 CBSE NCERT solutions for class 11 chapter 3 Trigonometric Functions chapter-4 NCERT solutions for class 11 chapter 4 Principle of Mathematical Induction chapter-5 Solutions of NCERT for class 11 chapter 5 Complex Numbers and Quadratic equations chapter-6 CBSE NCERT solutions for class 11 maths chapter 6 Linear Inequalities chapter-7 NCERT solutions for class 11 maths chapter 7 Permutation and Combinations chapter-8 Solutions of NCERT for class 11 maths chapter 8 Binomial Theorem chapter-9 CBSE NCERT solutions for class 11 maths chapter 9 Sequences and Series chapter-10 NCERT solutions for class 11 maths chapter 10 Straight Lines chapter-11 Solutions of NCERT for class 11 maths chapter 11 Conic Section chapter-12 CBSE NCERT solutions for class 11 maths chapter 12 Introduction to Three Dimensional Geometry chapter-13 NCERT solutions for class 11 maths chapter 13 Limits and Derivatives chapter-14 Solutions of NCERT for class 11 maths chapter 14 Mathematical Reasoning chapter-15 CBSE NCERT solutions for class 11 maths chapter 15 Statistics chapter-16 NCERT solutions for class 11 maths chapter 16 Probability

## NCERT solutions for class 11- Subject wise

 Solutions of NCERT for class 11 biology CBSE NCERT solutions for class 11 maths NCERT solutions for class 11 chemistry Solutions of NCERT for Class 11 physics

## Benefits of NCERT solutions

• NCERT solutions for class 11 maths chapter 6 linear inequalities will build your fundamentals which will be helpful in solving many real-life problems like maximizing the profit, minimizing the expenditure, allocating the resources with given constraints.

• As all the above questions are prepared and explained in a step-by-step manner with the help of the graphs, it can be understood and visualize the problem easily.

• CBSE NCERT solutions for class 11 maths chapter 6 linear inequalities will some innovative ways of solving the problems which become very important to solve some specific problems in an easy way.

• This chapter also useful in the prediction of future events based on the past data which is the fundamentals of machine learning