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NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities

Edited By Ramraj Saini | Updated on Sep 23, 2023 06:01 PM IST

Linear Inequalities Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities are provided here. These NCERT Solutions are created by expert team at Careers360 keeping in mind of latest syllabus of CBSE 2023-24. 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 linear inequalities class 11 NCERT solutions. Class 11 Mathematics NCERT book will help you understand the concepts in a much easier way. Here you will get NCERT solutions for class 11 also.

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This Story also Contains
  1. Linear Inequalities Class 11 Questions And Answers
  2. Linear Inequalities Class 11 Questions And Answers PDF Free Download
  3. Linear Inequalities Class 11 Solutions - Important Formulae And Points
  4. Linear Inequalities Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. Linear Inequality Example
  6. NCERT Solutions For Class 11 Mathematics - Chapter Wise
  7. NCERT Solutions For Class 11- Subject Wise
NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities
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. T he concept of inequality is used in formulating the constraints. In NCERT solutions for class 11 maths chapter 6 linear inequalities you will understand questions based on inequalities in one variable and two variables.

Linear Inequalities Class 11 Questions And Answers PDF Free Download

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Linear Inequalities Class 11 Solutions - Important Formulae And Points

Inequation (Inequality):

An inequation or inequality is a statement involving variables and the sign of inequality like >, <, ≥, or ≤.

Symbols used in inequalities:

The symbol < means less than.

The symbol > means greater than.

The symbol < with a bar underneath ≤ means less than or equal to.

The symbol > with a bar underneath ≥ means greater than or equal to.

The symbol ≠ means the quantities on the left and right sides are not equal to.

Algebraic Solutions for Linear Inequalities in One Variable:

Linear inequalities involve expressions with variables and inequality symbols like <, >, ≤, or ≥.

The solution to a linear inequality can be determined using algebraic methods.

Important rules to follow when solving linear inequalities:

  • Rule 1: Don’t change the sign of an inequality by adding or subtracting the same integer on both sides of an equation.

  • Rule 2: Add or subtract the same positive integer from both sides of an inequality equation.

Graphical Representation of Linear Inequalities:

Linear inequalities can also be represented graphically on a number line.

For example, x > 3 represents all real numbers greater than 3, which can be shaded on the number line to the right of 3.

Similarly, x ≤ -2 represents all real numbers less than or equal to -2, which can be shaded on the number line to the left of -2.

Free download NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities for CBSE Exam.

Linear Inequalities Class 11 NCERT Solutions (Intext Questions and Exercise)

Class 11 maths chapter 6 question answer - Exercise 6.1

Question:1(i) Solve 24x<100 , when
x i s a natural number.

Answer:

Given : 24x<100

24x<100

Divide by 24 from both sides

2424x<10024

x<256

x<4.167

x i s a natural number which is less than 4.167.

Hence, values of x can be {1,2,3,4}

Question:1(ii) Solve 24x<100 , when

x is an integer.

Answer:

Given : 24x<100

24x<100

Divide by 24 from both sides

2424x<10024

x<256

x<4.167


x i s are integers which are less than 4.167.

Hence, values of x can be {..........3,2,1,0,1,2,3,4}

Question:2(i) Solve 12x>30 , when
x is a natural number.

Answer:

Given : 12x>30

12x>30

Divide by -12 from both side

1212x<3012

x<3012

x<2.5

x i s a natural number which is less than - 2.5.

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

Question:2(ii) Solve 12x>30 , when

x is an integer.

Answer:

Given : 12x>30

12x>30

Divide by -12 from both side

1212x<3012

x<3012

x<2.5

x are integers less than - 2.5 .

Hence, values of x can be {.............,6,5,4,3}

Question:3(i) Solve 5x3<7 , when

x is an integer.

Answer:

Given : 5x3<7

5x3<7

5x<10

Divide by 5 from both sides

55x<105

x<2

x are integers less than 2

Hence, values of x can be {.........3,21,0,1,}

Question:3(ii) Solve 5x3<7 , when

x is a real number.

Answer:

Given : 5x3<7

5x3<7

5x<10

Divide by 5 from both sides

55x<105

x<2

x are real numbers less than 2

i.e. x(,2)

Question:4(i) Solve 3x+8>2 , when
x is an integer.

Answer:

Given : 3x+8>2

3x+8>2

3x>6

Divide by 3 from both sides

33x>63

x>2


x are integers greater than -2

Hence, the values of x can be {1,0,1,2,3,4...............} .

Question:4(ii) Solve 3x+8>2 , when ) x is a real number.

Answer:

Given : 3x+8>2

3x+8>2

3x>6

Divide by 3 from both side

33x>63

x>2


x are real numbers greater than -2

Hence , values of x can be as x(2,)

Question:5 Solve the inequality for real x . 4x+3<5x+7

Answer:

Given : 4x+3<5x+7

4x+3<5x+7

4x5x<73

x>4

x are real numbers greater than -4.

Hence, values of x can be as x(4,)

Question:6 Solve the inequality for real x 3x7>5x1

Answer:

Given : 3x7>5x1

3x7>5x1

2x>6

x<62

x<3

x are real numbers less than -3.

Hence, values of x can be x(,3)

Question:7 Solve the inequality for real x . 3(x1)2(x3)

Answer:

Given : 3(x1)2(x3)

3(x1)2(x3)

3x32x6

3x2x6+3

x3

x are real numbers less than equal to -3

Hence , values of x can be as , x(,3]

Question:8 Solve the inequality for real x 3(2x)2(1x)

Answer:

Given : 3(2x)2(1x)

3(2x)2(1x)

63x22x

623x2x

4x

x are real numbers less than equal to 4

Hence, values of x can be as x(,4]

Question:9 Solve the inequality for real x x+x2+x3<11

Answer:

Given : x+x2+x3<11

x+x2+x3<11

x(1+12+13)<11

x(116)<11

11x<11×6

x<6

x are real numbers less than 6

Hence, values of x can be as x(,6)

Question:10 Solve the inequality for real x . x3>x2+1

Answer:

Given : x3>x2+1

x3>x2+1

x3x2>1

x(1312)>1

x(16)>1

x>6

x<6

x are real numbers less than -6

Hence, values of x can be as x(,6)

Question:11 Solve the inequality for real x 3(x2)55(2x)3

Answer:

Given : 3(x2)55(2x)3

3(x2)55(2x)3

9(x2)25(2x)

9x185025x

9x+25x50+18

34x68

x2

x are real numbers less than equal to 2.

Hence, values of x can be as x(,2]

Question:12 Solve the inequality for real x 12(3x5+4)13(x6)

Answer:

Given : 12(3x5+4)13(x6)

12(3x5+4)13(x6)

3(3x5+4)2(x6)

9x5+122x12

12+122x9x5

24x5

120x

x are real numbers less than equal to 120.

Hence, values of x can be as x(,120] .

Question:13 Solve the inequality for real x 2(2x+3)10<6(x2)

Answer:

Given : 2(2x+3)10<6(x2)

2(2x+3)10<6(x2)

4x+610<6x12

610+12<6x4x

8<2x

4<x

x are real numbers greater than 4

Hence , values of x can be as x(4,)

Question:14 Solve the inequality for real x 37(3x+5)9x8(x3)

Answer:

Given : 37(3x+5)9x8(x3)

37(3x+5)9x8(x3)

373x59x8x+24

323xx+24

3224x+3x

84x

2x

x are real numbers less than equal to 2.

Hence , values of x can be as x(,2]

Question:15 Solve the inequality for real x x4<(5x2)3(7x3)5

Answer:

Given : x4<(5x2)3(7x3)5

x4<(5x2)3(7x3)5

15x<20(5x2)12(7x3)

15x<100x4084x+36

15x<16x4

4<x

x are real numbers greater than 4.

Hence, values of x can be as x(4,)

Question:16 Solve the inequality for real x (2x1)33x24(2x)5

Answer:

Given : (2x1)33x24(2x)5

(2x1)33x24(2x)5

20(2x1)15(3x2)12(2x)

40x2045x3024+12x

30+242045x40x+12x

3417x

2x

x are real numbers less than equal 2.

Hence, values of x can be as x(,2] .

Question:17 Solve the inequality and show the graph of the solution on number line 3x2<2x+1

Answer:

Given : 3x2<2x+1

3x2<2x+1

3x2x<2+1

x<3

x are real numbers less than 3

Hence, values of x can be as x(,3)

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

1635762263704

Question:18 Solve the inequality and show the graph of the solution on number line 5x33x5

Answer:

Given : 5x33x5

5x33x5

5x3x35

2x2

x1

x are real numbers greater than equal to -1.

Hence, values of x can be as x[1,)

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

1635762285990

Question:19 Solve the inequality and show the graph of the solution on number line 3(1x)<2(x+4)

Answer:

Given : 3(1x)<2(x+4)

3(1x)<2(x+4)

33x<2x+8

38<2x+3x

5<5x

1<x

x are real numbers greater than -1

Hence, values of x can be as x(1,)

The graphical representation of solutions of given inequality is as :

1635762322180

Question:20 Solve the inequality and show the graph of the solution on number line x2(5x2)3(7x3)5

Answer:

Given : x2(5x2)3(7x3)5

x2(5x2)3(7x3)5

15x10(5x2)6(7x3)

15x50x2042x+18

15x+42x50x1820

7x2

x27

x are real numbers greater than equal to =27

Hence, values of x can be as x(27,)

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

1635762357828

Question:21 Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Answer:

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

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

70+75+x360

145+x180

x180145

x35

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

Question:22 To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

Answer:

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.

87+92+94+95+x590

368+x590

368+x450

x450368

x82

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

Question:23 Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Answer:

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

Both integers are smaller than 10.

x+2<10

x<102

x<8

Sum of both integers is more than 11.

x+(x+2)>11

(2x+2)>11

2x>112

2x>9

x>92

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) .

Question:24 Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Answer:

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

Both integers are larger than 5.

x>5

Sum of both integers is less than 23.

x+(x+2)<23

(2x+2)<23

2x<232

2x<21

x<212

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) .

Question:25 The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Answer:

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.

x+3x+(3x2)61

7x261

7x61+2

7x63

x637

x9

Minimum length of the shortest side is 9 cm.

Question:26 A man wants to cut three lengths from a single piece of board of length 91cm. The second length is to be 3cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5cm longer than the second?

[ 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)+2x91 and 2x(x+3)+5 ].

Answer:

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)+2x91

4x+391

4x913

4x88

x884

x22

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

2x(x+3)+5

2xx+8

2xx8

x8

We conclude that x8 and x22 .

Thus , 8x22 .

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


Class 11 maths chapter 6 question answer - Exercise: 6.2

Question:1 Solve the following inequality graphically in two-dimensional plane:

x+y<5

Answer:

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:

1635762532191

Question:2 Solve the following inequality graphically in two-dimensional plane: 2x+y6

Answer:

2x+y6

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+26 i.e. 86 , which is true.

Therefore, half plane II is not a solution region of given inequality i.e. 2x+y6

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:

1635762558839

Question:3 Solve the following inequality graphically in two-dimensional plane: 3x+4y12

Answer:

3x+4y12

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+212 i.e. 312 , which is true.

Therefore, the half plane I(above the line) is not a solution region of given inequality i.e. 3x+4y12 .

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:

1635762572337

Question:4 Solve the following inequality graphically in two-dimensional plane: y+82x

Answer:

y+82x

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+82×1 i.e. 102 , which is true.

Therefore, half plane II is not solution region of given inequality i.e. y+82x .

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:

1635762587304

Question:5 Solve the following inequality graphically in two-dimensional plane: xy2

Answer:

xy2

Graphical representation of xy=2 is given in the graph below.

The line xy=2 divides plot in two half planes.

Select a point (not on the line xy=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

122 i.e. 12 , which is true.

Therefore, half plane Ii is not solution region of given inequality i.e. xy2 .

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:

1635762601555

Question:6 Solve the following inequality graphically in two-dimensional plane: 2x3y>6

Answer:

2x3y>6

Graphical representation of 2x3y=6 is given in the graph below.

The line 2x3y=6 divides plot in two half planes.

Select a point (not on the line 2x3y=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

26>6 i.e. 4>6 , which is false .

Therefore, half plane I is not solution region of given inequality i.e. 2x3y>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:

1635762622339

Question:7 Solve the following inequality graphically in two-dimensional plane: 3x+2y6

Answer:

3x+2y6

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+46 i.e. 16 , which is true.

Therefore, half plane II is not solution region of given inequality i.e. 3x+2y6 .

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:

1635762636448

Question:8 Solve the following inequality graphically in two-dimensional plane: 3y5x<30

Answer:

3y5x<30

Graphical representation of 3y5x=30 is given in graph below.

The line 3y5x=30 divides plot in two half planes.

Select a point (not on the line 3y5x=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

65<30 i.e. 1<30 , which is true.

Therefore, half plane II is not solution region of given inequality i.e. 3y5x<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:

1635762649943

Question:9 Solve the following inequality graphically in two-dimensional plane: y<2

Answer:

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:

1635762675271

Question:10 Solve the following inequality graphically in two-dimensional plane: x>3

Answer:

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:

1635762691903


Class 11 maths chapter 6 question answer - Exercise 6.3

Question:1 Solve the following system of inequalities graphically:

x3, y2

Answer:

x3, y2

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 x3 ,

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 y2 ,

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 x3, y2 is common region of graph i.e. region III.

Thus, solution of x3, y2 is region III.

This can be represented as follows:

1654687280859

The below green colour represents the solution

1654687281313

Question:2 Solve the following system of inequalities graphically: 3x+2y12, x1, y2

Answer:

3x+2y12, x1, y2

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

For x1 ,

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 y2 ,

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+2y12

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:

1635762856044

Question:3 Solve the following system of inequalities graphically: 2x+y6,3x+4y12

Answer:

2x+y6,3x+4y12

Graphical representation of 2x+y=6and3x+4y=12 is given in the graph below.

For 2x+y6 ,

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+4y12 ,

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:

1635762935162

Question:4 Solve the following system of inequalities graphically: x+y4,2xy<0

Answer:

x+y4,2xy<0

Graphical representation of x+y=4and2xy=0 is given in the graph below.

For x+y4, ,

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 2xy<0 ,

The solution to this inequality is half plane corresponding to the line (2xy=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 (2xy=0) .

This can be represented as follows:

1635763100696

Question:5 Solve the following system of inequalities graphically: 2xy>1, x2y<1

Answer:

2xy>1, x2y<1

Graphical representation of x2y=1and2xy=1 is given in graph below.

For 2xy>1,

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

For  x2y<1 ,

The solution to this inequality is region above the line (x2y=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:

cz1635763130508

Question:6 Solve the following system of inequalities graphically: x+y6,x+y4

Answer:

x+y6,x+y4

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

For x+y6,

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

For x+y4 ,

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:

1635763165801

Question:7 Solve the following system of inequalities graphically: 2x+y8,x+2y10

Answer:

2x+y8,x+2y10

Graphical representation of 2x+y=8andx+2y=10 is given in graph below.

For 2x+y8,

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+2y10 ,

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:

1635763179656

Question:8 Solve the following system of inequalities graphically: x+y9,y>x,x0

Answer:

x+y9,y>x,x0

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

For x+y9 ,

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 x0 ,

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:

1635763196635

Question:9 Solve the following system of inequalities graphically: 5x+4y20, x1, y2

Answer:

5x+4y20, x1, y2

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

For 5x+4y20, ,

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  x1, ,

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  y2,

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:

1635763228699

Question:10 Solve the following system of inequalities graphically: 3x+4y60, x+3y30, x0, y0

Answer:

3x+4y60, x+3y30, x0, y0

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

For 3x+4y60 ,

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+3y30 ,

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  x0,

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  y0,

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:

1635763315671

Question:11 Solve the following system of inequalities graphically: 2x+y4, x+y3, 2x3y6

Answer:

2x+y4, x+y3, 2x3y6

Graphical representation of 2x+y=4,x+y=3 and 2x3y=6 is given in graph below.

For 2x+y4, ,

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+y3, ,

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  2x3y6,

The solution to this inequality is region above the line (2x3y=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:

1635763400655

Question:12 Solve the following system of inequalities graphically: x2y3,3x+4y12,x0,y1

Answer:

x2y3,3x+4y12,x0,y1

Graphical representation of x2y=3,3x+4y=12,x=0andy=1 is given in graph below.

For x2y3 ,

The solution to this inequality is region above the line (x2y=3) including points on this line because points on line also satisfy the inequality.

For 3x+4y12 ,

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  x0,

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  y1,

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:

1635763420814

Question:13 Solve the following system of inequalities graphically: 4x+3y60, y2x, x3, x,y0

Answer:

4x+3y60, y2x, x3, x,y0

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

For 4x+3y60,

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 y2x ,

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 x3 ,

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  x0,

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  y0,

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:

1635763491115

Question:14 Solve the following system of inequality graphically: 3x+2y150, x+4y80, x15 y0, x0

Answer:

3x+2y150, x+4y80, x15 y0, x0

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

For 3x+2y150,

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+4y80 ,

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 x15 ,

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  x0,

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  y0,

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:

1635763512601

Question:15 Solve the following system of inequality graphically: x+2y10, x+y1, xy0,x0, y0

Answer:

x+2y10, x+y1, xy0,x0, y0

Graphical representation of x+2y=10,x+y=1,xy=0,x=0andy=0 is given in graph below.

For x+2y10,

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+y1, ,

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  xy0, ,

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

For  x0,

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  y0,

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:

1635763543684


Linear inequalities equations ncert solutions - Miscellaneous Exercise

Question:1 Solve the inequality 23x45

Answer:

Given : 23x45


23x45

2+43x5+4

63x9

63x93

2x3

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

Solution set is Missing or unrecognized delimiter for \left

Question:2 Solve the inequality 63(2x4)<12

Answer:

Given 63(2x4)<12

63(2x4)<12

 63(2x4)<123

 2(2x4)>4

 2+42x>4+4

 22x>0

 1x>0

Solution set is (01]

Question:3 Solve the inequality 347x218

Answer:

Given 347x218


347x218

347x2184

77x214

77x214

7×27x14×2

147x28

147x287

2x4

Solution set is [4,2]

Question:4 Solve the inequality 15<3(x2)50

Answer:

Given The inequality

15<3(x2)50


15<3(x2)50

 15×5<3(x2)0×5

 75<3(x2)0

 753<(x2)03

 25<(x2)0

 25+2<x0+2

 23<x2

The solution set is (23,2]

Question:5 Solve the inequality 12<43x52

Answer:

Given the inequality

12<43x52


12<43x52

124<3x524

16<3x52

16<3x52

16×5<3x2×5

80<3x10

803<3x103

Solution set is (803,103]

Question:6 Solve the inequality 7(3x+11)211

Answer:

Given the linear inequality

7(3x+11)211

7(3x+11)211

7×2(3x+11)11×2

14(3x+11)22

1411(3x)2211

33x11

1x113

The solution set of the given inequality is [1,113]

Question:7 Solve the inequality and represent the solution graphically on number line. 5x+1>24, 5x1<24

Answer:

Given : 5x+1>24, 5x1<24


5x+1>24and 5x1<24

5x>241and 5x<24+1

5x>25and 5x<25

x>255and x<255

x>5and x<5

(5,5)

The solution graphically on the number line is as shown :

1635763658335

Question:8 Solve the inequality and represent the solution graphically on number line. 2(x1)<x+5, 3(x+2)>2x

Answer:

Given : 2(x1)<x+5, 3(x+2)>2x


2(x1)<x+5and 3(x+2)>2x

2x2<x+5and 3x+6>2x

2xx<2+5and 3x+x>26

x<7and 4x>4

x<7and x>1

(1,7)

The solution graphically on the number line is as shown :

1635763681053

Question:9 Solve the inequality and represent the solution graphically on number line. 3x7>2(x6), 6x>112x

Answer:

Given : 3x7>2(x6), 6x>112x


3x7>2(x6)and 6x>112x

3x7>2x12and 6x>112x

3x2x>712and 2xx>116

x>5and x>5

x(5,)

The solution graphically on the number line is as shown :

1635763701201

Question:10 Solve the inequality and represent the solution graphically on number line.

5(2x7)3(2x+3)0,2x+196x+47

Answer:

Given : 5(2x7)3(2x+3)0,2x+196x+47


5(2x7)3(2x+3)0and2x+196x+47

10x356x90and2x6x4719

4x440and4x28

4x44and4x28

x11andx7

x[7,11]

The solution graphically on the number line is as shown :

1635763730673

Question:11 A solution is to be kept between 68° F and 77° F. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by F=95C+32

Answer:

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

68<F<77

Putting the value of F=95C+32 , we have

68<95C+32<77

6832<95C<7732

36<95C<45

36×5<9C<45×5

180<9C<225

1809<C<2259

20<C<25

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

Question:12 A solution of 8% boric acid is to be diluted by adding a 2% boric acid solution to it. The resulting mixture is to be more than 4% but less than 6% boric acid. If we have 640 litres of the 8% solution, how many litres of the 2% solution will have to be added?

Answer:

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.

2%x+8%of640>4%of(640+x) and 2%x+8%of640<6%of(x+640)

2%x+8%of640>4%of(640+x) and 2%x+8%of640<6%of(x+640)

2100x+(8100)640>4100(640+x) 2100x+(8100)640<6100(640+x)

2x+5120>4x+2560 2x+5120<6x+3840

51202560>4x2x 51203840<6x2x

2560>2x 1280<4x

1280>x 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.

Question:13 How many litres of water will have to be added to 1125 litres of the 45% solution of acid so that the resulting mixture will contain more than 25% but less than 30% acid content?

Answer:

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.

30%of(1125+x)>45%of(1125) and 25%of(1125+x)<45%of1125

30%of(1125+x)>45%of(1125) and 25%of(1125+x)<45%of1125

30100(1125+x)>45100(1125) (25100)(1125+x)<45100(1125)

30×1125+30x>45×(1125) 25(1125+x)<45(1125)

30x>(4530)×(1125) 25x<(4525)1125

30x>(15)×(1125) 25x<(20)1125

x>15×112530 x<20×112525

x>562.5 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.

Question:14 IQ of a person is given by the formula IQ=MACA×100 where MA is mental age and CA is chronological age. If 80IQ140 for a group of 12 years old children, find the range of their mental age.

Answer:

Given that group of 12 years old children.

80IQ140

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

IQ=MACA×100

Putting the value of IQ, we obtain

80IQ140

80MACA×100140

80MA12×100140

80×12MA×100140×12

80×12100MA140×12100

9.6MA16.8

Thus, the range of mental age of the group of 12 years old children is 9.6MA16.8

Class 11 maths chapter 6 NCERT solutions - Symmary

Definition of Inequality: An inequality is a statement that two values are not equal. In mathematics, inequalities are used to compare values and to represent constraints in real-world problems.

Linear Inequalities: Linear inequalities are a type of inequality where the variables appear only in the first degree, that is, raised to the power of 1.

Solving Linear Inequalities: The process of finding all the possible values of the variable that satisfy a given linear inequality is called solving the inequality. In this chapter, various methods of solving linear inequalities are discussed.

Graphical Representation: Graphical representation is an important tool to visualize the solution of a linear inequality. The chapter 6 class 11 maths discusses how to plot linear inequalities on a coordinate plane.

Solution of System of Linear Inequalities: The NCERT solution for class 11 maths chapter 6 also discusses the solution of a system of linear inequalities, which involves finding the region on the coordinate plane that satisfies all the inequalities in the system.

Application in Real-World Problems: Linear inequalities are widely used in real-world problems, such as optimizing production, minimizing costs, and maximizing profits. The chapter provides various examples of real-world problems that can be solved using linear inequalities.

Linear Inequality 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+12y180\(fabricating constraint)x+3y30 (finishing constraint)

Linear Inequalities Exercise Wise Solutions

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NCERT Solutions For Class 11 Mathematics - Chapter Wise

Key Features Of Linear Inequalities Class 11 Maths NCERT Chapter

Conceptual Clarity: The chapter begins by introducing the basic concepts of linear inequalities, ensuring that students understand the fundamental principles.

Real-Life Applications: Linear inequalities are explained with reference to real-life scenarios, helping students relate mathematical concepts to practical situations.

Inequality Notations: The chapter covers different types of inequality notations, such as "less than," "greater than," "less than or equal to," and "greater than or equal to".

NCERT Solutions For Class 11- Subject Wise

Benefits of NCERT Solutions

  • Linear inequalities equations ncert solutions 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 class 11 maths ch 6 question answer 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.
  • NCERT solutions for maths chapter 6 class 11 will some innovative ways of solving the problems which become very important to solve some specific problems in an easy way.
  • This ch 6 maths class 11 also useful in the prediction of future events based on the past data which is the fundamentals of machine learning

NCERT Books and NCERT Syllabus

Happy Reading !!!

Frequently Asked Questions (FAQs)

1. What are important topics of the chapter Linear Inequalities ?

Inequalities class 11 includes the important topics such as Basic concept of inequalities, algebraic solutions of linear inequalities in one variable and their graphical representation, graphical solution of linear inequalities in two variables, and solution of system of linear inequalities in two variables. students should practice these concepts to get good a hold of the concepts discussed in class 11 chapter 6.

2. Explain the steps to plot a graph of linear inequality covered in NCERT Solutions for Class 11 Maths Chapter 6.

The steps to plot a graph of a linear inequality covered in linear inequalities class 11 ncert solutions are as follows:

  • Write the inequality in the form of a linear equation

  • Solve the equation for y

  • Identify the boundary line

  • Choose a test point

  • Substitute the test point

  • Shade the region

  • Identify the solution set

  • Label the graph

Students can find NCERT solutions for class 11 maths  by clicking on the link.

3. List out the number of exercises present in NCERT solutions for class 11 maths chapter 6.

The linear inequalities class 11 solutions includes there three exercises and one miscellaneous exercise.

Exercise 6.1 – 26 Questions
Exercise 6.2 – 10 Questions
Exercise 6.3 – 15 Questions
Miscellaneous Exercise – 14 Questions

4. Which is the official website of NCERT ?

NCERT official is the official website of the NCERT where you can get NCERT textbooks and syllabus from class 1 to 12.

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Get answers from students and experts

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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