Class 11 - Maths - Linear Inequalities

                                                                        Exercise 6.1

Question 1:

Solve 24x < 100, when           (i) x is a natural number               (ii) x is an integer

Answer:

The given inequality is 24x < 100

=> x < 100/24

=> x < 25/6

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than 25/6

Thus, when x is a natural number, the solutions of the given inequality are 1, 2, 3, and 4.

Hence, in this case, the solution set is {1, 2, 3, 4}.

(ii) The integers less than are ...–3, –2, –1, 0, 1, 2, 3, 4.

Thus, when x is an integer, the solutions of the given inequality are ...–3, –2, –1, 0, 1, 2, 3, 4.

Hence, in this case, the solution set is {......,–3, –2, –1, 0, 1, 2, 3, 4}.

Question 2:

Solve –12x > 30, when (i) x is a natural number               (ii) x is an integer

Answer:

The given inequality is –12x > 30

=> -x > 30/12

=> -x > 5/2

=> x < -5/2

(i) There is no natural number less than (-5/2).

Thus, when x is a natural number, there is no solution of the given inequality.

(ii) The integers less than -5/2 are ..., –5, –4, –3.

Thus, when x is an integer, the solutions of the given inequality are ..., –5, –4, –3.

Hence, in this case, the solution set is {..., –5, –4, –3}.

Question 3:

Solve 5x– 3 < 7, when (i) x is an integer                                        (ii) x is a real number

Answer:

The given inequality is 5x– 3 < 7

=> 5x – 3 + 3 < 7 + 3                             [3 is added both sides]

=> 5x < 10

=> x < 10/5

=> x < 2

(i)The integers less than 2 are ..., –4, –3, –2, –1, 0, 1.

Thus, when x is an integer, the solutions of the given inequality are ..., –4, –3, –2, –1, 0, 1.

Hence, in this case, the solution set is {..., –4, –3, –2, –1, 0, 1}.

(ii) When x is a real number, the solutions of the given inequality are given by x < 2, i.e., all real

numbers x which are less than 2.

Thus, the solution set of the given inequality is x ϵ (–∞, 2).

Question 4:

Solve 3x + 8 > 2, when (i) x is an integer                            (ii) x is a real number

Answer:

The given inequality is 3x + 8 > 2

=> 3x + 8 – 8 > 2 – 8                           [8 is subtracted both sides]

=> 3x > -6

=> x > -6/3

=> x > -2

(i)The integers greater than –2 are –1, 0, 1, 2, ...

Thus, when x is an integer, the solutions of the given inequality are –1, 0, 1, 2 ...

Hence, in this case, the solution set is {–1, 0, 1, 2, ...}.

(ii) When x is a real number, the solutions of the given inequality are all the real numbers,

which are greater than –2.

Thus, in this case, the solution set is (– 2, ∞).

Question 5:

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

Answer:

Given, 4x + 3 < 5x + 7

=> 4x + 3 – 7 < 5x + 7 – 7              [7 is subtracted on both sides]

=> 4x – 4 < 5x

=> 4x – 4 – 4x < 5x – 4x                [4x is subtracted on both sides]

=> –4 < x

=> x > -4

Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–4, ∞).

Question 6:

Solve the given inequality for real x: 3x – 7 > 5x – 1

Answer:

Given, 3x – 7 > 5x – 1

=> 3x – 7 + 7 > 5x – 1 + 7                  [7 is added on both sides]

=> 3x > 5x + 6

=> 3x – 5x > 5x + 6 – 5x                    [5x is subtracted on both sides]   

=> – 2x > 6

=> -x > 6/2

=>-x > 3

=> x < -3

Thus, all real numbers x, which are less than –3, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, –3).

Question 7:

Solve the given inequality for real x: 3(x – 1) ≤ 2 (x – 3)

Answer:

Given, 3(x – 1) ≤ 2(x – 3)

=> 3x – 3 ≤ 2x – 6

=> 3x – 3 + 3 ≤ 2x – 6 + 3                 [3 is added on both sides]

=> 3x ≤ 2x – 3

=> 3x – 2x ≤ 2x – 3 – 2x                   [2x is subtracted on both sides]

=> x ≤ – 3

Thus, all real numbers x, which are less than or equal to –3, are the solutions of the given

inequality. Hence, the solution set of the given inequality is (–∞, –3].

Question 8:

Solve the given inequality for real x: 3(2 – x) ≥ 2(1 – x)

Answer:

Given, 3(2 – x) ≥ 2(1 – x)

=> 6 – 3x ≥ 2 – 2x

=> 6 – 3x + 2x ≥ 2 – 2x + 2x             [2x is added on both sides]

=> 6 – x ≥ 2

=> 6 – x – 6 ≥ 2 – 6                            [6 is subtracted on both sides]

=> –x ≥ –4

=> x ≤ 4

Thus, all real numbers x, which are less than or equal to 4, are the solutions of the given

inequality. Hence, the solution set of the given inequality is (–∞, 4].

Question 9:

Solve the given inequality for real x: x + x/2 + x/3 < 11

Answer:

Given, x + x/2 + x/3 < 11

=> (6x + 3x + 2x)/6 < 11

=> 11x/6 < 11

=> x/6 < 1

=> x < 6

Thus, all real numbers x, which are less than 6, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, 6).

Question 10:

Solve the given inequality for real x: x/3 > x/2 + 1

Answer:

Given, x/3 > x/2 + 1

=> x/3 – x/2 > 1

=> (2x – 3x)/6 > 1

=> -x/6 > 1

=> -x > 6

=> x < -6

Thus, all real numbers x, which are less than –6, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (–∞, –6).

Question 11:

Solve the given inequality for real x: 3(x - 2)/5 ≤ 5(2 - x)/3

Answer:

Given, 3(x - 2)/5 ≤ 5(2 - x)/3

=> 3(x - 2) * 3 ≤ 5(2 - x) * 5

=> 9(x - 2) ≤ 25(2 - x)

=> 9x - 18 ≤ 50 - 25x

=> 9x - 18 + 25x ≤ 50

=> 34x - 18 ≤ 50

=> 34x ≤ 50 + 18

=> 34x ≤ 68

=> x ≤ 68/34

=> x ≤ 2

Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given

inequality. Hence, the solution set of the given inequality is (–∞, 2].

Question 12:

Solve the given inequality for real x: (1/2)(3x/5 + 4) ≥ (x - 6)/3

Answer:

Given, (1/2)(3x/5 + 4) ≥ (x - 6)/3

=> 3(3x/5 + 4) ≥ 2(x - 6)

=> 9x/5 + 12 ≥ 2x – 12

=> 9x/5 – 2x ≥ -12 – 12

=> (9x – 10x)/5 ≥ -24

=> -x/5 ≥ -24

=> -x ≥ -24 * 5

=> -x ≥ -120

=> x ≤ 120

Thus, all real numbers x, which are less than or equal to 120, are the solutions of the given

inequality. Hence, the solution set of the given inequality is (–∞, 120].

Question 13:

Solve the given inequality for real x: 2(2x + 3) – 10 < 6(x – 2)

Answer:

Given, 2(2x + 3) – 10 < 6(x – 2)

=> 4x + 6 – 10 < 6x – 12

=> 4x – 4 < 6x – 12

=> 4x – 6x < -12 + 4

=> -2x < -8

=> 2x > 8

=> x > 4

Thus, all real numbers x, which are greater than or equal to 4, are the solutions of the given

inequality. Hence, the solution set of the given inequality is [4, ∞).

Question 14:

Solve the given inequality for real x: 37 – (3x + 5) ≥ 9x – 8(x – 3)

Answer:

Given, 37 – (3x + 5) ≥ 9x – 8(x – 3)

=> 37 – 3x - 5 ≥ 9x – 8x + 24

=> 32 – 3x ≥ x + 24

=> 32 – 24 ≥ x + 3x

=> 8 ≥ 4x

=> 2 ≥ x

=> x ≤ 2

Thus, all real numbers x, which is less than or equal to 2, are the solutions of the given

inequality. Hence, the solution set of the given inequality is (–∞, 2].

Question 15:

Solve the given inequality for real x: x/4 < (5x - 2)/3 – (7x - 3)/5

Answer:

Given, x/4 < (5x - 2)/3 – (7x - 3)/5

=> x/4 < {5(5x - 2) – 3(7x - 3)}/15

=> x/4 < (25x - 10 – 21x + 9)/15

=> x/4 < (4x - 1)/15

=> 15x < 4(4x - 1)

=> 15x < 16x - 4

=> 15x – 16x < -4

=> -x < -4

=> x > 4

Thus, all real numbers x, which are greater than 4, are the solutions of the given inequality.

Hence, the solution set of the given inequality is (4, ∞).

Question 16:

Solve the given inequality for real x: (2x - 1)/3 ≥ (3x - 2)/4 – (2 - x)/5

Answer:

Given, (2x - 1)/3 ≥ (3x - 2)/4 – (2 - x)/5

=> (2x - 1)/3 ≥ {5(3x - 2) – 4(2 - x)}/20

=> (2x - 1)/3 ≥ (15x - 10 – 8 + 4x)/20

=> (2x - 1)/3 ≥ (19x - 18)/20

=> 20(2x - 1) ≥ 3(19x - 18)

=> 40x – 20 ≥ 57x - 54

=> 40x – 57x ≥ -54 + 20

=> -17x ≥ -34

=> -x ≥ -2

=> x ≤ 2

Thus, all real numbers x, which are less than or equal to 2, are the solutions of the given

inequality. Hence, the solution set of the given inequality is (–∞, 2].

Question 17:

Solve the given inequality and show the graph of the solution on number line:

3x – 2 < 2x +1

 

Answer:

Given, 3x – 2 < 2x +1

=> 3x – 2x < 1 + 2

=> x < 3

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

          Class_11_Maths_Linear_InEqualities_NumberLine_7                           

Question 18:

Solve the given inequality and show the graph of the solution on number line:

5x – 3 ≥ 3x – 5

Answer:

Given, 5x – 3 ≥ 3x – 5

=> 5x – 3x ≥ -5 + 3

=> 2x ≥ -2

=> x ≥ -1

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

            Class_11_Maths_Linear_InEqualities_NumberLine_6                            

Question 19:

Solve the given inequality and show the graph of the solution on number line:

3(1 – x) < 2(x + 4)

Answer:

Given, 3(1 – x) < 2(x + 4)

=> 3 – 3x < 2x + 8

=> 3 – 8 < 2x + 3x

=> -5 < 5x

=> -1 < x

=> x > -1

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

           Class_11_Maths_Linear_InEqualities_NumberLine_5                   

Question 20:

Solve the given inequality and show the graph of the solution on number line:

x/2 ≥ (5x - 2)/3 – (7x - 3)/5

Answer:

Given, x/2 ≥ (5x - 2)/3 – (7x - 3)/5 

=> x/2 ≥ {5(5x - 2) – 3(7x - 3)}/15

=> x/2 ≥ (25x - 10 – 21x + 9)/15

=> x/2 ≥ (4x – 1)/15

=> 15x ≥ 2(4x – 1)

=> 15x ≥ 8x – 2

=> 15x – 8x ≥ -2

=> 7x ≥ -2

=> x ≥ -2/7

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

 Class_11_Maths_Linear_InEqualities_NumberLine_4

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 the marks obtained by Ravi in the third unit test.

Since the student should have an average of at least 60 marks,

      (70 + 75 + x)/3 ≥ 60

=> (145 + x)/3 ≥ 60

=> 145 + x ≥ 60 * 3

=> 145 + x ≥ 180

=> x ≥ 180 – 145

=> x ≥ 35

Thus, the student must obtain a minimum of 35 marks to have an average of at least 60 marks.

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:

Let x be the marks obtained by Sunita in the fifth examination.

In order to receive grade ‘A’ in the course, she must obtain an average of 90 marks or more in

five examinations.

Therefore,

      (87 + 92 + 94 + 95 + x)/5 ≥ 90

=> (368 + x)/5 ≥ 90

=> 368 + x ≥ 90 * 5

=> 368 + x ≥ 450

=> x ≥ 450 – 368

=> x ≥ 82

Hence, Sunita must obtain greater than or equal to 82 marks in the fifth examination.

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 the smaller of the two consecutive odd positive integers.

Then, the other integer is x + 2

Since both the integers are smaller than 10,

     x + 2 < 10

=> x < 10 – 2

=> x < 8        ........1

Also, the sum of the two integers is more than 11.

So, x + (x + 2) > 11

=> 2x + 2 > 11

=> 2x > 11 – 2

=> 2x > 9

=> x > 9/2

=> x > 4.5    …………….2

From equation 1 and 2, we get

4.5 < x < 8

Since x is an odd number, x can take the values, 5 and 7.

Hence, the required possible pairs 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 the smaller of the two consecutive even positive integers.

Then, the other integer is x + 2

Since both the integers are larger than 5,

So, x > 5     ...........1

Also, the sum of the two integers is less than 23.

      x + (x + 2) < 23

=> 2x + 2 < 23

=> 2x < 23 – 2

=> 2x < 21

=> x < 21/2

=> x < 10.5    ……………2

From equation 1 and 2, we get

5 < x < 10.5

Since x is an even number, x can take the values, 6, 8, and 10.

Thus, the required possible pairs are (6, 8), (8, 10), and (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 shortest side of the triangle be x cm.

Then, length of the longest side = 3x cm

Length of the third side = (3x – 2) cm

Since the perimeter of the triangle is at least 61 cm,

So, x + 3x + (3x - 2) ≥ 61

=> 7x - 2 ≥ 61

=> 7x ≥ 61 + 2

=> 7x ≥ 63

=> x ≥ 63/7

=> x ≥ 9

Thus, the 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 91 cm.

The second length is to be 3 cm 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 5 cm 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) + 2x ≤ 91 and 2x ≥ (x + 3) + 5]

Answer:

Let the length of the shortest piece be x cm. Then, length of the second piece and the third

piece are (x + 3) cm and 2x cm respectively.

Since the three lengths are to be cut from a single piece of board of length 91 cm,

So, x + (x + 3) + 2x ≤ 91

=> 4x + 3 ≤ 91

=> 4x ≤ 91 – 3

=> 4x ≤ 88

=> x ≤ 88/4

=> x ≤ 22    …………….1

Also, the third piece is at least 5 cm longer than the second piece.

So, 2x ≥ (x + 3) + 5

=> 2x ≥ x + 8

=> x ≥ 8      .............2

From equation 1 and 2, we get

8 ≤ x ≤ 22

Thus, the possible length of the shortest board is greater than or equal to 8 cm but less than or

equal to 22 cm.

 

 

 

                                                                        Exercise 6.2

Question 1:

Solve the given inequality graphically in two-dimensional plane: x + y < 5

Answer:

The graphical representation of x + y = 5 is given as dotted line in the figure below.

This line divides the xy-plane in two half planes, I and II.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

0 + 0 < 5 or, 0 < 5, which is true.

Therefore, half plane II is not the solution region of the given inequality. Also, it is evident that

any point on the line does not satisfy the given strict inequality.

Thus, the solution region of the given inequality is the shaded half plane I excluding the points

on the line. This can be represented as follows:

 Class_11_Maths_Linear_InEqualities_Graph_25

 

Question 2:

Solve the given inequality graphically in two-dimensional plane: 2x + y ≥ 6

Answer:

The graphical representation of 2x + y = 6 is given in the figure below.

This line divides the xy-plane in two half planes, I and II.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

2(0) + 0 ≥ 6 or 0 ≥ 6, which is false

Therefore, half plane I is not the solution region of the given inequality. Also, it is evident that

any point on the line satisfies the given inequality.

Thus, the solution region of the given inequality is the shaded half plane II including the points

on the line. This can be represented as follows:

 

 Class_11_Maths_Linear_InEqualities_Graph_24

 

Question 3:

Solve the given inequality graphically in two-dimensional plane: 3x + 4y ≤ 12

Answer 3:

3x + 4y ≤ 12

The graphical representation of 3x + 4y = 12 is given in the figure below.

This line divides the xy-plane in two half planes, I and II.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

3(0) + 4(0) ≤ 12 or 0 ≤ 12, which is true.

Therefore, half plane II is not the solution region of the given inequality. Also, it is evident that

any point on the line satisfies the given inequality.

Thus, the solution region of the given inequality is the shaded half plane I including the points

on the line. This can be represented as follows:

 Class_11_Maths_Linear_InEqualities_Graph_23

 

Question 4:

Solve the given inequality graphically in two-dimensional plane: y + 8 ≥ 2x

Answer:

The graphical representation of y + 8 = 2x is given in the figure below.

This line divides the xy-plane in two half planes.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

0 + 8 ≥ 2(0) or 8 ≥ 0, which is true

Therefore, lower half plane is not the solution region of the given inequality. Also, it is evident

that any point on the line satisfies the given inequality.

Thus, the solution region of the given inequality is the half plane containing the point (0, 0)

including the line.

The solution region is represented by the shaded region as follows:

 Class_11_Maths_Linear_InEqualities_Graph_22

 

Question 5:

Solve the given inequality graphically in two-dimensional plane: x – y ≤ 2

Answer:

The graphical representation of x – y = 2 is given in the figure below.

This line divides the xy-plane in two half planes.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

0 – 0 ≤ 2 or 0 ≤ 2, which is true.

Therefore, the lower half plane is not the solution region of the given inequality. Also, it is

clear that any point on the line satisfies the given inequality.

Thus, the solution region of the given inequality is the half plane containing the point (0, 0)

including the line.

The solution region is represented by the shaded region as follows:

 Class_11_Maths_Linear_InEqualities_Graph_21

 

  

Question 6:

Solve the given inequality graphically in two-dimensional plane: 2x – 3y > 6

Answer:

The graphical representation of 2x – 3y = 6 is given as dotted line in the figure below.

This line divides the xy-plane in two half planes.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

2(0) – 3(0) > 6 or 0 > 6, which is false

Therefore, the upper half plane is not the solution region of the given inequality. Also, it is

clear that any point on the line does not satisfy the given inequality.

Thus, the solution region of the given inequality is the half plane that does not contain the

point (0, 0) excluding the line.

The solution region is represented by the shaded region as follows:

 

 Class_11_Maths_Linear_InEqualities_Graph_19

 

Question 7:

Solve the given inequality graphically in two-dimensional plane: –3x + 2y ≥ -6

Answer:

The graphical representation of – 3x + 2y = – 6 is given in the figure below.

This line divides the xy-plane in two half planes.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

– 3(0) + 2(0) ≥ – 6 or 0 ≥ –6, which is true

Therefore, the lower half plane is not the solution region of the given inequality. Also, it is

evident that any point on the line satisfies the given inequality.

Thus, the solution region of the given inequality is the half plane containing the point (0, 0)

including the line.

The solution region is represented by the shaded region as follows:

 Class_11_Maths_Linear_InEqualities_Graph_19

 

Question 8:

Solve the given inequality graphically in two-dimensional plane: 3y – 5x < 30

Answer:

The graphical representation of 3y – 5x = 30 is given as dotted line in the figure below.

This line divides the xy-plane in two half planes.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

3(0) – 5(0) < 30 or 0 < 30, which is true

Therefore, the upper half plane is not the solution region of the given inequality. Also, it is

evident that any point on the line does not satisfy the given inequality.

Thus, the solution region of the given inequality is the half plane containing the point (0, 0)

excluding the line.

The solution region is represented by the shaded region as follows:

 Class_11_Maths_Linear_InEqualities_Graph_18

 

Question 9:

Solve the given inequality graphically in two-dimensional plane: y < –2

Answer:

The graphical representation of y = –2 is given as dotted line in the figure below. This line

divides the xy-plane in two half planes.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

0 < –2, which is false

Also, it is evident that any point on the line does not satisfy the given inequality.

Hence, every point below the line, y = –2 (excluding all the points on the line), determines the

solution of the given inequality.

The solution region is represented by the shaded region as follows:

 Class_11_Maths_Linear_InEqualities_Graph_17

 

 

Question 10:

Solve the given inequality graphically in two-dimensional plane: x > –3

Answer:

The graphical representation of x = –3 is given as dotted line in the figure below. This line

divides the xy-plane in two half planes.

Select a point (not on the line), which lies in one of the half planes, to determine whether the

point satisfies the given inequality or not.

We select the point as (0, 0).

It is observed that,

0 > –3, which is true

Also, it is evident that any point on the line does not satisfy the given inequality.

Hence, every point on the right side of the line, x = –3 (excluding all the points on the line),

determines the solution of the given inequality.

The solution region is represented by the shaded region as follows:

 Class_11_Maths_Linear_Inequalities_Graph_10

 

 

                                                                      Exercise 6.3

Question 1:

Solve the following system of inequalities graphically: x ≥ 3, y ≥ 2.

Answer:

x ≥ 3      .............1

y ≥ 2      .............2

The graph of the lines, x = 3 and y = 2, are drawn in the figure below.

Inequality 1 represents the region on the right hand side of the line,

x = 3 (including the line x = 3), and inequality 2 represents the region above the line, y = 2

(including the line y = 2).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as follows:

 Class_11_Maths_Linear_InEqualities_Graph_16

 

 

Question 2:

Solve the following system of inequalities graphically: 3x + 2y ≤ 12, x ≥ 1, y ≥ 2

Answer 2:

3x + 2y ≤ 12     ..............1

x ≥ 1      .........................2

y ≥ 2      .........................3

The graphs of the lines, 3x + 2y = 12, x = 1, and y = 2, are drawn in the figure below:

Inequality 1 represents the region below the line, 3x + 2y = 12 ( including the line 3x + 2y =

12). Inequality 2 represents the region on the right side of the line, x = 1 (including the line x =

1). Inequality 3 represents the region above the line, y = 2 (including the line y = 2).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as above.

 

 Class_11_Maths_Linear_InEqualities_Graph_15

 

Question 3:

Solve the following system of inequalities graphically: 2x + y≥ 6, 3x + 4y ≤ 12

Answer:

2x + y ≥ 6     ..............1

3x + 4y ≤ 12   ...........2

The graph of the lines, 2x + y= 6 and 3x + 4y = 12, are drawn in the figure below.

Inequality 1 represents the region above the line, 2x + y= 6 ( including the line 2x + y= 6), and

inequality 2 represents the region below the line, 3x + 4y =12 ( including the line 3x + 4y =12).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as follows:

Class_11_Maths_Linear_InEqualities_Graph_14

Question 4:

Solve the following system of inequalities graphically: x + y ≥ 4, 2x – y > 0

Answer:

x + y ≥ 4        ................1

2x – y > 0     ................2

The graph of the lines, x + y = 4 and 2x – y = 0, are drawn in the figure below.

Inequality 1 represents the region above the line, x + y = 4 (including the line x + y = 4). It is

observed that (1, 0) satisfies the inequality, 2x – y > 0. [2(1) – 0 = 2 > 0]

Therefore, inequality 2 represents the half plane corresponding to the line, 2x – y = 0,

containing the point (1, 0) [excluding the line 2x – y > 0].

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on line x + y = 4 and excluding the points on line 2x – y = 0

as follows:

 Class_11_Maths_Linear_InEqualities_Graph_13

 

Question 5:

Solve the following system of inequalities graphically: 2x – y > 1, x – 2y < –1

Answer:

2x – y > 1      ..............1

x – 2y < –1    ..............2

The graph of the lines, 2x – y = 1 and x – 2y = –1, are drawn in the figure below.

Inequality 1 represents the region below the line, 2x – y = 1 (excluding the line 2x – y = 1), and

inequality 2 represents the region above the line, x – 2y = –1 (excluding the line x – 2y = –1).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region excluding the points on the respective lines as follows:

 Class_11_Maths_Linear_InEqualities_Graph_12

 

Question 6:

Solve the following system of inequalities graphically: x + y ≤ 6, x + y ≥ 4

Answer:

x + y ≤ 6     ............1

x + y ≥ 4     ............2

The graph of the lines, x + y = 6 and x + y = 4, are drawn in the figure below.

Inequality 1 represents the region below the line, x + y = 6 (including the line x + y = 6), and

inequality 2 represents the region above the line, x + y = 4 (including the line x + y = 4).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as follows:

 

 Class_11_Maths_Linear_InEqualities_Graph_11

Question 7:

Solve the following system of inequalities graphically: 2x + y ≥ 8, x + 2y ≥ 10

Answer:

2x + y = 8     ..............1

x + 2y = 10   .............2

The graph of the lines, 2x + y = 8 and x + 2y = 10, are drawn in the figure below.

Inequality 1 represents the region above the line, 2x + y = 8, and inequality 2 represents the

region above the line, x + 2y = 10.

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as follows.

 Class_11_Maths_Linear_InEqualities_Graph_10

 

Question 8:

Solve the following system of inequalities graphically: x + y ≤ 9, y > x, x ≥ 0

Answer:

X + y ≤ 9    .............1

y > x          ..............2

x ≥ 0          ..............3

The graph of the lines, x + y= 9 and y = x, are drawn in the figure below.

Inequality 1 represents the region below the line, x + y = 9 (including the line x + y = 9). It is

observed that (0, 1) satisfies the inequality, y > x. [1 > 0].

Therefore, inequality 2 represents the half plane corresponding to the line, y = x, containing

the point (0, 1) [excluding the line y = x]. Inequality 3 represents the region on the right hand

side of the line, x = 0 or y-axis (including y-axis).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the lines, x + y = 9 and x = 0, and excluding the points on

line y = x as follows:

 Class_11_Maths_Linear_InEqualities_Graph_9

 

Question 9:

Solve the following system of inequalities graphically: 5x + 4y ≤ 20, x ≥ 1, y ≥ 2

Answer:

5x + 4y ≤ 20      ................1

x ≥ 1                  .................2

y ≥ 2                  .................3

The graph of the lines, 5x + 4y = 20, x = 1, and y = 2, are drawn in the figure below.

Inequality 1 represents the region below the line, 5x + 4y = 20 (including the line 5x + 4y = 20).

Inequality 2 represents the region on the right hand side of the line, x = 1 (including the line x =

1). Inequality 3 represents the region above the line, y = 2 (including the line y = 2).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as follows:

 Class_11_Maths_Linear_InEqualities_Graph_8

 

Question 10:

Solve the following system of inequalities graphically: 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0

Answer:

3x + 4y ≤ 60    ..............1

x + 3y ≤ 30      ..............2

The graph of the lines, 3x + 4y = 60 and x + 3y = 30, are drawn in the figure below.

Inequality 1 represents the region below the line, 3x + 4y = 60 (including the line 3x + 4y = 60),

and inequality 1 represents the region below the line, x + 3y = 30 (including the line x + 3y =

30).

Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including

the points on the respective line and the axes represents the solution of the given system of

linear inequalities.

 Class_11_Maths_Linear_InEqualities_Graph_7

 

 

Question 11:

Solve the following system of inequalities graphically: 2x + y≥ 4, x + y ≤ 3, 2x – 3y ≤ 6

Answer:

2x + y≥ 4    .............1

x + y ≤ 3     .............2

2x – 3y ≤ 6 ............3

The graph of the lines, 2x + y = 4, x + y = 3, and 2x – 3y = 6, are drawn in the figure below.

Inequality 1 represents the region above the line, 2x + y = 4 (including the line 2x + y = 4).

Inequality 2 represents the region below the line, x + y = 3 (including the line x + y = 3).

Inequality 3 represents the region above the line, 2x – 3y = 6 (including the line 2x – 3y = 6).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as follows:

 Class_11_Maths_Linear_InEqualities_Graph_6

 

 

Question 12:

Solve the following system of inequalities graphically: x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1

Answer:

x – 2y ≤ 3         ......................1

3x + 4y ≥ 12    ......................2

y ≥ 1                 ......................3

The graph of the lines, x – 2y = 3, 3x + 4y = 12, and y = 1, are drawn in the figure below.

Inequality 1 represents the region above the line, x – 2y = 3 (including the line x – 2y = 3).

Inequality 2 represents the region above the line, 3x + 4y = 12 (including the line 3x + 4y = 12).

Inequality 3 represents the region above the line, y = 1 (including the line y = 1). The

inequality, x ≥ 0, represents the region on the right hand side of y-axis (including y – axis).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines and y- axis as follows:

    Class_11_Maths_Linear_InEqualities_Graph_5                                 

 

 

Question 13:

Solve the following system of inequalities graphically:  4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0

Answer:

4x + 3y ≤ 60     ..................1

y ≥ 2x                ..................2

x ≥ 3                  ..................3

The graph of the lines, 4x + 3y = 60, y = 2x, and x = 3, are drawn in the figure below.

Inequality 1 represents the region below the line, 4x + 3y = 60 (including the line 4x + 3y = 60).

Inequality 2 represents the region above the line, y = 2x (including the line y = 2x).

Inequality 3 represents the region on the right hand side of the line, x = 3 (including the line x =

3).

Hence, the solution of the given system of linear inequalities is represented by the common

shaded region including the points on the respective lines as follows:

 Class_11_Maths_Linear_InEqualities_Graph_4

 

Question 14:

Solve the following system of inequalities graphically:

3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0

Answer:

3x + 2y ≤ 150    ……………1

x + 4y ≤ 80        ……………2

x ≤ 15                ……………3

The graph of the lines, 3x + 2y = 150, x + 4y = 80, and x = 15, are drawn in the figure below.

Inequality 1 represents the region below the line, 3x + 2y = 150 (including the line 3x + 2y =

150). Inequality 2 represents the region below the line, x + 4y = 80 (including the line x + 4y =

80). Inequality 3 represents the region on the left hand side of the line, x = 15 (including the

line x = 15).

Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including

the points on the respective lines and the axes represents the solution of the given system of

linear inequalities.

 Class_11_Maths_Linear_InEqualities_Graph_3

 

Question 15:

Solve the following system of inequalities graphically:

x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0

Answer:

x + 2y ≤ 10     .....................1

x + y ≥ 1         ......................2

x – y ≤ 0        .......................3

The graph of the lines, x + 2y = 10, x + y = 1, and x – y = 0, are drawn in the figure below.

Inequality 1 represents the region below the line, x + 2y = 10 (including the line x + 2y = 10).

Inequality 2 represents the region above the line, x + y = 1 (including the line x + y = 1).

Inequality 3 represents the region above the line, x – y = 0 (including the line x – y = 0).

Since x ≥ 0 and y ≥ 0, every point in the common shaded region in the first quadrant including

the points on the respective lines and the axes represents the solution of the given system of

linear inequalities.

 

 Class_11_Maths_Linear_InEqualities_Graph_1

 

                                                        Miscellaneous Exercise on Chapter 6

Question 1:

Solve the inequality 2 ≤ 3x – 4 ≤ 5

Answer:

Given, 2 ≤ 3x – 4 ≤ 5

=> 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

=> 6 ≤ 3x ≤ 9

=> 2 ≤ x ≤ 3

Thus, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3,

are the solutions of the given inequality. The solution set for the given inequality is [2, 3].

Question 2:

Solve the inequality 6 ≤ –3(2x – 4) < 12

Answer:

Given, 6 ≤ – 3(2x – 4) < 12

=> 2 ≤ –(2x – 4) < 4

=> –2 ≥ 2x – 4 > –4

=> 4 – 2 ≥ 2x > 4 – 4

=> 2 ≥ 2x > 0

=> 1 ≥ x > 0

=> 0 < x ≤ 1

Thus, the solution set for the given inequality is (0, 1].

Question 3:

Solve the inequality -3 ≤ 4 – 7x/2 ≤ 18

Answer:

Given, -3 ≤ 4 – 7x/2 ≤ 18

=> -3 – 4 ≤ -7x/2 ≤ 18 – 4

=> -7 ≤ -7x/2 ≤ 14

=> 7 ≥ 7x/2 ≥ -14

=> 1 ≥ x/2 ≥ -2

=> 2 ≥ x ≥ -4

=> -4 ≤ x ≤ 2

Thus, the solution set for the given inequality is [–4, 2].

Question 4:

Solve the inequality -15 < 3(x - 2)/5 ≤ 0

Answer:

Given, -15 < 3(x - 2)/5 ≤ 0

=> –75 < 3(x – 2) ≤ 0

=> –25 < x – 2 ≤ 0

=> – 25 + 2 < x ≤ 2

=> –23 < x ≤ 2

Thus, the solution set for the given inequality is (–23, 2].

Question 5:

Solve the inequality -12 < 4 – 3x/(-5) ≤ 2

Answer:

Given, -12 < 4 – 3x/(-5) ≤ 2

=> –12 - 4 < -3x/(-5) ≤ 2 - 4

=> –16 < 3x/5 ≤ -2

=> -80 < 3x ≤ -10

=> -80/3 < x ≤ -10/3

Thus, the solution set for the given inequality is (-80/3, -10/3].

Question 6:

Solve the inequality 7 ≤ (3x + 11)/2 ≤ 11

Answer:

Given, 7 ≤ (3x + 11)/2 ≤ 11

=> 14 ≤ 3x + 11 ≤ 22

=> 14 – 11 ≤ 3x ≤ 22 - 11

=> 3 ≤ 3x ≤ 11

=> 1 ≤ x ≤ 11/3

Thus, the solution set for the given inequality is [1, 11/3].

Question 7:

Solve the inequalities and represent the solution graphically on number line:

5x + 1 > – 24, 5x – 1 < 24

Answer:

Given inequality is:

      5x + 1 > - 24

=> 5x > -24 - 1

=> 5x > -25

=> x > -25/5

=> x > -5

Again given inequality is:

      5x - 1 < 24

=> 5x < 24 + 1

=> 5x < 25

=> x < 25/5

=> x < 5

So, the value of x lies between (-5, 5). The solution of the given system of inequalities can be

Represented on number line as:

             Class_11_Maths_Linear_InEqualities_NumberLine_3               

Question 8:

Solve the inequalities and represent the solution graphically on number line:

2(x – 1) < x + 5, 3(x + 2) > 2 – x

Answer:

Given inequality is:

      2(x - 1) < x + 5

=> 2x - 2 < x + 5

=> 2x - x < 5 + 2

=> x < 7

Again given inequality is:

      3(x + 2) > 2 - x

=> 3x + 6 > 2 - x

=> 3x + x > 2 - 6

=> 4x > -4

=> x > -4/4

=> x > -1

So, the value of x lies between (-1, 7). The solution of the given system of inequalities can be

represented on number line as:

               Class_11_Maths_Linear_InEqualities_NumberLine_2        

Question 9:

Solve the following inequalities and represent the solution graphically on number line:

3x – 7 > 2(x – 6), 6 – x > 11 – 2x

Answer:

Given, 3x – 7 > 2(x – 6)

=> 3x – 7 > 2x – 12 

=> 3x – 2x > – 12 + 7

=> x > –5     .............1

Again, 6 – x > 11 – 2x

=> –x + 2x > 11 – 6

=> x > 5       ..............2

From equation 1 and 2, it can be concluded that the solution set for the given system of

inequalities is (5, ∞). The solution of the given system of inequalities can be represented

on number line as:

 Class_11_Maths_Linear_InEqualities_NumberLine_1

Question 10:

Solve the inequalities and represent the solution graphically on number line:

5(2x – 7) – 3(2x + 3) ≤ 0, 2x + 19 ≤ 6x + 47

Answer 10:

5(2x – 7) – 3(2x + 3) ≤ 0

=> 10x – 35 – 6x – 9 ≤ 0

=> 4x – 44 ≤ 0

=> 4x ≤ 44

=> x ≤ 11     ..................1

Again, 2x + 19 ≤ 6x + 47

=> 19 – 47 ≤ 6x – 2x

=> –28 ≤ 4x

=> –7 ≤ x

=> x ≥ -7      .................2

From (1) and (2), it can be concluded that the solution set for the given system of inequalities

is [–7, 11]. The solution of the given system of inequalities can be represented on number line

as:

 Class_11_Maths_Linear_InEqualities_NumberLine

 

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 = 9C/5 + 32

Answer:

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

So, 68 < F < 77

Put F = 9C/5 + 32, we get

=> 68 < 9C/5 + 32 < 77

=> 68 - 32 < 9C/5 + 32 - 32 < 77 – 32

=> 36 < 9C/5 < 45

=> 36 * (5/9) < 9C/5 * (5/9) < 45 * (5/9)

=> 20 < 9C/5 < 25

Hence, the required range of temperature in degree Celsius is between 20°C and 25°C

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:

Given mixture = 640 litres

Let 2% boric acid solution is x litre.

So, total amount of mixture = 640 + x.

Now given in the question

       2% of x + 8% of 640 > 4% of (640 + x)

=> 2x/100 + (8*640)/100 > {4(640 + x)}/100

=> 2x + 5120 > 2560 + 4x

=> 2x - 4x > 2560 - 5120

=> -2x > -2560

=> 2x < 2560

=> x < 2560/2

=> x < 1280

Again from the question

      2% of x + 8% of 640 < 6% of (640 + x)

=> 2x/100 + (8 * 640)/100 < {6 * (640 + x)}/100

=> 2x + 5120 < 3840 + 6x

=> 2x - 6x < 3840 - 5120

=> -4x < -1280

=> 4x > 1280

=> x > 1280/4

=> x > 320

So, 320 < x < 1280

Thus, the number of litres of 2% of boric acid solution that is to be added will have to be more

than 320 litres but 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 amount of water is added = x litre

Given solution is 1125 litre

Percentage of acid in solution = 45%

So percentage of water in solution = 55%

Case 1: Resulting mixture will contain more than 25% acid

      x + 1125 * 55% = (x + 1125) * 75%

=> x + (1125 * 55)/100 = (x + 1125) * (75/100)

=> 100x + 1125 * 55 =(x + 1125) * 75

=> 100x + 1125 * 55 = 75x + 1125 * 75

=> 100x - 75x = 1125 * 75 – 1125 * 55

=> 25x = 1125 * (75 - 55)

=> 25x = 1125 * 20

=> x = (1125 * 20)/25

=> x = (1125 * 4)/5

=> x = 225 * 4

=> x = 900

Case 2: Resulting mixture will contain more than 30% acid

      x + 1125 * 55% = (x + 1125) * 70%

=> x + (1125 * 55)/100 =(x + 1125) * (70/100)

=> 100x + 1125 * 55 =(x + 1125) * 70

=> 100x + 1125 * 55 = 70x + 1125 * 70

=> 100x - 70x = 1125 * 70 – 1125 * 55

=> 30x = 1125 * (70-55)

=> 30x = 1125 * 15

=> x = (1125 * 15)/30

=> x = (1125 * 5)/10

=> x = (1125)/2

=> x = 562.5

Now, 562.5 < x < 900

Thus, the required number of litres of water that is to be added will have to be more than

562.5 but less than 900.

Question 14:

IQ of a person is given by the formula IQ = (MA/CA) * 100

Where MA is mental age and CA is chronological age. If 80 ≤ IQ ≤ 140 for a group of 12 years old children, find the range of their mental age.

Answer:

It is given that for a group of 12 years old children,

80 ≤ IQ ≤ 140    .................1

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

IQ = (MA/CA) * 100

Putting this value of IQ in 1, we get

      80 ≤ (MA/12) * 100 ≤ 140

=> 80 * (12/100) ≤ MA ≤ 140 * (12/100)

=> 960/100 ≤ MA ≤ 1680/100

=> 9.6 ≤ MA ≤ 16.8

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

 

 

 

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