NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities

ANSWER:

The given inequality is –12x > 30.

(i) There is no natural number less than.

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

(ii) The integers less than  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}.

ANSWER:

The given inequality is 5x– 3 < 7.

(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, that is, all real numbers x which are less than 2.

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

ANSWER:

The given inequality is 3x + 8 > 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, ∞).

ANSWER:

4x + 3 < 5x + 7

⇒ 4x + 3 – 7 < 5x + 7 – 7

⇒ 4x – 4 < 5x

⇒ 4x – 4 – 4x < 5x – 4x

⇒ –4 < x

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, ∞).

ANSWER:

3x – 7 > 5x – 1

⇒ 3x – 7 + 7 > 5x – 1 + 7

⇒ 3x > 5x + 6

⇒ 3x – 5x > 5x + 6 – 5x

⇒ – 2x > 6

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).

ANSWER:

3(x – 1) ≤ 2(x – 3)

⇒ 3x – 3 ≤ 2x – 6

⇒ 3x – 3 + 3 ≤ 2x – 6 + 3

⇒ 3x ≤ 2x – 3

⇒ 3x – 2x ≤ 2x – 3 – 2x

⇒ 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].

ANSWER:

3(2 – x) ≥ 2(1 – x)

⇒ 6 – 3x ≥ 2 – 2x

⇒ 6 – 3x + 2x ≥ 2 – 2x + 2x

⇒ 6 – x ≥ 2

⇒ 6 – x – 6 ≥ 2 – 6

⇒ –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].

ANSWER:

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).

ANSWER:

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).

ANSWER:

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].

ANSWER:

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].

ANSWER:

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, ∞).

ANSWER:

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, ∞).

ANSWER:

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].

ANSWER:

3x – 2 < 2x +1

⇒ 3x – 2x < 1 + 2

⇒ x < 3

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

ANSWER:

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

ANSWER:

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

ANSWER:

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

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,

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

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,

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

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 … (i)

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

∴x + (x + 2) > 11

⇒ 2x + 2 > 11

⇒ 2x > 11 – 2

⇒ 2x > 9

From (i) and (ii), we obtain 4.5<x<8.4.5<x<8.

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

Thus, the required possible pairs are (5, 7) and (7, 9).

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,

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

From (1) and (2), we obtain 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).

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,

Thus, the minimum length of the shortest side is 9 cm.

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,

x cm + (x + 3) cm + 2x cm ≤ 91 cm

⇒ 4x + 3 ≤ 91

⇒ 4x ≤ 91 ­– 3

⇒ 4x ≤ 88

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

∴2x ≥ (x + 3) + 5

⇒ 2x ≥ x + 8

⇒ x ≥ 8 … (2)

From (1) and (2), we obtain

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.

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.

ANSWER:

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.

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.

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.

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.

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.

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.

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.

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.

ANSWER:

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 follows.

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.

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.

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.

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.

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.

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.

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.

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 (2) 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.

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.

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.

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.

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.

ANSWER:

6 ≤ – 3(2x – 4) < 12

⇒ 2 ≤ –(2x – 4) < 4

⇒ –2 ≥ 2x – 4 > –4

⇒ 4 – 2 ≥ 2x > 4 – 4

⇒ 2 ≥ 2x > 0

⇒1 ≥ x > 0

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

ANSWER:

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

ANSWER:

⇒ –75 < 3(x – 2) ≤ 0

⇒ –25 < x – 2 ≤ 0

⇒ – 25 + 2 < x ≤ 2

⇒ –23 < x ≤ 2

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

ANSWER:

Thus, the solution set for the given inequalityis.

ANSWER:

Thus, the solution set for the given inequalityis.

ANSWER:

5x + 1 > –24

⇒ 5x > –25

⇒ x > –5 … (1)

5x – 1 < 24

⇒ 5x < 25

⇒ x < 5 … (2)

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (–5, 5). The solution of the given system of inequalities can be represented on number line as

ANSWER:

2(x – 1) < x + 5

⇒ 2x – 2 < x + 5

⇒ 2x – x < 5 + 2

⇒ x < 7 … (1)

3(x + 2) > 2 – x

⇒ 3x + 6 > 2 – x

⇒ 3x + x > 2 – 6

⇒ 4x > – 4

⇒ x > – 1 … (2)

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is (–1, 7). The solution of the given system of inequalities can be represented on number line as

ANSWER:

3x – 7 > 2(x – 6)

⇒ 3x – 7 > 2x – 12

⇒ 3x – 2x > – 12 + 7

⇒ x > –5 … (1)

6 – x > 11 – 2x

⇒ –x + 2x > 11 – 6

⇒ x > 5 … (2)

From (1) and (2), it can be concluded that the solution set for the given system of inequalities is. The solution of the given system of inequalities can be represented on number line as

ANSWER:

5(2x – 7) – 3(2x + 3) ≤ 0

⇒ 10x – 35 – 6x – 9 ≤ 0

⇒ 4x – 44 ≤ 0

⇒ 4x ≤ 44

⇒ x ≤ 11 … (1)

2x + 19 ≤ 6x + 47

⇒ 19 – 47 ≤ 6x – 2x

⇒ –28 ≤ 4x

⇒ –7 ≤ x … (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

ANSWER:

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

68 < F < 77

Putting  we obtain

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

ANSWER:

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

Then, total mixture = (x + 640) litres

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

∴2%x + 8% of 640 > 4% of (x + 640)

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

2%x + 8% of 640 > 4% of (x + 640)

⇒ 2x + 5120 > 4x + 2560

⇒ 5120 – 2560 > 4x – 2x

⇒ 5120 – 2560 > 2x

⇒ 2560 > 2x

⇒ 1280 > x

2% x + 8% of 640 < 6% of (x + 640)

⇒ 2x + 5120 < 6x + 3840

⇒ 5120 – 3840 < 6x – 2x

⇒ 1280 < 4x

⇒ 320 < x

∴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.

ANSWER:

Let x litres of water is required to be added.

Then, total mixture = (x + 1125) litres

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

This resulting mixture will contain more than 25% but less than 30% acid content.

∴30% of (1125 + x) > 45% of 1125

And, 25% of (1125 + x) < 45% of 1125

30% of (1125 + x) > 45% of 1125

25% of (1125 + x) < 45% of 1125

⇒25100(1125 + x)<45100×1125 ⇒25(1125 + x)<45×1125

⇒25×1125 +25x<45×1125 ⇒25x<45×1125 −25×1125 ⇒25x<22500

⇒x<900⇒251001125 + x<45100×1125 ⇒251125 + x<45×1125⇒25×1125 +25x<45×1125 

⇒25x<45×1125 -25×1125 ⇒25x<22500⇒x<900

∴ 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.