Page 1 :
Mathematics, (Chapter – 6) (Linear Inequalities), (Class – XI), , Exercise 6.1, Question 1:, Solve 24x < 100, when (i) x is a natural number, , (ii) x is an integer, , Answer 1:, The given inequality is 24x < 100., , (i), , It is evident that 1, 2, 3, and 4 are the only natural numbers less than, , 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 2:, The given inequality is –12x > 30., , (i) There is no natural number less than, , ., , 1

Page 2 :
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}., Question 3:, Solve 5x– 3 < 7, when, (i) x is an integer, , (ii) x is a real number, , Answer 3:, 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)., , Question 4:, Solve 3x + 8 > 2, when, (i) x is an integer, , (ii) x is a real number, , Answer 4:, The given inequality is 3x + 8 > 2., , 2

Page 3 :
(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 5:, 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, ∞)., Question 6:, Solve the given inequality for real x: 3x – 7 > 5x – 1, Answer 6:, 3x – 7 > 5x – 1, ⇒ 3x – 7 + 7 > 5x – 1 + 7, ⇒ 3x > 5x + 6, ⇒ 3x – 5x > 5x + 6 – 5x, ⇒ – 2x > 6, , 3

Page 4 :
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 7:, 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]., Question 8:, Solve the given inequality for real x: 3(2 – x) ≥ 2(1 – x), Answer 8:, 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]., , 4

Page 5 :
Question 9:, Solve the given inequality for real x:, Answer 9:, , 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:, Answer 10:, , 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)., , 5

Page 6 :
Question 11:, Solve the given inequality for real x:, Answer 11:, , 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:, Answer 12:, , 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]., , 6

Page 7 :
Question 13:, Solve the given inequality for real x: 2(2x + 3) – 10 < 6 (x – 2), Answer 13:, , 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 14:, , 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 15:, Solve the given inequality for real x:, , 7

Page 8 :
Answer 15:, , 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:, Answer 16:, , 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]., , 8

Page 9 :
Question 17:, Solve the given inequality and show the graph of the solution on number line:, 3x – 2 < 2x +1, , Answer 17:, 3x – 2 < 2x +1, ⇒ 3x – 2x < 1 + 2, ⇒x<3, The graphical representation of the solutions of the given inequality is as follows., , Question 18:, Solve the given inequality and show the graph of the solution on number line:, 5x – 3 ≥ 3x – 5, , Answer 18:, , The graphical representation of the solutions of the given inequality is as follows., , 9

Page 10 :
Question 19:, Solve the given inequality and show the graph of the solution on number line:, 3(1 – x) < 2 (x + 4), Answer 19:, , The graphical representation of the solutions of the given inequality is as follows., , Question 20:, Solve the given inequality and show the graph of the solution on number line:, , Answer 20:, , The graphical representation of the solutions of the given inequality is as follows., , 10

Page 11 :
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 21:, 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., , 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 22:, 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., , 11

Page 12 :
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 23:, 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, , ., , 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)., 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 24:, 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, , 12

Page 13 :
⇒ 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)., 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 25:, 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., , 13

Page 14 :
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 26:, 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., , 14

Page 15 :
Mathematics, (Chapter – 6) (Linear Inequalities), (Class – XI), , Exercise 6.2, Question 1:, Solve the given inequality graphically in two-dimensional plane: x + y < 5, Answer 1:, 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., , 1

Page 16 :
Question 2:, Solve the given inequality graphically in two-dimensional plane: 2x + y ≥ 6, Answer 2:, 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., , 2

Page 17 :
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., , 3

Page 18 :
Question 4:, Solve the given inequality graphically in two-dimensional plane: y + 8 ≥ 2x, Answer 4:, 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., , 4

Page 19 :
Question 5:, Solve the given inequality graphically in two-dimensional plane: x – y ≤ 2, Answer 5:, 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., , 5

Page 20 :
Question 6:, Solve the given inequality graphically in two-dimensional plane: 2x – 3y > 6, Answer 6:, 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., , 6

Page 21 :
Question 7:, Solve the given inequality graphically in two-dimensional plane: –3x + 2y ≥ –6, , Answer 7:, 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., , 7

Page 22 :
Question 8:, Solve the given inequality graphically in two-dimensional plane: 3y – 5x < 30, Answer 8:, 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., , 8

Page 23 :
Question 9:, Solve the given inequality graphically in two-dimensional plane: y < –2, , Answer 9:, 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., , 9

Page 24 :
Question 10:, Solve the given inequality graphically in two-dimensional plane: x > –3, , Answer 10:, 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., , 10

Page 25 :
Mathematics, (Chapter – 6) (Linear Inequalities), (Class – XI), , Exercise 6.3, Question 1:, Solve the following system of inequalities graphically: x ≥ 3, y ≥ 2., Answer 1:, 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., , 1

Page 26 :
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 follows., , 2

Page 27 :
Question 3:, Solve the following system of inequalities graphically: 2x + y≥ 6, 3x + 4y ≤ 12, , Answer 3:, 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, , 2x + y= 6), and inequality (2), , above the line, 2x + y= 6 ( including the line, , 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., , 3

Page 28 :
Question 4:, Solve the following system of inequalities graphically: x + y≥ 4, 2x – y > 0, Answer 4:, 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., , 4

Page 29 :
Question 5:, Solve the following system of inequalities graphically: 2x – y > 1, x – 2y < –1, Answer 5:, 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., , 5

Page 30 :
Question 6:, Solve the following system of inequalities graphically: x + y ≤ 6,, , x+y≥4, , Answer 6:, 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., , 6

Page 31 :
Question 7:, Solve the following system of inequalities graphically: 2x + y≥ 8, x + 2y ≥ 10, Answer 7:, 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., , 7

Page 32 :
Question 8:, Solve the following system of inequalities graphically: x + y ≤ 9, y > x, x ≥ 0, Answer 8:, 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., , 8

Page 33 :
Question 9:, Solve the following system of inequalities graphically: 5x + 4y ≤ 20, x ≥ 1, y ≥ 2, , Answer 9:, 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, 4y = 20)., (including, (including, , (1) represents the region below the line, 5x + 4y = 20 (including the line 5x +, Inequality (2) represents the region on the right hand side of the line, x = 1, the line x = 1). Inequality (3) represents the region above the line, y = 2, 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., , 9

Page 34 :
Question 10:, Solve the following system of inequalities graphically: 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0,, y≥0, Answer 10:, 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., , 10

Page 35 :
Question 11:, Solve the following system of inequalities graphically: 2x + y≥ 4, x + y ≤ 3, 2x – 3y ≤ 6, Answer 11:, 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., , 11

Page 36 :
Question 12:, Solve the following system of inequalities graphically:, x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1, Answer 12:, 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., , 12

Page 37 :
Question 13:, Solve the following system of inequalities graphically:, 4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, x, y ≥ 0, Answer 13:, 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., , 13

Page 38 :
Question 14:, Solve the following system of inequalities graphically:, 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0, , Answer 14:, 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., , 14

Page 39 :
Question 15:, Solve the following system of inequalities graphically:, x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0, Answer 15:, 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., , 15

Page 40 :
Mathematics, (Chapter – 6) (Linear Inequalities), (Class – XI), , Miscellaneous Exercise on Chapter 6, Question 1:, Solve the inequality 2 ≤ 3x – 4 ≤ 5, , Answer 1:, 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 2:, 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 inequality is (0, 1]., , 1

Page 41 :
Question 3:, Solve the inequality, , Answer 3:, , Thus, the solution set for the given inequality is [–4, 2]., , Question 4:, Solve the inequality, Answer 4:, , ⇒ –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]., , 2

Page 42 :
Question 5:, Solve the inequality, , Answer 5:, , Thus, the solution set for the given inequality is., , Question 6:, Solve the inequality, Answer 6:, , Thus, the solution set for the given inequality is, , 3

Page 43 :
Question 7:, Solve the inequalities and represent the solution graphically on number line:, 5x + 1 > – 24,, , 5x – 1 < 24, , Answer 7:, ⇒, , 5x + 1 > –24, ⇒, , x > –5 ……………………………. (1), , 5x – 1 < 24, ⇒, , 5x > –25, , ⇒, , 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, , Question 8:, Solve the inequalities and represent the solution graphically on number line:, 2(x – 1) < x + 5,, , 3(x + 2) > 2 – x, , Answer 8:, 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, , 4

Page 44 :
Question 9:, Solve the following inequalities and represent the solution graphically on number line:, 3x – 7 > 2(x – 6), 6 – x > 11 – 2x, , Answer 9:, 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 (5, ∞). The solution of the given system of inequalities can be represented, on number line as, , 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), 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, , 5

Page 45 :
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, , Answer 11:, 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., , 6

Page 46 :
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 12:, 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., , 7

Page 47 :
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 13:, 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, , 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., , 8

Page 48 :
Question 14:, IQ of a person is given by the formula, 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 14:, It is given that for a group of 12 years old children,, 80 ≤ IQ ≤ 140 …………………………………. (i), For a group of 12 years old children, CA = 12 years, , Putting this value of IQ in (i), we obtain, , Thus, the range of mental age of the group of 12 years old children is, , 9