NCERT Solutions for Class 11 Maths Chapter 6
Linear Inequalities Class 11
Chapter 6 Linear Inequalities Exercise 6.1, 6.2, 6.3, miscellaneous Solutions
Exercise 6.1 : Solutions of Questions on Page Number : 122
Solve 24x < 100, when (i) x is a natural number (ii) x is an integer
(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 β3, β2, β1, 0, 1, 2, 3, 4.
Thus, when x is an integer, the solutions of the given inequality are
β3, β2, β1, 0, 1, 2, 3, 4.
Solve -12x> 30, when
Answer :
The given inequality is β12x> 30.
(i) There is no natural number less than .
Thus, when xis a natural number, there is no solution of the given inequality.
(ii) The integers less than are , β5, β4, β3.
Thus, when xis an integer, the solutions of the given inequality are
, β5, β4, β3.
Solve 5x- 3 < 7, when
Answer :
(i) The integers less than 2are , β4, β3, β2, β1, 0, 1. Thus, when xis an integer, the solutions of the given inequality are
, β4, β3, β2, β1, 0, 1.
Hence, in this case, the solution set is { , β4, β3, β2, β1, 0, 1}.
(ii) When xis a real number, the solutions of the given inequality are given by x< 2, that is, all real numbers xwhich are less than 2.
Solve 3x+ 8 > 2, when
Answer :
(i) The integers greater than β2are β1, 0, 1, 2,
Thus, when xis an integer, the solutions of the given inequality are β1, 0, 1, 2
Hence, in this case, the solution set is {β1, 0, 1, 2, }.
(ii) When xis a real number, the solutions of the given inequality are all the real numbers, which are greater than β2. Thus, in this case, the solution set is (β 2, ∞).
Solve the given inequality for real x: 4x + 3 < 5x + 7
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, ∠ΕΎ).
Solve the given inequality for real x: 3x - 7 > 5x - 1
3xβ 7 > 5xβ 1
⇒ 3xβ 7 + 7 > 5xβ 1 + 7
⇒ 3x> 5x+ 6
⇒ 3xβ 5x> 5x+ 6 β 5x
Thus, all real numbers x,which are less than β3, are the solutions of the given inequality. Hence, the solution set of the given inequality is (β∞, β3).
Solve the given inequality for real x: 3(x - 1) Γ’°Β€ 2 (x - 3)
3(x- 1) Γ’°Β€2(x- 3)
⇒ 3x- 3 Γ’°Β€ 2x- 6
⇒ 3x- 3 + 3 Γ’°Β€ 2x- 6 + 3
⇒ 3x Γ’°Β€ 2x- 3
⇒ 3x - 2x Γ’°Β€2x- 3 - 2x
⇒ x Γ’°Β€- 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 (-∠ΕΎ, -3].
Solve the given inequality for real x: 3(2 - x) Γ’°Β₯ 2(1 - x)
3(2 - x) Γ’°Β₯2(1 - x)
⇒ 6 - 3x Γ’°Β₯2 - 2x
⇒ 6 - 3x+ 2x Γ’°Β₯2 - 2x + 2x
⇒ 6 - x Γ’°Β₯2
⇒ 6 - x- 6 Γ’°Β₯2 - 6
⇒ -x Γ’°Β₯-4
⇒ x Γ’°Β€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 (-∠ΕΎ, 4].
Solve the given inequality for real x:
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 (β∞, 6).
Solve the given inequality for real x: Answer :
Thus, all real numbers x,which are less than β6, are the solutions of the given inequality. Hence, the solution set of the given inequality is (β∞, β6).
Solve the given inequality for real x:
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 (β∞, 2].
Solve the given inequality for real x:
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 (β∞, 120].
Solve the given inequality for real x: 2(2x + 3) - 10 < 6 (x - 2)
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, ∞).
Solve the given inequality for real x: 37 - (3x + 5) Γ’°Β₯ 9x - 8(x - 3)
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 (β∞, 2].
Solve the given inequality for real x: 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, ∞).
Solve the given inequality for real x:
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 (β∞, 2].
Solve the given inequality and show the graph of the solution on number line: 3x - 2 < 2x +1
3x- 2 < 2x+1
⇒ 3x - 2x< 1 + 2
⇒ x< 3
Solve the given inequality and show the graph of the solution on number line: 5x - 3 Γ’°Β₯ 3x - 5
The graphical representation of the solutions of the given inequality is as follows.
Solve the given inequality and show the graph of the solution on number line: 3(1 - x) < 2 (x + 4)
The graphical representation of the solutions of the given inequality is as follows.
Solve the given inequality and show the graph of the solution on number line: Answer :
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.
Let xbe the marks obtained by Ravi in the third unit test.
Thus, the student must obtain a minimum of 35 marks to have an average of at least 60 marks.
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.
Let xbe 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,
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
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 β 2
⇒ x< 8 (i)
Also, the sum of the two integers is more than 11.
∴x+ (x + 2) > 11
⇒ 2x+ 2 > 11
⇒ 2x> 11 β 2
From (i) and (ii), we obtain .
Since x is an odd number, xcan take the values, 5 and 7. Thus, the required possible pairs are (5, 7) and (7, 9).
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.
Let the length of the shortest side of the triangle be x cm. Then, length of the longest side = 3xcm
Length of the third side = (3xβ 2) cm
Thus, the minimum length of the shortest side is 9 cm.
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?
Answer :
Let the length of the shortest piece be xcm. Then, length of the second piece and the third piece are (x+ 3) cm and 2xcm respectively.
Since the three lengths are to be cut from a single piece of board of length 91 cm,
xcm + (x+ 3) cm + 2xcm ≤91 cm
⇒ 4x+ 3 ≤91
⇒ 4x ≤91 β 3
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
Exercise 6.2 : Solutions of Questions on Page Number : 127
Solve the given inequality graphically in two-dimensional plane: x + y < 5
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, Iand 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 Iexcluding the points on the line. This can be represented as follows.
Solve the given inequality graphically in two-dimensional plane: 2x + y Γ’°Β₯ 6
The graphical representation of 2x+ y= 6 is given in the figure below. This line divides the xy-plane in two half planes, Iand 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 Γ’°Β≦ 6 or 0 Γ’°Β≦ 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 IIincluding the points on the line. This can be represented as follows.
Solve the given inequality graphically in two-dimensional plane: 3x + 4y Γ’°Β€ 12
3x+ 4y Γ’°Β€12
The graphical representation of 3x+ 4y= 12 is given in the figure below. This line divides the xy-plane in two half planes, Iand 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) Γ’°Β€12 or 0 Γ’°Β€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 Iincluding the points on the line. This can be represented as follows.
Solve the given inequality graphically in two-dimensional plane: y + 8 Γ’°Β₯ 2x
The graphical representation of y+ 8 = 2xis 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 Γ’°Β₯2(0) or 8 Γ’°Β₯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.
Solve the given inequality graphically in two-dimensional plane: x - y Γ’°Β€ 2
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 Γ’°Β€2 or 0 Γ’°Β€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.
Solve the given inequality graphically in two-dimensional plane: 2x - 3y > 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.
Solve the given inequality graphically in two-dimensional plane: -3x + 2y Γ’°Β₯ -6
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) Γ’°Β₯- 6 or 0 Γ’°Β₯-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.
Solve the given inequality graphically in two-dimensional plane: 3y - 5x < 30
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.
Solve the given inequality graphically in two-dimensional plane: y < -2
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.
Solve the given inequality graphically in two-dimensional plane: x > -3
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.
Exercise 6.3 : Solutions of Questions on Page Number : 129
Solve the following system of inequalities graphically: x Γ’°Β₯ 3, y Γ’°Β₯ 2
x Γ’°Β₯3 ... (1)
y Γ’°Β₯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.
Solve the following system of inequalities graphically: 3x + 2y Γ’°Β€ 12, x Γ’°Β₯ 1, y Γ’°Β₯ 2
3x+ 2y Γ’°Β€12 ... (1)
x Γ’°Β₯1 ... (2)
y Γ’°Β₯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.
Solve the following system of inequalities graphically: 2x + yΓ’°Β₯ 6, 3x + 4y Γ’°Β€ 12
2x+ yΓ’°Β₯6 ... (1)
3x+ 4y Γ’°Β€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.
Solve the following system of inequalities graphically: x + yΓ’°Β₯ 4, 2x - y > 0
x+ yΓ’°Β₯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.
Solve the following system of inequalities graphically: 2x - y > 1, x - 2y < -1
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 linesas follows.
Solve the following system of inequalities graphically: x + y Γ’°Β€ 6, x + y Γ’°Β₯ 4
x+ y Γ’°Β€6 ... (1)
x+ y Γ’°Β₯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.
Solve the following system of inequalities graphically: 2x + yΓ’°Β₯ 8, x + 2y Γ’°Β₯ 10
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.
Solve the following system of inequalities graphically: x + y Γ’°Β€ 9, y > x, x Γ’°Β₯ 0
x + y Γ’°Β€ 9 ... (1)
y > x ... (2)
x Γ’°Β₯ 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.
Solve the following system of inequalities graphically: 5x + 4y Γ’°Β€ 20, x Γ’°Β₯ 1, y Γ’°Β₯ 2
5x+ 4y Γ’°Β€20 ... (1)
x Γ’°Β₯1 ... (2)
y Γ’°Β₯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.
Solve the following system of inequalities graphically: 3x + 4y Γ’°Β€ 60, x + 3y Γ’°Β€ 30, x Γ’°Β₯ 0, y Γ’°Β₯ 0
3x+ 4y Γ’°Β€60 ... (1)
x+ 3y Γ’°Β€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 Γ’°Β₯0 and y Γ’°Β₯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.
Solve the following system of inequalities graphically: 2x + yΓ’°Β₯ 4, x + y Γ’°Β€ 3, 2x - 3y Γ’°Β€ 6
2x+ yΓ’°Β₯4 ... (1)
x+ y Γ’°Β€3 ... (2)
2x- 3y Γ’°Β€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.
Solve the following system of inequalities graphically:
x - 2y ≤ 3, 3x + 4y Γ’β°Β₯ 12, x Γ’β°Β₯ 0, y Γ’β°Β₯ 1
x - 2y Γ’°Β€ 3 ... (1)
3x + 4y Γ’°Β≦ 12 ... (2)
y Γ’°Β≦ 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 Γ’°Β≦ 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.
[[Q]]
4x + 3y ≤ 60, y Γ’β°Β₯ 2x, x Γ’β°Β₯ 3, x, y Γ’β°Β₯ 0
4x + 3y Γ’°Β€ 60 ... (1)
y Γ’°Β≦ 2x ... (2)
x Γ’°Β≦ 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.
Solve the following system of inequalities graphically: 3x + 2y Γ’°Β€ 150, x + 4y Γ’°Β€ 80, x Γ’°Β€ 15, yΓ’°Β₯ 0, x Γ’°Β₯ 0
3x+ 2y Γ’°Β€150 ... (1)
x+ 4y Γ’°Β€80 ... (2)
x Γ’°Β€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 Γ’°Β₯0 and y Γ’°Β₯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.
Solve the following system of inequalities graphically: x + 2y Γ’°Β€ 10, x + y Γ’°Β₯ 1, x -
Answer :
x+ 2y Γ’°Β€10 ... (1)
x+ y Γ’°Β₯1 ... (2)
x- y Γ’°Β€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 Γ’°Β₯0 and y Γ’°Β₯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.
Exercise Miscellaneous : Solutions of Questions on Page Number : 132
Solve the inequality 2 Γ’°Β€ 3x - 4 Γ’°Β€ 5
2 Γ’°Β€3x - 4 Γ’°Β€5
⇒ 2 + 4 Γ’°Β€3x - 4 + 4 Γ’°Β€5 + 4
⇒ 6 Γ’°Β€3x Γ’°Β€9
⇒ 2 Γ’°Β€ x Γ’°Β€3
Solve the inequality 6 Γ’°Β€ -3(2x - 4) < 12
6 Γ’°Β€- 3(2x- 4) < 12
⇒ 2 Γ’°Β€-(2x- 4) < 4
⇒ -2 Γ’°Β₯2x - 4 > -4
⇒ 4 - 2 Γ’°Β₯2x > 4 - 4
⇒ 2 Γ’°Β₯2x > 0
⇒1 Γ’°Β₯ x > 0
Solve the inequality Answer :
Solve the inequality Answer :
⇒ β75 < 3(xβ 2) ≤0
⇒ β25 < xβ 2 ≤0
⇒ β 25 + 2 < x ≤2
⇒ β23 < x ≤2
Solve the inequality Answer :
Solve the inequality Answer :
Solve the inequalities and represent the solution graphically on number line: 5x + 1 > -24, 5x - 1 < 24
5x+ 1 > -24
⇒ 5x > -25
⇒ x > -5 ... (1)
5x- 1 < 24
⇒ 5x < 25
⇒ x< 5 ... (2)
Solve the inequalities and represent the solution graphically on number line: 2(x - 1) < x + 5, 3(x + 2) > 2 - x
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
Solve the following inequalities and represent the solution graphically on number line: 3x - 7 > 2(x - 6), 6 - x > 11 - 2x
3x β 7 > 2(x β 6)
⇒ 3x β 7 > 2x β 12
⇒ 3x β 2x > β 12 + 7
⇒ x > β5 (1) 6 β x > 11 β 2x
⇒ βx + 2x > 11 β 6
⇒ x > 5 (2)
Solve the inequalities and represent the solution graphically on number line: 5(2x - 7) - 3(2x + 3) Γ’°Β€ 0, 2x + 19 Γ’°Β€ 6x + 47
5(2x- 7) - 3(2x+ 3) Γ’°Β€0
⇒ 10x- 35 - 6x- 9 Γ’°Β€0
⇒ 4x- 44 Γ’°Β€0
⇒ 4x Γ’°Β€44
⇒ x Γ’°Β€11 ... (1) 2x+ 19 Γ’°Β€6x + 47
⇒ 19 - 47 Γ’°Β€6x - 2x
⇒ -28 Γ’°Β€4x
⇒ -7 Γ’°Β€ x ... (2)
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
Since the solution is to be kept between 68°F and 77°F, 68 < F < 77
Thus, the required range of temperature in degree Celsius is between 20°C and 25°C.
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?
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)
⇒ 2x+ 5120 > 4x+ 2560
⇒ 5120 β 2560 > 4xβ 2x
⇒ 5120 β 2560 > 2x
⇒ 2560 > 2x
⇒ 1280 > x
⇒ 2x+ 5120 < 6x+ 3840
⇒ 5120 β 3840 < 6xβ 2x
⇒ 1280 < 4x
⇒ 320 < x
∴320 < x < 1280
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?
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
∴562.5 < x < 900
Answer :
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 .