Chapter 5 Linear Inequalities

NCERT Solutions Class 11 Maths Chapter 5 Linear Inequalities: Download PDF

NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities helps the students to study the mathematical expressions which contain the symbols <, >, \leq and \geq. Equations depict perfect balance, but inequalities describe ranges and regions, which are far more common in real-life situations like budgeting, resource allocation and optimisation.

ALLEN NCERT Solutions for Class 11 Maths Chapter 5 have been prepared by the expert faculty to make the rules of inequality easily understandable with clear explanations and systematic problem solving. The solutions emphasise the important rules for multiplying and dividing negative numbers and the graphical representation of solutions on line numbers and coordinate planes.

Linear Programming is a chapter in Class 12 and to understand it you need to have a good grasp on linear inequalities. It is also an important chapter for various competitive exams like JEE Main, NDA etc. The solutions give a strong logical basis for helping students grasp the idea of the "shaded region" and interval notation, which are often used in calculus and coordinate geometry.

1.0Class 11 Maths Chapter 5 Linear Inequalities: Key Concepts

Class 11 Maths Chapter 5, Linear Inequalities, explains how to find all values that satisfy an inequality. The chapter covers linear inequalities in one and two variables, their solutions, and graphical representation.

Inequality Symbols: Understand the symbols < (less than), > (greater than), ≤ (less than or equal to), and ≥ (greater than or equal to).
Linear Inequalities in One Variable: Learn how to solve inequalities by adding, subtracting, multiplying, or dividing both sides. When both sides are multiplied or divided by a negative number, the inequality sign is reversed.
Solution on a Number Line: Represent solutions on a number line using an open circle for < or > and a closed circle for ≤ or ≥.
Linear Inequalities in Two Variables: Learn how to represent the solution of an inequality as a region on the Cartesian plane. Use a dashed line for strict inequalities and a solid line when equality is included.
System of Linear Inequalities: Find the common region that satisfies two or more linear inequalities at the same time.
Graphical Representation: Learn how to identify, plot, and shade the correct solution region for linear inequalities on a graph.

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 5

EXERCISE - 5.1

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

Sol. The given inequality is $24 x < 100$

$24 x < 100$

$\Rightarrow \frac{24 x}{24} < \frac{100}{24}$
[Dividing both sides by same positive number]

$\Rightarrow x < \frac{25}{6}$

(i) It is evident that 1, 2, 3, and 4 are the only natural numbers less than $\frac{25}{6}$ .
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 $\frac{25}{6}$ are $\dots - 3, - 2, - 1, 0$ , 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 ${\dots - 3, - 2, - 1, 0, 1, 2, 3, 4}$ .
2. Solve $- 12 x > 30$ , when

$(i) x is a natural number (ii) x is an integer$

Sol. The given inequality is $- 12 x > 30$ .

$- 12 x > 30$

$\Rightarrow \frac{- 12 x}{- 12} < \frac{30}{- 12}$
[Dividing both sides by same positive number]

$\Rightarrow x < - \frac{5}{2}$

(i) There is no natural number less than $(- \frac{5}{2})$ .
Thus, when x is a natural number, there is no solution of the given inequality.
(ii) The integers less than $(- \frac{5}{2})$ are $\dots, - 5$ , $- 4, - 3$ .
Thus, when x is an integer, the solutions of the given inequality are ..., -5, -4, -3.
Hence, in this case, the solution set is ${\dots, - 5$ , $- 4, - 3}$ .
3. Solve $5 x - 3 < 7$ , when

$(i) x is an integer (ii) x is a real number$

Sol. The given inequality is $5 x - 3 < 7$

$\Rightarrow 5 x - 3 + 3 < 7 + 3 \Rightarrow \frac{5 x}{5} < \frac{10}{5} \Rightarrow 5 x < 10 \Rightarrow x < 2$

(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 $\dots, - 4, - 3, - 2, - 1, 0, 1$ .
Hence, in this case, the solution set is ${\dots, - 4, - 3, - 2, - 1, 0, 1}$ .
(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 $x \in (- \infty, 2)$ .
4. Solve $3 x + 8 > 2$ , when

$(i) x is an integer (ii) x is a real number$

Sol. The given inequality is $3 x + 8 > 2$ ,

$\Rightarrow 3 x + 8 - 8 > 2 - 8 \Rightarrow 3 x > - 6 \Rightarrow \frac{3 x}{3} > \frac{- 6}{3} \Rightarrow x > - 2$

(i) The integers greater than -2 are -1, 0, 1, 2, .....
Thus, when x is an integer, the solutions of the given inequality are $- 1, 0, 1, 2 \dots$
Hence, in this case, the solution set is ${- 1, 0, 1, 2, \dots}$ .
(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 $(- 2, \infty)$ .

Solve the inequalities in Q. 5 to 16 for real x.

5. $4 x + 3 < 5 x + 7$

Sol. The given inequality is $4 x + 3 < 5 x + 7$

$\Rightarrow 4 x + 3 - 7 < 5 x + 7 - 7 \Rightarrow 4 x - 4 < 5 x \Rightarrow 4 x - 4 - 4 x < 5 x - 4 x \Rightarrow - 4 < x$

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 $(- 4, \infty)$ .
6. $3 x - 7 > 5 x - 1$

Sol. The given inequality is $3 x - 7 > 5 x - 1$

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow 3 x - 7 + 7 > 5 x - 1 + 7 3 x > 5 x + 6 3 x - 5 x > 5 x + 6 - 5 x - 2 x > 6 \frac{- 2 x}{- 2} < \frac{6}{- 2} x < - 3$

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 $(- \infty, - 3)$ .
7. $3 (x - 1) \leq 2 (x - 3)$

Sol. The given inequality is $3 (x - 1) \leq 2 (x - 3)$

$\Rightarrow 3 x - 3 \leq 2 x - 6 \Rightarrow 3 x - 3 + 3 \leq 2 x - 6 + 3 \Rightarrow 3 x \leq 2 x - 3 \Rightarrow 3 x - 2 x \leq 2 x - 3 - 2 x \Rightarrow x \leq - 3$

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 $(- \infty, - 3]$ .
8. $3 (2 - x) \geq 2 (1 - x)$

Sol. The given inequality is $3 (2 - x) \geq 2 (1 - x)$

$\Rightarrow 6 - 3 x \geq 2 - 2 x \Rightarrow 6 - 3 x + 2 x \geq 2 - 2 x + 2 x \Rightarrow 6 - x \geq 2 \Rightarrow 6 - x - 6 \geq 2 - 6 \Rightarrow - x \geq - 4 \Rightarrow x \leq 4$

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 $(- \infty, 4]$ .
9. $x + \frac{x}{2} + \frac{x}{3} < 11$

Sol. The given inequality is $x + \frac{x}{2} + \frac{x}{3} < 11$

$\Rightarrow x (1 + \frac{1}{2} + \frac{1}{3}) < 11 \Rightarrow x (\frac{6 + 3 + 2}{6}) < 11 \Rightarrow \frac{11 x}{6} < 11 \Rightarrow \frac{11 x}{6 \times 11} < \frac{11}{11} \Rightarrow \frac{x}{6} < 1 \Rightarrow x < 6$

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 $(- \infty, 6)$ .
10. $\frac{x}{3} > \frac{x}{2} + 1$

Sol. The given inequality is $\frac{x}{3} > \frac{x}{2} + 1$

$\Rightarrow \frac{x}{3} - \frac{x}{2} > 1 \Rightarrow - \frac{x}{6} > 1 \Rightarrow x < - 6 \Rightarrow \frac{2 x - 3 x}{6} > 1 \Rightarrow - x > 6$

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 $(- \infty, - 6)$ .

11. $\frac{3 ( x - 2 )}{5} \leq \frac{5 ( 2 - x )}{3}$

Sol. The given inequality is $\frac{3 ( x - 2 )}{5} \leq \frac{5 ( 2 - x )}{3}$
$\Rightarrow 9 (x - 2) \leq 25 (2 - x)$
$\Rightarrow 9 x - 18 \leq 50 - 25 x$
$\Rightarrow 9 x - 18 + 25 x \leq 50$
$\Rightarrow 34 x - 18 \leq 50 \Rightarrow 34 x \leq 50 + 18$
$\Rightarrow 34 x \leq 68 \Rightarrow \frac{34 x}{34} \leq \frac{68}{34}$
$\Rightarrow x \leq 2$

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 $(- \infty, 2]$ .

12. $\frac{1}{2} (\frac{3 x}{5} + 4) \geq \frac{1}{3} (x - 6)$

Sol. The given inequality is

$\frac{1}{2} (\frac{3 x}{5} + 4) \geq \frac{1}{3} (x - 6)$

$\Rightarrow 3 (\frac{3 x}{5} + 4) \geq 2 (x - 6)$
$\Rightarrow \frac{9 x}{5} + 12 \geq 2 x - 12$
$\Rightarrow 12 + 12 \geq 2 x - \frac{9 x}{5}$
$\Rightarrow 24 \geq \frac{10 x - 9 x}{5}$
$\Rightarrow 24 \geq \frac{x}{5} \Rightarrow 120 \geq x$
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 $(- \infty, 120]$ .
13. $2 (2 x + 3) - 10 < 6 (x - 2)$

Sol. The given inequality is

$2 (2 x + 3) - 10 < 6 (x - 2)$

$\Rightarrow 4 x + 6 - 10 < 6 x - 12$
$\Rightarrow 4 x - 4 < 6 x - 12$

$\Rightarrow - 4 + 12 < 6 x - 4 x$

$\Rightarrow 8 < 2 x$

$\Rightarrow 4 < x$

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 $[4, \infty)$ .
14. $37 - (3 x + 5) \geq 9 x - 8 (x - 3)$

Sol. The given inequality is

$37 - (3 x + 5) \geq 9 x - 8 (x - 3)$

$\Rightarrow 37 - 3 x - 5 \geq 9 x - 8 x + 24$
⇒

$32 - 3 x \geq x + 24$

$\Rightarrow 32 - 24 \geq x + 3 x$

⇒

$8 \geq 4 x$

$\Rightarrow 2 \geq x$

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 $(- \infty, 2]$ .
15. $\frac{x}{4} < \frac{( 5 x - 2 )}{3} - \frac{( 7 x - 3 )}{5}$

Sol. The given inequality is

$\frac{x}{4} < \frac{( 5 x - 2 )}{3} - \frac{( 7 x - 3 )}{5}$

⇒

$\frac{x}{4} < \frac{5 ( 5 x - 2 ) - 3 ( 7 x - 3 )}{15}$

⇒

$\frac{x}{4} < \frac{25 x - 10 - 21 x + 9}{15}$

⇒

$\frac{x}{4} < \frac{4 x - 1}{15}$

$\Rightarrow 15 x < 4 (4 x - 1)$

⇒

$15 x < 16 x - 4$

$\Rightarrow 4 < 16 x - 15 x$

⇒

$4 < x$

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 $(4, \infty)$ .

16. $\frac{( 2 x - 1 )}{3} \geq \frac{( 3 x - 2 )}{4} - \frac{( 2 - x )}{5}$

Sol. The given inequality is

$\frac{( 2 x - 1 )}{3} \geq \frac{( 3 x - 2 )}{4} - \frac{( 2 - x )}{5}$

$\Rightarrow \frac{( 2 x - 1 )}{3} \geq \frac{5 ( 3 x - 2 ) - 4 ( 2 - x )}{20}$
$\Rightarrow \frac{( 2 x - 1 )}{3} \geq \frac{15 x - 10 - 8 + 4 x}{20}$
$\Rightarrow \frac{( 2 x - 1 )}{3} \geq \frac{19 x - 18}{20}$
$\Rightarrow 20 (2 x - 1) \geq 3 (19 x - 18)$
$\Rightarrow 40 x - 20 \geq 57 x - 54$
$\Rightarrow - 20 + 54 \geq 57 x - 40 x$
$\Rightarrow 34 \geq 17 x \Rightarrow 2 \geq x$
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 $(- \infty, 2]$ .

Solve the given inequalities in Q. 17 to 20 and show the graph of the solution in each case on number line.

17. $3 x - 2 < 2 x + 1$

Sol. The given inequality is $3 x - 2 < 2 x + 1$

$3 x - 2 < 2 x + 1$

$\Rightarrow 3 x - 2 x < 1 + 2 \Rightarrow x < 3$
The graphical representation of the solutions of the given inequality is as follows.

18. $5 x - 3 \geq 3 x - 5$

Sol. The given inequality is $5 x - 3 \geq 3 x - 5$

$5 x - 3 \geq 3 x - 5$

$\Rightarrow 5 x - 3 x \geq - 5 + 3 \Rightarrow 2 x \geq - 2$
$\Rightarrow \frac{2 x}{2} \geq \frac{- 2}{2} \Rightarrow x \geq - 1$
The graphical representation of the solutions of the given inequality is as follows.

19. $3 (1 - x) < 2 (x + 4)$

Sol. The given inequality is $3 (1 - x) < 2 (x + 4)$

$\Rightarrow 3 - 3 x < 2 x + 8 \Rightarrow - 5 < 5 x \Rightarrow - 1 < x \Rightarrow 3 - 8 < 2 x + 3 x \Rightarrow \frac{- 5}{5} < \frac{5 x}{5}$

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

20. $\frac{x}{2} \geq \frac{( 5 x - 2 )}{3} - \frac{( 7 x - 3 )}{5}$

Sol. The given inequality is

$\frac{x}{2} \geq \frac{( 5 x - 2 )}{3} - \frac{( 7 x - 3 )}{5}$

$\Rightarrow \frac{x}{2} \geq \frac{5 ( 5 x - 2 ) - 3 ( 7 x - 3 )}{15}$
$\Rightarrow \frac{x}{2} \geq \frac{25 x - 10 - 21 x - 9}{15}$
$\Rightarrow \frac{x}{2} \geq \frac{4 x - 1}{15}$
$\Rightarrow 15 x \geq 2 (4 x - 1)$
$\Rightarrow 15 x \geq 8 x - 2$
$\Rightarrow 15 x - 8 x \geq 8 x - 2 - 8 x$
$\Rightarrow 7 x \geq - 2 \Rightarrow x \geq - \frac{2}{7}$
The graphical representation of the solutions of the given inequality is as follows

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.

Sol. 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,

$\frac{70 + 75 + x}{3} \geq 60 \Rightarrow 145 + x \geq 180$

$\Rightarrow x \geq 180 - 145 \Rightarrow x \geq 35$

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

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.

Sol. 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,

$\frac{87 + 92 + 94 + 95 + x}{5} \geq 90$

$\Rightarrow \frac{368 + x}{5} \geq 90 \Rightarrow x \geq 450 - 368 \Rightarrow 368 + x \geq 450 \Rightarrow x \geq 82$

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

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

Sol. 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,

$x + 2 < 10$

$\Rightarrow x < 10 - 2$
$\Rightarrow x < 8$
Also, the sum of the two integers is more than 11.
$∴ x + (x + 2) > 11$
$\Rightarrow 2 x + 2 > 11 \Rightarrow 2 x > 11 - 2$
$\Rightarrow 2 x > 9 \Rightarrow x > \frac{9}{2}$
$\Rightarrow x > 4.5$
From (1) and (2), 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).
24. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23 .

Sol. 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,

$x > 5$

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

$x + (x + 2) < 23$

$\Rightarrow 2 x + 2 < 23 \Rightarrow 2 x < 23 - 2$
$\Rightarrow 2 x < 21 \Rightarrow x > \frac{21}{2}$

$\Rightarrow x < 10.5$

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).
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.

Sol. Let the length of the shortest side of the triangle be x cm.
Then, length of the longest side $= 3 x cm$
Length of the third side $= (3 x - 2) cm$
Since the perimeter of the triangle is at least 61 cm,

$\Rightarrow \Rightarrow \Rightarrow x cm + 3 x cm + (3 x - 2) cm \geq 61 cm 7 x - 2 \geq 61 \Rightarrow 7 x \geq 61 + 2 7 x \geq 63 \Rightarrow \frac{7 x}{7} \geq \frac{63}{7} x \geq 9$

Thus, the minimum length of the shortest side is 9 cm.
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, $x =$ $(x + 3) + 2 x \leq 91$ and $2 x \geq (x + 3) + 5$ ]

Sol. 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 2 x cm respectively. Since the three lengths are to be cut from a single piece of board of length 91 cm,

$\Rightarrow \Rightarrow \Rightarrow x cm + (x + 3) cm 4 x + 3 \leq 91 4 x \leq 88 x \leq 22 + 2 x cm \leq 91 cm \Rightarrow \Rightarrow 4 x \leq 91 - 3 \frac{4 x}{4} \leq \frac{88}{4}$

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

$∴ 2 x \geq (x + 3) + 5 \Rightarrow 2 x \geq x + 8 \Rightarrow x \geq 8$

From (1) and (2), we obtain $8 \leq x \leq 22$
Thus, the possible length of the shortest board is greater than or equal to 8 cm but less than or equal to 22 cm.

3.0MISCELLANEOUS EXERCISE

Solve the inequalities Q. 1 to 6

1. $2 \leq 3 x - 4 \leq 5$

Sol. The given inequality is $2 \leq 3 x - 4 \leq 5$

$\Rightarrow 2 + 4 \leq 3 x - 4 + 4 \leq 5 + 4$
$\Rightarrow 6 \leq 3 x \leq 9$
$\Rightarrow 2 \leq x \leq 3$

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].

2. $6 \leq - 3 (2 x - 4) < 12$

Sol. The given inequality is

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow \Rightarrow 6 \leq - 3 (2 x - 4) < 1 2 \leq - (2 x - 4) < 4 - 2 \geq 2 x - 4 > - 4 4 - 2 \geq 2 x > 4 - 4 2 \geq 2 x > 0 1 \geq x > 0$

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

3. $- 3 \leq 4 - \frac{7 x}{2} \leq 18$

Sol. The given inequality is

$- 3 \leq 4 - \frac{7 x}{2} \leq 18 \Rightarrow - 3 - 4 \leq \frac{7 x}{2} \leq 18 - 4$

$\Rightarrow - 7 \leq - \frac{7 x}{2} \leq 14 \Rightarrow 7 \geq \frac{7 x}{2} \geq - 14$
$\Rightarrow 1 \geq \frac{x}{2} \geq - 2 \Rightarrow 2 \geq x \geq - 4$

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

4. $- 15 < \frac{3 ( x - 2 )}{5} \leq 0$

Sol. The given inequality is

$- 15 < \frac{3 ( x - 2 )}{5} \leq 0$

$\Rightarrow \Rightarrow \Rightarrow \Rightarrow - 75 < 3 (x - 2) \leq 0 - 25 < x - 2 \leq 0 - 25 + 2 < x \leq 2 - 23 < x \leq 2$

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

5. $- 12 < 4 - \frac{3 x}{- 5} \leq 2$

Sol. The given inequality is

$\Rightarrow \Rightarrow \Rightarrow - 12 < 4 - \frac{3 x}{- 5} \leq 2 - 12 - 4 < \frac{- 3 x}{- 5} \leq 2 - 4 - 16 < \frac{3 x}{5} \leq - 2 - 80 < 3 x \leq - 10 \Rightarrow \frac{- 80}{3} < x \leq \frac{- 10}{3}$

Thus, the solution set for the given inequality is $(\frac{- 80}{3}, \frac{- 10}{3}]$
6. $7 \leq \frac{( 3 x + 11 )}{2} \leq 11$

Sol. The given inequality is

$7 \leq \frac{( 3 x + 11 )}{2} \leq 11$

$\Rightarrow 14 \leq 3 x + 11 \leq 22$

$\Rightarrow 14 - 11 \leq 3 x \leq 22 - 11$

$\Rightarrow 3 \leq 3 x \leq 11$

$\Rightarrow 1 \leq x \leq \frac{11}{3}$

Thus, the solution set for the given inequality is $[1, \frac{11}{3}]$ .

Solve the inequalities in Q. 7 to 10 and represent the solution graphically on number line.

7. $5 x + 1 > - 24, 5 x - 1 < 24$

Sol. The given inequality is $5 x + 1 > - 24$ and

$\Rightarrow 5 x - 1 < 24 5 x + 1 > - 24 5 x > - 25 5 x - 1 < 24 \Rightarrow x > - 5$

$\Rightarrow 5 x < 25 \Rightarrow x < 5$

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

8. $2 (x - 1) < x + 5, 3 (x + 2) > 2 - x$

Sol. The given inequality is $2 (x - 1) < x + 5$ and

$\Rightarrow \Rightarrow 3 (x + 2) > 2 - x 2 (x - 1) < x + 5 2 x - 2 < x + 5 2 x - x < 5 + 2 \Rightarrow x < 7 3 (x + 2) > 2 - x$

$\Rightarrow 3 x + 6 > 2 - x \Rightarrow 3 x + x > 2 - 6 \Rightarrow 4 x > - 4 \Rightarrow x > - 1$

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

9. $3 x - 7 > 2 (x - 6)$ and $6 - x > 11 - 2 x$

Sol. The given inequality is $3 x - 7 > 2 (x - 6)$ and

$\Rightarrow \Rightarrow \Rightarrow 6 - x > 11 - 2 x 3 x - 7 > 2 (x - 6) 3 x - 7 > 2 x - 12 3 x - 2 x > - 12 + 7 \Rightarrow x > - 5 6 - x > 11 - 2 x - x + 2 x > 11 - 6 \Rightarrow x > 5$

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

10. $5 (2 x - 7) - 3 (2 x + 3) \leq 0, 2 x + 19 \leq 6 x + 47$

Sol. The given inequality is $5 (2 x - 7) - 3 (2 x + 3) \leq 0$ and $2 x + 19 \leq 6 x + 47$

$5 (2 x - 7) - 3 (2 x + 3) \leq 0$

$\Rightarrow 10 x - 35 - 6 x - 9 \leq 0$

$\Rightarrow 4 x - 44 \leq 0$

$\Rightarrow 4 x \leq 44$

$\Rightarrow x \leq 11$

$2 x + 19 \leq 6 x + 47$

$\Rightarrow 19 - 47 \leq 6 x - 2 x$

$\Rightarrow - 28 \leq 4 x$

$\Rightarrow - 7 \leq x$

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

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 $F = \frac{9}{8} C + 32$ ?

Sol. Since the solution is to be kept between 68°F and $7 7^{\circ} F, 68 < F < 77$ Putting $F = \frac{9}{8} C + 32$ , we obtain

$\Rightarrow 68 < \frac{9}{8} C + 32 < 77 \Rightarrow 68 - 32 < \frac{9}{5} C < 77 - 32 36 < \frac{9}{5} C < 45 \Rightarrow 36 \times \frac{5}{9} C < 45 \times \frac{5}{9}$

Thus, the required range of temperature in degree Celsius is between $2 0^{\circ} C$ and $2 5^{\circ} C$ .
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?

Sol. 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 $(x + 640)$
$\Rightarrow \frac{2}{100} x + \frac{8}{100} (640) > \frac{4}{100} (x + 640)$
$\Rightarrow 2 x + 5120 > 4 x + 2560$
$\Rightarrow 5120 - 2560 > 4 x - 2 x$
$\Rightarrow$ 5120-2560 > 2x
$\Rightarrow 2560 > 2 x$
$\Rightarrow 1280 > x$
$2% x + 8%$ of $640 < 6%$ of $(x + 640)$
$\frac{2}{100} x + \frac{8}{100} (640) < \frac{6}{100} (x + 640)$
$\Rightarrow 2 x + 5120 < 6 x + 3840$
$\Rightarrow$ 5120-3840<6x-2x
$\Rightarrow 1280 < 4 x \Rightarrow 320 < x$
$∴ 320 < x < 1280$
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.

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?

Sol. 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
$\Rightarrow \frac{30}{100} (1125 + x) > \frac{45}{100} \times 1125$ $\Rightarrow 30 (1125 + x) > 45 \times 1125$ $\Rightarrow 30 \times 1125 + 30 x > 45 \times 1125$ $\Rightarrow 30 x > 45 \times 1125 - 30 \times 1125$ $\Rightarrow 30 x > (45 - 30) \times 1125$ $\Rightarrow x > \frac{15 \times 1125}{30}$
$\Rightarrow x > 562.5$ $25%$ of $(1125 + x) < 45%$ of 1125
$\Rightarrow \frac{25}{100} (1125 + x) < \frac{45}{100} \times 1125$ $\Rightarrow 25 (1125 + x) < 45 \times 1125$ $\Rightarrow 25 \times 1125 + 25 x < 45 \times 1125$ $\Rightarrow 25 x < 45 \times 1125 - 25 \times 1125$ $\Rightarrow 25 x < (45 - 25) \times 1125$ $\Rightarrow x < \frac{20 \times 1125}{25}$
$\Rightarrow x < 900$ $∴ 562.5 < x < 900$

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.

14.IQ of a person is given by the formula $IQ = \frac{MA}{CA} \times 100$
Where MA is mental age and CA is chronological age. If $80 \leq IQ \leq 140$ for a group of 12 years old children, find the range of their mental age.

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

$80 \leq IQ \leq 140$

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

$IQ = \frac{MA}{12} \times 100$

Putting this value of IQ in (1), we obtain

$80 \leq \frac{MA}{12} \times 100 \leq 140$

$\Rightarrow 80 \times \frac{12}{100} \leq MA \leq 140 \times \frac{12}{100}$ $\Rightarrow 9.6 \leq MA \leq 16.8$
Thus, the range of mental age of the group of 12 years old children is $9.6 \leq MA \leq 16.8$

4.0Class 11 Maths NCERT Solutions – Chapter-wise Links

Get NCERT Solutions for Class 11 Maths with chapter-wise exercise answers, important formulas, and easy step-by-step explanations to solve questions confidently.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Sets
Chapter 2	Relations and Functions
Chapter 3	Trigonometric Functions
Chapter 4	Complex Numbers and Quadratic Equations
Chapter 6	Permutations and Combinations
Chapter 7	Binomial Theorem
Chapter 8	Sequence and Series
Chapter 9	Straight Lines
Chapter 10	Conic Sections
Chapter 11	Introduction to Three-dimensional Geometry
Chapter 12	Limits and Derivatives
Chapter 13	Statistics
Chapter 14	Probability

5.0Class 11 Maths Chapter 5 Linear Inequalities: Exercise-wise Questions and Topics

Exercise	Number of Questions	Important Topics Covered
Exercise 5.1	26 Questions with Solutions	Linear inequalities, solution sets, number line representation, and graphical solutions
Miscellaneous Exercise	14 Questions with Solutions	Absolute value inequalities, interval method, and word problems based on linear inequalities

6.0Key Features of NCERT Solutions for Class 11 Maths Chapter 5 (Linear Inequalities)

Clear Inequality Rules: The solutions explain when and why the inequality sign changes, especially when multiplying or dividing by a negative number. This helps students avoid common mistakes.
Easy Interval Notation: Students learn how to write solution sets using parentheses ( ) for open intervals and brackets [ ] for closed intervals.
Step-by-Step Graphical Solutions: For inequalities in two variables, the solutions explain how to choose test points and identify the correct region on the graph.
Simple Word Problem Solutions: Real-life problems involving mixtures, marks, and other conditions are converted into mathematical inequalities through clear and easy steps.
Prepared by ALLEN Subject Experts: The solutions are prepared by ALLEN subject experts with accurate methods and alignment with the latest NCERT syllabus.
Clear Graphs and Methods: Graph-based questions include easy-to-follow diagrams that help students understand solution regions and common solutions.
Complete NCERT Exercise Coverage: All NCERT exercises, including the miscellaneous exercise, are covered with detailed solutions to help students solve different types of questions.

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Frequently Asked Questions

Why are NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities important for students?

How do Class 11 Maths NCERT Solutions for Chapter 5 help in board, JEE preparation?

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Are NCERT Solutions for Class 11 Maths Chapter 5 by ALLEN useful for exam preparation?

Why are linear inequalities important in mathematics?

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NCERT Solutions Class 11 Maths Chapter 5 Linear Inequalities: Download PDF

1.0Class 11 Maths Chapter 5 Linear Inequalities: Key Concepts

2.0Detailed NCERT Textbook Solutions : Class 11 Maths Chapter 5

3.0MISCELLANEOUS EXERCISE

4.0Class 11 Maths NCERT Solutions – Chapter-wise Links

5.0Class 11 Maths Chapter 5 Linear Inequalities: Exercise-wise Questions and Topics

6.0Key Features of NCERT Solutions for Class 11 Maths Chapter 5 (Linear Inequalities)

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