Last Updated on November 24, 2025 by Aman Singh

Get detailed, step-by-step solutions for NCERT Class 11 Maths Chapter 5 Exercise 5.1 . Learn to solve linear inequalities for $\mathbf{x}$ as a natural number, integer, or real number (Q.1-4). Practice complex algebraic inequalities (Q.5-16), including those involving fractions and graphing solutions on the number line (Q.17-20). Also includes solving application problems involving averages, consecutive integers, and geometric constraints (Q.21-26).

Ex. 5.1

Ex. 5.Misc.

This exercise involves solving linear inequalities and applying them to various word problems.

Rbse Solutions for Class 11 maths Chapter 4 Miscellaneous | Complex Numbers and Quadratic Equations

Rbse Solutions for Class 11 maths Chapter 5 Exercise 5.1 | Linear Inequalities

Solving Simple Inequalities (Exercises 1-4)

1. Solve $24x < 100$

Divide by 24: $x < \frac{100}{24} \implies x < \frac{25}{6} \implies x < 4.166…$

(i) $x$ is a natural number.

Natural numbers are $\{1, 2, 3, \dots\}$.

$$\mathbf{\text{Solution: } \{1, 2, 3, 4\}}$$

(ii) $x$ is an integer.

Integers are $\{\dots, -2, -1, 0, 1, 2, 3, \dots\}$.

$$\mathbf{\text{Solution: } \{\dots, -2, -1, 0, 1, 2, 3, 4\}}$$

2. Solve $-12x > 30$

Divide by $-12$ and reverse the inequality sign: $x < \frac{30}{-12} \implies x < -\frac{5}{2} \implies x < -2.5$.

(i) $x$ is a natural number.

Since natural numbers are positive, there is no natural number less than $-2.5$.

$$\mathbf{\text{Solution: } \emptyset \text{ (No solution)}}$$

(ii) $x$ is an integer.

$$\mathbf{\text{Solution: } \{\dots, -6, -5, -4, -3\}}$$

3. Solve $5x – 3 < 7$

Add 3: $5x < 7 + 3 \implies 5x < 10$.

Divide by 5: $x < 2$.

(i) $x$ is an integer.

$$\mathbf{\text{Solution: } \{\dots, -1, 0, 1\}}$$

(ii) $x$ is a real number.

$$\mathbf{\text{Solution: } (-\infty, 2)}$$

4. Solve $3x + 8 > 2$

Subtract 8: $3x > 2 – 8 \implies 3x > -6$.

Divide by 3: $x > -2$.

(i) $x$ is an integer.

$$\mathbf{\text{Solution: } \{-1, 0, 1, 2, \dots\}}$$

(ii) $x$ is a real number.

$$\mathbf{\text{Solution: } (-2, \infty)}$$

Solving Inequalities for Real $x$ (Exercises 5-16)

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

Subtract $4x$: $3 < 5x – 4x + 7 \implies 3 < x + 7$.

Subtract 7: $3 – 7 < x \implies -4 < x$.

$$\mathbf{\text{Solution: } (-4, \infty)}$$

6. $3x – 7 > 5x – 1$

Subtract $3x$: $-7 > 5x – 3x – 1 \implies -7 > 2x – 1$.

Add 1: $-7 + 1 > 2x \implies -6 > 2x$.

Divide by 2: $-3 > x$ or $x < -3$.

$$\mathbf{\text{Solution: } (-\infty, -3)}$$

7. $3(x – 1) \le 2 (x – 3)$

Expand: $3x – 3 \le 2x – 6$.

Subtract $2x$: $3x – 2x – 3 \le -6 \implies x – 3 \le -6$.

Add 3: $x \le -6 + 3 \implies x \le -3$.

$$\mathbf{\text{Solution: } (-\infty, -3]}$$

8. $3 (2 – x) \ge 2 (1 – x)$

Expand: $6 – 3x \ge 2 – 2x$.

Add $3x$: $6 \ge 2 – 2x + 3x \implies 6 \ge 2 + x$.

Subtract 2: $6 – 2 \ge x \implies 4 \ge x$ or $x \le 4$.

$$\mathbf{\text{Solution: } (-\infty, 4]}$$

9. $x + \frac{x}{2} + \frac{x}{3} < 11$

Find a common denominator (6):

$$\frac{6x}{6} + \frac{3x}{6} + \frac{2x}{6} < 11$$

$$\frac{11x}{6} < 11$$

Multiply by $\frac{6}{11}$: $x < 11 \cdot \frac{6}{11} \implies x < 6$.

$$\mathbf{\text{Solution: } (-\infty, 6)}$$

10. $\frac{x}{3} > \frac{x}{2} + 1$

Subtract $\frac{x}{2}$: $\frac{x}{3} – \frac{x}{2} > 1$.

Find a common denominator (6): $\frac{2x – 3x}{6} > 1 \implies \frac{-x}{6} > 1$.

Multiply by $-6$ and reverse the inequality sign: $-x < 6 \implies x < -6$.

$$\mathbf{\text{Solution: } (-\infty, -6)}$$

11. $\frac{3(x – 2)}{5} \le \frac{5(2 – x)}{3}$

Multiply by the LCM of 5 and 3 (which is 15):

$$15 \cdot \frac{3(x – 2)}{5} \le 15 \cdot \frac{5(2 – x)}{3}$$

$$3 \cdot 3(x – 2) \le 5 \cdot 5(2 – x)$$

$$9(x – 2) \le 25(2 – x)$$

Expand: $9x – 18 \le 50 – 25x$.

Add $25x$: $9x + 25x – 18 \le 50 \implies 34x – 18 \le 50$.

Add 18: $34x \le 50 + 18 \implies 34x \le 68$.

Divide by 34: $x \le 2$.

$$\mathbf{\text{Solution: } (-\infty, 2]}$$

12. $\frac{1}{2} \left(\frac{3x}{5} + 4\right) \ge \frac{1}{3} (x – 6)$

Multiply by the LCM of 2 and 3 (which is 6):

$$6 \cdot \frac{1}{2} \left(\frac{3x}{5} + 4\right) \ge 6 \cdot \frac{1}{3} (x – 6)$$

$$3 \left(\frac{3x}{5} + 4\right) \ge 2 (x – 6)$$

$$3 \cdot \frac{3x}{5} + 12 \ge 2x – 12$$

$$\frac{9x}{5} + 12 \ge 2x – 12$$

Subtract $2x$: $\frac{9x}{5} – 2x + 12 \ge -12$.

$\frac{9x – 10x}{5} + 12 \ge -12 \implies -\frac{x}{5} + 12 \ge -12$.

Subtract 12: $-\frac{x}{5} \ge -12 – 12 \implies -\frac{x}{5} \ge -24$.

Multiply by $-5$ and reverse the inequality sign: $x \le (-24) \cdot (-5) \implies x \le 120$.

$$\mathbf{\text{Solution: } (-\infty, 120]}$$

13. $2 (2x + 3) – 10 < 6 (x – 2)$

Expand: $4x + 6 – 10 < 6x – 12 \implies 4x – 4 < 6x – 12$.

Subtract $4x$: $-4 < 6x – 4x – 12 \implies -4 < 2x – 12$.

Add 12: $-4 + 12 < 2x \implies 8 < 2x$.

Divide by 2: $4 < x$ or $x > 4$.

$$\mathbf{\text{Solution: } (4, \infty)}$$

14. $37 – (3x + 5) > 9x – 8 (x – 3)$

Expand: $37 – 3x – 5 > 9x – 8x + 24$.

Simplify both sides: $32 – 3x > x + 24$.

Add $3x$: $32 > x + 3x + 24 \implies 32 > 4x + 24$.

Subtract 24: $32 – 24 > 4x \implies 8 > 4x$.

Divide by 4: $2 > x$ or $x < 2$.

$$\mathbf{\text{Solution: } (-\infty, 2)}$$

15. $\frac{x}{4} < \frac{5x – 2}{3} – \frac{7x – 3}{5}$

Multiply by the LCM of 4, 3, and 5 (which is 60):

$$60 \cdot \frac{x}{4} < 60 \cdot \frac{5x – 2}{3} – 60 \cdot \frac{7x – 3}{5}$$

$$15x < 20(5x – 2) – 12(7x – 3)$$

$$15x < (100x – 40) – (84x – 36)$$

$$15x < 100x – 40 – 84x + 36$$

$$15x < (100x – 84x) + (-40 + 36)$$

$$15x < 16x – 4$$

Subtract $16x$: $15x – 16x < -4 \implies -x < -4$.

Multiply by $-1$ and reverse the inequality sign: $x > 4$.

$$\mathbf{\text{Solution: } (4, \infty)}$$

16. $\frac{2x – 1}{3} \ge \frac{3x – 2}{4} – \frac{2 – x}{5}$

Multiply by the LCM of 3, 4, and 5 (which is 60):

$$60 \cdot \frac{2x – 1}{3} \ge 60 \cdot \frac{3x – 2}{4} – 60 \cdot \frac{2 – x}{5}$$

$$20(2x – 1) \ge 15(3x – 2) – 12(2 – x)$$

Expand: $40x – 20 \ge (45x – 30) – (24 – 12x)$.

$$40x – 20 \ge 45x – 30 – 24 + 12x$$

$$40x – 20 \ge (45x + 12x) + (-30 – 24)$$

$$40x – 20 \ge 57x – 54$$

Subtract $40x$: $-20 \ge 57x – 40x – 54 \implies -20 \ge 17x – 54$.

Add 54: $-20 + 54 \ge 17x \implies 34 \ge 17x$.

Divide by 17: $2 \ge x$ or $x \le 2$.

$$\mathbf{\text{Solution: } (-\infty, 2]}$$

Inequalities and Graphing on Number Line (Exercises 17-20)

17. $3x – 2 < 2x + 1$

Subtract $2x$: $x – 2 < 1$.

Add 2: $x < 3$.

$$\mathbf{\text{Solution: } (-\infty, 3)}$$

Graph: An open circle at 3, with shading to the left.

18. $5x – 3 > 3x – 5$

Subtract $3x$: $2x – 3 > -5$.

Add 3: $2x > -5 + 3 \implies 2x > -2$.

Divide by 2: $x > -1$.

$$\mathbf{\text{Solution: } (-1, \infty)}$$

Graph: An open circle at $-1$, with shading to the right.

19. $3 (1 – x) < 2 (x + 4)$

Expand: $3 – 3x < 2x + 8$.

Add $3x$: $3 < 2x + 3x + 8 \implies 3 < 5x + 8$.

Subtract 8: $3 – 8 < 5x \implies -5 < 5x$.

Divide by 5: $-1 < x$ or $x > -1$.

$$\mathbf{\text{Solution: } (-1, \infty)}$$

Graph: An open circle at $-1$, with shading to the right.

20. $\frac{x}{2} \ge \frac{5x – 2}{3} – \frac{7x – 3}{5}$

(Note: The question provided is $\frac{x}{2} \ge \frac{5x – 2}{3} – \frac{7x – 3}{5}$. Let’s solve the inequality as written in the text: $\frac{5x – 2}{3} – \frac{7x – 3}{5} \ge \frac{x}{2}$)

Multiply by the LCM of 3, 5, and 2 (which is 30):

$$30 \cdot \frac{5x – 2}{3} – 30 \cdot \frac{7x – 3}{5} \ge 30 \cdot \frac{x}{2}$$

$$10(5x – 2) – 6(7x – 3) \ge 15x$$

$$50x – 20 – 42x + 18 \ge 15x$$

$$(50x – 42x) + (-20 + 18) \ge 15x$$

$$8x – 2 \ge 15x$$

Subtract $8x$: $-2 \ge 15x – 8x \implies -2 \ge 7x$.

Divide by 7: $-\frac{2}{7} \ge x$ or $x \le -\frac{2}{7}$.

$$\mathbf{\text{Solution: } \left(-\infty, -\frac{2}{7}\right]}$$

Graph: A closed circle at $-\frac{2}{7}$, with shading to the left.

Word Problems (Exercises 21-26)

21. Minimum marks for an average of at least 60.

Let $x$ be the marks in the third test. The average of three tests must be $\ge 60$.

$$\frac{70 + 75 + x}{3} \ge 60$$

$$145 + x \ge 60 \times 3$$

$$145 + x \ge 180$$

$$x \ge 180 – 145$$

$$x \ge 35$$

The minimum marks Ravi should get is 35.

22. Minimum marks for Grade ‘A’ (average $\ge 90$).

Let $x$ be the marks in the fifth examination. The average of five tests must be $\ge 90$.

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

$$\frac{368 + x}{5} \ge 90$$

$$368 + x \ge 90 \times 5$$

$$368 + x \ge 450$$

$$x \ge 450 – 368$$

$$x \ge 82$$

The minimum marks Sunita must obtain is 82.

23. Pairs of consecutive odd positive integers $< 10$ and sum $> 11$.

Let the two consecutive odd positive integers be $x$ and $x + 2$.

Both are smaller than 10: $x + 2 < 10 \implies x < 8$.
$x$ is a positive odd integer: $x \in \{1, 3, 5, 7, 9, \dots\}$
Their sum is more than 11: $x + (x + 2) > 11 \implies 2x + 2 > 11 \implies 2x > 9 \implies x > 4.5$.

Combining the conditions: $4.5 < x < 8$, where $x$ is an odd positive integer.

The possible values for $x$ are $5$ and $7$.

If $x=5$, the pair is $(5, 7)$. Sum $5+7=12 > 11$.
If $x=7$, the pair is $(7, 9)$. Sum $7+9=16 > 11$.$$\mathbf{\text{Pairs: } (5, 7) \text{ and } (7, 9)}$$

24. Pairs of consecutive even positive integers $> 5$ and sum $< 23$.

Let the two consecutive even positive integers be $x$ and $x + 2$.

Both are larger than 5: $x > 5$.
$x$ is an even positive integer: $x \in \{2, 4, 6, 8, 10, \dots\}$. Combining with $x>5$: $x \in \{6, 8, 10, \dots\}$.
Their sum is less than 23: $x + (x + 2) < 23 \implies 2x + 2 < 23 \implies 2x < 21 \implies x < 10.5$.

Combining the conditions: $x \in \{6, 8, 10, \dots\}$ and $x < 10.5$.

The possible values for $x$ are $6$, $8$, and $10$.

If $x=6$, the pair is $(6, 8)$. Sum $6+8=14 < 23$.
If $x=8$, the pair is $(8, 10)$. Sum $8+10=18 < 23$.
If $x=10$, the pair is $(10, 12)$. Sum $10+12=22 < 23$.$$\mathbf{\text{Pairs: } (6, 8), (8, 10), \text{ and } (10, 12)}$$

25. Minimum length of the shortest side of a triangle.

Let $x$ be the length of the shortest side.

Shortest side: $x$
Longest side: $3x$
Third side: $3x – 2$

The perimeter is at least 61 cm: $x + 3x + (3x – 2) \ge 61$.

$$7x – 2 \ge 61$$

$$7x \ge 63$$

$$x \ge 9$$

Also, side lengths must be positive, which is satisfied if $x \ge 9$.

And the Triangle Inequality must be satisfied: the sum of any two sides must be greater than the third side. The most restrictive is:

Shortest side + Third side $>$ Longest side

$$x + (3x – 2) > 3x$$

$$4x – 2 > 3x$$

$$x > 2$$

Since $x \ge 9$ already satisfies $x > 2$, the minimum length of the shortest side is 9 cm.

26. Possible lengths of the shortest board.

Let $x$ be the length of the shortest board.

Shortest length: $x$
Second length: $x + 3$
Third length: $2x$

Total length constraint: The sum of the three lengths cannot exceed the total board length (91 cm).$$x + (x + 3) + 2x \le 91$$$$4x + 3 \le 91$$$$4x \le 88$$$$x \le 22 \quad (*1)$$
Third piece constraint: The third piece must be at least 5 cm longer than the second piece.$$2x \ge (x + 3) + 5$$$$2x \ge x + 8$$$$x \ge 8 \quad (*2)$$
Physical constraint: Lengths must be positive.$$x > 0$$

Combining (*1) and (*2):

$$8 \le x \le 22$$

The possible lengths of the shortest board are between 8 cm and 22 cm, inclusive.

Ex. 5.1

Ex. 5 Misc.

Author

Aman Singh

Aman Singh | M.Sc. Mathematics, RRBMU University Alwar

A seasoned Mathematics Educator with 7 years of dedicated experience in the field of education. Specializing in simplifying complex mathematical concepts, Aman has a proven track record of helping students master advanced topics. Holds an M.Sc. in Mathematics from RRBMU University, Alwar. Passionate about leveraging conceptual clarity and effective teaching methodologies to drive student success and achievement.

"Transforming mathematical complexity into conceptual clarity."

For the past 7 years, Aman Singh has been on a mission to redefine math education. Armed with an M.Sc. in Mathematics from RRBMU University Alwar, Aman brings a deep well of knowledge and seven years of classroom insight to every lesson. Specializing in turning student struggle into genuine mastery, Aman believes math isn't just about numbers—it's about building confidence and problem-solving muscle.