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).
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
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.