Anand Classes presents accurate and detailed Linear Inequalities NCERT Solutions Exercise 5.1 Class 11 Maths Chapter 5 to help students understand and solve inequality problems with confidence. These step-by-step solutions are prepared according to the latest NCERT and CBSE syllabus, ensuring conceptual clarity and effective exam preparation. Perfect for Class 11 students, these solutions simplify complex topics like linear inequalities in one and two variables, making learning easy and efficient. Click the print button to download study material and notes.
NCERT Question 1. Solve $24x < 100$, when
(i) $x$ is a natural number
(ii) $x$ is an integer
Solution :
(i) When $x$ is a natural number
Given:
$$24x < 100$$
Dividing both sides by $24$:
$$x < \dfrac{25}{6}$$
Since $x$ is a natural number $\{1, 2, 3, 4, 5, \ldots\}$,
The natural numbers less than $\dfrac{25}{6}$ are:
$$x = \{1, 2, 3, 4\}$$
Hence, the solution set is:
$$\{1, 2, 3, 4\}$$
(ii) When $x$ is an integer
Again:
$$24x < 100 \Rightarrow x < \frac{25}{6}$$
As $x$ is an integer, the integers less than $\dfrac{25}{6}$ are:
$$\{\ldots, -2, -1, 0, 1, 2, 3, 4\}$$
Hence, the solution set is:
$$\{\ldots, -2, -1, 0, 1, 2, 3, 4\}$$
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NCERT Question 2. Solve $-12x > 30$, when
(i) $x$ is a natural number
(ii) $x$ is an integer
Solution :
(i) When $x$ is a natural number
Given:
$$-12x > 30$$
Dividing both sides by $-12$
(Note: inequality sign reverses when dividing by a negative number)
$$x < -\dfrac{5}{2}$$
Since natural numbers are positive:
$$\mathbb{N} = \{1, 2, 3, 4, \ldots\}$$
There is no natural number less than $-\dfrac{5}{2}$.
Therefore, the solution set is: $varnothing$
(ii) When $x$ is an integer
Again:
$$-12x > 30 \Rightarrow x < -\dfrac{5}{2}$$
The integers less than $-\dfrac{5}{2}$ are:
$$\{\ldots, -6, -5, -4, -3\}$$
Hence, the solution set is:
$$\{\ldots, -6, -5, -4, -3\}$$
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NCERT Question 3. Solve $5x – 3 < 7$, when
(i) $x$ is an integer
(ii) $x$ is a real number
Solution :
(i) When $x$ is an integer
Given:
$$5x – 3 < 7$$
Add $3$ to both sides:
$$5x < 10$$
Divide both sides by $5$:
$$x < 2$$
The integer values less than $2$ are:
$$\{\ldots, -3, -2, -1, 0, 1\}$$
Therefore, the solution set is:
$$\{\ldots, -3, -2, -1, 0, 1\}$$
(ii) When $x$ is a real number
Same steps:
$$5x – 3 < 7$$
$$5x < 10$$
$$x < 2$$
Thus the solution set in interval form is:
$$(-\infty, 2)$$
All real numbers less than $2$ satisfy the inequality.
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NCERT Question 4. Solve $3x + 8 > 2$, when
(i) $x$ is an integer
(ii) $x$ is a real number
Solution :
(i) When $x$ is an integer
Given:
$$3x + 8 > 2$$
Subtract $8$ from both sides:
$$3x > -6$$
Divide by $3$:
$$x > -2$$
The integers greater than $-2$ are:
$$\{-1, 0, 1, 2, 3, \ldots\}$$
Therefore, the solution set is:
$$\{-1, 0, 1, 2, 3, \ldots\}$$
(ii) When $x$ is a real number
Again from the inequality:
$$3x + 8 > 2 \Rightarrow 3x > -6 \Rightarrow x > -2$$
So the solution set in interval notation is:
$$(-2, \infty)$$
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NCERT Question 5. Solve $4x + 3 < 5x + 7$ for real $x$
Solution :
Given:
$$4x + 3 < 5x + 7$$
Subtract $7$ from both sides:
$$4x + 3 – 7 < 5x + 7 – 7$$
$$4x – 4 < 5x$$
Subtract $4x$ from both sides:
$$4x – 4 – 4x < 5x – 4x$$
$$-4 < x$$
Rewriting:
$$x > -4$$
Thus, the solution includes all real numbers greater than $-4$.
Required solution set:
$$(-4, \infty)$$
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Question 6. Solve $3x – 7 > 5x – 1$ for real $x$.
Solution :
Given:
$$3x – 7 > 5x – 1$$
Add $7$ to both sides:
$$3x – 7 + 7 > 5x – 1 + 7$$
$$3x > 5x + 6$$
Subtract $5x$ from both sides:
$$3x – 5x > 5x + 6 – 5x$$
$$-2x > 6$$
Divide both sides by $-2$
(Note: the inequality sign reverses when dividing by a negative number)
$$x < -3$$
Thus, the solutions are all real numbers less than $-3$.
Required solution set:
$$(-\infty, -3)$$
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Question 7. Solve $3(x – 1) \le 2(x – 3)$ for real $x$.
Solution :
Given:
$$3(x – 1) \le 2(x – 3)$$
Expand both sides:
$$3x – 3 \le 2x – 6$$
Add $3$ to both sides:
$$3x – 3 + 3 \le 2x – 6 + 3$$
$$3x \le 2x – 3$$
Subtract $2x$ from both sides:
$$3x – 2x \le 2x – 3 – 2x$$
$$x \le -3$$
Therefore, the solution includes all real numbers less than or equal to $-3$.
Required solution set:
$$(-\infty, -3]$$
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Question 8. Solve $3(2 – x) \ge 2(1 – x)$ for real $x$.
Solution :
Given:
$$3(2 – x) \ge 2(1 – x)$$
Expand:
$$6 – 3x \ge 2 – 2x$$
Add $2x$ to both sides:
$$6 – 3x + 2x \ge 2 – 2x + 2x$$
$$6 – x \ge 2$$
Subtract $6$ from both sides:
$$6 – x – 6 \ge 2 – 6$$
$$-x \ge -4$$
Multiply by $-1$ on both sides
(Note: the inequality sign reverses):
$$x \le 4$$
Therefore, the solution includes all real numbers less than or equal to 4.
Required solution set:
$$(-\infty, 4]$$
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Question 9. Solve : $x + \dfrac{x}{2} + \dfrac{x}{3} < 11$ for real $x$.
Solution :
Given:
$$x + \dfrac{x}{2} + \dfrac{x}{3} < 11$$
Take LCM of $1, 2, 3$ which is $6$:
$$\frac{6x + 3x + 2x}{6} < 11$$
$$\frac{11x}{6} < 11$$
Multiply both sides by $6$:
$$11x < 66$$
Divide both sides by $11$:
$$x < 6$$
Thus, the solution includes all real numbers less than $6$.
Required solution set:
$$(-\infty, 6)$$
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Question 10. Solve $\dfrac{x}{3} > \dfrac{x}{2} + 1$ for real $x$.
Solution :
Given:
$$\frac{x}{3} > \frac{x}{2} + 1$$
Move all $x$-terms to the left-hand side:
$$\frac{x}{3} – \frac{x}{2} > 1$$
Take LCM $6$:
$$\frac{2x – 3x}{6} > 1$$
$$\frac{-x}{6} > 1$$
Multiply both sides by $6$:
$$-x > 6$$
Multiply both sides by $-1$
(Note: inequality sign reverses):
$$x < -6$$
Thus, the solution includes all real numbers less than $-6$.
Required solution set:
$$(-\infty, -6)$$
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