Inequalities

(i) 2x > 5

We have :

2x > 5

⇒ x > $\frac{5}{2} \quad \text{[Dividing both sides by 2]}$

⇒ x > 2.5

From the set, values greater than 2.5 are {3, 4, 5}.

∴ Solution set = {3, 4, 5}

(ii) 3x - 8 < 1

We have:

3x - 8 < 1

⇒ 3x < 1 + 8 $\quad$ [Adding 8 on both sides]

⇒ 3x < 9

⇒ x < $\frac{9}{3} \quad \text{[Dividing both sides by 3]}$

⇒ x < 3

From the set, values less than 3 are {-2, -1, 0, 1, 2}.

∴ Solution set = {-2, -1, 0, 1, 2}

(iii) 3 - 12x > -21

We have:

3 - 12x > -21

⇒ -12x > -21 - 3 $\quad$ [Subtracting 3 from both sides]

⇒ -12x > -24

Dividing by a negative number reverses the sign:

⇒ x < $\frac{-24}{-12} \quad \text{[Dividing both sides by -12]}$

⇒ x < 2

From the set, values less than 2 are {-2, -1, 0, 1}.

∴ Solution set = {-2, -1, 0, 1}

(iv) 7 - x > 0

We have:

7 - x > 0

⇒ 7 > x

⇒ x < 7

All values in the set are less than 7.

∴ Solution set = {-2, -1, 0, 1, 2, 3, 4, 5}

(v) 3 - 4x > -2

We have:

3 - 4x > -2

⇒ -4x > -2 - 3 $\quad$ [Subtracting 3 from both sides]

⇒ -4x > -5

Dividing by a negative number reverses the sign:

⇒ x < $\frac{-5}{-4} \quad \text{[Dividing both sides by -4]}$

⇒ x < 1.25

From the set, values less than 1.25 are {-2, -1, 0, 1}.

∴ Solution set = {-2, -1, 0, 1}

(vi) 3x + 4 < 15

We have:

3x + 4 < 15

⇒ 3x < 15 - 4 $\quad$ [Subtracting 4 from both sides]

⇒ 3x < 11

⇒ x < $\frac{11}{3} \quad \text{[Dividing both sides by 3]}$

⇒ x < 3.66...

From the set, values less than 3.66... are {-2, -1, 0, 1, 2, 3}.

∴ Solution set = {-2, -1, 0, 1, 2, 3}

(vii) $\frac{3}{4}x$ > - 1

We have:

$\frac{3}{4}x$ > - 1

⇒ 3x > -1 x 4 $\quad$ [Multiplying 4 on both sides]

⇒ 3x > -4

⇒ x > $\frac{-4}{3} \quad \text{[Dividing both sides by 3]}$

⇒ x > -1.33...

From the set, values greater than -1.33... are {-1, 0, 1, 2, 3, 4, 5}.

∴ Solution set = {-1, 0, 1, 2, 3, 4, 5}

(viii) $\frac{2}{3} + x$ < - $\frac{1}{6}$

We have:

$$\begin{aligned}\phantom{=} \frac{2}{3} + x \lt - \frac{1}{6} \\[1em] \text{Subtract} \frac{2}{3} \text{ from both sides} \\[1em] \Rightarrow x \lt - \frac{1}{6} - \frac{2}{3} \\[1em] \Rightarrow x \lt \frac{-1 - 4}{6} \\[1em] \Rightarrow x \lt \frac{-5}{6} \\[1em] \Rightarrow x \lt -0.833...\end{aligned}$$

From the set, values less than -0.833 are {-2 , -1}.

∴ Solution set = {-2, -1}

(ix) $\frac{7}{4} - 3x$ < $\frac{5}{6}$

We have:

$$\begin{aligned}\phantom{=} \frac{7}{4} - 3x \lt \frac{5}{6} \\[1em] \text{Subtract } \frac{7}{4} \text{ from both sides} \\[1em] \Rightarrow -3x \lt \frac{5}{6} - \frac{7}{4} \\[1em] \Rightarrow -3x \lt \frac{5}{6} - \frac{7}{4} \\[1em] \Rightarrow -3x \lt \frac{10 - 21}{12} \\[1em] \Rightarrow -3x \lt \frac{-11}{12} \\[1em] \text{Divide both sides by -3 and reverse the sign} \\[1em] \Rightarrow x \gt \frac{-11}{12 \times (-3)} \\[1em] \Rightarrow x \gt \frac{-11}{-36} \\[1em] \Rightarrow x \gt 0.305...\end{aligned}$$

From the set, values greater than 0.305... are {1, 2, 3, 4, 5}.

∴ Solution set = {1, 2, 3, 4, 5}

(i) 4x < 13

We have:

⇒ 4x < 13

⇒ x < $\frac{13}{4} \quad \text{[Dividing both sides by 4]}$

⇒ x < 3.25

Natural numbers less than 3.25 are {1, 2, 3}.

∴ Solution set = {1, 2, 3}

(ii) 2x - 9 < -1

We have:

2x - 9 < -1

⇒ 2x < -1 + 9 $\quad$ [Adding 9 on both sides]

⇒ 2x < 8

⇒ x < $\frac{8}{2} \quad \text{[Dividing both sides by 2]}$

⇒ x < 4

Natural numbers less than 4 are {1, 2, 3}.

∴ Solution set = {1, 2, 3}

(iii) 3 - x < -2

We have:

3 - x < -2

⇒ -x < -2 - 3 $\quad$ [Subtracting 3 from both sides]

⇒ -x < -5

Multiply -1 on both sides and reverse the sign

⇒ x > 5

Natural numbers greater than 5 are {6, 7, 8, 9, ...}

∴ Solution set = {6, 7, 8, 9, ...}

(iv) 5 - 7x > - 16

We have:

5 - 7x > - 16

⇒ -7x > -16 - 5 $\quad$ [Subtracting 5 from both sides]

⇒ -7x > -21

Dividing by a negative number reverses the inequality:

⇒ x < $\frac{-21}{-7} \quad \text{[Dividing both sides by -7]}$

⇒ x < 3

Natural numbers less than 3 are {1, 2}

∴ Solution set = {1, 2}

(v) $\frac{4}{7}-\frac{x}{4}$ > - 2

We have:

$$\begin{aligned}\phantom{=} \frac{4}{7}-\frac{x}{4} \gt - 2 \\[1em] \text{Subtract } \frac{4}{7} \text{ from both sides} \\[1em] \Rightarrow -\frac{x}{4} \gt -2 - \frac{4}{7} \\[1em] \Rightarrow -\frac{x}{4} \gt \frac{-14 - 4}{7} \\[1em] \Rightarrow -\frac{x}{4} \gt \frac{-18}{7} \\[1em] \text{Multiply 4 on both sides} \\[1em] \Rightarrow -x \gt \frac{-18}{7} \times 4 \\[1em] \Rightarrow -x \gt \frac{-72}{7} \\[1em] \Rightarrow -x \gt -10.28... \\[1em] \text{[Multiply -1 on both sides and reverse the sign]} \\[1em] \Rightarrow x \lt 10.28...\end{aligned}$$

Natural numbers less than 10.28 are {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}.

∴ Solution set = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vi) $-\frac{1}{2}$ > $\frac{1}{4} - \frac{x}{3}$

We have:

$$\begin{aligned}\phantom{=} -\frac{1}{2} \gt \frac{1}{4} - \frac{x}{3} \\[1em] \text{Subtract } \frac{1}{4} \text{ from both sides} \\[1em] \Rightarrow -\frac{1}{2} - \frac{1}{4} \gt - \frac{x}{3} \\[1em] \Rightarrow \frac{-2 - 1}{4} \gt - \frac{x}{3} \\[1em] \Rightarrow \frac{-3}{4} \gt - \frac{x}{3} \\[1em] \text{Multiplying 3 on both sides} \\[1em] \Rightarrow \frac{-3}{4} \times 3 \gt -x \\[1em] \Rightarrow \frac{-9}{4} \gt -x \\[1em] \Rightarrow -2.25 \gt -x \\[1em] \text{Multiply -1 on both sides and reverse the sign} \\[1em] \Rightarrow 2.25 \lt x \\[1em]\end{aligned}$$

Natural numbers greater than 2.25 are {3, 4, 5, ...}

∴ Solution set = {3, 4, 5, ...}

(i) 7x < 17

We have:

7x < 17

⇒ x < $\frac{17}{7} \quad \text{[Dividing both sides by 7]}$

⇒ x < 2.42...

Positive integers less than 2.42... are {1, 2}

∴ Solution set = {1, 2}

(ii) 4x - 11 < 5

We have:

4x - 11 < 5

⇒ 4x < 5 + 11 $\quad$ [Adding 11 on both sides]

⇒ 4x < 16

⇒ x < $\frac{16}{4} \quad \text{[Dividing both sides by 4]}$

⇒ x < 4

Positive integers less than 4 are {1, 2, 3}

∴ Solution set = {1, 2, 3}

(iii) 8 - x > $\frac{1}{3}$

We have:

$$\begin{aligned}\phantom{=} 8 - x \gt \frac{1}{3} \\[1em] \text{Subtract 8 from both sides} \\[1em] \Rightarrow -x \gt \frac{1}{3} - 8 \\[1em] \Rightarrow -x \gt \frac{1 - 24}{3} \\[1em] \Rightarrow -x \gt \frac{-23}{3} \\[1em] \Rightarrow -x \gt -7.66... \\[1em] \text{Multiply -1 on both sides and reverse the sign} \\[1em] \Rightarrow x \lt 7.66...\end{aligned}$$

Positive integers less than 7.66.. are {1, 2, 3, 4, 5, 6, 7}

∴ Solution set = {1, 2, 3, 4, 5, 6, 7}

(iv) 4(x + 5) < 29

We have:

4(x + 5) < 29

⇒ 4x + 20 < 29

⇒ 4x < 29 - 20 $\quad$ [Subtracting 20 from both sides]

⇒ 4x < 9

⇒ x < $\frac{9}{4} \quad \text{[Dividing both sides by 4]}$

⇒ x < 2.25

Positive integers less than 2.25 are {1, 2}

∴ Solution set = {1, 2}

(v) 5 > $\frac{2}{3}$x

We have:

5 > $\frac{2}{3}$x

⇒ 5 x 3 > 2x $\quad$ [Multiplying 3 on both sides]

⇒ 15 > 2x

⇒ $\frac{15}{2}$ > x $\quad$ [Dividing both sides by 2]

⇒ 7.5 > x

⇒ x < 7.5

Positive integers less than 7.5 are {1, 2, 3, 4, 5, 6, 7}

∴ Solution set = {1, 2, 3, 4, 5, 6, 7}

(vi) 2 - $\frac{7x}{29}$ < $\frac{5}{3}$

We have:

$$\begin{aligned}\phantom{=} 2 - \frac{7x}{29} \lt \frac{5}{3} \\[1em] \text{Subtracting 2 from both sides} \\[1em] \Rightarrow -\frac{7x}{29} \lt \frac{5}{3} - 2 \\[1em] \Rightarrow -\frac{7x}{29} \lt \frac{5 - 6}{3} \\[1em] \Rightarrow -\frac{7x}{29} \lt \frac{-1}{3} \\[1em] \text{Multiplying} \frac{29}{7} \text{ on both sides} \\[1em] \Rightarrow -x \lt \frac{-1}{3} \times \frac{29}{7} \\[1em] \Rightarrow -x \lt \frac{-29}{21} \\[1em] \Rightarrow -x \lt - 1.38... \\[1em] \text{Multiplying -1 on both sides and reversing the sign} \\[1em] \Rightarrow x \gt 1.38\end{aligned}$$

Positive integers greater than 1.38 are {2, 3, 4, 5, ...}

∴ Solution set = {2, 3, 4, 5, ...}

(i) 3x > - 14

We have:

3x > - 14

⇒ x > $\frac{-14}{3} \quad \text{[Dividing both sides by 3]}$

⇒ x > -4.66...

Negative integers greater than -4.66.. are {-4, -3, -2, -1}

∴ Solution set = {-4, -3, -2, -1}

(ii) -29 < 9x - 2

We have:

-29 < 9x - 2

⇒ -29 + 2 < 9x $\quad$ [Adding 2 on both sides]

⇒ -27 < 9x

⇒ $\frac{-27}{9}$ < x $\quad$ [Dividing both sides by 9]

⇒ -3 < x

⇒ x > -3

Negative integers greater than -3 are {-2, -1}

∴ Solution set = {-2, -1}

(iii) -4(x + 5) < 9

We have:

-4(x + 5) < 9

⇒ -4x - 20 < 9

⇒ -4x < 9 + 20 $\quad$ [Adding 20 on both sides]

⇒ -4x < 29

⇒ -x < $\frac{29}{4} \quad \text{[Dividing both sides by 4]}$

⇒ -x < 7.25

⇒ x > -7.25 $\quad$ [Multiplying -1 on both sides and reversing the sign]

Negative integers greater than -7.25 are {-6, -5, -4, -3, -2, -1}

∴ Solution set = {-6, -5, -4, -3, -2, -1}

(iv) 5 + 6x > x - 10

We have:

5 + 6x > x - 10

Subtract x and 5 from both sides

⇒ 6x - x > - 10 - 5

⇒ 5x > -15

⇒ x > $\frac{-15}{5} \quad \text{[Dividing both sides by 5]}$

⇒ x > -3

Negative integers greater than -3 are {-2, -1}

∴ Solution set = {-2, -1}

(v) 10 - 2(1 + 4x) < 26

We have:

10 - 2(1 + 4x) < 26

⇒ 10 - 2 - 8x < 26

⇒ 8 - 8x < 26

⇒ -8x < 26 - 8 $\quad$ [Subtracting 8 from both sides]

⇒ -8x < 18

⇒ -x < $\frac{18}{8} \quad \text{[Dividing both sides by 8]}$

⇒ -x < 2.25

Multiply -1 on both sides and reverse the sign

⇒ x > -2.25

Negative integers greater than -2.25 are {-2, -1}

∴ Solution set = {-2, -1}

(vi) $\frac{1}{3}$ > $\frac{6}{7}x + 4$

We have:

$$\begin{aligned}\phantom{=} \frac{1}{3} \gt \frac{6}{7}x + 4 \\[1em] \text{Subtract 4 from both sides} \\[1em] \Rightarrow \frac{1}{3} - 4 \gt \frac{6}{7}x \\[1em] \Rightarrow \frac{1 - 12}{3} \gt \frac{6}{7}x \\[1em] \Rightarrow \frac{-11}{3} \gt \frac{6}{7}x \\[1em] \text{Multiply } \frac{7}{6} \text{ on both sides} \\[1em] \Rightarrow \frac{7}{6} \times \frac{-11}{3} \gt x \\[1em] \Rightarrow \frac{-77}{18} \gt x \\[1em] \Rightarrow -4.27... \gt x \\[1em] \Rightarrow x \lt -4.27...\end{aligned}$$

Negative integers less than -4.27... are {..., -6, -5}

∴ Solution set = {..., -6, -5}

(i) 2 < x - 3 < 7, x ∈ N

We have:

2 < x - 3 < 7

Let's separate the inequalities:

Case 1:

2 < x - 3

⇒ 2 + 3 < x $\quad$ [Adding 3 on both sides]

⇒ 5 < x

⇒ x > 5

Natural numbers greater than 5 are {6, 7, 8, 9, 10, ...}

∴ Solution set A = {6, 7, 8, 9, 10, ...}

Case 2:

x - 3 < 7

⇒ x < 7 + 3 $\quad$ [Adding 3 on both sides]

⇒ x < 10

Natural numbers less than 10 are {1, 2, 3, 4, 5, 6, 7, 8, 9}

∴ Solution set B = {1, 2, 3, 4, 5, 6, 7, 8, 9}

Final solution set = A ∩ B = {6, 7, 8, 9}

(ii) 10 < 4x - 5 < 21, x ∈ N

We have:

10 < 4x - 5 < 21

Let's separate the inequalities:

Case 1:

10 < 4x - 5

⇒ 10 + 5 < 4x $\quad$ [Adding 5 on both sides]

⇒ 15 < 4x

⇒ $\frac{15}{4}$ < x $\quad$ [Dividing both sides by 4]

⇒ 3.75 < x

⇒ x > 3.75

Natural numbers greater than 3.75 are {4, 5, 6, 7, ...}

∴ Solution set A = {4, 5, 6, 7, ...}

Case 2:

4x - 5 < 21

⇒ 4x < 21 + 5 $\quad$ [Adding 5 on both sides]

⇒ 4x < 26

⇒ x < $\frac{26}{4} \quad \text{[Dividing both sides by 4]}$

⇒ x < 6.5

Natural numbers less than 6.5 are {1, 2, 3, 4, 5, 6}

∴ Solution set B = {1, 2, 3, 4, 5, 6}

Final solution set = A ∩ B = {4, 5, 6}

(iii) 2 - x < 4x - 7 < 11 - 2x, x ∈ Z

We have:

2 - x < 4x - 7 < 11 - 2x

Let's separate the inequalities:

Case 1:

2 - x < 4x - 7

⇒ 2 + 7 < 4x + x $\quad$ [Adding 7 and x on both sides]

⇒ 9 < 5x

⇒ $\frac{9}{5}$ < x $\quad$ [Dividing both sides by 5]

⇒ 1.8 < x

⇒ x > 1.8

Integers greater than 1.8 are {2, 3, 4, 5, ...}

∴ Solution set A = {2, 3, 4, 5, ...}

Case 2:

4x - 7 < 11 - 2x

⇒ 4x + 2x < 11 + 7 $\quad$ [Adding 7 and 2x on both sides]

⇒ 6x < 18

⇒ x < $\frac{18}{6} \quad \text{[Dividing both sides by 6]}$

⇒ x < 3

Integers less than 3 are {..., -1, 0, 1, 2}

∴ Solution set B = {..., -1, 0, 1, 2}

Final solution set = A ∩ B = {2}

(iv) 4 - 2x < 3x + 19 < 42 - 5x, x ∈ Z

We have:

4 - 2x < 3x + 19 < 42 - 5x

Let's separate the inequalities:

Case 1:

4 - 2x < 3x + 19

Subtract 19 and add 2x on both sides

⇒ 4 - 19 < 3x + 2x

⇒ -15 < 5x

⇒ $\frac{-15}{5}$ < x $\quad$ [Dividing both sides by 5]

⇒ -3 < x

⇒ x > -3

Integers greater than -3 are {-2, -1, 0, 1, 2, ...}

∴ Solution set A = {-2, -1, 0, 1, 2, ...}

Case 2:

3x + 19 < 42 - 5x

Subtract 19 and add 5x on both sides

⇒ 3x + 5x < 42 - 19

⇒ 8x < 23

⇒ x < $\frac{23}{8} \quad \text{[Dividing both sides by 8]}$

⇒ x < 2.875

Integers less than 2.875 are {..., -2, -1, 0, 1, 2}

∴ Solution set B = {..., -2, -1, 0, 1, 2}

Final solution set = A ∩ B = {-2, -1, 0, 1, 2}

(v) -5 < $\frac{x}{2}$ - 3 < $\frac{5}{2}$, x ∈ Z

We have:

-5 < $\frac{x}{2}$ - 3 < $\frac{5}{2}$

Let's separate the inequalities:

Case 1:

-5 < $\frac{x}{2}$ - 3

⇒ -5 + 3 < $\frac{x}{2} \quad \text{[Adding 3 on both sides]}$

⇒ -2 < $\frac{x}{2}$

⇒ -2 x 2 < x $\quad$ [Multiplying 2 on both sides]

⇒ -4 < x

⇒ x > -4

Integers greater than -4 are {-3, -2, -1, 0, ...}

∴ Solution set A = {-3, -2, -1, 0, ...}

Case 2:

$$\begin{aligned}\phantom{=} \frac{x}{2} - 3 \lt \frac{5}{2} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{5}{2} + 3 \quad \text{[Adding 3 on both sides]} \\[1em] \Rightarrow \frac{x}{2} \lt \frac{5 + 6}{2} \\[1em] \text{[Since both denominators are same, we cancel them]} \\[1em] \Rightarrow x \lt 5 + 6 \\[1em] \Rightarrow x \lt 11\end{aligned}$$

Integers less than 11 are {..., 8, 9, 10}

∴ Solution set B = {..., 8, 9, 10}

Final solution set = A ∩ B = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vi) 9 - $\frac{2}{3}$x < 5x - 11 < 17 - $\frac{x}{4}$, x ∈ Z

We have:

9 - $\frac{2}{3}$x < 5x - 11 < 17 - $\frac{x}{4}$

Let's separate the inequalities:

Case 1:

$$\begin{aligned}\phantom{=} 9 - \frac{2}{3}x \lt 5x - 11 \\[1em] \text{Add 11 on both sides} \\[1em] \Rightarrow 9 + 11 \lt 5x + \frac{2}{3}x \\[1em] \Rightarrow 20 \lt \frac{15x + 2x}{3} \\[1em] \Rightarrow 20 \lt \frac{17x}{3} \\[1em] \text{Multiply 3 on both sides} \\[1em] \Rightarrow 20 \times 3 \lt 17x \\[1em] \Rightarrow 60 \lt 17x \\[1em] \Rightarrow \frac{60}{17} \lt x \quad \text{[Dividing 17 on both sides]} \\[1em] \Rightarrow 3.52.. \lt x \\[1em] \Rightarrow x \gt 3.52..\end{aligned}$$

Integers greater than 3.52.. are {4, 5, 6, ...}

∴ Solution set A = {4, 5, 6, ...}

Case 2:

$$\begin{aligned}\phantom{=} 5x - 11 \lt 17 - \frac{x}{4} \\[1em] \text{Add 11 and} \frac{x}{4} \text{ on both sides} \\[1em] \Rightarrow 5x + \frac{x}{4} \lt 17 + 11 \\[1em] \Rightarrow \frac{20x + x}{4} \lt 28 \\[1em] \Rightarrow \frac{21x}{4} \lt 28 \\[1em] \text{Multiply 4 on both sides} \\[1em] \Rightarrow 21x \lt 28 \times 4 \\[1em] \Rightarrow 21x \lt 112 \\[1em] \Rightarrow x \lt \frac{112}{21} \quad \text{[Dividing 21 on both sides]} \\[1em] \Rightarrow x \lt 5.33..\end{aligned}$$

Integers less than 5.33.. are {..., 3, 4, 5}

∴ Solution set B = {..., 3, 4, 5}

Final solution set = A ∩ B = {4, 5}

Given:

a > b

m < 0 means m is negative.

Let's evaluate each:

1. am < bm

Multiplying an inequality by a negative number (m < 0) requires reversing the inequality sign. It is correct.

2. am = bm

Multiplying by a non-zero number (m < 0) maintains a difference between unequal values; they cannot become equal. It is incorrect.

3. am > bm

This is incorrect because, m is negative number and multiplying an inequality by a negative number requires reversing the inequality sign. But here the sign remains same.

4. am and bm cannot be compared

Since the relationship between a, b and the sign of m is known, their products are strictly comparable. It is incorrect.

Among all four options, option 1 is correct.

Hence, option 1 is the correct option.

Given:

3x - 5 < 6

⇒ 3x < 6 + 5 $\quad$ [Adding 5 on both sides]

⇒ 3x < 11

⇒ x < $\frac{11}{3} \quad \text{[Dividing both sides by 3]}$

⇒ x < 3.66

Among all four options only 3 is less than 3.66

Hence, option 1 is the correct option.

Given:

17 - 4x < 7

⇒ 17 - 7 < 4x $\quad$ [Subtracting 7 and adding 4x on both sides]

⇒ 10 < 4x

⇒ $\frac{10}{4}$ < x $\quad$ [Dividing both sides by 4]

⇒ 2.5 < x

⇒ x > 2.5

Integers greater than 2.5 are {3, 4, 5, ...}

Hence, option 3 is the correct option.

Given:

2x > 18 - 5x

⇒ 2x + 5x > 18 $\quad$ [Adding 5x on both sides]

⇒ 7x > 18

⇒ x > $\frac{18}{7} \quad \text{[Dividing both sides by 7]}$

⇒ x > 2.57

The solution must be greater than 2.57. Since 1 is not greater than 2.57, therefore 1 is not a solution.

Hence, option 4 is the correct option.

Let's evaluate each:

1. If a < b, then a - m < b - m
   Adding/Subtracting doesn't change the sign. So, it is correct.

2. If a > b and m > 0, then am > bm
   Multiplying by a positive number keeps the sign. So, it is correct.

3. If a < b and m > 0, then $\frac{a}{m}$ > $\frac{b}{m}$.
   Incorrect. If we divide by a positive number, the sign should remain the same. It should be $\frac{a}{m}$ < $\frac{b}{m}$.

4. If a ≠ 0 and b ≠ 0, then a > b ⇒ $\frac{1}{a}$ < $\frac{1}{b}$.
   Generally correct for positive numbers (Reciprocal rule).

Option 3 is the incorrect statement.

Hence, option 3 is the correct option.

(i) A statement of inequality between two expressions is called an inequation.

(ii) The set from which the values of the variable satisfying a given inequality are chosen, is called the replacement set.

(iii) Subset of the replacement set, consisting of all those values of the variable which satisfy the given inequation is called the solution set.

(iv) Multiplying each side of an inequality by a negative number, reverses the inequality.

(v) If a ≠ 0, b ≠ 0 and a < b, then $\frac{1}{a}$ > $\frac{1}{b}$.

Explanation

(i) While an "equation" uses an equals sign (=), a statement using symbols like <, >, ≤, or ≥ is termed an inequation.

(ii) Replacement set is the "universe" of numbers (like Natural numbers N or Integers Z) from which you are allowed to pick potential answers.

(iii) Solution set is the specific group of numbers that actually make the inequation true. It is always a subset of the replacement set.

(iv) Multiplying by a negative value changes the direction of the sign (> becomes <). This is the most important rule in inequalities.

(v) For non-zero numbers, taking the reciprocal reverses the inequality. This is the Reciprocal Rule.

For example, if 2 < 5, then $\frac{1}{2}$ > $\frac{1}{5}$ (0.5 > 0.2).

(i) True
Reason — Any mathematical statement that uses inequality symbols like <, >, ≤, or ≥ to compare two expressions is defined as an inequation.

(ii) True
Reason — This follows the Negative Division Rule. When we divide both sides of an inequality by a negative number (m < 0), the direction of the inequality sign must be reversed (< becomes >).

(iii) False
Reason — Adding or subtracting any number (whether positive or negative) from both sides of an inequality never changes the direction of the sign.

If a < b, then a - m < b - m remains true regardless of the value of m.

(iv) True
Reason — Similar to the division rule, multiplying by a negative number (m < 0) requires flipping the sign.

(v) False
Reason — When we divide by a positive number (m > 0), the inequality sign stays the same. Therefore, if a > b, the result should be $\frac{a}{m} \gt \frac{b}{m}$.

(1)

Total Rent = (Fixed Daily Charge) + (Charge per Kilometre x distance)

Let the distance driven be x km.

Total Rent = ₹ 250 + 15x

The total rent is more than 500:

250 + 15x > 500

15x > 500 - 250 $\quad$ [Subtracting 250 from both sides]

15x > 250

Divide both sides by 5:

3x > 50

Hence, option 3 is the correct option.

(2)

Let's solve 3x > 50:

3x > 50

x > $\frac{50}{3}$ $\quad$ [Dividing 3 from both sides]

x > 16.66..

The solution must be greater than 16.66..

Solution set = {17, 18, 19, ....}

Hence, option 2 is the correct option.

(3)

Let the distance driven be y km. The total rent is less than 600:

250 + 15y < 600

15y < 600 - 250 $\quad$ [Subtracting 250 from both sides]

15y < 350

Divide both sides by 5:

3y < 70

Hence, option 3 is the correct option.

(4)

Let's solve 3y < 70:

3y < 70

y < $\frac{70}{3}$ $\quad$ [Dividing 3 from both sides]

y < 23.33..

The distance must be 23 km or less.

Solution set = {..............., 20, 21, 22, 23}

Hence, option 1 is the correct option.

Assertion (A) is true but Reason (R) is false.

Explanation

Assertion:

Given inequality:

2x + 3 > 8

Subtract 3 from both sides:

2x + 3 - 3 > 8 - 3

2x > 5

∴ Assertion is true.

Subtracting or adding a number never reverses an inequality. Only multiplying or dividing by a negative number does that. Therefore, the Reason is false.

Hence, option 3 is the correct option.

Assertion (A) is false but Reason (R) is true.

Explanation

Assertion:

Let's solve the given inequality:

2x - 1 > 7

2x > 7 + 1 $\quad$ [Adding 1 on both sides]

2x > 8

x > $\frac{8}{2} \quad \text{[Dividing both sides by 2]}$

x > 4

Since x must be a Natural number (N) greater than 4, the solution set should be {5, 6, 7, ...}. So, Assertion is false.

The statement in reason is a standard rule of inequalities. For example, if 2 < 4, then taking the reciprocal gives $\frac{1}{2}$ > $\frac{1}{4}$ (0.5 > 0.25). The sign reverses. Therefore, the Reason is True.

Hence, option 4 is the correct option.

Let the number of books purchased by a student be x.

Given:

At least 3 books and no more than 10 books, so :

3 ≤ x ≤ 10

Cost per book = ₹ 250.

Total amount spent = ₹ 250x.

When x = 3:
Minimum amount = ₹ 250 × 3 = ₹ 750 $\quad$

When x = 10:
Maximum amount = ₹ 250 × 10 = ₹ 2500 $\quad$

Hence, option 1 is the correct option.

Given:

3x - 5 > 7

⇒ 3x > 7 + 5 $\quad$ [Adding 5 to both sides]

⇒ 3x > 12

⇒ x > $\frac{12}{3} \quad \text{[Dividing both sides by 3]}$

⇒ x > 4

From the replacement set {2, 3, 4, 5}, the value greater than 4 is {5}.

Since the solution set is {5}, it does not match any of the given options (1, 2 or 3).

Hence, option 4 is the correct option.

Given:

Number of students = 40

Space occupied by each student = x sq. ft.

Total space occupied = 40x sq. ft.

Total space available is 120 sq. ft, so:

40x < 120

⇒ x < $\frac{120}{40} \quad \text{[Dividing both sides by 40]}$

⇒ x < 3

The maximum space per student is less than 3 sq. ft.

Hence, option 2 is the correct option.

Given:

Marbles collected by Friend A = 2x - 3

Marbles collected by Friend B = 3x + 5

Total marbles collected by both friends together is fewer than 50 :

(2x - 3) + (3x + 5) < 50

⇒ 5x + 2 < 50

⇒ 5x < 50 - 2 $\quad$ [Subtracting 2 from both sides]

⇒ 5x < 48

⇒ x < $\frac{48}{5} \quad \text{[Dividing both sides by 5]}$

⇒ x < 9.6

Since the numbers of marbles collected by both friends must be whole numbers, we take the greatest whole-number value of x satisfying x < 9.6, which is 9.

∴ Maximum marbles collected by Friend A = 2(9) - 3 = 18 - 3 = 15.

Hence, option 1 is the correct option.

Let the number of classes attended in a month be n.

Given:

At least 5 classes but fewer than 15 classes, so :

5 ≤ n < 15

Cost per class = ₹ 200.

Total amount spent y = 200n.

Multiplying each side of the inequality by 200 :

5 × 200 ≤ 200n < 15 × 200

⇒ 1000 ≤ y < 3000

So, the amount y lies between ₹ 1,000 and ₹ 3,000.

Hence, option 2 is the correct option.