Question 1: If the replacement set is the set of natural numbers (N), find the solution set of:
(i) $3x+ 4 < 16$      (ii) $8 - x \leq 4x -2$

 $\text{ (i) } 3x+ 4 < 16$ $\Rightarrow 3x < 16 - 4$ $\Rightarrow 3x < 12$ $\Rightarrow x < 4$ Since, the replacement set = N (set of natural numbers) $\text{Solution set} = \{1, 2, 3 \}$ $\text{ (ii) } 8 - x \leq 4x -2$ $\Rightarrow -x - 4x \leq -2 -8$ $\Rightarrow -5x \leq - 10$ $\Rightarrow x \geq 2$ Since, the replacement set = N (set of natural numbers) $\text{Solution set} = \{ 2, 3, 4, 5, 6, \cdots \}$

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Question 2: If the replacement set is the set of whole numbers (W), find the solution set of:
(i) $5x + 4 \leq 24$      (ii) $4x - 2 < 2x + 10$

 $\text{ (i) } 5x+4 \leq 24$ $\Rightarrow 5x \leq 24-4$ $\Rightarrow 5x \leq 20$ $\Rightarrow x \leq 4$ Since, the replacement set = W (set of whole numbers) $\text{Solution set} = \{0, 1, 2, 3, 4 \}$ $\text{ (ii) } 4x-2 < 2x+10$ $\Rightarrow 4x-2x < 10+2$ $\Rightarrow 2x < 12$ $\Rightarrow x < 6$ Since, the replacement set = W (set of whole numbers) $\text{Solution set} = \{0, 1, 2, 3, 4 , 5 \}$

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Question 3: If the replacement set is the set of integers, (I or Z), between -6 and 8, find the solution set of: (i) $6x-1 \geq 9+x$     (ii) $15-3x > x - 3$

 $\text{ (i) } 6x-1 \geq 9+x$ $\Rightarrow 6x-x \geq 9+1$ $\Rightarrow 5x \geq 10$ $\Rightarrow x \geq 2$ Since, the replacement set = I (set of whole numbers between -6 and 8 ) $\text{Solution set} = \{2, 3, 4, 5, 6,7\}$ $\text{ (ii) } 15-3x > x-3$ $\Rightarrow -3x-x > -3 - 15$ $\Rightarrow -4x > - 18$ $\displaystyle \Rightarrow x < 4\frac{1}{2}$ Since, the replacement set = I (set of whole numbers between -6 and 8 ) $\text{Solution set} = \\ \{-5, -4, -3, -2, -1, 0, 1, 2, ,3, 4\}$

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Question 4: f the replacement set is the set of real numbers (R), find the solution set of :
(i) $5 - 3x < 11$       (ii) $8 + 3x \geq 28 -2x$

 $\text{ (i) } 5-3x < 11$ $\Rightarrow -3x < 11-5$ $\Rightarrow -3x < 6$ $\Rightarrow x > -2$ Since, the replacement set is the set of real numbers (R) $\text{Solution set} \\ = \{ x:x > -2 \text{ and } x \in R \}$ $\text{ (i) } 8+3x \geq 28-2x$ $\Rightarrow 3x+2x \geq 28-8$ $\Rightarrow 5x \geq 20$ $\Rightarrow x \geq 4$ Since, the replacement set is the set of real numbers (R) $\text{Solution set} \\ = \{ x:x \geq 4 \text{ and } x \in R \}$

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$\displaystyle \text{Question 5: Solve } \frac{x}{2} - 5 \leq \frac{x}{3} - 4 , \text{ where } x \text{ is a positive odd integer. }$

$\displaystyle \frac{x}{2} - 5 \leq \frac{x}{3} - 4$

$\displaystyle \Rightarrow \frac{x}{2} - \frac{x}{3} \leq -4 + 5$

$\displaystyle \Rightarrow \frac{3x-2x}{6} \leq 1$

$\displaystyle \Rightarrow x \leq 6$

$\displaystyle \text{Since, } x \text{ is a positive odd integer, the Solution set } = \{ 1, 3, 5 \}$

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Question 6: Solve the following equation:  $2y - 3 < y+1 \leq 4y + 7 :$

(i) $y \in \{ \text{Integers} \}$   (ii) $y \in R \text{ (Real Numbers} )$

$2y - 3 < y+1 \leq 4y + 7$

$\Rightarrow 2y - 3 < y + 1 \text{ and } \hspace{1.0cm} y+1 \leq 4y+7$

$\Rightarrow y < 4 \hspace{1.7cm} \text{ and } \hspace{0.9cm} -6 \leq 3y$

$\Rightarrow y < 4 \hspace{1.7cm} \text{ and } \hspace{1.0cm} y \geq -2$

$\Rightarrow - 2 \leq y < 4$

(i) When $y \in \{ \text{Integer} \}$

Therefore Solution set $= \{ -2, -1, 0, 1, 2, 3 \}$

(ii) When $y \in R \text{ (Real Numbers} )$

Therefore Solution set $= \{ y : - 2 \leq y < 4 \text{ and } y \in R \}$

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Question 7: Given that $x \in R$, solve the following inequality and graph the solution on the number line: $-1 \leq 3 + 4x<23$  [ICSE Board 2006]

$-1 \leq 3 + 4x<23 ; x \in R$

$\Rightarrow -1 \leq 3 + 4x \hspace{1.0cm} \text{ and } \hspace{1.0cm} 3+4x < 23$

$\Rightarrow -4 \leq 4x \hspace{1.7cm} \text{ and } \hspace{0.9cm} 4x < 20$

$\Rightarrow -1 \leq x \hspace{2.0cm} \text{ and } \hspace{1.0cm} x < 5$

$\Rightarrow -1 \leq x < 5; x \in R$

Therefore Solution $= \{ -1 \leq x < 5 : x \in R \}$

Solution on the number line is :

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$\displaystyle \text{Question 8: Simplify: } -\frac{1}{3} \leq \frac{x}{2} - 1\frac{1}{3} < \frac{1}{6} : x \in R .$

Graph the values of $x$ on the real number line.

$\displaystyle -\frac{1}{3} \leq \frac{x}{2} - 1\frac{1}{3} < \frac{1}{6} : x \in R .$

The given inequation has two parts :

$\displaystyle -\frac{1}{3} \leq \frac{x}{2} - 1\frac{1}{3} \hspace{1.0cm} \text{ and } \hspace{1.0cm} \frac{x}{2} - 1\frac{1}{3} < \frac{1}{6}$

$\displaystyle \Rightarrow -\frac{1}{3} + 1\frac{1}{3} \leq \frac{x}{2} \hspace{1.0cm} \text{ and } \hspace{0.9cm} \Rightarrow \frac{x}{2} < \frac{1}{6} + 1\frac{1}{3}$

$\displaystyle \Rightarrow 1 \leq \frac{x}{2} \hspace{2.0cm} \text{ and } \hspace{1.0cm} \Rightarrow \frac{x}{2} < \frac{9}{6}$

$\displaystyle \Rightarrow 2 \leq x \hspace{2.0cm} \text{ and } \hspace{1.0cm} \Rightarrow x < 3$

On simplifying, the given inequation reduces to $2 \leq x < 3$ and the required graph on number line is:

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Question 9: List the solution set of $50 - 3 (2x - 5) < 25,$ given that $x \in W.$ Also represent the solution set obtained on a number line.

$\displaystyle 50 - 3 (2x - 5) < 25$

$\displaystyle \Rightarrow 50-6x+15 < 25$

$\displaystyle \Rightarrow -6x < 25 - 65$

$\displaystyle \Rightarrow -6x < -40$

$\displaystyle \Rightarrow \frac{-6x}{-6} > \frac{-40}{-6}$

$\displaystyle \Rightarrow x > 6\frac{2}{3}$

The required solution set $\displaystyle = \{ 7, 8, 9, \cdots \}$

And the required no. line is

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Question 10: Solve and graph the solution set of $3x + 6 \geq 9$ and $-5x > -15;$ where $x \in R.$

$\displaystyle 3x + 6 \geq 9 \hspace{1.0cm}\Rightarrow 3x \geq 9-6 \hspace{1.0cm}\Rightarrow 3x \geq 3 \hspace{1.0cm}\Rightarrow x \geq 1$

And

$\displaystyle -5x > -15 \hspace{1.0cm}\Rightarrow \frac{-5x}{-5} < \frac{-15}{-5} \hspace{1.0cm}\Rightarrow x < 3$

Graph for $\displaystyle x \geq 1:$

Graph for $\displaystyle x < 3:$

Therefore graph of solution set of $\displaystyle x \geq 1$ and $\displaystyle x < 3$

= Graph of points common to both $\displaystyle x \geq$ and  $\displaystyle x < 3$

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Question 11:  Solve and graph the solution set of $-2 < 2x - 6$ or $-2x + 5 \geq 13;$ where $x \in R.$

$\displaystyle -2 < 2x - 6 \hspace{1.0cm} \Rightarrow 2x - 6 > - 2 \hspace{1.0cm} \Rightarrow 2x > 4 \hspace{1.0cm} \Rightarrow x > 2$

$\displaystyle 2x + 5 \geq 13 \hspace{1.0cm} \Rightarrow -2x \geq 8 \hspace{1.0cm} \Rightarrow x \leq -4$

Graph for $\displaystyle x > 2 :$

Graph for $\displaystyle x \leq - 4 :$

Therefore graph of solution set of $\displaystyle x > 2$ or $\displaystyle x \leq - 4$

= Graph of points which belong to $\displaystyle x > 2$ or  $\displaystyle x \leq - 4$ or both

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Question 12: Given: $P = \{ x : 5 < 2x - 1 \leq 11, x \in R \}$

$Q = \{ x : -1 \leq 3 + 4 x < 23, x \in I \}$

$\text{Where } R= \{ \text{real numbers} \} \text{ and } I = \{ \text{integers} \} .$

Represent P and Q on two different number lines. Write down the elements of $P \cap Q.$

$\displaystyle \text{For P: } \ 5 < 2x - 1 \leq 11, \ x \in R$

$\displaystyle \Rightarrow 5 < 2x - 1 \hspace{1.0cm} \text{ and } \hspace{1.0cm} 2x - 1 \leq 11$

$\displaystyle \Rightarrow 3 < x \hspace{1.0cm} \text{ and } \hspace{1.0cm} x \leq 6$

$\displaystyle P = 3 < x \leq 6; \ \ \ x \in R$

$\displaystyle \text{For Q: } \ -1 \leq 3 + 4 x < 23, \ x \in I$

$\displaystyle \Rightarrow -1 \leq 3 + 4x \hspace{1.0cm} \text{ and } \hspace{1.0cm} 3 + 4x < 23$

$\displaystyle \Rightarrow -1 \leq x \hspace{1.0cm} \text{ and } \hspace{1.0cm} x < 5$

$\displaystyle Q = -1 \leq x < 5 ; \ \ \ x \in I$

$\displaystyle \text{Hence,} P \cap Q = \{ \text{elements common to both P and Q} \} = \{ 4 \}$

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Question 13: Write down the range of values of $x$ for which both the inequations $x > 2$ and $-1 \leq x \leq 4$ are true.

Both the given inequations are true in the range; where their graphs on the real number lines overlap.

The graphs for the given inequalities are drawn alongside.

It is clear from both the graphs that their common range is $2 < x \leq 4.$

Therefore required range is $2 < x \leq 4$

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Question 14: The diagram, given below, represents two inequations p and e on real number lines:

(i) Write down P and Q in set builder notation.
(ii) Represent each of the following sets on different number lines :

$\text{(a)} P \cup Q \hspace{0.5cm} \text{(b)} P \cap Q \hspace{0.5cm} \text{(c)} P - Q \hspace{0.5cm} \text{(d)} Q - P \hspace{0.5cm} \text{(e)} P \cap Q' \hspace{0.5cm} \text{(f)} P' \cap Q$

$\text{(i) } P = \{ x : - 2 < x \leq 6 \text{ and } x \in R \} \text{ and } Q = \{ x : 2 \leq x < 7 \text{ and } x \in R \}$

(ii)

$\displaystyle \text{(a) } P \cup Q = \text{ Numbers which belong to P or Q or both }$

$\displaystyle \text{(b) } P \cap Q = \text{ Numbers common to both P and Q }$

$\displaystyle \text{(c) } P - Q = \text{ Numbers which belong to P but do not belong to Q }$

$\displaystyle \text{(d) }Q - P = \text{ Numbers which belong to Q but do not belong to P }$

$\displaystyle \text{(e) } P \cap Q'= \text{ Numbers which belong to P but do not belong to } Q = P - Q$

$\displaystyle \text{(f) } P' \cap Q = \text{ Numbers which do not belong to P but belong to Q } = Q - P$

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Question 15: Find three smallest consecutive whole numbers such that the difference between one-fourth of the largest and one-fifth of the smallest is at least 3

Let the required whole numbers be $x, x + 1 \text{ and } x + 2.$

According to the given statement :

$\displaystyle \frac{x+2}{4} - \frac{x}{5} \geq 3$

$\displaystyle \Rightarrow \frac{5x+10-4x}{20} \geq 3$

$\displaystyle \Rightarrow x + 10 \geq 60$

$\displaystyle \Rightarrow x \geq 50$

Since, the smallest value of $x = 50$ that satisfies the inequation $x \geq 50.$

Therefore Required smallest consecutive whole numbers are : $50, 51, \text{ and } 52$