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  

Answer:

\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

Answer:

\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 

Answer:

\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

Answer:

\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. }

Answer:

\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} ) 

Answer:

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]

Answer:

-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 :

7

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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.

Answer:

\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:

8

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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.

Answer:

\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

9

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

Answer:

\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:

101

Graph for \displaystyle x < 3:

102

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

103

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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.

Answer:

\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 :

111

Graph for \displaystyle x \leq - 4 :

112

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

113

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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.

Answer:

\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

121

\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

122

\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.

Answer:

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.

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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:

q14

14 2

(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

Answer:

\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 }

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

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

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

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

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

2022-02-27_10-45-39

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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

Answer:

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