Note: Refer to the following if you need clarifications. Reference on Numbers

Question 1: State True or False:

$\displaystyle \text{i) } x < -y \Rightarrow -x> y \text{ : True }$

$\displaystyle \text{ii) } -5x \geq 15 \Rightarrow x \geq -3 \text{ : False }$

$\displaystyle \text{iii) } 2x \leq -7 \Rightarrow \frac{2x}{-4} \geq \frac{-7}{-4} \text{ : True }$

$\displaystyle \text{iv) } 7 > 5 \Rightarrow \frac{1}{7} < \frac{1}{5} \text{ : True }$

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Question 2: State True or False: Given that $\displaystyle a, b, c \text{ and } d$ , are real numbers and $\displaystyle c \neq 0$

$\displaystyle \text{i) If } a < b \text{ , then } a-c

$\displaystyle \text{ii) If } a > b \text{ , then } a+c > b+c \text{ : True }$

$\displaystyle \text{iii) If } a < b \text{ , then } ac > bc \text{ : False }$

$\displaystyle \text{iv) If } a > b \text{ , then } \frac{a}{c} < \frac{b}{c} \text{ : False }$

$\displaystyle \text{v) If } a-c > b-d$ ; then $\displaystyle a+d > b+c \text{ : True }$

$\displaystyle \text{vi) If } a < b, \text{ and } c>0 \text{ , then } a-c > b-c \text{ : False }$

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Question 3: If $\displaystyle x \in N$ , find the solution set of the inequations:

$\displaystyle \text{i) } 5x+3 \leq 2x+18$    $\displaystyle \text{ii) } 3x-2 < 19-4x$

$\displaystyle \text{i) } 5x+3 \leq 2x+18$

$\displaystyle \Rightarrow 3x \leq 15$

$\displaystyle \Rightarrow x \leq 5 \text{ or } x \in \{1, 2, 3, 4, 5 \}$

$\displaystyle \text{ii) } 3x-2 < 19-4x$

$\displaystyle 3x-2 < 19-4x$

$\displaystyle \Rightarrow 7x <21$

$\displaystyle \Rightarrow x < 3 \text{ or } x \in \{1, 2 \}$

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Question 4: If the replacement set is a set of whole numbers, solve:

$\displaystyle \text{i) } x+7 \leq 11$          $\displaystyle \text{ii) } 3x-1 > 8$          $\displaystyle \text{iii) } x- \frac{3}{2} < \frac{3}{2} - x$          $\displaystyle \text{iv) } 18 \leq 3x-2$

$\displaystyle \text{i) } x+7 \leq 11$

$\displaystyle x+7 \leq 11$

$\displaystyle \Rightarrow x \leq 4 \text{ or } x \in \{0, 1 2, 3, 4 \}$

$\displaystyle \text{ii) } 3x-1 > 8$

$\displaystyle 3x-1 > 8$

$\displaystyle \Rightarrow 3x > 9$

$\displaystyle \Rightarrow x > 3 \text{ or } x \in \{4, 5, 6, ... \}$

$\displaystyle \text{iii) } x- \frac{3}{2} < \frac{3}{2} - x$

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

$\displaystyle \Rightarrow 2x < \frac{3}{2} + \frac{3}{2}$

$\displaystyle \Rightarrow 2x < 3 \text{ or } x \in \{0, 1 \}$

$\displaystyle \text{iv) } 18 \leq 3x-2$

$\displaystyle 18 \leq 3x-2$

$\displaystyle \Rightarrow 3x \geq 20 \text{ or } x \in \{7, 8, 9, ... \}$

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Question 5: Solve the inequation: $\displaystyle 3-2x \geq x-12$ given that $\displaystyle x \in N$ [1987]

$\displaystyle 3-2x \geq x-12$

$\displaystyle \Rightarrow 3x \leq 15$

$\displaystyle \Rightarrow x \leq 5 \text{ or } x \in \{1, 2, 3, 4, 5 \}$

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Question 6: If $\displaystyle 25-4x \leq 16$ , find: i) the smallest value of $\displaystyle x$ , when $\displaystyle x$ is a real number ii) smallest value of $\displaystyle x$ when $\displaystyle x$ is an integer

$\displaystyle 25-4x \leq 16$

$\displaystyle \Rightarrow 4x \geq 9$

Therefore if $\displaystyle x$ is a real number the $\displaystyle x=2.25$ and if $\displaystyle x$ is an integer then $\displaystyle x = 3$

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Question 7: If the replacement set is a set of real numbers, solve

$\displaystyle \text{i) } -4x \geq -16$          $\displaystyle \text{ii) } 8-3x \leq 20$      $\displaystyle \text{iii) } 5+ \frac{x}{4} > \frac{x}{5} +9$          $\displaystyle \text{iv) } \frac{x+3}{8} < \frac{x-3}{5}$

$\displaystyle \text{i) } -4x \geq -16$

$\displaystyle -4x \geq -16$

$\displaystyle \Rightarrow 4x \leq 16$

$\displaystyle \Rightarrow x \leq 4 \text{ or } \{x: x\in R \text{ and } x \leq 4 \}$

$\displaystyle \text{ii) } 8-3x \leq 20$

$\displaystyle 8-3x \leq 20$

$\displaystyle \Rightarrow 3x \geq -12$

$\displaystyle \Rightarrow x \geq -4 \text{ or } \{x: x\in R \text{ and } x \geq 4 \}$

$\displaystyle \text{iii) } 5+ \frac{x}{4} > \frac{x}{5} +9$

$\displaystyle 5+ \frac{x}{4} > \frac{x}{5} +9$

$\displaystyle \Rightarrow \frac{x}{4} - \frac{x}{5} >4$

$\displaystyle \Rightarrow x > 80 \text{ or } \{x: x\in R \text{ and } x \geq 80 \}$

$\displaystyle \text{iv) } \frac{x+3}{8} < \frac{x-3}{5}$

$\displaystyle \frac{x+3}{8} < \frac{x-3}{5}$

$\displaystyle \Rightarrow 5x+15< 8x-24$

$\displaystyle \Rightarrow 39 < 3x$

$\displaystyle \Rightarrow x>13 \text{ or } \{x: x\in R \text{ and } x \geq 13 \}$

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Question 8: Find the smallest value of $\displaystyle x$ for which $\displaystyle 5-2x < 5 \frac{1}{2} - \frac{5}{3} x$ , where $\displaystyle x \in I$ .

$\displaystyle 5-2x<5 \frac{1}{2} - \frac{5}{3} x$

$\displaystyle \Rightarrow 5-2x < \frac{11}{2} - \frac{5}{3} x$

$\displaystyle \Rightarrow 30-12x<33-10x$

$\displaystyle \Rightarrow -3<2x$

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

Therefore $\displaystyle x = -1$

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Question 9: Find the largest value of $\displaystyle x$ for which $\displaystyle 2(x-1) \leq (9-x)$ and $\displaystyle x \in W$ .

$\displaystyle 2(x-1) \leq (9-x)$

$\displaystyle \Rightarrow 2x-2 \leq 9-x$

$\displaystyle \Rightarrow 3x \leq 11$

Therefore $\displaystyle x =3$

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Question 10: Solve the inequation: $\displaystyle 12+1 \frac{5}{6} x \leq 5+3x$ and $\displaystyle x \in R$ . [1999]

$\displaystyle 12+1 \frac{5}{6} x \leq 5+3x$

$\displaystyle \Rightarrow 12+ \frac{11}{6} x \leq 5+3x$

$\displaystyle \Rightarrow 7 \leq \frac{7}{6} x$

$\displaystyle \Rightarrow x \geq 6 \text{ or } \{x: x\in R \text{ and } x \geq 6 \}$

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Question 11: Given $\displaystyle x \in I$ , find the solution set for $\displaystyle -5 \leq 2x-3 < x+2$

$\displaystyle -5 \leq 2x-3 < x+2$

Equation 1:$\displaystyle -5 \leq 2x-3$

$\displaystyle \Rightarrow -2 \leq 2x$

$\displaystyle \Rightarrow -1 \leq x$

Equation 2: $\displaystyle 2x-3 < x+2$

$\displaystyle \Rightarrow x < 5$

Therefore $\displaystyle \{ x : x \in I \text{ and } -1 \leq x < 5 \} \text{ or } x \in \{-1, 0, 1, 2, 3, 4 \}$

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Question 12: $\displaystyle x \in W$ , find the solution set for $\displaystyle -1 \leq 3+4x < 23$

$\displaystyle -1 \leq 3+4x < 23$

Equation 1: $\displaystyle -1 \leq 3+4x$

$\displaystyle \Rightarrow -4 \leq 4x$

$\displaystyle \Rightarrow -1 \leq x$

Equation 2: $\displaystyle 3x+4 < 23$

$\displaystyle \Rightarrow 4x < 20$

$\displaystyle \Rightarrow x < 5$

Therefore $\displaystyle \{ x: x \in W \text{ and } -1 \leq x < 5 \} \text{ or } x \in \{0, 1, 2, 3, 4 \}$