Question 1: Solve the following inequation and graph the solution on a number line \displaystyle  2x- 5 \leq 5x+4 < 11 , where \displaystyle  x \in I .      [ICSE 2011]
\displaystyle \text{Answer:}
\displaystyle  2x-5 \leq 5x+4 < 11
\displaystyle  2x-5 \leq 5x+4 \text{  or  } -9 \leq 3x \text{  or  } -3 \leq x
\displaystyle  5x+4 < 11 \text{  or  } 5x < 7 \text{  or  } x < \frac{7}{5}
\displaystyle  -3 \leq x <\frac{7}{5}
Therefore \displaystyle  x \in \{-3, -2, -1, 0, 1 \}


\displaystyle  \\

Question 2: Given that \displaystyle  x \in I , solve the inequation and graph it on a number line: \displaystyle  3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2 .       [ICSE 2004]
\displaystyle \text{Answer:}
\displaystyle  3 \geq \frac{x-4}{2}+\frac{x}{3} \geq 2
\displaystyle  18 \geq 3(x-4)+2x \geq 12
\displaystyle  30 \geq 5x \geq 24
\displaystyle  6 \geq x \geq 4.8
Therefore \displaystyle  x \in \{5, 6 \}
\displaystyle  \\

Question 3: Given \displaystyle  A = \{x: 11x-5 > 7x + 3, x \in R \} , \displaystyle  B = \{x: 18x-9 \geq 15+12x , x \in R \} . Find the range of the set \displaystyle  A \cap B and represent it on a number line.       [ICSE 2005]
\displaystyle \text{Answer:}
\displaystyle  A: 11x-5 > 7x+3
\displaystyle  4x >8 \text{  or  } x >2
\displaystyle  B: 18x-9 \geq 15+12x
\displaystyle  6x \geq 24 \text{  or  } x \geq 4
\displaystyle  A \cap B = \{ x: x \geq 4, x \in R \}
\displaystyle  \\

Question 4: Solve the given inequation and graph it on a number line: \displaystyle  2y-3 < y+1 \leq 4y+7, y \in R .       [ICSE 2008]
\displaystyle \text{Answer:}
\displaystyle  2y-3 < y+1 \leq 4y+7
\displaystyle  2y-3 < y+1 \text{  or  } y < 4
\displaystyle  y+1 \leq 4y+7 \text{  or  } -6 \leq 3y \text{  or  } -2 \leq y
Hence \displaystyle  \{ x: -2 \leq y < 4, x \in R \}
\displaystyle  \\

Question 5: Solve the given inequation and graph it on a number line: \displaystyle  -3 < -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6}, x \in R .      [ICSE 2010]
\displaystyle \text{Answer:}
\displaystyle  -3 < -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6}
\displaystyle  -3 < -\frac{1}{2}-\frac{2x}{3}
\displaystyle  -18 < -3 -4x
\displaystyle  4x < 15 \text{  or  } x < \frac{15}{4}
\displaystyle  -\frac{1}{2}-\frac{2x}{3} \leq \frac{5}{6} \text{  or  } -3-4x \leq 5
\displaystyle  -8 \leq 4x \text{  or  } -2 \leq x
Therefore \displaystyle  \{ x: -2 \leq x < \frac{15}{4}, x \in R \}


\displaystyle  \\

Question 6: Solve the given inequation and graph it on a number line: \displaystyle  4x-19 < \frac{3x}{5}-2 \leq -\frac{2}{5}+x, x \in R .       [ICSE 2012]
\displaystyle \text{Answer:}
\displaystyle  4x-19 < \frac{3x}{5}-2 \leq -\frac{2}{5}+x
\displaystyle  4x-19 < \frac{3x}{5}-2 \text{  or  } 20x-95 < 3x-10 \text{  or  } 17x < 85 \text{  or  } x < 5
\displaystyle  \frac{3x}{5}-2 \leq -\frac{2}{5}+x \text{  or  } 3x-10 \leq -2 +5x \text{  or  } -8 \leq 2x \text{  or  } -4 \leq x
Therefore \displaystyle  \{x : -4 \leq x < 5, x \in R \}


\displaystyle  \\

Question 7: Solve the given inequation and graph it on a number line: \displaystyle  -\frac{x}{3} \leq \frac{x}{2}-1\frac{1}{3} <\frac{1}{6}.x \in R .       [ICSE 2013]
\displaystyle \text{Answer:}
\displaystyle  -\frac{x}{3} \leq \frac{x}{2}-1\frac{1}{3} < \frac{1}{6}
\displaystyle  -\frac{x}{3} \leq \frac{x}{2}-1\frac{1}{3} < \frac{1}{6}
\displaystyle  -2x \leq 3x-8 < 1 \text{  or  } -2x \leq 3x-8 \text{  or  } 8 \leq 5x
\displaystyle  \frac{8}{5} \leq x \text{  or  } 3x-8 < 1 \text{  or  } 3x < 9 \text{  or  } x < 3
Therefore \displaystyle  \{ x : \frac{8}{5} \leq x < 3, x \in R \}


\displaystyle  \\

Question 8: Find the value of \displaystyle  x which satisfies the inequation: \displaystyle  -2\frac{5}{6} < \frac{1}{2} - \frac{2x}{3} \leq 2, x \in W .       [ICSE 2014]
\displaystyle \text{Answer:}
\displaystyle  -2\frac{5}{6} < \frac{1}{2} - \frac{2x}{3} \leq 2
\displaystyle  -\frac{17}{6} < \frac{1}{2} -\frac{2x}{3} \leq 2
\displaystyle  -17 < 3-4x \leq 12 \text{  or  } -17 < 3-4x \text{  or  } 4x < 20 \text{  or  } x < 5
\displaystyle  3-4x \leq 12 \text{  or  } -9 \leq 4x \text{  or  } -2.25 \leq x
Therefore \displaystyle  \{x : -2.25 \leq x < 5, x \in W \}
\displaystyle  x \in \{0, 1, 2, 3, 4\}

Question 9: Solve the inequation: \displaystyle  3-2x \geq x-12 given that \displaystyle  x \in N       [ICSE 1987]
\displaystyle \text{Answer:}
\displaystyle  3-2x \geq x-12
\displaystyle  \Rightarrow 3x \leq 15
\displaystyle  \Rightarrow x \leq 5 \ or \ x \in \{1, 2, 3, 4, 5 \}
\displaystyle  \\

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

\displaystyle \textbf{Question 11.}\ \text{The solution set for the inequation }2x+4\le 14,\ x\in W\text{ is}\qquad \text{ICSE 2023}
\displaystyle (a)\ \{1,2,3,4,5\}\qquad (b)\ \{0,1,2,3,4,5\}
\displaystyle (c)\ \{1,2,3,4\}\qquad (d)\ \{0,1,2,3,4\}
\displaystyle \text{Answer:}
\displaystyle  (b)\ \text{Given, expression is }2x+4\le 14
\displaystyle \Rightarrow 2x\le 14-4
\displaystyle \Rightarrow 2x\le 10\Rightarrow x\le 5,\ x\in W
\displaystyle \therefore\ \text{Solution set}=\{0,1,2,3,4,5\}

\displaystyle \textbf{Question 12.}\ \text{The solution set for the given inequation is}
\displaystyle -8\le 2x<8,\ x\in W\qquad \text{ICSE 2023}
\displaystyle (a)\ \{-4,-3,-2,-1,0,1,2,3,4\}
\displaystyle (b)\ \{-4,-3,-2,-1\}
\displaystyle (c)\ \{0,1,2,3\}
\displaystyle (d)\ \{-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8\}
\displaystyle \text{Answer:}
\displaystyle  (c)\ \text{Given, }-8\le 2x<8
\displaystyle \Rightarrow -\frac{8}{2}\le x<\frac{8}{2}
\displaystyle \Rightarrow -4\le x<4
\displaystyle \text{But }x\in W
\displaystyle \therefore\ \text{Solution set}=\{0,1,2,3\}

\displaystyle \textbf{Question 13.}\ \text{The solution set of the inequation }x-3\ge -5,\ x\in R\text{ is } \\ \text{ICSE Semester I 2022}
\displaystyle (a)\ \{x:x>-2,\ x\in R\}
\displaystyle (b)\ \{x:x\le -2,\ x\in R\}
\displaystyle (c)\ \{x:x\ge -2,\ x\in R\}
\displaystyle (d)\ \{-2,-1,0,1,2\}
\displaystyle \text{Answer:}
\displaystyle  (c)\ \text{Given, }x-3\ge -5,\ x\in R
\displaystyle \text{Adding }3\text{ on both sides}
\displaystyle \Rightarrow x-3+3\ge -5+3
\displaystyle \Rightarrow x\ge -2,\ x\in R
\displaystyle \therefore\ \text{Solution set}=\{x:x\ge -2,\ x\in R\}

\displaystyle \textbf{Question 14.}\ \text{The solution set on the number line of the linear inequation}
\displaystyle 2y-6<y+2\le 2y,\ y\in N\qquad \text{ICSE Semester I 2022}
\displaystyle (a)\ \text{(number line as shown)}


\displaystyle \text{Answer:}
\displaystyle (b)\ \text{Given, }2y-6<y+2\le 2y,\ y\in N
\displaystyle \Rightarrow 2y-6<y+2
\displaystyle \Rightarrow 2y-y<2+6
\displaystyle \Rightarrow y<8\quad (i)
\displaystyle \text{Also, }y+2\le 2y
\displaystyle \Rightarrow 2\le 2y-y
\displaystyle \Rightarrow 2\le y\quad (ii)
\displaystyle \text{From (i) and (ii), }2\le y<8
\displaystyle \therefore\ \text{Solution set}=\{y:2\le y<8,\ y\in N\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 15.}\ \text{The solution set representing the following number line is } \\ \text{ICSE Semester I 2022}

\displaystyle (a)\ \{x:x\in R,\ -3\le x<2\}
\displaystyle (b)\ \{x:x\in R,\ -3<x<2\}
\displaystyle (c)\ \{x:x\in R,\ -3<x\le 2\}
\displaystyle (d)\ \{x:x\in R,\ -3\le x\le 2\}
\displaystyle \text{Answer:}
\displaystyle  (a)\ \text{Given number line from }x=-3\ (\text{included})\ \text{to }x=2\ (\text{not included})
\displaystyle \therefore\ \text{Solution set}=\{x:x\in R,\ -3\le x<2\}

\displaystyle \textbf{Question 16.}\ \text{If }x\in W,\ \text{then the solution set of the inequation }-x>-7\text{ is } \\ \text{ICSE Semester I 2022}
\displaystyle (a)\ \{8,9,10,\dots\}\qquad (b)\ \{0,1,2,3,4,5,6\}
\displaystyle (c)\ \{0,1,2,3,\dots\}\qquad (d)\ \{-8,-9,-10,\dots\}
\displaystyle \text{Answer:}
\displaystyle  \ (b)\ \text{Given, inequation is }-x>-7
\displaystyle \Rightarrow x<7,\ x\in W
\displaystyle \therefore\ \text{Solution set}=\{0,1,2,3,4,5,6\}

\displaystyle \textbf{Question 17.}\ \text{The solution set for the linear inequation}
\displaystyle -8\le x-7<-4,\ x\in I\qquad \text{ICSE Semester I 2022}
\displaystyle (a)\ \{x:x\in R,\ -1\le x\le 3\}
\displaystyle (b)\ \{0,1,2,3\}
\displaystyle (c)\ \{-1,0,1,2,3\}
\displaystyle (d)\ \{-1,0,1,2\}
\displaystyle \text{Answer:}
\displaystyle  \ (d)\ \text{Given, }-8\le x-7<-4
\displaystyle \text{Adding }7\text{ on both sides}
\displaystyle \Rightarrow -8+7\le x-7+7
\displaystyle \Rightarrow -1\le x
\displaystyle \text{Also, }x-7<-4
\displaystyle \Rightarrow x-7+7<-4+7
\displaystyle \Rightarrow x<3
\displaystyle \therefore\ -1\le x<3
\displaystyle \text{Hence, solution set}=\{-1,0,1,2\}

\displaystyle \textbf{Question 18.}\ \text{Solve the following inequation and represent the solution set on} \\ \text{the number line.}
\displaystyle \frac{3x}{5}+2<x+4\le \frac{x}{2}+5,\ x\in R\qquad \text{ICSE 2020}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }\frac{3x}{5}+2<x+4\le \frac{x}{2}+5,\ x\in R
\displaystyle \Rightarrow \frac{3x}{5}+2<x+4\ \text{and}\ x+4\le \frac{x}{2}+5
\displaystyle \text{Now, }\frac{3x}{5}+2<x+4
\displaystyle \Rightarrow \frac{3x}{5}+2-x<x+4-x
\displaystyle \Rightarrow \frac{3x-5x}{5}+2<4
\displaystyle \Rightarrow -\frac{2x}{5}+2<4
\displaystyle \Rightarrow -\frac{2x}{5}<2
\displaystyle \Rightarrow \frac{2x}{5}>-2
\displaystyle \Rightarrow x>-5\quad (i)
\displaystyle \text{Also, }x+4\le \frac{x}{2}+5
\displaystyle \Rightarrow x+4-\frac{x}{2}\le 5
\displaystyle \Rightarrow \frac{x}{2}+4\le 5
\displaystyle \Rightarrow \frac{x}{2}\le 1
\displaystyle \Rightarrow x\le 2\quad (ii)
\displaystyle \text{From (i) and (ii), }-5<x\le 2
\displaystyle \therefore\ \text{Solution set}=\{x:-5<x\le 2,\ x\in R\}=(-5,2]

\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 19.}\ \text{Solve the following inequation and write down the solution set.} \\ \text{Represent the solution set on a real number line.}
\displaystyle 11x-4<15x+4\le 13x+14,\ x\in W\qquad \text{ICSE 2019}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }11x-4<15x+4\le 13x+14,\ x\in W
\displaystyle \Rightarrow 11x-4<15x+4\ \text{and}\ 15x+4\le 13x+14
\displaystyle \text{Now, }11x-4<15x+4
\displaystyle \Rightarrow 11x-4-4<15x
\displaystyle \Rightarrow 11x-8<15x
\displaystyle \Rightarrow -8<4x
\displaystyle \Rightarrow -2<x\quad (i)
\displaystyle \text{Also, }15x+4\le 13x+14
\displaystyle \Rightarrow 15x-13x\le 14-4
\displaystyle \Rightarrow 2x\le 10
\displaystyle \Rightarrow x\le 5\quad (ii)
\displaystyle \text{From (i) and (ii), }-2<x\le 5
\displaystyle \text{Since }x\in W
\displaystyle \therefore\ \text{Solution set}=\{0,1,2,3,4,5\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 20.}\ \text{Solve the following inequation, write down the solution set and} \\ \text{represent it on the real number line.}
\displaystyle -3+x\le \frac{7x}{2}+2<8+2x,\ x\in I\qquad \text{ICSE 2024}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }-3+x\le \frac{7x}{2}+2<8+2x,\ x\in I
\displaystyle \Rightarrow -3+x\le \frac{7x}{2}+2\ \text{and}\ \frac{7x}{2}+2<8+2x
\displaystyle \text{First, }-3+x\le \frac{7x}{2}+2
\displaystyle \Rightarrow -3+x-2\le \frac{7x}{2}
\displaystyle \Rightarrow x-5\le \frac{7x}{2}
\displaystyle \Rightarrow 2x-10\le 7x
\displaystyle \Rightarrow -10\le 5x
\displaystyle \Rightarrow -2\le x\quad (i)
\displaystyle \text{Now, }\frac{7x}{2}+2<8+2x
\displaystyle \Rightarrow \frac{7x}{2}-2x<8-2
\displaystyle \Rightarrow \frac{3x}{2}<6
\displaystyle \Rightarrow x<4\quad (ii)
\displaystyle \text{From (i) and (ii), }-2\le x<4
\displaystyle \text{Since }x\in I
\displaystyle \therefore\ \text{Solution set}=\{-2,-1,0,1,2,3\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 21.}\ \text{Solve the following inequation. Write down the solution set and} \\ \text{represent it on the real number line.}
\displaystyle -5(x-9)\ge 17-9x>x+2,\ x\in R\qquad \text{ICSE 2023}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }-5(x-9)\ge 17-9x>x+2,\ x\in R
\displaystyle \Rightarrow -5x+45\ge 17-9x\ \text{and}\ 17-9x>x+2
\displaystyle \text{First, }-5x+45\ge 17-9x
\displaystyle \Rightarrow -5x+9x\ge 17-45
\displaystyle \Rightarrow 4x\ge -28
\displaystyle \Rightarrow x\ge -7\quad (i)
\displaystyle \text{Now, }17-9x>x+2
\displaystyle \Rightarrow 17-2> x+9x
\displaystyle \Rightarrow 15>10x
\displaystyle \Rightarrow x<\frac{3}{2}\quad (ii)
\displaystyle \text{From (i) and (ii), }-7\le x<\frac{3}{2}
\displaystyle \therefore\ \text{Solution set}=\{x:-7\le x<\frac{3}{2},\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 22.}\ \text{Solve the following inequation and represent the solution set on a number line.}
\displaystyle -\frac{x}{3}-4\le \frac{x}{2}-\frac{7}{3}<-\frac{7}{6},\ x\in R\qquad \text{ICSE 2023}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }-\frac{x}{3}-4\le \frac{x}{2}-\frac{7}{3}<-\frac{7}{6},\ x\in R
\displaystyle \Rightarrow -\frac{x}{3}-4\le \frac{x}{2}-\frac{7}{3}\ \text{and}\ \frac{x}{2}-\frac{7}{3}<-\frac{7}{6}
\displaystyle \text{First, }-\frac{x}{3}-4\le \frac{x}{2}-\frac{7}{3}
\displaystyle \Rightarrow -\frac{x}{3}-4+\frac{7}{3}\le \frac{x}{2}
\displaystyle \Rightarrow -\frac{x}{3}-\frac{5}{3}\le \frac{x}{2}
\displaystyle \Rightarrow -\frac{2x}{6}-\frac{10}{6}\le \frac{3x}{6}
\displaystyle \Rightarrow -\frac{10}{6}\le \frac{5x}{6}
\displaystyle \Rightarrow -2\le x\quad (i)
\displaystyle \text{Now, }\frac{x}{2}-\frac{7}{3}<-\frac{7}{6}
\displaystyle \Rightarrow \frac{x}{2}<-\frac{7}{6}+\frac{7}{3}
\displaystyle \Rightarrow \frac{x}{2}<\frac{7}{6}
\displaystyle \Rightarrow x<\frac{7}{3}\quad (ii)
\displaystyle \text{From (i) and (ii), }-2\le x<\frac{7}{3}
\displaystyle \therefore\ \text{Solution set}=\{x:-2\le x<\frac{7}{3},\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 23.}\ \text{Solve the following inequation and represent your solution on the real number line.}
\displaystyle -5\frac{1}{2}-x\le \frac{1}{2}-3x\le \frac{1}{2}-x,\ x\in R\qquad \text{ICSE 2020}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }-5\frac{1}{2}-x\le \frac{1}{2}-3x\le \frac{1}{2}-x,\ x\in R
\displaystyle \Rightarrow -\frac{11}{2}-x\le \frac{1}{2}-3x\ \text{and}\ \frac{1}{2}-3x\le \frac{1}{2}-x
\displaystyle \text{First, }-\frac{11}{2}-x\le \frac{1}{2}-3x
\displaystyle \Rightarrow -\frac{11}{2}+2x\le \frac{1}{2}
\displaystyle \Rightarrow 2x\le 6
\displaystyle \Rightarrow x\le 3\quad (i)
\displaystyle \text{Now, }\frac{1}{2}-3x\le \frac{1}{2}-x
\displaystyle \Rightarrow -3x\le -x
\displaystyle \Rightarrow -2x\le 0
\displaystyle \Rightarrow x\ge 0\quad (ii)
\displaystyle \text{From (i) and (ii), }0\le x\le 3
\displaystyle \therefore\ \text{Solution set}=\{x:0\le x\le 3,\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 24.}\ \text{Solve the following equation, write down the solution set and} \\ \text{represent it on the number line.}
\displaystyle -2+10x\le 13x+10<24+10x,\ x\in Z\qquad \text{ICSE 2018}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }-2+10x\le 13x+10<24+10x,\ x\in Z
\displaystyle \Rightarrow -2+10x\le 13x+10\ \text{and}\ 13x+10<24+10x
\displaystyle \text{First, }-2+10x\le 13x+10
\displaystyle \Rightarrow -2-10\le 13x-10x
\displaystyle \Rightarrow -12\le 3x
\displaystyle \Rightarrow -4\le x\quad (i)
\displaystyle \text{Now, }13x+10<24+10x
\displaystyle \Rightarrow 13x-10x<24-10
\displaystyle \Rightarrow 3x<14
\displaystyle \Rightarrow x<\frac{14}{3}\quad (ii)
\displaystyle \text{From (i) and (ii), }-4\le x<\frac{14}{3}
\displaystyle \text{Since }x\in Z
\displaystyle \therefore\ \text{Solution set}=\{-4,-3,-2,-1,0,1,2,3,4\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 25.}\ \text{Solve the following inequation, write the solution set and} \\ \text{represent it on the number line.}
\displaystyle -3(x-7)\ge 15-7x-\frac{x+1}{3},\ x\in R\qquad \text{ICSE 2016}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-3(x-7)\ge 15-7x>\frac{x+1}{3},\ x\in R
\displaystyle \Rightarrow -3(x-7)\ge 15-7x\ \text{and}\ 15-7x>\frac{x+1}{3}
\displaystyle \text{Consider }-3(x-7)\ge 15-7x
\displaystyle \Rightarrow -3x+21\ge 15-7x
\displaystyle \Rightarrow -3x+7x\ge 15-21
\displaystyle \Rightarrow 4x\ge -6\Rightarrow x\ge -\frac{6}{4}
\displaystyle \Rightarrow x\ge -\frac{3}{2}\quad (i)
\displaystyle \text{Also, }15-7x>\frac{x+1}{3}\Rightarrow 3(15-7x)>x+1
\displaystyle \Rightarrow 45-21x>x+1
\displaystyle \Rightarrow -21x-x>1-45
\displaystyle \Rightarrow -22x>-44\Rightarrow 22x<44
\displaystyle \Rightarrow x<2\quad (ii)
\displaystyle \text{From (i) and (ii), }-\frac{3}{2}\le x<2
\displaystyle \therefore\ \text{Solution set}=\{x:-\frac{3}{2}\le x<2,\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 26.}\ \text{Solve the following inequation and represent the solution on} \\ \text{a real number line.}
\displaystyle 13x-5<15x+4<7x+12,\ x\in R\qquad \text{ICSE 2015}
\displaystyle \text{Answer:}
\displaystyle  \text{Given inequation is }13x-5<15x+4<7x+12,\ x\in R
\displaystyle \Rightarrow 13x-5<15x+4\ \text{and}\ 15x+4<7x+12
\displaystyle \text{Solution of }13x-5<15x+4
\displaystyle \Rightarrow 13x-15x<4+5
\displaystyle \Rightarrow -2x<9\Rightarrow x>-\frac{9}{2}\quad (iii)
\displaystyle \text{Solution of }15x+4<7x+12
\displaystyle \Rightarrow 15x-7x<12-4
\displaystyle \Rightarrow 8x<8\Rightarrow x<1\quad (iv)
\displaystyle \text{From (iii) and (iv), }-\frac{9}{2}<x<1
\displaystyle \therefore\ \text{Solution set}=\{x:-\frac{9}{2}<x<1,\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 27.}\ \text{Find the value of }x,\ \text{which satisfies the inequation. Graph} \\ \text{the solution set on the number line.}
\displaystyle -2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\le 2,\ x\in W\qquad \text{ICSE 2014}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\le 2,\ x\in W
\displaystyle \Rightarrow -\frac{17}{6}<\frac{1}{2}-\frac{2x}{3}\le 2
\displaystyle \Rightarrow -\frac{17}{6}<\frac{3-4x}{6}\le 2
\displaystyle \text{On multiplying throughout by }6,\ \text{we get}
\displaystyle \Rightarrow -17<3-4x\le 12
\displaystyle \Rightarrow -17<3-4x\ \text{and}\ 3-4x\le 12
\displaystyle \Rightarrow 4x<3+17\ \text{and}\ 3-12\le 4x
\displaystyle \Rightarrow 4x<20\ \text{and}\ -9\le 4x
\displaystyle \Rightarrow x<\frac{20}{4}\ \text{and}\ -\frac{9}{4}\le x
\displaystyle \Rightarrow x<5\ \text{and}\ x\ge -\frac{9}{4}
\displaystyle \therefore\ -\frac{9}{4}\le x<5
\displaystyle \text{In set notation form, }\{x:x\in W,\ -\frac{9}{4}\le x<5\}
\displaystyle \therefore\ \text{Required set is }\{0,1,2,3,4\}\ [\text{as }x\in W]
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 28.}\ \text{Solve the following inequation and write the solution set and} \\ \text{represent it on the number line.}
\displaystyle -\frac{x}{3}\le \frac{x}{2}-1\frac{1}{3}<\frac{1}{6},\ x\in R\qquad \text{ICSE 2013}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-\frac{x}{3}\le \frac{x}{2}-1\frac{1}{3}<\frac{1}{6},\ x\in R
\displaystyle \Rightarrow -\frac{x}{3}\le \frac{x}{2}-\frac{4}{3}<\frac{1}{6}
\displaystyle \Rightarrow -\frac{x}{3}\le \frac{3x-8}{6}<\frac{1}{6}
\displaystyle \text{Taking LCM of }(3,6)\ \text{i.e. }6
\displaystyle \text{On multiplying throughout by }6,\ \text{we get}
\displaystyle \Rightarrow -\frac{x}{3}\times 6\le \frac{3x-8}{6}\times 6<\frac{1}{6}\times 6
\displaystyle \Rightarrow -2x\le 3x-8<1
\displaystyle \Rightarrow -2x\le 3x-8\ \text{and}\ 3x-8<1
\displaystyle \text{Now, }-2x\le 3x-8
\displaystyle \Rightarrow -2x+2x+8\le 3x-8+2x+8
\displaystyle \Rightarrow 8\le 5x\Rightarrow x\ge \frac{8}{5}\quad (i)
\displaystyle \text{Also, }3x-8<1
\displaystyle \Rightarrow 3x-8+8<1+8
\displaystyle \Rightarrow 3x<9\Rightarrow x<3\quad (ii)
\displaystyle \text{From (i) and (ii), }\frac{8}{5}\le x<3
\displaystyle \therefore\ \text{Solution set}=\{x:\frac{8}{5}\le x<3,\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 29.}\ \text{Solve the following inequation and represent the solution set on} \\ \text{the number line.}
\displaystyle 4x-19<\frac{3x}{5}-2\le -\frac{2}{5}+x,\ x\in R\qquad \text{ICSE 2012}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }4x-19<\frac{3x}{5}-2\le -\frac{2}{5}+x,\ x\in R
\displaystyle \Rightarrow 4x-19<\frac{3x}{5}-2\ \text{and}\ \frac{3x}{5}-2\le -\frac{2}{5}+x
\displaystyle \text{Consider, }4x-19<\frac{3x}{5}-2
\displaystyle \Rightarrow 4x-19+(2-4x)<\frac{3x}{5}-2+(2-4x)
\displaystyle \Rightarrow -17<\frac{3x}{5}-4x
\displaystyle \Rightarrow -17<\frac{3x-20x}{5}=-\frac{17x}{5}
\displaystyle \Rightarrow -17\times \frac{-5}{17}>-\frac{17x}{5}\times \frac{-5}{17}
\displaystyle \Rightarrow 5>x\Rightarrow x<5\quad (i)
\displaystyle \text{Also, }\frac{3x}{5}-2\le -\frac{2}{5}+x
\displaystyle \Rightarrow \frac{3x}{5}-2+(2-x)\le -\frac{2}{5}+x+(2-x)
\displaystyle \Rightarrow \frac{3x}{5}-x\le -\frac{2}{5}+2
\displaystyle \Rightarrow \frac{3x-5x}{5}\le \frac{-2+10}{5}
\displaystyle \Rightarrow -\frac{2x}{5}\le \frac{8}{5}
\displaystyle \Rightarrow x\ge -4\quad (ii)
\displaystyle \text{From (i) and (ii), }-4\le x<5
\displaystyle \therefore\ \text{Solution set}=\{x:-4\le x<5,\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 30.}\ \text{Solve the following inequation and represent the solution set} \\ \text{on the number line.}
\displaystyle 2x-5\le 5x+4<11,\ x\in I\qquad \text{ICSE 2011, 02}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }2x-5\le 5x+4<11,\ x\in I
\displaystyle \Rightarrow 2x-5\le 5x+4\ \text{and}\ 5x+4<11
\displaystyle \text{Consider, }2x-5\le 5x+4
\displaystyle \Rightarrow (2x-5)+(-2x-4)\le 5x+4+(-2x-4)
\displaystyle \Rightarrow -9\le 3x\Rightarrow x\ge -\frac{9}{3}
\displaystyle \Rightarrow x\ge -3\quad (i)
\displaystyle \text{Also, }5x+4<11
\displaystyle \Rightarrow 5x+4-4<11-4
\displaystyle \Rightarrow 5x<7\Rightarrow x<\frac{7}{5}\quad (ii)
\displaystyle \text{From (i) and (ii), }-3\le x<\frac{7}{5}
\displaystyle \text{In set notation, }\{x:-3\le x<\frac{7}{5},\ x\in I\}
\displaystyle \therefore\ \text{Required solution set is }\{-3,-2,-1,0,1\}\ [x\in I]
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 31.}\ \text{Solve the following inequation and represent the solution set} \\ \text{on the number line.}
\displaystyle -3<\frac{1}{2}-\frac{2x}{3}\le \frac{5}{6},\ x\in R\qquad \text{ICSE 2010}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-3<-\frac{1}{2}-\frac{2x}{3}\le \frac{5}{6},\ x\in R
\displaystyle \Rightarrow -3<-\frac{1}{2}-\frac{2x}{3}\ \text{and}\ -\frac{1}{2}-\frac{2x}{3}\le \frac{5}{6}
\displaystyle \text{Consider, }-3<-\frac{1}{2}-\frac{2x}{3}=-\frac{3+4x}{6}
\displaystyle \Rightarrow -3<-\frac{3+4x}{6}
\displaystyle \Rightarrow -18<-3-4x
\displaystyle \Rightarrow -18+3<-4x
\displaystyle \Rightarrow -15<-4x\Rightarrow 15>4x
\displaystyle \Rightarrow x<\frac{15}{4}\quad (i)
\displaystyle \text{Also, }-\frac{1}{2}-\frac{2x}{3}\le \frac{5}{6}
\displaystyle \Rightarrow -\frac{3+4x}{6}\le \frac{5}{6}
\displaystyle \Rightarrow -3-4x\le 5
\displaystyle \Rightarrow -3-4x+3\le 5+3
\displaystyle \Rightarrow -4x\le 8\Rightarrow x\ge -2\quad (ii)
\displaystyle \text{From (i) and (ii), }-2\le x<\frac{15}{4}
\displaystyle \therefore\ \text{Solution set}=\{x:-2\le x<\frac{15}{4},\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 32.}\ \text{Solve the inequation and represent the solution set on the number line.}
\displaystyle -3+x\le \frac{8x}{3}+2\le \frac{14}{3}+2x,\ x\in I\qquad \text{ICSE 2009}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-3+x\le \frac{8x}{3}+2\le \frac{14}{3}+2x,\ x\in I
\displaystyle \Rightarrow -3+x\le \frac{8x}{3}+2\ \text{and}\ \frac{8x}{3}+2\le \frac{14}{3}+2x
\displaystyle \text{Consider, }-3+x\le \frac{8x}{3}+2=\frac{8x+6}{3}
\displaystyle \Rightarrow -9+3x\le 8x+6
\displaystyle \Rightarrow -9+3x-3x-6\le 8x+6-3x-6
\displaystyle \Rightarrow -15\le 5x\ \text{or}\ 5x\ge -15
\displaystyle \Rightarrow x\ge \frac{-15}{5}
\displaystyle \Rightarrow x\ge -3\quad (i)
\displaystyle \text{Also, }\frac{8x}{3}+2\le \frac{14}{3}+2x
\displaystyle \Rightarrow \frac{8x+6}{3}\le \frac{14+6x}{3}
\displaystyle \Rightarrow 8x+6\le 14+6x
\displaystyle \Rightarrow 8x+6-(6x+6)\le 14+6x-(6x+6)
\displaystyle \Rightarrow 2x\le 8
\displaystyle \Rightarrow x\le 4\quad (ii)
\displaystyle \text{From (i) and (ii), }-3\le x\le 4
\displaystyle \text{In set notation, }\{x:-3\le x\le 4,\ x\in I\}
\displaystyle \therefore\ \text{Required solution set is }\{-3,-2,-1,0,1,2,3,4\}\ [x\in I]
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 33.}\ \text{Solve the given inequation and graph the solution set on the number line.}
\displaystyle 2y-3<y+1\le 4y+7,\ y\in R\qquad \text{ICSE 2008}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }2y-3<y+1\le 4y+7,\ y\in R
\displaystyle \Rightarrow 2y-3<y+1\ \text{and}\ y+1\le 4y+7
\displaystyle \text{Consider, }2y-3<y+1
\displaystyle \Rightarrow 2y-3+(3-y)<y+1+(3-y)
\displaystyle \Rightarrow y<4\quad (i)
\displaystyle \text{Also, }y+1\le 4y+7
\displaystyle \Rightarrow y+1-(7+y)\le 4y+7-(7+y)
\displaystyle \Rightarrow -6\le 3y\ \text{or}\ 3y\ge -6
\displaystyle \Rightarrow y\ge -\frac{6}{3}
\displaystyle \Rightarrow y\ge -2\quad (ii)
\displaystyle \text{From (i) and (ii), }-2\le y<4
\displaystyle \therefore\ \text{Solution set}=\{y:-2\le y<4,\ y\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 34.}\ \text{Solve the following inequation and graph the solution set on the} \\ \text{number line.}
\displaystyle -2\frac{2}{3}\le x+\frac{1}{3}<3\frac{1}{3},\ x\in R\qquad \text{ICSE 2007}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-2\frac{2}{3}\le x+\frac{1}{3}<3\frac{1}{3},\ x\in R
\displaystyle \Rightarrow -\frac{8}{3}\le x+\frac{1}{3}<\frac{10}{3}
\displaystyle \Rightarrow -\frac{8}{3}\le x+\frac{1}{3}\ \text{and}\ x+\frac{1}{3}<\frac{10}{3}
\displaystyle \text{Consider, }-\frac{8}{3}\le x+\frac{1}{3}=\frac{3x+1}{3}
\displaystyle \Rightarrow -8\le 3x+1
\displaystyle \Rightarrow -8-1\le 3x+1-1
\displaystyle \Rightarrow -9\le 3x
\displaystyle \Rightarrow -3\le x\Rightarrow x\ge -3\quad (i)
\displaystyle \text{Also, }x+\frac{1}{3}<3\frac{1}{3}=\frac{10}{3}
\displaystyle \Rightarrow \frac{3x+1}{3}<\frac{10}{3}
\displaystyle \Rightarrow 3x+1<10
\displaystyle \Rightarrow 3x+1-1<10-1
\displaystyle \Rightarrow 3x<9
\displaystyle \Rightarrow x<3\quad (ii)
\displaystyle \text{From (i) and (ii), }-3\le x<3
\displaystyle \therefore\ \text{Solution set}=\{x:-3\le x<3,\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 35.}\ \text{Given that }x\in R,\ \text{solve the following inequality and graph} \\ \text{the solution set on the number line.}
\displaystyle -1\le 3+4x<23\qquad \text{ICSE 2006}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-1\le 3+4x<23,\ x\in R
\displaystyle \Rightarrow -1\le 3+4x\ \text{and}\ 3+4x<23
\displaystyle \text{Consider, }-1\le 3+4x
\displaystyle \Rightarrow -1-3\le 3+4x-3
\displaystyle \Rightarrow -4\le 4x
\displaystyle \Rightarrow -1\le x\Rightarrow x\ge -1\quad (i)
\displaystyle \text{Also, }3+4x<23
\displaystyle \Rightarrow 3+4x-3<23-3
\displaystyle \Rightarrow 4x<20
\displaystyle \Rightarrow x<5\quad (ii)
\displaystyle \text{From (i) and (ii), }-1\le x<5
\displaystyle \therefore\ \text{Solution set}=\{x:-1\le x<5,\ x\in R\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 36.}\ \text{If }A=\{x:1x-5>7x+3,\ x\in R\}\ \text{and }B=\{x:18x-9\ge 15+12x,\ x\in R\},\ \text{find }A\cap B\ \text{and represent it on the number line.}\qquad \text{ICSE 2005}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, }A=\{x:11x-5>7x+3,\ x\in R\}\ \text{and }B=\{x:18x-9\ge 15+12x,\ x\in R\}
\displaystyle \text{Now, }11x-5>7x+3
\displaystyle \Rightarrow 11x-5+(-7x+5)>7x+3+(-7x+5)
\displaystyle \Rightarrow 4x>8\Rightarrow x>\frac{8}{4}
\displaystyle \Rightarrow x>2
\displaystyle \text{Also, }18x-9\ge 15+12x
\displaystyle \Rightarrow 18x-9+(9-12x)\ge 15+12x+(9-12x)
\displaystyle \Rightarrow 6x\ge 24\Rightarrow x\ge \frac{24}{6}
\displaystyle \Rightarrow x\ge 4
\displaystyle \therefore\ A\cap B=\{x:x>2\ \text{and}\ x\ge 4\}=\{x:x\ge 4\}
\displaystyle \therefore\ \text{Hence, the range of the set is }x\ge 4.
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 37.}\ \text{Given that }x\in I,\ \text{solve the inequation and graph the solution on the} \\ \text{number line.}
\displaystyle 3\ge \frac{x-4}{2}+\frac{x}{3}\ge 2\qquad \text{ICSE 2004}
\displaystyle \text{Answer:}
\displaystyle 3\ge \frac{3(x-4)+2x}{6}\ge 2
\displaystyle 3\ge \frac{5x-12}{6}\ge 2
\displaystyle \Rightarrow 18\ge 5x-12\ge 12
\displaystyle \Rightarrow 30\ge 5x\ge 24
\displaystyle \Rightarrow 6\ge x\ge \frac{24}{5}
\displaystyle \Rightarrow \frac{24}{5}\le x\le 6
\displaystyle \text{Since }x\in I,\ \text{solution set}=\{5,6\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 38.}\ \text{Solve the following inequation and represent the solution set on a} \\ \text{number line.}
\displaystyle -8\frac{1}{2}<-\frac{1}{2}-4x\le 7\frac{1}{2},\ x\in I\qquad \text{ICSE 2018}
\displaystyle \text{Answer:}
\displaystyle  \text{We have, }-8\frac{1}{2}<-\frac{1}{2}-4x\le 7\frac{1}{2}
\displaystyle \text{By adding throughout by }\frac{1}{2},\ \text{we get}
\displaystyle -8<-4x\le 8
\displaystyle \text{By dividing throughout by }-4,\ \text{we get}
\displaystyle -2\le x<2
\displaystyle \text{It is given that }x\in I
\displaystyle \therefore\ \text{Solution set is }\{x:-2\le x<2,\ x\in I\}=\{-2,-1,0,1\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 39.}\ \text{Given: }A=\{x:3<2x-1<9,\ x\in R\},\ B=\{x:11\le 3x+2\le 23,\ x\in R\}\ \text{where }R\ \text{is the set of real numbers.}
\displaystyle (i)\ \text{Represent }A\ \text{and }B\ \text{on number lines.}\qquad (ii)\ \text{On the number line also mark }A\cap B.\qquad \textbf{ICSE 2016}
\displaystyle \text{Answer:}
\displaystyle  \ (i)\ A=\{x:3<2x-1<9,\ x\in R\}
\displaystyle \text{Adding }1\text{ on both sides and after that dividing by }2,\ \text{we get}
\displaystyle -2<x<5
\displaystyle \text{And }B=\{x:11\le 3x+2\le 23,\ x\in R\}
\displaystyle \Rightarrow 11\le 3x+2\le 23
\displaystyle \text{Subtracting }2\text{ from both sides and then dividing by }3
\displaystyle \Rightarrow 3\le x\le 7
\displaystyle \text{Hence, on number lines}
\displaystyle A\Rightarrow -2<x<5
\displaystyle B\Rightarrow 3\le x\le 7
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}


\displaystyle (ii)\ \therefore\ A\cap B=3\le x<5
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

\displaystyle \textbf{Question 40.}\ \text{Find the value of }x,\ \text{which satisfies the inequation. Graph} \\ \text{the solution set on the number line.}
\displaystyle -2\frac{5}{2}-\frac{2x}{3}\le \frac{5}{6},\ x\in N\qquad \text{ICSE 2001}
\displaystyle \text{Answer:}
\displaystyle  \text{Given, inequation is }-2\le \frac{1}{2}-\frac{2x}{3}\le \frac{5}{6},\ x\in N
\displaystyle \Rightarrow -2\le \frac{1}{2}-\frac{2x}{3}\ \text{and}\ \frac{1}{2}-\frac{2x}{3}\le \frac{5}{6}
\displaystyle \text{Consider, }-2\le \frac{1}{2}-\frac{2x}{3}=\frac{3-4x}{6}
\displaystyle \Rightarrow -2\le \frac{3-4x}{6}
\displaystyle \Rightarrow -2\times 6\le 3-4x
\displaystyle \Rightarrow -12-3\le 3-4x-3
\displaystyle \Rightarrow -15\le -4x\Rightarrow \frac{15}{4}\ge x\ \text{or}\ x\le \frac{15}{4}\quad (i)
\displaystyle \text{Also, }\frac{1}{2}-\frac{2x}{3}\le \frac{5}{6}\Rightarrow \frac{3-4x}{6}\le \frac{5}{6}
\displaystyle \Rightarrow 3-4x\le 5
\displaystyle \Rightarrow 3-4x-3\le 5-3
\displaystyle \Rightarrow -4x\le 2\Rightarrow x\ge \frac{8}{-4}
\displaystyle \Rightarrow x\ge -2\quad (ii)
\displaystyle \text{From (i) and (ii), }-2\le x\le \frac{15}{4}
\displaystyle \text{But }x\in N,\ \text{solution set is }\{1,2,3\}
\displaystyle \text{The graph of the soution set on the number line is shown by the dark line}

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