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

\displaystyle \textbf{Question 1: }\text{State True or False:}

\displaystyle \text{Answer:}

\displaystyle \text{(i) }x<-y\Rightarrow -x>y\hfill\text{True} 

\displaystyle \text{(ii) }-5x\geq15\Rightarrow x\geq-3\hfill\text{False} 

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

\displaystyle \text{(iv) }7>5\Rightarrow\frac{1}{7}<\frac{1}{5}\hfill\text{True} 

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\displaystyle \textbf{Question 2: }\text{State True or False: Given that }a,b,c\text{ and }d
\displaystyle \text{are real numbers and }c\neq0.
\displaystyle \text{Answer:}
\displaystyle \text{(i) If }a<b,\text{ then }a-c<b-c\hfill\text{True}

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

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

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

\displaystyle \text{(v) If }a-c>b-d,\text{ then }a+d>b+c\hfill\text{True}

\displaystyle \text{(vi) If }a<b\text{ and }c>0,\text{ then }a-c>b-c\hfill\text{False}
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\displaystyle \textbf{Question 3: }\text{If }x\in N,\text{ find the solution set of the inequations:}
\displaystyle \text{(i) }5x+3\leq2x+18\qquad\qquad\text{(ii) }3x-2<19-4x
\displaystyle \text{Answer:}
\displaystyle \text{(i) }5x+3\leq2x+18
\displaystyle \Rightarrow 3x\leq15
\displaystyle \Rightarrow x\leq5
\displaystyle \therefore x\in\{1,2,3,4,5\}
\displaystyle \text{(ii) }3x-2<19-4x
\displaystyle \Rightarrow 7x<21
\displaystyle \Rightarrow x<3
\displaystyle \therefore x\in\{1,2\}
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\displaystyle \textbf{Question 4: }\text{If the replacement set is a set of whole numbers, solve:}
\displaystyle \text{(i) }x+7\leq11\qquad\text{(ii) }3x-1>8
\displaystyle \text{(iii) }x-\frac{3}{2}<\frac{3}{2}-x\qquad\text{(iv) }18\leq3x-2
\displaystyle \text{Answer:}
\displaystyle \text{(i) }x+7\leq11
\displaystyle \Rightarrow x\leq4
\displaystyle \therefore x\in\{0,1,2,3,4\}
\displaystyle \text{(ii) }3x-1>8
\displaystyle \Rightarrow 3x>9
\displaystyle \Rightarrow x>3
\displaystyle \therefore x\in\{4,5,6,\ldots\}
\displaystyle \text{(iii) }x-\frac{3}{2}<\frac{3}{2}-x
\displaystyle \Rightarrow 2x<\frac{3}{2}+\frac{3}{2}
\displaystyle \Rightarrow 2x<3
\displaystyle \therefore x\in\{0,1\}
\displaystyle \text{(iv) }18\leq3x-2
\displaystyle \Rightarrow 3x\geq20
\displaystyle \Rightarrow x\geq\frac{20}{3}
\displaystyle \therefore x\in\{7,8,9,\ldots\}
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\displaystyle \textbf{Question 5: }\text{Solve the inequation }3-2x\geq x-12,\text{ given that }x\in N.\hfill\text{[ICSE 1987]}
\displaystyle \text{Answer:}
\displaystyle 3-2x\geq x-12
\displaystyle \Rightarrow 3+12\geq x+2x
\displaystyle \Rightarrow 15\geq3x
\displaystyle \Rightarrow x\leq5
\displaystyle \therefore x\in\{1,2,3,4,5\}
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\displaystyle \textbf{Question 6: }\text{If }25-4x\leq16,\text{ find:}
\displaystyle \text{(i) the smallest value of }x\text{, when }x\text{ is a real number}
\displaystyle \text{(ii) the smallest value of }x\text{, when }x\text{ is an integer}
\displaystyle \text{Answer:}
\displaystyle 25-4x\leq16
\displaystyle \Rightarrow -4x\leq-9
\displaystyle \Rightarrow x\geq\frac{9}{4}
\displaystyle \text{(i) Smallest real value of }x=\frac{9}{4}=2.25
\displaystyle \text{(ii) Smallest integer value of }x=3
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\displaystyle \textbf{Question 7: }\text{If the replacement set is a set of real numbers, solve:}
\displaystyle \text{(i) }-4x\geq-16\qquad\text{(ii) }8-3x\leq20
\displaystyle \text{(iii) }5+\frac{x}{4}>\frac{x}{5}+9\qquad\text{(iv) }\frac{x+3}{8}<\frac{x-3}{5}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }-4x\geq-16
\displaystyle \Rightarrow 4x\leq16
\displaystyle \Rightarrow x\leq4
\displaystyle \therefore \{x:x\in R\text{ and }x\leq4\}
\displaystyle \text{(ii) }8-3x\leq20
\displaystyle \Rightarrow -3x\leq12
\displaystyle \Rightarrow x\geq-4
\displaystyle \therefore \{x:x\in R\text{ and }x\geq-4\}
\displaystyle \text{(iii) }5+\frac{x}{4}>\frac{x}{5}+9
\displaystyle \Rightarrow \frac{x}{4}-\frac{x}{5}>4
\displaystyle \Rightarrow \frac{x}{20}>4
\displaystyle \Rightarrow x>80
\displaystyle \therefore \{x:x\in R\text{ and }x>80\}
\displaystyle \text{(iv) }\frac{x+3}{8}<\frac{x-3}{5}
\displaystyle \Rightarrow 5x+15<8x-24
\displaystyle \Rightarrow 39<3x
\displaystyle \Rightarrow x>13
\displaystyle \therefore \{x:x\in R\text{ and }x>13\}
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\displaystyle \textbf{Question 8: }\text{Find the smallest value of }x\text{ for which}
\displaystyle 5-2x<5\frac{1}{2}-\frac{5}{3}x,\text{ where }x\in I.
\displaystyle \text{Answer:}
\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}
\displaystyle \therefore \text{Smallest integral value of }x=-1
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\displaystyle \textbf{Question 9: }\text{Find the largest value of }x\text{ for which }2(x-1)\leq(9-x),
\displaystyle \text{where }x\in W.
\displaystyle \text{Answer:}
\displaystyle 2(x-1)\leq(9-x)
\displaystyle \Rightarrow 2x-2\leq9-x
\displaystyle \Rightarrow 3x\leq11
\displaystyle \Rightarrow x\leq\frac{11}{3}
\displaystyle \therefore \text{Largest whole number satisfying the inequation is }x=3
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\displaystyle \textbf{Question 10: }\text{Solve the inequation }12+1\frac{5}{6}x\leq5+3x,\text{ where }x\in R.\hfill\text{[ICSE 1999]}
\displaystyle \text{Answer:}
\displaystyle 12+1\frac{5}{6}x\leq5+3x
\displaystyle \Rightarrow 12+\frac{11}{6}x\leq5+3x
\displaystyle \Rightarrow 7\leq\frac{7}{6}x
\displaystyle \Rightarrow x\geq6
\displaystyle \therefore \{x:x\in R\text{ and }x\geq6\}
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\displaystyle \textbf{Question 11: }\text{Given }x\in I,\text{ find the solution set for }-5\leq2x-3<x+2.
\displaystyle \text{Answer:}
\displaystyle -5\leq2x-3<x+2
\displaystyle \text{Equation 1: }-5\leq2x-3
\displaystyle \Rightarrow -2\leq2x
\displaystyle \Rightarrow -1\leq x
\displaystyle \text{Equation 2: }2x-3<x+2
\displaystyle \Rightarrow x<5
\displaystyle \therefore \{x:x\in I\text{ and }-1\leq x<5\}
\displaystyle \text{or }x\in\{-1,0,1,2,3,4\}
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\displaystyle \textbf{Question 12: }\text{Given }x\in W,\text{ find the solution set for }-1\leq3+4x<23.
\displaystyle \text{Answer:}
\displaystyle -1\leq3+4x<23
\displaystyle \text{Equation 1: }-1\leq3+4x
\displaystyle \Rightarrow -4\leq4x
\displaystyle \Rightarrow -1\leq x
\displaystyle \text{Equation 2: }3+4x<23
\displaystyle \Rightarrow 4x<20
\displaystyle \Rightarrow x<5
\displaystyle \therefore \{x:x\in W\text{ and }-1\leq x<5\}
\displaystyle \text{or }x\in\{0,1,2,3,4\}
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