Class 10: Linear Inequations

$\displaystyle \textbf{Question 18: }P\text{ is the solution set of }7x-2>4x+1\text{ and }Q\text{ is the}$
$\displaystyle \text{solution set of }9x-45\leq5(x-5),\text{ where }x\in R.\text{ Represent }$
$\displaystyle \text{(i) }P\cap Q\qquad\text{(ii) }P-Q\qquad\text{(iii) }P\cap Q'\text{ on different number lines.}$
$\displaystyle \text{Answer:}$
$\displaystyle P:7x-2>4x+1$
$\displaystyle \Rightarrow 3x>3$
$\displaystyle \Rightarrow x>1$
$\displaystyle \therefore P=\{x:x\in R\text{ and }x>1\}$
$\displaystyle Q:9x-45\leq5(x-5)$
$\displaystyle \Rightarrow 9x-45\leq5x-25$
$\displaystyle \Rightarrow 4x\leq20$
$\displaystyle \Rightarrow x\leq5$
$\displaystyle \therefore Q=\{x:x\in R\text{ and }x\leq5\}$
$\displaystyle \text{(i) }P\cap Q=\{x:x\in R\text{ and }1<x\leq5\}$
$\displaystyle \text{(ii) }P-Q=\{x:x\in R\text{ and }x>5\}$
$\displaystyle \text{(iii) }Q'=\{x:x\in R\text{ and }x>5\}$
$\displaystyle \therefore P\cap Q'=\{x:x\in R\text{ and }x>5\}$

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$\displaystyle \textbf{Question 19: }\text{If }P=\{x:7x-4>5x+2,\ x\in R\}\text{ and }Q=\{x:x-19\geq1-3x,\ x\in R\},$
$\displaystyle \text{find the range of the set }P\cap Q\text{ and represent it on the number line.}$
$\displaystyle \text{Answer:}$
$\displaystyle P:7x-4>5x+2$
$\displaystyle \Rightarrow 2x>6$
$\displaystyle \Rightarrow x>3$
$\displaystyle \therefore P=\{x:x\in R\text{ and }x>3\}$
$\displaystyle Q:x-19\geq1-3x$
$\displaystyle \Rightarrow 4x\geq20$
$\displaystyle \Rightarrow x\geq5$
$\displaystyle \therefore Q=\{x:x\in R\text{ and }x\geq5\}$
$\displaystyle P\cap Q=\{x:x\in R\text{ and }x\geq5\}$
$\displaystyle \text{Hence, the range of }P\cap Q\text{ is }[5,\infty).$
$\displaystyle \text{Number line: Closed circle at }5\text{ and shade the region to the right.}$
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$\displaystyle \textbf{Question 20: }\text{Find the range of values of }x\text{ which satisfy}$
$\displaystyle -\frac{1}{3}\leq\frac{x}{2}+1\frac{2}{3}<5\frac{1}{6}.$
$\displaystyle \text{Graph the values of }x\text{ for each of the following cases:}$
$\displaystyle \text{(i) }x\in W\qquad\text{(ii) }x\in Z\qquad\text{(iii) }x\in R$
$\displaystyle \text{Answer:}$
$\displaystyle -\frac{1}{3}\leq\frac{x}{2}+1\frac{2}{3}<5\frac{1}{6}$
$\displaystyle \Rightarrow -\frac{1}{3}\leq\frac{x}{2}+\frac{5}{3}<\frac{31}{6}$
$\displaystyle \Rightarrow -2\leq\frac{x}{2}<\frac{7}{2}$
$\displaystyle \Rightarrow -4\leq x<7$
$\displaystyle \text{(i) If }x\in W,\text{ then }x\in\{0,1,2,3,4,5,6\}$
$\displaystyle \text{(ii) If }x\in Z,\text{ then }x\in\{-4,-3,-2,-1,0,1,2,3,4,5,6\}$
$\displaystyle \text{(iii) If }x\in R,\text{ then }\{x:x\in R\text{ and }-4\leq x<7\}$
$\displaystyle \text{Number lines:}$

i) $x \in W$ : $0 \leq x < 7$

ii) $x \in Z$ : $-4 \leq x < 7$

iii) $x \in R$ : $-4 \leq x < 7$

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$\displaystyle \textbf{Question 21: }\text{Given }A=\{x:-8<5x+2\leq17,\ x\in I\},$
$\displaystyle B=\{x:-2\leq7+3x<17,\ x\in R\},\text{ where }R\text{ is the set of real}$
$\displaystyle \text{numbers and }I\text{ is the set of integers. Represent }A\text{ and }B$
$\displaystyle \text{on two different number lines. Also write down the elements of }A\cap B.$
$\displaystyle \text{Answer:}$
$\displaystyle A:-8<5x+2\leq17$
$\displaystyle \Rightarrow -10<5x\leq15$
$\displaystyle \Rightarrow -2<x\leq3$
$\displaystyle \therefore A=\{-1,0,1,2,3\}$
$\displaystyle B:-2\leq7+3x<17$
$\displaystyle \Rightarrow -9\leq3x<10$
$\displaystyle \Rightarrow -3\leq x<\frac{10}{3}$
$\displaystyle \therefore B=\{x:x\in R\text{ and }-3\leq x<\frac{10}{3}\}$
$\displaystyle A\cap B=\{-1,0,1,2,3\}$
$\displaystyle \text{Number line for }A:\text{ Plot the points }-1,0,1,2,3.$
$\displaystyle \text{Number line for }B:\text{ Draw a closed circle at }-3,$
$\displaystyle \text{an open circle at }\frac{10}{3}\text{ and shade the interval between them.}$
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$\displaystyle \textbf{Question 22: }\text{Solve the inequation and graph the solution on a number line }$
$\displaystyle 2x-5\leq5x+4<11,\text{ where }x\in I.\hfill\text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2x-5\leq5x+4<11$
$\displaystyle \text{Equation 1: }2x-5\leq5x+4$
$\displaystyle \Rightarrow -9\leq3x$
$\displaystyle \Rightarrow -3\leq x$
$\displaystyle \text{Equation 2: }5x+4<11$
$\displaystyle \Rightarrow 5x<7$
$\displaystyle \Rightarrow x<\frac{7}{5}$
$\displaystyle \therefore \{x:x\in I\text{ and }-3\leq x<\frac{7}{5}\}$
$\displaystyle \text{or }x\in\{-3,-2,-1,0,1\}$
$\displaystyle \text{Number line: Plot the points }-3,-2,-1,0\text{ and }1\text{ on the number line.}$
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$\displaystyle \textbf{Question 23: }\text{Given that }x\in I,\text{ solve the inequation and graph it on}$
$\displaystyle \text{a number line: }3\geq\frac{x-4}{2}+\frac{x}{3}\geq2.\hfill\text{[ICSE 2004]}$
$\displaystyle \text{Answer:}$
$\displaystyle 3\geq\frac{x-4}{2}+\frac{x}{3}\geq2$
$\displaystyle \Rightarrow 2\leq\frac{x-4}{2}+\frac{x}{3}\leq3$
$\displaystyle \Rightarrow 2\leq\frac{3(x-4)+2x}{6}\leq3$
$\displaystyle \Rightarrow 2\leq\frac{5x-12}{6}\leq3$
$\displaystyle \Rightarrow 12\leq5x-12\leq18$
$\displaystyle \Rightarrow 24\leq5x\leq30$
$\displaystyle \Rightarrow \frac{24}{5}\leq x\leq6$
$\displaystyle \therefore \{x:x\in I\text{ and }\frac{24}{5}\leq x\leq6\}$
$\displaystyle \text{or }x\in\{5,6\}$
$\displaystyle \text{Number line: Plot the points }5\text{ and }6\text{ on the number line.}$
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$\displaystyle \textbf{Question 24: }\text{Given }A=\{x:11x-5>7x+3,\ x\in R\},$
$\displaystyle B=\{x:18x-9\geq15+12x,\ x\in R\}.\text{ Find the range of the set }A\cap B$
$\displaystyle \text{and represent it on a number line.}\hfill\text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle A:11x-5>7x+3$
$\displaystyle \Rightarrow 4x>8$
$\displaystyle \Rightarrow x>2$
$\displaystyle \therefore A=\{x:x\in R\text{ and }x>2\}$
$\displaystyle B:18x-9\geq15+12x$
$\displaystyle \Rightarrow 6x\geq24$
$\displaystyle \Rightarrow x\geq4$
$\displaystyle \therefore B=\{x:x\in R\text{ and }x\geq4\}$
$\displaystyle \therefore A\cap B=\{x:x\in R\text{ and }x\geq4\}$
$\displaystyle \text{Hence, the range of }A\cap B\text{ is }[4,\infty).$
$\displaystyle \text{Number line: Draw a closed circle at }4\text{ and shade the region to the right.}$
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$\displaystyle \textbf{Question 25: }\text{Find the set of values of }x\text{ satisfying }7x+3\geq3x-5$
$\displaystyle \text{and }\frac{x}{4}-5\leq\frac{5}{4}-x,\text{ where }x\in N.$
$\displaystyle \text{Answer:}$
$\displaystyle 7x+3\geq3x-5$
$\displaystyle \Rightarrow 4x\geq-8$
$\displaystyle \Rightarrow x\geq-2$
$\displaystyle \frac{x}{4}-5\leq\frac{5}{4}-x$
$\displaystyle \Rightarrow x-20\leq5-4x$
$\displaystyle \Rightarrow 5x\leq25$
$\displaystyle \Rightarrow x\leq5$
$\displaystyle \therefore -2\leq x\leq5$
$\displaystyle \text{Since }x\in N,\text{ the solution set is }\{1,2,3,4,5\}.$
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$\displaystyle \textbf{Question 26: }\text{Solve:}$
$\displaystyle \text{(i) }\frac{x}{2}+5\leq\frac{x}{3}+6,\text{ where }x\text{ is a positive odd integer.}$
$\displaystyle \text{(ii) }\frac{2x+3}{3}\geq\frac{3x-1}{4},\text{ where }x\text{ is a positive even integer.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }\frac{x}{2}+5\leq\frac{x}{3}+6$
$\displaystyle \Rightarrow \frac{x}{2}-\frac{x}{3}\leq1$
$\displaystyle \Rightarrow \frac{x}{6}\leq1$
$\displaystyle \Rightarrow x\leq6$
$\displaystyle \text{Since }x\text{ is a positive odd integer, }x\in\{1,3,5\}$
$\displaystyle \text{(ii) }\frac{2x+3}{3}\geq\frac{3x-1}{4}$
$\displaystyle \Rightarrow 4(2x+3)\geq3(3x-1)$
$\displaystyle \Rightarrow 8x+12\geq9x-3$
$\displaystyle \Rightarrow x\leq15$
$\displaystyle \text{Since }x\text{ is a positive even integer, }x\in\{2,4,6,8,10,12,14\}$
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$\displaystyle \textbf{Question 27: }\text{Solve the inequation }-2\frac{1}{2}+2x\leq\frac{4x}{5}\leq\frac{4}{3}+2x,$
$\displaystyle \text{where }x\in W.\text{ Graph the solution on a number line.}$
$\displaystyle \text{Answer:}$
$\displaystyle -2\frac{1}{2}+2x\leq\frac{4x}{5}\leq\frac{4}{3}+2x$
$\displaystyle \Rightarrow -\frac{5}{2}+2x\leq\frac{4x}{5}\leq\frac{4}{3}+2x$
$\displaystyle \text{Equation 1: }-\frac{5}{2}+2x\leq\frac{4x}{5}$
$\displaystyle \Rightarrow -25+20x\leq8x$
$\displaystyle \Rightarrow 12x\leq25$
$\displaystyle \Rightarrow x\leq\frac{25}{12}$
$\displaystyle \text{Equation 2: }\frac{4x}{5}\leq\frac{4}{3}+2x$
$\displaystyle \Rightarrow 12x\leq20+30x$
$\displaystyle \Rightarrow -18x\leq20$
$\displaystyle \Rightarrow x\geq-\frac{10}{9}$
$\displaystyle \therefore -\frac{10}{9}\leq x\leq\frac{25}{12}$
$\displaystyle \text{Since }x\in W,\text{ the solution set is }\{0,1,2\}.$
$\displaystyle \text{Number line: Plot the points }0,1\text{ and }2\text{ on the number line.}$

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$\displaystyle \textbf{Question 28: }\text{Find three consecutive largest positive integers such that}$
$\displaystyle \text{the sum of one third of the first, one fourth of the second and}$
$\displaystyle \text{one fifth of the third is less than or equal to }20.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three consecutive positive integers be }x,\ x+1,\ x+2.$
$\displaystyle \frac{x}{3}+\frac{x+1}{4}+\frac{x+2}{5}\leq20$
$\displaystyle \Rightarrow 20x+15(x+1)+12(x+2)\leq1200$
$\displaystyle \Rightarrow 20x+15x+15+12x+24\leq1200$
$\displaystyle \Rightarrow 47x+39\leq1200$
$\displaystyle \Rightarrow 47x\leq1161$
$\displaystyle \Rightarrow x\leq\frac{1161}{47}$
$\displaystyle \Rightarrow x\leq24.70$
$\displaystyle \text{Since }x\text{ is a positive integer, the largest possible value of }x=24.$
$\displaystyle \therefore \text{The three consecutive largest positive integers are }24,\ 25,\ 26.$
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$\displaystyle \textbf{Question 29: }\text{Solve the given inequation and graph it on a number line:}$
$\displaystyle 2y-3<y+1\leq4y+7,\text{ where }y\in R.\hfill\text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2y-3<y+1\leq4y+7$
$\displaystyle \text{Equation 1: }2y-3<y+1$
$\displaystyle \Rightarrow y<4$
$\displaystyle \text{Equation 2: }y+1\leq4y+7$
$\displaystyle \Rightarrow -6\leq3y$
$\displaystyle \Rightarrow -2\leq y$
$\displaystyle \therefore -2\leq y<4$
$\displaystyle \therefore \{y:y\in R\text{ and }-2\leq y<4\}$
$\displaystyle \text{Number line: Draw a closed circle at }-2,\text{ an open circle at }4,$
$\displaystyle \text{and shade the interval between them.}$
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$\displaystyle \textbf{Question 30: }\text{Solve the inequation }3z-5\leq z+3<5z-9,\text{ where }z\in R.$
$\displaystyle \text{Graph the solution set on a number line.}$
$\displaystyle \text{Answer:}$
$\displaystyle 3z-5\leq z+3<5z-9$
$\displaystyle \text{Equation 1: }3z-5\leq z+3$
$\displaystyle \Rightarrow 2z\leq8$
$\displaystyle \Rightarrow z\leq4$
$\displaystyle \text{Equation 2: }z+3<5z-9$
$\displaystyle \Rightarrow 12<4z$
$\displaystyle \Rightarrow z>3$
$\displaystyle \therefore 3<z\leq4$
$\displaystyle \therefore \{z:z\in R\text{ and }3<z\leq4\}$
$\displaystyle \text{Number line: Draw an open circle at }3,\text{ a closed circle at }4,$
$\displaystyle \text{and shade the interval between them.}$
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$\displaystyle \textbf{Question 31: }\text{Solve the given inequation and graph it on a number line:}$
$\displaystyle -3<-\frac{1}{2}-\frac{2x}{3}\leq\frac{5}{6},\text{ where }x\in R.\hfill\text{[ICSE 2010]}$
$\displaystyle \text{Answer:}$
$\displaystyle -3<-\frac{1}{2}-\frac{2x}{3}\leq\frac{5}{6}$
$\displaystyle \text{Equation 1: }-3<-\frac{1}{2}-\frac{2x}{3}$
$\displaystyle \Rightarrow -\frac{5}{2}<-\frac{2x}{3}$
$\displaystyle \Rightarrow -15<-4x$
$\displaystyle \Rightarrow x<\frac{15}{4}$
$\displaystyle \text{Equation 2: }-\frac{1}{2}-\frac{2x}{3}\leq\frac{5}{6}$
$\displaystyle \Rightarrow -\frac{2x}{3}\leq\frac{4}{3}$
$\displaystyle \Rightarrow -2x\leq4$
$\displaystyle \Rightarrow x\geq-2$
$\displaystyle \therefore -2\leq x<\frac{15}{4}$
$\displaystyle \therefore \{x:x\in R\text{ and }-2\leq x<\frac{15}{4}\}$
$\displaystyle \text{Number line: Draw a closed circle at }-2,\text{ an open circle at }\frac{15}{4},$
$\displaystyle \text{and shade the interval between them.}$

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$\displaystyle \textbf{Question 32: }\text{Solve the given inequation and graph it on a number line:}$
$\displaystyle 4x-19<\frac{3x}{5}-2\leq-\frac{2}{5}+x,\text{ where }x\in R.\hfill\text{[ICSE 2012]}$
$\displaystyle \text{Answer:}$
$\displaystyle 4x-19<\frac{3x}{5}-2\leq-\frac{2}{5}+x$
$\displaystyle \text{Equation 1: }4x-19<\frac{3x}{5}-2$
$\displaystyle \Rightarrow 20x-95<3x-10$
$\displaystyle \Rightarrow 17x<85$
$\displaystyle \Rightarrow x<5$
$\displaystyle \text{Equation 2: }\frac{3x}{5}-2\leq-\frac{2}{5}+x$
$\displaystyle \Rightarrow 3x-10\leq-2+5x$
$\displaystyle \Rightarrow -8\leq2x$
$\displaystyle \Rightarrow x\geq-4$
$\displaystyle \therefore -4\leq x<5$
$\displaystyle \therefore \{x:x\in R\text{ and }-4\leq x<5\}$
$\displaystyle \text{Number line: Draw a closed circle at }-4,\text{ an open circle at }5,$
$\displaystyle \text{and shade the interval between them.}$

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$\displaystyle \textbf{Question 33: }\text{Solve the given inequation and graph it on a number line:}$
$\displaystyle -\frac{x}{3}\leq\frac{x}{2}-1\frac{1}{3}<\frac{1}{6},\text{ where }x\in R.\hfill\text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle -\frac{x}{3}\leq\frac{x}{2}-1\frac{1}{3}<\frac{1}{6}$
$\displaystyle \Rightarrow -\frac{x}{3}\leq\frac{x}{2}-\frac{4}{3}<\frac{1}{6}$
$\displaystyle \text{Equation 1: }-\frac{x}{3}\leq\frac{x}{2}-\frac{4}{3}$
$\displaystyle \Rightarrow -2x\leq3x-8$
$\displaystyle \Rightarrow 8\leq5x$
$\displaystyle \Rightarrow x\geq\frac{8}{5}$
$\displaystyle \text{Equation 2: }\frac{x}{2}-\frac{4}{3}<\frac{1}{6}$
$\displaystyle \Rightarrow 3x-8<1$
$\displaystyle \Rightarrow 3x<9$
$\displaystyle \Rightarrow x<3$
$\displaystyle \therefore \frac{8}{5}\leq x<3$
$\displaystyle \therefore \{x:x\in R\text{ and }\frac{8}{5}\leq x<3\}$
$\displaystyle \text{Number line: Draw a closed circle at }\frac{8}{5},\text{ an open circle at }3,$
$\displaystyle \text{and shade the interval between them.}$
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$\displaystyle \textbf{Question 34: }\text{Find the value of }x\text{ which satisfies the inequation:}$
$\displaystyle -2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\leq2,\text{ where }x\in W.\hfill\text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle -2\frac{5}{6}<\frac{1}{2}-\frac{2x}{3}\leq2$
$\displaystyle \Rightarrow -\frac{17}{6}<\frac{1}{2}-\frac{2x}{3}\leq2$
$\displaystyle \text{Equation 1: }-\frac{17}{6}<\frac{1}{2}-\frac{2x}{3}$
$\displaystyle \Rightarrow -\frac{20}{6}<-\frac{2x}{3}$
$\displaystyle \Rightarrow -10<-2x$
$\displaystyle \Rightarrow x<5$
$\displaystyle \text{Equation 2: }\frac{1}{2}-\frac{2x}{3}\leq2$
$\displaystyle \Rightarrow -\frac{2x}{3}\leq\frac{3}{2}$
$\displaystyle \Rightarrow -4x\leq9$
$\displaystyle \Rightarrow x\geq-\frac{9}{4}$
$\displaystyle \therefore -\frac{9}{4}\leq x<5$
$\displaystyle \text{Since }x\in W,\text{ the solution set is }\{0,1,2,3,4\}.$
$\displaystyle \text{Hence, the values of }x\text{ are }0,\ 1,\ 2,\ 3\text{ and }4.$