Class 10: Linear Inequations -Miscellaneous Problems

$\displaystyle \textbf{Question 1: }\text{If the replacement set is the set of natural numbers }(N),$
$\displaystyle \text{find the solution set of:}$
$\displaystyle \text{(i) }3x+4<16\qquad\qquad\text{(ii) }8-x\leq4x-2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }3x+4<16$
$\displaystyle \Rightarrow 3x<16-4$
$\displaystyle \Rightarrow 3x<12$
$\displaystyle \Rightarrow x<4$
$\displaystyle \text{Since the replacement set is }N,\text{ the solution set is }\{1,2,3\}.$
$\displaystyle \text{(ii) }8-x\leq4x-2$
$\displaystyle \Rightarrow -x-4x\leq-2-8$
$\displaystyle \Rightarrow -5x\leq-10$
$\displaystyle \Rightarrow x\geq2$
$\displaystyle \text{Since the replacement set is }N,\text{ the solution set is }\{2,3,4,5,6,\cdots\}.$
$\\$

$\displaystyle \textbf{Question 2: }\text{If the replacement set is the set of whole numbers }(W),$
$\displaystyle \text{find the solution set of:}$
$\displaystyle \text{(i) }5x+4\leq24\qquad\qquad\text{(ii) }4x-2<2x+10$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }5x+4\leq24$
$\displaystyle \Rightarrow 5x\leq24-4$
$\displaystyle \Rightarrow 5x\leq20$
$\displaystyle \Rightarrow x\leq4$
$\displaystyle \text{Since the replacement set is }W,\text{ the solution set is }\{0,1,2,3,4\}.$
$\displaystyle \text{(ii) }4x-2<2x+10$
$\displaystyle \Rightarrow 4x-2x<10+2$
$\displaystyle \Rightarrow 2x<12$
$\displaystyle \Rightarrow x<6$
$\displaystyle \text{Since the replacement set is }W,\text{ the solution set is }\{0,1,2,3,4,5\}.$
$\\$

$\displaystyle \textbf{Question 3: }\text{If the replacement set is the set of integers }(I\text{ or }Z),$
$\displaystyle \text{between }-6\text{ and }8,\text{ find the solution set of:}$
$\displaystyle \text{(i) }6x-1\geq9+x\qquad\qquad\text{(ii) }15-3x>x-3$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }6x-1\geq9+x$
$\displaystyle \Rightarrow 6x-x\geq9+1$
$\displaystyle \Rightarrow 5x\geq10$
$\displaystyle \Rightarrow x\geq2$
$\displaystyle \text{Since the replacement set is }\{-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7\},$
$\displaystyle \text{the solution set is }\{2,3,4,5,6,7\}.$
$\displaystyle \text{(ii) }15-3x>x-3$
$\displaystyle \Rightarrow -3x-x>-3-15$
$\displaystyle \Rightarrow -4x>-18$
$\displaystyle \Rightarrow x<\frac{9}{2}$
$\displaystyle \text{Since the replacement set is }\{-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7\},$
$\displaystyle \text{the solution set is }\{-5,-4,-3,-2,-1,0,1,2,3,4\}.$
$\\$

$\displaystyle \textbf{Question 4: }\text{If the replacement set is the set of real numbers }(R),$
$\displaystyle \text{find the solution set of:}$
$\displaystyle \text{(i) }5-3x<11\qquad\qquad\text{(ii) }8+3x\geq28-2x$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }5-3x<11$
$\displaystyle \Rightarrow -3x<11-5$
$\displaystyle \Rightarrow -3x<6$
$\displaystyle \Rightarrow x>-2$
$\displaystyle \text{Since the replacement set is }R,\text{ the solution set is }$
$\displaystyle \{x:x\in R\text{ and }x>-2\}.$
$\displaystyle \text{(ii) }8+3x\geq28-2x$
$\displaystyle \Rightarrow 3x+2x\geq28-8$
$\displaystyle \Rightarrow 5x\geq20$
$\displaystyle \Rightarrow x\geq4$
$\displaystyle \text{Since the replacement set is }R,\text{ the solution set is }$
$\displaystyle \{x:x\in R\text{ and }x\geq4\}.$
$\\$

$\displaystyle \textbf{Question 5: }\text{Solve }\frac{x}{2}-5\leq\frac{x}{3}-4,\text{ where }x\text{ is a positive odd integer.}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x}{2}-5\leq\frac{x}{3}-4$
$\displaystyle \Rightarrow \frac{x}{2}-\frac{x}{3}\leq-4+5$
$\displaystyle \Rightarrow \frac{3x-2x}{6}\leq1$
$\displaystyle \Rightarrow x\leq6$
$\displaystyle \text{Since }x\text{ is a positive odd integer, the solution set is }\{1,3,5\}.$
$\\$

$\displaystyle \textbf{Question 6: }\text{Solve the following inequation: }2y-3<y+1\leq4y+7.$
$\displaystyle \text{(i) }y\in\{\text{Integers}\}\qquad\text{(ii) }y\in R\text{ (Real Numbers)}$
$\displaystyle \text{Answer:}$
$\displaystyle 2y-3<y+1\leq4y+7$
$\displaystyle \Rightarrow 2y-3<y+1\qquad\text{and}\qquad y+1\leq4y+7$
$\displaystyle \Rightarrow y<4\qquad\qquad\qquad\text{and}\qquad -6\leq3y$
$\displaystyle \Rightarrow y<4\qquad\qquad\qquad\text{and}\qquad y\geq-2$
$\displaystyle \Rightarrow -2\leq y<4$
$\displaystyle \text{(i) When }y\in\{\text{Integers}\},\text{ the solution set is }\{-2,-1,0,1,2,3\}.$
$\displaystyle \text{(ii) When }y\in R,\text{ the solution set is }\{y:y\in R\text{ and }-2\leq y<4\}.$
$\\$

$\displaystyle \textbf{Question 7: }\text{Given that }x\in R,\text{ solve the following inequation and graph}$
$\displaystyle \text{the solution on the number line: }-1\leq3+4x<23.\hfill\text{[ICSE 2006]}$
$\displaystyle \text{Answer:}$
$\displaystyle -1\leq3+4x<23$
$\displaystyle \Rightarrow -1\leq3+4x\qquad\text{and}\qquad3+4x<23$
$\displaystyle \Rightarrow -4\leq4x\qquad\qquad\text{and}\qquad4x<20$
$\displaystyle \Rightarrow -1\leq x\qquad\qquad\text{and}\qquad x<5$
$\displaystyle \Rightarrow -1\leq x<5$
$\displaystyle \therefore \{x:x\in R\text{ and }-1\leq x<5\}$
$\displaystyle \text{Number line: Draw a closed circle at }-1,\text{ an open circle at }5,$
$\displaystyle \text{and shade the interval between them.}$
Solution on the number line is :

$\\$

$\displaystyle \textbf{Question 8: }\text{Simplify: }-\frac{1}{3}\leq\frac{x}{2}-1\frac{1}{3}<\frac{1}{6},\text{ where }x\in R.$
$\displaystyle \text{Graph the values of }x\text{ on the real number line.}$
$\displaystyle \text{Answer:}$
$\displaystyle -\frac{1}{3}\leq\frac{x}{2}-1\frac{1}{3}<\frac{1}{6}$
$\displaystyle \text{The given inequation has two parts:}$
$\displaystyle -\frac{1}{3}\leq\frac{x}{2}-1\frac{1}{3}\qquad\text{and}\qquad\frac{x}{2}-1\frac{1}{3}<\frac{1}{6}$
$\displaystyle \Rightarrow -\frac{1}{3}+1\frac{1}{3}\leq\frac{x}{2}\qquad\text{and}\qquad\frac{x}{2}<\frac{1}{6}+1\frac{1}{3}$
$\displaystyle \Rightarrow 1\leq\frac{x}{2}\qquad\qquad\qquad\text{and}\qquad\frac{x}{2}<\frac{9}{6}$
$\displaystyle \Rightarrow 2\leq x\qquad\qquad\qquad\text{and}\qquad x<3$
$\displaystyle \therefore 2\leq x<3$
$\displaystyle \therefore \{x:x\in R\text{ and }2\leq x<3\}$
$\displaystyle \text{Number line: Draw a closed circle at }2,\text{ an open circle at }3,$
$\displaystyle \text{and shade the interval between them.}$
On simplifying, the given inequation reduces to $2 \leq x < 3$ and the required graph on number line is:

$\\$

$\displaystyle \textbf{Question 9: }\text{List the solution set of }50-3(2x-5)<25,\text{ given that }x\in W.$
$\displaystyle \text{Also represent the solution set obtained on a number line.}$
$\displaystyle \text{Answer:}$
$\displaystyle 50-3(2x-5)<25$
$\displaystyle \Rightarrow 50-6x+15<25$
$\displaystyle \Rightarrow -6x<25-65$
$\displaystyle \Rightarrow -6x<-40$
$\displaystyle \Rightarrow \frac{-6x}{-6}>\frac{-40}{-6}$
$\displaystyle \Rightarrow x>\frac{20}{3}=6\frac{2}{3}$
$\displaystyle \text{Since }x\in W,\text{ the solution set is }\{7,8,9,\cdots\}.$
$\displaystyle \text{Number line: Plot the points }7,8,9,\ldots\text{ on the number line.}$
And the required no. line is

$\\$

$\displaystyle \textbf{Question 10: }\text{Solve and graph the solution set of }3x+6\geq9\text{ and }-5x>-15,$
$\displaystyle \text{where }x\in R.$
$\displaystyle \text{Answer:}$
$\displaystyle 3x+6\geq9$
$\displaystyle \Rightarrow 3x\geq9-6$
$\displaystyle \Rightarrow 3x\geq3$
$\displaystyle \Rightarrow x\geq1$
$\displaystyle -5x>-15$
$\displaystyle \Rightarrow \frac{-5x}{-5}<\frac{-15}{-5}$
$\displaystyle \Rightarrow x<3$
$\displaystyle \therefore 1\leq x<3$
$\displaystyle \therefore \{x:x\in R\text{ and }1\leq x<3\}$
$\displaystyle \text{Number line: Draw a closed circle at }1,\text{ an open circle at }3,$
$\displaystyle \text{and shade the interval between them.}$
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

$\\$

$\displaystyle \textbf{Question 11: }\text{Solve and graph the solution set of }-2<2x-6\text{ or }-2x+5\geq13,$
$\displaystyle \text{where }x\in R.$
$\displaystyle \text{Answer:}$
$\displaystyle -2<2x-6$
$\displaystyle \Rightarrow 2x-6>-2$
$\displaystyle \Rightarrow 2x>4$
$\displaystyle \Rightarrow x>2$
$\displaystyle -2x+5\geq13$
$\displaystyle \Rightarrow -2x\geq8$
$\displaystyle \Rightarrow x\leq-4$
$\displaystyle \therefore x\leq-4\text{ or }x>2$
$\displaystyle \therefore \{x:x\in R\text{ and }(x\leq-4\text{ or }x>2)\}$
$\displaystyle \text{Number line: Draw a closed circle at }-4\text{ and shade to the left.}$
$\displaystyle \text{Draw an open circle at }2\text{ and shade to the right.}$
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

$\\$

$\displaystyle \textbf{Question 12: }\text{Given }P=\{x:5<2x-1\leq11,\ x\in R\}$
$\displaystyle Q=\{x:-1\leq3+4x<23,\ x\in I\}.$
$\displaystyle \text{Where }R=\{\text{real numbers}\}\text{ and }I=\{\text{integers}\}.$
$\displaystyle \text{Represent }P\text{ and }Q\text{ on two different number lines.}$
$\displaystyle \text{Write down the elements of }P\cap Q.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }P:5<2x-1\leq11,\ x\in R$
$\displaystyle \Rightarrow 5<2x-1\qquad\text{and}\qquad2x-1\leq11$
$\displaystyle \Rightarrow 3<x\qquad\qquad\text{and}\qquad x\leq6$
$\displaystyle \therefore P=\{x:x\in R\text{ and }3<x\leq6\}$
$\displaystyle \text{For }Q:-1\leq3+4x<23,\ x\in I$
$\displaystyle \Rightarrow -1\leq3+4x\qquad\text{and}\qquad3+4x<23$
$\displaystyle \Rightarrow -1\leq x\qquad\qquad\text{and}\qquad x<5$
$\displaystyle \therefore Q=\{-1,0,1,2,3,4\}$
$\displaystyle \therefore P\cap Q=\{4\}$
$\displaystyle \text{Number line for }P:\text{ Draw an open circle at }3,\text{ a closed circle at }6,$
$\displaystyle \text{and shade the interval between them.}$
$\displaystyle \text{Number line for }Q:\text{ Plot the points }-1,0,1,2,3\text{ and }4.$

121

122

$\displaystyle \text{Hence,} P \cap Q = \{ \text{elements common to both P and Q} \} = \{ 4 \}$

$\\$

$\displaystyle \textbf{Question 13: }\text{Write down the range of values of }x\text{ for which both the}$
$\displaystyle \text{inequations }x>2\text{ and }-1\leq x\leq4\text{ are true.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given }x>2\text{ and }-1\leq x\leq4$
$\displaystyle \text{The common values satisfying both inequations are those greater than }2$
$\displaystyle \text{and at the same time less than or equal to }4.$
$\displaystyle \therefore 2<x\leq4$
$\displaystyle \therefore \{x:x\in R\text{ and }2<x\leq4\}$
2022-02-27_10-58-06

It is clear from both the graphs that their common range is $2 < x \leq 4.$

Therefore required range is $2 < x \leq 4$

$\\$

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)

$\displaystyle \text{(a) } P \cup Q = \text{ Numbers which belong to P or Q or both }$

2022-02-27_10-44-26

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

2022-02-27_10-44-42

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

2022-02-27_10-45-26

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

2022-02-27_10-45-39

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

2022-02-27_10-45-26

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

2022-02-27_10-45-39

$\\$

$\displaystyle \textbf{Question 15: }\text{Find three smallest consecutive whole numbers such that the}$
$\displaystyle \text{difference between one-fourth of the largest and one-fifth of the}$
$\displaystyle \text{smallest is at least }3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required consecutive whole numbers be }x,\ x+1\text{ and }x+2.$
$\displaystyle \text{According to the given statement,}$
$\displaystyle \frac{x+2}{4}-\frac{x}{5}\geq3$
$\displaystyle \Rightarrow \frac{5x+10-4x}{20}\geq3$
$\displaystyle \Rightarrow x+10\geq60$
$\displaystyle \Rightarrow x\geq50$
$\displaystyle \text{Since the smallest value of }x\text{ is }50,$
$\displaystyle \therefore \text{the required smallest consecutive whole numbers are }50,\ 51\text{ and }52.$
$\\$