Class 10, Exercise 6(a), Linear Inequations (In One Variable)

Class 10: Linear Inequations – Exercise 6(a)

Date: July 13, 2017Author: ICSE CBSE ISC Board Mathematics Portal for Students 0 Comments

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

Question 1: State True or False:

i) $x < -y \Rightarrow -x> y$ : True

ii) $-5x \geq 15 \Rightarrow x \geq -3$ : False

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

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

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

i) If $a < b$ , then $a-c <b-c$ : True

ii) If $a > b$ , then $a+c > b+c$ : True

iii) If $a < b$ , then $ac > bc$ : False

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

v) If $a-c > b-d$ ; then $a+d > b+c$ : True

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

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

i) $5x+3 \leq 2x+18$

Answer

$5x+3 \leq 2x+18$

$\Rightarrow 3x \leq 15$

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

ii) $3x-2 < 19-4x$

Answer:

$3x-2 < 19-4x$

$\Rightarrow 7x <21$

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

$\\$

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

i) $x+7 \leq 11$

Answer:

$x+7 \leq 11$

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

ii) $3x-1 > 8$

Answer:

$3x-1 > 8$

$\Rightarrow 3x > 9$

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

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

Answer:

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

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

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

iv) $18 \leq 3x-2$

Answer:

$18 \leq 3x-2$

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

$\\$

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

Answer:

$3-2x \geq x-12$

$\Rightarrow 3x \leq 15$

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

$\\$

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

Answer:

$25-4x \leq 16$

$\Rightarrow 4x \geq 9$

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

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

i) $-4x \geq -16$

Answer:

$-4x \geq -16$

$\Rightarrow 4x \leq 16$

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

ii) $8-3x \leq 20$

Answer:

$8-3x \leq 20$

$\Rightarrow 3x \geq -12$

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

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

Answer:

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

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

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

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

Answer:

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

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

$\Rightarrow 39 < 3x$

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

$\\$

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

Answer:

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

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

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

$\Rightarrow -3<2x$

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

Therefore $x = -1$

$\\$

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

Answer:

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

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

$\Rightarrow 3x \leq 11$

Therefore $x =3$

$\\$

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

Answer:

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

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

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

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

$\\$

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

Answer:

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

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

$\Rightarrow -2 \leq 2x$

$\Rightarrow -1 \leq x$

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

$\Rightarrow x < 5$

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

$\\$

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

Answer:

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

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

$\Rightarrow -4 \leq 4x$

$\Rightarrow -1 \leq x$

Equation 2: $3x+4 < 23$

$\Rightarrow 4x < 20$

$\Rightarrow x < 5$

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

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