Exercise 5.1 (Revised) - Chapter 6 - Linear Inequalities - Ncert Solutions class 11 - Maths

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NCERT Solutions for Class 11 Maths Chapter 5 - Linear Inequalities

Ex 5.1 Question 1.

Solve $24 x<100$ when:
(i) $x$ is a natural number
(ii) $x$ is an integer

Answer.

Given: $24 \mathrm{x}<100$

Divide both sides by 24 ,
$
\begin{aligned}
& =>\frac{24 x}{24}<\frac{100}{24} \\
& =>x<\frac{25}{6}
\end{aligned}
$
(i) When $x$ is a natural number then values of $x$ that make statement true are 1, 2, 3, 4. The solution set of inequality is $\{1,2,3,4\}$.
(ii) When $\mathbf{x}$ is an integer, then values of $\mathbf{x}$ that make statement true for all (-ve number and $0,1,2,3,4)$. So, The solution set of inequality is $\{\ldots . . .,-4,-3-2,-1,0,1,2,3,4\}$.
Ex 5.1 Question 2.

Solve $-12 x>30$ when:
(i) $x$ is a natural number
(ii) $x$ is an integer

Answer.

Given: $-12 \mathrm{x}>30$

Divide both sides by -12 , then we get,
$
\begin{aligned}
& =>\frac{-12 x}{-12}<\frac{30}{-12} \\
& =>x<-\frac{5}{2}
\end{aligned}
$

(i) There is no natural number less than $-\frac{5}{2}$, so when $\mathrm{x}$ is natural number, there is no solution for the given inequality.
(ii) When $x$ is an integer then values of $x$ that make statement true for \{ $\qquad$ $,-5,-4,-3\}$. The solution set of inequality is \{ $\qquad$ $,-5,-4,-3\}$

Ex 5.1 Question 3.

Solve $5 x-3<7$ when
(i) $x$ is an integer
(ii) $x$ is a real number

Answer.

Given: $5 x-3<7$
$
\begin{aligned}
& \Rightarrow 5 x<7+3 \\
& \Rightarrow 5 x<10 \\
& \Rightarrow x<2
\end{aligned}
$

(i) When $x$ is an integer then values of $x$ that make statement true are $\ldots,-3,-2,-1,0,1$. The solution set of inequality is $\{\ldots,-3,-2,-1,0,1\}$.
(ii) When $x$ is a real number then solution set of inequality is $x \in(-\infty, 2)$.
Ex 5.1 Question 4.

Solve $3 x+8>2$ when:
(i) $x$ is an integer
(ii) $x$ is a real number

Answer.

Given: $3 x+8>2$
$
\begin{aligned}
& \Rightarrow 3 x>2-8 \\
& \Rightarrow 3 x<-6 \\
& \Rightarrow x>-2
\end{aligned}
$
(i) When $x$ is an integer then values of $x$ that make statement true are $-1,0,1,2,3, \ldots \ldots$ The solution set of inequality is $\{-1,0,1,2,3, \ldots \ldots$.$\} .$
(ii) When $x$ is a real number then solution set of inequality is $x \in(-2, \infty)$.

Solve the inequalities in Exercises 5 to 16 for real $x$ :
Ex 5.1 Question 5.

$4 x+3<5 x+7$

Answer.

Here $4 x+3<5 x+7$
$
\begin{aligned}
& \Rightarrow 4 x-5 x<7-3 \\
& \Rightarrow-x<4 \\
& \Rightarrow x>-4
\end{aligned}
$

Therefore, the solution set is $(-4, \infty)$.

Ex 5.1 Question 6.

$3 x-7>5 x-1$

Answer.

Here $3 x-7>5 x-1$
$
\begin{aligned}
& \Rightarrow 3 x-5 x>-1+7 \\
& \Rightarrow-2 x>6 \\
& \Rightarrow x<-3
\end{aligned}
$

Therefore, the solution set is $(-\infty,-3)$.
Ex 5.1 Question 7.

$3(x-1) \leq 2(x-3)$

Answer.

Here $3(x-1) \leq 2(x-3)$
$
\begin{aligned}
& \Rightarrow 3 x-3 \leq 2 x-6 \\
& \Rightarrow 3 x-2 x \leq-6+3 \\
& \Rightarrow x \leq-3
\end{aligned}
$

Therefore, the solution set is $(-\infty,-3)$.
Ex 5.1 Question 8.

$3(2-x) \geq 2(1-x)$

Answer.

Here $3(2-x) \geq 2(1-x)$

$
\begin{aligned}
& \Rightarrow 6-3 x \geq 2-2 x \\
& \Rightarrow-3 x+2 x \geq 2-6 \\
& \Rightarrow-x \leq-4 \\
& \Rightarrow x \leq 4
\end{aligned}
$

Therefore, the solution set is $(-\infty, 4)$.

Ex 5.1 Question 9.

$x+\frac{x}{2}+\frac{x}{3}<11$

Answer.

Here $x+\frac{x}{2}+\frac{x}{3}<11$
$
\begin{aligned}
& \Rightarrow \frac{6 x+3 x+2 x}{6}<11 \\
& \Rightarrow \frac{11 x}{6}<11 \\
& \Rightarrow 11 x<66 \\
& \Rightarrow x<6
\end{aligned}
$

Therefore, the solution set is $(-\infty, 6)$.
Ex 5.1 Question 10.

$\frac{x}{3}>\frac{x}{2}+1$

Answer.

Here $\frac{x}{3}>\frac{x}{2}+1$
$
\Rightarrow \frac{x}{3}-\frac{x}{2}>1
$

$
\begin{aligned}
& \Rightarrow \frac{2 x-3 x}{6}>1 \\
& \Rightarrow \frac{-x}{6}>1 \\
& \Rightarrow x<-6
\end{aligned}
$

Therefore, the solution set is $(-\infty,-6)$.

Ex 5.1 Question 11.

$\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$

Answer.

Here $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$
$
\begin{aligned}
& \Rightarrow \frac{3 x-6}{5} \leq \frac{10-5 x}{3} \\
& \Rightarrow 9 x-18 \leq 50-25 x \\
& \Rightarrow 9 x+25 x \leq 50+18 \\
& \Rightarrow 34 x \leq 68 \\
& \Rightarrow x \leq 2
\end{aligned}
$

Therefore, the solution set is $(-\infty, 2)$.
Ex 5.1 Question 12.

$\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$

Answer.

Here $\frac{1}{2}\left(\frac{3 x}{5}+4\right) \geq \frac{1}{3}(x-6)$

$\begin{aligned}
& \Rightarrow \frac{3 x}{10}+2 \geq \frac{x}{3}-2 \\
& \Rightarrow \frac{3 x}{10}-\frac{x}{3} \geq-2-2 \\
& \Rightarrow \frac{9 x-10 x}{30} \geq-4 \\
& \Rightarrow \frac{-x}{30} \geq-4 \\
& \Rightarrow-x \geq-120
\end{aligned}$

$
\Rightarrow x \leq 120
$

Therefore, the solution set is $(-\infty, 120)$.
Ex 5.1 Question 13.

$2(2 x+3)-10<6(x-2)$

Answer.

Here $2(2 x+3)-10<6(x-2)$
$
\begin{aligned}
& \Rightarrow 4 x+6-10<6 x-12 \\
& \Rightarrow 4 x-4<6 x-12 \\
& \Rightarrow 4 x-6 x<-12+4 \\
& \Rightarrow-2 x<-8 \\
& \Rightarrow x>4
\end{aligned}
$

Therefore, the solution set is $(4, \infty)$.
Ex 5.1 Question 14.

$37-(3 x+5) \geq 9 x-8(x-3)$

Answer.

Here $37-(3 x+5) \geq 9 x-8(x-3)$
$
\Rightarrow 37-3 x-5 \geq 9 x-8 x+24
$

$
\begin{aligned}
& \Rightarrow 32-3 x \geq x+24 \\
& \Rightarrow-3 x-x \geq 24-32 \\
& \Rightarrow-4 x \geq-8 \\
& \Rightarrow x \leq 2
\end{aligned}
$

Therefore, the solution set is $(-\infty, 2)$.

Ex 5.1 Question 15.

$\frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$

Answer.

Here $\frac{x}{4}<\frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$
$
\begin{aligned}
& \Rightarrow \frac{x}{4}<\frac{5 x}{3}-\frac{2}{3}-\frac{7 x}{5}+\frac{3}{5} \\
& \Rightarrow \frac{x}{4}-\frac{5 x}{3}+\frac{7 x}{5}<-\frac{2}{3}+\frac{3}{5} \\
& \Rightarrow \frac{15 x-100 x+84 x}{60}<\frac{-10+9}{15} \\
& \Rightarrow \frac{-x}{60}<\frac{-1}{15} \Rightarrow x>4
\end{aligned}
$

Therefore, the solution set is $(4, \infty)$.
Ex 5.1 Question 16.

$\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$

Answer.

Here $\frac{(2 x-1)}{3} \geq \frac{(3 x-2)}{4}-\frac{(2-x)}{5}$

$\begin{aligned}
& \Rightarrow \frac{2 x}{3}-\frac{1}{3} \geq \frac{3 x}{4}-\frac{2}{4}-\frac{2}{5}+\frac{x}{5} \\
& \Rightarrow \frac{2 x}{3}-\frac{3 x}{4}-\frac{x}{5} \geq-\frac{2}{4}-\frac{2}{5}+\frac{1}{3} \\
& \Rightarrow \frac{40 x-45 x-12 x}{60} \geq \frac{-30-24+20}{60} \\
& \Rightarrow \frac{-17 x}{60} \geq \frac{-34}{60}
\end{aligned}$

$
\Rightarrow x \leq 2
$

Therefore, the solution set is $(-\infty, 2)$.

Solve the inequalities in Exercises 17 to 20 and show the graph of the solution in each case on number line:
Ex 5.1 Question 17.

$3 x-2<2 x+1$

Answer. Here $3 x-2<2 x+1$
$
\begin{aligned}
& \Rightarrow 3 x-2 x<1+2 \\
& \Rightarrow x<3
\end{aligned}
$
$\therefore$ The solution set is $(-\infty, 3)$.

Ex 5.1 Question 18.

$5 x-3 \geq 3 x-5$

Answer.

Here $5 x-3 \geq 3 x-5$
$
\begin{aligned}
& \Rightarrow 5 x-3 x \geq-5+3 \\
& \Rightarrow 2 x \geq-2 \\
& \Rightarrow x \geq-1
\end{aligned}
$
$\therefore$ The solution set is $(-1, \infty)$.

Ex 5.1 Question 19.

$3(1-x)<2(x+4)$

Answer.

Here $3(1-x)<2(x+4)$

$
\begin{aligned}
& \Rightarrow 3-3 x<2 x+8 \\
& \Rightarrow-3 x-2 x<8-3 \\
& \Rightarrow-5 x<5 \\
& \Rightarrow x>-1
\end{aligned}
$
$\therefore$ The solution set is $(-1, \infty)$.

Ex 5.1 Question 20.

$\frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$

Answer.

Here $\frac{x}{2} \geq \frac{(5 x-2)}{3}-\frac{(7 x-3)}{5}$
$
\begin{aligned}
& \Rightarrow \frac{x}{2} \geq \frac{5 x}{3}-\frac{2}{3}-\frac{7 x}{5}+\frac{3}{5} \\
& \Rightarrow \frac{x}{2}-\frac{5 x}{3}+\frac{7 x}{5} \geq-\frac{2}{3}+\frac{3}{5} \\
& \Rightarrow \frac{15 x-50 x+42 x}{30} \geq \frac{-10+9}{15} \\
& \Rightarrow \frac{7 x}{30} \geq \frac{-1}{15} \\
& \Rightarrow 7 x \geq-2 \\
& \Rightarrow x \geq \frac{-2}{7}
\end{aligned}
$
$\therefore$ The solution set is $\left(\frac{-2}{7}, \infty\right)$.