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

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NCERT Solutions for Class 11 Maths Chapter 5 - Linear Inequalities

Example 1

Solve $30 x<200$ when
(i) $x$ is a natural number,
(ii) $x$ is an integer.

Solution

We are given $30 x<200$
or $\frac{30 x}{30}<\frac{200}{30}$ (Rule 2), i.e., $x<20 / 3$.
(i) When $x$ is a natural number, in this case the following values of $x$ make the statement true.
$
1,2,3,4,5,6 \text {. }
$

The solution set of the inequality is $\{1,2,3,4,5,6\}$.
(ii) When $x$ is an integer, the solutions of the given inequality are
$
\text { ..., }-3,-2,-1,0,1,2,3,4,5,6
$

The solution set of the inequality is $\{\ldots,-3,-2,-1,0,1,2,3,4,5,6\}$
Example 2

Solve $5 x-3<3 x+1$ when
(i) $x$ is an integer,
(ii) $x$ is a real number.

Solution We have, $5 x-3<3 x+1$
or $\quad 5 x-3+3<3 x+1+3$
(Rule 1)

or $\quad 5 x<3 x+4$
or $\quad 5 x-3 x<3 x+4-3 x$
(Rule 1)

or $\quad 2 x<4$
or $\quad x<2$
(Rule 2)
(i) When $x$ is an integer, the solutions of the given inequality are
$
\ldots,-4,-3,-2,-1,0,1
$
(ii) When $x$ is a real number, the solutions of the inequality are given by $x<2$, i.e., all real numbers $x$ which are less than 2 . Therefore, the solution set of the inequality is $x \in(-\infty, 2)$.

We have considered solutions of inequalities in the set of natural numbers, set of integers and in the set of real numbers. Henceforth, unless stated otherwise, we shall solve the inequalities in this Chapter in the set of real numbers.

Example 3

Solve $4 x+3<6 x+7$.
Solution

We have, $\quad 4 x+3<6 x+7$
or $4 x-6 x<6 x+4-6 x$
or $\quad-2 x<4 \quad$ or $x>-2$
i.e., all the real numbers which are greater than -2 , are the solutions of the given inequality. Hence, the solution set is $(-2, \infty)$.

Example 4

Solve $\frac{5-2 x}{3} \leq \frac{x}{6}-5$.
Solution

We have
$
\begin{gathered}
\frac{5-2 x}{3} \leq \frac{x}{6}-5 \\
2(5-2 x) \leq x-30 . \\
10-4 x \leq x-30 \\
-5 x \leq-40, \text { i.e., } x \geq 8
\end{gathered}
$

Thus, all real numbers $x$ which are greater than or equal to 8 are the solutions of the given inequality, i.e., $x \in[8, \infty)$.
Example 5

Solve $7 x+3<5 x+9$. Show the graph of the solutions on number line.
Solution

We have $7 x+3<5 x+9$ or
$
2 x<6 \text { or } x<3
$

The graphical representation of the solutions are given in Fig 5.1.

Example 6

Solve $\frac{3 x-4}{2} \geq \frac{x+1}{4}-1$. Show the graph of the solutions on number line.
Solution

We have
$
\frac{3 x-4}{2} \geq \frac{x+1}{4}-1
$
or
$
\frac{3 x-4}{2} \geq \frac{x-3}{4}
$
or
$
2(3 x-4) \geq(x-3)
$

or
$
6 x-8 \geq x-3
$
or
$
5 x \geq 5 \text { or } x \geq 1
$

The graphical representation of solutions is given in Fig 5.2.

Example 7

The marks obtained by a student of Class XI in first and second terminal examination are 62 and 48 , respectively. Find the minimum marks he should get in the annual examination to have an average of at least 60 marks.
Solution

Let $x$ be the marks obtained by student in the annual examination. Then
$
\frac{62+48+x}{3} \geq 60
$
or
$
110+x \geq 180
$
or
$
x \geq 70
$

Thus, the student must obtain a minimum of 70 marks to get an average of at least 60 marks.

Example 8

Find all pairs of consecutive odd natural numbers, both of which are larger than 10 , such that their sum is less than 40 .

Solution

Let $x$ be the smaller of the two consecutive odd natural number, so that the other one is $x+2$. Then, we should have
$
\text { and } \begin{aligned}
& x>10 \\
& x+(x+2)<40
\end{aligned}
$
and $x+(x+2)<40$
Solving (2), we get
$
\begin{array}{ll} 
& 2 x+2<40 \\
\text { i.e., } & x<19
\end{array}
$

From (1) and (3), we get
$
10<x<19
$

Since $x$ is an odd number, $x$ can take the values $11,13,15$, and 17. So, the required possible pairs will be
$
(11,13),(13,15),(15,17),(17,19)
$