Miscellaneous Example (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 $-8 \leq 5 x-3<7$.
Solution

In this case, we have two inequalities, $-8 \leq 5 x-3$ and $5 x-3<7$, which we will solve simultaneously. We have $-8 \leq 5 x-3<7$
or $-5 \leq 5 x<10 \quad$ or $-1 \leq x<2$
Example 2

Solve $-5 \leq \frac{5-3 x}{2} \leq 8$.
Solution

We have $\quad-5 \leq \frac{5-3 x}{2} \leq 8$
or $\quad-10 \leq 5-3 x \leq 16 \quad$ or $\quad-15 \leq-3 x \leq 11$
or $\quad 5 \geq x \geq-\frac{11}{3}$
which can be written as $\frac{-11}{3} \leq x \leq 5$
Example 3

Solve the system of inequalities:
$
\begin{aligned}
& 3 x-7<5+x \\
& 11-5 x \leq 1
\end{aligned}
$
and represent the solutions on the number line.
Solution

From inequality (1), we have
$
\begin{aligned}
& 3 x-7<5+x \\
& \text { or } \quad x<6
\end{aligned}
$

Also, from inequality (2), we have
$
11-5 x \leq 1
$
or
$
-5 x \leq-10
$
i.e., $x \geq 2$

If we draw the graph of inequalities (3) and (4) on the number line, we see that the values of $x$, which are common to both, are shown by bold line in Fig 5.3.

Thus, solution of the system are real numbers $x$ lying between 2 and 6 including 2, i.e.,
$
2 \leq x<6
$

Example 4

In an experiment, a solution of hydrochloric acid is to be kept between $30^{\circ}$ and $35^{\circ}$ Celsius. What is the range of temperature in degree Fahrenheit if conversion formula is given by $\mathrm{C}=\frac{5}{9} \quad(\mathrm{~F}-32)$, where $\mathrm{C}$ and $\mathrm{F}$ represent temperature in degree Celsius and degree Fahrenheit, respectively.
Solution

It is given that $30<\mathrm{C}<35$.
Putting
$
\begin{aligned}
& \mathrm{C}=\frac{5}{9}(\mathrm{~F}-32), \text { we get } \\
& 30<\frac{5}{9}(\mathrm{~F}-32)<35,
\end{aligned}
$
or
$
\begin{aligned}
\frac{9}{5} & \times(30)<(F-32)<\frac{9}{5} \times \\
54 & <(F-32)<63 \\
86 & <\mathrm{F}<95 .
\end{aligned}
$

Thus, the required range of temperature is between $86^{\circ} \mathrm{F}$ and $95^{\circ} \mathrm{F}$.
Example 5

A manufacturer has 600 litres of a $12 \%$ solution of acid. How many litres of a $30 \%$ acid solution must be added to it so that acid content in the resulting mixture will be more than $15 \%$ but less than $18 \%$ ?
Solution

Let $x$ litres of $30 \%$ acid solution is required to be added. Then Total mixture $=(x+600)$ litres

Therefore $30 \% x+12 \%$ of $600>15 \%$ of $(x+600)$
and
$
30 \% x+12 \% \text { of } 600<18 \% \text { of }(x+600)
$

$\text { or } \quad \frac{30 x}{100}+\frac{12}{100}(600)>\frac{15}{100}(x+600)$

$\begin{array}{ll}
\text { and } & \frac{30 x}{100}+\frac{12}{100}(600)<\frac{18}{100}(x+600) \\
\text { or } & 30 x+7200>15 x+9000 \\
\text { and } & 30 x+7200<18 x+10800 \\
\text { or } & 15 x>1800 \text { and } 12 x<3600 \\
\text { or } & x>120 \text { and } x<300 \text {, } \\
\text { i.e. } & 120<x<300
\end{array}$

Thus, the number of litres of the $30 \%$ solution of acid will have to be more than 120 litres but less than 300 litres.