# Miscellaneous Exercise Linear Inequalities Solution - NCERT Class 11

## Chapter 6 Ex.6.ME Question 1

Solve the inequality $$2 \le 3x - 4 \le 5$$

### Solution

\begin{align}&\quad\;\; 2 \le 3x - 4 \le 5\\&\Rightarrow 2 + 4 \le 3x - 4 + 4 \le 5 + 4\\&\Rightarrow 6 \le 3x \le 9\\&\Rightarrow 2 \le x \le 3\end{align}

Thus, all the real numbers, $$x$$, which are greater than or equal to $$2$$ but less than or equal to $$3$$, are the solutions of the given inequality.

The solution set for the given inequality is $$\left[ {{\rm{2}},{\rm{3}}} \right]$$.

## Chapter 6 Ex.6.ME Question 2

Solve the inequality  $$6 \le - 3(2x - 4) < 12$$

### Solution

$\begin{array}{l} 6 \le - 3(2x - 4) < 12\\ \Rightarrow 2 \le - (2x - 4) < 4\\ \Rightarrow - 2 \ge 2x - 4 > - 4\\ \Rightarrow 4 - 2 \ge 2x > 4 - 4\\ \Rightarrow 2 \ge 2x > 0\\ \Rightarrow 1 \ge x > 0 \end{array}$

Thus, the solution set for the given inequality is [0,1].

## Chapter 6 Ex.6.ME Question 3

Solve the inequality $$- 3 \le 4 - \frac{{7x}}{2} \le 18$$

### Solution

\begin{align}\;\;\;\; &- 3 \le 4 - \frac{{7x}}{2} \le 18\\&\Rightarrow - 3 - 4 \le - \frac{{7x}}{2} \le 18 - 4\\&\Rightarrow - 7 \le - \frac{{7x}}{2} \le 14\\&\Rightarrow 7 \ge \frac{{7x}}{2} \ge - 14\\&\Rightarrow 1 \ge \frac{x}{2} \ge - 2\\&\Rightarrow 2 \ge x \ge - 4\\&\Rightarrow - 4 \le x \le 2\end{align}

Thus, the solution set for the given inequality is $$\left[ { - {\rm{4}},{\rm{2}}} \right]$$.

## Chapter 6 Ex.6.ME Question 4

Solve the inequality $$- 15 < \frac{{3\left( {x - 2} \right)}}abc \le 0$$

### Solution

\begin{align}\;\;\;\;& - 15 < \frac{{3\left( {x - 2} \right)}}{5} \le 0\\&\Rightarrow - 75 < 3(x - 2) \le 0\\&\Rightarrow - 25 < x - 2 \le 0\\&\Rightarrow - 25 + 2 < x \le 2\\&\Rightarrow - 23 < x \le 2\end{align}

Thus, the solution set for the given inequality is $$\left( { - {\rm{23}},{\rm{2}}} \right]$$

## Chapter 6 Ex.6.ME Question 5

Solve the inequality $$- 12 < 4 - \frac{{3x}}{{ - 5}} \le 2$$

### Solution

\begin{align}\;\;\;\; &- 12 < 4 - \frac{{3x}}{{ - 5}} \le 2\\&\Rightarrow - 12 - 4 < \frac{{3x}}{5} \le 2 - 4\\&\Rightarrow - 16 < \frac{{3x}}{5} \le - 2\\&\Rightarrow - 80 < 3x \le - 10\\&\Rightarrow \frac{{ - 80}}{3} < x \le \frac{{ - 10}}{3}\end{align}

Thus, the solution set for the given inequality is $$\left( {\frac{{ - 80}}{3},\frac{{ - 10}}{3}} \right]$$.

## Chapter 6 Ex.6.ME Question 6

Solve the inequality $$7 \le \frac{{\left( {3x + 11} \right)}}{2} \le 11$$

### Solution

\begin{align}\;\;\;\;&7 \le \frac{{\left( {3x + 11} \right)}}{2} \le 11\\&\Rightarrow 14 \le 3x + 11 \le 22\\&\Rightarrow 14 - 11 \le 3x \le 22 - 11\\&\Rightarrow 3 \le 3x \le 11\\&\Rightarrow 1 \le x \le \frac{{11}}{3}\end{align}

Thus, the solution set for the given inequality is $$\left[ {1,\frac{{11}}{3}} \right]$$ .

## Chapter 6 Ex.6.ME Question 7

Solve the inequalities and represent the solution graphically on number line.

$5x + 1 > - 24,\;5x - 1 < 24$

### Solution

\begin{align}\;\;\;\;&5x + 1 > - 24\\&\Rightarrow 5x > - 25\\&\Rightarrow x > - 5\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\\\\;\;\;\;&5x - 1 < 24\\&\Rightarrow 5x < 25\\&\Rightarrow x < 5\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From ($$1$$) and ($$2$$), we get, $$- 5 < x < 5$$

Hence, it can be concluded that the solution set for the given system of inequalities is $$\left( { - 5,5} \right)$$.

The solution of the given system of inequalities can be represented on number line as

## Chapter 6 Ex.6.ME Question 8

Solve the inequalities and represent the solution graphically on number line: $2\left( {x - 1} \right) < x + 5,\;3\left( {x + 2} \right) > 2 - x$

### Solution

\begin{align} & \ \ \ \ 2\left( x-1 \right)<x+5 \\ & \Rightarrow 2x-2<x+5 \\ & \Rightarrow 2x-x<5+2 \\ & \Rightarrow x<7\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \;\;\; \ldots \left( 1 \right) \\ & \\ & \ \ \ \ 3\left( x+2 \right)>2-x \\ & \Rightarrow 3x+6>2-x \\ & \Rightarrow 3x+x>2-6 \\ & \Rightarrow 4x>-4 \\ & \Rightarrow x>-1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) \\ \end{align}

From ($$1$$) and ($$2$$), we get, $$- 1 < x < 7$$

Hence, it can be concluded that the solution set for the given system of inequalities is $$\left( { - 1,7} \right)$$.

The solution of the given system of inequalities can be represented on number line as

## Chapter 6 Ex.6.ME Question 9

Solve the inequalities and represent the solution graphically on number line: $$3x - 7 > 2(x - 6),\;6 - x > 11 - 2x$$

### Solution

\begin{align}\;\;\;\;&3x - 7 > 2(x - 6)\\&\Rightarrow 3x - 7 > 2x - 12\\&\Rightarrow 3x - 2x > - 12 + 7\\&\Rightarrow x > - 5\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

\begin{align}\;\;\;\;&6 - x > 11 - 2x\\&\Rightarrow - x + 2x > 11 - 6\\&\Rightarrow x > 5\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From ($$1$$) and ($$2$$), we get, $$- 5 < x > 5$$

Hence, it can be concluded that the solution set for the given system of inequalities is $$\left( {5,\infty } \right)$$.

The solution of the given system of inequalities can be represented on number line as

## Chapter 6 Ex.6.ME Question 10

Solve the inequalities and represent the solution graphically on number line: $5\left( {2x - 7} \right) - 3\left( {2x + 3} \right) \le 0,\;2x + 19 \le 6x + 47$

### Solution

\begin{align} & \ 5\left( 2x-7 \right)-3\left( 2x+3 \right)\le 0 \\ & \Rightarrow 10x-35-6x-9\le 0 \\ & \Rightarrow 4x-44\le 0 \\ & \Rightarrow 4x\le 44 \\ & \Rightarrow x\le 11\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 1 \right) \\ & \\ & \ 2x+19\le 6x+47 \\ & \Rightarrow 19-47\le 6x-2x \\ & \Rightarrow -28\le 4x \\ & \Rightarrow -7\le x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots \left( 2 \right) \\ \end{align}

From ($$1$$) and ($$2$$), we get, $$- 7 \le x \le 11$$

Hence, it can be concluded that the solution set for the given system of inequalities is $$\left[ { - 7,11} \right]$$.

The solution of the given system of inequalities can be represented on number line as

## Chapter 6 Ex.6.ME Question 11

A solution is to be kept between $$~{{68}^{\circ }}F$$ and $${77^ \circ }F$$. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by $$F = \frac{9}abcC + 32?$$

### Solution

Since the solution is to be kept between $$~{{68}^{\circ }}F$$ and $${77^ \circ }F$$, $$68^\circ < F < 77^\circ$$

Putting $$F = \frac{9}{5}C + 32$$, we obtain

\begin{align}\;\;\;\;&68 < \frac{9}{5}C + 32 < 77\\&\Rightarrow 68 - 32 < \frac{9}{5}C < 77 - 32\\&\Rightarrow 36 < \frac{9}{5}C < 45\\&\Rightarrow 36 \times \frac{5}{9} < C < 45 \times \frac{5}{9}\\&\Rightarrow 20 < C < 25\\\end{align}

Thus, the required range of temperature in degree Celsius is between $$~{{20}^{\circ }}C$$ and $$~{{25}^{\circ }}C$$.

## Chapter 6 Ex.6.ME Question 12

A solution of $$8\%$$ boric acid is to be diluted by adding a $$2\%$$ boric acid solution to it. The resulting mixture is to be more than $$4\%$$ but less than $$6\%$$ boric acid. If we have $$640$$ litres of the $$8\%$$ solution, how many litres of the $$2\%$$ solution will have to be added?

### Solution

Let $$x$$ litres of $$2\%$$ boric acid solution is required to be added.

Then, total mixture $$= \left( {x + 640} \right)$$ litres

This resulting mixture is to be more than $$4\%$$ but less than $$6\%$$ boric acid.

Therefore,

$$2\% x + 8\% \;{\rm{of}}\;640 > 4\% \;{\rm{of}}\;\left( {x + 640} \right)$$ and $$\\ 2\% x + 8\% \;{\rm{of}}\;640 < 6\% \;{\rm{of}}\;\left( {x + 640} \right)$$

\begin{align}&2\% x + 8\% \;{\rm{of}}\;640 > 4\% \;{\rm{of}}\;\left( {x + 640} \right)\\&\Rightarrow \frac{2}{{100}}x + \frac{8}{{100}}\left( {640} \right) > \frac{4}{{100}}\left( {x + 640} \right)\\&\Rightarrow 2x + 5120 > 4x + 2560\\&\Rightarrow 5120 - 2560 > 4x - 2x\\&\Rightarrow 2560 > 2x\\&\Rightarrow x < 1280 \qquad \quad\quad \ldots \left( 1 \right)\end{align}

\begin{align}\;\;\;\;&2\% x + 8\% \;{\rm{of}}\;640 < 6\% \;{\rm{of}}\;\left( {x + 640} \right)\\&\Rightarrow \frac{2}{{100}}x + \frac{8}{{100}}\left( {640} \right) < \frac{6}{{100}}\left( {x + 640} \right)\\&\Rightarrow 2x + 5120 < 6x + 2560\\&\Rightarrow 5120 - 2560 < 6x - 2x\\&\Rightarrow 2560 < 3x\\&\Rightarrow x > 320\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From ($$1$$) and ($$2$$), we get, $$320 < x < 1280$$

Thus, the number of litres of $$2\%$$ of boric acid solution that is to be added will have to be more than $$320$$ litres but less than $$1280$$ litres.

## Chapter 6 Ex.6.ME Question 13

How many litres of water will have to be added to $$1125$$ litres of the $$45\%$$ solution of acid so that the resulting mixture will contain more than $$25\%$$ but less than $$30\%$$ acid content?

### Solution

Let $$x$$ litres of water is required to be added.

Then, total mixture $$= \left( {1125 + x} \right)$$ litres

It is evident that the amount of acid contained in the resulting mixture is $$45\%$$ of $$1125$$ litres. This resulting mixture will contain more than $$25\%$$ but less than $$30\%$$ acid content.

Therefore, $$30\% \;{\rm{of}}\;\left( {1125 + x} \right) > 45\% \;{\rm{of}}\;1125$$ and $$25\% \;{\rm{of}}\;\left( {1125 + x} \right) < 45\% \;{\rm{of}}\;1125$$

\begin{align}\;\;\;\;&30\% \;{\rm{of}}\;\left( {1125 + x} \right) > 45\% \;{\rm{of}}\;1125\\&\Rightarrow \frac{{30}}{{100}}\left( {1125 + x} \right) > \frac{{45}}{{100}} \times 1125\\&\Rightarrow 30\left( {1125 + x} \right) > 45 \times 1125\\&\Rightarrow 30 \times 1125 + 30x > 45 \times 1125\\&\Rightarrow 30x > 45 \times 1125 - 30 \times 1125\\&\Rightarrow 30x > \left( {45 - 30} \right) \times 1125\\&\Rightarrow x > \frac{{15 \times 1125}}{{30}}\\&\Rightarrow x > 562.5\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}

\begin{align}\;\;\;\;&25\% \;{\rm{of}}\;\left( {1125 + x} \right) < 45\% \;{\rm{of}}\;1125\\&\Rightarrow \frac{{25}}{{100}}\left( {1125 + x} \right) < \frac{{45}}{{100}} \times 1125\\&\Rightarrow 25\left( {1125 + x} \right) < 45 \times 1125\\&\Rightarrow 25 \times 1125 + 25x < 45 \times 1125\\&\Rightarrow 25x < 45 \times 1125 - 25 \times 1125\\&\Rightarrow 25x < \left( {45 - 25} \right) \times 1125\\&\Rightarrow x < \frac{{20 \times 1125}}{{25}}\\&\Rightarrow x < 900\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}

From ($$1$$) and ($$2$$), we get, $$562.5 < x < 900$$

Thus, the required number of litres of water that is to be added will have to be more than $$562.5$$ litres but less than $$900$$ litres.

## Chapter 6 Ex.6.ME Question 14

$$IQ$$ of a person is given by the formula

$$IQ = \frac{{MA}}{{CA}} \times 100$$

where $$MA$$ is mental age and $$CA$$ is chronological age. If $$80 \le IQ \le 140$$ for a group of $$12$$ years old children, find the range of their mental age.

### Solution

It is given that for a group of $$12$$ years old children, $$80 \le IQ \le 140\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)$$

For a group of $$12$$ years old children, $$CA = 12$$ years

Therefore, $$IQ = \frac{{MA}}{{12}} \times 100$$

Putting this value of $$IQ$$ in ($$1$$), we obtain

\begin{align}\;\;\;\;\;\;\;\;\;\; &80 \le \frac{{MA}}{{12}} \times 100 \le 140\\&\Rightarrow 80 \times \frac{{12}}{{100}} \le MA \le 140 \times \frac{{12}}{{100}}\\&\Rightarrow 9.6 \le MA \le 16.8\end{align}

Thus, the range of mental age of the group of $$12$$ years old children is $$9.6 \le MA \le 16.8$$

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