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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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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