NCERT Solutions for Class 11 Maths Chapter 5 - Linear Inequalities Miscellaneous Exercise

Miscellaneous Exercise

1. Solve the inequality $2 \leqslant 3x - 4 \leqslant 5$

Ans: $2 \leqslant 3x - 4 \leqslant 5$

$\Rightarrow 2 + 4 \leqslant 3x - 4 + 4 \leqslant 5 + 4$

$\Rightarrow 6 \leqslant 3x \leqslant 9$

$\Rightarrow 2 \leqslant x \leqslant 3$

As a result, the solutions of the following inequality are all real values higher than or equal to 2 but less than or equal to 3. For the given inequality, the solution set is [2,3].

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

Ans: $6 \leqslant  - 3(2x - 4) < 12$

$\Rightarrow 2 \leqslant  - (2x - 4) < 4$

$\Rightarrow  - 2 \geqslant 2x - 4 >  - 4$

$\Rightarrow 4 - 2 \geqslant 2x > 4 - 4$

$\Rightarrow 2 \geqslant 2x > 0$

$\Rightarrow 1 \geqslant x > 0$

As a result, the set of solutions for the given inequality is [1 ,0 ).

3. Solve the inequality $ - 3 \leqslant 4 - \dfrac{{7x}}{2} \leqslant 18$

Ans: $ - 3 \leqslant 4 - \dfrac{{7x}}{2} \leqslant 18$

$\Rightarrow  - 3 - 4 \leqslant  - \dfrac{{7x}}{2} \leqslant 18 - 4$

$\Rightarrow  - 7 \leqslant  - \dfrac{{7x}}{2} \leqslant 14$

$\Rightarrow 7 \geqslant \dfrac{{7x}}{2} \geqslant  - 14$

$\Rightarrow 1 \geqslant \dfrac{x}{2} \geqslant  - 2$

$\Rightarrow 2 \geqslant x \geqslant  - 4$

As a result, the set of solutions for the given inequality is $[ - 4,2]$.

4. Solve the inequality $ - 15 < \dfrac{{3(x - 2)}}{5} \leqslant 0$

Ans: $ - 15 < \dfrac{{3(x - 2)}}{5} \leqslant 0$

$\Rightarrow  - 75 < 3(x - 2) \leqslant 0$

$\Rightarrow  - 25 < x - 2 \leqslant 0$

$\Rightarrow  - 25 + 2 < x \leqslant 2$

$\Rightarrow  - 23 < x \leqslant 2$

As a result, the set of solutions for the given inequality is (-23, 2]

5. Solve the inequality $ - 12 < 4 - \dfrac{{3x}}{{ - 5}} \leqslant 2$

Ans: \[ - 12 < 4 - \dfrac{{3x}}{{ - 5}} \leqslant 2\]

$\Rightarrow  - 12 - 4 < \dfrac{{ - 3x}}{{ - 5}} \leqslant 2 - 4$

$\Rightarrow  - 16 < \dfrac{{3x}}{5} \leqslant  - 2$

$\Rightarrow  - 80 < 3x \leqslant  - 10$

$\Rightarrow \dfrac{{ - 80}}{3} < x \leqslant \dfrac{{ - 10}}{3}$

As a result, the set of solutions for the given inequality is $\left( {\dfrac{{ - 80}}{3},\dfrac{{ - 10}}{3}} \right]$.

6. Solve the inequality $7 \leqslant \dfrac{{(3x + 11)}}{2} \leqslant 11$

Ans: $7 \leqslant \dfrac{{(3x + 11)}}{2} \leqslant 11$

$\Rightarrow 14 \leqslant 3x + 11 \leqslant 22$

$\Rightarrow 14 - 11 \leqslant 3x \leqslant 22 - 11$

$\Rightarrow 3 \leqslant 3x \leqslant 11$

$\Rightarrow 1 \leqslant x \leqslant \dfrac{{11}}{3}$

As a result, the set of solutions for the given inequality is $\left[ {1,\dfrac{{11}}{3}} \right]$.

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

$5x + 1 >  - 24,5x - 1 < 24$

Ans: $5x + 1 >  - 24 \Rightarrow 5x >  - 25$

$\Rightarrow x >  - 5 \ldots .(1)$

$5x - 1 < 24 \Rightarrow 5x < 25$

$\Rightarrow x < 5$

From (1) and ( 2 ), The solution set for the given system of inequalities can be deduced to be$( - 5,5)$. On a number line, the solution to the above system of inequalities can be expressed as

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

$2(x - 1) < x + 5,3(x + 2) > 2 - x$

Ans: $2(x - 1) < x + 5 \Rightarrow 2x - 2 < x + 5 \Rightarrow 2x - x < 5 + 2$

$\Rightarrow x < 7$

(1) $3(x + 2) > 2 - x \Rightarrow 3x + 6 > 2 - x \Rightarrow 3x + x > 2 - 6$

$\Rightarrow 4x >  - 4$

$\Rightarrow x >  - 1 \ldots  \ldots (2)$

From (1) and (2), The solution set for the given system of inequalities can be deduced to be $( - 1,7)$. On a number line, the solution to the above system of inequalities can be expressed as

9. Solve the following inequalities and represent the solution graphically on number line:

$3x - 7 > 2(x - 6),6 - x > 11 - 2x$

Ans: $3x - 7 > 2(x - 6) \Rightarrow 3x - 7 > 2x - 12 \Rightarrow 3x - 2x >  - 12 + 7$

$\Rightarrow x >  - 5 \ldots  \ldots  \ldots .(1)$

$ 6 - x > 11 - 2x \Rightarrow  - x + 2x > 11 - 6$

$\Rightarrow x > 5$

From (1) and (2), The solution set for the given system of inequalities can be deduced to be

$(5,\infty )$. On a number line, the solution to the above system of inequalities can be expressed as

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

$5(2x - 7) - 3(2x + 3) \leqslant 0,2x + 19 \leqslant 6x + 47$

Ans: $5(2x - 7) - 3(2x + 3) \leqslant 0 \Rightarrow 10x - 35 - 6x - 9 \leqslant 0 \Rightarrow 4x - 44 \leqslant 0 \Rightarrow 4x \leqslant 44$

$\Rightarrow x \leqslant 11$

$2x + 19 \leqslant 6x + 47 \Rightarrow 19 - 47 \leqslant 6x - 2x \Rightarrow  - 28 \leqslant 4x$

$\Rightarrow  - 7 \leqslant x$

From (1) and (2), The solution set for the given system of inequalities can be deduced to be $[ - 7,11]$. On a number line, the solution to the above system of inequalities can be expressed as

11. A solution is to be kept between ${68^\circ }{\text{F}}$ and ${77^\circ }{\text{F}}$. What is the range in temperature in degree Celsius (C) if the Celsius/Fahrenheit (F) conversion formula is given by $F = \dfrac{9}{5}C + 32?$

Ans: Because the solution must be preserved somewhere in the middle, ${68^\circ }{\text{F}}$ and ${77^\circ }{\text{F}},68 < F < 77$ Putting $F = \dfrac{9}{5}C + 32$, we obtain $68 < \dfrac{9}{5}C + 32 < 77$

$\Rightarrow 68 - 32 < \dfrac{9}{5}C < 77 - 32$

$\Rightarrow 36 < \dfrac{9}{5}C < 45$

$\Rightarrow 36 \times \dfrac{5}{9} < C < 45 \times \dfrac{5}{9}$

$\Rightarrow 20 < C < 25$

As a result, the required temperature range in degrees Celsius is between ${20^\circ }C$ and ${25^\circ }C$.

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?

Ans: Let $2\% $ of $x$ litres of boric acid solution is required to be added. Then, total mixture $ = (x + 640)$ litres.

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

$\therefore \quad 2 \%$ of $x+8 \%$ of $640>4 \%$ of $(x+640)$

And, $2 \%$ of $x+8 \%$ of $640<6 \%$ of $(x+640)$`

$\Rightarrow 2x + 5120 > 4x + 2560$

$\Rightarrow 5120 - 2560 > 4x - 2x$

$\Rightarrow 5120 - 2560 > 2x$

$\Rightarrow 2560 > 2x$

$\Rightarrow 1280 > x$

$2\% x + 8\% $ of $640 < 6\% $ of $(x + 640)$ $\dfrac{2}{{100}}x + \dfrac{8}{{100}}(640) < \dfrac{6}{{100}}(x + 640)$

$\Rightarrow 2x + 5120 < 6x + 3840$

$\Rightarrow 5120 - 3840 < 6x - 2x$

$\Rightarrow 5120 - 3840 < 6x - 2x$

$\Rightarrow 1280 < 4x$

$\Rightarrow 320 < x$

$\therefore 320 < x < 1280$

As a result, the total amount of boric acid solution to be added must be greater than 320 litres but less than 1280 litres.

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?

Ans: Allow for the addition of x litres of water. The entire mixture is then calculated $ = (x + 1125)$litres It is clear that the amount of acid in the final mixture is excessive. $45\% $ of 1125 litres. The resulting mixture will have a higher concentration of $25\% $ but less than $30\% $ acid content.

$\therefore 30\% $ of $(1125 + x) > 45\% $ of 1125

And, $25\% $ of $(1125 + x) < 45\% $ of 1125 

$\Rightarrow \dfrac{{30}}{{100}}(1125 + x) > \dfrac{{45}}{{100}} \times 1125$

$\Rightarrow 30(1125 + x) > 45 \times 1125$

$\Rightarrow 30 \times 1125 + 30x > 45 \times 1125$

$\Rightarrow 30 > 45 \times 1125 - 30 \times 1125$

$\Rightarrow 30x > (45 - 30) \times 1125$

$\Rightarrow x > \dfrac{{15 \times 1125}}{{30}} = 562.5$

$25\% $ of $(1125 + x) < 45\% $ of 1125 $\Rightarrow \dfrac{{25}}{{100}}(1125 + x) < \dfrac{{45}}{{100}} \times 1125$

$\Rightarrow 25(1125 + x) > 45 \times 1125$

$\Rightarrow 25 \times 1125 + 25x > 45 \times 1125$

$\Rightarrow 25x > 45 \times 1125 - 25 \times 1125$

$\Rightarrow 25x > (45 - 25) \times 1125$

$\Rightarrow x > \dfrac{{20 \times 1125}}{{25}} = 900$

$\therefore 562.5 < x < 900$

As a result, the required number of litres of water must be greater than 562.5 but less than 900.

14. IQ of a person is given by the formula $IQ = \dfrac{{MA}}{{CA}} \times 100$, Where MA is mental age and CA is chronological age. If $80 \leqslant 1Q \leqslant 140$ for a group of 12 years old children, find the range of their mental age.

Ans: It is reported that for a group of twelve-year-olds $80 \leqslant IQ \leqslant 140 \ldots  \ldots (i)$

For a group of 12 years old children, ${\text{CA}} = 12$ years ${\text{IQ}} = \dfrac{{{\text{MA}}}}{{12}} \times 100$

Putting this value of IQ in (i), we obtain $80 \leqslant \dfrac{{{\text{MA}}}}{{12}} \times 100 \leqslant 140$

$\Rightarrow 80 \times \dfrac{{12}}{{100}} \leqslant {\text{MA}} \leqslant 140 \times \dfrac{{12}}{{100}}$

$\Rightarrow 9.6 \leqslant {\text{MA}} \leqslant 16.8$

As a result, the mental age range of the 12-year-olds has widened.$\Rightarrow 9.6 \leqslant {\text{MA}} \leqslant 16.8$.

Conclusion

NCERT Miscellaneous Exercise in Chapter 5 of Class 11 Maths helps you practice and understand linear inequalities better. These solutions have been developed by Vedantu's experienced teachers to simplify and make learning clear. Solving and representing these problems can help you become more skilled in solving inequality problems. It is important to practise this to absorb the material completely and perform well on examinations.

Class 11 Maths Chapter 5: Exercises Breakdown

CBSE Class 11 Maths Chapter 5 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for CBSE Class 11 Maths