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

Last updated date: 02nd Aug 2024
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## NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities Miscellaneous Exercise - Free PDF Download

NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities includes solutions to all Miscellaneous Exercise problems. The NCERT Solutions for Class 11 Miscellaneous Exercises are based on the concepts presented in Maths Chapter 5. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the Class 11 Maths NCERT Solutions in PDF format and practice them offline. Get the latest CBSE Class 11 Maths Syllabus here.

Table of Content
1. NCERT Solutions for Class 11 Maths Chapter 5 Linear Inequalities Miscellaneous Exercise - Free PDF Download
2. Access NCERT Solutions for Class 11 Maths Chapter 13 Statistics
2.1Miscellaneous Exercise
3. Class 11 Maths Chapter 5: Exercises Breakdown
4. CBSE Class 11 Maths Chapter 5 Other Study Materials
5. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs
Competitive Exams after 12th Science

## Access NCERT Solutions for Class 11 Maths Chapter 13 Statistics

### 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

 Exercise Number of Questions Exercise 5.1 26 Questions & Solutions

## 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.

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

1. What is the importance of Miscellaneous Exercise Class 11 Chapter 5 Linear Inequalities?

In NCERT solutions of Miscellaneous Exercise Class 11 Chapter 5, Linear Inequalities are crucial for understanding more complex mathematical concepts. They form the basis for solving problems involving constraints and optimization. Mastery of linear inequalities is essential for higher studies in mathematics and other fields.

2. What should I focus on in the NCERT solutions of Miscellaneous Exercise Class 11 Chapter 5?

In NCERT solutions of Miscellaneous Exercise Class 11 Chapter 5, focus on understanding the graphical representation of inequalities. Practice solving systems of linear inequalities and interpreting their solutions. This exercise helps reinforce your conceptual knowledge and problem-solving skills.

3. What types of questions are commonly asked in the NCERT Class 11 Maths Chapter 5 Miscellaneous Exercise Solutions?

Common questions include solving linear inequalities algebraically and graphically. You will also encounter problems requiring the graphical representation of solutions. Real-life applications of inequalities are another frequent topic. Word problems that translate real-world scenarios into inequalities are common. Questions may also involve solving systems of inequalities and finding their solutions graphically.

4. How can I efficiently solve graphical representation problems in NCERT Class 11 Maths Chapter 5 Miscellaneous Exercise Solutions?

In NCERT Class 11 Maths Chapter 5 Miscellaneous Exercise Solutions, start by practising plotting inequalities on a graph accurately. Learn to identify and shade the feasible regions correctly. Understanding the intersection points of inequalities is crucial. Pay attention to the boundaries and open or closed nature of inequality lines. Familiarize yourself with different types of lines, such as dashed for '<' or '>' and solid for '≤' or '≥'. Practice various problems to gain confidence and speed. Review your graphs carefully to ensure accuracy. Consistent practice will enhance your graphing skills.

5. Are there any shortcuts to solving linear inequalities in Class 11 Maths Chapter 5 Miscellaneous Exercise Solutions?

In NCERT Class 11 Maths Chapter 5 Miscellaneous Exercise Solutions while there are no real shortcuts, a strong grasp of fundamental principles is crucial. Practice different types of problems to build speed and accuracy. Understanding the logic behind each step can make the process faster. Always double-check your solutions to ensure they meet the inequality conditions.

6. What topics are covered in the Linear Inequalities Class 11 Miscellaneous Exercise?

The Linear Inequalities Class 11 Miscellaneous Exercise includes a variety of problems related to the concepts of complex numbers and quadratic equations.

7. How can NCERT Solutions help me with the Linear Inequalities Class 11 Miscellaneous Exercise?

NCERT Linear Inequalities Class 11 Miscellaneous Exercise solutions provides clear, step-by-step answers to each question. This helps you understand how to solve each problem and learn the correct methods.

8. Why is practising the Linear Inequalities Class 11 Miscellaneous Exercise important?

Practising the Miscellaneous Exercise is important because it helps you apply what you’ve learned in the chapter. It covers different types of questions that can appear in exams, giving you good practice.

9. What is a good way to approach solving these Linear Inequalities Class 11 Miscellaneous Exercise problems?

A good way to solve these problems is to start by trying them on your own. Then, use NCERT Solutions to check your work and understand any mistakes. This helps reinforce your learning.

10. Are the questions in the Linear Inequalities Class 11 Miscellaneous Exercise likely to appear in exams?

Yes, the questions from the Miscellaneous Exercise are often similar to those asked in exams. Practising these problems can help you prepare effectively for your tests.