# NCERT Solutions Class 11 Maths Chapter 6

## NCERT Solutions Class 11 Mathematics Chapter 6- Linear Inequalities

Class 11 is the foundation for Class 12. The NCERT Solutions Class 11 Mathematics Chapter 6 is dependable and has been curated in a very simple and precise format. The solutions cover all the answers to the textual questions including all the concepts of the chapter. With the help of the Class 11 Mathematics Chapter 6 solutions , students fetch excellent marks in their exams as well as competitive entrance examinations such as JEE Mains and JEE Advanced. NCERT Solutions offer thorough and quick revision before the exams.

The NCERT Solutions Class 11 Mathematics Chapter 6 discusses the various concepts related to Linear Inequalities. To clearly understand all concepts of this chapter, students must refer to the NCERT Solutions Class 7 and NCERT Solutions Class 8 to recall all the related concepts. The NCERT Solutions Class 11 Mathematics Chapter 6 provides solved key to textual questions which clears the conceptual understanding of the students which finally helps them to gain high scores in the examinations. Using these solutions, they can comprehend all theorems, sums and formulas from this chapter. Students must consider using Chapter 6 Class 11 Mathematics solutions provided by Extramarks for effective preparation.

### Key Topics Covered in NCERT Solutions Class 11 Mathematics Chapter 6

The subject matter experts at Extramarks have covered the following key topics in the Chapter 6 Mathematics Class 11 solutions through detailed research and analysis:

 Exercise Topic 6.1 Introduction 6.2 Inequalities 6.3 Algebraic Solutions of Linear Inequalities and Graphical Representation 6.4 Solutions of the Linear Inequalities in Two Variables using graphs 6.5 Solution of System of Linear Inequalities

Introduction:

In this Chapter 6, Mathematics Class 11, students will study linear equalities. The previous knowledge about equations in one and two variables studied introduced in the NCERT solutions Class 9 and NCERT solutions Class 10 is used in this Chapter. In NCERT Solutions Class 11 Mathematics Chapter 6, students will learn to translate statement problems in the equation format, using signs of inequalities such as less than (<), greater than (>), less than equal to (), and greater than equal to ().

The study of NCERT Solutions Class 11 Mathematics Chapter 6 is very useful in the field of science, optimisation problems, statistics, etc.

Inequalities:

Students will learn about the term ‘inequality’, which is the relation between two real numbers or algebraic expressions. They will also gain understanding about the different types of inequalities, such as numerical inequalities, literal inequalities and double inequalities. A complete explanation is given in the NCERT Solutions Class 11 Mathematics Chapter 6 with the help of several examples of inequalities. Students will also understand the difference between strict inequalities and slack inequalities.

Algebraic Solutions of Linear Inequalities and Graphical Representation:

In this section, students gain knowledge about the left-hand side (LHS) and right-hand side (RHS) of the inequalities. They will also learn about the solutions and the solution set of the inequalities. A true statement is also defined in this section. It also recalls the rules for solving a linear equation. Similarly, rules for solving linear inequalities with solved examples are also mentioned in this part. In the Chapter 6 Class 11 Mathematics, students will also learn about visual representation using graphs of inequalities in one variable.

Students must refer to the NCERT Solutions Class 11 Mathematics Chapter 6 for practising several questions.

Graphical Solution of Linear Inequalities with Two Variables:

In this section, students will learn to draw a graph and represent the relation between inequalities of two variables. Concepts such as the half-planes and their types (lower and upper half-planes) are discussed in this chapter. Students will learn to solve sums based on the three possibilities, namely:

(i) ax + by = c

(ii) ax + by > c

(iii) ax + by < c

The solution region is defined in this section. Several examples are included to explain the procedure for solving linear inequalities in two variables.

1. The solution region contains all the solutions to the linear inequality.
2. To identify the half-plane, which is represented by the linear inequality, it is necessary to take the point (a, b) and check if it satisfies the inequality or not. If it satisfies, then the inequality will represent the half-plane and shade the solution region which contains the point.
3. If the linear inequality is of the form ax + by > c or of the form ax + by < c, then any points on the line ax + by = c are also included in the solution region. Draw a dark  line in the solution region.
4. If the linear inequality is of the form ax + by > c or of the form ax + by < c, then any points on the line ax + by = c will not be included in the solution region. Draw a broken or dotted line.

Visit the Extramarks platform to get access to the NCERT Solutions Class 11 Mathematics Chapter 6

NOTE:

1. Any point which lies in the half-plane II satisfies ax + by > c.
2. Any point satisfying the inequality ax + by > c lies in half-plane II.
3. If b < 0, then the point which satisfies ax + by > c lies in the half-plane I.
4. All points satisfying ax + by > c lies in half-planes II or I as b > 0 or b < 0.
5. The graph of inequality ax + by > c will be the half-plane (solution region). It is represented by shading.

Solution of System of Linear Inequalities:

In the previous section, students learnt to solve linear inequalities involving one or two variables using graphs. In this section, they will learn to illustrate the method for solving a system of linear inequalities in two variables.

### NCERT Solutions Class 11 Mathematics Chapter 6: Exercise & Solutions

The NCERT Solutions Class 11 Mathematics Chapter 6 Linear Inequalities are available for students on the Extramarks’ mobile application and website . The notes comprehensively explain and provide detailed answers to all the textual questions included in the Chapter 6 Class 11 Mathematics. Several solved examples are used to explain the practical applications of linear inequalities. The NCERT Solutions for Class 11 Mathematics Chapter 6 help students to revise all important formulas, derivations and key points. With knowledge of Class 11 Mathematics Chapter 6, students will be able to tackle all difficult questions asked in the examinations.

Click on the below links to learn and practice the Extramarks NCERT Solutions Class 11 Mathematics Chapter 6:

Students can also gain access to the other study materials like the revision notes, CBSE sample papers and mock tests. Students who are preparing for the board exams are advised to use the NCERT Solutions Class 11 Mathematics Chapter 6 from our Extramarks’ website. They can also go through several NCERT reference books for preparing efficiently.

• NCERT Solutions Class 1
• NCERT Solutions Class 2
• NCERT Solutions Class 3
• NCERT Solutions Class 4
• NCERT Solutions Class 5
• NCERT Solutions Class 6

### NCERT Exemplar Class 11 Mathematics

The NCERT Exemplar Mathematics notes are an important study material which students must refer to. It includes all detailed explanations and helps students to gain an in-depth understanding of all complex topics. Students can also practice several challenging questions from the  chapter. Using the Exemplar notes along with NCERT Solutions Class 11 Mathematics Chapter 6 will help students to enhance their problem-solving and thinking abilities. The study materials on the Extramarks platform will help students prepare for board exams such as JEE Mains, IIT, etc.

The platform of Extramarks provides the best learning experience to all students. Through the best study notes, we provide guidance and support to enable academic progress. Students are advised to practice and study using the all-inclusive NCERT Solutions Class 11 Mathematics Chapter 6 and solve many sample papers and past  years’ question papers.

### Key Features of NCERT Solutions Class 11 Mathematics Chapter 6

• The NCERT Solutions Class 11 Mathematics Chapter 6 explains all concepts in a detailed manner using proper logic and reasoning.
• Students will learn tips, tricks and methods to solve problems quickly and easily.
• The NCERT Solutions Class 11 Mathematics Chapter 6 is based on the CBSE syllabus and follow the latest guidelines by the CBSE.
• Students can prepare for board exams and annual examinations effectively.
• These notes help to clarify all doubts by providing detailed and accurate answers to the chapter end exercises. This enables students to solve the questions and cross-check their answers with an authentic source. By practising through NCERT Solutions Class 11 Mathematics Chapter 6, students will gain confidence and problem-solving abilities.

Q.1 Solve 24x < 100, when
(i) x is a natural number.
(ii) x is an integer.

Ans

$\begin{array}{l}\text{We are given}\\ \text{\hspace{0.17em}24x}<\text{1}00\\ \mathrm{or}\frac{\text{24x}}{24}<\frac{100}{24}\left[\mathrm{Dividing}\text{both sides by 24}\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<\frac{25}{6}\\ \left(\mathrm{i}\right)\mathrm{When}\text{x is a natural number, in this case following values}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of x make the statement true.}\\ \text{x}=1,2,3,4.\\ \mathrm{The}\text{solution set of the inequality is}\left\{1,2,3,4\right\}.\\ \left(\mathrm{ii}\right)\mathrm{When}\text{x is an integer, the solutions of the given inequlity are}\\ ...\text{,}-3,-2,-1\text{, 0, 1, 2, 3,4.}\\ \mathrm{The}\text{solution set of the inequality is}\left\{...\text{,}-3,-2,-1\text{, 0, 1, 2, 3,4}\right\}.\end{array}$

Q.2 Solve – 12x > 30, when
(i) x is a natural number.
(ii) x is an integer.

Ans

$\begin{array}{l}\text{We are given}\\ \text{\hspace{0.17em}}-12\mathrm{x}>30\\ \mathrm{or}\frac{-12\mathrm{x}}{-12}<\frac{30}{-12}\left[\mathrm{Dividing}\text{both sides by}-12\right]\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<-\frac{5}{2}\\ \left(\mathrm{i}\right)\mathrm{When}\text{x is a natural number, in this case following no value}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}of x make the statement true.}\\ \therefore \mathrm{There}\text{is no solution.}\\ \left(\mathrm{ii}\right)\mathrm{When}\text{x is an integer, the solutions of the given inequality are}\\ ...\text{,}-4,-3.\\ \mathrm{The}\text{solution set of the inequality is}\left\{...\text{,}-4,-3\right\}.\end{array}$

Q.3 Solve 5x – 3 < 7, when
(i) x is an integer.
(ii) x is a real number.

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5x}-\text{3}<\text{7}\\ \text{or 5x}-\text{3}+\text{3}<\text{7}+3\left[\mathrm{Adding}\text{3 both sides}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}5x}<10\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{5x}}{5}<\frac{10}{5}\left[\mathrm{Divide}\text{both sides by 5}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<2\\ \left(\mathrm{i}\right)\mathrm{When}\text{}\mathrm{x}\text{is an integer, the solutions of given inequality are}\\ ...,-3,-2,-1,0,1.\\ \mathrm{The}\text{solution set of the inequality is}\left\{...,-3,-2,-1,0,1\right\}.\\ \left(\mathrm{ii}\right)\mathrm{When}\text{}\mathrm{x}\text{is a real number, the solutions of given inequality are}\\ \text{given by x}<\text{2, i.e., all real numbers x which are less than 2.\hspace{0.17em}}\\ \text{Therefore, the solution set of the inequality is x}\in \left(-\mathrm{\infty },2\right).\end{array}$

Q.4 Solve 3x + 8 > 2, when
(i) x is an integer.
(ii) x is a real number.

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x}+\text{8}>\text{2}\\ \text{or 3x}+\text{8}-\text{8}>\text{2}-8\left[\mathrm{Adding}\left(-8\right)\text{both sides}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}3x}>-\text{\hspace{0.17em}}6\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{\text{3x}}{3}>\frac{-\text{\hspace{0.17em}}6}{3}\left[\mathrm{Divide}\text{both sides by 3}\right]\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}>-2\\ \left(\mathrm{i}\right)\mathrm{When}\text{}\mathrm{x}\text{is an integer, the solutions of given inequality are}\\ -1,0, 1,2,3,...\\ \mathrm{The}\text{solution set of the inequality is}\left\{-1,0, 1,2,3,...\right\}.\\ \left(\mathrm{ii}\right)\mathrm{When}\text{}\mathrm{x}\text{is a real number, the solutions of given inequality are}\\ \text{given by x}>-\text{2, i.e., all real numbers x which are greater than\hspace{0.17em}\hspace{0.17em}}-\text{2.\hspace{0.17em}}\\ \text{Therefore, the solution set of the inequality is x}\in \left(-2,\mathrm{\infty }\right).\end{array}$

Q.5 4x + 3 < 5x + 7

Ans

$\begin{array}{l}\text{We are given}\\ \text{4x}+\text{3}<\text{5x}+\text{7}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}4x}+\text{3}-\text{4x}<\text{5x}+\text{7}-\text{4x}\left[\mathrm{Subtracting}\text{4x from both sides}\right]\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3<\mathrm{x}+7\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3-7<\mathrm{x}+7-7\left[\mathrm{Subtracting}\text{7 from both sides}\right]\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-4<\mathrm{x}\\ \text{i.e., all the real numbers which are greater than}-\text{4, are the}\\ \text{solutions of the given inequality. Hence, the solution set is}\\ \text{(}-\text{4,}\mathrm{\infty }\text{).}\end{array}$

Q.6 3x – 7 > 5x – 1

Ans

$\begin{array}{l}\text{We are given}\\ \text{ 3x}-\text{7}>\text{5x}-\text{1}\\ \mathrm{or}\text{ 3x}-\text{7}-\text{3x}>\text{5x}-\text{1}-\text{3x }\left[\mathrm{Subtracting}\text{3x from both sides}\right]\\ \mathrm{or}\text{ }-\text{7}>\text{2x}-\text{1}\\ \mathrm{or}\text{ }-\text{7}+\text{1}>\text{2x}-\text{1}+1\text{ }\left[\mathrm{Adding}\text{1 both sides}\right]\\ \mathrm{or}\text{ }-6>\text{2x}\\ \mathrm{or}\text{ }-\frac{6}{2}>\frac{\text{2x}}{2}\text{ }\left[\mathrm{Dividing}\text{both sides by 2}\right]\\ \mathrm{or}\text{ }-3>\text{x}\\ \text{i.e., all the real numbers which are less than}-\text{3, are the}\\ \text{solutions of the given inequality. Hence, the solution set is}\\ \text{(}-\infty \text{,}-\text{3).}\end{array}$

Q.7 3(x – 1) ≤ 2 (x – 3)

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{ 3}\left(\text{x}-\text{1}\right)\le \text{2}\left(\text{x}-\text{3}\right)\\ \mathrm{or}\text{ }3\text{x}-3\le 2\text{x}-6\\ \mathrm{or}\text{ }3\text{x}-3-2\text{x}\le 2\text{x}-6-2\text{x}\left[\mathrm{Subtracting}\text{2x from both sides}\right]\\ \mathrm{or}\text{ x}-3\le -6\\ \mathrm{or}\text{ x}-3+3\le -6+3\left[\mathrm{Adding}\text{3 both sides}\right]\\ \mathrm{or}\text{ x}\le -3\\ \text{Thus, all real numbers x which are less than or equal to}-3\\ \text{are the solutions of the given inequality, i.e., x}\in \left(-\infty \text{,}-\text{3].}\end{array}$

Q.8 3(2 – x) ≥ 2 (1 – x)

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{3}\left(\text{2}-\text{x}\right)\ge \text{2}\left(\text{1}-\text{x}\right)\\ \mathrm{or}6-3\mathrm{x}\ge 2-2\mathrm{x}\\ \mathrm{or}6-3\mathrm{x}+3\mathrm{x}\ge 2-2\mathrm{x}+3\mathrm{x}\left[\mathrm{Adding}\text{3x both sides}\right]\\ \mathrm{or}6\ge 2+\mathrm{x}\\ \mathrm{or}6-2\ge 2+\mathrm{x}-2\left[\mathrm{Subtracting}\text{2 from both sides}\right]\\ \mathrm{or}4\ge \mathrm{x}\\ \text{Thus, all real numbers x which are less than or equal to}4\\ \text{are the solutions of the given inequality, i.e., x}\in \left(-\mathrm{\infty }\text{,\hspace{0.17em}}4\text{].}\end{array}$

Q.9

$\mathrm{x}+\frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{3}<11$

Ans

$\begin{array}{l}\mathrm{We}\mathrm{are}\mathrm{given}\\ \mathrm{x}+\frac{\mathrm{x}}{2}+\frac{\mathrm{x}}{3}<11\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{6\mathrm{x}+3\mathrm{x}+2\mathrm{x}}{6}<11\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{11\mathrm{x}}{6}<11\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{11\mathrm{x}}{6}×\frac{6}{11}<11×\frac{6}{11}\left[\mathrm{Multliplying}\text{a both sides by}\frac{6}{11}\right]\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<6\\ \text{Thus, all real numbers x which are less than}6\text{are the solutions}\\ \text{of the given inequality, i.e., x}\in \left(-\mathrm{\infty }\text{,\hspace{0.17em}}6\right)\text{.}\end{array}$

Q.10

$\frac{\mathbf{x}}{\mathbf{3}}\mathbf{>}\frac{x}{2}\mathbf{+}\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{\mathrm{x}}{3}>\frac{\mathrm{x}}{2}+1\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{\mathrm{x}}{3}-\frac{\mathrm{x}}{2}>\frac{\mathrm{x}}{2}+1-\frac{\mathrm{x}}{2}\left[\mathrm{Subtracting}\text{​}\frac{\mathrm{x}}{2}\text{from both sides}\right]\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{\mathrm{x}}{6}>1\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{\mathrm{x}}{6}×-\text{\hspace{0.17em}}6<1×-6\left[\mathrm{Multiplying}\text{both sides by}-\text{6}\right]\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<-6\\ \text{Thus, all real numbers x which are greater than}6\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in \left(-\mathrm{\infty },-6\right)\text{.}\end{array}$

Q.11

$\frac{3\left(\mathrm{x}-2\right)}{5}\le \frac{5\left(2-\mathrm{x}\right)}{3}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{3\left(\mathrm{x}-2\right)}{5}\le \frac{5\left(2-\mathrm{x}\right)}{3}\\ \mathrm{or}9\left(\mathrm{x}-2\right)\le 25\left(2-\mathrm{x}\right)\\ \mathrm{or}9\mathrm{x}-18\le 50-25\mathrm{x}\\ \mathrm{Adding}\text{25x both sides, we get}\\ 9\mathrm{x}-18+25\mathrm{x}\le 50-25\mathrm{x}+25\mathrm{x}\\ \mathrm{or}34\mathrm{x}-18\le 50\\ \mathrm{Adding}\text{18 both sides, we get}\\ 34\mathrm{x}-18+18\le 50+18\\ \mathrm{or}34\mathrm{x}\le 68\\ \mathrm{Dividing}\text{both sides by 34, we get}\\ \frac{34\mathrm{x}}{34}\le \frac{68}{34}\\ ⇒\mathrm{x}\le 2\\ \mathrm{i}.\mathrm{e}.,\text{\hspace{0.17em}all real numbers x which are less than or equal to}2\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in \left(-\mathrm{\infty },2\right]\text{.}\end{array}$

Q.12

$\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\left(}\frac{\mathbf{3}\mathbf{x}}{\mathbf{5}}\mathbf{+}\mathbf{4}\mathbf{\right)}\mathbf{\ge }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{\left(}\mathbf{x}\mathbf{–}\mathbf{6}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{1}{2}\left(\frac{3\mathrm{x}}{5}+4\right)\ge \frac{1}{3}\left(\mathrm{x}-6\right)\\ \mathrm{or}\frac{1}{2}\left(\frac{3\mathrm{x}+20}{5}\right)\ge \frac{1}{3}\left(\mathrm{x}-6\right)\\ \mathrm{or}\left(\frac{3\mathrm{x}+20}{10}\right)\ge \frac{1}{3}\left(\mathrm{x}-6\right)\\ \mathrm{or}3\left(3\mathrm{x}+20\right)\ge 10\left(\mathrm{x}-6\right)\\ \mathrm{or}9\mathrm{x}+60\ge 10\mathrm{x}-60\\ ⇒60+60\ge 10\mathrm{x}-9\mathrm{x}\\ ⇒120\ge \mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than or equal to 1}20\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in \left(-\mathrm{\infty },120\right]\text{.}\end{array}$

Q.13 2 (2x + 3) – 10 < 6 (x – 2)

Ans

$\begin{array}{l}\text{We are given}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2}\left(\text{2x}+\text{3}\right)-\text{1}0<\text{6}\left(\text{x}-\text{2}\right)\\ \mathrm{or}4\mathrm{x}+6-10<6\mathrm{x}-12\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}-4<6\mathrm{x}-12\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}12-4<6\mathrm{x}-4\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8<2\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4<\mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than}4\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in \left(4,\mathrm{\infty }\right).\end{array}$

Q.14

$\mathbf{37}\mathbf{–}\mathbf{\left(}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{5}\mathbf{\right)}\mathbf{\ge }\mathbf{9}\mathbf{x}\mathbf{–}\mathbf{8}\mathbf{\left(}\mathbf{x}\mathbf{–}\mathbf{3}\mathbf{\right)}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ 37–\left(3\mathrm{x}+5\right)\ge 9\mathrm{x}–8\left(\mathrm{x}–3\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}37-3\mathrm{x}-5\ge 9\mathrm{x}-8\mathrm{x}+24\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}32-3\mathrm{x}\ge \mathrm{x}+24\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}32-24\ge \mathrm{x}+3\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\ge 4\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\ge \mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than or equal to}2\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in \left(-\mathrm{\infty },2\right]\text{.}\end{array}$

Q.15

$\frac{\mathbf{x}}{\mathbf{4}}\mathbf{<}\frac{\mathbf{\left(}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\right)}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{\left(}\mathbf{7}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{\right)}}{\mathbf{5}}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{\mathrm{x}}{4}<\frac{\left(5\mathrm{x}-2\right)}{3}-\frac{\left(7\mathrm{x}-3\right)}{5}\\ \mathrm{or}\frac{\mathrm{x}}{4}<\frac{5\left(5\mathrm{x}-2\right)-3\left(7\mathrm{x}-3\right)}{15}\\ \mathrm{or}\frac{\mathrm{x}}{4}<\frac{25\mathrm{x}-10-21\mathrm{x}+9}{15}\\ \mathrm{or}\frac{\mathrm{x}}{4}<\frac{4\mathrm{x}-1}{15}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}15\mathrm{x}<16\mathrm{x}-4\\ \mathrm{or}4<16\mathrm{x}-15\mathrm{x}\\ \mathrm{or}4<\mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than}4\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in \left(4,\mathrm{\infty }\right)\text{.}\end{array}$

Q.16

$\frac{\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}\mathbf{\right)}}{\mathbf{3}}\mathbf{\ge }\frac{\mathbf{\left(}\mathbf{3}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\right)}}{\mathbf{4}}\mathbf{-}\frac{\mathbf{\left(}\mathbf{2}\mathbf{-}\mathbf{x}\mathbf{\right)}}{\mathbf{5}}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{\left(2\mathrm{x}-1\right)}{3}\ge \frac{\left(3\mathrm{x}-2\right)}{4}-\frac{\left(2-\mathrm{x}\right)}{5}\\ \mathrm{or}\frac{\left(2\mathrm{x}-1\right)}{3}\ge \frac{5\left(3\mathrm{x}-2\right)-4\left(2-\mathrm{x}\right)}{20}\\ \mathrm{or}\frac{\left(2\mathrm{x}-1\right)}{3}\ge \frac{15\mathrm{x}-10-8+4\mathrm{x}}{20}\\ \mathrm{or}\frac{\left(2\mathrm{x}-1\right)}{3}\ge \frac{19\mathrm{x}-18}{20}\\ \mathrm{or}20\left(2\mathrm{x}-1\right)\ge 3\left(19\mathrm{x}-18\right)\\ \mathrm{or}40\mathrm{x}-20\ge 57\mathrm{x}-54\\ \mathrm{or}54-20\ge 57\mathrm{x}-40\mathrm{x}\\ \mathrm{or}34\ge 17\mathrm{x}\\ \mathrm{or}2\ge \mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than or equal to}2\text{}\\ \text{are the solutions of the given inequality, i.e., x}\in \left(-\mathrm{\infty },2\right]\text{.}\end{array}$

Q.17 3x – 2 < 2x + 1

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.17em} \hspace{0.17em}3x}-\text{2}<\text{2x}+\text{1}\\ \text{or 3x}-2\mathrm{x}<1+2\\ \text{or \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<3\\ \text{The graphical representation of solutions is given in Fig}\mathrm{ure}.\end{array}$

Q.18

$5\mathrm{x}-3\ge 3\mathrm{x}-5$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}5\mathrm{x}-3\ge 3\mathrm{x}-5\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}5\mathrm{x}-3\mathrm{x}\ge -5+3\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}\ge -2\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge -1\\ \mathrm{The}\mathrm{graphical}\mathrm{representation}\mathrm{of}\mathrm{solutions}\mathrm{is}\mathrm{given}\mathrm{in}\mathrm{Figure}.\end{array}$

Q.19 3(1 – x) < 2(x + 4)

Ans

We are given
3(1 – x) < 2(x + 4)
or 3 – 3x < 2x + 8
or 3 – 8 < 2x + 3x
or – 5 < 5x
or – 1 < x
The graphical representation of solution is given below in the figure:

Q.20

$\frac{\mathbf{x}}{\mathbf{2}}\mathbf{\ge }\frac{\mathbf{\left(}\mathbf{5}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{\right)}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{\left(}\mathbf{7}\mathbf{x}\mathbf{-}\mathbf{3}\mathbf{\right)}}{\mathbf{5}}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \frac{\mathrm{x}}{2}\ge \frac{\left(5\mathrm{x}-2\right)}{3}-\frac{\left(7\mathrm{x}-3\right)}{5}\\ \mathrm{or}\frac{\mathrm{x}}{2}\ge \frac{5\left(5\mathrm{x}-2\right)-3\left(7\mathrm{x}-3\right)}{15}\\ \mathrm{or}\frac{\mathrm{x}}{2}\ge \frac{25\mathrm{x}-10-21\mathrm{x}+9}{15}\\ \mathrm{or}\frac{\mathrm{x}}{2}\ge \frac{4\mathrm{x}-1}{15}\\ \mathrm{or}15\mathrm{x}\ge 8\mathrm{x}-2\\ \mathrm{or}15\mathrm{x}-8\mathrm{x}\ge -2\\ \mathrm{or}7\mathrm{x}\ge -2\\ \mathrm{or}\mathrm{x}\ge -\frac{2}{7}\end{array}$

The graphical representation of solution is given below in the figure:

Q.21 Ravi obtained 70 and 75 marks in first two unit test. Find the minimum marks he should get in the third test to have an average of at least 60 marks.

Ans

$\begin{array}{l}\mathrm{Let}\text{Ravi obtained marks in third test}=\text{x}\\ \text{Ravi obtained marks in first two tests}=70\text{and 75}\\ \text{The average marks obtained by Ravi}=\frac{70+75+\mathrm{x}}{3}\\ \mathrm{According}\text{​​ to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{70+75+\mathrm{x}}{3}\ge 60\\ \text{\hspace{0.17em}\hspace{0.17em}}145+\mathrm{x}\ge 180\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 180-145\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 35\\ \mathrm{Thus},\text{​ Ravi should get greater or equal to 35 marks in}\\ \text{third unit test.}\end{array}$

Q.22 To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations (each of 100 marks). If Sunita’s marks in first four examinations are 87, 92, 94 and 95, find minimum marks that Sunita must obtain in fifth examination to get grade ‘A’ in the course.

Ans

$\begin{array}{l}\mathrm{Let}\text{the minimum marks obtained by Sunita in}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}fifth examination}=\text{x}\\ \text{Marks obtained by Sunita in first four}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}examinations}=87,92,94,95\\ \mathrm{Average}\text{marks obtained by Sunita}=\frac{87+92+94+95+\mathrm{x}}{5}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\frac{87+92+94+95+\mathrm{x}}{5}\ge 90\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{368+\mathrm{x}}{5}\ge 90\\ \mathrm{or}\text{\hspace{0.17em}}368+\mathrm{x}\ge 450\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 450-368\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 82\\ \mathrm{Thus},\text{​ Sunita will get grade ‘A’ in examination if she will get}\\ \text{marks greater than or equal to 82.}\end{array}$

Q.23 Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}x be the smaller number of the two consecutive}\\ \text{odd positive integers, so that the other one is x+2. Then}\\ \text{we should have}\\ \text{x}+\text{2}<\text{10}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<8\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\left(\text{x}+\text{2}\right)>11\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}>11-2\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}>\frac{9}{2}\\ \mathrm{x}>4.5\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4.5<\mathrm{x}<8\\ \mathrm{Since},\text{x is odd number. So, x will be 5 and 7.}\\ \text{Second number is x}+\text{2, so it wil be 7 and 9.}\\ \text{Thus the required pair of possible numbers will}\\ \text{be\hspace{0.17em}\hspace{0.17em}}\left(5,7\right)\text{and}\left(7,9\right).\end{array}$

Q.24 Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.

Ans

$\begin{array}{l}\mathrm{Let}\text{\hspace{0.17em}x be the smaller number of the two consecutive}\\ \text{even positive integers, so that the other one is x+2.}\\ \text{Then we should have}\\ \text{x}+\text{2}>\text{5}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}>3\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\left(\text{x}+\text{2}\right)<23\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}<23-2\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<\frac{21}{2}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}<10.5\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3<\mathrm{x}<10.5\\ \mathrm{Since},\text{x is even number. So, x will be 6, 8 and 10.}\\ \text{Second number is x}+\text{2, so it wil be 8, 10 and 12.}\\ \text{Thus the required pair of possible numbers will}\\ \text{be\hspace{0.17em}}\left(6,8\right)\text{,}\left(8,10\right)\text{and}\left(10,12\right).\end{array}$

Q.25 The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest side.

Ans

$\begin{array}{l}\mathrm{Let}\text{the shortest side of triangle}=\text{x\hspace{0.17em}cm}\\ \text{The longest side of triangle}=3\mathrm{x}\text{\hspace{0.17em}cm}\\ \mathrm{Third}\text{side of triangle}=\left(3\mathrm{x}-2\right)\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{The}\text{perimeter of the triangle}=\mathrm{x}+3\mathrm{x}+\left(3\mathrm{x}-2\right)\\ =\left(7\mathrm{x}-2\right)\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left(7\mathrm{x}-2\right)\ge 61\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}7\mathrm{x}\ge 63\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge \frac{63}{7}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 9\\ \mathrm{Thus},\text{the minimum length of the shortest side is 9 cm.}\end{array}$

Q.26 A man wants to cut three lengths from a single piece of board of length 91cm.
The second length is to be 3cm longer than the shortest and the third length is to twice as long as the shortest. What are the possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?

Ans

$\begin{array}{l}\text{The length of the single piece of board}=91\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{Let}\text{t}\mathrm{he}\text{length of the smaller piece of board}=\mathrm{x}\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{The}\text{​ length of the second piece of board}=\left(\mathrm{x}+3\right)\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{The}\text{length of the third piece of board}=2\mathrm{x}\text{\hspace{0.17em}}\mathrm{cm}\\ \mathrm{According}\text{to given conditions:}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2x}\ge \left(\mathrm{x}+3\right)+\text{5}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 8...\left(1\right)\\ \mathrm{Since},\text{these the total length of these three pieces will at most}\\ \text{equal to 91 cm. Then,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}+\left(\mathrm{x}+3\right)+2\mathrm{x}\le 91\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}\le 91-3\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}4\mathrm{x}\le 88\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\le 22...\left(2\right)\\ \mathrm{From}\text{​}\left(1\right)\text{​ and}\left(2\right),\text{we have}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}8}\le \text{x}\le \text{22}\\ \mathrm{Thus},\text{the length of the shortest piece of wood is greater or equal}\\ \text{to 8 cm but less than or equal to 22 cm.}\end{array}$

Q.27 x + y < 5

Ans

Graph of x + y = 5 is given as dotted line in the figure.

This line divides the xy-plane in two half planes I and II.
We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that
(0) + (0) < 5
or 0 < 5 , which is true.
Hence, half plane I is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane I excluding the points on the line is the solution region of the inequality.

Q.28

$2\mathrm{x}+\mathrm{y}\ge 6$

Ans

Graph of 2x + y = 6 is given as line in the figure.

This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that 2(0) + (0) = 6 or 0 = 6 , which is false.
Hence, half plane II is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that
2(0) + (0) ≥ 6
or 0 ≥ 6 , which is false.
Hence, half plane II is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

Q.29

$3\mathrm{x}+4\mathrm{y}\le 12$

Ans

Graph of 3x + 4y = 12 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

3(0) + 4(0) ≤ 12
or 0 ≤ 12, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.30

$\mathrm{y}+8\ge 2\mathrm{x}$

Ans

Graph of y + 8 = 2x is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

(0) + 8 ≥ 2(0)
or 8 ≥ 0, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.31

$\mathrm{x}-\mathrm{y}\le 2$

Ans

Graph of x – y = 2 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

(0) – (0) ≤ 2
or 0 ≤ 2, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.32

$2\mathrm{x}-3\mathrm{y}>6$

Ans

Graph of 2x – 3y = 6 is given as dotted line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality, we note that

2(0) – 3(0) > 6
or 0 > 6, which is false.
Hence, half plane II is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

Q.33

$-3\mathrm{x}+2\mathrm{y}\ge \text{\hspace{0.17em}}-6$

Ans

Graph of –3x+2y = – 6 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0),
which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

– 3 (0) + 2(0) ≥ – 6
or 0 ≥ – 6, which is true.
Hence, half plane I is the solution region of the given inequality including all the points on line.
In other words, the shaded half plane I including the points on the line is the solution region of the inequality.

Q.34

$3\mathrm{y}-5\mathrm{x}<30$

Ans

Graph of 3y – 5x = 30 is given as line in the figure.
This line divides the xy-plane in two half planes I and II. We select a point (not on the line), say (0, 0),
which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

3(0) – 5(0) < 30
or 0 < 30, which is true.
Hence, half plane I is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane I excluding the points on the line is the solution region of the inequality.

Q.35 y < – 2

Ans

Graph of y = – 2 is given as line in the figure.
This line divides the xy-plane in two half planes I and II.
We select a point (not on the line), say (0, 0), which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

(0) < – 2, which is false.
Hence, half plane II is the solution region of the given inequality. Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane II excluding the points on the line is the solution region of the inequality.

Q.36 x > – 3

Ans

Graph of x = – 3 is given as line in the figure.
This line divides the xy-plane in two half planes I and II.
We select a point (not on the line), say (0, 0),
which lies in one of the half planes and determine if this point satisfies the given inequality,
we note that

0 > – 2, which is true.
Hence, half plane I is the solution region of the given inequality.
Clearly, any point on the line does not satisfy the given strict inequality.
In other words, the shaded half plane I excluding the points on the line is the solution region of the inequality.

Q.37 Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{\ge }\mathbf{3}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{2}$

Ans

$\begin{array}{l}\mathrm{We}\text{are given}\\ \mathrm{x}\ge 3...\left(\mathrm{i}\right)\\ \mathrm{y}\ge 2...\left(\mathrm{ii}\right)\\ \mathrm{We}\text{draw the graph of linear equations x}=\text{3 and y}=\text{2 in xy-plane}\\ \text{as shown in figure. Solution of inequality}\left(\text{1}\right)\text{is represented by}\\ \text{the}\text{shaded region right the line\hspace{0.17em}\hspace{0.17em}x}=\text{3},\text{including the points on}\\ \text{the line and solution of inequality}\left(\text{2}\right)\text{is represented by the}\\ \text{shaded region above the line y}=\text{2},\text{including the points on the}\\ \text{line}.\text{\hspace{0.17em}The common region of these two lines is shaded and it is the}\\ \text{required solution region of the given system of inequalities}.\end{array}$

Q.38 Solve the following system of inequalities graphically:

$\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\le }\mathbf{12}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{The}\text{given inequalities are:}\\ \text{3x}+\text{2y}\le \text{12 …}\left(1\right)\\ \text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}x}\ge \text{1 …}\left(2\right)\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}y}\ge \text{1 …}\left(3\right)\\ \mathrm{The}\text{graphs of following lines are drawn in the figure:}\\ \text{3x}+2\mathrm{y}=1\text{2,\hspace{0.17em}\hspace{0.17em}x}=\text{1 and y}=\text{1.}\end{array}$ $\begin{array}{l}\text{The inequality}\left(1\right)\text{shows shaded region below the line}\\ \text{3x + 2y = 12, inequality}\left(2\right)\text{shows the shaded region right of}\\ \text{the line x = 1 and inequality}\left(3\right)\text{shows the shaded region}\\ \text{above the line y}=\text{1.Hence, the common region of all these}\\ \text{lines is the solution of the given system of the linear inequalities.}\end{array}$

Q.39

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{6}\mathbf{,}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\le }\mathbf{12}$

Ans

$\begin{array}{l}\text{The given inequalities are:}\\ \mathrm{x}+\mathrm{y}\ge 4\text{ }...\left(1\right)\\ 2\mathrm{x}-\mathrm{y}\ge 0\text{ }...\left(2\right)\\ \text{The graphs of linear equations}\mathrm{x}+\mathrm{y}=4\text{and}2\mathrm{x}-\mathrm{y}=0\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x + y = 6, including the points on the line.
On the same set of axes, we draw the graph of the equation 3x + 4y = 12 as shown in Figure.
Then we see that inequality (2) represents the shaded region below the line 3x + 4y = 12, including the points on the line.

In this way, the double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.40

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{\ge }\mathbf{0}$

Ans

$\begin{array}{l}\text{The given inequalities are:}\\ \mathrm{x}+\mathrm{y}\ge 4\text{ }...\text{(1)}\\ 2\mathrm{x}-\mathrm{y}\ge 0\text{ }...\left(2\right)\\ \text{The graphs of linear equations}\mathrm{x}+\mathrm{y}=4\text{and}2\mathrm{x}-\mathrm{y}=0\\ \text{are drawn in}\mathrm{xy}\text{-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line x + y = 4, including the points on the line.
On the same set of axes, we draw the graph of the equation 2x – y = 0 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line 2x – y =0, including the points on the line.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.41 Solve the following system of inequalities graphically:

$\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{y}\mathbf{>}\mathbf{1}\mathbf{,}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{<}\mathbf{-}\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ 2\text{x}-\text{y}>1\text{ }...\left(1\right)\\ \text{x}-2\text{y}<-1\text{ }...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations 2x}-\text{y}=1\text{and x}-2\text{y}=-1\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x – y = 1, excluding the points on the line.
On the same set of axes, we draw the graph of the equation 2x – y = 1 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x – 2y = –1.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.42 Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le }\mathbf{6}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{4}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ \text{x}+\text{y}\le 6\text{ }...\left(1\right)\\ \text{x}+\text{y}\ge 4\text{ }...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations x}+\text{y}=6\text{and x}+\text{y}=4\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line x + y = 6, including the points on the line.
On the same set of axes, we draw the graph of the equation x + y = 4 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x + y = 4, including the points on the line.

In this way, the green shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.43 Solve the following system of inequalities graphically:

$\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{8}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{2}\mathbf{y}\mathbf{\ge }\mathbf{10}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ 2\text{x}+\text{y}\ge 8\text{ }...\left(1\right)\\ \text{x}+2\text{y}\ge 10\text{ }...\left(2\right)\\ \mathrm{The}\text{graphs of linear equations 2x}+\text{y}=8\text{and x}+2\text{y}=10\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region above the line 2x + y = 8,
including the points on the line.
On the same set of axes, we draw the graph of the equation x + 2y = 10 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x + 2y = 10,
including the points on the line.

In this way, the double shaded region, common to the above two shaded regions, is the required solution region of the given system of inequalities.

Q.44 Solve the following system of inequalities graphically:

$\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le }\mathbf{9}\mathbf{,}\mathbf{y}\mathbf{>}\mathbf{x}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{0}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ \text{x}+\text{y}\le 9\text{ }...\left(1\right)\\ \text{y}>\text{x }...\left(2\right)\\ \text{x}\ge 0\text{ }...\left(3\right)\\ \mathrm{The}\text{graphs of linear equations 2x}+\text{y}=8\text{and x}+2\text{y}=10\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line x + y = 9, including the points on the line.
On the same set of axes, we draw the graph of the equation y = x and x = 0 as shown in Figure.
Then we see that inequality (2) and (3) represent the shaded region above the line y = x and right of the line and
x = 0 excluding the points on the line respectively.

In this way, the triple shaded region, common to the above three shaded regions, is the required solution region of the given system of inequalities.

Q.45

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{5}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\le }\mathbf{60}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{1}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{2}$

Ans

$\begin{array}{l}\text{The given inequalities are:.}\\ 5\mathrm{x}+4\mathrm{y}\le 60\text{ }...\left(1\right)\\ \text{ }\mathrm{x}\ge 1\text{ }...\left(2\right)\\ \text{ }\mathrm{y}\ge 2\text{ }...\left(3\right)\\ \text{The graphs of linear equations}5\mathrm{x}+4\mathrm{y}=60,\mathrm{x}=1\text{and}\mathrm{y}=2\\ \text{are drawn in xy-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line 5x + 4y = 60, including the points on the line.
On the same set of axes, we draw the graph of the equation x = 1 and y = 2 as shown in Figure.
Then we see that inequality (2) represents the shaded region above the line x = 1 on the line including the points on the line and (3) represents the shaded region above the line y = 2 on the line including the points on the line.

In this way, green region is the common region. Which is the required solution region of the given system of inequalities.

Q.46

$\begin{array}{l}\mathbf{\text{Solve the following system of inequalities graphically:}}\\ \mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\le }\mathbf{60}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\le }\mathbf{30}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{0}\end{array}$

Ans

$\begin{array}{l}\text{The given inequalities are:}\\ 3\mathrm{x}+4\mathrm{y}\le 60\text{ }...\left(1\right)\\ \text{ }\mathrm{x}+3\mathrm{y}\le 30\text{ }...\left(2\right)\\ \text{ }\mathrm{x}\ge 0\text{ }...\left(3\right)\\ \text{ }\mathrm{y}\ge 0\text{ }...\left(4\right)\\ \text{The graphs of linear equations}3\mathrm{x}+4\mathrm{y}=60,\mathrm{x}+3\mathrm{y}=30\text{,}\\ \mathrm{x}=0\text{and}\mathrm{y}=0\text{are drawn in}\mathrm{xy}\text{-plane.}\end{array}$

Here, the solution of inequality (1) is represented by the shaded region below the line 3x + 4y = 60, including the points on the line.
On the same set of axes, we draw the graph of the equation x + 3y = 30, x = 0 and y = 0 as shown in Figure. Then we see that inequality (2), (3) and (4) represent the shaded region below the line x + 3y = 30,
x = 0 and y = 0 including the points on the line.

In this way, the green shaded region is the required solution region of the given system of inequalities.

Q.47

$\mathbf{\text{Solve the following system of inequalities graphically:}}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\ge }\mathbf{4}\mathbf{,}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{\le }\mathbf{3}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{–}\mathbf{3}\mathbf{y}\mathbf{\le }\mathbf{6}$

Ans

$\begin{array}{l}\text{The given inequalities are:.}\\ 2\mathrm{x}+\mathrm{y}\ge 4\text{ }...\left(1\right)\\ \text{ }\mathrm{x}+\mathrm{y}\le 3\text{ }...\left(2\right)\\ 2\mathrm{x}-3\mathrm{y}\le 6\text{ }...\left(3\right)\\ \text{The graphs of linear equations}2\mathrm{x}+\mathrm{y}=4,\mathrm{x}+\mathrm{y}=3,\\ \mathrm{and}\text{ }2\mathrm{x}-3\mathrm{y}=6\text{ are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality }\left(1\right)\text{is false for the point}\left(0,0\right),\text{so its solution area is above the line of equation}\\ \text{2x}+\text{y}=4\text{including the points on the line.}\\ \text{Inequality }\left(2\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is below the line of equation}\\ \text{x}+\text{y}=3\text{including the points on the line.}\\ \text{Inequality }\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is below the line of equation}\\ \text{2x}-\text{3y = 6 including the points on the line.}\\ \text{The common shaded area of the given three inequalities is the solution area.}\end{array}$

Q.48

$\mathbf{Solve}\mathbf{}\mathbf{the}\mathbf{}\mathbf{following}\mathbf{}\mathbf{system}\mathbf{}\mathbf{of}\mathbf{}\mathbf{inequalities}\mathbf{}\mathbf{graphically}\mathbf{:}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{2}\mathbf{y}\mathbf{\le }\mathbf{3}\mathbf{,}\mathbf{\text{\hspace{0.17em}}}\mathbf{3}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{y}\mathbf{\ge }\mathbf{12}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{0}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{1}$

Ans

$\begin{array}{l}\mathrm{The}\text{ given inequalities are:}\\ \text{ x}-2\text{y}\le 3\text{ }...\left(1\right)\\ 3\text{x}+4\text{y}\ge 12\text{ }...\left(2\right)\\ \text{ x}\ge 0\text{ }...\left(3\right)\\ \text{ y}\ge 1\text{ }...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations x}-2\text{y}=3\text{, }3\text{x}+4\text{y}=12,\text{x}=0\\ \text{and y}=\text{1 are drawn in xy-plane.}\\ \text{Inequality }\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is left of the line of equation}\\ \text{x}-2\text{y}=3\text{including the points on }\mathrm{the}\text{ }\mathrm{line}.\\ \text{Inequality }\left(2\right)\text{is false for the point}\left(0,0\right),\text{so its solution area is above the line of equation}\\ 3\text{x}+4\text{y}=12\text{including the points on the line.}\\ \text{Inequality }\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is right of the line of equation}\\ \text{x}=0\text{including the points on the line.}\end{array}$

$\begin{array}{l}\text{Inequality }\left(4\right)\text{is false for the point}\left(0,0\right),\text{so its solution area is above the line of equation}\\ \text{2x}-3\text{y}=6\text{including the points on }\mathrm{the}\text{ }\mathrm{line}.\\ \text{Thus, the common green area of the given three inequalities is the solution area.}\end{array}$

Q.49

$\mathbf{Solve}\mathbf{}\mathbf{the}\mathbf{}\mathbf{following}\mathbf{}\mathbf{system}\mathbf{}\mathbf{of}\mathbf{}\mathbf{inequalities}\mathbf{}\mathbf{graphically}\mathbf{:}\mathbf{}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{y}\mathbf{\le }\mathbf{60}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{2}\mathbf{x}\mathbf{,}\mathbf{x}\mathbf{\ge }\mathbf{3}\mathbf{,}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{\ge }\mathbf{0}$

Ans

$\begin{array}{l}\mathrm{The}\text{​ given inequalities are:}\\ 4\mathrm{x}+3\mathrm{y}\le 60...\left(1\right)\\ \mathrm{y}\ge 2\mathrm{x}...\left(2\right)\\ \mathrm{x}\ge 3...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations 4}\mathrm{x}+3\mathrm{y}=60\text{,\hspace{0.17em}}\mathrm{y}=2\mathrm{x}\text{,x}=\text{3,}\mathrm{x}=0\\ \text{and y}=\text{0 are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality }\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is below of the line of equation}\\ \text{4x}+3\text{y}=60\text{including the points on the line.}\\ \text{Inequality }\left(2\right)\text{is true for the point}\left(1,3\right),\text{so its solution area is above the line of equation}\\ \text{y}=2\text{x including the points on the line.}\\ \text{Inequality }\left(3\right)\text{is false for the point}\left(0,0\right),\text{so its solution area is right of the line of equation}\\ \text{x}=3\text{including the points on the line.}\\ \text{Inequality }\left(4\right)\text{is true for the values of x and y equal to zero or greater than zero.}\\ \text{So, its solution area is right of the x}-\mathrm{axis}\text{ and above to the y-axis.}\\ \text{Thus, the green shaded area of the given three inequalities is the solution area.}\end{array}$

Q.50

$\begin{array}{l}\text{Solve the following system of inequalities graphically:}\\ 3\text{x}+2\text{y}\le 150,\text{ x}+4\text{y}\le 80,\text{x}\le 15,\text{x}\ge 0,\text{ y}\ge 0\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{​ given inequalities are:}\\ 3\mathrm{x}+2\mathrm{y}\le 150...\left(1\right)\\ \mathrm{x}+4\mathrm{y}\le 80...\left(2\right)\\ \mathrm{x}\le 15...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations}3\mathrm{x}+2\mathrm{y}=150\text{,\hspace{0.17em}}\mathrm{x}+4\mathrm{y}=80\text{,\hspace{0.17em}x}=\text{15,}\\ \mathrm{x}=0\text{\hspace{0.17em}\hspace{0.17em}and y}=\text{0 are drawn in xy-plane.}\end{array}$

$\begin{array}{l}\text{Inequality }\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is below of the line of equation}\\ \text{3x}+2\text{y}=150\text{including the points on the line.}\\ \text{Inequality }\left(2\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is below the line of equation}\\ \text{x}+4\text{y}=80\text{including the points on the line.}\\ \text{Inequality }\left(3\right)\text{is true for the point}\left(0,0\right),\text{so its solution area is left of the line of equation}\\ \text{x}=15\text{including the points on the }\mathrm{line}.\\ \text{Inequality }\left(4\right)\text{is true for the values of x and y equal to zero or greater than zero.}\\ \text{So, its solution area is right of the x}-\mathrm{axis}\text{ and above to the y-axis.}\\ \text{Thus, the green shaded area of the given four inequalities is }\mathrm{the}\text{ }\mathrm{solution}\text{ }\mathrm{area}.\end{array}$

Q.51

$\begin{array}{l}\text{Solve the following system of inequalities graphically:}\\ \text{}\mathrm{x}+2\mathrm{y}\le 10,\mathrm{x}+\mathrm{y}\ge 1,\mathrm{x}-\mathrm{y}\le 0,\mathrm{x}\ge 0,\mathrm{y}\ge 0\end{array}$

Ans

$\begin{array}{l}\mathrm{The}\text{​ given inequalities are:}\\ \mathrm{x}+2\mathrm{y}\le 10...\left(1\right)\\ \mathrm{x}+\mathrm{y}\ge 1...\left(2\right)\\ \mathrm{x}-\mathrm{y}\le 0...\left(3\right)\\ \mathrm{x},\mathrm{y}\ge 0...\left(4\right)\\ \mathrm{The}\text{graphs of linear equations}\mathrm{x}+2\mathrm{y}=10\text{, \hspace{0.17em}}\mathrm{x}+\mathrm{y}=1\text{,\hspace{0.17em}\hspace{0.17em}}\mathrm{x}-\mathrm{y}=\text{0,}\\ \mathrm{x}=0\text{\hspace{0.17em}\hspace{0.17em}and y}=\text{0 are drawn in xy-plane.}\\ \text{Inequality\hspace{0.17em}}\left(1\right)\text{is true for the point}\left(0,0\right),\text{so its solution area}\\ \text{is below of the line of equation}\mathrm{x}+2\mathrm{y}=10\text{including the points}\\ \text{on the line.}\\ \text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(2\right)\text{is false for the point}\left(0,0\right),\text{so its solution area}\\ \text{is above the line of equation}\mathrm{x}+\mathrm{y}=1\text{including the points on}\\ \text{the line.}\end{array}$

$\begin{array}{l}\text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(3\right)\text{is true for the point}\left(2,1\right),\text{so its solution area}\\ \text{is above the line of equation}\mathrm{x}-\mathrm{y}=0\text{including the points on the}\\ \text{line.}\\ \text{Inequality\hspace{0.17em}\hspace{0.17em}}\left(4\right)\text{is true for the values of x and y equal to zero}\\ \text{or greater than zero. So, its solution area is right of the}\mathrm{x}-\mathrm{axis}\\ \text{and above to the y-axis.}\\ \text{Thus, the blue shaded area of the given four inequalities is}\\ \text{the solution area.}\end{array}$

Q.52

$2\le 3\mathrm{x}–4\le 5$

Ans

$\begin{array}{l}\mathrm{We}\text{have}\\ 2\le 3\mathrm{x}-4\le 5\end{array}$ $\begin{array}{l}\mathrm{or}6\le 3\mathrm{x}\le 9\\ \mathrm{or}\frac{6}{3}\le \mathrm{x}\le \frac{9}{3}\\ \mathrm{or}\text{\hspace{0.17em}}2\le \mathrm{x}\le 3\\ ⇒\mathrm{x}\in \left[2,3\right]\end{array}$

Q.53

$6\le -\mathbf{3}\left(2\mathrm{x}–4\right)<12$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{\hspace{0.17em}\hspace{0.17em}}6\le -3\left(2\mathrm{x}-4\right)<12\\ \mathrm{or}\text{\hspace{0.17em}}\frac{6}{3}\le -\left(2\mathrm{x}-4\right)<\frac{12}{3}\\ \mathrm{or}2\le -\text{\hspace{0.17em}}2\mathrm{x}+4<4\\ \mathrm{or}-2\le -\text{\hspace{0.17em}}2\mathrm{x}<0\\ \mathrm{or}1\le \mathrm{x}<0\\ ⇒\mathrm{x}\in \left(0,1\right]\end{array}$

Q.54

$-3\le 4-\frac{7\mathrm{x}}{2}\le 18$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ -3\le 4-\frac{7\mathrm{x}}{2}\le 18\\ \mathrm{or}\text{\hspace{0.17em}}-3-4\le -\frac{7\mathrm{x}}{2}\le 18-4\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-7\le -\frac{7\mathrm{x}}{2}\le 14\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-14\le -7\mathrm{x}\le 28\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{14}{7}\le -\mathrm{x}\le \frac{28}{7}\end{array}$ $\begin{array}{l}\mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2\le -\mathrm{x}\le 4\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\ge \mathrm{x}\ge -4\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\in \left[-4,2\right]\end{array}$

Q.55

$-15<\frac{3\left(\mathrm{x}-2\right)}{5}\le 0$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ -15<\frac{3\left(\mathrm{x}-2\right)}{5}\le 0\\ \mathrm{or}-75<3\left(\mathrm{x}-2\right)\le 0\\ \mathrm{or}-\frac{75}{3}>\left(\mathrm{x}-2\right)\le \frac{0}{3}\\ \mathrm{or}-25>\mathrm{x}-2\le 0\\ \mathrm{or}-25+2>\mathrm{x}\text{\hspace{0.17em}}\le 0+2\\ \mathrm{or}-23>\mathrm{x}\text{\hspace{0.17em}}\le 2\\ ⇒\mathrm{x}\in \left(-23,2\right]\end{array}$

Q.56

$-12<\mathbf{4}-\frac{3\mathbf{x}}{-5}\le 2$

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-12<4-\frac{3\mathrm{x}}{-5}\le 2\\ \mathrm{or}-12-4<\frac{3\mathrm{x}}{5}\le 2-4\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-16<\frac{3\mathrm{x}}{5}\le -2\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-16×5<3\mathrm{x}\le -2×5\end{array}$ $\begin{array}{l}\mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-80<3\mathrm{x}\le -10\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-\frac{80}{3}<\mathrm{x}\le -\frac{10}{3}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\in \left(-\frac{80}{3},-\frac{10}{3}\right]\end{array}$

Q.57

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

Ans

$\begin{array}{l}\mathrm{We}\text{have,}\\ 7\le \frac{\left(3\mathrm{x}+11\right)}{2}\le 11\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}14\le 3\mathrm{x}+11\le 22\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}14-11\le 3\mathrm{x}\le 22-11\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}3\le 3\mathrm{x}\le 11\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\le \mathrm{x}\le \frac{11}{3}\\ ⇒\mathrm{x}\in \left[1,\frac{11}{3}\right]\end{array}$

Q.58 Solve the inequalities in Exercises 7 to 10 and represent the solution graphically on number line.

7. 5x + 1 > – 24, 5x – 1 < 24
8. 2 (x – 1) < x + 5, 3 (x + 2) > 2 – x
9. 3x – 7 > 2(x – 6), 6 – x >11–2x
10.

$\mathbf{}\mathbf{5}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{–}\mathbf{7}\mathbf{\right)}\mathbf{–}\mathbf{3}\mathbf{\left(}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{3}\mathbf{\right)}\mathbf{\le }\mathbf{0}\mathbf{,}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{19}\mathbf{\le }\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{47}\mathbf{.}$

Ans

7.

The given inequalities are:
5x + 1 > – 24 …(1)
5x – 1 < 24 …(2)
From inequality (1), we have
5x + 1 > – 24
or 5x > – 25
or x > – 5 …(3)
From inequality (2), we have
5x – 1 < 24
or 5x < 25
or x < 5 …(4)
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 Figure.

Thus, solution of the system is real numbers x lying between – 5 and 5,
i.e., – 5 < x < 5,
or x ∈ (– 5, 5).

8.

The given inequalities are:
2(x – 1) < x + 5 …(1)
3(x + 2) > 2 – x …(2)
From inequality (1), we have
2(x – 1) < x + 5
or 2x – 2 < x + 5
or x < 7 …(3)
From inequality (2), we have
3(x + 2) > 2 – x
or 3x + 6 > 2 – x
or 4x > – 4
or x > – 1 …(4)
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 Figure.

Thus, solution of the system is real numbers x lying between – 1 and 7,
i.e., –1< x < 7,
or x ∈ (– 1, 7).

9.

3x – 7 > 2(x – 6) …(1)
6 – x > 11–2x …(2)
From inequality (1), we have
3x – 7 > 2(x – 6)
or 3x – 7 > 2x – 12
or x > – 5 …(3)
From inequality (2), we have
6 – x > 11–2x
or x > 5 …(4)
From equation (3) and equation (4), we can say that
x > 5

Thus, solution of the system is real numbers x lying between 5 and infinity,
i.e., 5 < x < ∞,
or x ∈ (5, ∞).

10.

$\begin{array}{l}5\left(2\text{x}-7\right)-3\left(2\text{x}+3\right)\le 0,2\text{x}+19\le 6\text{x}+47.\\ 5\left(2\text{x}-7\right)-3\left(2\text{x}+3\right)\le 0\text{ }...\left(1\right)\\ \text{ }2\text{x}+19\le 6\text{x}+47\text{ }...\left(2\right)\\ \text{From inequality}\left(\text{1}\right),\text{we have}\\ 10\text{x}-35-6\text{x}-9\le 0\\ 4\text{x}-44\le 0\\ 4\text{x}\le 44\\ \text{From inequality}\left(\text{1}\right),\text{we have}\\ 10\text{x}-35-6\text{x}-9\le 0\\ 4\text{x}-44\le 0\\ 4\text{x}\le 44\\ \text{x}\le 11\text{ }...\left(3\right)\\ \text{From inequality}\left(\text{2}\right),\text{we have}\\ 19-47\le 6\text{x}-2\text{x}\\ \text{ }-28\le 4\text{x}\\ \text{ }-7\le \text{x }...\left(4\right)\end{array}$

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 Figure.

$\begin{array}{l}\text{Thus},\text{solution of the system is real numbers x lying between}–\text{7 and 11},\\ \text{i}.\text{e}.,–\text{7}\le \text{x}\le \text{11},\\ \text{or}\text{x}\in \left[–\text{7},\text{11}\right].\end{array}$

Q.59

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

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{solution}\mathrm{is}\mathrm{to}\mathrm{be}\mathrm{kept}\mathrm{between}68\mathrm{°}\mathrm{F}\mathrm{}\mathrm{and}77\mathrm{°}\mathrm{F}.\\ \mathrm{So},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}68<\mathrm{F}<77\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}68<\frac{9}{5}\mathrm{C}+32<77\\ ⇒\text{\hspace{0.17em}}68-32<\frac{9}{5}\mathrm{C}<77-32\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}36<\frac{9}{5}\mathrm{C}<45\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}}36×\frac{5}{9}<\mathrm{C}<45×\frac{5}{9}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}20<\mathrm{C}<25\\ ⇒\mathrm{C}\in \left(20,25\right)\\ \mathrm{Thus},\text{range of temperature is 20°C to 25°C.}\end{array}$

Q.60 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

$\begin{array}{l}\mathrm{Let}\text{added 2% boric acid solution}=\text{x litres}\\ \text{Volume of 8% boric acid solution}=\text{640 litres}\\ \text{Total solution of boric acid}=\left(640+\mathrm{x}\right)\text{\hspace{0.17em}litres}\\ \mathrm{Therefore},\\ 2\mathrm{%}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{x}+8\mathrm{%}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}640>4\mathrm{%}\left(640+\mathrm{x}\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}2\mathrm{%}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}\mathrm{x}+8\mathrm{%}\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}640<6\mathrm{%}\left(640+\mathrm{x}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{2}{100}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\mathrm{x}+\frac{8}{100}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}640>\frac{4}{100}\text{\hspace{0.17em}}×\left(640+\mathrm{x}\right)\end{array}$ $\begin{array}{l}\mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}}\frac{2}{100}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\mathrm{x}+\frac{8}{100}\text{\hspace{0.17em}}×\text{\hspace{0.17em}}\mathrm{of}\text{\hspace{0.17em}}640<\frac{6}{100}\text{\hspace{0.17em}}×\left(640+\mathrm{x}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+8\text{\hspace{0.17em}}×\text{\hspace{0.17em}}640>4\left(640+\mathrm{x}\right)\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+8\text{\hspace{0.17em}}×\text{\hspace{0.17em}}640<6\text{\hspace{0.17em}}×\left(640+\mathrm{x}\right)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+5120>2560+4\mathrm{x}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\mathrm{x}+5120<3840+6\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2560>2\mathrm{x}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1280<4\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1280>\mathrm{x}\\ \mathrm{and}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}320<\mathrm{x}\\ ⇒\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\in \left(320,1280\right)\\ \mathrm{Thus},\text{2% solution of boric acid should be more than}\\ \text{320 litres but less than 1280 litres.}\end{array}$

Q.61 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

$\begin{array}{l}\mathrm{Let}\text{Volume of added water to acid}=\text{x litres}\\ \text{ Volume of 45% solution of acid}=\text{1125 litres}\\ \text{ Total solution of acid}=\left(1125+\text{x}\right)\text{ litres}\\ \mathrm{Therefore},\\ 45\text{% }\mathrm{of}\text{ }1125>25\text{%}\left(1125+\text{x}\right)\\ \mathrm{and}\text{ }45\text{% }\mathrm{of}\text{ }1125<30\text{%}\left(1125+\text{x}\right)\\ \mathrm{or}\text{ }\frac{45}{100}\text{ }×1125>\frac{25}{100}\text{ }×\left(1125+\text{x}\right)\\ \mathrm{and}\text{ }\frac{45}{100}\text{ }×\text{ }1125<\frac{30}{100}\text{ }×\left(1125+\text{x}\right)\\ \mathrm{or}\text{ }45\text{ }×1125>25\text{ }×\left(1125+\text{x}\right)\\ \mathrm{and}\text{ }445\text{ }×\text{ }1125<30\text{ }×\left(1125+\text{x}\right)\\ \mathrm{or}\text{ }50625>28125+25\text{x}\\ \mathrm{and}\text{ }50625<33750+30\text{x}\\ \mathrm{or}\text{ }50625-28125>25\text{x}\\ \mathrm{and}\text{ }50625-33750<30\text{x}\\ \mathrm{or}\text{ }22500>25\text{x}\\ \mathrm{and}\text{ }16875<30\text{x}\\ \mathrm{or}\text{ }900>\text{x}\\ \mathrm{and}\text{ }562.5<\text{x}\\ ⇒\text{ x}\in \left(562.5,900\right)\\ \mathrm{Thus},\text{added water should be more than}562.5\text{litres but less than 900 litres.}\end{array}$

Q.62

$\begin{array}{l}\text{IQ of a person is given by the formula}\\ \text{IQ=}\frac{\text{MA}}{\text{CA}}\text{×100,}\\ \begin{array}{l}\text{where MA is mental age and CA is chronological age.}\\ \text{If 80}\le \text{IQ}\le \text{140 for a group of 12 years old children,}\\ \text{find the range of their mental age.}\end{array}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{IQ}\mathrm{of}\text{a}\mathrm{person}\mathrm{is}\mathrm{given}\mathrm{by}\mathrm{the}\mathrm{formula}.\\ \mathrm{IQ}=\frac{\mathrm{MA}}{\mathrm{CA}}×100,\\ \mathrm{For}\text{a}\mathrm{group}\mathrm{of}\text{}12\text{}\mathrm{years}\mathrm{old}\mathrm{children}\\ 80\le \mathrm{IQ}\le 140\\ 80\le \frac{\mathrm{MA}}{\mathrm{CA}}×100\le 140\\ 80\le \frac{\mathrm{MA}}{12}×100\le 140\\ 80×\frac{12}{100}\le \mathrm{MA}\le 140×\frac{12}{100}\\ \text{ }\frac{48}{5}\le \mathrm{MA}\le \frac{84}{5}\\ \text{ }9.6\le \mathrm{MA}\le 16.8\\ \text{Thus, the range of mental age of a group of 12 years old}\\ \text{children is 9.6 to 16.8 years.}\end{array}$

## FAQs (Frequently Asked Questions)

### 1. Are the important concepts covered in the NCERT Solutions Class 11 Mathematics Chapter 6?

Yes. Chapter 6 Mathematics Class 11 starts with the introduction to linear inequalities. It defines and explains all the terms related to them. It also includes some important algebraic concepts to solve the linear inequalities in one or two variables with the help of their graphical representation. With regular practice of these NCERT Solutions Class 11 Mathematics Chapter 6, students can accomplish any desired learning goal like – studying a concept/chapter from scratch, revising before examination, etc.

### 2. Why should I practice the NCERT Solutions Class 11 Mathematics Chapter 6?

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