# NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities (Ex 6.1) Exercise 6.1

In order to improve the quality of Indian education, the National Council of Educational Research and Training was established by the government. A key responsibility of this department is the development and publication of model textbooks, supplements, newsletters, journals, educational kits, and multimedia digital goods. There are both private and public secondary schools served by the Central Board of Secondary Education in India. It is the responsibility of the Government of India to govern the CBSE Board. In order to be promoted to Class 11, students must achieve a minimum score of 33% on both the theory and practical examinations during Class 11. A compartment exam is required in July for students who perform poorly in a particular subject. Those students who fail more than two compartments or two topics in the following year will be required to retake them all the following year.

The understanding of the world is based on Mathematics, which is among the most essential components of the human intellect and logic. Mathematics not only encourages logical reasoning and mental clarity but also develops mental discipline. Additionally, it is important to possess a good understanding of Mathematics in order to comprehend topics in preparation for the Class 11 final examinations. It is often necessary to use mathematical concepts to simplify Linear Inequalities Class 11 Exercise 6.1 in order to reduce their complexity. For students interested in pursuing careers in Science, Engineering, Commerce, Astronomy, and Economics, among other fields, an understanding of mathematical concepts and terminologies is essential.

The previous classes have covered equations with one variable and two variables as well as solving some statement problems through the translation of statements into equations. The term inequality refers to the case in which an unequal comparison is made between two mathematical expressions or two numbers. As a general rule, inequalities can either be numerical in nature or algebraic in nature, or they can be both. The concept of linear inequalities describes inequalities involving at least one linear algebraic expression. Different kinds of linear inequalities may be represented in a variety of ways. The purpose of this chapter is to examine linear inequalities in one and two variables. There are numerous disciplines in which inequalities are studied to solve problems, including science, Mathematics, Statistics, Economics, Psychology, and Economic Theory. Using the NCERT Solutions For Class 11 Maths Chapter 6 Exercise 6.1, students may be able to solve all the questions presented in the exercises in this chapter at a steady pace. Students can also comprehend the real-world applications of the chapter as a result of the concepts discussed in the NCERT solutions. A variety of competitive examinations, such as JEE, BITSAT, and other college entrance exams, can also be prepared using these provided solutions.

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## NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities (Ex 6.1) Exercise 6.1

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### What are Linear Inequalities?

An inequality involving a linear function is known as a linear inequality in Mathematics. There is a symbol of inequality associated with linear inequality. A graph is presented in the chapter that illustrates the data that is not equal.

### Rules to be Followed While Solving Inequalities

Students must pay attention when solving an inequality, as they can add the same quantity to each side and can subtract the same quantity from each side as well. They can multiply or divide each side by the same positive quantity. If they multiply or divide each side by a negative quantity, the inequality symbol must be reversed.

### Access NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities

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### NCERT Solutions for Class 11 Maths Chapters

For all the chapters of Class 11 Mathematics, NCERT Solutions can be found on Extramarks. Below are the chapter-wise solutions:

### NCERT Solution Class 11 Maths of Chapter 6 Exercise

Those students who find it difficult to locate the Class 11 Mathematics Chapter 6 questions can refer to the table below, which includes questions from all the exercises.

### Maths Chapter 6 Class 11 CBSE NCERT Solution

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**Q.1 ** Solve 24x < 100, when

(i) x is a natural number.

(ii) x is an integer.

**Ans.**

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

**Q.2 ** Solve – 12x > 30, when

(i) x is a natural number.

(ii) x is an integer.

**Ans.**

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

**Q.3 ** Solve 5x – 3 < 7, when

(i) x is an integer.

(ii) x is a real number.

**Ans.**

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

**Q.4 ** Solve 3x + 8 > 2, when

(i) x is an integer.

(ii) x is a real number.

**Ans.**

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

**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.**

$\begin{array}{l}\text{We are given}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3x}-\text{7}>\text{5x}-\text{1}\\ \mathrm{or}\text{\hspace{0.33em}3x}-\text{7}-\text{3x}>\text{5x}-\text{1}-\text{3x\hspace{0.33em}\hspace{0.33em}}\left[\mathrm{Subtracting}\text{3x from both sides}\right]\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}-\text{7}>\text{2x}-\text{1}\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}-\text{7}+\text{1}>\text{2x}-\text{1}+1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\left[\mathrm{Adding}\text{1 both sides}\right]\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}-6>\text{2x}\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}-\frac{6}{2}>\frac{\text{2x}}{2}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\left[\mathrm{Dividing}\text{both sides by 2}\right]\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}-3>\text{x}\\ \text{i.e., all the real numbers which are less than}-\text{3, are the}\\ \text{solutions of the given inequality. Hence, the solution set is}\\ \text{(}-\infty \text{,}-\text{3).}\end{array}$

**Q.7 ** 3(x – 1) ≤ 2 (x – 3)

**Ans.**

$\begin{array}{l}\mathrm{We}\text{are given}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}3}\left(\text{x}-\text{1}\right)\le \text{2}\left(\text{x}-\text{3}\right)\\ \mathrm{or}\text{\hspace{0.33em}}3\text{x}-3\le 2\text{x}-6\\ \mathrm{or}\text{\hspace{0.33em}}3\text{x}-3-2\text{x}\le 2\text{x}-6-2\text{x}\left[\mathrm{Subtracting}\text{2x from both sides}\right]\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}x}-3\le -6\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}x}-3+3\le -6+3\left[\mathrm{Adding}\text{3 both sides}\right]\\ \mathrm{or}\text{\hspace{0.33em}\hspace{0.33em}x}\le -3\\ \text{Thus, all real numbers x which are less than or equal to}-3\\ \text{are the solutions of the given inequality, i.e., x}\in (-\infty \text{,}-\text{3].}\end{array}$

**Q.8 ** 3(2 – x) ≥ 2 (1 – x)

**Ans.**

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

**Q.9 **

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

**Ans.**

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

**Q.10 **

**Ans.**

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

**Q.11 **

**Ans.**

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

**Q.12 **

**Ans.**

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

**Q.13 ** 2 (2x + 3) – 10 < 6 (x – 2)

**Ans.**

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

**Q.14 **

**Ans.**

$\begin{array}{l}\mathrm{We}\text{are given}\\ 37\u2013(3\mathrm{x}+5)\ge 9\mathrm{x}\u20138(\mathrm{x}\u20133)\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}37-3\mathrm{x}-5\ge 9\mathrm{x}-8\mathrm{x}+24\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}32-3\mathrm{x}\ge \mathrm{x}+24\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}32-24\ge \mathrm{x}+3\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}8\ge 4\mathrm{x}\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}2\ge \mathrm{x}\\ \mathrm{i}.\mathrm{e},\text{\hspace{0.17em}all real numbers x which are less than or equal to}2\text{are the}\\ \text{solutions of the given inequality, i.e., x}\in (-\mathrm{\infty},2]\text{.}\end{array}$

**Q.15 **

**Ans.**

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

**Q.16 **

**Ans.**

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

**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 **

**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 **

**Ans.**

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

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.**

$\begin{array}{l}\mathrm{Let}\text{the minimum marks obtained by Sunita in}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}fifth examination}=\text{x}\\ \text{Marks obtained by Sunita in first four}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}examinations}=87,92,94,95\\ \mathrm{Average}\text{marks obtained by Sunita}=\frac{87+92+94+95+\mathrm{x}}{5}\\ \mathrm{According}\text{to given condition,}\\ \text{\hspace{0.17em}\hspace{0.17em}}\frac{87+92+94+95+\mathrm{x}}{5}\ge 90\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}}\frac{368+\mathrm{x}}{5}\ge 90\\ \mathrm{or}\text{\hspace{0.17em}}368+\mathrm{x}\ge 450\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 450-368\\ \mathrm{or}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\mathrm{x}\ge 82\\ \mathrm{Thus},\text{ Sunita will get grade \u2018A\u2019 in examination if she will get}\\ \text{marks greater than or equal to 82.}\end{array}$

**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.**

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

**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.**

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

**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.**

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

**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}=(\mathrm{x}+3)\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 (\mathrm{x}+3)+\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}+(\mathrm{x}+3)+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}$

## FAQs (Frequently Asked Questions)

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