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

NCERT Solutions For Class 11 Maths Chapter 6 - Linear Inequalities can make your preparation for exams much easier. Maths is one of the more challenging subjects in the NCERT syllabus. So, these solved problems will help you understand and solve difficult questions effortlessly. Vedantu’s solutions for Chapter 6 Maths Class 11 cover all the NCERT questions on Linear Inequalities and gives you the ready reference that you need to succeed. Download the Free PDF on Class 11 Maths Chapter 6 NCERT Solutions.

NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.1  based on the following topics:

1. Introduction to Linear Inequalities

2. Inequalities

3. Algebraic Solutions of Linear Inequalities in One Variable and their Graphical Representation

4. Graphical Solution of Linear Inequalities in Two Variables

5. Solution of System of Linear Inequalities in Two Variables

### What are Linear Inequalities?

The expressions where two values are compared by the inequalities symbols such as '<', '>', '≤' or '≥ 'are termed linear inequalities. These values can be numerical, algebraic, or both.

### Rules to be Followed While Solving Inequalities

• Equal numbers may be subtracted or added from both sides of an equation without affecting the sign of inequality.

• When both sides of an inequality are added or subtracted by the same positive number, their sign remains unchanged. But when both sides of an equation are multiplied or divided by a negative number, then the sign of inequality is reversed.

Last updated date: 29th Sep 2023
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## Access NCERT Solutions for Class 11 Maths Chapter 6 – Linear Inequalities

Exercise 6.1

1. Solve $\text{24x}<\text{100}$ when

(i) $\text{x}$ is a natural number

Ans: We are given $\text{24x}<\text{100}$.

$24\text{x}<100$

$\Rightarrow \frac{\text{24x}}{\text{24}}\text{}<\frac{\text{100}}{\text{24}}$

$\Rightarrow \text{x}<\frac{25}{6}$

Therefore, we can say that only $1,2,3,4$ are only natural numbers less than $\frac{25}{6}$.

Therefore, the solution set is $\left\{ 1,2,3,4 \right\}$

(ii) $\text{x}$ is an integer.

Ans: We are given $\text{24x}<\text{100}$.

$24\text{x}<100$

$\Rightarrow \frac{\text{24x}}{\text{24}}<\frac{\text{100}}{\text{24}}$

$\Rightarrow \text{x}<\frac{\text{25}}{\text{6}}$

Therefore, we can say that only $..................-3,-2,-1,0,1,2,3,4.$ the integers less than $\frac{25}{6}$.

So, the solution set is $\left\{ ...............-3,-2,-1,0,1,2,3,4 \right\}$.

2. Solve $\mathrm{-12x}>\mathrm{30}$, when

(i) $\text{x}$ is a natural number

Ans: We are given $\text{-12x}>\text{30}$.

$\text{-12x}>\text{30}$

$\Rightarrow \frac{-12}{12}\text{x}>\frac{30}{12}$

$\Rightarrow -\text{x}>\frac{5}{2}$

$\Rightarrow \text{x}<-\frac{5}{2}$

Therefore, we can say that there is no natural number less than $\left( -\frac{5}{2} \right)$

So, there is no solution set for the given condition.

(ii) $\mathrm{x}$ is an integer.

We are given $\text{-12x}>\text{30}$.

$\text{-12x}>\text{30}$

$\Rightarrow \frac{-12}{12}\text{x}>\frac{30}{12}$

$\Rightarrow -\text{x}>-\frac{5}{2}$

$\Rightarrow \text{x}<\frac{5}{2}$

Therefore, we can say that $.........-5,-4,-3$ are the numbers that satisfy the given inequality.

Hence, the solution set is $\left\{ .....-5,-4,-3 \right\}$.

3. Solve $\mathrm{5x-3}<\mathrm{7}$, when

(i) $\text{x}$ is an integer.

Ans: We are given $\text{5x-3}<7$.

$\text{5x-3}<7$

$\Rightarrow \text{5x-3+3}<7+3$

$\Rightarrow \text{5x}<10$

$\Rightarrow \frac{\text{5}}{\text{5}}\text{x}<\frac{10}{5}$

$\Rightarrow \text{x}<2$

Therefore, we can say that $........-4,-3,-2,-1,0,1$ are the integers that satisfy the given inequality.

So, the solution set is $\left\{ ........-4,-3,-2,-1,0,1 \right\}$

(ii) $\mathrm{x}$ is a real number.

We are given $\text{5x-3}<7$.

$\text{5x-3}<7$

$\Rightarrow \text{5x-3+3}<7+3$

$\Rightarrow \text{5x}<10$

$\Rightarrow \frac{\text{5}}{\text{5}}\text{x}<\frac{10}{5}$

$\Rightarrow \text{x}<2$

Therefore, we can say that the real numbers which are less than $2$ will satisfy the inequality.

Hence, the solution set is $\text{x}\in \left( -\infty ,2 \right)$.

4. Solve $\mathrm{3x+8}>\mathrm{2}$, when

(i) $\text{x}$ is an integer.

Ans: We are given $\text{3x+8}>2$.

$3\text{x}+8>2$

$\Rightarrow 3\text{x}+8-8>2-8$

$\Rightarrow 3\text{x}>-6$

$\Rightarrow \frac{3}{3}\text{x}>\frac{-6}{3}$

$\Rightarrow \text{x}>-2$

Therefore, we can say that $-1,0,1,2,3.........$ are the integers that satisfy the given inequality.

So, the solution set is $\left\{ -1,0,1,2,3......... \right\}$

(ii) $\mathrm{x}$ is a real number.

$\text{3x+8}>\text{2}$

$\Rightarrow 3\text{x}+8-8>2-8$

$\Rightarrow 3\text{x}>-6$

$\Rightarrow \frac{3}{3}\text{x}>\frac{-6}{3}$

$\Rightarrow \text{x}>-2$

Therefore, we can say that the real numbers which are greater than $-2$ will satisfy the inequality.

Hence, the solution set is $\text{x}\in \left( -2,\infty\right)$.

5. Solve the given inequality for real $\mathrm{x:4x+3}<\mathrm{5x+7}$

Ans: Let us rewrite the given inequality, we get

$\text{4x+3}<\text{5x+7}$

$\Rightarrow \text{4x+3-7}<\text{5x+7-7}$

$\Rightarrow \text{4x-4}<\text{5x}$

$\Rightarrow \text{4x-4-4x}<\text{5x-4x}$

$\Rightarrow \text{-4}<\text{x}$

Therefore, we can say that the numbers greater than $\text{-4}$ will be the solutions that satisfy the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -4,\infty\right)$.

6. Solve the given inequality for real $\mathrm{x:3x-7}>\mathrm{5x-1}$

Ans: Let us rewrite the given inequality, we get

$3\text{x}-7>5x-1$

$\Rightarrow 3\text{x}-7\text{+75x-1+7}$

$\Rightarrow 3\text{x}>\text{5x+6}$

$\Rightarrow 3\text{x}-5\text{x}>5\text{x}-5\text{x}+6$

$\Rightarrow -2\text{x}>6$

$\Rightarrow \frac{-2}{-2}\text{x}<-\frac{6}{-2}$

$\Rightarrow \text{x}<-3$

Therefore, we can say that the numbers less than $\text{-3}$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,-3 \right)$.

7. Solve the given inequality for real $\mathrm{x:3}\left( \mathrm{x-1} \right)\leq\mathrm{2}\left( \mathrm{x-3} \right)$

Ans: Let us rewrite the given inequality, we get

$\text{3}\left( \text{x-1} \right)\leq\text{2}\left( \text{x-3} \right)$

$\Rightarrow \text{3x-3}\leq\text{2x-6}$

$\Rightarrow \text{3x-3+3}\leq\text{2x-6+3}$

$\Rightarrow \text{3x}\leq\text{2x-3}$

$\Rightarrow 3\text{x-2x}\leq\text{2x-2x-3}$

$\Rightarrow \text{x}\leq\text{-3}$

Therefore, we can say that the numbers less than or equal to $\text{-3}$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,-3 \right]$.

8. Solve the given inequality for real $\mathrm{x:3}\left( \mathrm{2-x} \right)\geq\mathrm{2}\left( \mathrm{1-x} \right)$

Ans: Let us rewrite the given inequality, we get

$\text{3}\left( \text{2-x} \right)\geq\text{2}\left( \text{1-x} \right)$

$\Rightarrow \text{6-3x}\geq\text{2-2x}$

$\Rightarrow \text{6-6-3x}\geq\text{2-6-2x}$

$\Rightarrow \text{-3x}\geq\text{-4-2x}$

$\Rightarrow \text{-3x+2x}\geq\text{-4-2x+2x}$

$\Rightarrow \text{-x}\geq\text{-4}$

$\Rightarrow \text{x}\leq\text{4}$

Therefore, we can say that the numbers less than or equal to $4$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,4 \right]$.

9. Solve the given inequality for real $\mathrm{x:x+}\frac{\mathrm{x}}{\mathrm{2}}\mathrm{+}\frac{\mathrm{x}}{\mathrm{3}}<\mathrm{11}$

Ans: Let us rewrite the given inequality, we get

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

$\Rightarrow \text{x}\left( \text{1+}\frac{\text{1}}{\text{2}}\text{+}\frac{\text{1}}{\text{3}} \right)<\text{11}$

$\Rightarrow \text{x}\left( \frac{\text{6+3+2}}{\text{6}} \right)<\text{11}$

$\Rightarrow \text{x}\left( \frac{\text{11}}{\text{6}} \right)<\text{11}$

$\Rightarrow \text{x}\left( \frac{\text{11}}{\text{6}} \right)\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{6}}{\text{11}} \right)\text{11 }\!\!\times\!\!\text{ }\frac{\text{6}}{\text{11}}$

$\Rightarrow \text{x}<\text{6}$

Therefore, we can say that the numbers less than $6$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,6 \right)$.

10. Solve the given inequality for real $\mathrm{x:}\frac{\mathrm{x}}{\mathrm{3}}>\frac{\mathrm{x}}{\mathrm{2}}\mathrm{+1}$

Ans: Let us rewrite the given inequality, we get

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

$\Rightarrow \frac{\text{x}}{\text{3}}\text{-}\frac{\text{x}}{\text{2}}>\frac{\text{x}}{\text{2}}\text{-}\frac{\text{x}}{\text{2}}\text{+1}$

$\Rightarrow \text{x}\left( \frac{\text{1}}{\text{3}}\text{-}\frac{\text{1}}{\text{2}} \right)>\text{1}$

$\Rightarrow \text{x}\left( \frac{\text{-1}}{6} \right)>\text{1}$

$\Rightarrow \text{x}\left( \frac{\text{-1}}{6} \right)\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{-6}}{1} \right)<\text{-6}$

$\Rightarrow \text{x}<\text{-6}$

Therefore, we can say that the numbers less than $-6$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,-6 \right)$.

11. Solve the given inequality for real $\mathrm{x:}\frac{\mathrm{3}\left( \mathrm{x-2} \right)}{\mathrm{5}}\leq\frac{\mathrm{5}\left( \mathrm{2-x} \right)}{\mathrm{3}}$

Ans: Let us rewrite the given inequality, we get

$\frac{\text{3}\left( \text{x-2} \right)}{\text{5}}\leq\frac{\text{5}\left( \text{2-x} \right)}{\text{3}}$

$\Rightarrow \text{9}\left( \text{x-2} \right)\leq\text{25}\left( \text{2-x} \right)$

$\Rightarrow \text{9x-18}\leq\text{50-25x}$

$\Rightarrow \text{9x+25x-18}\leq\text{50-25x+25x}$

$\Rightarrow \text{34x-18}\leq\text{50}$

$\Rightarrow \text{34x-18+18}\leq\text{50+18}$

$\Rightarrow \text{34x}\leq\text{68}$

$\Rightarrow \text{x}\leq\frac{\text{68}}{\text{34}}$

$\Rightarrow \text{x}\leq\text{2}$

Therefore, we can say that the numbers less than or equal to $2$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,2 \right]$.

12. Solve the given inequality for real $\mathrm{x:}\frac{\mathrm{1}}{\mathrm{2}}\left( \frac{\mathrm{3x}}{\mathrm{5}}\mathrm{+4} \right)\geq\frac{\mathrm{1}}{\mathrm{3}}\mathrm{x-2}$

Ans: Let us rewrite the given inequality, we get

$\frac{\text{1}}{\text{2}}\left( \frac{\text{3x}}{\text{5}}\text{+4} \right)\geq\frac{\text{1}}{\text{3}}\left( \text{x-6} \right)$

$\Rightarrow \text{3}\left( \frac{\text{3x}}{\text{5}}\text{+4} \right)\geq\text{2}\left( \text{x-6} \right)$

$\Rightarrow \frac{\text{9}}{\text{5}}\text{x+12}\geq\text{2x-12}$

$\Rightarrow \frac{\text{9}}{\text{5}}\text{x+12-12}\geq\text{2x-12-12}$

$\Rightarrow \frac{\text{9}}{\text{5}}\text{x}\geq\text{2x-24}$

$\Rightarrow \text{9x}\geq\text{10x-120}$

$\Rightarrow \text{9x-10x}\geq\text{10x-10x-120}$

$\Rightarrow \text{-x}\geq\text{-120}$

$\Rightarrow \text{x}\leq\text{120}$

Therefore, we can say that the numbers less than or equal to $120$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,120 \right]$.

13. Solve the given inequality for real $\mathrm{x:2}\left( \mathrm{2x+3} \right)\mathrm{-10}<\mathrm{6}\left( \mathrm{x-2} \right)$

Ans: Let us rewrite the given inequality, we get

$\text{2}\left( \text{2x+3} \right)\text{-10}<\text{6}\left( \text{x-2} \right)$

$\Rightarrow \text{4x+6-10}<\text{6x-12}$

$\Rightarrow \text{4x-4}<\text{6x-12}$

$\Rightarrow \text{4x-4+4}<\text{6x-12+4}$

$\Rightarrow \text{4x}<\text{6x-8}$

$\Rightarrow \text{4x-6x}<\text{6x-6x-8}$

$\Rightarrow \text{-2x}<\text{-8}$

$\Rightarrow \text{x}>\text{4}$

Therefore, we can say that the numbers greater than $4$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( 4,\infty\right)$.

14. Solve the given inequality for real $\mathrm{x:37-}\left( \mathrm{3x+5} \right)\geq\mathrm{9x-8}\left( \mathrm{x-3} \right)$

Ans: Let us rewrite the given inequality, we get

$\text{37-}\left( \text{3x+5} \right)\geq\text{9x-8}\left( \text{x-3} \right)$

$\Rightarrow \text{37-3x-5}\geq\text{9x-8x+24}$

$\Rightarrow \text{-3x+32}\geq\text{x+24}$

$\Rightarrow \text{-3x-x+32}\geq\text{x-x+24}$

$\Rightarrow \text{-4x+32}\geq\text{24}$

$\Rightarrow \text{-4x+32-32}\geq\text{24-32}$

$\Rightarrow \text{-4x\geq-8}$

$\Rightarrow \text{x\leq2}$

Therefore, we can say that the numbers less than or equal to $2$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,2 \right]$.

15. Solve the given inequality for real $\mathrm{x:}\frac{\mathrm{x}}{\mathrm{4}}<\frac{\left( \mathrm{5x-2} \right)}{\mathrm{3}}\mathrm{-}\frac{\left( \mathrm{7x-3} \right)}{\mathrm{5}}$

Ans: Let us rewrite the given inequality, we get

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

$\Rightarrow \frac{\text{x}}{\text{4}}<\frac{\text{5}\left( \text{5x-2} \right)\text{-3}\left( \text{7x-3} \right)}{\text{15}}$

$\Rightarrow \frac{\text{x}}{\text{4}}<\frac{\text{25x-10-21x+9}}{\text{15}}$

$\Rightarrow \frac{\text{x}}{\text{4}}<\frac{\text{4x-1}}{\text{15}}$

$\Rightarrow \text{15x}<\text{4}\left( \text{4x-1} \right)$

$\Rightarrow \text{15x-16x}<\text{16x-16x-4}$

$\Rightarrow \text{-x}<\text{-4}$

$\text{x}>\text{4}$

Therefore, we can say that the numbers greater $4$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( 4,\infty\right)$.

16. Solve the given inequality for real $\mathrm{x:}\frac{\left( \mathrm{2x-1} \right)}{\mathrm{3}}\geq\frac{\left( \mathrm{3x-2} \right)}{\mathrm{4}}\mathrm{-}\frac{\left( \mathrm{2-x} \right)}{\mathrm{5}}$

Ans: Let us rewrite the given inequality, we get

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

$\Rightarrow \frac{\left( \text{2x-1} \right)}{\text{3}}\geq\frac{\text{5}\left( \text{3x-2} \right)\text{-4}\left( \text{2-x} \right)}{\text{20}}$

$\Rightarrow \frac{\left( \text{2x-1} \right)}{\text{3}}\text{=}\frac{\text{15x-10-8+4x}}{\text{20}}$

$\Rightarrow \frac{\left( \text{2x-1} \right)}{\text{3}}\geq\frac{\text{19x-18}}{\text{20}}$

$\Rightarrow \text{20}\left( \text{2x-1} \right)\geq\text{3}\left( \text{19x-18} \right)$

$\Rightarrow \text{40x-20}\geq\text{57x-54}$

$\Rightarrow \text{-20+54}\geq\text{57x-40x}$

$\Rightarrow \text{34}\geq\text{17x}$

$\Rightarrow \text{2}\geq\text{x}$

Therefore, we can say that the numbers less than or equal to $4$ will be in the solution set for satisfying the given condition.

Therefore, we can say that the solution set for the given inequality is $\left( -\infty ,2 \right]$.

17. Solve the inequality and show the graph of the solution in number line:

$\mathrm{3x-2}<\mathrm{2x+1}$

Ans: Let us rewrite the given inequality, we get

$\text{3x-2}<\text{2x+1}$

$\Rightarrow \text{3x-2x}<\text{1+2}$

$\Rightarrow \text{x}<\text{3}$

Therefore, the graphical representation for the inequality in number line is: 18. Solve the inequality and show the graph of the solution in number line:

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

Ans: Let us rewrite the given inequality, we get

$\text{5x-3}\geq\text{3x-5}$

$\Rightarrow \text{5x-3x}\geq\text{-5+3}$

$\Rightarrow \text{2x}\geq\text{-2}$

$\Rightarrow \frac{\text{2x}}{\text{2}}\geq\frac{\text{-2}}{\text{2}}$

$\Rightarrow \text{x}\geq\text{-1}$

Therefore, the graphical representation for the inequality in number line is: 19. Solve the inequality and show the graph of the solution in number line:

$\mathrm{3}\left( \mathrm{1-x} \right)<\mathrm{2}\left( \mathrm{x+4} \right)$

Ans: Let us rewrite the given inequality, we get

$\text{3}\left( \text{1-x} \right)<\text{2}\left( \text{x+4} \right)$

$\Rightarrow 3-3\text{x}<\text{2x+8}$

$\Rightarrow \text{3-8}<\text{2x+3x}$

$\Rightarrow \text{-5}<\text{5x}$

$\Rightarrow \frac{-5}{5}<\frac{5\text{x}}{5}$

$\Rightarrow -1<\text{x}$

Therefore, the graphical representation for the inequality in number line is: 20. Solve the inequality and show the graph of the solution in number line:

$\frac{\mathrm{x}}{\mathrm{2}}\geq\frac{\left( \mathrm{5x-2} \right)}{\mathrm{3}}\mathrm{-}\frac{\left( \mathrm{7x-3} \right)}{\mathrm{5}}$

Ans: Let us rewrite the given inequality, we get

$\frac{\text{x}}{\text{2}}\geq\frac{\left( \text{5x-2} \right)}{\text{3}}\text{-}\frac{\left( \text{7x-3} \right)}{\text{5}}$

$\Rightarrow \frac{\text{x}}{\text{2}}\geq\frac{\text{5}\left( \text{5x-2} \right)\text{-3}\left( \text{7x-3} \right)}{\text{15}}$

$\Rightarrow \frac{\text{x}}{\text{2}}\geq\frac{\text{25x-10-21x+9}}{\text{15}}$

$\Rightarrow \frac{\text{x}}{\text{2}}\geq\frac{\text{4x-1}}{\text{15}}$

$\Rightarrow \text{15x}\geq\text{2}\left( \text{4x-1} \right)$

$\Rightarrow \text{15x}\geq\text{8x-2}$

$\Rightarrow \text{15x-8x}\geq\text{8x-2-8x}$

$\Rightarrow \text{7x}\geq\text{-2}$

$\Rightarrow \text{x}\geq\text{-}\frac{\text{2}}{\text{7}}$

Therefore, the graphical representation for the inequality in number line is: 21. Ravi obtained $\mathrm{70}$ and $\mathrm{75}$ marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least $\mathrm{60}$ marks.

Ans: Let us assume $\text{x}$ as the marks obtained by Ravi in the third unit test.

As it is given that the student must have an average of $\text{60}$ marks,

Therefore, the expression is:

$\frac{\text{70+75+x}}{\text{3}}\geq\text{60}$

$\Rightarrow \text{145+x}\geq\text{180}$

$\Rightarrow \text{x}\geq\text{180-145}$

$\Rightarrow \text{x}\geq\text{35}$

Therefore, we can say that the student should obtain greater than or equal

to $35$ marks.

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

Ans: Let us assume $\text{x}$ as the fifth examination marked by Sunita.

As it is given that to obtain ‘A’ one should obtain an average of $\text{90}$ marks or

more in five examinations.

So, the expression is:

$\frac{87+92+94+95+\text{x}}{5}\geq90$

$\Rightarrow \frac{368+\text{x}}{5}\geq90$

$\Rightarrow 368+\text{x}\geq450$

$\Rightarrow \text{x}\geq450-368$

$\Rightarrow \text{x}\geq82$

Therefore, we can say that sunita must obtain $\text{82}$ or greater than it to obtain

23. Find all pairs of consecutive odd positive integers both of which are smaller than $\mathrm{10}$ such that their sum is more than $\mathrm{11}$.

Ans: Let us assume $\text{x}$ as the smallest number in two consecutive odd integers. Then, the other odd integer will be $\text{x+2}$.

As it is given that both the integers are smaller than $\text{10}$, we get

$\text{x+2}<10$

$\Rightarrow \text{x}<\text{10-2}$

$\Rightarrow \text{x}8$

Let us assume it as inequality $\left( 1 \right)$, we get

$\text{x}<8.............\left( 1 \right)$

Also, it is given that the sum of the two integers is more than $\text{11}$.

So,

$\text{x+}\left( \text{x}+2 \right)>11$

$\Rightarrow 2\text{x}+2>11$

$\Rightarrow 2\text{x}>11-2$

$\Rightarrow 2\text{x}>\text{9}$

$\Rightarrow \text{x}>\frac{9}{2}$

$\Rightarrow \text{x}>\text{4}\text{.5}$

Let us assume this as inequality $\left( 2 \right)$

$\text{x}>4.5........\left( 2 \right)$

Therefore, from inequality $\left( 1 \right)$ and $\left( 2 \right)$, we get that

$\text{x}$ can take the values of

$5$ and $7$.

Therefore, the required pairs are $\left( 5,7 \right)$ and $\left( 7,9 \right)$.

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

Ans: Let us assume $\text{x}$ as the smallest number in two consecutive even integers.

Then, the other odd integer will be $\text{x+2}$.

As it is given that both the integers are greater than $5$, we get

$\text{x}>\text{5}..........\left( 1 \right)$

Also, it is given that the sum of the two integers is less than $23$.

So,

$\text{x+}\left( \text{x}+2 \right)<23$

$\Rightarrow 2\text{x}+2<23$

$\Rightarrow 2\text{x}<\text{23}-2$

$\Rightarrow 2\text{x}<\text{21}$

$\Rightarrow \text{x}<\frac{21}{2}$

$\Rightarrow \text{x}<\text{10}\text{.5}$

Let us assume this as inequality $\left( 2 \right)$

$\text{x}<10.5........\left( 2 \right)$

From $\left( 1 \right)$ and $\left( 2 \right)$, we obtain

$5<\text{x}<\text{10}\text{.5}$.

Therefore, we can say that $\text{x}$satisfies $\text{6,8}$ and $\text{10}$ values.

Therefore, the pairs that are possible are $\left( 6,8 \right)$, $\left( 8,10 \right)$ and $\left( 10,12 \right)$.

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

Ans: Let us assume the shortest side of the given triangle as $\text{x}$.

As it is given that the longest side is the three times the shortest side, we

get longest side$\text{=3x}$

Therefore, the length of the other side will be $\left( 3x-2 \right)$

As it is given that the perimeter should be at least $61$ cm, we can write it as:

$\text{xcm+3xcm+}\left( 3\text{x}-2 \right)\text{cm=61cm}$

$\Rightarrow \text{7x-2}\geq\text{61}$

$\Rightarrow \text{7x}\geq\text{61+2}$

$\Rightarrow \text{7x}\geq\text{63}$

$\Rightarrow \text{x}\geq\text{9}$

Therefore, we can say that the shortest side of the given triangle should

satisfy the inequality$\text{x}\geq\text{9}$.

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

Ans: Let us assume the length of the shortest piece be $\text{x}$ cm.

From the question we can write that the second and third piece will be $\left( \text{x+3} \right)$ cm and $\text{2x}$ respectively.

As it is given that the three lengths are single piece of board of length $\text{91}$ cm,

$\text{x cm+}\left( \text{x+3} \right)\text{cm+2xcm=91cm}$

$\Rightarrow \text{4x+3}\leq\text{91}$

$\Rightarrow \text{4x}\leq\text{91-3}$

$\Rightarrow \text{4x}\leq\text{88}$

$\Rightarrow \frac{4}{4}\text{x}\leq\frac{88}{4}$

$\Rightarrow \text{x}\leq22........\left( 1 \right)$

Also, the third piece is at least $5\text{cm}$ longer than the second piece.

$\therefore \text{2x}\geq\left( \text{x+3} \right)+5$

$\Rightarrow 2\text{x}\geq\text{x+8}$

$\Rightarrow \text{x}\geq\text{8}..........\left( 2 \right)$

Therefore, from $\left( 1 \right)$and $\left( 2 \right)$, we get

$8\leq\text{x}\leq22$

So, we can say that the shortest board should be greater than or equal to

$8$ cm but less than or equal to $22$ cm.

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

 Chapter 6 - Linear Inequalities Exercises in PDF Format Exercise 6.1 26 Questions & Solutions Exercise 6.2 10 Questions & Solutions Exercise 6.3 15 Questions & Solutions Miscellaneous Exercise 14 Questions & Solutions

## Maths Chapter 6 Class 11 CBSE NCERT Solution

Maths Chapter 6 Class 11 CBSE NCERT Solution by Vedantu includes all the top NCERT questions. Each solution comes with detailed diagrams and in-depth instructions on linear inequalities. This way, you can follow the solutions step by step and score big even if you are faced with difficult questions.

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## FAQs on NCERT Solutions for Class 11 Maths Chapter 6 - Exercise

1. Why must I refer to online NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Exercise 6.1?

Class 11 CBSE students must refer to NCERT Solutions for Maths Chapter 6 Linear Inequalities for Exercise 6.1 and other exercises to have a thorough understanding of the chapter. Provided by ed-tech solutions like Vedantu, these are the best study material to avail answers to all the exercise questions at one place. With the help of NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Ex 6.1, students will be able to solve the questions conveniently on their own. Students can refer to the solutions provided by subject matter experts for easy doubt clearance and revision during exams. The solutions are provided in a step-by-step manner for an easy understanding. These are designed as per the latest guidelines and syllabus. Students can refer to exercise-wise NCERT Solutions for Class 11 Maths Chapter 6 to score well in the exams.

2. Where can Class 11 students find NCERT Solutions for Maths Chapter 6 Linear Inequalities Exercise 6.1?

Vedantu provides exercise-wise NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities. These solutions are in a free to download PDF format. Chapter 6 of CBSE Class 11 Maths is important to score well in Class 11 exams as well as competitive exams. Students can find NCERT Solutions for Exercise 6.1 of Chapter 6 on Vedantu’s site provided by expert tutors. Students must solve every problem given in the exercise and can refer to Vedantu’s site for the needed solutions. They can also register for masterclasses by experts to understand the chapter and clear any further doubts.

3. What are some of the key aspects of Vedantu’s NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities Exercise 6.1?

Designed by Vedantu, the exercise-wise NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities are the most comprehensive and well-designed study material available online. NCERT Solutions for Class 11 Mathematics Chapter 6 Linear Inequalities Exercise 6.1 include step-by-step explanations for each problem. Vedantu’s NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.1 allow quick revision, a thorough understanding of the chapter and easy doubt resolution. The Key aspects of NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.1 are that these are easy to access, available for free of cost and designed by subject matter experts.

4. What are other study materials available on Vedantu for Class 11 Maths Chapter 6 Linear Inequalities?

Apart from exercise-wise NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequalities, students can also find important questions for the chapter, revision notes, previous year papers and solved sample question papers. All these study materials are designed by expert teachers at Vedantu who have years of teaching experience. They are also well versed in NCERT guidelines and exam pattern. Students can download the free PDFs of these materials to practice the chapter and score well in the exam. Students can also avail online LIVE Classes on Vedantu by experts to clear any doubts regarding the chapter.

5. How many questions are covered in NCERT Solutions for Chapter 6 of Class 11 Maths?

Chapter 6 of NCERT Solutions for Class 11 Maths is called ‘Linear Inequalities’. There are a total of three exercises in the chapter with an additional miscellaneous exercise. NCERT Solutions for Class 11 Maths Chapter 6 cover the following number of questions with solutions in each exercise:

• Exercise 6.1 - 26 Questions

• Exercise 6.2 - 10 Questions

• Exercise 6.3 - 15 Questions

• Miscellaneous Exercise - 14 Questions

6. What is Chapter 6 of Class 11 Maths NCERT about?

Chapter 6 - Linear Inequalities of Class 11 Maths NCERT talks about the concepts, techniques, and theories involved in solving linear inequalities. This chapter will also help students understand the one major difference between solving inequality and a linear equation. Students will be able to learn more about these through solving various subjective and objective questions given in the exercises. Some questions will also require students to draw graphical representations to solve them.

7. Do I need to practise all the questions provided in NCERT Solutions for Chapter 6 of Class 11 Maths?

All questions that have been covered in the NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.1 have equal importance and should be prepared with seriousness. Questions that may be asked in your Class 11 Maths Exams cannot be predicted so prepare accordingly. Hence, students must make sure they practice all questions and their solutions to avoid losing any marks in their exams. To get the solutions of all the NCERT questions for Chapter 6 free of cost students can refer to the Vedantu app or website.

8. Is Exercise 6.1 of Class 11 Maths easy?

With regular practice Class 11 Exercise, 6.1 becomes easy. Before a student proceeds, it becomes important that they are clear with the concepts. When they do that, the exercise becomes very easy. However, it is also important to come back to it every now and then and at least solve a couple of questions from the exam point of view. For a complete explanation of the exercise, you can visit Vedantu.

9. Should I practise the examples provided in Exercise 6.1 of Chapter 6 of NCERT Class 11 Maths?

Examples are an important part of practice when preparing for your Class 11 Maths exams. Questions in each exercise are based on the examples provided before them. Understanding how these examples have been solved will help you solve the questions using the same concepts and techniques. Often, examples are given as questions in the exams, and practising them will give you an advantage in achieving better marks. Make sure you do not miss practising all examples provided Exercise 6.1 of Chapter 6 of NCERT Class 11 Maths.