NCERT Solutions for Exercise 6.1 Class 11 Maths Chapter 6 - Linear Inequalities

# NCERT Solutions for Exercise 6.1 Class 11 Maths Chapter 6 - Linear Inequalities

Edited By Vishal kumar | Updated on Nov 06, 2023 12:01 PM IST

## NCERT Solutions for Class 11 Maths Chapter 6 - Linear Inequalities Exercise 6.1- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 6: Linear Inequalities Exercise 6.1- NCERT Solutions for Exercise 6.1 Class 11 Maths Chapter 6 is the very first exercise of the chapter Linear Inequalities. The NCERT has covered linear equations in the lower classes. The concepts of linear inequalities are introduced in the NCERT Class 11 Maths book and these concepts are important in solving problems in various fields like economics, optimisation problems, statistics, mathematics etc. Class 11 maths ex 6.1 gives practice problems on solving inequalities in one variable. Also, NCERT solutions for exercise 6.1 Class 11 Maths chapter 6 covers forming an inequality in one variable and solving them.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

NCERT book Class 11 Maths chapter 6 exercise 6.1 mainly deals with solutions of inequalities in one variable. A few more exercises are coming after NCERT syllabus 11th class maths exercise 6.1 answers. Those are listed below, which are prepared by subject matter experts at Careers360. They are presented in a straightforward language that is easy to comprehend. Additionally, these resources are available in PDF format, enabling students to access them at their convenience, free of cost, and without requiring an internet connection.

Linear Inequalities Exercise 6.2

Linear Inequalities Exercise 6.3

Linear Inequalities Miscellaneous Exercise

**As per the new CBSE Syllabus for 2023-24, this chapter has been assigned a different number, and it is now referred to as Chapter 5.

## Access Linear Inequalities Class 11 Chapter 6- Exercise: 6.1

Given : $24x < 100$

$\Rightarrow$ $24x < 100$

Divide by 24 from both sides

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

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

$\Rightarrow \, \, \, x< 4.167$

$x$ i s a natural number which is less than 4.167.

Hence, values of x can be $\left \{ 1,2,3,4 \right \}$

$x$ is an integer.

Given : $24x < 100$

$\Rightarrow$ $24x < 100$

Divide by 24 from both sides

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

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

$\Rightarrow \, \, \, x< 4.167$

$x$ i s are integers which are less than 4.167.

Hence, values of x can be $\left \{..........-3,-2,-1,0, 1,2,3,4 \right \}$

Given : $-12x>30$

$\Rightarrow$ $-12x>30$

Divide by -12 from both side

$\Rightarrow \, \, \, \frac{-12}{-12}x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< -2.5$

$x$ i s a natural number which is less than - 2.5.

Hence, the values of x do not exist for given inequality.

$x$ is an integer.

Given : $-12x>30$

$\Rightarrow$ $-12x>30$

Divide by -12 from both side

$\Rightarrow \, \, \, \frac{-12}{-12}x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< \frac{30}{-12}$

$\Rightarrow \, \, \, x< -2.5$

$x$ are integers less than - 2.5 .

Hence, values of x can be $\left \{ .............,-6,-5,-4,-3 \right \}$

$x$ is an integer.

Given : $5x - 3 < 7$

$\Rightarrow$ $5x - 3 < 7$

$\Rightarrow \, \, \, 5x< 10$

Divide by 5 from both sides

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

$\Rightarrow \, \, \, x< 2$

$x$ are integers less than 2

Hence, values of x can be $\left \{.........-3,-2-1,0,1,\right \}$

$x$ is a real number.

Given : $5x - 3 < 7$

$\Rightarrow$ $5x - 3 < 7$

$\Rightarrow \, \, \, 5x< 10$

Divide by 5 from both sides

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

$\Rightarrow \, \, \, x< 2$

$x$ are real numbers less than 2

i.e. $x\in (-\infty ,2)$

Given : $3x + 8 >2$

$\Rightarrow$ $3x + 8 >2$

$\Rightarrow \, \, \, 3x> -6$

Divide by 3 from both sides

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

$\Rightarrow \, \, \, x> - 2$

$x$ are integers greater than -2

Hence, the values of x can be $\left \{-1,0,1,2,3,4...............\right \}$ .

) $x$ is a real number.

Given : $3x + 8 >2$

$\Rightarrow$ $3x + 8 >2$

$\Rightarrow \, \, \, 3x> -6$

Divide by 3 from both side

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

$\Rightarrow \, \, \, x> - 2$

$x$ are real numbers greater than -2

Hence , values of x can be as $x\in (-2,\infty )$

Given : $4x + 3 < 5x + 7$

$\Rightarrow$$4x + 3 < 5x + 7$

$\Rightarrow \, \, \, 4x-5x< 7-3$

$\Rightarrow \, \, \, x> -4$

$x$ are real numbers greater than -4.

Hence, values of x can be as $x\in (-4 ,\infty )$

Given : $3x - 7 > 5x -1$

$\Rightarrow$$3x - 7 > 5x -1$

$\Rightarrow \, \, \, -2x> 6$

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

$\Rightarrow \, \, \, x< -3$

$x$ are real numbers less than -3.

Hence, values of x can be $x\in (-\infty ,-3)$

Given : $3(x-1) \leq 2(x-3)$

$\Rightarrow$$3(x-1) \leq 2(x-3)$

$\Rightarrow \, \, \, 3x-3\leq 2x-6$

$\Rightarrow \, \, \, 3x-2x\leq -6+3$

$\Rightarrow \, \, \, x\leq -3$

$x$ are real numbers less than equal to -3

Hence , values of x can be as , $x\in (-\infty ,-3]$

Given : $3(2- x) \geq 2(1-x)$

$\Rightarrow$$3(2- x) \geq 2(1-x)$

$\Rightarrow \, \, \, 6-3x\geq 2-2x$

$\Rightarrow \, \, \, 6-2\geq 3x-2x$

$\Rightarrow \, \, \, 4\geq x$

$x$ are real numbers less than equal to 4

Hence, values of x can be as $x\in (-\infty ,4]$

Given : $x + \frac{x}{2} + \frac{x}{3} < 11$

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

$\Rightarrow \, \, \, x(1+\frac{1}{2}+\frac{1}{3})< 11$

$\Rightarrow \, \, \, x(\frac{11}{6})< 11$

$\Rightarrow \, \, \, 11 x< 11\times 6$

$\Rightarrow \, \, \, x< 6$

$x$ are real numbers less than 6

Hence, values of x can be as $x\in (-\infty ,6)$

Given : $\frac{x}{3} > \frac{x}{2} + 1$

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

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

$\Rightarrow \, \, \,x (\frac{1}{3}-\frac{1}{2})> 1$

$\Rightarrow \, \, \,x (-\frac{1}{6})> 1$

$\Rightarrow \, \, \, -x > 6$

$\Rightarrow \, \, \, x< -6$

$x$ are real numbers less than -6

Hence, values of x can be as $x\in (-\infty ,-6)$

Given : $\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$

$\Rightarrow$$\frac{3(x-2)}{5} \leq \frac{5(2-x)}{3}$

$\Rightarrow \, \, \, 9(x-2)\leq 25(2-x)$

$\Rightarrow \, \, \, 9x-18\leq 50-25x$

$\Rightarrow \, \, \, 9x+25x\leq 50+18$

$\Rightarrow \, \, \, 34x\leq 68$

$\Rightarrow \, \, \, x\leq 2$

$x$ are real numbers less than equal to 2.

Hence, values of x can be as $x\in (-\infty ,2]$

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

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

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

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

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

$\Rightarrow \, \, 24 \geq \frac{x}{5}$

$\Rightarrow \, \, 120 \geq x$

$x$ are real numbers less than equal to 120.

Hence, values of x can be as $x\in (-\infty,120 ]$ .

Given : $2(2x + 3) - 10 < 6(x-2)$

$\Rightarrow$$2(2x + 3) - 10 < 6(x-2)$

$\Rightarrow \, \, \, 4x+6-10 < 6x-12$

$\Rightarrow \, \, \, 6-10+12 < 6x-4x$

$\Rightarrow \, \, \, 8 < 2x$

$\Rightarrow \, \, \, 4 < x$

$x$ are real numbers greater than 4

Hence , values of x can be as $x\in (4,\infty )$

Given : $37 - (3x + 5) \geq 9x - 8(x-3)$

$\Rightarrow$$37 - (3x + 5) \geq 9x - 8(x-3)$

$\Rightarrow \, \, \, 37 - 3x - 5 \geq 9x - 8x+24$

$\Rightarrow \, \, \, 32 - 3x \geq x+24$

$\Rightarrow \, \, \, 32 - 24 \geq x+3x$

$\Rightarrow \, \, \, 8 \geq 4x$

$\Rightarrow \, \, \, 2\geq x$

$x$ are real numbers less than equal to 2.

Hence , values of x can be as $x\in (-\infty ,2]$

Given : $\frac{x}{4}< \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

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

$\Rightarrow \, \, \, \, 15x< 20(5x-2)-12(7x-3)$

$\Rightarrow \, \, \, \, 15x< 100x-40-84x+36$

$\Rightarrow \, \, \, \, 15x< 16x-4$

$\Rightarrow \, \, \, \, 4< x$

$x$ are real numbers greater than 4.

Hence, values of x can be as $x\in (4,\infty)$

Given : $\frac{(2x - 1)}{3} \geq \frac{3x-2}{4} - \frac{(2-x)}{5}$

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

$\Rightarrow \, \, \, 20(2x - 1) \geq 15(3x-2) - 12(2-x)$

$\Rightarrow \, \, \, 40x - 20 \geq 45x-30 - 24+12x$

$\Rightarrow \, \, \, 30+24 - 20 \geq 45x-40x+12x$

$\Rightarrow \, \, \, 34 \geq 17x$

$\Rightarrow \, \, \, 2 \geq x$

$x$ are real numbers less than equal 2.

Hence, values of x can be as $x\in (-\infty,2 ]$ .

Given : $3x - 2 < 2x + 1$

$\Rightarrow$$3x - 2 < 2x + 1$

$\Rightarrow \, \, \, 3x - 2x< 2 + 1$

$\Rightarrow \, \, \, x< 3$

$x$ are real numbers less than 3

Hence, values of x can be as $x\in (-\infty ,3)$

The graphical representation of solutions of the given inequality is as :

Given : $5x - 3 \geq 3x -5$

$\Rightarrow$$5x - 3 \geq 3x -5$

$\Rightarrow \, \, \, 5x - 3x \geq 3 -5$

$\Rightarrow \, \, \, 2x \geq -2$

$\Rightarrow \, \, \, x \geq -1$

$x$ are real numbers greater than equal to -1.

Hence, values of x can be as $x\in [-1,\infty )$

The graphical representation of solutions of the given inequality is as :

Given : $3(1-x) < 2 (x +4)$

$\Rightarrow$$3(1-x) < 2 (x +4)$

$\Rightarrow \, \, \, 3- 3x< 2x + 8$

$\Rightarrow \, \, \, 3- 8< 2x + 3x$

$\Rightarrow \, \, \, -5< 5 x$

$\Rightarrow \, \, \, -1< x$

$x$ are real numbers greater than -1

Hence, values of x can be as $x\in (-1,\infty )$

The graphical representation of solutions of given inequality is as :

Given : $\frac{x}{2} \geq \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

$\Rightarrow$$\frac{x}{2} \geq \frac{(5x-2)}{3} - \frac{(7x-3)}{5}$

$\Rightarrow \, \, \, 15x \geq 10(5x-2) - 6(7x-3)$

$\Rightarrow \, \, \, 15x \geq 50x-20 - 42x+18$

$\Rightarrow \, \, \, 15x+42x-50x \geq 18-20$

$\Rightarrow \, \, \, 7x \geq -2$

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

$x$ are real numbers greater than equal to $= \frac{-2}{7}$

Hence, values of x can be as $x\in (-\frac{2}{7},\infty )$

The graphical representation of solutions of the given inequality is as :

Let x be marks obtained by Ravi in the third test.

The student should have an average of at least 60 marks.

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

$\, \, \, 145+x\geq 180$

$x\geq 180-145$

$x\geq 35$

the student should have minimum marks of 35 to have an average of 60

Sunita’s marks in the first four examinations are 87, 92, 94 and 95.

Let x be marks obtained in the fifth examination.

To receive Grade ‘A’ in a course, one must obtain an average of 90 marks or more in five examinations.

$\therefore \, \, \, \frac{87+92+94+95+x}{5}\geq 90$

$\Rightarrow \, \, \, \frac{368+x}{5}\geq 90$

$\Rightarrow \, \, \, 368+x\geq 450$

$\Rightarrow \, \, \, x\geq 450-368$

$\Rightarrow \, \, \, x\geq 82$

Thus, Sunita must obtain 82 in the fifth examination to get grade ‘A’ in the course.

Let x be smaller of two consecutive odd positive integers. Then the other integer is x+2.

Both integers are smaller than 10.

$\therefore \, \, \, x+2< 10$

$\Rightarrow \, \, \, \, x< 10-2$

$\Rightarrow \, \, \, \, x< 8$

Sum of both integers is more than 11.

$\therefore \, \, \, x+(x+2)> 11$

$\Rightarrow \, \, \, (2x+2)> 11$

$\Rightarrow \, \, \, 2x> 11-2$

$\Rightarrow \, \, \, 2x> 9$

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

$\Rightarrow \, \, \, x> 4.5$

We conclude $\, \, \, \, x< 8$ and $\, \, \, x> 4.5$ and x is odd integer number.

x can be 5,7.

The two pairs of consecutive odd positive integers are $(5,7)\, \, \, and\, \, \, (7,9)$ .

Let x be smaller of two consecutive even positive integers. Then the other integer is x+2.

Both integers are larger than 5.

$\therefore \, \, \, x> 5$

Sum of both integers is less than 23.

$\therefore \, \, \, x+(x+2)< 23$

$\Rightarrow \, \, \, (2x+2)< 23$

$\Rightarrow \, \, \, 2x< 23-2$

$\Rightarrow \, \, \, 2x< 21$

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

$\Rightarrow \, \, \, x< 10.5$

We conclude $\, \, \, \, x< 10.5$ and $\, \, \, x> 5$ and x is even integer number.

x can be 6,8,10.

The pairs of consecutive even positive integers are $(6,8),(8,10),(10,12)$ .

Let the length of the smallest side be x cm.

Then largest side = 3x cm.

Third side = 3x-2 cm.

Given: The perimeter of the triangle is at least 61 cm.

$\therefore\, \, \, x+3x+(3x-2)\geq 61$

$\Rightarrow \, \, \, 7x-2\geq 61$

$\Rightarrow \, \, \, 7x\geq 61+2$

$\Rightarrow \, \, \, 7x\geq 63$

$\Rightarrow \, \, \, x\geq \frac{63}{7}$

$\Rightarrow \, \, \, x\geq 9$

Minimum length of the shortest side is 9 cm.

[ Hint : If x is the length of the shortest board, then $x$ , $(x + 3)$ and $2x$ are the lengths of the second and third piece, respectively. Thus, $x + (x + 3) + 2x \leq 91$ and $2x \geq (x + 3) + 5$ ].

Let x is the length of the shortest board,

then $(x + 3)$ and $2x$ are the lengths of the second and third piece, respectively.

The man wants to cut three lengths from a single piece of board of length 91cm.

Thus, $x + (x + 3) + 2x \leq 91$

$4x+3\leq 91$

$\Rightarrow \, \, \, \, 4x\leq 91-3$

$\Rightarrow \, \, \, \, 4x\leq 88$

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

$\Rightarrow \, \, \, \, x\leq 22$

if the third piece is to be at least 5cm longer than the second, than

$2x \geq (x + 3) + 5$

$\Rightarrow \, \, \, \, 2x\geq x+8$

$\Rightarrow \, \, \, \, 2x-x\geq 8$

$\Rightarrow \, \, \, \, x\geq 8$

We conclude that $\, \, \, \, x\geq 8$ and $\, \, \, \, x\leq 22$ .

Thus , $8\leq x\leq 22$ .

Hence, the length of the shortest board is greater than equal to 8 cm and less than equal to 22 cm.

## More About NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.1

A total of twenty-six practice problems are given in the Class 11 Maths ch 6 ex 6.1. Solutions to exercise 6.1 Class 11 Maths are written in detail in this page for NCERT solutions for Class 11 Maths chapter 6 exercise 6.1. Questions 21 to 26 of 11th class maths exercise 6.1 answers are to form Linear Inequalities in one variable from the given statements. After Class 11th Maths chapter 6 exercise 6.1 linear inequalities in two variables are introduced in the NCERT book.

Also Read| Linear Inequalities Class 11th Notes

Benefits of NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.1

• The NCERT syllabus Class 11 chapter Linear Inequalities gives an insight into the basic concepts of linear inequalities and the exercise 6.1 Class 11 Maths gives practice questions on this.

• NCERT Solutions for Class 11 Maths chapter 6 exercise 6.1 give an idea of forming linear inequalities in one variable and solving them.

## Key Features of NCERT 11th Class Maths Exercise 6.1 Answers

1. Step-by-step explanations: Detailed, step-by-step ex 6.1 class 11 solutions for each problem to help students understand the concepts and problem-solving techniques.

2. Clarity and accuracy: Clear and precise presentation, ensuring students can confidently prepare for their exams and improve their understanding.

3. Variety of practice problems: Class 11 maths ex 6.1 typically includes a range of practice problems to help students reinforce their knowledge and problem-solving skills.

4. Accessibility: These class 11 ex 6.1 solutions are often available for free, making them easily accessible to students.

5. Format options: PDF versions of the solutions may be provided, allowing students to download and use them conveniently, both online and offline.

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## Subject Wise NCERT Exampler Solutions

1. Give an example of numerical inequality.

3<5 represents a numerical inequality

2. Write some examples of literal inequalities

x<5, y>2 etc...

3. Can you write a double inequality?

Yes, an example for double inequality is 3<5<7

4. List two inequalities in one variable

2x+3<0

2x+3>0

5. Write two inequalities in two variable

3x+4y>7

5x-2y<3

6. Write a quadratic inequality in one variable

2x^2+2x-2>0

7. Whether quadratic inequalities are solved in the exercise 6.1 Class 11 Maths

No, the NCERT solutions for Class 11 Maths chapter 6 exercise 6.1 deal with linear inequalities in one variable.

8. Give examples of strict inequalities

2x+3<0

5x+2y<0

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