# NCERT Solutions for Class 11 Maths Chapter 6

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequations (Raikhik Asamikaen) Exercise 6.1, Exercise 6.2, Exercise 6.3 and Miscellaneous in English to view online or download in PDF format free.

All UP Board Solutions and NCERT Solutions 2020-21 are based on new NCERT Books 2020-2021 following latest CBSE Syllabus 2020-2021. Ask your questions through discussion forum and share your knowledge with the other uses.

## NCERT Solutions for Class 11 Maths Chapter 6

Class: | 11 |

Subject: | Maths |

Chapter 6: | Linear Inequations |

### 11th Maths Chapter 6 Solutions

NCERT Solutions for Class 11 Maths Chapter 6 Linear Inequations Exercise 6.1 to 6.3 and miscellaneous exercise are given below for new academic session 2020-21. Visit to Class 11 Maths main page to download solutions of other chapters. All NCERT solutions are in accordance with Latest CBSE Curriculum 2020-21 for CBSE, UP Board, MP Board, Gujrat Board and all other board following NCERT Books 2020-2021.

### 11th Maths Chapter 6 Solutions in English Medium

### 11th Maths Chapter 6 Solutions in PDF

#### Class 11 Maths Chapter 6 Exercise 6.1 Solutions in Videos

#### Class 11 Maths Chapter 6 Exercise 6.2 Solutions in Videos

#### Class 11 Maths Chapter 6 Exercise 6.3 Solutions in Videos

#### Class 11 Maths Chapter 6 Miscellaneous Exercise 6 in Videos

##### What is meant by the solution set of the inequality?

Solution Set: A solution to an inequality is a number which when substituted for the variable, makes the inequality true. The set of all solutions of an inequality is called the solution set of the inequality.

##### What is a replacement set?

Replacement Set: The set from which values of the variable (involved in the inequality) are chosen is called replacement set.

#### Procedure to solve a linear inequality

1. Simplify both sides by removing graph symbols and collecting like terms.

2. Remove fractions (or decimals) by multiplying both sides by appropriate factor (L.C.M of denomination or a power of 10 in case of decimals.)

3. Isolate the variable on one side and all constants on the other side. Collect like terms whenever possible.

4. Make the coefficient of the variable.

5. Choose the solution set from the replacement set.

##### Important Terms Related to Linear Inequations

1. The graph of the inequality ax + by > c is one of the half planes and is called the solution region.

2. When the inequality involves the sign ≤ or ≥ then the points on the line are included in the solution region but if it has the sign < or > then the points on the line are not included in the solution region and it has to be drawn as a dotted line.

3. The common values of the variable form the required solution of the given system of linear inequalities in one variable.

4. The common part of coordinate plane is the required solution of the system of linear inequations in two variables when solved by graphical method.

### Important Questions on 11th Maths Chapter 6

⇒3x>-6

⇒x>(-6)/3

⇒x>-2

x is an integer.

The integers greater than –2 are –1, 0, 1, 2, …

Thus, when x is an integer, the solutions of the given inequality are –1, 0, 1, 2 …

Hence, in this case, the solution set is {–1, 0, 1, 2, …}.

⇒ 3x – 7 + 7 > 5x – 1 + 7

⇒ 3x > 5x + 6

⇒ 3x – 5x > 5x + 6 – 5x

⇒ – 2x > 6

⇒-x>6/2

⇒-x>3

⇒x<-3 Thus, all real numbers x, which are less than –3, are the solutions of the given inequality. Hence, the solution set of the given inequality is (–∞, –3).

Since both the integers are smaller than 10, therefore

x + 2 < 10 ⇒ x < 10 – 2 ⇒ x < 8 … (i) Also, the sum of the two integers is more than 11. ∴ x + (x + 2) > 11

⇒ 2x + 2 > 11

⇒ 2x > 11 – 2

⇒ 2x > 9

⇒x>9/2

⇒x>4.5 …(ii)

From (i) and (ii), we obtain that the value of x can be 4,5,6 or 7. .

Since x is an odd number, x can take the values, 5 and 7.

Hence, the required possible pairs of numbers are (5, 7) and (7, 9).

Since both the integers are larger than 5,

x > 5 … (1)

Also, the sum of the two integers is less than 23.

x + (x + 2) < 23 ⇒ 2x + 2 < 23 ⇒ 2x < 23 – 2 ⇒2x<21 ⇒x<21/2 ⇒x<10.5 … (2) From (1) and (2), we obtain 5 < x < 10.5. Since x is an even number, so x can take the values, 6, 8 and 10. Hence, the required possible pairs are (6, 8), (8, 10) and (10, 12).

Then, length of the longest side = 3x cm and the length of the third side = (3x – 2) cm

Since the perimeter of the triangle is at least 61 cm,

⇒x+3x+(3x-2)≥61

⇒7x-2≥61

⇒7x≥63

⇒x≥9

Hence, the minimum length of the shortest side is 9 cm.

Since the three lengths are to be cut from a single piece of board of length 91 cm,

x + (x + 3) + 2x≤ 91

⇒ 4x + 3 ≤ 91

⇒ 4x ≤ 91 – 3

⇒ 4x ≤ 88

⇒x≤22 … (1)

Also, the third piece is at least 5 cm longer than the second piece.

∴ 2x ≥ (x + 3) + 5

⇒ 2x ≥ x + 8

⇒ x ≥ 8 … (2)

From (1) and (2), we obtain, 8 ≤ x ≤ 22

Thus, the possible length of the shortest board is greater than or equal to 8 cm but less than or equal to 22 cm.

⇒ 2 + 4 ≤ 3x – 4 + 4 ≤ 5 + 4

⇒ 6 ≤ 3x ≤ 9

⇒ 2 ≤ x ≤ 3

Therefore, all the real numbers, x, which are greater than or equal to 2 but less than or equal to 3, are the solutions of the given inequality. The solution set for the given inequality is [2, 3].

It is evident that the amount of acid contained in the resulting mixture is 45% of 1125 litres.

This resulting mixture will contain more than 25% but less than 30% acid content.

∴ 25% of (1125 + x) < 45% of 1125 < 30% of (1125 + x) ⇒25/100 (1125+x) < 45/100×1125<30/100 (1125+x) ⇒25(1125+x) < 45×1125<30(1125+x) ⇒5(1125+x) < 9×1125<6(1125+x) ⇒5625+5x < 10125<6750+6x ⇒5x<10125-5625 < 6750-5625+6x ⇒5x<4500 < 1125+6x ⇒5x<4500 and 4500<1125+6x ⇒x<900 and 3375<6x ⇒x<900 and 562.5