# 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.

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## 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 PDF

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

Class 11 Maths Chapter 6 Exercise 6.1 Solution
Class 11 Maths Exercise 6.1 Solution in Hindi

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

Class 11 Maths Chapter 6 Exercise 6.2 Solution
Class 11 Maths Exercise 6.2 Solution in Hindi

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

Class 11 Maths Chapter 6 Exercise 6.3 Solution
Class 11 Maths Exercise 6.3 Solution in Hindi

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

Class 11 Maths Chapter 6 Miscellaneous Solutions
Class 11 Maths Chapter 6 Miscellaneous in Hindi
##### 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

Solve 3x + 8 >2, when x is an integer.
Given that: 3x + 8 > 2
⇒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, …}.
Solve the following inequality for real x: 4x+3<6x+7
4x + 3 < 5x + 7 ⇒ 4x + 3 – 7 < 5x + 7 – 7 ⇒ 4x – 4 < 5x ⇒ 4x – 4 – 4x < 5x – 4x ⇒ –4 < x Thus, all real numbers x, which are greater than –4, are the solutions of the given inequality. Hence, the solution set of the given inequality is (–4, ∞).
Solve the following inequality for real x: 3x-7>5x-1.
3x – 7 > 5x – 1
⇒ 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).
Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11.
Let x be the smaller of the two consecutive odd positive integers. Then, the other integer is x + 2.
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).
Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is less than 23.
Let x be the smaller of the two consecutive even positive integers. Then, the other integer is x + 2.
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).
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.
Let the length of the shortest side of the triangle be x cm.
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.
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 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 5cm longer than the second?
Let the length of the shortest piece be x cm. Then, length of the second piece and the third piece are (x + 3) cm and 2x cm respectively.
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.
Solve the following inequality: 2 ≤ 3x – 4 ≤ 5.
2 ≤ 3x – 4 ≤ 5
⇒ 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].
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?
Let x litres of water is required to be added. Then, the total mixture = (x + 1125) litres
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                    