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.

We are given  24x<100or 24x24<10024[Dividing both sides by 24]     x<256(i) When x is a natural number, in this case following values      of x make the statement true.x=1,2,3,4.The solution set of the inequality is {1,2,3,4}.(ii) When x is an integer, the solutions of the given inequlity are...,−3,−2,−1, 0, 1, 2, 3,4.The solution set of the inequality is {...,−3,−2,−1, 0, 1, 2, 3,4}.

Q.2 Solve – 12x > 30, when
(i) x is a natural number.
(ii) x is an integer.

Ans.

We are given  −12x>30or −12x−12<30−12 [Dividing both sides by −12]       x<−52(i) When x is a natural number, in this case following no value      of x make the statement true.∴There is no solution.(ii) When x is an integer, the solutions of the given inequality are...,−4,−3.The solution set of the inequality is {...,−4,−3}.

Q.3 Solve 5x – 3 < 7, when
(i) x is an integer.
(ii) x is a real number.

Ans.

We are given       5x−3<7or 5x−3+3<7+3 [Adding 3 both sides]or    5x<10or    5x5<105 [Divide both sides by 5]or      x<2(i) When x is an integer, the solutions of given inequality are...,−3,−2,−1,0,1.The solution set of the inequality is {...,−3,−2,−1,0,1}.(ii) When x is a real number, the solutions of given inequality are given by x<2, i.e., all real numbers x which are less than 2. Therefore, the solution set of the inequality is x∈(−∞,2).

Q.4 Solve 3x + 8 > 2, when
(i) x is an integer.
(ii) x is a real number.

Ans.

We are given        3x+8>2or 3x+8−8>2−8 [Adding (−8) both sides]or    3x>− 6or    3x3>− 63 [Divide both sides by 3]or      x>−2(i) When x is an integer, the solutions of given inequality are−1,0, 1,2,3, ...The solution set of the inequality is {−1,0, 1,2,3,...}.(ii) When x is a real number, the solutions of given inequality are given by x>−2, i.e., all real numbers x which are greater than  −2. Therefore, the solution set of the inequality is x∈(−2,∞).

Q.5 4x + 3 < 5x + 7

Ans.

We are given 4x+3<5x+7or     4x+3−4x<5x+7−4x [Subtracting 4x from both sides]or   3<x+7or     3−7<x+7−7[Subtracting 7 from both sides]or        −4<xi.e., all the real numbers which are greater than −4, are the solutions of the given inequality. Hence, the solution set is (−4,∞).

Q.6 3x – 7 > 5x – 1

Ans.

We are given    3x−7>5x−1or 3x−7−3x>5x−1−3x  Subtracting 3x from both sidesor     −7>2x−1or     −7+1>2x−1+1      Adding 1 both sidesor     −6>2xor     −62>2x2                  Dividing both sides by 2or     −3>xi.e., all the real numbers which are less than −3, are thesolutions of the given inequality. Hence, the solution set is(−∞,−3).

Q.7 3(x – 1) ≤ 2 (x – 3)

Ans.

We are given    3x−1≤2x−3or 3x−3≤2x−6or 3x−3−2x≤2x−6−2xSubtracting 2x from both sidesor  x−3≤−6or  x−3+3≤−6+3Adding 3 both sidesor  x≤−3Thus, all real numbers x which are less than or equal to −3 are the solutions of the given inequality, i.e., x∈(−∞,−3].

Q.8 3(2 – x) ≥ 2 (1 – x)

Ans.

We are given3(2−x)≥2(1−x)or 6−3x≥2−2xor 6−3x+3x≥2−2x+3x[Adding 3x both sides]or 6≥2+xor 6−2≥2+x−2[Subtracting 2 from both sides]or 4≥xThus, all real numbers x which are less than or equal to 4 are the solutions of the given inequality, i.e., x∈(−∞, 4].

Q.9

x+x2+x3<11

Ans.

We are givenx+x2+x3<11or  6x+3x+2x6<11or  11x6<11or         11x6×611<11×611[Multliplying a both sides by 611]or   x<6Thus, all real numbers x which are less than 6 are the solutions of the given inequality, i.e., x∈(−∞, 6).

Q.10

x3>x2+1

Ans.

We are givenx3>x2+1or  x3−x2>x2+1−x2[Subtracting​ x2 from both sides]or      −x6>1or          −x6×− 6<1×−6[Multiplying both sides by−6]or         x<−6Thus, all real numbers x which are greater than 6 are the solutions of the given inequality, i.e., x∈(−∞,−6).

Q.11

3(x−2)5≤5(2−x)3

Ans.

We are given3(x−2)5≤5(2−x)3or 9(x−2)≤25(2−x)or 9x−18≤50−25xAdding 25x both sides, we get 9x−18+25x≤50−25x+25xor 34x−18≤50Adding 18 both sides, we get 34x−18+18≤50+18or 34x≤68Dividing both sides by 34, we get 34x34≤6834⇒ x≤2i.e., all real numbers x which are less than or equal to 2 are the solutions of the given inequality, i.e., x∈(−∞,2].

Q.12

12(3x5+4)≥13(x–6)

Ans.

We are given 12(3x5+4)≥13(x−6)or 12(3x+205)≥13(x−6)or (3x+2010)≥13(x−6)or 3(3x+20)≥10(x−6)or 9x+60≥10x−60⇒ 60+60≥10x−9x⇒ 120≥xi.e, all real numbers x which are less than or equal to 120 are the solutions of the given inequality, i.e., x∈(−∞,120].

Q.13 2 (2x + 3) – 10 < 6 (x – 2)

Ans.

We are given     2(2x+3)−10<6(x−2)or4x+6−10<6x−12or        4x−4<6x−12or       12−4<6x−4xor               8<2xor               4<xi.e, all real numbers x which are less than 4 are the solutions of the given inequality, i.e., x∈(4,∞).

Q.14

37–(3x+5)≥9x–8(x–3)

Ans.

We are given37–(3x+5)≥9x–8(x–3)or   37−3x−5≥9x−8x+24or   32−3x≥x+24or   32−24≥x+3xor        8≥4xor        2≥xi.e, all real numbers x which are less than or equal to 2 are the solutions of the given inequality, i.e., x∈(−∞,2].

Q.15 

x4<(5x−2)3−(7x−3)5

Ans.

We are givenx4<(5x−2)3−(7x−3)5orx4<5(5x−2)−3(7x−3)15orx4<25x−10−21x+915orx4<4x−115or      15x<16x−4or4<16x−15xor4<xi.e, all real numbers x which are less than 4 are the solutions of the given inequality, i.e., x∈(4,∞).

Q.16 

(2x−1)3≥(3x−2)4−(2−x)5

Ans.

We are given (2x−1)3≥(3x−2)4−(2−x)5or (2x−1)3≥5(3x−2)−4(2−x)20or (2x−1)3≥15x−10−8+4x20or (2x−1)3≥19x−1820or 20(2x−1)≥3(19x−18)or 40x−20≥57x−54or 54−20≥57x−40xor 34≥17xor 2≥xi.e, all real numbers x which are less than or equal to 2 are the solutions of the given inequality, i.e., x∈(−∞,2].

Q.17 3x – 2 < 2x + 1

Ans.

We are given   3x−2<2x+1or 3x−2x<1+2or             x<3The graphical representation of solutions is given in Figure.

Q.18 

5x−3≥3x−5

Ans.

We are given    5x−3≥3x−5or  5x−3x≥−5+3or          2x≥−2or            x≥−1The graphical representation of solutions is given in Figure.

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 

x2≥(5x−2)3−(7x−3)5

Ans.

We are given x2≥(5x−2)3−(7x−3)5or x2≥5(5x−2)−3(7x−3)15or x2≥25x−10−21x+915or x2≥4x−115or 15x≥8x−2or 15x−8x≥−2or 7x≥−2or x ≥−27

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.

Let Ravi obtained marks in third test=xRavi obtained marks in first two tests=70 and 75The average marks obtained by Ravi=70+75+x3According​​ to given condition,    70+75+x3≥60  145+x≥180              x≥180−145              x≥35Thus,​ Ravi should get greater or equal to 35 marks inthird unit test.

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.

Let the minimum marks obtained by Sunita in                     fifth examination=xMarks obtained by Sunita in first four                                              examinations=87,92,94,95Average marks obtained by Sunita=87+92+94+95+x5According to given condition,  87+92+94+95+x5≥90or  368+x5≥90or 368+x≥450or    x≥450−368or    x≥82Thus,​ Sunita will get grade ‘A’ in examination if she will get marks greater than or equal to 82.

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.

Let x be the smaller number of the two consecutive odd positive integers, so that the other one is x+2. Then we should havex+2<10or       x<8and  x+(x+2)>11or     2x>11−2or      x>92 x>4.5So,      4.5<x<8Since, x is odd number. So, x will be 5 and 7. Second number is x+2, so it wil be 7 and 9.Thus the required pair of possible numbers will be  (5,7) and (7,9).

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.

Let x be the smaller number of the two consecutive even positive integers, so that the other one is x+2. Then we should havex+2>5or       x>3and  x+(x+2)<23or     2x<23−2or      x<212      x<10.5So,           3<x<10.5Since, x is even number. So, x will be 6, 8 and 10.Second number is x+2, so it wil be 8, 10 and 12.Thus the required pair of possible numbers will be (6,8),(8,10) and (10,12).

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.

Let the shortest side of triangle=x cmThe longest side of triangle=3x cmThird side of triangle=(3x−2) cmThe perimeter of the triangle=x+3x+(3x−2)=(7x−2) cmAccording to given condition,   (7x−2)≥61or    7x≥63or     x≥637or     x≥9Thus, the minimum length of the shortest side is 9 cm.

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.

The length of the single piece of board=91 cmLet the length of the smaller piece of board=x cmThe​ length of the second piece of board=(x+3) cmThe length of the third piece of board=2x cmAccording to given conditions:         2x≥(x+3)+5or            x≥8 ...(1)Since, these the total length of these three pieces will at most equal to 91 cm. Then,    x+(x+3)+2x≤91             4x≤91−3             4x≤88                x≤22 ...(2)From​ (1)​ and (2), we have    8≤x≤22Thus, the length of the shortest piece of wood is greater or equal to 8 cm but less than or equal to 22 cm.

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