Class 11 Maths Chapter 1 Sets
NCERT Solutions for Class 11 Maths Chapter 1 Sets includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 11 Maths Chapter 1 Sets. Going through the solutions provided on this page will help you to know how to approach and solve the problems.
Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Maths Chapter 1 Sets NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time. Students can also download NCERT Solutions for Class 11 Maths Chapter 1 Sets PDF to access them even in offline mode.
Here we are providing the Chapter-wise NCERT Book for "Class 11 Maths Subject". Students can easily download and access the chapters of Class 11 Maths. Digital edition of "NCERT Books Class 11 Maths pdf" are always handy to use when you do not have access to a physical copy. Here you can get "Class 11 Maths" NCERT Books. The "NCERT for Class 11 Maths" Chapters wise relevant topic covered on this page.
Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 11 Maths Chapter 1 Sets NCERT Solutions will be most helpful to the students to solve their Homeworks and Assignments on time. Students can also download NCERT Solutions for Class 11 Maths Chapter 1 Sets PDF to access them even in offline mode.
Here we are providing the Chapter-wise NCERT Book for "Class 11 Maths Subject". Students can easily download and access the chapters of Class 11 Maths. Digital edition of "NCERT Books Class 11 Maths pdf" are always handy to use when you do not have access to a physical copy. Here you can get "Class 11 Maths" NCERT Books. The "NCERT for Class 11 Maths" Chapters wise relevant topic covered on this page.
UP Board Solutions for Class 11 Maths Chapter 1 Sets (set)
Questionnaire 1.1
Question 1.
Which of the following sets are there? Justify your answer.
(i) Collection of all months of the year starting with the letter j.
(ii) Collection of the ten most talented writers in India.
(iii) Collection of the best eleven balloons in the world.
(iv) Collection of all the children in your class.
(v) Collection of all natural numbers less than 100.
(vi) A collection of novels written by author Premchand.
(vii) Collection of Sarbhisam Purnaka.
(viii) A collection of questions appearing in this chapter.
(ix) Collection of the most dangerous animals in the world.
Solution:
(i) Names of months starting with j: January, June and July. Hence, it is a set.
(ii) A talented writer cannot be defined. That is why it is not a set.
(iii) Can not define the best batsman. (Vidfom.com) So this is not a set.
(iv) The number of all the students in the class is fixed. Hence, it is a set.
(v) Prakrit numbers less than 100 are 1, 2, 3, …… .. 99. Hence, it is a set.
(vi) The collection of novels written by author Prem Chandra is defined by embezzlement, Godan etc. Hence, it is a set.
(vii) integer {… .. -6, -4. There are 2, 4, 6,….}. So it is a set.
(viii) The questions in this chapter are defined. Hence, it is a set.
(ix) The collection of the world's most dangerous animals cannot be defined. Therefore it is not a set.
Which of the following sets are there? Justify your answer.
(i) Collection of all months of the year starting with the letter j.
(ii) Collection of the ten most talented writers in India.
(iii) Collection of the best eleven balloons in the world.
(iv) Collection of all the children in your class.
(v) Collection of all natural numbers less than 100.
(vi) A collection of novels written by author Premchand.
(vii) Collection of Sarbhisam Purnaka.
(viii) A collection of questions appearing in this chapter.
(ix) Collection of the most dangerous animals in the world.
Solution:
(i) Names of months starting with j: January, June and July. Hence, it is a set.
(ii) A talented writer cannot be defined. That is why it is not a set.
(iii) Can not define the best batsman. (Vidfom.com) So this is not a set.
(iv) The number of all the students in the class is fixed. Hence, it is a set.
(v) Prakrit numbers less than 100 are 1, 2, 3, …… .. 99. Hence, it is a set.
(vi) The collection of novels written by author Prem Chandra is defined by embezzlement, Godan etc. Hence, it is a set.
(vii) integer {… .. -6, -4. There are 2, 4, 6,….}. So it is a set.
(viii) The questions in this chapter are defined. Hence, it is a set.
(ix) The collection of the world's most dangerous animals cannot be defined. Therefore it is not a set.
Question 2.
Suppose A = {1, 2, 3, 4, 5, 6}. Fill in the blank with the appropriate symbol थ or वा.
(i) 5 …… A
(ii) 8 …… .. A
(iii) 0 ……… A
(iv) 4 ……. A
(v) 2 …… .. A
(vi) 10 ……. A
solution:
(I) 5 ∈ A
(Ii) 8 ∉ A
(Iii) 0 ∉ 4
(Iv) 4 ∈ A
(V) 2 ∈ A
(Vi) L0 ∉ A
Suppose A = {1, 2, 3, 4, 5, 6}. Fill in the blank with the appropriate symbol थ or वा.
(i) 5 …… A
(ii) 8 …… .. A
(iii) 0 ……… A
(iv) 4 ……. A
(v) 2 …… .. A
(vi) 10 ……. A
solution:
(I) 5 ∈ A
(Ii) 8 ∉ A
(Iii) 0 ∉ 4
(Iv) 4 ∈ A
(V) 2 ∈ A
(Vi) L0 ∉ A
Question 3.
Write the following sets in roster form:
(i) A = {x: x is an integer and -3 <x <7}
(ii) B = {x: x is a natural number less than 6.}
(iii) C = {x: x is a natural number of two digits whose sum of digits is 8.}
(iv) D = {x: x is a prime number which is a divisor of 60.
(v) E = set of all letters of the word TRIGONOMETRY
(vi) F = BETTER The set of all letters of the word is
solved:
(i) A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}
(ii) B = {1, 2, 3, 4, 5} (Vidfom.com)
(iii) C = {17, 26, 35, 44, 53, 62, 71, 80}
(iv) D = {2, 3, 5}
(v) E = {T, R, I, G, O, N, M, E, Y}
(vi) F = {B, E, T, R}
Write the following sets in roster form:
(i) A = {x: x is an integer and -3 <x <7}
(ii) B = {x: x is a natural number less than 6.}
(iii) C = {x: x is a natural number of two digits whose sum of digits is 8.}
(iv) D = {x: x is a prime number which is a divisor of 60.
(v) E = set of all letters of the word TRIGONOMETRY
(vi) F = BETTER The set of all letters of the word is
solved:
(i) A = {-2, -1, 0, 1, 2, 3, 4, 5, 6}
(ii) B = {1, 2, 3, 4, 5} (Vidfom.com)
(iii) C = {17, 26, 35, 44, 53, 62, 71, 80}
(iv) D = {2, 3, 5}
(v) E = {T, R, I, G, O, N, M, E, Y}
(vi) F = {B, E, T, R}
Question 4.
Express the following sets in set form:
(i) {3, 6, 9, 12}
(i) {2, 4, 8, 16, 32}
(iii) {5, 25, 125, 625 }
(iv) {2, 4, 6,….}
(v) {1, 4, 9, …… 100}
Solution:
(i) {x: x = 3n and 1 ≤ n ≤ 4}
(ii) { x: x = 2n and 1 ≤ n ≤ 5}
(iii) {x: x = 5 and 1 ≤ n ≤ 4}
(iv) {x: x is an even natural number.}
(v) {x: x = n², 1 ≤ n ≤ 10}
Express the following sets in set form:
(i) {3, 6, 9, 12}
(i) {2, 4, 8, 16, 32}
(iii) {5, 25, 125, 625 }
(iv) {2, 4, 6,….}
(v) {1, 4, 9, …… 100}
Solution:
(i) {x: x = 3n and 1 ≤ n ≤ 4}
(ii) { x: x = 2n and 1 ≤ n ≤ 5}
(iii) {x: x = 5 and 1 ≤ n ≤ 4}
(iv) {x: x is an even natural number.}
(v) {x: x = n², 1 ≤ n ≤ 10}
Question 5.
List all the components (members) of the following sets.
(i) A = {x: x is an odd natural number.
(i) B = x: x is an integer,
}
List all the components (members) of the following sets.
(i) A = {x: x is an odd natural number.
(i) B = x: x is an integer,
}
(iii) C = {x: x is an integer, x² ≤ 4}
(iv) D = {x: x is a letter of the word LOYAL.}
(v) F = {x: x is one such month of the year. Is, which does not contain 31 days.}
(Vi) F = {x: x is a consonant in the English alphabet, which comes before k.}
Solution:
(i) A = {1, 3, 5, 7, ……….}
(Ii) B = {0, 1, 2, 3, 4}
(iii) C = {-2, - 1, 0, 1, 2}
(iv) D = {L, O, Y , A}
(v) E = {February, April, June, (Vidfom.com) September, November}
(vi) F = {b, c, d, f, g, h, j}
(iv) D = {x: x is a letter of the word LOYAL.}
(v) F = {x: x is one such month of the year. Is, which does not contain 31 days.}
(Vi) F = {x: x is a consonant in the English alphabet, which comes before k.}
Solution:
(i) A = {1, 3, 5, 7, ……….}
(Ii) B = {0, 1, 2, 3, 4}
(iii) C = {-2, - 1, 0, 1, 2}
(iv) D = {L, O, Y , A}
(v) E = {February, April, June, (Vidfom.com) September, November}
(vi) F = {b, c, d, f, g, h, j}
Question 6.
Correctly match the sets written in the roster form on the left side and the form set as the right.
(i) {1, 2, 3, 6} (a) {x: x is a prime number and a divisor of 6.
(ii) {2, 3} (b) {x: x is an odd natural number less than 10.
(iii) {M, A, T, H, E, I, C, S} (c) {x: x is a natural number and a divisor of 6.
(iv) {1, 3, 5, 7, 9} (d) {x: x MATHEMATICS is a letter of the word:
Solution:
(i) → (c)
(ii) → (a)
(iii) → (d) )
(iv) → (b)
Correctly match the sets written in the roster form on the left side and the form set as the right.
(i) {1, 2, 3, 6} (a) {x: x is a prime number and a divisor of 6.
(ii) {2, 3} (b) {x: x is an odd natural number less than 10.
(iii) {M, A, T, H, E, I, C, S} (c) {x: x is a natural number and a divisor of 6.
(iv) {1, 3, 5, 7, 9} (d) {x: x MATHEMATICS is a letter of the word:
Solution:
(i) → (c)
(ii) → (a)
(iii) → (d) )
(iv) → (b)
Questionnaire 1.2
Question 1.
Which of the following are examples of blank sets?
(i) The set of odd natural numbers divisible by 2
(ii) The set of even prime numbers
(iii) {x: x is a natural number, x <5 as well as x> 7}
(iv) {y: y any Is also the common point of two parallel lines.}
Solution:
(i) There are no odd natural numbers divisible by 2. Hence, it is a blank set.
(ii) The set of even prime numbers is {2}. This is not a (Vidfom.com) empty set.
(iii) x <5 and x> 7 is not a natural number. Hence, it is a blank set.
(iv) Parallel lines are not found anywhere. Hence, it is a blank set.
Which of the following are examples of blank sets?
(i) The set of odd natural numbers divisible by 2
(ii) The set of even prime numbers
(iii) {x: x is a natural number, x <5 as well as x> 7}
(iv) {y: y any Is also the common point of two parallel lines.}
Solution:
(i) There are no odd natural numbers divisible by 2. Hence, it is a blank set.
(ii) The set of even prime numbers is {2}. This is not a (Vidfom.com) empty set.
(iii) x <5 and x> 7 is not a natural number. Hence, it is a blank set.
(iv) Parallel lines are not found anywhere. Hence, it is a blank set.
Question 2.
Which of the following sets are finite and which are infinite?
(i) A set of months of the year.
(ii) {1, 2, 3,… ..}
(ii) {1, 2, 3,… .. 99, 100}
(iv) The set of positive integers greater than 100
(v) The prime integers smaller than 99 Set of
solutions:
(i) There are 12 months in a year.
Hence, it is a finite set.
(ii) The set {1, 2, 3, ……} has infinite elements.
In the end it is an infinite set.
(iii) The set {1, 2, 3, …… 99, 100} has a total of 100 components.
Hence, it is a finite set.
(iv) The set of integers greater than 100 is {101, 102, 103,….} which has infinite elements.
Hence, it is an infinite set.
(v) The set of prime integers smaller than 99 is {2, 3, 5, 7, …… 97} with a fixed number of elements.
Hence, it is a finite set.
Which of the following sets are finite and which are infinite?
(i) A set of months of the year.
(ii) {1, 2, 3,… ..}
(ii) {1, 2, 3,… .. 99, 100}
(iv) The set of positive integers greater than 100
(v) The prime integers smaller than 99 Set of
solutions:
(i) There are 12 months in a year.
Hence, it is a finite set.
(ii) The set {1, 2, 3, ……} has infinite elements.
In the end it is an infinite set.
(iii) The set {1, 2, 3, …… 99, 100} has a total of 100 components.
Hence, it is a finite set.
(iv) The set of integers greater than 100 is {101, 102, 103,….} which has infinite elements.
Hence, it is an infinite set.
(v) The set of prime integers smaller than 99 is {2, 3, 5, 7, …… 97} with a fixed number of elements.
Hence, it is a finite set.
Question 3. For
each of the following sets, state who is finite and who is infinite?
(i) A set of lines parallel to the x-axis.
(ii) A set of letters of the English alphabet.
(iii) A set of numbers that are multiples of 5.
(iv) A set of animals living on the earth
(v) A set of circles passing through the point (0, 0).
Solution:
(i) Infinite lines can be drawn parallel to the x-axis. Hence, it is an infinite set.
(ii) The English alphabet consists of a total of 26 letters. The set formed by these letters will be finite.
(iii) The set of numbers divided by 5 is {5, 10, 15, 20,….}, which has infinite elements. Hence, it is an infinite set.
(iv) The set of animals living on the earth will be finite.
(v) After considering the origin as the center, one can go to the eternal circle (Vidfom.com). Therefore, it will be infinite.
each of the following sets, state who is finite and who is infinite?
(i) A set of lines parallel to the x-axis.
(ii) A set of letters of the English alphabet.
(iii) A set of numbers that are multiples of 5.
(iv) A set of animals living on the earth
(v) A set of circles passing through the point (0, 0).
Solution:
(i) Infinite lines can be drawn parallel to the x-axis. Hence, it is an infinite set.
(ii) The English alphabet consists of a total of 26 letters. The set formed by these letters will be finite.
(iii) The set of numbers divided by 5 is {5, 10, 15, 20,….}, which has infinite elements. Hence, it is an infinite set.
(iv) The set of animals living on the earth will be finite.
(v) After considering the origin as the center, one can go to the eternal circle (Vidfom.com). Therefore, it will be infinite.
Question 4. In the
following, state whether A = B or not.
(i) A = {a, b, c, a}, B = {a, c, b, a}
(ii) A = {4, 8, 12, 16}, B = {8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}, B = {x: x is even integer and x ≤ 10}
(iv) A = {x: x is a multiple of 10} , B = {10, 15,20, 25, 30,…}
Solution:
(i) The components of both sets A and B are a, b, c, d so A = B.
(ii) A has element 12 but B does not contain, hence A ≠ B.
(iii) Both A and B sets have elements 2, 4, 6, 8 and 10. Hence A = B.
(iv) A = {10, 20, 30, 40,… ..}, B = {10, 15, 25, 30,….}
Multiples of 10 do not contain 5, 15, 25. . Hence A ≠ B.
following, state whether A = B or not.
(i) A = {a, b, c, a}, B = {a, c, b, a}
(ii) A = {4, 8, 12, 16}, B = {8, 4, 16, 18}
(iii) A = {2, 4, 6, 8, 10}, B = {x: x is even integer and x ≤ 10}
(iv) A = {x: x is a multiple of 10} , B = {10, 15,20, 25, 30,…}
Solution:
(i) The components of both sets A and B are a, b, c, d so A = B.
(ii) A has element 12 but B does not contain, hence A ≠ B.
(iii) Both A and B sets have elements 2, 4, 6, 8 and 10. Hence A = B.
(iv) A = {10, 20, 30, 40,… ..}, B = {10, 15, 25, 30,….}
Multiples of 10 do not contain 5, 15, 25. . Hence A ≠ B.
Question 5.
Are the following set of pairs the same? Explain with reasons.
(i) A = {2, 3}
B = {x: x is a solution to the equation x² + 5x + 6 = 0.}
(ii) A = {k: x is a letter of the word 'FOLLOW'.}
B = {y: y is a letter of the word 'WOLF.}
Solution:
(i) A = {2, 3}, B = x: x Equation x² + 5x + 6 = 0} = {-2, -3}
Explicit. Is that the components of sets A and B are different.
Hence A ≠ B.
(ii) A = {F, O, L, W}, B = {W, O, L, F}
The components of the set A and B are the same. Hence A = B.
Are the following set of pairs the same? Explain with reasons.
(i) A = {2, 3}
B = {x: x is a solution to the equation x² + 5x + 6 = 0.}
(ii) A = {k: x is a letter of the word 'FOLLOW'.}
B = {y: y is a letter of the word 'WOLF.}
Solution:
(i) A = {2, 3}, B = x: x Equation x² + 5x + 6 = 0} = {-2, -3}
Explicit. Is that the components of sets A and B are different.
Hence A ≠ B.
(ii) A = {F, O, L, W}, B = {W, O, L, F}
The components of the set A and B are the same. Hence A = B.
Question 6.
Select the same sets from the sets given below:
A = {2, 4, 8, 12}
B = {1, 2, 3, 4}
C = {4, 8, 12, 14}
D = {3, 1, 4, 2}
E = {- 1, 1}
F = {0, a}
G = {1, -1}
H = {0, 1}
Solution:
Here the elements of sets B and D are 1, 2, 3, 4, are
B = D
and -1 and 1 elements in the set E and G are same.
E = G
Select the same sets from the sets given below:
A = {2, 4, 8, 12}
B = {1, 2, 3, 4}
C = {4, 8, 12, 14}
D = {3, 1, 4, 2}
E = {- 1, 1}
F = {0, a}
G = {1, -1}
H = {0, 1}
Solution:
Here the elements of sets B and D are 1, 2, 3, 4, are
B = D
and -1 and 1 elements in the set E and G are same.
E = G
Questionnaire 1.3
Question 1.
Make the correct statement by filling the symbol ⊂ or ⊄ in the blank spaces:
(i) {2, 3, 4}…. {1, 2, 3, 4, 5}
(ii) {a, b, c}… .. {b, c, d}
(iii) {x: x is a student of class XI in your school}…. {x: x is a student of your school.}
(iv) {x: x is a circle in a plane}… .. {x: x is a circle in the same plane. Whose radius is 1 unit.}
(V) {x: x is a triangle in a plane}…. {x: x is a rectangle in a plane.}
(vi) {x: x is an equilateral triangle located in a plane} …… {x: x is a triangle located in a plane.}
(vii) {x: x is an even natural number} ……. {x: x is an integer}
Solution:
(i) Components 2, 3, 4 ∈ {1, 2, 3, 4, 5}
Hence {2, 3, 4} ⊂ {1, 2, 3, 4, 5 }
(ii) Component of {a, b, c} a ∉ {b, c, d}
Hence {a, b, c} ⊄ {b, c, d}
(iii) Students who are in class XI of the school Are also in school.
Hence {x: x student of class XI of school} ⊂ {x: x student of your school}
(iv) The radius of a set (Vidfom.com) of an element (Vidfom.com) of a set {x: x is a circle} Can.
Hence, {x: x circle in the plane} ⊄ {x: x radius of the circle is 1 unit}
(v) The set of triangles is completely different from the set of rectangles.
Hence {x: a triangle in the plane} ⊄ {x: x a rectangle in the plane}
(vi) Each equilateral triangle is a triangle.
Hence {x: x is an equilateral triangle in plane} ⊂ {x: x is a triangle in plane}
(vii) Every even natural number is a whole number.
Hence {x: x is an even natural number} ⊂ {x: x is an integer}
Make the correct statement by filling the symbol ⊂ or ⊄ in the blank spaces:
(i) {2, 3, 4}…. {1, 2, 3, 4, 5}
(ii) {a, b, c}… .. {b, c, d}
(iii) {x: x is a student of class XI in your school}…. {x: x is a student of your school.}
(iv) {x: x is a circle in a plane}… .. {x: x is a circle in the same plane. Whose radius is 1 unit.}
(V) {x: x is a triangle in a plane}…. {x: x is a rectangle in a plane.}
(vi) {x: x is an equilateral triangle located in a plane} …… {x: x is a triangle located in a plane.}
(vii) {x: x is an even natural number} ……. {x: x is an integer}
Solution:
(i) Components 2, 3, 4 ∈ {1, 2, 3, 4, 5}
Hence {2, 3, 4} ⊂ {1, 2, 3, 4, 5 }
(ii) Component of {a, b, c} a ∉ {b, c, d}
Hence {a, b, c} ⊄ {b, c, d}
(iii) Students who are in class XI of the school Are also in school.
Hence {x: x student of class XI of school} ⊂ {x: x student of your school}
(iv) The radius of a set (Vidfom.com) of an element (Vidfom.com) of a set {x: x is a circle} Can.
Hence, {x: x circle in the plane} ⊄ {x: x radius of the circle is 1 unit}
(v) The set of triangles is completely different from the set of rectangles.
Hence {x: a triangle in the plane} ⊄ {x: x a rectangle in the plane}
(vi) Each equilateral triangle is a triangle.
Hence {x: x is an equilateral triangle in plane} ⊂ {x: x is a triangle in plane}
(vii) Every even natural number is a whole number.
Hence {x: x is an even natural number} ⊂ {x: x is an integer}
Question 2.
Check whether the following statements are true or false:
(i) {a, b} ⊄ {b, c, a}
(ii) {a, e} ⊂ {x: x is a vowel in the English alphabet.
(iii) {1, 2, 3} ⊂ {1, 3, 5}
(iv) {a} ⊂ {a, b, c}
(v) {a} ⊂ {a, b, c}
(vi) { x: x is an even natural number less than 6. ⊂ {x: x is a natural number that divides the number 36.
Solution:
(i) The components of a set {a, b} are in a, b dasamuchya {b, c, a}.
{a, b} ⊄ {b, c, a}
Hence the above statement is false.
(ii) Both a, e are vowels.
{a, e} = {x: x, is a vowel in the English alphabet.
Hence this statement is true.
(iii) The set {1, 2, 3} and {1, 3, 5} does not contain the element 2 set {1, 3, 5}.
{1, 2, 3} ⊂ {1, 3, 5} The statement is untrue.
(iv) a ∈ {a, b, c}
{a} ⊂ {a, b, c} This statement is true.
(v) {4} is a set, not an element.
{a}} ∈ {a, b, c} The statement is false.
(vi) Even natural number 2, 4 is less than 6 and divides 36.
{x: x is an even natural number that is less than 6} ⊂ {x: x is a (Vidfom.com) divided by even natural number 36. So this statement is true.}
Check whether the following statements are true or false:
(i) {a, b} ⊄ {b, c, a}
(ii) {a, e} ⊂ {x: x is a vowel in the English alphabet.
(iii) {1, 2, 3} ⊂ {1, 3, 5}
(iv) {a} ⊂ {a, b, c}
(v) {a} ⊂ {a, b, c}
(vi) { x: x is an even natural number less than 6. ⊂ {x: x is a natural number that divides the number 36.
Solution:
(i) The components of a set {a, b} are in a, b dasamuchya {b, c, a}.
{a, b} ⊄ {b, c, a}
Hence the above statement is false.
(ii) Both a, e are vowels.
{a, e} = {x: x, is a vowel in the English alphabet.
Hence this statement is true.
(iii) The set {1, 2, 3} and {1, 3, 5} does not contain the element 2 set {1, 3, 5}.
{1, 2, 3} ⊂ {1, 3, 5} The statement is untrue.
(iv) a ∈ {a, b, c}
{a} ⊂ {a, b, c} This statement is true.
(v) {4} is a set, not an element.
{a}} ∈ {a, b, c} The statement is false.
(vi) Even natural number 2, 4 is less than 6 and divides 36.
{x: x is an even natural number that is less than 6} ⊂ {x: x is a (Vidfom.com) divided by even natural number 36. So this statement is true.}
Question 3.
Suppose that A = {1, 2, 3, 4, 5}, which of the following statements is not correct and why?
(i) {3, 4} ⊂ A
(i) {3, 4} ∈ A
(iii) {{3, 4}} ⊂ A
(iv) 1 ∈ A
(v) 1 ⊂ A
(vi) {1, 2, 5} ⊂ A
(vii) {1, 2, 5} ∈ A
(viii) {1, 2, 3} ⊂ A
(ix) Φ ∈ A
(x) Φ ⊂A
(xi) {Φ} ⊂ A
Solution:
(i) is not correct. The set {3, 4} is an element.
(ii) is correct. Because {3, 4} is a component of the set A.
(iii) is correct. The components of A are a subset of {3, 4}.
(iv) 1 ∈ A is correct.
(v) 1 ⊂ A is not correct because 1 is not a set.
(vi) {1, 2, 5} ⊂ A is correct. The components of the set {1, 2,5} are 1, 2, 5 in the set A.
(vii) {1, 2, 5} ∈ is not correct. {12, 5} is not an element. This is a set.
(viii) {1, 2, 3} ⊂ A is not correct. Element 3 is not in the set.
(ix) Φ ∈ A is not correct. है is a set, not an element.
(x) {Φ} ⊂ A is correct. (Vidfom.com) is a subset of all sets.
(xi) {Φ} ⊂ A is not correct. {Φ} is a set of sets.
Suppose that A = {1, 2, 3, 4, 5}, which of the following statements is not correct and why?
(i) {3, 4} ⊂ A
(i) {3, 4} ∈ A
(iii) {{3, 4}} ⊂ A
(iv) 1 ∈ A
(v) 1 ⊂ A
(vi) {1, 2, 5} ⊂ A
(vii) {1, 2, 5} ∈ A
(viii) {1, 2, 3} ⊂ A
(ix) Φ ∈ A
(x) Φ ⊂A
(xi) {Φ} ⊂ A
Solution:
(i) is not correct. The set {3, 4} is an element.
(ii) is correct. Because {3, 4} is a component of the set A.
(iii) is correct. The components of A are a subset of {3, 4}.
(iv) 1 ∈ A is correct.
(v) 1 ⊂ A is not correct because 1 is not a set.
(vi) {1, 2, 5} ⊂ A is correct. The components of the set {1, 2,5} are 1, 2, 5 in the set A.
(vii) {1, 2, 5} ∈ is not correct. {12, 5} is not an element. This is a set.
(viii) {1, 2, 3} ⊂ A is not correct. Element 3 is not in the set.
(ix) Φ ∈ A is not correct. है is a set, not an element.
(x) {Φ} ⊂ A is correct. (Vidfom.com) is a subset of all sets.
(xi) {Φ} ⊂ A is not correct. {Φ} is a set of sets.
Question 4.
Write all the subsets of the following sets.
(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) Φ
Solution:
(i) Φ, {a}
(ii) Φ, {a}, {b} , {a, b}
(iii) Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
(iv ) Φ
Write all the subsets of the following sets.
(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) Φ
Solution:
(i) Φ, {a}
(ii) Φ, {a}, {b} , {a, b}
(iii) Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
(iv ) Φ
Question 5.
How many components of P (A), if A = Φ
Solution:
A = Φ, P (A) = Φ Thus P (A) has 2 ° = 1 element.
How many components of P (A), if A = Φ
Solution:
A = Φ, P (A) = Φ Thus P (A) has 2 ° = 1 element.
Question 6.
Write the following in interval form:
(i) {x: x ∈R, -4 <x ≤ 6}
(ii) {x: x ∈R, -12 <x <-10}
(iii) {x : x ∈ R, 0 ≤ x <7}
(iv) {x: x ∈ R, 3 ≤ x ≤ 4}
Solution: The
desired intervals are as follows.
(i) (-4, 6]
(ii) (-12, - 10)
(iii) [0, 7)
(iv) [3, 4]
Write the following in interval form:
(i) {x: x ∈R, -4 <x ≤ 6}
(ii) {x: x ∈R, -12 <x <-10}
(iii) {x : x ∈ R, 0 ≤ x <7}
(iv) {x: x ∈ R, 3 ≤ x ≤ 4}
Solution: The
desired intervals are as follows.
(i) (-4, 6]
(ii) (-12, - 10)
(iii) [0, 7)
(iv) [3, 4]
Question 7.
Write the following intervals in a set form:
(i) (-3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [-23, 5]
Solution:
(i) ) (-3, 0) = {x: x ∈ R, -3 <x <0}
(ii) [6, 12] = {x: x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12] = {x: x ∈ R, 6 <x ≤ 12}.
(Iv) [-23, 5] = {x: x ∈ R, -23 ≤ x ≤ 5}
Write the following intervals in a set form:
(i) (-3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [-23, 5]
Solution:
(i) ) (-3, 0) = {x: x ∈ R, -3 <x <0}
(ii) [6, 12] = {x: x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12] = {x: x ∈ R, 6 <x ≤ 12}.
(Iv) [-23, 5] = {x: x ∈ R, -23 ≤ x ≤ 5}
Question 8.
Which universal set would you propose for each of the following?
(i) the set of right-angled triangles
(ii) the aggregate of isosceles triangles
solution:
universal for both sets aggregates:
{X: X is a triangle in the plane}
Which universal set would you propose for each of the following?
(i) the set of right-angled triangles
(ii) the aggregate of isosceles triangles
solution:
universal for both sets aggregates:
{X: X is a triangle in the plane}
Question 9. The
set A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8} are given. Which of the following universal sets can be taken for these three sets A, B and C?
(i) {0,1, 2, 3, 4, 5, 6}
(ii) Φ
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) ) {1, 2, 3, 4, 5, 6, 7, 8}
Solution: For
set (iii),
all three sets A, B, C {0, 1, 2, 3, 4, 5, 6, 7 , 8, 9, 10} are universal sets.
set A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8} are given. Which of the following universal sets can be taken for these three sets A, B and C?
(i) {0,1, 2, 3, 4, 5, 6}
(ii) Φ
(iii) {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) ) {1, 2, 3, 4, 5, 6, 7, 8}
Solution: For
set (iii),
all three sets A, B, C {0, 1, 2, 3, 4, 5, 6, 7 , 8, 9, 10} are universal sets.
Questionnaire 1.4
Question 1. Find the
combination of each of the following sets:
(i) X = {1, 3, 5}, Y = {1, 2, 3}
(i) A = {a, e, i, 0, u}, B = {a, b, c}
(iii) A = {x: is a natural number and is a multiple of 3.
B = {x: x is a natural number less than 6.
(iv) A = {x: x is a natural number and 1 <x ≤ 6}
B = {x: x is a natural number and 6 <x <10}
(v) A = {1, 2, 3}, B = Φ
Solution:
(i) X ∪ Y = {1, 3, 5} ∪ {1, 2, 3} = {1, 2, 3, 5}
(ii) A ∪ B = {a, e, i , 0, u} ∪ {a, b, c} = {a, b, c, e, i, 0, u}
(iii) A ∪ B = {3, 6, 9….} ∪ {1, 2 , 3, 4, 5} = {1, 2, 4, 5 or multiples of number 3}
(iv) A = {2, 3, 4, 5, 6}, B = {7, 8, 9}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9} ie (Vidfom.com) {x: 1 <x <10, x ∈ N}
(v) A ∪B = {1, 2, 3} ∪ Φ = {1, 2, 3}
combination of each of the following sets:
(i) X = {1, 3, 5}, Y = {1, 2, 3}
(i) A = {a, e, i, 0, u}, B = {a, b, c}
(iii) A = {x: is a natural number and is a multiple of 3.
B = {x: x is a natural number less than 6.
(iv) A = {x: x is a natural number and 1 <x ≤ 6}
B = {x: x is a natural number and 6 <x <10}
(v) A = {1, 2, 3}, B = Φ
Solution:
(i) X ∪ Y = {1, 3, 5} ∪ {1, 2, 3} = {1, 2, 3, 5}
(ii) A ∪ B = {a, e, i , 0, u} ∪ {a, b, c} = {a, b, c, e, i, 0, u}
(iii) A ∪ B = {3, 6, 9….} ∪ {1, 2 , 3, 4, 5} = {1, 2, 4, 5 or multiples of number 3}
(iv) A = {2, 3, 4, 5, 6}, B = {7, 8, 9}
A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9} ie (Vidfom.com) {x: 1 <x <10, x ∈ N}
(v) A ∪B = {1, 2, 3} ∪ Φ = {1, 2, 3}
Question 2.
Suppose that A = {a, b}, B = {a, b, c} whether A ⊂ B? Find A ∪ B
Solution:
A = {a, b}, B = {a, b, c}. The components of set A are also in a, b set B. A ⊂ B = A ∪ B = B and A ∪ B = {a, b} ∪ {a, b, c} = {a, b, c}
Suppose that A = {a, b}, B = {a, b, c} whether A ⊂ B? Find A ∪ B
Solution:
A = {a, b}, B = {a, b, c}. The components of set A are also in a, b set B. A ⊂ B = A ∪ B = B and A ∪ B = {a, b} ∪ {a, b, c} = {a, b, c}
Question 3.
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution:
All elements of A ⊂ B set A are in set B. A ⊂ B = B.
If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Solution:
All elements of A ⊂ B set A are in set B. A ⊂ B = B.
Question 4.
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10 }, Then find the following:
(i) A ∪ B
(ii) A ∪ C
(ii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪D
(vii) B ∪ C ∪ D
Solution:
(i) A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3 , 4, 5, 6} ∪ {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6} ∪ {7 , 8, 9, 10} = {3,4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B ∪C = ({1, 2, 3, 4} {3, 4, 5 , 6}) ∪ {5, 6, 7, 8} = (Vidfom.com) {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8} = {1, 2, 3 , 4, 5, 6, 7, 8}.
(vi) A ∪ B ∪ D = ({1, 2, 3, 4} ∪ {3, 4, 5, 6}) ∪ {7, 8, 9, 10} = {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = ({3, 4 , 5, 6} ∪ {5, 6, 7, 8}) ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8} ∪ {7, 8, 9, 10) = {3, 4, 5, 6, 7, 8, 9, 10}
If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10 }, Then find the following:
(i) A ∪ B
(ii) A ∪ C
(ii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪D
(vii) B ∪ C ∪ D
Solution:
(i) A ∪ B = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4} ∪ {5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3 , 4, 5, 6} ∪ {5, 6, 7, 8} = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6} ∪ {7 , 8, 9, 10} = {3,4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B ∪C = ({1, 2, 3, 4} {3, 4, 5 , 6}) ∪ {5, 6, 7, 8} = (Vidfom.com) {1, 2, 3, 4, 5, 6} ∪ {5, 6, 7, 8} = {1, 2, 3 , 4, 5, 6, 7, 8}.
(vi) A ∪ B ∪ D = ({1, 2, 3, 4} ∪ {3, 4, 5, 6}) ∪ {7, 8, 9, 10} = {1, 2, 3, 4, 5, 6} ∪ {7, 8, 9, 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = ({3, 4 , 5, 6} ∪ {5, 6, 7, 8}) ∪ {7, 8, 9, 10} = {3, 4, 5, 6, 7, 8} ∪ {7, 8, 9, 10) = {3, 4, 5, 6, 7, 8, 9, 10}
Question 5. Find the
common set of each pair of pairs in Question 1:
Solution:
(i) X ∩ Y = {1, 3, 5} ∩ {1, 2, 3} = {1, 3}
(ii) A ∩ B = {a, e, i, o, u} ∩ {a, b, c} = {a}.
(iii) A ∩ B = {3, 6, 9… ..} ∩ {1, 2, 3, 4, 5} = {3}.
(iv) A ∩ B = {2, 3, 4, 5, 6} ∩ {7, 8, 9} = Φ
(v) A ∩ B = {1, 2, 3} ∩ Φ = 0
common set of each pair of pairs in Question 1:
Solution:
(i) X ∩ Y = {1, 3, 5} ∩ {1, 2, 3} = {1, 3}
(ii) A ∩ B = {a, e, i, o, u} ∩ {a, b, c} = {a}.
(iii) A ∩ B = {3, 6, 9… ..} ∩ {1, 2, 3, 4, 5} = {3}.
(iv) A ∩ B = {2, 3, 4, 5, 6} ∩ {7, 8, 9} = Φ
(v) A ∩ B = {1, 2, 3} ∩ Φ = 0
Question 6.
If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; Then find the following:
(i) A ∩ B
(ii) B ∩C
(iii) A ∩ C ∩D
(iv) A ∩C
(v) B ∩ D
(vi) Á ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪D)
(ix) (A ∩ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Solution:
(i) A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
(ii) B ∩ C = {7, 9, 11, 13} ∩ {11, 13 , 15} = {11, 13}
(iii) Á ∩ C ∩ D = ({3, 5, 7, 9, 11} ∩ {11, 13, 15}) ∩ {15, 17} = {11} ∩ {15, 17} = Φ
(iv) A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}.
(v) B ∩ D = {7, 9, 11, 13} ∩ {15, 17} = Φ
(vi) A ∩ (B ∪ C) = {3, 5, 7, 9, 11} ∩ ({7, 9, 11, 13} ∪ {11, 13, 15}) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}.
(vii) A ∩ D = {3, 5, 7, 9, 11} ∩ {15, 17} = Φ
(viii) A ∩ (B ∪D) = {3, 5, 7, 9, 11} ∩ { 7, 9, 11, 13} ∪ {15, 17}) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17} = {7, 9, 11).
(ix) A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
B ∪ C = {7, 9, 11, 13} ∪ {11, 13, 15) = {7, 9, 11, 13, 15).
(A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}.
(x) A ∪ D = {3, 5, 7, 9, 11} ∪ {15, 17} = {3, 5, 7, 9, 11, 15, 17}
B ∪ C = {7, 9, 11 , 13} ∪ {11, 13, 15} = {7, 9, 11, 13, 15}
(A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15} = {7, 9, 11, 15}
If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; Then find the following:
(i) A ∩ B
(ii) B ∩C
(iii) A ∩ C ∩D
(iv) A ∩C
(v) B ∩ D
(vi) Á ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪D)
(ix) (A ∩ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Solution:
(i) A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
(ii) B ∩ C = {7, 9, 11, 13} ∩ {11, 13 , 15} = {11, 13}
(iii) Á ∩ C ∩ D = ({3, 5, 7, 9, 11} ∩ {11, 13, 15}) ∩ {15, 17} = {11} ∩ {15, 17} = Φ
(iv) A ∩ C = {3, 5, 7, 9, 11} ∩ {11, 13, 15} = {11}.
(v) B ∩ D = {7, 9, 11, 13} ∩ {15, 17} = Φ
(vi) A ∩ (B ∪ C) = {3, 5, 7, 9, 11} ∩ ({7, 9, 11, 13} ∪ {11, 13, 15}) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}.
(vii) A ∩ D = {3, 5, 7, 9, 11} ∩ {15, 17} = Φ
(viii) A ∩ (B ∪D) = {3, 5, 7, 9, 11} ∩ { 7, 9, 11, 13} ∪ {15, 17}) = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13, 15, 17} = {7, 9, 11).
(ix) A ∩ B = {3, 5, 7, 9, 11} ∩ {7, 9, 11, 13} = {7, 9, 11}
B ∪ C = {7, 9, 11, 13} ∪ {11, 13, 15) = {7, 9, 11, 13, 15).
(A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}.
(x) A ∪ D = {3, 5, 7, 9, 11} ∪ {15, 17} = {3, 5, 7, 9, 11, 15, 17}
B ∪ C = {7, 9, 11 , 13} ∪ {11, 13, 15} = {7, 9, 11, 13, 15}
(A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17} ∩ {7, 9, 11, 13, 15} = {7, 9, 11, 15}
Question 7.
If A = {x: x is a natural number}, B = {x: x is an even natural number} C = {x: x is an odd natural number}, D = {x: is a prime number } Then find the following:
(i) A ∩ B
(ii) A ∩C
(iii) A ∩ D
(iv) B ∩C
(v) B ∩D
(vi) C ∩ D
Solution:
A = {x: x is a natural number} = {1, 2, 3, 4 ……} B = {x: x is an even natural number} = {2, 4, 6, 8…} C = {x: x is an odd natural The number is} = {1, 3, 5, 7…} D = {x: x is a prime number} = {2, 3, 5, 7, 11….}
(I) A ∩ B = {1, 2 , 3, 4….} ∩ {2, 4, 6, 8….} = {2, 4, 6, 8….} = B
(ii) A ∩C = {1, 2, 3, 4 …… } ∩ {1, 3, 5, 7….} = {1, 3, 5, 7….} = C
(iii) A ∩D = {1, 2, 3, 4…} ∩ {2, 3, 5, 7… ..} = {2, 3, 5, 7 ……} = D
(iv) B ∩ C = {2, 4, 6, 8…} ∩ {1, 3, 5, 7 ……} = Φ
(v) B ∩D = {2, 4, 6, 8… ..} ∩ {2, 3, 5, 7… ..} = {2}
(vi) C ∩D = {1, 3, 5, 7… ..} ∩ {2, 3, 5, 7, 11 ……. } = {3, 5, 7, 11, 13 ……} = (Vidfom.com) {x: x an odd prime number}
If A = {x: x is a natural number}, B = {x: x is an even natural number} C = {x: x is an odd natural number}, D = {x: is a prime number } Then find the following:
(i) A ∩ B
(ii) A ∩C
(iii) A ∩ D
(iv) B ∩C
(v) B ∩D
(vi) C ∩ D
Solution:
A = {x: x is a natural number} = {1, 2, 3, 4 ……} B = {x: x is an even natural number} = {2, 4, 6, 8…} C = {x: x is an odd natural The number is} = {1, 3, 5, 7…} D = {x: x is a prime number} = {2, 3, 5, 7, 11….}
(I) A ∩ B = {1, 2 , 3, 4….} ∩ {2, 4, 6, 8….} = {2, 4, 6, 8….} = B
(ii) A ∩C = {1, 2, 3, 4 …… } ∩ {1, 3, 5, 7….} = {1, 3, 5, 7….} = C
(iii) A ∩D = {1, 2, 3, 4…} ∩ {2, 3, 5, 7… ..} = {2, 3, 5, 7 ……} = D
(iv) B ∩ C = {2, 4, 6, 8…} ∩ {1, 3, 5, 7 ……} = Φ
(v) B ∩D = {2, 4, 6, 8… ..} ∩ {2, 3, 5, 7… ..} = {2}
(vi) C ∩D = {1, 3, 5, 7… ..} ∩ {2, 3, 5, 7, 11 ……. } = {3, 5, 7, 11, 13 ……} = (Vidfom.com) {x: x an odd prime number}
Question 8.
Which of the following sets of pairs are disjoint?
(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}
(ii) {a, e, i, 0, u} and {c, d, e, f}
(iii) {x: x is an even integer. And {x: x is an odd integer.
Solution:
(i) Suppose E = {1, 2, 3, 4} F = {x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6} components 4, both E and F sets. Is in Hence, both pairs are not disjoint.
(ii) In a given set, the component is common. Hence, it is not a disjoint set.
(iii) Suppose A = {x: x is an even integer. = {….-4, -2, 0, 2, 4…} B = {x: x is an odd integer} = {….-5, -3, -1, 1, 3, 5… ..} No elements are common in sets A and B. Therefore, this set is disjoint.
Which of the following sets of pairs are disjoint?
(i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}
(ii) {a, e, i, 0, u} and {c, d, e, f}
(iii) {x: x is an even integer. And {x: x is an odd integer.
Solution:
(i) Suppose E = {1, 2, 3, 4} F = {x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6} components 4, both E and F sets. Is in Hence, both pairs are not disjoint.
(ii) In a given set, the component is common. Hence, it is not a disjoint set.
(iii) Suppose A = {x: x is an even integer. = {….-4, -2, 0, 2, 4…} B = {x: x is an odd integer} = {….-5, -3, -1, 1, 3, 5… ..} No elements are common in sets A and B. Therefore, this set is disjoint.
Question 9.
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12 , 14, 16}, D = {5, 10, 15, 20}, then find the following:
(i) A - B
(ii) A - C
(iii) A - D
(iv) B - A
(v) ) C - A
(vi) D - A
(vii) B - C
(viii) B - D
(ix) C - B
(x) D - B
(xi) C - D
(xii) D - C
Solution:
(i ) A - B = {3, 6, 9, 12, 15, 18, 21} - {4, 8, 12, 16, 20} = {3, 6, 9, 15, 18, 21}
(ii) A - C = {3, 6, 9, 12, 15, 18, 21} - {2, 4, 6, 8, 10, 12, 14, 16} = {3, 9, 15, 18, 21}
(iii ) A - D = {3, 6, 9, 12, 15, 18, 21} - {5, 10, 15, 20} = {3, 6, 9, 12, 18, 21}
(iv) B - A = {4, 8, 12, 16, 20} - {3, 6, 9, 12, 15, 18, 21} = {4, 8, 16, 20}
(v) C - A = {2, 4, 6, 8, 10, 12, 14, 16} - {3, 6, 9, 12, 15, 18, 21} = {2, 4, 8, 10, 14, 16}
(vi) D - A = {5, 10, 15, 20} - {3, 6, 9, 12, 15, 18, 21} = {5, 10, 20}
(vii) B - C = {4, 8, 12, 16, 20} - {2, 4, 6, 8, 10, 12, 14, 16} = {20}
(viii) B - D = {4, 8, 12, 16, 20} - {5, 10, 15, 20} = {4, 8, 12, 16}
(ix) C - B = {2, 4, 6, 8, 10, 12, 14, 16} - {4, 8, 12, 16, 20} = {2, 6, 10, 14}
(x) D - B = {5, 10, 15, 20} - {4, 8, 12, 16, 20} = {5, 10, 15}
(xi) C - D = {2, 4, 6, 8, 10, 12, 14, 16} - {5, 10, 15, 20} = {2, 4, 6, 8, 12, 14, 16}
(xii) D - C = {5, 10, 15, 20} - {2, 4, 6, 8, 10, 12, 14, 16} = (Vidfom.com) {5, 15, 20 }
If A = {3, 6, 9, 12, 15, 18, 21}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12 , 14, 16}, D = {5, 10, 15, 20}, then find the following:
(i) A - B
(ii) A - C
(iii) A - D
(iv) B - A
(v) ) C - A
(vi) D - A
(vii) B - C
(viii) B - D
(ix) C - B
(x) D - B
(xi) C - D
(xii) D - C
Solution:
(i ) A - B = {3, 6, 9, 12, 15, 18, 21} - {4, 8, 12, 16, 20} = {3, 6, 9, 15, 18, 21}
(ii) A - C = {3, 6, 9, 12, 15, 18, 21} - {2, 4, 6, 8, 10, 12, 14, 16} = {3, 9, 15, 18, 21}
(iii ) A - D = {3, 6, 9, 12, 15, 18, 21} - {5, 10, 15, 20} = {3, 6, 9, 12, 18, 21}
(iv) B - A = {4, 8, 12, 16, 20} - {3, 6, 9, 12, 15, 18, 21} = {4, 8, 16, 20}
(v) C - A = {2, 4, 6, 8, 10, 12, 14, 16} - {3, 6, 9, 12, 15, 18, 21} = {2, 4, 8, 10, 14, 16}
(vi) D - A = {5, 10, 15, 20} - {3, 6, 9, 12, 15, 18, 21} = {5, 10, 20}
(vii) B - C = {4, 8, 12, 16, 20} - {2, 4, 6, 8, 10, 12, 14, 16} = {20}
(viii) B - D = {4, 8, 12, 16, 20} - {5, 10, 15, 20} = {4, 8, 12, 16}
(ix) C - B = {2, 4, 6, 8, 10, 12, 14, 16} - {4, 8, 12, 16, 20} = {2, 6, 10, 14}
(x) D - B = {5, 10, 15, 20} - {4, 8, 12, 16, 20} = {5, 10, 15}
(xi) C - D = {2, 4, 6, 8, 10, 12, 14, 16} - {5, 10, 15, 20} = {2, 4, 6, 8, 12, 14, 16}
(xii) D - C = {5, 10, 15, 20} - {2, 4, 6, 8, 10, 12, 14, 16} = (Vidfom.com) {5, 15, 20 }
Question 10.
If X = {a, b, c, d} and Y = {f, b, d, g} then find the following:
(i) X - Y
(ii) Y - X
(iii) X ∩. Y
Solution:
(i) X - Y = {a, b, c, d} - {f, b, d, g} = {a, c}
(ii) Y - X = {f, b, d, g } - {a, b, c, d} = {f, g}
(iii) X ∩Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}
If X = {a, b, c, d} and Y = {f, b, d, g} then find the following:
(i) X - Y
(ii) Y - X
(iii) X ∩. Y
Solution:
(i) X - Y = {a, b, c, d} - {f, b, d, g} = {a, c}
(ii) Y - X = {f, b, d, g } - {a, b, c, d} = {f, g}
(iii) X ∩Y = {a, b, c, d} ∩ {f, b, d, g} = {b, d}
Question 11.
If R is a set of real numbers and Q rational numbers, what will R - Q be?
Solution:
R = {x: x is a real number.} P = {x: x is a rational number.}
R - Q = {x: x is an irrational number.} Hence it is a set of irrational numbers.
If R is a set of real numbers and Q rational numbers, what will R - Q be?
Solution:
R = {x: x is a real number.} P = {x: x is a rational number.}
R - Q = {x: x is an irrational number.} Hence it is a set of irrational numbers.
Question 12. State
whether each of the following statements is true or false? Justify your answer.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, 0, u} and {a, b, c, 4} are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.
Solution:
(i) This statement is not true because the elements in the set {2, 3, 4, 5} and {3, 6} are common.
(ii) This statement is not true because the components {a, e, i, o, u} and {a, b, c, d} are common in the set.
(iii) This statement is true because the set {2, 6, 10, 14} and {3, 7, 11, 15} does not have an element common. Therefore, this set is disjoint.
(iv) This statement is true because the set {2, 6, 10} and {3, 7, 11} has no (Vidfom.com) element in common. Hence, this set is disjoint.
whether each of the following statements is true or false? Justify your answer.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, 0, u} and {a, b, c, 4} are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.
Solution:
(i) This statement is not true because the elements in the set {2, 3, 4, 5} and {3, 6} are common.
(ii) This statement is not true because the components {a, e, i, o, u} and {a, b, c, d} are common in the set.
(iii) This statement is true because the set {2, 6, 10, 14} and {3, 7, 11, 15} does not have an element common. Therefore, this set is disjoint.
(iv) This statement is true because the set {2, 6, 10} and {3, 7, 11} has no (Vidfom.com) element in common. Hence, this set is disjoint.
Questionnaire 1.5
Question 1.
Suppose U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8 } And C = {3, 4, 5, 6} then find the following:
(i) A '
(ii) B'
(iii) (A ∪ C) '
(iv) (A ∪ B)'
(v) (A ')'
(vi) (B - C) '
Solution:
(i) A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2 , 3, 4} = {5, 6, 7, 8, 9}
(ii) B '= U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8} = {1, 3, 5, 7, 9)
(iii) A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6} (A ∪ C) '= U - (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 5, 6} = {7, 8, 9}
(iv) A ∪ B = {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8} (A ∪ B) '= U - ( A ∪B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 8} = {5, 7, 9} (v) ( A) '= U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4} = {5, 6, 7, 8, 9} ( A ')' = U- A '= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {5, 6, 7, 8, 9} = {1, 2, 3, 4 }
(vi) B - C = {2, 4, 6, 8} - {3, 4, 5, 6} = {2, 8} (B - C) '= U - (B - C) = {1 , 2, 3, 4, 5, 6, 7, 8, 9} - {2,8} = {1, 3, 4, 5, 6, 7, 9}
Suppose U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8 } And C = {3, 4, 5, 6} then find the following:
(i) A '
(ii) B'
(iii) (A ∪ C) '
(iv) (A ∪ B)'
(v) (A ')'
(vi) (B - C) '
Solution:
(i) A' = U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2 , 3, 4} = {5, 6, 7, 8, 9}
(ii) B '= U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8} = {1, 3, 5, 7, 9)
(iii) A ∪ C = {1, 2, 3, 4} ∪ {3, 4, 5, 6} = {1, 2, 3, 4, 5, 6} (A ∪ C) '= U - (A ∪ C) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 5, 6} = {7, 8, 9}
(iv) A ∪ B = {1, 2, 3, 4} ∪ {2, 4, 6, 8} = {1, 2, 3, 4, 6, 8} (A ∪ B) '= U - ( A ∪B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 8} = {5, 7, 9} (v) ( A) '= U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4} = {5, 6, 7, 8, 9} ( A ')' = U- A '= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {5, 6, 7, 8, 9} = {1, 2, 3, 4 }
(vi) B - C = {2, 4, 6, 8} - {3, 4, 5, 6} = {2, 8} (B - C) '= U - (B - C) = {1 , 2, 3, 4, 5, 6, 7, 8, 9} - {2,8} = {1, 3, 4, 5, 6, 7, 9}
Question 2.
If U = {a, b, c, d, e, f, g, h}, then find the complement of the following sets:
(i) A = {a, b, c}
(ii) B = { d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
Solution:
(i) A '= U - A = {a , b, c, d, e, f, g, h} - {a, b, c} = {d, e, f, g, h}
(ii) B '= U - B = {a, b, c, d, e, f, g, h} - {d, e, f, g} = {a, b, c, h}
(iii) C = U - C = {a, b, c, d, e, f, g, h} - {a, c, e, g} = {b, d, f, h}
(iv) D '= U - D = {a, b, c, d, e, f , g, h} - {f, g, h, a} = {b, c, d, e}.
If U = {a, b, c, d, e, f, g, h}, then find the complement of the following sets:
(i) A = {a, b, c}
(ii) B = { d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
Solution:
(i) A '= U - A = {a , b, c, d, e, f, g, h} - {a, b, c} = {d, e, f, g, h}
(ii) B '= U - B = {a, b, c, d, e, f, g, h} - {d, e, f, g} = {a, b, c, h}
(iii) C = U - C = {a, b, c, d, e, f, g, h} - {a, c, e, g} = {b, d, f, h}
(iv) D '= U - D = {a, b, c, d, e, f , g, h} - {f, g, h, a} = {b, c, d, e}.
Question 3.
Assuming the set of natural numbers as a universal set, write the complement of the following sets:
(i) {x: x is a natural even number.}
(Ii) {x: x is a natural odd number.}
(Iii) {x: x is a positive multiple to the number 3.}
(iv) {x: x is a prime number.}
(v) {x: x is a number divided by 3 and 5.}
(vi) { x: x is a perfect square number.}
(vii) {x: x is a perfect cube number.}
(viii) {x: x + 5 = 8}
(ix) {x: 2x + 5 = 9}
(x ) {x: x ≥ 7}
(xi) {x: x ∈ N and 2x + 1> 10}
Solution:
(i) {x: x is an odd natural number.}
(ii) {x: is an even number .}
(Iii) {x: x is not a money multiple of ∈ N and x number 3.}
(iv) {x: x = 1 and x is a positive divisional number.}
(v) {x: x ∈ N and x, the number 3 and 5 are not divisible by any one.}
(vi) {x: x ∈ N and x are not a perfect square number.}
(Vii) {x: x ∈ N and x is not a perfect square cube number.}
(Viii) {x: x ∈ N and x ≠ 3}
(ix) {x : x ∈ N and x ≠ 2}
(x) {x: x ∈ N and x <7}
(xi) {x: x ∈ N and x <}
Assuming the set of natural numbers as a universal set, write the complement of the following sets:
(i) {x: x is a natural even number.}
(Ii) {x: x is a natural odd number.}
(Iii) {x: x is a positive multiple to the number 3.}
(iv) {x: x is a prime number.}
(v) {x: x is a number divided by 3 and 5.}
(vi) { x: x is a perfect square number.}
(vii) {x: x is a perfect cube number.}
(viii) {x: x + 5 = 8}
(ix) {x: 2x + 5 = 9}
(x ) {x: x ≥ 7}
(xi) {x: x ∈ N and 2x + 1> 10}
Solution:
(i) {x: x is an odd natural number.}
(ii) {x: is an even number .}
(Iii) {x: x is not a money multiple of ∈ N and x number 3.}
(iv) {x: x = 1 and x is a positive divisional number.}
(v) {x: x ∈ N and x, the number 3 and 5 are not divisible by any one.}
(vi) {x: x ∈ N and x are not a perfect square number.}
(Vii) {x: x ∈ N and x is not a perfect square cube number.}
(Viii) {x: x ∈ N and x ≠ 3}
(ix) {x : x ∈ N and x ≠ 2}
(x) {x: x ∈ N and x <7}
(xi) {x: x ∈ N and x <
}
Question 4.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}, Then verify that:
(i) (A ∪ B) '= A' ∩ B '
(ii) (A ∩ B)' = A '∪ B'
Solution:
(i) A ∪ B = {2, 4, 6 , 8} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8}
left side = (A ∪B) '= U - (A ∪ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 4, 5, 6, 7, 8} = {1, 9}
A '= U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8} = {1, 3, 5, 7, 9}
B '= U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7} = {1, 4, 6, 8, 9}
right side = A '∩ B' = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}
Hence (A ∪ B) '= A' ∩ B '.
(ii) Left side = (A ∩B) '
(A ∩B) = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2}
(A ∩ B) '= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2} = {1, 3, 4, 5, 6, 7, 8, 9}
right sides. : A '∪ B' = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}
Hence (A ∩ B) '= A' ∪ B '.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}, Then verify that:
(i) (A ∪ B) '= A' ∩ B '
(ii) (A ∩ B)' = A '∪ B'
Solution:
(i) A ∪ B = {2, 4, 6 , 8} ∪ {2, 3, 5, 7} = {2, 3, 4, 5, 6, 7, 8}
left side = (A ∪B) '= U - (A ∪ B) = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 4, 5, 6, 7, 8} = {1, 9}
A '= U - A = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8} = {1, 3, 5, 7, 9}
B '= U - B = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7} = {1, 4, 6, 8, 9}
right side = A '∩ B' = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}
Hence (A ∪ B) '= A' ∩ B '.
(ii) Left side = (A ∩B) '
(A ∩B) = {2, 4, 6, 8} ∩ {2, 3, 5, 7} = {2}
(A ∩ B) '= {1, 2, 3, 4, 5, 6, 7, 8, 9} - {2} = {1, 3, 4, 5, 6, 7, 8, 9}
right sides. : A '∪ B' = {1, 3, 5, 7, 9} ∪ {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}
Hence (A ∩ B) '= A' ∪ B '.
Question 5.
Draw the appropriate Venn diagram for each of the following.
(i) (A ∪ B) '
(ii) A' ∩ B '
(iii) (A ∩B)'
(iv) (A '∪ B')
Solution:
Represent the shaded area by the following sets (Vidfom.com) Huh:
Draw the appropriate Venn diagram for each of the following.
(i) (A ∪ B) '
(ii) A' ∩ B '
(iii) (A ∩B)'
(iv) (A '∪ B')
Solution:
Represent the shaded area by the following sets (Vidfom.com) Huh:
Question 6.
Suppose the set of all triangles in a plane is a universal set. If A is the set of all triangles in which at least one angle is different from 60 °, then what is A?
Solution:
U = {x: is a triangle in the plane.}
A = {x: x is a triangle with at least one angle of 60 °.}
A '= {is the set of all equilateral triangles.}
Solution:
U = {x: is a triangle in the plane.}
A = {x: x is a triangle with at least one angle of 60 °.}
A '= {is the set of all equilateral triangles.}
Question 7.
Fill in the blanks to make the following statements true:
(i) A 'A' = ……… ..
(ii) Φ '∩ A = ……… ..
(iii) A ∩A' = …… …….
(iv) U '∩ A = …………
Solution:
(i) A ∪ A' = U
(ii) Φ '∩ A = U ∩ A = A
(iii) A ∩A' = Φ
(iv) U ' ∩ A = Φ ∩ A = Φ
Fill in the blanks to make the following statements true:
(i) A 'A' = ……… ..
(ii) Φ '∩ A = ……… ..
(iii) A ∩A' = …… …….
(iv) U '∩ A = …………
Solution:
(i) A ∪ A' = U
(ii) Φ '∩ A = U ∩ A = A
(iii) A ∩A' = Φ
(iv) U ' ∩ A = Φ ∩ A = Φ
Questionnaire 1.6
Question 1.
If X and Y are two sets such that n (X) = 17, n (Y) = 23 and n (X ∪ Y) = 38, then find n (X ∩Y).
Solution:
Given. n (X) = 17, n (Y) = 23
n (X ∪ Y) = 38
n (X ∪ Y) = n (X) + n (Y) - n (X ∩ Y)
38 = 17 + 23 - n (K ∩ Y) = 40 - n (X ∩ Y).
n (X ∩ Y) = 40 - 38 = 2.
If X and Y are two sets such that n (X) = 17, n (Y) = 23 and n (X ∪ Y) = 38, then find n (X ∩Y).
Solution:
Given. n (X) = 17, n (Y) = 23
n (X ∪ Y) = 38
n (X ∪ Y) = n (X) + n (Y) - n (X ∩ Y)
38 = 17 + 23 - n (K ∩ Y) = 40 - n (X ∩ Y).
n (X ∩ Y) = 40 - 38 = 2.
Question 2.
If X and Y are two sets such that X ∪ Y has 18 components, X in 8 and Y in 15, then how many elements will X ∩Y have?
Solution:
n (X) = 8, n (Y) = 15, n (X ∪ Y) = 18
We know that,
n (X ∪ Y) = n (X) + n (Y) - n (X ∩ Y )
18 = 8 + 15 - n (X ∩ Y) = 23 - n (X ∩ Y)
n (X ∩ Y) = 23 - 18 = 5.
If X and Y are two sets such that X ∪ Y has 18 components, X in 8 and Y in 15, then how many elements will X ∩Y have?
Solution:
n (X) = 8, n (Y) = 15, n (X ∪ Y) = 18
We know that,
n (X ∪ Y) = n (X) + n (Y) - n (X ∩ Y )
18 = 8 + 15 - n (X ∩ Y) = 23 - n (X ∩ Y)
n (X ∩ Y) = 23 - 18 = 5.
Question 3. In
a group of 400 people, 250 can speak Hindi and 200 English. How many people can speak both Hindi and English?
Solution:
Suppose H and E are the sets of Hindi and English speakers respectively, then
n (H) = 250, n (E) = 200 and
n (H ∪ E) = 400
n (H ∪E) = n ( H) + n (E) - n (H ∩E)
400 = 250 + 200 - n (H ∩E) = 450 - n (H ∩E)
n (H ∩E) = 450 - 400 = 50.
a group of 400 people, 250 can speak Hindi and 200 English. How many people can speak both Hindi and English?
Solution:
Suppose H and E are the sets of Hindi and English speakers respectively, then
n (H) = 250, n (E) = 200 and
n (H ∪ E) = 400
n (H ∪E) = n ( H) + n (E) - n (H ∩E)
400 = 250 + 200 - n (H ∩E) = 450 - n (H ∩E)
n (H ∩E) = 450 - 400 = 50.
Question 4.
If S and T are two sets such that S has 21 elements in T, 32 in T and 11 in S ∩ T, then how many elements will S ∪T have?
Solution:
Here n (S) = 21, n (T) = 32, n (S ∩T) = 11
n (S∪T) = n (S) + n (T) - n (S ∩ T) = 21 + 32 - 11 = 53 - 11 = 42.
If S and T are two sets such that S has 21 elements in T, 32 in T and 11 in S ∩ T, then how many elements will S ∪T have?
Solution:
Here n (S) = 21, n (T) = 32, n (S ∩T) = 11
n (S∪T) = n (S) + n (T) - n (S ∩ T) = 21 + 32 - 11 = 53 - 11 = 42.
Question 5.
If X and two are such sets that X contains 40, X ∪Y contains 60, and X ∩ Y contains 10 elements, then? How many components will I have?
Solution:
n (X) = 40, n (X ∪Y) = 60, n (X ∩ Y) = 10, n (Y) =?
n (X ∪ Y) = n (X) + n (Y) - n (X ∩Y)
60 = 40 + n (Y) - 10
n (Y) = 60 - 40 + 10 = 30.
If X and two are such sets that X contains 40, X ∪Y contains 60, and X ∩ Y contains 10 elements, then? How many components will I have?
Solution:
n (X) = 40, n (X ∪Y) = 60, n (X ∩ Y) = 10, n (Y) =?
n (X ∪ Y) = n (X) + n (Y) - n (X ∩Y)
60 = 40 + n (Y) - 10
n (Y) = 60 - 40 + 10 = 30.
Question 6. In
a group of 70 people, 37 likes coffee, 52 tea and each person likes at least one of the two drinks, then how many people like both coffee and tea?
Solution:
Let C be the set of people who drink coffee and T is the set of people who drink tea, then
n (C ∪T) = 70, n (C) = 37, n (T) = 52
n (C). ∪T) = n (C) + n (T) - n (C ∩ T)
70 = 37 + 52 - n (C ∩T)
n (C ∩ T) = 37 + 52 -70 = 89 - 70 = 19 .
a group of 70 people, 37 likes coffee, 52 tea and each person likes at least one of the two drinks, then how many people like both coffee and tea?
Solution:
Let C be the set of people who drink coffee and T is the set of people who drink tea, then
n (C ∪T) = 70, n (C) = 37, n (T) = 52
n (C). ∪T) = n (C) + n (T) - n (C ∩ T)
70 = 37 + 52 - n (C ∩T)
n (C ∩ T) = 37 + 52 -70 = 89 - 70 = 19 .
Question 7. In
a group of 65 people, 40 persons like cricket and 10 persons like both cricket and tennis, then how many people like only tennis but not cricket? How many people like tennis?
Solution:
Suppose C, cricket is the set of people who appreciate and T is a set of quality tennis choice, then
n (C ∪T) = 65, n (C) = 40, n (C ∩T) = 10
we Know that
n (C ∪ T) = n (C) + n (T) - n (C ∩ T)
65 = 40 + n (T) - 10 = 30 + n (T)
n (T) = 65 - 30 = 35
the number of members to only tennis choice = n (T) - n ( C ∩T) = 35 - 10 = 25.
the number of this type of tennis appreciate those who do not appreciate cricket (Vidfom.com) = 25
therefore, : Number of people who like tennis = 35.
a group of 65 people, 40 persons like cricket and 10 persons like both cricket and tennis, then how many people like only tennis but not cricket? How many people like tennis?
Solution:
Suppose C, cricket is the set of people who appreciate and T is a set of quality tennis choice, then
n (C ∪T) = 65, n (C) = 40, n (C ∩T) = 10
we Know that
n (C ∪ T) = n (C) + n (T) - n (C ∩ T)
65 = 40 + n (T) - 10 = 30 + n (T)
n (T) = 65 - 30 = 35
the number of members to only tennis choice = n (T) - n ( C ∩T) = 35 - 10 = 25.
the number of this type of tennis appreciate those who do not appreciate cricket (Vidfom.com) = 25
therefore, : Number of people who like tennis = 35.
Question 8. In
a committee, 50 people can speak French 20 people Spanish and 10 people can speak both Spanish and French languages. How many people can speak at least one of these two languages?
Solution:
Suppose the set of French speaking people is represented by F and the set of Spanish speaking people is represented by S, then
n (F) = 50, n (S) = 20, n (F ∩S) = 10.
Now, n (F ∪S) = n (F) + n (S) - n (F ∩S) = 50 + 20 - 10 = 60
Number of people speaking at least one language = 60.
a committee, 50 people can speak French 20 people Spanish and 10 people can speak both Spanish and French languages. How many people can speak at least one of these two languages?
Solution:
Suppose the set of French speaking people is represented by F and the set of Spanish speaking people is represented by S, then
n (F) = 50, n (S) = 20, n (F ∩S) = 10.
Now, n (F ∪S) = n (F) + n (S) - n (F ∩S) = 50 + 20 - 10 = 60
Number of people speaking at least one language = 60.
Miscellaneous Questionnaire on Chapter 1
Question 1.
Decide which of the following sets is a subset of:
All real numbers satisfying A = x: x ∈R and x² - 8x + 12 = 0 = x}, B = {2, 4, 6 }, C = {2, 4, 6, 8….}, D = {6}.
Solution:
A = {x: x ∈ R, x satisfies the equation x² - 8x + 12 = 0. That is,
A = {2, 6}
B = {2, 4, 6}.
C = {2, 4, 6, 8….}
D = {6}
(i) The components of set A are also in the 2, 6 set B.
A ⊂ B.
(ii) Thus the elements of the set A are also in the 2, 6 set C.
A ⊂ C.
(iii) The elements of set B are 2, 4, 6 in the set C.
B ⊂C.
(iv) The components of set D are 6, the sets A, B and C are all three,
D ⊂ A, D ⊂ B, D ⊂C.
Decide which of the following sets is a subset of:
All real numbers satisfying A = x: x ∈R and x² - 8x + 12 = 0 = x}, B = {2, 4, 6 }, C = {2, 4, 6, 8….}, D = {6}.
Solution:
A = {x: x ∈ R, x satisfies the equation x² - 8x + 12 = 0. That is,
A = {2, 6}
B = {2, 4, 6}.
C = {2, 4, 6, 8….}
D = {6}
(i) The components of set A are also in the 2, 6 set B.
A ⊂ B.
(ii) Thus the elements of the set A are also in the 2, 6 set C.
A ⊂ C.
(iii) The elements of set B are 2, 4, 6 in the set C.
B ⊂C.
(iv) The components of set D are 6, the sets A, B and C are all three,
D ⊂ A, D ⊂ B, D ⊂C.
Question 2. Find out
whether each of the following statements is true or false. If true, prove it. If false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂C, then A ⊂C
(iv ) If A ⊄ BB ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Solution:
(i) False : Suppose A = {1}, B = {{1}, 2} is
clear that 1 ∈ A, A ∈ B but 1 ∉ set B because 1 is not in B. Thus, the statement given is not true
(ii) False: Suppose A = {1}, B = {1, 2} and C = {{1, 2}, 3} the
elements of set A are in the set B of A ∈. B
components {1, 2} are in the set C B ∈ C
but A = {1} is not in the set C.
The statement A ∈ C is not true.
(iii) True: A ⊂ B ⇒ if x ∈ A and x ∈ B
but B ⊂ C ⇒ if x ∈ B then x ∈ C
if x ∈ A then x ∈ A then x ∈ C ⇒ A ⊂ C
(iv) False : Suppose A = {1, 2}, B = {2, 3}, C = {1, 2, 5} not
all elements of set A are in 1, 2 set B. All the components of
A ⊄ D
set B are not in 2, 3 sets (Vidfom.com) C.
A के C
but all the elements of set A are in 1, 2 set C.
A ⊂ C
Thus the statement given is not true.
(v) The set A = {1, 2}, B = {2, 3, 4, 5}
is not a component of the set A, 2 in the set B. The components of
A ⊄ B
set A are not in 1 set B.
x ∉ B
The statement given thus is not true.
(vi) True: A ⊂ B = if x ∈ A then x ∈ B if x ∉ B and x ∉ A
Thus the statements A ⊂ B, x ∉ B then x ∉ A are true.
whether each of the following statements is true or false. If true, prove it. If false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂C, then A ⊂C
(iv ) If A ⊄ BB ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Solution:
(i) False : Suppose A = {1}, B = {{1}, 2} is
clear that 1 ∈ A, A ∈ B but 1 ∉ set B because 1 is not in B. Thus, the statement given is not true
(ii) False: Suppose A = {1}, B = {1, 2} and C = {{1, 2}, 3} the
elements of set A are in the set B of A ∈. B
components {1, 2} are in the set C B ∈ C
but A = {1} is not in the set C.
The statement A ∈ C is not true.
(iii) True: A ⊂ B ⇒ if x ∈ A and x ∈ B
but B ⊂ C ⇒ if x ∈ B then x ∈ C
if x ∈ A then x ∈ A then x ∈ C ⇒ A ⊂ C
(iv) False : Suppose A = {1, 2}, B = {2, 3}, C = {1, 2, 5} not
all elements of set A are in 1, 2 set B. All the components of
A ⊄ D
set B are not in 2, 3 sets (Vidfom.com) C.
A के C
but all the elements of set A are in 1, 2 set C.
A ⊂ C
Thus the statement given is not true.
(v) The set A = {1, 2}, B = {2, 3, 4, 5}
is not a component of the set A, 2 in the set B. The components of
A ⊄ B
set A are not in 1 set B.
x ∉ B
The statement given thus is not true.
(vi) True: A ⊂ B = if x ∈ A then x ∈ B if x ∉ B and x ∉ A
Thus the statements A ⊂ B, x ∉ B then x ∉ A are true.
Question 3.
Suppose A, B and C that are aggregates that A ∪ B = A ∪ C and A ∩ B = A ∩ C, then Drshaia B = C
solution:
have:
A ∪ B = A ∪ C
(A ∪ B) ∩ C = (A ∪ C) ∩ C = C [(A ∪ C) ∩ C = C]
(A ∩ C) ∩ (B ∩ C) = C
(A ∩ B) ∪ (B ∩ C) = C…. (I) [Given A ∩ C = A ∩ B =.]
A ∪ B = A ∪ C
(A ∪ B) ∩ B = (A ∪ C) ∩ B
B = (A ∪C) ∩ B = (A ∩ B) ∪ (C ∩ B)
or (A ∩) ∪ (B ∩ C) = B …… (ii)
(i) and (ii) gives B = C.
Suppose A, B and C that are aggregates that A ∪ B = A ∪ C and A ∩ B = A ∩ C, then Drshaia B = C
solution:
have:
A ∪ B = A ∪ C
(A ∪ B) ∩ C = (A ∪ C) ∩ C = C [(A ∪ C) ∩ C = C]
(A ∩ C) ∩ (B ∩ C) = C
(A ∩ B) ∪ (B ∩ C) = C…. (I) [Given A ∩ C = A ∩ B =.]
A ∪ B = A ∪ C
(A ∪ B) ∩ B = (A ∪ C) ∩ B
B = (A ∪C) ∩ B = (A ∩ B) ∪ (C ∩ B)
or (A ∩) ∪ (B ∩ C) = B …… (ii)
(i) and (ii) gives B = C.
Question 4.
Show that the following four restrictions are equivalent:
(i) A ⊂ B
(ii) A - B = Φ
(iii) A ∪ B = B
Solution:
(i) A ⊂ B means that all the elements of the set A are in B. .
A - B = Φ ie (i) ⇔ (ii)
(ii) A - B = Φ ⇔ All the elements of the set A are in B.
A ∪ B = B
i.e. (ii) ⇔ (iii)
(iii) A ∪ B = B ⇔ All elements of the set A are in B.
The elements of set A are common in the set A and B.
A ∩B = A is
clear from it All statements are same.
Show that the following four restrictions are equivalent:
(i) A ⊂ B
(ii) A - B = Φ
(iii) A ∪ B = B
Solution:
(i) A ⊂ B means that all the elements of the set A are in B. .
A - B = Φ ie (i) ⇔ (ii)
(ii) A - B = Φ ⇔ All the elements of the set A are in B.
A ∪ B = B
i.e. (ii) ⇔ (iii)
(iii) A ∪ B = B ⇔ All elements of the set A are in B.
The elements of set A are common in the set A and B.
A ∩B = A is
clear from it All statements are same.
Question 5.
Show that if A ⊂ B then C - B ⊂ C - A.
Solution:
Suppose x ∈ C - B ⇒ x ∈ C but
given x ∈ B : A ⊂ B ⇒ If x ∉ B ⇒ x ∉ A
That is, x ∈ C and x ∉ A ⇒ x ∈ C - A
Here we find that
if x ∈ C - B then x ∈ C - A
⇒ C - B ⊂ C - A.
Show that if A ⊂ B then C - B ⊂ C - A.
Solution:
Suppose x ∈ C - B ⇒ x ∈ C but
given x ∈ B : A ⊂ B ⇒ If x ∉ B ⇒ x ∉ A
That is, x ∈ C and x ∉ A ⇒ x ∈ C - A
Here we find that
if x ∈ C - B then x ∈ C - A
⇒ C - B ⊂ C - A.
Question 6.
Suppose P (A) = P (B), prove that A = B.
Solution:
Suppose x is an element of the set A.
Then a subset X (assume) will be such that x ∈ A according to which
X ⊂ A ⇒ X ∈ P (A)
X ∈ P (B) [P (A) = P (B)]
X ⊂ B or x ∈ B
That is, if
x ∈ A then x ∈ B ⇒ A ∈ B… .. (i)
y is an element of set B, then
a subset of set B will be Y (Vidfom.com) (assume) so that y ∈ Y
Y ⊂. B ⇒ Y ∈P (B)
Y ∈P (A) [P (A) = P (B)]
Y ⊂ A if y ∈ B then y ∈ A
B ⊂ A ……… (ii)
Equation (i) and From (ii), we get.
A = B.
Suppose P (A) = P (B), prove that A = B.
Solution:
Suppose x is an element of the set A.
Then a subset X (assume) will be such that x ∈ A according to which
X ⊂ A ⇒ X ∈ P (A)
X ∈ P (B) [P (A) = P (B)]
X ⊂ B or x ∈ B
That is, if
x ∈ A then x ∈ B ⇒ A ∈ B… .. (i)
y is an element of set B, then
a subset of set B will be Y (Vidfom.com) (assume) so that y ∈ Y
Y ⊂. B ⇒ Y ∈P (B)
Y ∈P (A) [P (A) = P (B)]
Y ⊂ A if y ∈ B then y ∈ A
B ⊂ A ……… (ii)
Equation (i) and From (ii), we get.
A = B.
Question 7. For
any set of A's, is it true that P (A) तक P (B) = P (A ∪ B)? Justify your answer.
Solution:
Suppose.
A = {a}, B = {b}, and A ∪ B = {a, b}
P (A) = {Φ, {a}}, P (B) = {Φ, {b}}
P (A) ) ∪ P (B) = {Φ, {4}, {5}}… (i)
Now A ∪ B = {a, b}
P (A ∪ B) = {Φ, {a}, {b}, {a, b}}
review From (i) and (ii) we see that
hence P (A) ∪ P (B) ≠ P (A ∪ B)
any set of A's, is it true that P (A) तक P (B) = P (A ∪ B)? Justify your answer.
Solution:
Suppose.
A = {a}, B = {b}, and A ∪ B = {a, b}
P (A) = {Φ, {a}}, P (B) = {Φ, {b}}
P (A) ) ∪ P (B) = {Φ, {4}, {5}}… (i)
Now A ∪ B = {a, b}
P (A ∪ B) = {Φ, {a}, {b}, {a, b}}
review From (i) and (ii) we see that
hence P (A) ∪ P (B) ≠ P (A ∪ B)
Question 8. For
any two sets A and B, prove that
A = (A ∩ B) ∪ (A - B) and A ∪ (B - A) = A ∪ B.
Solution:
(i) Right side = (A) ∩ B) ∪ (A - B)
= (A ∩ B) ∪ (A - B) [A - B = A ∩ B ']
= (A ∩ (B ∪ B') (from the distribution property)
= A ∩ U (Here U universal set)
= A
Hence (A ∩ B) ∪ (A - B) = A.
(ii) Left side = A ∪ (B - A)
= A ∪ (B ∩ A ') [B - A = B ∩ A ']
= (A ∪ B) ∩ (A ∪ A') (from the distribution property)
= (A ∪ B) ∩ U [here U is the universal set]
= A ∪ B
Hence: A ∪ (B - A) = A ∪ B
any two sets A and B, prove that
A = (A ∩ B) ∪ (A - B) and A ∪ (B - A) = A ∪ B.
Solution:
(i) Right side = (A) ∩ B) ∪ (A - B)
= (A ∩ B) ∪ (A - B) [A - B = A ∩ B ']
= (A ∩ (B ∪ B') (from the distribution property)
= A ∩ U (Here U universal set)
= A
Hence (A ∩ B) ∪ (A - B) = A.
(ii) Left side = A ∪ (B - A)
= A ∪ (B ∩ A ') [B - A = B ∩ A ']
= (A ∪ B) ∩ (A ∪ A') (from the distribution property)
= (A ∪ B) ∩ U [here U is the universal set]
= A ∪ B
Hence: A ∪ (B - A) = A ∪ B
Question 9.
Using the properties of the sets, prove that
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A.
Solution:
(i) Left side = A ∪ (A ∩ B)
= (A ∪ A) ∩ (A ∪ B) (by distribution property)
= A ∩ (A ∪ B) (A ∪ A = A)
= A [A ⊂ A ∪ B]
A ∪ (A ∩ B) = A.
(ii) Left-sided = A ∩ (A ∪ B)
= (A ∩ A) ∪ (A ∩ B) [from distribution property]
= A ∪ (A ∩ B) [A ∩ A = A]
= A [A ∩ B ⊂ A]
Hence A ∩ (A ∪ B) = A.
Using the properties of the sets, prove that
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A.
Solution:
(i) Left side = A ∪ (A ∩ B)
= (A ∪ A) ∩ (A ∪ B) (by distribution property)
= A ∩ (A ∪ B) (A ∪ A = A)
= A [A ⊂ A ∪ B]
A ∪ (A ∩ B) = A.
(ii) Left-sided = A ∩ (A ∪ B)
= (A ∩ A) ∪ (A ∩ B) [from distribution property]
= A ∪ (A ∩ B) [A ∩ A = A]
= A [A ∩ B ⊂ A]
Hence A ∩ (A ∪ B) = A.
Question 10.
Show that A ∩ B = A ∩ C does not necessarily mean B = C.
Solution:
Suppose A = {1, 2}, B = {1, 7} and C = {1, 4}, then
A ∩ B = {1, 2} ∩ {1, 7} = {1}
A ∩ C = {1, 2} ∩ {1, 4} = {1}
A ∩ B = A ∩ C
B ≠ C
If A ∩ B = A ∩ C then necessary (Vidfom.com) Noah is that B = C.
Show that A ∩ B = A ∩ C does not necessarily mean B = C.
Solution:
Suppose A = {1, 2}, B = {1, 7} and C = {1, 4}, then
A ∩ B = {1, 2} ∩ {1, 7} = {1}
A ∩ C = {1, 2} ∩ {1, 4} = {1}
A ∩ B = A ∩ C
B ≠ C
If A ∩ B = A ∩ C then necessary (Vidfom.com) Noah is that B = C.
Question 11.
Suppose A and B are sets. If A∪ X = B ∪ X = Φ and A ∪ X = B ∪ X for a set X, then prove that A = B.
Solution:
Given A ∪ X = B ∪ X, while X is a set.
A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ⊂ A ∪X, A ∩ (A ∪ X) = A]
A = A ∩ (B ∪ X)
= (A ∩ B) ∪ (A ∩ X) [from the distributive property]
= (A ∩ B) ∪ Φ (Given, A ∩ X = Φ
= A ∩ B
A ⊂ B …… (i)
A ∪ X = B ∪ X
B ∩ (A ∪ X ) = B ∩ (B ∪ X)
B ∩ (A ∪ X) = B [B ⊂ B ∪ X]
(B ∩ A) ∪ (B ∩ X) = B [from distribution property]
(B ∩ A) ∪ Φ = B [Given: B ∩ X = Φ]
(B ∩ A) = B
B ⊂ A… .. (ii) From
review (i) and (ii), we find that A = B.
Suppose A and B are sets. If A∪ X = B ∪ X = Φ and A ∪ X = B ∪ X for a set X, then prove that A = B.
Solution:
Given A ∪ X = B ∪ X, while X is a set.
A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ⊂ A ∪X, A ∩ (A ∪ X) = A]
A = A ∩ (B ∪ X)
= (A ∩ B) ∪ (A ∩ X) [from the distributive property]
= (A ∩ B) ∪ Φ (Given, A ∩ X = Φ
= A ∩ B
A ⊂ B …… (i)
A ∪ X = B ∪ X
B ∩ (A ∪ X ) = B ∩ (B ∪ X)
B ∩ (A ∪ X) = B [B ⊂ B ∪ X]
(B ∩ A) ∪ (B ∩ X) = B [from distribution property]
(B ∩ A) ∪ Φ = B [Given: B ∩ X = Φ]
(B ∩ A) = B
B ⊂ A… .. (ii) From
review (i) and (ii), we find that A = B.
Question 12. Find the
sets A, B and C so that A ∩ B, B ∩ C and A ∩ C are the reserved sets and A ∩ B ∩ C = Φ.
Solution:
Suppose. A = {1, 2}, B = {2, 3}, C = {1, 3}
A ∩ B = {1, 2} ∩ {2, 3} = {2},
B ∩ C = {2, 3} ∩ {1, 3} = {3}
C ∩ A = {1, 3} ∩ {1, 2} = {1}
Hence A ∩ B, B ∩ C, C ∩ A are not empty sets.
A ∩ B ∩ C = (A ∩ B) ∩ C = {2} ∩ {1, 3} = Φ
Iti Siddham
sets A, B and C so that A ∩ B, B ∩ C and A ∩ C are the reserved sets and A ∩ B ∩ C = Φ.
Solution:
Suppose. A = {1, 2}, B = {2, 3}, C = {1, 3}
A ∩ B = {1, 2} ∩ {2, 3} = {2},
B ∩ C = {2, 3} ∩ {1, 3} = {3}
C ∩ A = {1, 3} ∩ {1, 2} = {1}
Hence A ∩ B, B ∩ C, C ∩ A are not empty sets.
A ∩ B ∩ C = (A ∩ B) ∩ C = {2} ∩ {1, 3} = Φ
Iti Siddham
Question 13. From
a survey of 600 students of a school, it was found that 150 students drink tea, 225 students coffee and 100 students drink both tea and coffee. Find how many students neither drink tea nor drink coffee.
Solution:
Suppose 7 and C are a set of students who drink tea and coffee, then
n (T) = 150, n (C) = 225, n (T ∩ C) = 100
n (T ∪ C) = n (T ) + n (C) - n (T ∩ C) = 150 + 225 - 100 = 275
= Number of students who drink tea or coffee (Vidfom.com) or both tea and coffee.
Total number of students = 600
Number of students who do not drink tea or coffee = 600 - 275 = 325.
a survey of 600 students of a school, it was found that 150 students drink tea, 225 students coffee and 100 students drink both tea and coffee. Find how many students neither drink tea nor drink coffee.
Solution:
Suppose 7 and C are a set of students who drink tea and coffee, then
n (T) = 150, n (C) = 225, n (T ∩ C) = 100
n (T ∪ C) = n (T ) + n (C) - n (T ∩ C) = 150 + 225 - 100 = 275
= Number of students who drink tea or coffee (Vidfom.com) or both tea and coffee.
Total number of students = 600
Number of students who do not drink tea or coffee = 600 - 275 = 325.
Question 14. In
a group of students, 100 students know Hindi, 50 students English and 25 students know both languages. Each of the students knows either Hindi or English. How many students are there in the group?
Solution:
Find and have a set of people who know Hindi and English respectively, then
n (H) = 100, n (E) = 50, n (H ∩ E) = 25
n (H ∪ E) = n (H) + n (E) - n (H ∩E) = 100 + 50 - 25 = 125
Number of students who know Hindi or English = 125.
a group of students, 100 students know Hindi, 50 students English and 25 students know both languages. Each of the students knows either Hindi or English. How many students are there in the group?
Solution:
Find and have a set of people who know Hindi and English respectively, then
n (H) = 100, n (E) = 50, n (H ∩ E) = 25
n (H ∪ E) = n (H) + n (E) - n (H ∩E) = 100 + 50 - 25 = 125
Number of students who know Hindi or English = 125.
Question 15.
Survey of 60 people found that 25 people newspaper H, 26 people newspaper T, 26 people newspaper I, 9 people both H and I, 11 people both H and T, 8 people both T and I and 3 people read all three newspapers, then find the following:
(i) Number of readers who read at least one newspaper.
(ii) Exactly the number of people who read only one newspaper.
Solution:
Total number of people surveyed = 60
H Number of newspaper readers, n (H) = 25
T Number of newspaper readers, n (T) = 26
I Number of newspaper readers, n ( I) = 26
Survey of 60 people found that 25 people newspaper H, 26 people newspaper T, 26 people newspaper I, 9 people both H and I, 11 people both H and T, 8 people both T and I and 3 people read all three newspapers, then find the following:
(i) Number of readers who read at least one newspaper.
(ii) Exactly the number of people who read only one newspaper.
Solution:
Total number of people surveyed = 60
H Number of newspaper readers, n (H) = 25
T Number of newspaper readers, n (T) = 26
I Number of newspaper readers, n ( I) = 26
Number of readers of H and I newspapers, n (H ∩ I) = 9
Number of readers of H and I newspapers, n (H ∩ T) = 11
Number of readers of T and I newspapers, n (T ∩ I) = 8
Number of readers reading all three newspapers, n (H ∩ T ∩ I) = 3
H and I Number of newspaper readers and non-readers of T newspaper = 9 - 3 = 6
H and T reading newspapers Number of people who did not read newspapers and I = 11 - 3 = 8 Number of readers of
T and I newspapers and readers of H newspaper (Vidfom.com) = 8 - 3 = 5
of those who read only H newspapers Number = 25 - 8 - 6 - 3 = 8
Number of readers reading T only = 26 - 8 - 3 - 5 = 10
Number of readers reading only I = 26 - 6 - 3 - 5 = 12
Number of readers reading at least one newspaper = Number of readers reading only one newspaper + Number of readers of only two newspapers + Number of readers of all three newspapers = (8 + 10 + 12) + (8 + 6 + 5) + 3 = 30 + 19 + 3 = 52
Alternative method:
n (H ∪ T ∪ I) = n (H) + n (T) + n (I) - n (H ∩ T) - n (T ∩ I) - n (H∩I) + n (H ∩ T ∩ I)
= 25 + 26 + 26 - 11 - 8 - 9 + 3 = 77 - 28 + 3 = 80 - 28 = 52
Number of readers of H and I newspapers, n (H ∩ T) = 11
Number of readers of T and I newspapers, n (T ∩ I) = 8
Number of readers reading all three newspapers, n (H ∩ T ∩ I) = 3
H and I Number of newspaper readers and non-readers of T newspaper = 9 - 3 = 6
H and T reading newspapers Number of people who did not read newspapers and I = 11 - 3 = 8 Number of readers of
T and I newspapers and readers of H newspaper (Vidfom.com) = 8 - 3 = 5
of those who read only H newspapers Number = 25 - 8 - 6 - 3 = 8
Number of readers reading T only = 26 - 8 - 3 - 5 = 10
Number of readers reading only I = 26 - 6 - 3 - 5 = 12
Number of readers reading at least one newspaper = Number of readers reading only one newspaper + Number of readers of only two newspapers + Number of readers of all three newspapers = (8 + 10 + 12) + (8 + 6 + 5) + 3 = 30 + 19 + 3 = 52
Alternative method:
n (H ∪ T ∪ I) = n (H) + n (T) + n (I) - n (H ∩ T) - n (T ∩ I) - n (H∩I) + n (H ∩ T ∩ I)
= 25 + 26 + 26 - 11 - 8 - 9 + 3 = 77 - 28 + 3 = 80 - 28 = 52
(ii) Number of readers of H and T newspapers only = 11 - 3 = 8 Number of readers of
T and 1 newspapers only = 8 - 3 = 5 Number of readers of
only 1 and H newspapers = 9 - 3 = 6
Number of readers reading all three newspapers = 3 Number of readers reading
only one newspaper = 52 - (8 + 5 + 6 + 3) = 52 - 22 = 30.
T and 1 newspapers only = 8 - 3 = 5 Number of readers of
only 1 and H newspapers = 9 - 3 = 6
Number of readers reading all three newspapers = 3 Number of readers reading
only one newspaper = 52 - (8 + 5 + 6 + 3) = 52 - 22 = 30.
Question 16.
A survey found that 21 people like Product A, 26 people like Product B, 29 people like Product C. If 14 people like products A and B, 12 people produce products C and A, 14 people like products B and C and 8 people like all three products. Find out how many people like only Product C?
Solution:
Given:
n (A) = 21,
n (B) = 26,
n (C) = 29
n (A ∩ B) = 14,
n (A ∩ C) = 12
n (B ∩ C) = 14 ,
n (A ∩ B ∩ C) = 8
n (A ∩ C) = 12,
P (A ∩ B ∩ C) = 8
A survey found that 21 people like Product A, 26 people like Product B, 29 people like Product C. If 14 people like products A and B, 12 people produce products C and A, 14 people like products B and C and 8 people like all three products. Find out how many people like only Product C?
Solution:
Given:
n (A) = 21,
n (B) = 26,
n (C) = 29
n (A ∩ B) = 14,
n (A ∩ C) = 12
n (B ∩ C) = 14 ,
n (A ∩ B ∩ C) = 8
n (A ∩ C) = 12,
P (A ∩ B ∩ C) = 8
n (A and C only) = 12 - 8 = 4
n (B and C only) = 14 - 8 = 6
n (C only) = n (C) - n (A and C only) - n (B and only only) C) (Vidfom.com) - n (A ∩ B ∩ C) = 29 - 4 - 6 - 8 = 29 - 18 = 11..
n (B and C only) = 14 - 8 = 6
n (C only) = n (C) - n (A and C only) - n (B and only only) C) (Vidfom.com) - n (A ∩ B ∩ C) = 29 - 4 - 6 - 8 = 29 - 18 = 11..
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We suggest you go through NCERT Book for Class 11 Maths Chapter 1 Sets and get the distinguished study materials. Practising these study materials will help you a lot in your school exam & board exam. All NCERT Chapters for Class 11 Maths Chapter 1 Sets given here are as per the latest syllabus