Class 9th Mathematics Term 1 Tamilnadu Board Solution
Exercise 1.1- Which of the following are sets? Justify your answer. i. The collection of good…
- Let A = {0, 1, 2, 3, 4, 5}. Insert the appropriate i. 0 ____ A ii. 6 ____ A iii.…
- Write the following sets in Set-Builder form i. The set of all positive even…
- Write the following sets in Roster form i. a = x:x inn , 2x less than equal to…
- Write the following sets in Descriptive form i. A = {a, e, i, o, u} ii. B = {1,…
- Find the cardinal number of the following sets i. a = x:x = 5^n , n inn n5 ii. B…
- Identify the following sets as finite or infinite i. A = {4, 5, 6, ...} ii. B =…
- Which of the following sets are equivalent? i. a = 2 , 4 , 6 , 8 , 10 , b = 1 ,…
- Which of the following sets are equal? i. A = {1, 2, 3, 4}, B = {4, 3, 2, 1} ii.…
- From the sets given below, select equal sets. A = {12, 14, 18, 22}, B = {11,…
- Is Φ = {Φ}? Why?
- Which of the sets are equal sets? State the reason. Φ, {0}, {Φ}
- Fill in the blanks with sub set nsub set to make each statement true. i. {3}…
- Let X = {−3, −2, −1, 0, 1, 2} and Y = {x : x is an integer and −3 ≤ x 2} i. Is…
- Examine whether A = {x : x is a positive integer divisible by 3} is a subset of…
- Write down the power sets of the following sets. i. A = {x, y} ii. X = {a, b,…
- Find the number of subsets and the number of proper subsets of the following…
- i. If A = , find n[P(A)] ii. If n(A) = 3, find n[P(A)] iii. If n[P(A)] = 512…
- If n[P(A)] = 1, what can you say about the set A?
- Let A = {x : x is a natural number 11} B = {x : x is an even number and 1 x 21}…
Exercise 1.2- Find A ∪ B and A ∩ B for the following sets. i. A = {0, 1, 2, 4, 6} and B = {−3,…
- If A = {x : x is a multiple of 5, x ≤ 30 and x inn } B = {1, 3, 7, 10, 12, 15,…
- If X = {x : x = 2n, x ≤ 20 and n inn } and Y = {x : x = 4n, x ≤ 20 and n inw }…
- U = {1, 2, 3, 6, 7, 12, 17, 21, 35, 52, 56}, P = {numbers divisible by 7}, Q =…
- State which of the following sets are disjoint i. A = {2, 4, 6, 8}; B = {x : x…
- i. If U = {x : 0 ≤ x ≤ 10, x inw } and A = {x : x is a multiple of 3}, find A’…
- If U = {a, b, c, d, e, f, g, h}, A = {a, b, c, d} and B = {b, d, f, g}, Find i.…
- If U = x:1 less than equal to x less than equal to 10 , x inn , A = {1, 3, 5, 7,…
- Given that U = {3, 7, 9, 11, 15, 17, 18}, M = {3, 7, 9, 11} and N = {7, 11, 15,…
- If A = {3, 6, 9, 12, 15, 18}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12}…
- Let U = {x : x is a positive integer less than 50}, A = {x : x is divisible by…
- Find the symmetric difference between the following sets. i. X = {a, d, f, g,…
- i. List the elements of U, E, F, E ∪ F and E ∩ F ii. Find n(U), n(E ∪ F) and…
- i. List U, G and H ii. Find G’, H’, G’ ∩ H’, n(G ∪ H)’ and n(G ∩ H)’ Use the…
Exercise 1.3- Place the elements of the following sets in the proper location on the given…
- If A and B are two sets such that A has 50 elements, B has 65 elements and A ∪ B…
- If A and B are two sets containing 13 and 16 elements respectively, then find…
- If n(A ∩ B) = 5, n(A ∪ B) = 35, n(A) = 13, find n(B).
- If n(A) = 26, n(B) = 10, n (A ∪ B) = 30, n(A’) = 17, find n(A ∩ B) and n(U).…
- If n(U) = 38, n(A) = 16, n(A ∩ B) = 12, n(B’) = 20, find n(A ∪ B).…
- Let A and B be two finite sets such that n(A - B) = 30, n(A ∪ B) = 180. Find…
- The population of a town is 10000. Out of these 5400 persons read newspaper A…
- In a school, all the students play either Foot ball or Volley ball or both. 300…
- In an examination 150 students secured first class in English or Mathematics.…
- In a group of 30 persons, 10 take tea but not coffee 18 take tea. Find how many…
- In a village there are 60 families. Out of these 28 families speak only Tamil…
- In a school 150 students passed X standard Examination. 95 students applied for…
- Pradeep is a Section Chief for an electric utility company. The employees in…
- A and B are two sets such that n(A - B) = 32 + x, n (B - A) = 5x and n(A B) = x…
- The following table shows the percentage of the students of a school who…
- A village has total population of 2500 people. Out of which 1300 people use…
Exercise 1.4- If A = {5, {5, 6}, 7}, which of the following is correct?A. {5, 6} ∈ A B. {5} ∈…
- If X = {a, {b, c}, d}, which of the following is a subset of X?A. {a, b} B. {b,…
- Which of the following statements are true? i. For any set A, A is a proper…
- If a finite set A has m elements, then the number of non-empty proper subsets of…
- The number of subsets of the set {10, 11, 12} isA. 3 B. 8 C. 6 D. 7…
- Which of the following is correct?A. {x:x^2 = -1, x∈Z} = Φ B. Φ = 0 C. Φ = {0}…
- Which of the following is incorrect?A. Every subset of a finite set in finite B.…
- Which of the following is a correct statement?A. Φ ⊆ {a, b} B. Φ ∈ {a, b} C. {a}…
- Which one of the following is a finite set?A. {x: x ∈ Z, x 5} B. {x: x ∈ W, x ≥…
- Given A = {5, 6, 7, 8}. Which one of the following is incorrect?A. Φ ⊆ A B. A ⊆…
- If A = {3, 4, 5, 6} and B = {1, 2, 5, 6}, then A ∪ B =A. {1, 2, 3, 4, 5, 6} B.…
- The number of elements of the set {x: x ∈ Z, x^2 = 1}isA. 3 B. 2 C. 1 D. 0…
- If n(X) = m, n(Y) = n and n(X ∩ Y) = p then n(X ∪ Y) =A. m + n + p B. m + n - p…
- If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 5, 6, 9, 10} then A’ isA.…
- If A B, then A - B isA. B B. A C. Φ D. B - A
- If A is a proper subset of B, then A ∩ B =A. A B. B C. Φ D. A ∪ B…
- If A is a proper subset of B, then A ∪ BA. A B. Φ C. B D. A ∩ B
- The shaded region in the adjoint diagram represents (a)^there there eξ sts A. A…
- If A = {a, b, c}, B = {e, f, g}, then A ∩ B =A. Φ B. A C. B D. A ∪ B…
- The shaded region in the adjoining diagram represents A. A - B B. B - A C. A Δ…
- Which of the following are sets? Justify your answer. i. The collection of good…
- Let A = {0, 1, 2, 3, 4, 5}. Insert the appropriate i. 0 ____ A ii. 6 ____ A iii.…
- Write the following sets in Set-Builder form i. The set of all positive even…
- Write the following sets in Roster form i. a = x:x inn , 2x less than equal to…
- Write the following sets in Descriptive form i. A = {a, e, i, o, u} ii. B = {1,…
- Find the cardinal number of the following sets i. a = x:x = 5^n , n inn n5 ii. B…
- Identify the following sets as finite or infinite i. A = {4, 5, 6, ...} ii. B =…
- Which of the following sets are equivalent? i. a = 2 , 4 , 6 , 8 , 10 , b = 1 ,…
- Which of the following sets are equal? i. A = {1, 2, 3, 4}, B = {4, 3, 2, 1} ii.…
- From the sets given below, select equal sets. A = {12, 14, 18, 22}, B = {11,…
- Is Φ = {Φ}? Why?
- Which of the sets are equal sets? State the reason. Φ, {0}, {Φ}
- Fill in the blanks with sub set nsub set to make each statement true. i. {3}…
- Let X = {−3, −2, −1, 0, 1, 2} and Y = {x : x is an integer and −3 ≤ x 2} i. Is…
- Examine whether A = {x : x is a positive integer divisible by 3} is a subset of…
- Write down the power sets of the following sets. i. A = {x, y} ii. X = {a, b,…
- Find the number of subsets and the number of proper subsets of the following…
- i. If A = , find n[P(A)] ii. If n(A) = 3, find n[P(A)] iii. If n[P(A)] = 512…
- If n[P(A)] = 1, what can you say about the set A?
- Let A = {x : x is a natural number 11} B = {x : x is an even number and 1 x 21}…
- Find A ∪ B and A ∩ B for the following sets. i. A = {0, 1, 2, 4, 6} and B = {−3,…
- If A = {x : x is a multiple of 5, x ≤ 30 and x inn } B = {1, 3, 7, 10, 12, 15,…
- If X = {x : x = 2n, x ≤ 20 and n inn } and Y = {x : x = 4n, x ≤ 20 and n inw }…
- U = {1, 2, 3, 6, 7, 12, 17, 21, 35, 52, 56}, P = {numbers divisible by 7}, Q =…
- State which of the following sets are disjoint i. A = {2, 4, 6, 8}; B = {x : x…
- i. If U = {x : 0 ≤ x ≤ 10, x inw } and A = {x : x is a multiple of 3}, find A’…
- If U = {a, b, c, d, e, f, g, h}, A = {a, b, c, d} and B = {b, d, f, g}, Find i.…
- If U = x:1 less than equal to x less than equal to 10 , x inn , A = {1, 3, 5, 7,…
- Given that U = {3, 7, 9, 11, 15, 17, 18}, M = {3, 7, 9, 11} and N = {7, 11, 15,…
- If A = {3, 6, 9, 12, 15, 18}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12}…
- Let U = {x : x is a positive integer less than 50}, A = {x : x is divisible by…
- Find the symmetric difference between the following sets. i. X = {a, d, f, g,…
- i. List the elements of U, E, F, E ∪ F and E ∩ F ii. Find n(U), n(E ∪ F) and…
- i. List U, G and H ii. Find G’, H’, G’ ∩ H’, n(G ∪ H)’ and n(G ∩ H)’ Use the…
- Place the elements of the following sets in the proper location on the given…
- If A and B are two sets such that A has 50 elements, B has 65 elements and A ∪ B…
- If A and B are two sets containing 13 and 16 elements respectively, then find…
- If n(A ∩ B) = 5, n(A ∪ B) = 35, n(A) = 13, find n(B).
- If n(A) = 26, n(B) = 10, n (A ∪ B) = 30, n(A’) = 17, find n(A ∩ B) and n(U).…
- If n(U) = 38, n(A) = 16, n(A ∩ B) = 12, n(B’) = 20, find n(A ∪ B).…
- Let A and B be two finite sets such that n(A - B) = 30, n(A ∪ B) = 180. Find…
- The population of a town is 10000. Out of these 5400 persons read newspaper A…
- In a school, all the students play either Foot ball or Volley ball or both. 300…
- In an examination 150 students secured first class in English or Mathematics.…
- In a group of 30 persons, 10 take tea but not coffee 18 take tea. Find how many…
- In a village there are 60 families. Out of these 28 families speak only Tamil…
- In a school 150 students passed X standard Examination. 95 students applied for…
- Pradeep is a Section Chief for an electric utility company. The employees in…
- A and B are two sets such that n(A - B) = 32 + x, n (B - A) = 5x and n(A B) = x…
- The following table shows the percentage of the students of a school who…
- A village has total population of 2500 people. Out of which 1300 people use…
- If A = {5, {5, 6}, 7}, which of the following is correct?A. {5, 6} ∈ A B. {5} ∈…
- If X = {a, {b, c}, d}, which of the following is a subset of X?A. {a, b} B. {b,…
- Which of the following statements are true? i. For any set A, A is a proper…
- If a finite set A has m elements, then the number of non-empty proper subsets of…
- The number of subsets of the set {10, 11, 12} isA. 3 B. 8 C. 6 D. 7…
- Which of the following is correct?A. {x:x^2 = -1, x∈Z} = Φ B. Φ = 0 C. Φ = {0}…
- Which of the following is incorrect?A. Every subset of a finite set in finite B.…
- Which of the following is a correct statement?A. Φ ⊆ {a, b} B. Φ ∈ {a, b} C. {a}…
- Which one of the following is a finite set?A. {x: x ∈ Z, x 5} B. {x: x ∈ W, x ≥…
- Given A = {5, 6, 7, 8}. Which one of the following is incorrect?A. Φ ⊆ A B. A ⊆…
- If A = {3, 4, 5, 6} and B = {1, 2, 5, 6}, then A ∪ B =A. {1, 2, 3, 4, 5, 6} B.…
- The number of elements of the set {x: x ∈ Z, x^2 = 1}isA. 3 B. 2 C. 1 D. 0…
- If n(X) = m, n(Y) = n and n(X ∩ Y) = p then n(X ∪ Y) =A. m + n + p B. m + n - p…
- If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 5, 6, 9, 10} then A’ isA.…
- If A B, then A - B isA. B B. A C. Φ D. B - A
- If A is a proper subset of B, then A ∩ B =A. A B. B C. Φ D. A ∪ B…
- If A is a proper subset of B, then A ∪ BA. A B. Φ C. B D. A ∩ B
- The shaded region in the adjoint diagram represents (a)^there there eξ sts A. A…
- If A = {a, b, c}, B = {e, f, g}, then A ∩ B =A. Φ B. A C. B D. A ∪ B…
- The shaded region in the adjoining diagram represents A. A - B B. B - A C. A Δ…
Exercise 1.1
Question 1.Which of the following are sets? Justify your answer.
i. The collection of good books
ii. The collection of prime number less than 30
iii. The collection of ten most talented mathematics teachers.
iv. The collection of all students in your school
v. The collection of all even numbers.
Answer:i. The collection of good books is not a set, because determination of good books varies from person to person.
ii. The collection of prime number less than 30 is a set, because it contains the prime numbers which are less than 30.
iii. The collection of ten most talented mathematics teachers is not a set, because determination of ten most talented mathematics teachers varies from person to person.
iv. The collection of all students in your school is a set, because it contains all students in your school which is easily identifiable.
v. The collection of all even numbers is a set, because it contains all the even numbers and any number is easily identifiable from this set.
Question 2.Let A = {0, 1, 2, 3, 4, 5}. Insert the appropriate
i. 0 ____ A
ii. 6 ____ A
iii. 3 ____ A
iv. 4 ____ A
v. 7 ____ A
Answer:i. 0 ∈ A [∵ 0 is a member of set A]
ii. 6 ∉ A [∵ 6 is not a member of set A]
iii. 3 ∈ A [∵ 3 is a member of set A]
iv. 4 ∈ A [∵ 4 is a member of set A]
v. 7 ∉ A [∵ 7 is not a member of set A]
Question 3.Write the following sets in Set-Builder form
i. The set of all positive even numbers
ii. The set of all whole numbers less than 20
iii. The set of all positive integers which are multiples of 3
iv. The set of all odd natural numbers less than 15
v. The set of all letters in the word ‘computer’
Answer:i. {x : x is a even number and x>0}
ii. {x : x is a whole number and x<20}
iii. {x : x is a positive integer and multiple of 3}
iv. {x : x is a odd natural number and x<15}
v. {x : x is a letter in the word ‘computer’}
Question 4.Write the following sets in Roster form
i.
ii.
iii. C = {x : x is a prime number and a divisor of 6}
iv.
v.
vi.
Answer:i. The required natural numbers are : 3, 4, 5, 6, 7, 8, 9, 10.
∴ The set in roster form A = {3, 4, 5, 6, 7, 8, 9, 10}
ii. The required integers are : 0, 1, 2, 3, 4, 5.
∴ The set in roster form B = {0, 1, 2, 3, 4, 5}
iii. The required prime numbers are : 2, 3
∴ The set in roster form C = {2, 3}
iv. The possible values of n : 1, 2, 3, 4, 5
∴ The values of x : 2, 4, 8, 16, 32
∴ The set in roster form X = {2, 4, 8, 16, 32}
v. The possible values of y = 0, 1, 2, 3, 4, 5
∴ The values of x : -1, 1, 3, 5, 7, 9
∴ The set in roster form M = {-1, 1, 3, 5, 7, 9}
vi. The possible values of x : -4, -3, -2, -1, 0, 1, 2, 3, 4
∴ The set in roster form P = {-4, -3, -2, -1, 0, 1, 2, 3, 4}
Question 5.Write the following sets in Descriptive form
i. A = {a, e, i, o, u}
ii. B = {1, 3, 5, 7, 9, 11}
iii. C = {1, 4, 9, 16, 25}
iv. P = {x : x is a letter in the word ‘set theory’}
v. Q = {x : x is a prime number between 10 and 20}
Answer:i. A is the set of all vowels of English alphabet.
ii. B is the set of all positive odd numbers less than 12.
iii. C is the set of all square numbers less than 26.
iv. P is the set of all the letters in the word ‘set theory’
v. Q is the set of all prime numbers between 10 and 20.
Question 6.Find the cardinal number of the following sets
i.
ii. B = {x : x is a consonant in English Alphabet}
iii. X = {x: x is an even prime number}
iv.
v.
Answer:i. The possible values of n : 1, 2, 3, 4
∴ The possible values of x : 5, 25, 125, 625
∴ A = {5, 25, 125, 625}
∴ n(A) = 4
ii. B = {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}
∴n(B) = 21
iii. The only even prime number is : 2
∴ X = {2}
∴ n(X) = 1
iv. There are no possible values of x.
∴ P = ϕ
∴ n(P) = 0
v. The possible values of x : -3, -2, -1, 0, 1, 2, 3, 4, 5
∴ Q = {-3, -2, -1, 0, 1, 2, 3, 4, 5}
∴ n(Q) = 9
Question 7.Identify the following sets as finite or infinite
i. A = {4, 5, 6, ...}
ii. B = {0, 1, 2, 3, 4, ... 75}
iii. X = {x : x is an even natural number}
iv. Y = {x : x is a multiple of 6 and x > 0}
v. P = The set of letters in the word ‘freedom’
Answer:i. A = {4, 5, 6 ….}
A contains all the integers greater than 3.
∴ A is an infinite set.
ii. B = {0, 1, 2, 3, ……75}
B contains all the integers between 0 and 75.
∴ B is a finite set.
iii. X = {x : x is an even natural number}
∴ X = {2, 4, 6, ……}
∴ X is an infinite set.
iv. Y = {x : x is a multiple of 6 and x > 0}
∴ Y = {6, 12, 18, ……}
∴ Y is an infinite set.
v. P = The set of letters in the word ‘freedom’
P contains only the letters of word ‘freedom’.
∴ P is a finite set.
Question 8.Which of the following sets are equivalent?
i.
ii. Y = {x : x is a vowel in the English Alphabet}
iii. P = {x : x is a prime number and 5 < x < 23}
Answer:i. A = {2, 4, 6, 8, 10}
∴ n(A) = 5
B = {1, 3, 5, 7, 9}
∴ n(B) = 5
∴ n(A) = n(B)
∴ A ≈ B [≈ → equivalent]
ii. X = {x : x∈N, 1<x<6}
∴ The possible values of x : 2, 3, 4, 5
∴ X = {2, 3, 4, 5}
∴ n(X) = 4
Y = {x : x is a vowel in the English Alphabet}
∴ Y = {a, e, i, o, u}
∴ n(Y) = 5
∴ n(X) ≠ n(Y)
∴ The sets X and Y are not equivalent.
iii. P = {x : x is a prime number and 5 < x < 23}
The prime numbers between 5 and 23 are: 7, 11, 13, 17, 19
∴ P = {7, 11, 13, 17, 19, 21}
∴ n(P) = 6
Q = {x : x∈W, 0≤x<5}
∴ The possible values of x : 0, 1, 2, 3, 4
∴ Q = {0, 1, 2, 3, 4}
∴ n(Q) = 5
∴ n(P) ≠ n(Q)
∴ The sets P and Q are not equivalent.
Question 9.Which of the following sets are equal?
i. A = {1, 2, 3, 4}, B = {4, 3, 2, 1}
ii. A = {4, 8, 12, 16}, B = {8, 4, 16, 18}
iii. X = {2, 4, 6, 8}
Y = {x : x is a positive even integer 0 < x < 10}
iv. P = {x : x is a multiple of 10, }
Q = {10, 15, 20, 25, 30, ...}
Answer:i. A = {1, 2, 3, 4}
B = {4, 3, 2, 1}
∴ A and B have exactly same elements.
∴ A = B
ii. A = {4, 8, 12, 16}
B = {8, 4, 16, 18}
∴ A and B does not have exactly same elements.
∴ A ≠ B
iii. X = {2, 4, 6, 8}
Y = {x : x is a positive even integer 0 < x < 10}
Positive even numbers between 0 to 10 are : 2, 4, 6, 8
∴ Y = {2, 4, 6, 8}
∴ X and Y have exactly same elements.
∴ X = Y
iv. P = {x : x is a multiple of 10, }
Multiples of 10 which are natural numbers are : 10, 20, 30, 40 …
∴ P = {10, 20, 30, 40 ……}
Q = {10, 15, 20, 25, 30, ...}
∴ P and Q does not have exactly same elements.
∴ P ≠ Q
Question 10.From the sets given below, select equal sets.
A = {12, 14, 18, 22}, B = {11, 12, 13, 14}, C = {14, 18, 22, 24}
D = {13, 11, 12, 14}, E = {−11, 11}, F = {10, 19}, G = {11, −11}, H = {10, 11}
Answer:B = {11, 12, 13, 14}
D = {13, 11, 12, 14}
∴ B and D have exactly same elements.
∴ B = D
E = {−11, 11}
G = {11, −11}
∴ E and G have exactly same elements.
∴ E = G
Question 11.Is Φ = {Φ}? Why?
Answer:No, because Φ contains no element but {Φ} contains one element, which is a null element. That’s the reason they are not equal.
Question 12.Which of the sets are equal sets? State the reason.
Φ, {0}, {Φ}
Answer:Each one is different from each other as we can see below:
Φ - contains no element
{0} - contains one element which is 0.
{Φ} - contains one element, which is a null element.
Question 13.Fill in the blanks with to make each statement true.
i. {3} ---- {0, 2, 4, 6}
ii. {a} ---- {a, b, c}
iii. {8, 18} ---- {18, 8}
iv. {d} ---- {a, b, c}
Answer:i. {3} {0, 2, 4, 6}
because, here the first set contains only one element 3, which is not present in the second element. Hence we are using “does not belong to” symbol to represent it.
ii. {a} {a, b, c}
because, here the first set contains only one element a, which is present in the second element. Hence we are using “subset” symbol to represent it.
iii.{8, 18} {18, 8}
because, here the first set contains elements 8,18 which are present in the second element. Hence, they are equal sets, we are using “subset” symbol to represent it.
We know that all equal sets can be subsets of each other as well.
iv. {d} {a, b, c}
because, here the first set contains only one element d, which is not present in the second element. Hence we are using “does not belong to” symbol to represent it.
Question 14.Let X = {−3, −2, −1, 0, 1, 2} and Y = {x : x is an integer and −3 ≤ x < 2}
i. Is X a subset of Y?
ii. Is Y a subset of X?
Answer:X = {−3, −2, −1, 0, 1, 2}
Clearly Y= {−3, −2, −1, 0, 1} since ‘2’ is not included in the condition.
Here Y has less number of elements when compared to X. so, Y can be subset of X (or) may not be a subset.
On observing,
i. X is not a subset of Y
Because X has more number of elements than Y, it is not possible to form subset according to our theory.
ii. Y is a subset of X
iii. Because Y has less number of elements when compared to X. Also, X has all the elements of Y. therefore Y is a subset of X.
Question 15.Examine whether A = {x : x is a positive integer divisible by 3} is a subset of B = {x : x is a multiple of 5, }
Answer:Given A = {x : x is a positive integer divisible by 3}
= {3,6,9,12,15,18,21,24,……………….}
B = {x: x is a multiple of 5,x belongs to positive integer}
= {5,10,15,20,25,30,……………………..}
Here, if we observe carefully, we can see some common elements, but as a whole none of the sets are less than or subset of each other. Therefore A is not a subset of B
Question 16.Write down the power sets of the following sets.
i. A = {x, y}
ii. X = {a, b, c}
iii. A = {5, 6, 7, 8}
iv. A = Φ
Answer:Power set is nothing but all the possible combinations of the available set. Which can form a subset of that power set. For a set with ‘n’ elements, we will have 2n elements in the power set.
i. P(A) = {Φ {x}, {y}, {x, y}} - 22 =4 elements
ii. P(X)= {Φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} – 23 =8 elements
iii. P(A) = {Φ, {5}, {6}, {7}, {8}, {5, 6}, {5, 7}, {5, 8}, {6, 7}, {6, 8}, {7, 8}, {5, 6, 7}, {5, 6, 8}, {5, 7, 8}, {6, 7, 8}, {5, 6, 7, 8}} – 24 =16 elements
iv. P(A) = {Φ} – 20 =1=one element
Question 17.Find the number of subsets and the number of proper subsets of the following sets.
i. A = {13, 14, 15, 16, 17, 18}
ii. B = {a, b, c, d, e, f, g}
iii.
Answer:Number of subsets can be same as number of elements in its power set which is nothing but 2n elements. But for proper subset, there should be no element in the power set which is equal to our set itself, therefore 2n -1 elements.
i. number of subsets=26 =64,
number of proper subsets=26 -1= 63
ii. number of subsets=27 =128,
number of proper subsets=27 -1= 127
iii. number of subsets=21 =2,
number of proper subsets=21 -1= 1
Question 18.i. If A = Φ, find n[P(A)]
ii. If n(A) = 3, find n[P(A)]
iii. If n[P(A)] = 512 find n(A)
iv. If n[P(A)] = 1024 find n(A)
Answer:i. If A = Φ, find n[P(A)]
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
As Φ represents 0 elemets.
n(A) = cardinality of A = 0
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 20 = 1
ii) If n(A) = 3, find n[P(A)]
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A = 3
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 2n
where n of no of elements
N[P(A)] = 23 = 8
iii) If n[P(A)] = 512 find n(A)
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A and
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 2n
where n of no of elements
so 512 = 2n
so 29 = 2n
n = 9
so n(A) = 9
iv.If n[P(A)] = 1024 find n(A)
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A and n[ P(A) ] = cardinality of P(A)
let P(A) = n
then n[ P(A) ] = 2n = 1024 = 210
so n = n(A) = 10
Question 19.If n[P(A)] = 1, what can you say about the set A?
Answer:As we have seen in the previous problems, the number of elements in the power set are 2n elements, while there are n elements in the current set. Similarly, if there are n elements in the power set, then there will be Log2(n) elements in the current set.
Log2(1)=0, which implies A is the empty set.
Question 20.Let
A = {x : x is a natural number < 11}
B = {x : x is an even number and 1 < x < 21}
C = {x : x is an integer and 15 ≤ x ≤ 25}
i. List the element of A, B, C
ii. Find n(A), n(B), n(C)
iii. State whether the following are True (T) or False (F)
a)
b)
c)
d)
Answer:i. Given
A = {1, 2, 3, 4, 5, 6, 7, 8, 9,10},
B = {2, 4, 6, 8,10,12,14,16,18, 20},
C = {15,16,17,18,19, 20, 21, 22, 23, 24, 25}
ii. n(A)= number of elements = 10, n(B) = 10, n(C) = 11
iii. (a) F
since B set does not contain 7, the given statement is false.
(b) T
Since A set does not contain 16, the given statement is true
(c) T
Since 15,20,25 form a subset of C set, as C contains 15,20,25. Therefore it is true.
(d) T
Since 10,12 form a subset of B set, as B contains 10,12. Therefore it is true.
Which of the following are sets? Justify your answer.
i. The collection of good books
ii. The collection of prime number less than 30
iii. The collection of ten most talented mathematics teachers.
iv. The collection of all students in your school
v. The collection of all even numbers.
Answer:
i. The collection of good books is not a set, because determination of good books varies from person to person.
ii. The collection of prime number less than 30 is a set, because it contains the prime numbers which are less than 30.
iii. The collection of ten most talented mathematics teachers is not a set, because determination of ten most talented mathematics teachers varies from person to person.
iv. The collection of all students in your school is a set, because it contains all students in your school which is easily identifiable.
v. The collection of all even numbers is a set, because it contains all the even numbers and any number is easily identifiable from this set.
Question 2.
Let A = {0, 1, 2, 3, 4, 5}. Insert the appropriate
i. 0 ____ A
ii. 6 ____ A
iii. 3 ____ A
iv. 4 ____ A
v. 7 ____ A
Answer:
i. 0 ∈ A [∵ 0 is a member of set A]
ii. 6 ∉ A [∵ 6 is not a member of set A]
iii. 3 ∈ A [∵ 3 is a member of set A]
iv. 4 ∈ A [∵ 4 is a member of set A]
v. 7 ∉ A [∵ 7 is not a member of set A]
Question 3.
Write the following sets in Set-Builder form
i. The set of all positive even numbers
ii. The set of all whole numbers less than 20
iii. The set of all positive integers which are multiples of 3
iv. The set of all odd natural numbers less than 15
v. The set of all letters in the word ‘computer’
Answer:
i. {x : x is a even number and x>0}
ii. {x : x is a whole number and x<20}
iii. {x : x is a positive integer and multiple of 3}
iv. {x : x is a odd natural number and x<15}
v. {x : x is a letter in the word ‘computer’}
Question 4.
Write the following sets in Roster form
i.
ii.
iii. C = {x : x is a prime number and a divisor of 6}
iv.
v.
vi.
Answer:
i. The required natural numbers are : 3, 4, 5, 6, 7, 8, 9, 10.
∴ The set in roster form A = {3, 4, 5, 6, 7, 8, 9, 10}
ii. The required integers are : 0, 1, 2, 3, 4, 5.
∴ The set in roster form B = {0, 1, 2, 3, 4, 5}
iii. The required prime numbers are : 2, 3
∴ The set in roster form C = {2, 3}
iv. The possible values of n : 1, 2, 3, 4, 5
∴ The values of x : 2, 4, 8, 16, 32
∴ The set in roster form X = {2, 4, 8, 16, 32}
v. The possible values of y = 0, 1, 2, 3, 4, 5
∴ The values of x : -1, 1, 3, 5, 7, 9
∴ The set in roster form M = {-1, 1, 3, 5, 7, 9}
vi. The possible values of x : -4, -3, -2, -1, 0, 1, 2, 3, 4
∴ The set in roster form P = {-4, -3, -2, -1, 0, 1, 2, 3, 4}
Question 5.
Write the following sets in Descriptive form
i. A = {a, e, i, o, u}
ii. B = {1, 3, 5, 7, 9, 11}
iii. C = {1, 4, 9, 16, 25}
iv. P = {x : x is a letter in the word ‘set theory’}
v. Q = {x : x is a prime number between 10 and 20}
Answer:
i. A is the set of all vowels of English alphabet.
ii. B is the set of all positive odd numbers less than 12.
iii. C is the set of all square numbers less than 26.
iv. P is the set of all the letters in the word ‘set theory’
v. Q is the set of all prime numbers between 10 and 20.
Question 6.
Find the cardinal number of the following sets
i.
ii. B = {x : x is a consonant in English Alphabet}
iii. X = {x: x is an even prime number}
iv.
v.
Answer:
i. The possible values of n : 1, 2, 3, 4
∴ The possible values of x : 5, 25, 125, 625
∴ A = {5, 25, 125, 625}
∴ n(A) = 4
ii. B = {b, c, d, f, g, h, j, k, l, m, n, p, q, r, s, t, v, w, x, y, z}
∴n(B) = 21
iii. The only even prime number is : 2
∴ X = {2}
∴ n(X) = 1
iv. There are no possible values of x.
∴ P = ϕ
∴ n(P) = 0
v. The possible values of x : -3, -2, -1, 0, 1, 2, 3, 4, 5
∴ Q = {-3, -2, -1, 0, 1, 2, 3, 4, 5}
∴ n(Q) = 9
Question 7.
Identify the following sets as finite or infinite
i. A = {4, 5, 6, ...}
ii. B = {0, 1, 2, 3, 4, ... 75}
iii. X = {x : x is an even natural number}
iv. Y = {x : x is a multiple of 6 and x > 0}
v. P = The set of letters in the word ‘freedom’
Answer:
i. A = {4, 5, 6 ….}
A contains all the integers greater than 3.
∴ A is an infinite set.
ii. B = {0, 1, 2, 3, ……75}
B contains all the integers between 0 and 75.
∴ B is a finite set.
iii. X = {x : x is an even natural number}
∴ X = {2, 4, 6, ……}
∴ X is an infinite set.
iv. Y = {x : x is a multiple of 6 and x > 0}
∴ Y = {6, 12, 18, ……}
∴ Y is an infinite set.
v. P = The set of letters in the word ‘freedom’
P contains only the letters of word ‘freedom’.
∴ P is a finite set.
Question 8.
Which of the following sets are equivalent?
i.
ii. Y = {x : x is a vowel in the English Alphabet}
iii. P = {x : x is a prime number and 5 < x < 23}
Answer:
i. A = {2, 4, 6, 8, 10}
∴ n(A) = 5
B = {1, 3, 5, 7, 9}
∴ n(B) = 5
∴ n(A) = n(B)
∴ A ≈ B [≈ → equivalent]
ii. X = {x : x∈N, 1<x<6}
∴ The possible values of x : 2, 3, 4, 5
∴ X = {2, 3, 4, 5}
∴ n(X) = 4
Y = {x : x is a vowel in the English Alphabet}
∴ Y = {a, e, i, o, u}
∴ n(Y) = 5
∴ n(X) ≠ n(Y)
∴ The sets X and Y are not equivalent.
iii. P = {x : x is a prime number and 5 < x < 23}
The prime numbers between 5 and 23 are: 7, 11, 13, 17, 19
∴ P = {7, 11, 13, 17, 19, 21}
∴ n(P) = 6
Q = {x : x∈W, 0≤x<5}
∴ The possible values of x : 0, 1, 2, 3, 4
∴ Q = {0, 1, 2, 3, 4}
∴ n(Q) = 5
∴ n(P) ≠ n(Q)
∴ The sets P and Q are not equivalent.
Question 9.
Which of the following sets are equal?
i. A = {1, 2, 3, 4}, B = {4, 3, 2, 1}
ii. A = {4, 8, 12, 16}, B = {8, 4, 16, 18}
iii. X = {2, 4, 6, 8}
Y = {x : x is a positive even integer 0 < x < 10}
iv. P = {x : x is a multiple of 10, }
Q = {10, 15, 20, 25, 30, ...}
Answer:
i. A = {1, 2, 3, 4}
B = {4, 3, 2, 1}
∴ A and B have exactly same elements.
∴ A = B
ii. A = {4, 8, 12, 16}
B = {8, 4, 16, 18}
∴ A and B does not have exactly same elements.
∴ A ≠ B
iii. X = {2, 4, 6, 8}
Y = {x : x is a positive even integer 0 < x < 10}
Positive even numbers between 0 to 10 are : 2, 4, 6, 8
∴ Y = {2, 4, 6, 8}
∴ X and Y have exactly same elements.
∴ X = Y
iv. P = {x : x is a multiple of 10, }
Multiples of 10 which are natural numbers are : 10, 20, 30, 40 …
∴ P = {10, 20, 30, 40 ……}
Q = {10, 15, 20, 25, 30, ...}
∴ P and Q does not have exactly same elements.
∴ P ≠ Q
Question 10.
From the sets given below, select equal sets.
A = {12, 14, 18, 22}, B = {11, 12, 13, 14}, C = {14, 18, 22, 24}
D = {13, 11, 12, 14}, E = {−11, 11}, F = {10, 19}, G = {11, −11}, H = {10, 11}
Answer:
B = {11, 12, 13, 14}
D = {13, 11, 12, 14}
∴ B and D have exactly same elements.
∴ B = D
E = {−11, 11}
G = {11, −11}
∴ E and G have exactly same elements.
∴ E = G
Question 11.
Is Φ = {Φ}? Why?
Answer:
No, because Φ contains no element but {Φ} contains one element, which is a null element. That’s the reason they are not equal.
Question 12.
Which of the sets are equal sets? State the reason.
Φ, {0}, {Φ}
Answer:
Each one is different from each other as we can see below:
Φ - contains no element
{0} - contains one element which is 0.
{Φ} - contains one element, which is a null element.
Question 13.
Fill in the blanks with to make each statement true.
i. {3} ---- {0, 2, 4, 6}
ii. {a} ---- {a, b, c}
iii. {8, 18} ---- {18, 8}
iv. {d} ---- {a, b, c}
Answer:
i. {3} {0, 2, 4, 6}
because, here the first set contains only one element 3, which is not present in the second element. Hence we are using “does not belong to” symbol to represent it.
ii. {a} {a, b, c}
because, here the first set contains only one element a, which is present in the second element. Hence we are using “subset” symbol to represent it.
iii.{8, 18} {18, 8}
because, here the first set contains elements 8,18 which are present in the second element. Hence, they are equal sets, we are using “subset” symbol to represent it.
We know that all equal sets can be subsets of each other as well.
iv. {d} {a, b, c}
because, here the first set contains only one element d, which is not present in the second element. Hence we are using “does not belong to” symbol to represent it.
Question 14.
Let X = {−3, −2, −1, 0, 1, 2} and Y = {x : x is an integer and −3 ≤ x < 2}
i. Is X a subset of Y?
ii. Is Y a subset of X?
Answer:
X = {−3, −2, −1, 0, 1, 2}
Clearly Y= {−3, −2, −1, 0, 1} since ‘2’ is not included in the condition.
Here Y has less number of elements when compared to X. so, Y can be subset of X (or) may not be a subset.
On observing,
i. X is not a subset of Y
Because X has more number of elements than Y, it is not possible to form subset according to our theory.
ii. Y is a subset of X
iii. Because Y has less number of elements when compared to X. Also, X has all the elements of Y. therefore Y is a subset of X.
Question 15.
Examine whether A = {x : x is a positive integer divisible by 3} is a subset of B = {x : x is a multiple of 5, }
Answer:
Given A = {x : x is a positive integer divisible by 3}
= {3,6,9,12,15,18,21,24,……………….}
B = {x: x is a multiple of 5,x belongs to positive integer}
= {5,10,15,20,25,30,……………………..}
Here, if we observe carefully, we can see some common elements, but as a whole none of the sets are less than or subset of each other. Therefore A is not a subset of B
Question 16.
Write down the power sets of the following sets.
i. A = {x, y}
ii. X = {a, b, c}
iii. A = {5, 6, 7, 8}
iv. A = Φ
Answer:
Power set is nothing but all the possible combinations of the available set. Which can form a subset of that power set. For a set with ‘n’ elements, we will have 2n elements in the power set.
i. P(A) = {Φ {x}, {y}, {x, y}} - 22 =4 elements
ii. P(X)= {Φ, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}} – 23 =8 elements
iii. P(A) = {Φ, {5}, {6}, {7}, {8}, {5, 6}, {5, 7}, {5, 8}, {6, 7}, {6, 8}, {7, 8}, {5, 6, 7}, {5, 6, 8}, {5, 7, 8}, {6, 7, 8}, {5, 6, 7, 8}} – 24 =16 elements
iv. P(A) = {Φ} – 20 =1=one element
Question 17.
Find the number of subsets and the number of proper subsets of the following sets.
i. A = {13, 14, 15, 16, 17, 18}
ii. B = {a, b, c, d, e, f, g}
iii.
Answer:
Number of subsets can be same as number of elements in its power set which is nothing but 2n elements. But for proper subset, there should be no element in the power set which is equal to our set itself, therefore 2n -1 elements.
i. number of subsets=26 =64,
number of proper subsets=26 -1= 63
ii. number of subsets=27 =128,
number of proper subsets=27 -1= 127
iii. number of subsets=21 =2,
number of proper subsets=21 -1= 1
Question 18.
i. If A = Φ, find n[P(A)]
ii. If n(A) = 3, find n[P(A)]
iii. If n[P(A)] = 512 find n(A)
iv. If n[P(A)] = 1024 find n(A)
Answer:
i. If A = Φ, find n[P(A)]
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
As Φ represents 0 elemets.
n(A) = cardinality of A = 0
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 20 = 1
ii) If n(A) = 3, find n[P(A)]
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A = 3
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 2n
where n of no of elements
N[P(A)] = 23 = 8
iii) If n[P(A)] = 512 find n(A)
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A and
n[ P(A) ] = cardinality of P(A)
N[P(A)] = 2n
where n of no of elements
so 512 = 2n
so 29 = 2n
n = 9
so n(A) = 9
iv.If n[P(A)] = 1024 find n(A)
A is a set containing some elements.
P(A) = power set = the set of all subsets of A = power set of A.
n(A) = cardinality of A and n[ P(A) ] = cardinality of P(A)
let P(A) = n
then n[ P(A) ] = 2n = 1024 = 210
so n = n(A) = 10
Question 19.
If n[P(A)] = 1, what can you say about the set A?
Answer:
As we have seen in the previous problems, the number of elements in the power set are 2n elements, while there are n elements in the current set. Similarly, if there are n elements in the power set, then there will be Log2(n) elements in the current set.
Log2(1)=0, which implies A is the empty set.
Question 20.
Let
A = {x : x is a natural number < 11}
B = {x : x is an even number and 1 < x < 21}
C = {x : x is an integer and 15 ≤ x ≤ 25}
i. List the element of A, B, C
ii. Find n(A), n(B), n(C)
iii. State whether the following are True (T) or False (F)
a)
b)
c)
d)
Answer:
i. Given
A = {1, 2, 3, 4, 5, 6, 7, 8, 9,10},
B = {2, 4, 6, 8,10,12,14,16,18, 20},
C = {15,16,17,18,19, 20, 21, 22, 23, 24, 25}
ii. n(A)= number of elements = 10, n(B) = 10, n(C) = 11
iii. (a) F
since B set does not contain 7, the given statement is false.
(b) T
Since A set does not contain 16, the given statement is true
(c) T
Since 15,20,25 form a subset of C set, as C contains 15,20,25. Therefore it is true.
(d) T
Since 10,12 form a subset of B set, as B contains 10,12. Therefore it is true.
Exercise 1.2
Question 1.Find A ∪ B and A ∩ B for the following sets.
i. A = {0, 1, 2, 4, 6} and B = {−3, −1, 0, 2, 4, 5}
ii. A = {2, 4, 6, 8} and B = Φ
iii. and B = {x : x is a prime number less than 11}
iv. and
Answer:i. A = {0, 1, 2, 4, 6}
B = {-3, -1, 0, 2, 4, 5}
∴ A ∪ B = {-3, -1, 0, 1, 2, 4, 5, 6} [elements that are present in either A or B]
∴ A ∩ B = {0, 2, 4} [elements that are present in both A and B]
ii. A = {2, 4, 6, 8} and B = Φ
∴ A ∪ B = {2, 4, 6, 8} [elements that are present in either A or B]
∴ A ∩ B = Φ [elements that are present in both A and B]
iii. A = {x : x ∈ N, x ≤ 5}
∴ A contains the Natural numbers that are less than or equal to 5.
∴ A = {1, 2, 3, 4, 5}
B = {x : x is a prime number less than 11}
Prime numbers less than 11 are : 2, 3, 5, 7
∴ B = {2, 3, 5, 7}
∴ A ∪ B = {1, 2, 3, 4, 5, 7} [elements that are present in either A or B]
∴ A ∩ B = {2, 3, 5} [elements that are present in both A and B]
iv. A = {x : x ∈ N, 2 < x ≤ 7}
∴ A contains the Natural numbers that are greater than 2 and less than equal to 7
∴ A = {3, 4, 5, 6, 7}
B = {x : x ∈ W, 0≤ x ≤ 6}
∴ B contains the whole numbers that are greater than or equal to 0 and less than or equal to 6.
∴ B = {0, 1, 2, 3, 4, 5, 6}
∴ A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7} [elements that are present in either A or B]
∴ A ∩ B = {3, 4, 5, 6} [elements that are present in both A and B]
Question 2.If A = {x : x is a multiple of 5, x ≤ 30 and }
B = {1, 3, 7, 10, 12, 15, 18, 25},
Find (i) A ∪ B (ii) A ∩ B
Answer:A = {x : x is a multiple of 5, x ≤ 30 and x ∈}
A contains the natural numbers that are less than or equal to 30 and multiples of 5.
∴ A = {5, 10, 15, 20, 25, 30}
B = {1, 3, 7, 10, 12, 15, 18, 25}
∴ A ∪ B = {1, 3, 5, 7, 10, 12, 15, 18, 20, 25, 30} [elements that are present in either A or B]
∴ A ∩ B = {10, 15, 25} [elements that are present in both A and B]
Question 3.If X = {x : x = 2n, x ≤ 20 and } and Y = {x : x = 4n, x ≤ 20 and }
Find (i) X ∪ Y (ii) X ∩ Y
Answer:X = {x : x = 2n, x ≤ 20 and n ∈}
X is twice of a natural number(n ∈ N) and less than or equal to 20.
∴ X = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
Y = {x : x = 4n, x ≤ 20 and n ∈ }
Y is 4 times a positive whole number(n ∈ W) and less than equal to 20.
∴ Y = {0, 4, 8, 12, 16, 20}
∴ X ∪ Y = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20} [elements that are present in either X or Y]
∴ X ∩ Y = {4, 8, 12, 16, 20} [elements that are present in both X and Y]
Question 4.U = {1, 2, 3, 6, 7, 12, 17, 21, 35, 52, 56},
P = {numbers divisible by 7}, Q = {prime numbers}
List the elements of the set {x : x ∈ P ∩ Q}
Answer:U = {1, 2, 3, 6, 7, 12, 17, 21, 35, 52, 56}
Numbers divisible by 7 present in U are: 7, 21, 35, 56.
Prime numbers present in U are : 2, 3, 7, 17.
∴ P = {7, 21, 35, 56}
∴ Q = {2, 3, 7, 17}
∴ P ∩ Q = {7} [elements that are present in both P and Q]
∴ The set {x : x ∈ P ∩ Q} = {7}
Question 5.State which of the following sets are disjoint
i. A = {2, 4, 6, 8}; B = {x : x is an even number < 10, }
ii. X = {1, 3, 5, 7, 9}, Y = {0, 2, 4, 6, 8, 10}
iii. P = {x : x is a prime < 15}
Q = {x : x is a multiple of 2 and x < 16}
iv. R = {a, b, c, d, e}, S = {d, e, a, b, c}
Answer:i. A = {2, 4, 6, 8}
B = {x : x is an even number < 10, x ∈}
B contains even natural numbers that are less than 10.
∴ B = {2, 4, 6, 8}
∴ A ∩ B = {2, 4, 6, 8} [elements that are present in both A and B]
∵ A ∩ B ≠ Φ
∴ A and B are not disjoint sets.
ii. X = {1, 3, 5, 7, 9}
Y = {0, 2, 4, 6, 8, 10}
∴ X ∩ Y = Φ [elements that are present in both X and Y]
∴ X and Y are disjoint sets.
iii. P = {x : x is a prime < 15}
P contains the prime numbers less than 15.
∴ P = {2, 3, 5, 7, 11, 13}
Q = {x : x is a multiple of 2 and x < 16}
Q contains the numbers that are less than 16 and multiple of 2.
∴ Q = {…-2, 0, 2, 4, 6, 8, 10, 12, 14}
∴ P ∩ Q = {2} [elements that are present in both P and Q]
∵ P ∩ Q ≠ Φ
∴ P and Q are not disjoint sets.
iv. R = {a, b, c, d, e}
S = {d, e, a, b, c}
R ∩ S = {d, e} [elements that are present in both R and S]
∵ R ∩ S ≠ Φ
∴ R and S are not disjoint sets.
Question 6.i. If U = {x : 0 ≤ x ≤ 10, } and A = {x : x is a multiple of 3}, find A’
ii. If U is the set of natural numbers and A’ is the set of all composite numbers, then what is A?
Answer:i. U = {x : 0 ≤ x ≤ 10, x ∈}
U contains all the whole numbers that are greater than or equal to zero and less than or equal to 10.
∴ U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {x : x is a multiple of 3}
Multiples of 3 present in U are : 3, 6, 9 [U→ Universal set]
∴ A = {3, 6, 9}
∴ A’ = {0, 1, 2, 4, 5, 7, 8, 10} [element that are present in U but not in A]
ii. U is the set of natural numbers.
A’ is the set of all composite numbers.
⇒ (A’)’ = A
∴ A contains all the elements that are present in U but not in A’.
∴ A contains all the natural numbers that are not composite numbers.
∴ A is the set of all prime numbers and 1.
Question 7.If U = {a, b, c, d, e, f, g, h}, A = {a, b, c, d} and B = {b, d, f, g},
Find
i. A ∪ B
ii. (A ∪ B)’
iii. A ∩ B
iv. (A ∩ B)’
Answer:U = {a, b, c, d, e, f, g, h}
A = {a, b, c, d}
B = {b, d, f, g}
i. A ∪ B = {a, b, c, d, f, g} [elements that are present in either A or B]
ii. U = {a, b, c, d, e, f, g, h}
A ∪ B = {a, b, c, d, f, g}
∴ (A ∪ B)’ = {e, h} [elements that are present in U but not in A ∪ B]
iii. A ∩ B = {b, d} [elements that are present in both A and B]
iv. U = {a, b, c, d, e, f, g, h}
A ∩ B = {b, d}
∴ (A ∩ B)’ = {a, c, e, f, g, h} [elements that are present in U but not in A ∩ B]
Question 8.If U = , A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 9, 10},
Find
i. A’
ii. B’
iii. A’ ∪ B’
iv. A’ ∩ B’
Answer:U = {x : 1≤x≤10, x ∈ N}
U contains all the natural numbers that are greater than or equal to 1 and less than or equal to 10}
∴ U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
i. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 3, 5, 7, 9}
A’ = {2, 4, 6, 8, 10} [elements that are present in U but not in A]
ii. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {2, 3, 5, 9, 10}
B’ = {1, 4, 5, 6, 7, 8} [elements that are present in U but not in B]
iii. A’ = {2, 4, 6, 8, 10}
B’ = {1, 4, 5, 6, 7, 8}
∴ A’ ∪ B’ = {1, 2, 4, 5, 6, 7, 8, 10} [elements that are present in either A’ or B’]
iv. A’ = {2, 4, 6, 8, 10}
B’ = {1, 4, 5, 6, 7, 8}
∴ A’ ∩ B’ = {4, 6, 8} [elements that are present in both A’ and B’]
Question 9.Given that U = {3, 7, 9, 11, 15, 17, 18}, M = {3, 7, 9, 11} and N = {7, 11, 15, 17},
Find
i. M – N
ii. N – M
iii. N’ – M
iv. M’ – N
v. M ∩ (M – N)
vi. N ∪ (N – M)
vii. n(M – N)
Answer:U = {3, 7, 9, 11, 15, 17, 18}
M = {3, 7, 9, 11}
N = {7, 11, 15, 17}
i. M – N = {3, 9} [elements that are present in M but not in N]
ii. N – M = {15, 17} [elements that are present in N but not in M]
iii. U = {3, 7, 9, 11, 15, 17, 18}
M = {3, 7, 9, 11}
N = {7, 11, 15, 17}
∴ N’ = {3, 9, 18}
∴ N’ – M = {18} [elements that are present in N’ but not in M]
iv. U = {3, 7, 9, 11, 15, 17, 18}
M = {3, 7, 9, 11}
N = {7, 11, 15, 17}
∴ M’ = {15, 17, 18} [elements that are present in U but not in M]
∴ M’ – N = {18} [elements that are present in M’ but not in N]
v. M = {3, 7, 9, 11}
M – N = {3, 9}
∴ M ∩ (M – N) = {3, 9} [elements that are present in both in M and (M – N)]
vi. N = {7, 11, 15, 17}
N – M = {15, 17}
∴ N ∪ (N – M) = {7, 11, 15, 17} [elements that are present in either N or (N – M)]
vii. M – N = {3, 9}
∴ n(M – N) = 2 [no. of elements in (M – N)]
Question 10.If A = {3, 6, 9, 12, 15, 18}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12} and D = {5, 10, 15, 20, 25}, find
i. A – B
ii. B – C
iii. C – D
iv. D – A
v. n(A – C)
Answer:i. A = {3, 6, 9, 12, 15, 18}
B = {4, 8, 12, 16, 20}
∴ A – B = {3, 6, 9, 15, 18} [elements that are present in A but not in B]
ii. B = {4, 8, 12, 16, 20}
C = {2, 4, 6, 8, 10, 12}
∴ B – C = {16, 20} [elements that are present in B but not in C]
iii. C = {2, 4, 6, 8, 10, 12}
D = {5, 10, 15, 20, 25}
∴ C – D = {2, 4, 6, 8, 12} [elements that are present in C but not in D]
iv. D = {5, 10, 15, 20, 25}
A = {3, 6, 9, 12, 15, 18}
∴ D – A = {5, 10, 20, 25} [elements that are present in D but not in A]
v. A = {3, 6, 9, 12, 15, 18}
C = {2, 4, 6, 8, 10, 12}
∴ A – C = {3, 9, 15, 18} [elements that are present in A but not in C]
∴ n(A – C) = 4 [no. of elements in (A – C)]
Question 11.Let U = {x : x is a positive integer less than 50}, A = {x : x is divisible by 4} and B = {x : x leaves a remainder 2 when divided by 14}.
i. List the elements of U, A and B
ii. Find A ∪ B, A ∩ B, n(A ∪ B), n(A ∩ B)
Answer:i. U = {x : x is a positive integer less than 50}
U contains all the positive integers that are less than 50.
∴ U = {1, 2, 3, 4 ……… 49}
A = {x : x is divisible by 4}
A contains all the members of U that are divisible by 4. [∵U is universal set]
Numbers present in U that are divisible by 4 are : 4, 8, 12 …… 48.
∴ A = {4, 8, 12 …… 48}
B = {x : x leaves a remainder 2 when divided by 14}
B contains all the elements of U that leaves a remainder 2 when divided by 14.
Such numbers present in U are : 16, 30, 44
∴ B = {16, 30, 44}
ii. A = {4, 8, 12 …… 48}
B = {16, 30, 44}
∴ A ∪ B = {4, 8, 12, 16, 20, 24, 28, 30, 32, 36, 40, 44, 48} [elements that are present in either A or B]
∴ A ∩ B = {16, 44} [elements that are present in both A and B]
∴ n(A ∪ B) = 13 [no. of elements in A ∪ B]
∴ n(A ∩ B) = 2 [no. of elements in A ∩ B]
Question 12.Find the symmetric difference between the following sets.
i. X = {a, d, f, g, h}, Y = {b, e, g, h, k}
ii. P = {x : 3 < x < 9, }, Q = {x : x < 5, }
iii. A = {−3, −2, 0, 2, 3, 5}, B = {−4, −3, −1, 0, 2, 3}
Answer:i. X = {a, d, f, g, h}
Y = {b, e, g, h, k}
∴ XΔY = {a, b, d, e, f, k} [elements that are present either only in X or only in Y]
ii. P = {x : 3 < x < 9, x ∈}
P contains all the natural number between 3 and 9.
∴ P = {4, 5, 6, 7, 8}
Q = {x : x < 5, x ∈}
The possible values of Q are : 0, 1, 2, 3, 4
∴ Q = {0, 1, 2, 3, 4}
∴ PΔQ = {0, 1, 2, 3, 5, 6, 7, 8} [elements that are present either only in P or only in Q]
iii. A = {−3, −2, 0, 2, 3, 5}
B = {−4, −3, −1, 0, 2, 3}
∴ XΔY = {-4, -2, -1, 5} [elements that are present either only in A or only in B]
Question 13.Use the Venn diagram to answer the following questions
i. List the elements of
U, E, F, E ∪ F and E ∩ F
ii. Find n(U), n(E ∪ F) and n(E ∩ F)
Answer:i. From the Venn diagram we get,
U = {1, 2, 3, 4, 7, 9, 10, 11}
[U is the universal set, so it contains all the values of in the diagram]
E = {1, 2, 4, 7}
[The set of values in red Circle]
F = {4, 7, 9, 11}
[The set of values in red Circle]
∴ E ∪ F = {1, 2, 4, 7, 9, 11} [elements present either in E or in F]
∴ E ∩ F = {4, 7} [elements that are present in both E and F]
ii. U = {1, 2, 3, 4, 7, 9, 10, 11}
∴ n(U) = 8 [no. of elements in U]
E ∪ F = {1, 2, 4, 7, 9, 11}
∴ n(E ∪ F) = 6 [no. of elements in E ∪ F]
E ∩ F = {4, 7}
∴ n(E ∩ F) = 2 [no. of elements in E ∩ F]
Question 14.Use the Venn diagram to answer the following questions
i. List U, G and H
ii. Find G’, H’, G’ ∩ H’, n(G ∪ H)’ and n(G ∩ H)’
Answer:i. From the Venn diagram we get,
U = {1, 2, 3, 4, 5, 6, 8, 9, 10}
G = {1, 2, 4, 8}
H = {2, 6, 8, 10}
ii. U = {1, 2, 3, 4, 5, 6, 8, 9, 10}
G = {1, 2, 4, 8}
H = {2, 6, 8, 10}
∴ G’ = {3, 5, 6, 9, 10} [elements that are present in U but not in G]
∴ H’ = {1, 3, 4, 5, 9} [elements that are present in U but not in H]
∴ G’ ∩ H’ = {3, 5, 9} [elements that are present in both G’ and H’]
∴ G ∪ H = {1, 2, 4, 6, 8, 10} [elements that are present either in G or in H]
∴ (G ∪ H)’ = {3, 5, 9} [elements present in U but not in (G ∪ H)]
∴ n(G ∪ H)’ = 3 [no. of elements in (G ∪ H)’]
∴ G ∩ H = {2, 8} [elements present in both G and H]
∴ (G ∩ H)’ = {1, 3, 4, 5, 6, 9, 10} [elements present in U but not in (G ∩ H)]
∴ n(G ∩ H)’ = 7 [no. of elements in (G ∩ H)’]
Find A ∪ B and A ∩ B for the following sets.
i. A = {0, 1, 2, 4, 6} and B = {−3, −1, 0, 2, 4, 5}
ii. A = {2, 4, 6, 8} and B = Φ
iii. and B = {x : x is a prime number less than 11}
iv. and
Answer:
i. A = {0, 1, 2, 4, 6}
B = {-3, -1, 0, 2, 4, 5}
∴ A ∪ B = {-3, -1, 0, 1, 2, 4, 5, 6} [elements that are present in either A or B]
∴ A ∩ B = {0, 2, 4} [elements that are present in both A and B]
ii. A = {2, 4, 6, 8} and B = Φ
∴ A ∪ B = {2, 4, 6, 8} [elements that are present in either A or B]
∴ A ∩ B = Φ [elements that are present in both A and B]
iii. A = {x : x ∈ N, x ≤ 5}
∴ A contains the Natural numbers that are less than or equal to 5.
∴ A = {1, 2, 3, 4, 5}
B = {x : x is a prime number less than 11}
Prime numbers less than 11 are : 2, 3, 5, 7
∴ B = {2, 3, 5, 7}
∴ A ∪ B = {1, 2, 3, 4, 5, 7} [elements that are present in either A or B]
∴ A ∩ B = {2, 3, 5} [elements that are present in both A and B]
iv. A = {x : x ∈ N, 2 < x ≤ 7}
∴ A contains the Natural numbers that are greater than 2 and less than equal to 7
∴ A = {3, 4, 5, 6, 7}
B = {x : x ∈ W, 0≤ x ≤ 6}
∴ B contains the whole numbers that are greater than or equal to 0 and less than or equal to 6.
∴ B = {0, 1, 2, 3, 4, 5, 6}
∴ A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7} [elements that are present in either A or B]
∴ A ∩ B = {3, 4, 5, 6} [elements that are present in both A and B]
Question 2.
If A = {x : x is a multiple of 5, x ≤ 30 and }
B = {1, 3, 7, 10, 12, 15, 18, 25},
Find (i) A ∪ B (ii) A ∩ B
Answer:
A = {x : x is a multiple of 5, x ≤ 30 and x ∈}
A contains the natural numbers that are less than or equal to 30 and multiples of 5.
∴ A = {5, 10, 15, 20, 25, 30}
B = {1, 3, 7, 10, 12, 15, 18, 25}
∴ A ∪ B = {1, 3, 5, 7, 10, 12, 15, 18, 20, 25, 30} [elements that are present in either A or B]
∴ A ∩ B = {10, 15, 25} [elements that are present in both A and B]
Question 3.
If X = {x : x = 2n, x ≤ 20 and } and Y = {x : x = 4n, x ≤ 20 and }
Find (i) X ∪ Y (ii) X ∩ Y
Answer:
X = {x : x = 2n, x ≤ 20 and n ∈}
X is twice of a natural number(n ∈ N) and less than or equal to 20.
∴ X = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
Y = {x : x = 4n, x ≤ 20 and n ∈ }
Y is 4 times a positive whole number(n ∈ W) and less than equal to 20.
∴ Y = {0, 4, 8, 12, 16, 20}
∴ X ∪ Y = {0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20} [elements that are present in either X or Y]
∴ X ∩ Y = {4, 8, 12, 16, 20} [elements that are present in both X and Y]
Question 4.
U = {1, 2, 3, 6, 7, 12, 17, 21, 35, 52, 56},
P = {numbers divisible by 7}, Q = {prime numbers}
List the elements of the set {x : x ∈ P ∩ Q}
Answer:
U = {1, 2, 3, 6, 7, 12, 17, 21, 35, 52, 56}
Numbers divisible by 7 present in U are: 7, 21, 35, 56.
Prime numbers present in U are : 2, 3, 7, 17.
∴ P = {7, 21, 35, 56}
∴ Q = {2, 3, 7, 17}
∴ P ∩ Q = {7} [elements that are present in both P and Q]
∴ The set {x : x ∈ P ∩ Q} = {7}
Question 5.
State which of the following sets are disjoint
i. A = {2, 4, 6, 8}; B = {x : x is an even number < 10, }
ii. X = {1, 3, 5, 7, 9}, Y = {0, 2, 4, 6, 8, 10}
iii. P = {x : x is a prime < 15}
Q = {x : x is a multiple of 2 and x < 16}
iv. R = {a, b, c, d, e}, S = {d, e, a, b, c}
Answer:
i. A = {2, 4, 6, 8}
B = {x : x is an even number < 10, x ∈}
B contains even natural numbers that are less than 10.
∴ B = {2, 4, 6, 8}
∴ A ∩ B = {2, 4, 6, 8} [elements that are present in both A and B]
∵ A ∩ B ≠ Φ
∴ A and B are not disjoint sets.
ii. X = {1, 3, 5, 7, 9}
Y = {0, 2, 4, 6, 8, 10}
∴ X ∩ Y = Φ [elements that are present in both X and Y]
∴ X and Y are disjoint sets.
iii. P = {x : x is a prime < 15}
P contains the prime numbers less than 15.
∴ P = {2, 3, 5, 7, 11, 13}
Q = {x : x is a multiple of 2 and x < 16}
Q contains the numbers that are less than 16 and multiple of 2.
∴ Q = {…-2, 0, 2, 4, 6, 8, 10, 12, 14}
∴ P ∩ Q = {2} [elements that are present in both P and Q]
∵ P ∩ Q ≠ Φ
∴ P and Q are not disjoint sets.
iv. R = {a, b, c, d, e}
S = {d, e, a, b, c}
R ∩ S = {d, e} [elements that are present in both R and S]
∵ R ∩ S ≠ Φ
∴ R and S are not disjoint sets.
Question 6.
i. If U = {x : 0 ≤ x ≤ 10, } and A = {x : x is a multiple of 3}, find A’
ii. If U is the set of natural numbers and A’ is the set of all composite numbers, then what is A?
Answer:
i. U = {x : 0 ≤ x ≤ 10, x ∈}
U contains all the whole numbers that are greater than or equal to zero and less than or equal to 10.
∴ U = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {x : x is a multiple of 3}
Multiples of 3 present in U are : 3, 6, 9 [U→ Universal set]
∴ A = {3, 6, 9}
∴ A’ = {0, 1, 2, 4, 5, 7, 8, 10} [element that are present in U but not in A]
ii. U is the set of natural numbers.
A’ is the set of all composite numbers.
⇒ (A’)’ = A
∴ A contains all the elements that are present in U but not in A’.
∴ A contains all the natural numbers that are not composite numbers.
∴ A is the set of all prime numbers and 1.
Question 7.
If U = {a, b, c, d, e, f, g, h}, A = {a, b, c, d} and B = {b, d, f, g},
Find
i. A ∪ B
ii. (A ∪ B)’
iii. A ∩ B
iv. (A ∩ B)’
Answer:
U = {a, b, c, d, e, f, g, h}
A = {a, b, c, d}
B = {b, d, f, g}
i. A ∪ B = {a, b, c, d, f, g} [elements that are present in either A or B]
ii. U = {a, b, c, d, e, f, g, h}
A ∪ B = {a, b, c, d, f, g}
∴ (A ∪ B)’ = {e, h} [elements that are present in U but not in A ∪ B]
iii. A ∩ B = {b, d} [elements that are present in both A and B]
iv. U = {a, b, c, d, e, f, g, h}
A ∩ B = {b, d}
∴ (A ∩ B)’ = {a, c, e, f, g, h} [elements that are present in U but not in A ∩ B]
Question 8.
If U = , A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 9, 10},
Find
i. A’
ii. B’
iii. A’ ∪ B’
iv. A’ ∩ B’
Answer:
U = {x : 1≤x≤10, x ∈ N}
U contains all the natural numbers that are greater than or equal to 1 and less than or equal to 10}
∴ U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
i. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {1, 3, 5, 7, 9}
A’ = {2, 4, 6, 8, 10} [elements that are present in U but not in A]
ii. U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
B = {2, 3, 5, 9, 10}
B’ = {1, 4, 5, 6, 7, 8} [elements that are present in U but not in B]
iii. A’ = {2, 4, 6, 8, 10}
B’ = {1, 4, 5, 6, 7, 8}
∴ A’ ∪ B’ = {1, 2, 4, 5, 6, 7, 8, 10} [elements that are present in either A’ or B’]
iv. A’ = {2, 4, 6, 8, 10}
B’ = {1, 4, 5, 6, 7, 8}
∴ A’ ∩ B’ = {4, 6, 8} [elements that are present in both A’ and B’]
Question 9.
Given that U = {3, 7, 9, 11, 15, 17, 18}, M = {3, 7, 9, 11} and N = {7, 11, 15, 17},
Find
i. M – N
ii. N – M
iii. N’ – M
iv. M’ – N
v. M ∩ (M – N)
vi. N ∪ (N – M)
vii. n(M – N)
Answer:
U = {3, 7, 9, 11, 15, 17, 18}
M = {3, 7, 9, 11}
N = {7, 11, 15, 17}
i. M – N = {3, 9} [elements that are present in M but not in N]
ii. N – M = {15, 17} [elements that are present in N but not in M]
iii. U = {3, 7, 9, 11, 15, 17, 18}
M = {3, 7, 9, 11}
N = {7, 11, 15, 17}
∴ N’ = {3, 9, 18}
∴ N’ – M = {18} [elements that are present in N’ but not in M]
iv. U = {3, 7, 9, 11, 15, 17, 18}
M = {3, 7, 9, 11}
N = {7, 11, 15, 17}
∴ M’ = {15, 17, 18} [elements that are present in U but not in M]
∴ M’ – N = {18} [elements that are present in M’ but not in N]
v. M = {3, 7, 9, 11}
M – N = {3, 9}
∴ M ∩ (M – N) = {3, 9} [elements that are present in both in M and (M – N)]
vi. N = {7, 11, 15, 17}
N – M = {15, 17}
∴ N ∪ (N – M) = {7, 11, 15, 17} [elements that are present in either N or (N – M)]
vii. M – N = {3, 9}
∴ n(M – N) = 2 [no. of elements in (M – N)]
Question 10.
If A = {3, 6, 9, 12, 15, 18}, B = {4, 8, 12, 16, 20}, C = {2, 4, 6, 8, 10, 12} and D = {5, 10, 15, 20, 25}, find
i. A – B
ii. B – C
iii. C – D
iv. D – A
v. n(A – C)
Answer:
i. A = {3, 6, 9, 12, 15, 18}
B = {4, 8, 12, 16, 20}
∴ A – B = {3, 6, 9, 15, 18} [elements that are present in A but not in B]
ii. B = {4, 8, 12, 16, 20}
C = {2, 4, 6, 8, 10, 12}
∴ B – C = {16, 20} [elements that are present in B but not in C]
iii. C = {2, 4, 6, 8, 10, 12}
D = {5, 10, 15, 20, 25}
∴ C – D = {2, 4, 6, 8, 12} [elements that are present in C but not in D]
iv. D = {5, 10, 15, 20, 25}
A = {3, 6, 9, 12, 15, 18}
∴ D – A = {5, 10, 20, 25} [elements that are present in D but not in A]
v. A = {3, 6, 9, 12, 15, 18}
C = {2, 4, 6, 8, 10, 12}
∴ A – C = {3, 9, 15, 18} [elements that are present in A but not in C]
∴ n(A – C) = 4 [no. of elements in (A – C)]
Question 11.
Let U = {x : x is a positive integer less than 50}, A = {x : x is divisible by 4} and B = {x : x leaves a remainder 2 when divided by 14}.
i. List the elements of U, A and B
ii. Find A ∪ B, A ∩ B, n(A ∪ B), n(A ∩ B)
Answer:
i. U = {x : x is a positive integer less than 50}
U contains all the positive integers that are less than 50.
∴ U = {1, 2, 3, 4 ……… 49}
A = {x : x is divisible by 4}
A contains all the members of U that are divisible by 4. [∵U is universal set]
Numbers present in U that are divisible by 4 are : 4, 8, 12 …… 48.
∴ A = {4, 8, 12 …… 48}
B = {x : x leaves a remainder 2 when divided by 14}
B contains all the elements of U that leaves a remainder 2 when divided by 14.
Such numbers present in U are : 16, 30, 44
∴ B = {16, 30, 44}
ii. A = {4, 8, 12 …… 48}
B = {16, 30, 44}
∴ A ∪ B = {4, 8, 12, 16, 20, 24, 28, 30, 32, 36, 40, 44, 48} [elements that are present in either A or B]
∴ A ∩ B = {16, 44} [elements that are present in both A and B]
∴ n(A ∪ B) = 13 [no. of elements in A ∪ B]
∴ n(A ∩ B) = 2 [no. of elements in A ∩ B]
Question 12.
Find the symmetric difference between the following sets.
i. X = {a, d, f, g, h}, Y = {b, e, g, h, k}
ii. P = {x : 3 < x < 9, }, Q = {x : x < 5, }
iii. A = {−3, −2, 0, 2, 3, 5}, B = {−4, −3, −1, 0, 2, 3}
Answer:
i. X = {a, d, f, g, h}
Y = {b, e, g, h, k}
∴ XΔY = {a, b, d, e, f, k} [elements that are present either only in X or only in Y]
ii. P = {x : 3 < x < 9, x ∈}
P contains all the natural number between 3 and 9.
∴ P = {4, 5, 6, 7, 8}
Q = {x : x < 5, x ∈}
The possible values of Q are : 0, 1, 2, 3, 4
∴ Q = {0, 1, 2, 3, 4}
∴ PΔQ = {0, 1, 2, 3, 5, 6, 7, 8} [elements that are present either only in P or only in Q]
iii. A = {−3, −2, 0, 2, 3, 5}
B = {−4, −3, −1, 0, 2, 3}
∴ XΔY = {-4, -2, -1, 5} [elements that are present either only in A or only in B]
Question 13.
Use the Venn diagram to answer the following questions
i. List the elements of
U, E, F, E ∪ F and E ∩ F
ii. Find n(U), n(E ∪ F) and n(E ∩ F)
Answer:
i. From the Venn diagram we get,
U = {1, 2, 3, 4, 7, 9, 10, 11}
[U is the universal set, so it contains all the values of in the diagram]
E = {1, 2, 4, 7}
[The set of values in red Circle]
F = {4, 7, 9, 11}
[The set of values in red Circle]∴ E ∪ F = {1, 2, 4, 7, 9, 11} [elements present either in E or in F]
∴ E ∩ F = {4, 7} [elements that are present in both E and F]
ii. U = {1, 2, 3, 4, 7, 9, 10, 11}
∴ n(U) = 8 [no. of elements in U]
E ∪ F = {1, 2, 4, 7, 9, 11}
∴ n(E ∪ F) = 6 [no. of elements in E ∪ F]
E ∩ F = {4, 7}
∴ n(E ∩ F) = 2 [no. of elements in E ∩ F]
Question 14.
Use the Venn diagram to answer the following questions
i. List U, G and H
ii. Find G’, H’, G’ ∩ H’, n(G ∪ H)’ and n(G ∩ H)’
Answer:
i. From the Venn diagram we get,
U = {1, 2, 3, 4, 5, 6, 8, 9, 10}
G = {1, 2, 4, 8}
H = {2, 6, 8, 10}
ii. U = {1, 2, 3, 4, 5, 6, 8, 9, 10}
G = {1, 2, 4, 8}
H = {2, 6, 8, 10}
∴ G’ = {3, 5, 6, 9, 10} [elements that are present in U but not in G]
∴ H’ = {1, 3, 4, 5, 9} [elements that are present in U but not in H]
∴ G’ ∩ H’ = {3, 5, 9} [elements that are present in both G’ and H’]
∴ G ∪ H = {1, 2, 4, 6, 8, 10} [elements that are present either in G or in H]
∴ (G ∪ H)’ = {3, 5, 9} [elements present in U but not in (G ∪ H)]
∴ n(G ∪ H)’ = 3 [no. of elements in (G ∪ H)’]
∴ G ∩ H = {2, 8} [elements present in both G and H]
∴ (G ∩ H)’ = {1, 3, 4, 5, 6, 9, 10} [elements present in U but not in (G ∩ H)]
∴ n(G ∩ H)’ = 7 [no. of elements in (G ∩ H)’]
Exercise 1.3
Question 1.Place the elements of the following sets in the proper location on the given Venn diagram.
U = {5, 6, 7, 8, 9, 10, 11, 12, 13}
M = {5, 8, 10, 11},
N = {5, 6, 7, 9, 10}
Answer:Given, U = {5, 6, 7, 8, 9, 10, 11, 12, 13}
M = {5, 8, 10, 11}
N = {5, 6, 7, 9, 10}
Representing the elements in their respective sets, we get:
Question 2.If A and B are two sets such that A has 50 elements, B has 65 elements and A ∪ B has 100 elements, how many elements does A ∩ B have?
Answer:Given, n(A) = 50, n(B) = 65 and n(A ∪ B) = 100.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
100 = 50 + 65 – n(A ∩ B)
n(A ∩ B) = 50 + 65 – 100 = 115 – 100 = 15
So, A ∩ B has 15 elements.
Question 3.If A and B are two sets containing 13 and 16 elements respectively, then find the minimum and maximum number of elements in A ∪ B?
Answer:Let A and B be two sets with n(A) = 13, n(B) = 16.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Minimum number of elements in A ∪ B is possible if all the elements of A lie in B, i.e. A⊆ B (B cannot be a subset of A, obviously, as n(B) > n(A)).
In that case, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 13 + 16 – 13 = 16
Maximum number of elements in A ∪ B is possible if n(A ∩ B) = 0,
i.e. if A and B are disjoint sets.
In that case, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 13 + 16 – 0 = 29
So, minimum and maximum number of elements possible in A ∪ B are 16 and 29 respectively.
Question 4.If n(A ∩ B) = 5, n(A ∪ B) = 35, n(A) = 13, find n(B).
Answer:Given, n(A ∩ B) = 5, n(A ∪ B) = 35, n(A) = 13
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
35 = 13 + n(B) – 5
n(B) = 35 + 5 – 13 = 40 – 13 = 17
So, n(B) = 17
Question 5.If n(A) = 26, n(B) = 10, n (A ∪ B) = 30, n(A’) = 17, find n(A ∩ B) and n(U).
Answer:Given, n(A) = 26, n(B) = 10, n (A ∪ B) = 30 and n(A’) = 17.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
30 = 26 + 10 – n(A ∩ B)
n(A ∩ B) = 36 – 30 = 6
Using the result n(A) + n(A’) = n(U)
26 + 17 = n(U)
n(U) = 43
So, n(A ∩ B) = 6 and n(U) = 43.
Question 6.If n(U) = 38, n(A) = 16, n(A ∩ B) = 12, n(B’) = 20, find n(A ∪ B).
Answer:Using the result n(B) + n(B’) = n(U)
n(B) + 20 = 38
n(B) = 38 – 20 = 18
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 16 + 18 – 12
n(A ∪ B) = 34 – 12 = 22
Hence, n(A ∪ B) = 22.
Question 7.Let A and B be two finite sets such that n(A – B) = 30, n(A ∪ B) = 180. Find n(B).
Answer:
Given, n(A–B) = 30, n(A ∪ B) = 180.
It can be concluded that,
(A ∪ B) = (A–B) + B
∴ n(A ∪ B) = n(A–B) + n(B)
180 = 30 + n(B)
n(B) = 180 – 30 = 150.
Hence, n(B) = 150.
Question 8.The population of a town is 10000. Out of these 5400 persons read newspaper A and 4700 read newspaper B. 1500 persons read both the newspaper. Find the number of persons who do not read either of the two papers.
Answer:Let A denote the set of people who read newspaper A and B denote the set of people who read newspaper B.
Then, n(A) = 5400, n(B) = 4700, n(A ∩ B) = 1500, n(U) = 10000.
People who do not read either of the two papers are represented by (A ∪ B)’.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 5400 + 4700 – 1500
n(A ∪ B) = 10100 – 1500 = 8600
n(A ∪ B)’ = n(U) – n(A ∪ B) = 10000 – 8600 = 1400
∴ Number of persons who do not read either of the two papers are 1400.
Question 9.In a school, all the students play either Foot ball or Volley ball or both. 300 students play Foot ball, 270 students play Volley ball and 120 students play both games. Find
i. the number of students who play Foot ball only
ii. the number of students who play Volley ball only
iii. the total number of students in the school
Answer:Let A denote the set of students who play football and B denote the set of students who play volleyball.
Then, n(A) = 300, n(B) = 270, n(A ∩ B) = 120.
i. Number of students who play volleyball only are represented by (A – B).
n(A–B) = n(A) – n(A ∩ B)
n(A–B) = 300 – 120 = 180
ii. Number of students who play volleyball only are represented by (B – A).
n(B–A) = n(B) – n(A ∩ B)
n(A–B) = 270 – 120 = 150
iii. Number of students in the school are represented by (A ∪ B)
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 300 + 270 – 120
n(A ∪ B) = 570 – 120 = 450
Question 10.In an examination 150 students secured first class in English or Mathematics. 115 students secured first class in Mathematics. How many students secured first class in English only?
Answer:Number of students with first class in both maths & English are represented are given = 150.
And, the students who scored first class in maths = 115.
It means out of 150 students, 115 students scored 1st class in Maths but not may or may not in English.
So, the students who got only first class in English will be 150 – 115 = 35 students.
Question 11.In a group of 30 persons, 10 take tea but not coffee 18 take tea. Find how many take coffee but not tea, if each person takes atleast one of the drinks.
Answer:Given in a group of 30 persons, each one
takes at least one of the drinks.
So n(T∪C) = 30
n(T) = 18
Also given that number of persons drink only tea is 10.
There fore, number of persons drink both tea and coffee
n(T∩C) = n(T)–10 = 18–10 = 8
From the Venn diagram,
30 = 10 + 8 + x
x = 30 – 18
= 12
Number of persons drink only coffee not tea is 12.
Question 12.In a village there are 60 families. Out of these 28 families speak only Tamil and 20 families speak only Urdu. How many families speak both Tamil and Urdu.
Answer:Given
total no families in the village n(A∪B) = 60
No. of families speak only Tamil n(A) = 28
No. of families speak only Urdu n(B) = 20
To find families speak both Tamil and Urdu n(A∩B) let us use the result
n(A∪B) = n(A) + n(B) – n(A∩B)
n(A∩B) = n(A∪B)– [n(A) + n(B)]
= 60 – [28 + 20]
= 60 – 48
= 12
Number of families speak both Tamil and Urdu are 12
Question 13.In a school 150 students passed X standard Examination. 95 students applied for Group I and 82 students applied for Group II in the Higher Secondary course. If 20 students applied neither of the two, how many students applied for both groups?
Answer:Given
total no student passed in X standard exam n(U) = 150
No of Students applied for Group I only n(A) = 95
No of Student applied for Group II only n(B) = 82
No of student applied for neither of the exam n(A∪B)' = 20
So No of student applied for exam n(A∪B) = n(U) – n(A∪B)'
= 150 – 20
= 130
From the above Venn diagram
(95–x) + x + (82–x) = 130
No of student applied for both the group x = (95 + 82) – 130
= 177 – 130
= 47
Number of students applied for both the group = 47.
Question 14.Pradeep is a Section Chief for an electric utility company. The employees in his section cut down tall trees or climb polies recently reported the following information to the management of the utility.
Out of 100 employees in my section, 55 can cut tall trees,50 can climb poles, 11 can do both, 6 can’t do any of the two. Is this information correct?
Answer:Given
total no of employees n(U) = 100
No of Employees can climb and cut n(T∩P) = 11
No of employees can’t do any of the two n(T∪P)' = 6
No of employees can cut the tree n(T) = 55
No of employees climb the pole n(P) = 50
From the Venn diagram, we know that
Total no of employees n(U) = n(T∪P) + n(T∪P)'
= [44 + 11 + 39] + 6
= 100
So the given information is correct.
Question 15.A and B are two sets such that n(A – B) = 32 + x, n (B – A) = 5x and n(A B) = x Illustrate the information by means of a Venn diagram. Given that n(A) = n(B). Calculate (i) the value of x (ii) n(A B).
Answer:Given that
n(A–B) = 32 + x
n(B–A) = 5x
n(A∩B) = x
and n(A) = n(B)
using these results
n(A) = n(A–B) + n(A∩B)
n(B) = n(B–A) + n(A∩B)
n(A∪B) = n(A–B) + n(A∩B) + n(B–A)
But given that n(A) = n(B)
So, n(A–B) + n(A∩B) = n(B–A) + n(A∩B)
n(A–B) = n(B–A)
32 + x = 5x
5x–x = 32
4x = 32
(i) x = 32/4 = 8
(ii) n(A∪B) = n(A–B) + n(A∩B) + n(B–A)
= 32 + x + x + 5x
= 32 + 7x
= 32 + 7(8)
= 88
Question 16.The following table shows the percentage of the students of a school who participated in Elocution and Drawing competitions.
Draw a Venn diagram to represent this information and use it to find the percentage of the students who
i. participated in Elocution only
ii. participated in Drawing only
iii. did not participate in any one of the competitions.
Answer:
(i) Participated in Elocution only = 35
(ii) Participated in Drawing only = 25
(iii)Do not participated in any on of the competitions = 100–(35 + 20 + 25)
= 100–80
= 20
Question 17.A village has total population of 2500 people. Out of which 1300 people use brand A soap and 1050 people use brand B soap and 250 people use both brands. Find the percentage of population who use neither of these soaps.
Answer:Given
total population n(U) = 2500
no of people use brand A soap n(A) = 1300
no of people using brand B soap n(B) = 1050
no people using both brand n(A∩B) = 250
We know that
n(A∪B) = n(A) + n(B) – n(A∩B)
= 1300 + 1050–250
= 2100
n(A∪B)' = n(U) – n(A∪B)
= 2500–2100
= 400
Percentage of population who use neither of these soaps
= (400/2500)x100
= 0.16 x 100
= 16%
Place the elements of the following sets in the proper location on the given Venn diagram.
U = {5, 6, 7, 8, 9, 10, 11, 12, 13}
M = {5, 8, 10, 11},
N = {5, 6, 7, 9, 10}
Answer:
Given, U = {5, 6, 7, 8, 9, 10, 11, 12, 13}
M = {5, 8, 10, 11}
N = {5, 6, 7, 9, 10}
Representing the elements in their respective sets, we get:
Question 2.
If A and B are two sets such that A has 50 elements, B has 65 elements and A ∪ B has 100 elements, how many elements does A ∩ B have?
Answer:
Given, n(A) = 50, n(B) = 65 and n(A ∪ B) = 100.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
100 = 50 + 65 – n(A ∩ B)
n(A ∩ B) = 50 + 65 – 100 = 115 – 100 = 15
So, A ∩ B has 15 elements.
Question 3.
If A and B are two sets containing 13 and 16 elements respectively, then find the minimum and maximum number of elements in A ∪ B?
Answer:
Let A and B be two sets with n(A) = 13, n(B) = 16.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
Minimum number of elements in A ∪ B is possible if all the elements of A lie in B, i.e. A⊆ B (B cannot be a subset of A, obviously, as n(B) > n(A)).
In that case, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 13 + 16 – 13 = 16
Maximum number of elements in A ∪ B is possible if n(A ∩ B) = 0,
i.e. if A and B are disjoint sets.
In that case, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 13 + 16 – 0 = 29
So, minimum and maximum number of elements possible in A ∪ B are 16 and 29 respectively.
Question 4.
If n(A ∩ B) = 5, n(A ∪ B) = 35, n(A) = 13, find n(B).
Answer:
Given, n(A ∩ B) = 5, n(A ∪ B) = 35, n(A) = 13
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
35 = 13 + n(B) – 5
n(B) = 35 + 5 – 13 = 40 – 13 = 17
So, n(B) = 17
Question 5.
If n(A) = 26, n(B) = 10, n (A ∪ B) = 30, n(A’) = 17, find n(A ∩ B) and n(U).
Answer:
Given, n(A) = 26, n(B) = 10, n (A ∪ B) = 30 and n(A’) = 17.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
30 = 26 + 10 – n(A ∩ B)
n(A ∩ B) = 36 – 30 = 6
Using the result n(A) + n(A’) = n(U)
26 + 17 = n(U)
n(U) = 43
So, n(A ∩ B) = 6 and n(U) = 43.
Question 6.
If n(U) = 38, n(A) = 16, n(A ∩ B) = 12, n(B’) = 20, find n(A ∪ B).
Answer:
Using the result n(B) + n(B’) = n(U)
n(B) + 20 = 38
n(B) = 38 – 20 = 18
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 16 + 18 – 12
n(A ∪ B) = 34 – 12 = 22
Hence, n(A ∪ B) = 22.
Question 7.
Let A and B be two finite sets such that n(A – B) = 30, n(A ∪ B) = 180. Find n(B).
Answer:
Given, n(A–B) = 30, n(A ∪ B) = 180.
It can be concluded that,
(A ∪ B) = (A–B) + B
∴ n(A ∪ B) = n(A–B) + n(B)
180 = 30 + n(B)
n(B) = 180 – 30 = 150.
Hence, n(B) = 150.
Question 8.
The population of a town is 10000. Out of these 5400 persons read newspaper A and 4700 read newspaper B. 1500 persons read both the newspaper. Find the number of persons who do not read either of the two papers.
Answer:
Let A denote the set of people who read newspaper A and B denote the set of people who read newspaper B.
Then, n(A) = 5400, n(B) = 4700, n(A ∩ B) = 1500, n(U) = 10000.
People who do not read either of the two papers are represented by (A ∪ B)’.
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 5400 + 4700 – 1500
n(A ∪ B) = 10100 – 1500 = 8600
n(A ∪ B)’ = n(U) – n(A ∪ B) = 10000 – 8600 = 1400
∴ Number of persons who do not read either of the two papers are 1400.
Question 9.
In a school, all the students play either Foot ball or Volley ball or both. 300 students play Foot ball, 270 students play Volley ball and 120 students play both games. Find
i. the number of students who play Foot ball only
ii. the number of students who play Volley ball only
iii. the total number of students in the school
Answer:
Let A denote the set of students who play football and B denote the set of students who play volleyball.
Then, n(A) = 300, n(B) = 270, n(A ∩ B) = 120.
i. Number of students who play volleyball only are represented by (A – B).
n(A–B) = n(A) – n(A ∩ B)
n(A–B) = 300 – 120 = 180
ii. Number of students who play volleyball only are represented by (B – A).
n(B–A) = n(B) – n(A ∩ B)
n(A–B) = 270 – 120 = 150
iii. Number of students in the school are represented by (A ∪ B)
We have, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
n(A ∪ B) = 300 + 270 – 120
n(A ∪ B) = 570 – 120 = 450
Question 10.
In an examination 150 students secured first class in English or Mathematics. 115 students secured first class in Mathematics. How many students secured first class in English only?
Answer:
Number of students with first class in both maths & English are represented are given = 150.
And, the students who scored first class in maths = 115.
It means out of 150 students, 115 students scored 1st class in Maths but not may or may not in English.
So, the students who got only first class in English will be 150 – 115 = 35 students.
Question 11.
In a group of 30 persons, 10 take tea but not coffee 18 take tea. Find how many take coffee but not tea, if each person takes atleast one of the drinks.
Answer:
Given in a group of 30 persons, each one
takes at least one of the drinks.
So n(T∪C) = 30
n(T) = 18
Also given that number of persons drink only tea is 10.
There fore, number of persons drink both tea and coffee
n(T∩C) = n(T)–10 = 18–10 = 8
From the Venn diagram,
30 = 10 + 8 + x
x = 30 – 18
= 12
Number of persons drink only coffee not tea is 12.
Question 12.
In a village there are 60 families. Out of these 28 families speak only Tamil and 20 families speak only Urdu. How many families speak both Tamil and Urdu.
Answer:
Given
total no families in the village n(A∪B) = 60
No. of families speak only Tamil n(A) = 28
No. of families speak only Urdu n(B) = 20
To find families speak both Tamil and Urdu n(A∩B) let us use the result
n(A∪B) = n(A) + n(B) – n(A∩B)
n(A∩B) = n(A∪B)– [n(A) + n(B)]
= 60 – [28 + 20]
= 60 – 48
= 12
Number of families speak both Tamil and Urdu are 12
Question 13.
In a school 150 students passed X standard Examination. 95 students applied for Group I and 82 students applied for Group II in the Higher Secondary course. If 20 students applied neither of the two, how many students applied for both groups?
Answer:
Given
total no student passed in X standard exam n(U) = 150
No of Students applied for Group I only n(A) = 95
No of Student applied for Group II only n(B) = 82
No of student applied for neither of the exam n(A∪B)' = 20
So No of student applied for exam n(A∪B) = n(U) – n(A∪B)'
= 150 – 20
= 130
From the above Venn diagram
(95–x) + x + (82–x) = 130
No of student applied for both the group x = (95 + 82) – 130
= 177 – 130
= 47
Number of students applied for both the group = 47.
Question 14.
Pradeep is a Section Chief for an electric utility company. The employees in his section cut down tall trees or climb polies recently reported the following information to the management of the utility.
Out of 100 employees in my section, 55 can cut tall trees,50 can climb poles, 11 can do both, 6 can’t do any of the two. Is this information correct?
Answer:
Given
total no of employees n(U) = 100
No of Employees can climb and cut n(T∩P) = 11
No of employees can’t do any of the two n(T∪P)' = 6
No of employees can cut the tree n(T) = 55
No of employees climb the pole n(P) = 50
From the Venn diagram, we know that
Total no of employees n(U) = n(T∪P) + n(T∪P)'
= [44 + 11 + 39] + 6
= 100
So the given information is correct.
Question 15.
A and B are two sets such that n(A – B) = 32 + x, n (B – A) = 5x and n(A B) = x Illustrate the information by means of a Venn diagram. Given that n(A) = n(B). Calculate (i) the value of x (ii) n(A B).
Answer:
Given that
n(A–B) = 32 + x
n(B–A) = 5x
n(A∩B) = x
and n(A) = n(B)
using these results
n(A) = n(A–B) + n(A∩B)
n(B) = n(B–A) + n(A∩B)
n(A∪B) = n(A–B) + n(A∩B) + n(B–A)
But given that n(A) = n(B)
So, n(A–B) + n(A∩B) = n(B–A) + n(A∩B)
n(A–B) = n(B–A)
32 + x = 5x
5x–x = 32
4x = 32
(i) x = 32/4 = 8
(ii) n(A∪B) = n(A–B) + n(A∩B) + n(B–A)
= 32 + x + x + 5x
= 32 + 7x
= 32 + 7(8)
= 88
Question 16.
The following table shows the percentage of the students of a school who participated in Elocution and Drawing competitions.
Draw a Venn diagram to represent this information and use it to find the percentage of the students who
i. participated in Elocution only
ii. participated in Drawing only
iii. did not participate in any one of the competitions.
Answer:
(i) Participated in Elocution only = 35
(ii) Participated in Drawing only = 25
(iii)Do not participated in any on of the competitions = 100–(35 + 20 + 25)
= 100–80
= 20
Question 17.
A village has total population of 2500 people. Out of which 1300 people use brand A soap and 1050 people use brand B soap and 250 people use both brands. Find the percentage of population who use neither of these soaps.
Answer:
Given
total population n(U) = 2500
no of people use brand A soap n(A) = 1300
no of people using brand B soap n(B) = 1050
no people using both brand n(A∩B) = 250
We know that
n(A∪B) = n(A) + n(B) – n(A∩B)
= 1300 + 1050–250
= 2100
n(A∪B)' = n(U) – n(A∪B)
= 2500–2100
= 400
Percentage of population who use neither of these soaps
= (400/2500)x100
= 0.16 x 100
= 16%
Exercise 1.4
Question 1.If A = {5, {5, 6}, 7}, which of the following is correct?
A. {5, 6} ∈ A
B. {5} ∈ A
C. {7} ∈ A
D. {6} ∈ A
Answer:In the given set we have been given that 2 elements and 1 set
Elements are 5, 7 and {5,6} is the set
So, {5,6} ∈ A.
All the other options are not correct
Because {5} is a set not an element, so it does not belongs to A , but 5 ∈ A
In the same way {6}, {7} are sets not elements.
Question 2.If X = {a, {b, c}, d}, which of the following is a subset of X?
A. {a, b}
B. {b, c}
C. {c, d}
D. {a, d}
Answer:Given set X = {a, {b, c}, d}
Here a, d is the elements {b, c} is the set present in the X.
Subset:
A is subset to B if all the elements in the A presents in the B
E.g.,
A = {9,8,4,6,7} B = {1,2,3,4,5,6,7,8,9,}
Since all elements in A ⊆ B
{a, b} is not subset because a is present in the set X but b is from other set and it is not an element in the set X
In the same way {b, c}, {c, d} are not subset of X.
But a, d both are elements of set X, so {a, d} ⊆ X
Question 3.Which of the following statements are true?
i. For any set A, A is a proper subset of A
ii. For any set A, Φ is a subset of A
iii. For any set A, A is a subset of A
A. (i) and (ii)
B. (ii) and (iii)
C. (i) and (iii)
D. (i), (ii) and (iii)
Answer:i. For any set A, A is a proper subset of A
No set is proper subset of itself
(or)
Let us say A = {1,2,3,4}
For a proper subset n(A) ≠ n(A)
But here n(A) = n(A)
So, it is not proper subset to itself.
ii. For any set A, Φ is a subset of A
it is a property of null set, that it is subset to any other set except to itself (Φ)
So, it is true.
iii. For any set A, A is a subset of A
It is also property of subset that it is subset to itself and null set (Φ)
So, it is also true
Since (ii) & (iii) is correct statement. option (B) is the correct answer.
Question 4.If a finite set A has m elements, then the number of non-empty proper subsets of A is
A. 2m
B. 2m – 1
C. 2m – 1
D. 2(2m – 1 −1)
Answer:Given Set A with ‘m’ number of elements
So, the number of power subset for given set is given by 2m – 1
∴ option (B) is true.
Question 5.The number of subsets of the set {10, 11, 12} is
A. 3
B. 8
C. 6
D. 7
Answer:Given set A {10, 11, 12}
There are 3 elements in the set. ⇒ n(A) = 3
The number of subset for a set is given by n(p(A)) = 2m
Where m is the no. of elements.
So, no. of subsets = n(p(A)) = 23 = 8.
∴ option(B) is the correct answer.
Question 6.Which of the following is correct?
A. {x:x2 = -1, x∈Z} = Φ
B. Φ = 0
C. Φ = {0}
D. Φ = {Φ}
Answer:B. Φ = 0 is false, because Φ is a null set, it is not equal to any element.
C. Φ = {0} is false, because Φ should not contain any element in the set, but here it contains an element.
D. Φ = {Φ} is also false, because null set will be considered as an element in the set.
A. {x:x2 = -1, x∈Z} = Φ
For any value of x, x2 will not be going to -1. So, the set will be remained with no elements. So, this set will be considered as null set
∴ option(A) is true.
Question 7.Which of the following is incorrect?
A. Every subset of a finite set in finite
B. P = {x : x – 8 = − 8} is a singleton set
C. Every set has a proper subset
D. Every non – empty set has at least two subsets
Answer:A. For a set to be subset to any set it should have less than or same number of elements. If a set contains finite no. of elements its subset also gets number finite in the set.
So, this is true.
B. The set is defined by P = {x : x – 8 = − 8}
X -8 = -8
X = 0
∴ P = {0}
This is a singleton set.
This Is also true.
D. It is a rule that non-empty set has 2 subsets they are
They are subset to itself and null set is the subset to that non-empty set.
So, it is also true.
C. It is a property that null set is proper set to every other set except to itself. This means Some sets only has proper subset.
So, it is false.
∴ option(C) is correct answer.
Question 8.Which of the following is a correct statement?
A. Φ ⊆ {a, b}
B. Φ ∈ {a, b}
C. {a} ∈ {a, b}
D. a ⊆ {a, b}
Answer:A. Φ ⊆ {a, b}
We know that null set is subset to every other non-empty set. So, it is true.
B. Φ ∈ {a, b}
Null set consists of no elements, but here it contains elements. So, it is false
C. {a} ∈ {a, b}
In the above {a} ⊂ {a, b} not {a} ∈ {a, b} since they are given as set but not element.
D. a ⊆ {a, b}
In the above {a} ∈ {a, b} not {a} ⊆ {a, b} since they are given as element but not as set.
So, option(A) is the correct answer.
Question 9.Which one of the following is a finite set?
A. {x: x ∈ Z, x < 5}
B. {x: x ∈ W, x ≥ 5}
C. {x: x ∈ N, x > 10}
D. {x: x is an even prime number}
Answer:A. {x: x ∈ Z, x < 5}
The above expression says that x defines the set. While the x belongs to integers which are less than 5.
We know that Z contains -∞ to ∞
But her it will contain elements from -∞ to 5
So, this is an infinite set.
B. {x: x ∈ W, x ≥ 5}
The above expression says that x defines the set. While the x belongs to whole numbers(W) which is ≥ 5.
We know that W contains 0 to ∞
But her it will contain elements from 5 to ∞
So, this is an infinite set.
C. {x: x ∈ N, x > 10}
The above expression says that x defines the set. While the x belongs to natural numbers(N) which greater than 10.
We know that N contains 1 to ∞
But her it will contain elements from 10 to ∞
So, this is an infinite set.
D. {x: x is an even prime number}
The above expression says that x defines the set. While the x belongs to even prime number (prime numbers which are even). Fortunately, we have only one even prime number that is ‘2’
So, this is finite series.
Question 10.Given A = {5, 6, 7, 8}. Which one of the following is incorrect?
A. Φ ⊆ A
B. A ⊆ A
C. {7, 8, 9} ⊆ A
D. {5} ⊂ A
Answer:Given set A = {5, 6, 7, 8}
A. Φ ⊆ A
We know that null set will be subset to every other non-empty set.
So, it’s true
B. A ⊆ A
We know that a set is subset to itself. So, it’s true.
C. {7, 8, 9} ⊆ A
The above set will not be a subset to A because set A didn’t contain element ‘9’. But it contains. So, it is false.
D. {5} ⊂ A
The above set is power set to A because it has same element as in set A and the no. of elements in Set is less than set A. So, it is true.
Question 11.If A = {3, 4, 5, 6} and B = {1, 2, 5, 6}, then A ∪ B =
A. {1, 2, 3, 4, 5, 6}
B. {1, 2, 3, 4, 6}
C. {1, 2, 5, 6}
D. {3, 4, 5, 6}
Answer:A = {3, 4, 5, 6} & B = {1, 2, 5, 6}
Here we are asked to find A ∪ B
Union means cub the elements of the both the given two sets.
So,
A ∪ B = {3, 4, 5, 6} ∪ {1, 2, 5, 6}
= {1, 2, 3, 4, 5, 6}
∴ option A is the correct answer.
Question 12.The number of elements of the set {x: x ∈ Z, x2 = 1}is
A. 3
B. 2
C. 1
D. 0
Answer:The set defined by {x: x ∈ Z, x2 = 1}
This means the elements of the set is given by x2 = 1 and that x is belongs to integers(Z)
x = 1 ⇒ x2 = 1
x = -1 ⇒ x2 = (-1)2 = 1
For the x = 1 & -1 the statement holds true.
∴ 1, -1 are the elements of the set
The number of elements in the set is 2.
Option B is the correct answer.
Question 13.If n(X) = m, n(Y) = n and n(X ∩ Y) = p then n(X ∪ Y) =
A. m + n + p
B. m + n – p
C. m – p
D. m – n + p
Answer:Given
n(X) = m, n(Y) = n and n(X ∩ Y) = p
n(X ∪ Y) = ?
we know that n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
in the same way, n(X ∪ Y) = n(X) + n(Y) - n(X ∩ Y)
n(X ∪ Y) = m + n – p
∴ option B is the correct answer.
Question 14.If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 5, 6, 9, 10} then A’ is
A. {2, 5, 6, 9, 10}
B. Φ
C. {1, 3, 5, 10}
D. {1, 3, 4, 7, 8}
Answer:Given,
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 5, 6, 9, 10}
We have to find A’ (A compliment)
A’(A compliment) means we have to find the eliminate the elements of sets A in U.
∴ A’ = {1, 3, 4, 7, 8}
Option D is the correct answer.
Question 15.If A B, then A – B is
A. B
B. A
C. Φ
D. B – A
Answer:Given that A B that means B contains all the elements that A contains.
e.g., A = {9, 8, 4, 6} B = {9, 8, 4, 6, 7, 0, 1}
we can see that B contains all the elements of B
So, A B
Let us consider two sets which is A B
A = {9, 8, 4, 6} B = {9, 8, 4, 6, 7, 0, 1}
A – B means we eliminate all the elements of B from A.
A – B = {9, 8, 4, 6} - {9, 8, 4, 6, 7, 0, 1}
= {} ⇒ null set.
Option D is the correct answer.
Question 16.If A is a proper subset of B, then A ∩ B =
A. A
B. B
C. Φ
D. A ∪ B
Answer:Given, A is a proper subset of B
Let us consider two set as A is a proper subset of B
A = {9, 8, 7, 6, 5, 4} B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A ∩ B means we have write all the common elements of both the sets.
A ∩ B = {9, 8, 7, 6, 5, 4} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9}
= {4, 5, 6, 7, 8, 9}
The resultant set is same as A
Option A is the correct answer.
Question 17.If A is a proper subset of B, then A ∪ B
A. A
B. Φ
C. B
D. A ∩ B
Answer:Given, A is a proper subset of B
Let us consider two set as A is a proper subset of B
A = {9, 8, 7, 6, 5, 4} B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A ∪ B means we must combinedly write all the elements of both the sets.
A ∪ B = {9, 8, 7, 6, 5, 4} ∪ {1, 2, 3, 4, 5, 6, 7, 8, 9}
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
The resultant set is same as B
Option C is the correct answer.
Question 18.The shaded region in the adjoint diagram represents
A. A – B
B. A’
C. B’
D. B – A
Answer:The venn diagram contains 2 sets, and they are overlapped which means they have common elements in it.
In the venn diagram B is completely shaded Except the overlapped part with A that means “it shows the elements of B by eliminating the common elements of A”
Since we are eliminating the elements of A in B this can be written as B-A.
Option D is the correct answer.
Question 19.If A = {a, b, c}, B = {e, f, g}, then A ∩ B =
A. Φ
B. A
C. B
D. A ∪ B
Answer:Given, A = {a, b, c}, B = {e, f, g}
We have to find A ∩ B
A ∩ B = {a, b, c} ∩ {e, f, g} = Φ
Since there is no common elements in the both the sets the resultant is null set.
Option A is the correct answer.
Question 20.The shaded region in the adjoining diagram represents
A. A – B
B. B – A
C. A Δ B
D. A’
Answer:The green shaded represents A-B and the blue shaded part represents B-A
The overall shaded region will the union of A-B & B-A
∴ (A-B) ∪ (B-A) = A Δ B
⇒ The shaded region represents A Δ B of the venn diagram.
If A = {5, {5, 6}, 7}, which of the following is correct?
A. {5, 6} ∈ A
B. {5} ∈ A
C. {7} ∈ A
D. {6} ∈ A
Answer:
In the given set we have been given that 2 elements and 1 set
Elements are 5, 7 and {5,6} is the set
So, {5,6} ∈ A.
All the other options are not correct
Because {5} is a set not an element, so it does not belongs to A , but 5 ∈ A
In the same way {6}, {7} are sets not elements.
Question 2.
If X = {a, {b, c}, d}, which of the following is a subset of X?
A. {a, b}
B. {b, c}
C. {c, d}
D. {a, d}
Answer:
Given set X = {a, {b, c}, d}
Here a, d is the elements {b, c} is the set present in the X.
Subset:
A is subset to B if all the elements in the A presents in the B
E.g.,
A = {9,8,4,6,7} B = {1,2,3,4,5,6,7,8,9,}
Since all elements in A ⊆ B
{a, b} is not subset because a is present in the set X but b is from other set and it is not an element in the set X
In the same way {b, c}, {c, d} are not subset of X.
But a, d both are elements of set X, so {a, d} ⊆ X
Question 3.
Which of the following statements are true?
i. For any set A, A is a proper subset of A
ii. For any set A, Φ is a subset of A
iii. For any set A, A is a subset of A
A. (i) and (ii)
B. (ii) and (iii)
C. (i) and (iii)
D. (i), (ii) and (iii)
Answer:
i. For any set A, A is a proper subset of A
No set is proper subset of itself
(or)
Let us say A = {1,2,3,4}
For a proper subset n(A) ≠ n(A)
But here n(A) = n(A)
So, it is not proper subset to itself.
ii. For any set A, Φ is a subset of A
it is a property of null set, that it is subset to any other set except to itself (Φ)
So, it is true.
iii. For any set A, A is a subset of A
It is also property of subset that it is subset to itself and null set (Φ)
So, it is also true
Since (ii) & (iii) is correct statement. option (B) is the correct answer.
Question 4.
If a finite set A has m elements, then the number of non-empty proper subsets of A is
A. 2m
B. 2m – 1
C. 2m – 1
D. 2(2m – 1 −1)
Answer:
Given Set A with ‘m’ number of elements
So, the number of power subset for given set is given by 2m – 1
∴ option (B) is true.
Question 5.
The number of subsets of the set {10, 11, 12} is
A. 3
B. 8
C. 6
D. 7
Answer:
Given set A {10, 11, 12}
There are 3 elements in the set. ⇒ n(A) = 3
The number of subset for a set is given by n(p(A)) = 2m
Where m is the no. of elements.
So, no. of subsets = n(p(A)) = 23 = 8.
∴ option(B) is the correct answer.
Question 6.
Which of the following is correct?
A. {x:x2 = -1, x∈Z} = Φ
B. Φ = 0
C. Φ = {0}
D. Φ = {Φ}
Answer:
B. Φ = 0 is false, because Φ is a null set, it is not equal to any element.
C. Φ = {0} is false, because Φ should not contain any element in the set, but here it contains an element.
D. Φ = {Φ} is also false, because null set will be considered as an element in the set.
A. {x:x2 = -1, x∈Z} = Φ
For any value of x, x2 will not be going to -1. So, the set will be remained with no elements. So, this set will be considered as null set
∴ option(A) is true.
Question 7.
Which of the following is incorrect?
A. Every subset of a finite set in finite
B. P = {x : x – 8 = − 8} is a singleton set
C. Every set has a proper subset
D. Every non – empty set has at least two subsets
Answer:
A. For a set to be subset to any set it should have less than or same number of elements. If a set contains finite no. of elements its subset also gets number finite in the set.
So, this is true.
B. The set is defined by P = {x : x – 8 = − 8}
X -8 = -8
X = 0
∴ P = {0}
This is a singleton set.
This Is also true.
D. It is a rule that non-empty set has 2 subsets they are
They are subset to itself and null set is the subset to that non-empty set.
So, it is also true.
C. It is a property that null set is proper set to every other set except to itself. This means Some sets only has proper subset.
So, it is false.
∴ option(C) is correct answer.
Question 8.
Which of the following is a correct statement?
A. Φ ⊆ {a, b}
B. Φ ∈ {a, b}
C. {a} ∈ {a, b}
D. a ⊆ {a, b}
Answer:
A. Φ ⊆ {a, b}
We know that null set is subset to every other non-empty set. So, it is true.
B. Φ ∈ {a, b}
Null set consists of no elements, but here it contains elements. So, it is false
C. {a} ∈ {a, b}
In the above {a} ⊂ {a, b} not {a} ∈ {a, b} since they are given as set but not element.
D. a ⊆ {a, b}
In the above {a} ∈ {a, b} not {a} ⊆ {a, b} since they are given as element but not as set.
So, option(A) is the correct answer.
Question 9.
Which one of the following is a finite set?
A. {x: x ∈ Z, x < 5}
B. {x: x ∈ W, x ≥ 5}
C. {x: x ∈ N, x > 10}
D. {x: x is an even prime number}
Answer:
A. {x: x ∈ Z, x < 5}
The above expression says that x defines the set. While the x belongs to integers which are less than 5.
We know that Z contains -∞ to ∞
But her it will contain elements from -∞ to 5
So, this is an infinite set.
B. {x: x ∈ W, x ≥ 5}
The above expression says that x defines the set. While the x belongs to whole numbers(W) which is ≥ 5.
We know that W contains 0 to ∞
But her it will contain elements from 5 to ∞
So, this is an infinite set.
C. {x: x ∈ N, x > 10}
The above expression says that x defines the set. While the x belongs to natural numbers(N) which greater than 10.
We know that N contains 1 to ∞
But her it will contain elements from 10 to ∞
So, this is an infinite set.
D. {x: x is an even prime number}
The above expression says that x defines the set. While the x belongs to even prime number (prime numbers which are even). Fortunately, we have only one even prime number that is ‘2’
So, this is finite series.
Question 10.
Given A = {5, 6, 7, 8}. Which one of the following is incorrect?
A. Φ ⊆ A
B. A ⊆ A
C. {7, 8, 9} ⊆ A
D. {5} ⊂ A
Answer:
Given set A = {5, 6, 7, 8}
A. Φ ⊆ A
We know that null set will be subset to every other non-empty set.
So, it’s true
B. A ⊆ A
We know that a set is subset to itself. So, it’s true.
C. {7, 8, 9} ⊆ A
The above set will not be a subset to A because set A didn’t contain element ‘9’. But it contains. So, it is false.
D. {5} ⊂ A
The above set is power set to A because it has same element as in set A and the no. of elements in Set is less than set A. So, it is true.
Question 11.
If A = {3, 4, 5, 6} and B = {1, 2, 5, 6}, then A ∪ B =
A. {1, 2, 3, 4, 5, 6}
B. {1, 2, 3, 4, 6}
C. {1, 2, 5, 6}
D. {3, 4, 5, 6}
Answer:
A = {3, 4, 5, 6} & B = {1, 2, 5, 6}
Here we are asked to find A ∪ B
Union means cub the elements of the both the given two sets.
So,
A ∪ B = {3, 4, 5, 6} ∪ {1, 2, 5, 6}
= {1, 2, 3, 4, 5, 6}
∴ option A is the correct answer.
Question 12.
The number of elements of the set {x: x ∈ Z, x2 = 1}is
A. 3
B. 2
C. 1
D. 0
Answer:
The set defined by {x: x ∈ Z, x2 = 1}
This means the elements of the set is given by x2 = 1 and that x is belongs to integers(Z)
x = 1 ⇒ x2 = 1
x = -1 ⇒ x2 = (-1)2 = 1
For the x = 1 & -1 the statement holds true.
∴ 1, -1 are the elements of the set
The number of elements in the set is 2.
Option B is the correct answer.
Question 13.
If n(X) = m, n(Y) = n and n(X ∩ Y) = p then n(X ∪ Y) =
A. m + n + p
B. m + n – p
C. m – p
D. m – n + p
Answer:
Given
n(X) = m, n(Y) = n and n(X ∩ Y) = p
n(X ∪ Y) = ?
we know that n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
in the same way, n(X ∪ Y) = n(X) + n(Y) - n(X ∩ Y)
n(X ∪ Y) = m + n – p
∴ option B is the correct answer.
Question 14.
If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 5, 6, 9, 10} then A’ is
A. {2, 5, 6, 9, 10}
B. Φ
C. {1, 3, 5, 10}
D. {1, 3, 4, 7, 8}
Answer:
Given,
U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {2, 5, 6, 9, 10}
We have to find A’ (A compliment)
A’(A compliment) means we have to find the eliminate the elements of sets A in U.
∴ A’ = {1, 3, 4, 7, 8}
Option D is the correct answer.
Question 15.
If A B, then A – B is
A. B
B. A
C. Φ
D. B – A
Answer:
Given that A B that means B contains all the elements that A contains.
e.g., A = {9, 8, 4, 6} B = {9, 8, 4, 6, 7, 0, 1}
we can see that B contains all the elements of B
So, A B
Let us consider two sets which is A B
A = {9, 8, 4, 6} B = {9, 8, 4, 6, 7, 0, 1}
A – B means we eliminate all the elements of B from A.
A – B = {9, 8, 4, 6} - {9, 8, 4, 6, 7, 0, 1}
= {} ⇒ null set.
Option D is the correct answer.
Question 16.
If A is a proper subset of B, then A ∩ B =
A. A
B. B
C. Φ
D. A ∪ B
Answer:
Given, A is a proper subset of B
Let us consider two set as A is a proper subset of B
A = {9, 8, 7, 6, 5, 4} B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A ∩ B means we have write all the common elements of both the sets.
A ∩ B = {9, 8, 7, 6, 5, 4} ∩ {1, 2, 3, 4, 5, 6, 7, 8, 9}
= {4, 5, 6, 7, 8, 9}
The resultant set is same as A
Option A is the correct answer.
Question 17.
If A is a proper subset of B, then A ∪ B
A. A
B. Φ
C. B
D. A ∩ B
Answer:
Given, A is a proper subset of B
Let us consider two set as A is a proper subset of B
A = {9, 8, 7, 6, 5, 4} B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A ∪ B means we must combinedly write all the elements of both the sets.
A ∪ B = {9, 8, 7, 6, 5, 4} ∪ {1, 2, 3, 4, 5, 6, 7, 8, 9}
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
The resultant set is same as B
Option C is the correct answer.
Question 18.
The shaded region in the adjoint diagram represents
A. A – B
B. A’
C. B’
D. B – A
Answer:
The venn diagram contains 2 sets, and they are overlapped which means they have common elements in it.
In the venn diagram B is completely shaded Except the overlapped part with A that means “it shows the elements of B by eliminating the common elements of A”
Since we are eliminating the elements of A in B this can be written as B-A.
Option D is the correct answer.
Question 19.
If A = {a, b, c}, B = {e, f, g}, then A ∩ B =
A. Φ
B. A
C. B
D. A ∪ B
Answer:
Given, A = {a, b, c}, B = {e, f, g}
We have to find A ∩ B
A ∩ B = {a, b, c} ∩ {e, f, g} = Φ
Since there is no common elements in the both the sets the resultant is null set.
Option A is the correct answer.
Question 20.
The shaded region in the adjoining diagram represents
A. A – B
B. B – A
C. A Δ B
D. A’
Answer:
The green shaded represents A-B and the blue shaded part represents B-A
The overall shaded region will the union of A-B & B-A
∴ (A-B) ∪ (B-A) = A Δ B
⇒ The shaded region represents A Δ B of the venn diagram.