Theory Of Sets

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.

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]

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’}

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}

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.

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

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.

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.

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

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

No, because Φ contains no element but {Φ} contains one element, which is a null element. That’s the reason they are not equal.

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.

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.

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.

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

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

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

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

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.

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.

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]

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]

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]

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}

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.

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.

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]

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’]

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)]

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)]

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]

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]

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}

∴ 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]

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)’]

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:

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.

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.

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

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.

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.

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.

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.

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

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 1^st 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.

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.

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

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.

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.

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

(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

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%

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.

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

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.

Given Set A with ‘m’ number of elements

So, the number of power subset for given set is given by 2^m – 1

∴ option (B) is true.

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)) = 2^m

Where m is the no. of elements.

So, no. of subsets = n(p(A)) = 2³ = 8.

∴ option(B) is the correct 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.