Q.1. Find the union of each of the following pairs of sets:
(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = {a, e, i, o, u} B = {a, b, c}
(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}
(iv) A = {x : x is a natural number and 1 < x ≤ 6}
B = {x : x is a natural number and 6 < x < 10}
(v) A = {1, 2, 3}, B = Φ
Ans.
(i) X = {1, 3, 5} Y = {1, 2, 3}
X∪ Y= {1, 2, 3, 5}
(ii) A = {a, e, i, o, u} B = {a, b, c}
A∪ B = {a, b, c, e, i, o, u}
(iii) A = {x : x is a natural number and multiple of 3} = {3, 6, 9 …}
As B = {x : x is a natural number less than 6} = {1, 2, 3, 4, 5, 6}
A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 …}
∴ A ∪ B = {x : x = 1, 2, 4, 5 or a multiple of 3}
(iv) A = {x : x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x < 10} = {7, 8, 9}
A∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
∴ A∪ B = {x : x ∈ N and 1 < x < 10}
(v) A = {1, 2, 3}, B = Φ
A∪ B = {1, 2, 3}
Q.2. Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?
Ans. Here, A = {a, b} and B = {a, b, c}
Yes, A ⊂ B.
A∪ B = {a, b, c} = B
Q.3. If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Ans. If A and B are two sets such that A ⊂ B, then A ∪ B = B.
Q.4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}; find
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Ans. A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
(i) A ∪ B = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
Q.5. Find the intersection of each pair of sets:
(i) X = {1, 3, 5} Y = {1, 2, 3}
(ii) A = {a, e, i, o, u} B = {a, b, c}
(iii) A = {x: x is a natural number and multiple of 3}
B = {x: x is a natural number less than 6}
(iv) A = {x : x is a natural number and 1 < x ≤ 6}
B = {x : x is a natural number and 6 < x < 10}
(v) A = {1, 2, 3}, B = Φ
Ans.
(i) X = {1, 3, 5}, Y = {1, 2, 3}
X ∩ Y = {1, 3}
(ii) A = {a, e, i, o, u}, B = {a, b, c}
A ∩ B = {a}
(iii) A = {x : x is a natural number and multiple of 3} = (3, 6, 9 …}
B = {x : x is a natural number less than 6} = {1, 2, 3, 4, 5}
∴ A ∩ B = {3}
(iv) A = {x : x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x < 10} = {7, 8, 9}
A ∩ B = Φ
(v) A = {1, 2, 3}, B = Φ
A ∩ B = Φ
Q.6. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15} and D = {15, 17}; find
(i) A ∩ B
(ii) B ∩ C
(iii) A ∩ C ∩ D
(iv) A ∩ C
(v) B ∩ D
(vi) A ∩ (B ∪ C)
(vii) A ∩ D
(viii) A ∩ (B ∪ D)
(ix) (A ∩ B) ∩ (B ∪ C)
(x) (A ∪ D) ∩ (B ∪ C)
Ans.
(i) A ∩ B = {7, 9, 11}
(ii) B ∩ C = {11, 13}
(iii) A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ
(iv) A ∩ C = {11}
(v) B ∩ D = Φ
(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
= {7, 9, 11} ∪ {11} = {7, 9, 11}
(vii) A ∩ D = Φ
(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)
= {7, 9, 11} ∪ Φ = {7, 9, 11}
(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}
(x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}
= {7, 9, 11, 15}
Q.7. If A = {x : x is a natural number}, B ={x : x is an even natural number}
C = {x : x is an odd natural number} and D = {x : x is a prime number}, find
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Ans.
A = {x : x is a natural number} = {1, 2, 3, 4, 5 …}
B ={x : x is an even natural number} = {2, 4, 6, 8 …}
C = {x : x is an odd natural number} = {1, 3, 5, 7, 9 …}
D = {x : x is a prime number} = {2, 3, 5, 7 …}
(i) A ∩B = {x : x is a even natural number} = B
(ii) A ∩ C = {x : x is an odd natural number} = C
(iii) A ∩ D = {x : x is a prime number} = D
(iv) B ∩ C = Φ
(v) B ∩ D = {2}
(vi) C ∩ D = {x : x is odd prime number}
Q.8. Which of the following pairs of sets are disjoint
(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6}
(ii) {a, e, i, o, u} and {c, d, e, f}
(iii) {x : x is an even integer} and {x : x is an odd integer}
Ans.
(i) {1, 2, 3, 4}
{x : x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}
Therefore, this pair of sets is not disjoint.
(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}
Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.
(iii) {x : x is an even integer} ∩ {x : x is an odd integer} = Φ
Therefore, this pair of sets is disjoint.
Q.9. If A = {3, 6, 9, 12, 15, 18, 21},
B = {4, 8, 12, 16, 20},
C = {2, 4, 6, 8, 10, 12, 14, 16},
D = {5, 10, 15, 20}; find
(i) A – B
(ii) A – C
(iii) A – D
(iv) B – A
(v) C – A
(vi) D – A
(vii) B – C
(viii) B – D
(ix) C – B
(x) D – B
(xi) C – D
(xii) D – C
Ans.
(i) A – B = {3, 6, 9, 15, 18, 21}
(ii) A – C = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 16, 20}
(v) C – A = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 20}
(vii) B – C = {20}
(viii) B – D = {4, 8, 12, 16}
(ix) C – B = {2, 6, 10, 14}
(x) D – B = {5, 10, 15}
(xi) C – D = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C = {5, 15, 20}
Q.10. If X = {a, b, c, d} and Y = {f, b, d, g}, find
(i) X – Y
(ii) Y – X
(iii) X ∩ Y
Ans.
(i) X – Y = {a, c}
(ii) Y – X = {f, g}
(iii) X ∩ Y = {b, d}
Q.11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Ans.
R: set of real numbers
Q: set of rational numbers
Therefore, R – Q is a set of irrational numbers.
Q.12. State whether each of the following statement is true or false. Justify your answer.
(i) {2, 3, 4, 5} and {3, 6} are disjoint sets.
(ii) {a, e, i, o, u } and {a, b, c, d} are disjoint sets.
(iii) {2, 6, 10, 14} and {3, 7, 11, 15} are disjoint sets.
(iv) {2, 6, 10} and {3, 7, 11} are disjoint sets.
Ans.
(i) False
As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}
⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}
(ii) False
As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}
⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}
(iii) True
As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ
(iv) True
As {2, 6, 10} ∩ {3, 7, 11} = Φ
Q.1. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 3, 4}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6}. Find
(i) A'
(ii) B'
(iii) (A ∪ C)'
(iv) (A ∪ B)'
(v) (A')^{'}
(vi) (B  C)'
Ans.
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4}
B = {2, 4, 6, 8}
C = {3, 4, 5, 6}
(i) A' = {5, 6, 7, 8 ,9}
(ii) B' = {1, 3, 5, 7, 9}
(iii) A ∪ C = {1, 2, 3, 4, 5, 6} ∴ (A ∪ C)' = {7, 8, 9}
(iv) A ∪ B = {1, 2, 3, 4, 6, 8} (A ∪ B)' = {5, 7, 9}
(v) (A')' = A = {1, 2, 3, 4}
(vi) B  C = {2, 8} ∴ (B  C)' = {1, 3, 4, 5, 6, 7, 9}
Q.2. If U = {a, b, c, d, e, f, g, h}, find the complements of the following sets
(i) A = {a, b, c}
(ii) B = {d, e, f, g}
(iii) C = {a, c, e, g}
(iv) D = {f, g, h, a}
Ans. U = {a, b, c, d, e, f, g, h}
(i) A = {a, b, c} A' = {d, e, f, g, h}
(ii) B = {d, e, f, g} ∴ B' = {a, b, c, h}
(iii) C = {a, c, e, g} ∴ C' = {b, d, f, h}
(iv) D = {f, g, h, a} ∴ D' = {b, c, d, e}
Q.3. Taking the set of natural numbers as the universal set, write down the complements of the following sets:
(i) {x : x is an even natural number}
(ii) {x : x is an odd natural number}
(iii) {x : x is a positive multiple of 3}
(iv) {x : x is a prime number}
(v) {x : x is a natural number divisible by 3 and 5}
(vi) {x : x is a perfect square}
(vii) {x : x is perfect cube}
(viii) {x : x + 5 = 8}
(ix) {x : 2x + 5 = 9}
(x) {x : x ≥ 7}
(xi) {x : x ∈ N and 2x + 1 > 10}
Ans. U = N: Set of natural numbers
(i) {x : x is an even natural number}´ = {x : x is an odd natural number}
(ii) {x : x is an odd natural number}´ = {x : x is an even natural number}
(iii) {x : x is a positive multiple of 3}´ = {x : x ∈ N and x is not a multiple of 3}
(iv) {x : x is a prime number}´ = {x : x is a positive composite number and x = 1}
(v) {x : x is a natural number divisible by 3 and 5}´ = {x : x is a natural number that is not divisible by 3 or 5}
(vi) {x : x is a perfect square}´ = {x : x ∈ N and x is not a perfect square}
(vii) {x : x is a perfect cube}´ = {x : x ∈ N and x is not a perfect cube}
(viii) {x : x + 5 = 8}´ = {x : x ∈ N and x ≠ 3}
(ix) {x : 2x + 5 = 9}´ = {x : x ∈ N and x ≠ 2}
(x) {x : x ≥ 7}´ = {x : x ∈ N and x < 7}
(xi) {x : x ∈ N and 2x + 1 > 10}´ = {x : x ∈ N and x ≤ 9/2}
Q.4. If U = {1, 2, 3, 4, 5,6,7,8, 9}, A = {2, 4, 6, 8} and B = {2, 3, 5, 7}. Verify that
(i) (A U B)' = A' ∩ B'
(ii) (A ∩ B)' = A' U B'
Ans. It is given that
U = {1, 2, 3, 4, 5,6,7,8, 9}
A = {2, 4, 6, 8}
B = {2, 3, 5, 7}
(i) (A U B)' = {2, 3, 4, 5, 6, 7, 8}' = {1, 9}
A' ∩ B' = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}
Therefore, (A U B)' = A' ∩ B'.
(ii) (A ∩ B)' = {2}' = {1, 3, 4, 5, 6, 7, 8, 9}
A' U B' = {1, 3, 5, 7, 9} U {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}
Therefore, (A ∩ B)' = A' U B'.
Q.5. Draw appropriate Venn diagrams for each of the following:
(i) (A ∪ B)'
(ii) A' ∩ B'
(iii) (A ∩ B)'
(iv) A' ∪ B'
Ans.
(i) In the diagrams, shaded portion represents (A ∪ B)'
(ii) In the diagrams, shaded portion represents A' ∩ B'
(iii) In the diagrams, shaded portion represents (A ∩ B)'
(iv) In the diagrams, shaded portion represents A' ∪ B'
Q.6. Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60º what is A’?
Ans. Given: U = {x : x is a triangle}
A = {x : x is a triangle and has at least one angle different from 60º}
∴ A' = U – A = {x : x is a triangle and has all angles equal to 60º}
= Set of all equilateral triangles
Q.7. Fill in the blanks to make each of the following a true statement:
(i) A'∪A' = ____
(ii) ϕ'∩A = ____
(iii) A'∩A' = ____
(iv) U'∩A' = ____
Ans.
(i) A'∪ A' = U
(ii) ϕ'∩ A = U ∩ A = A
(iii) A'∩ A' = ϕ
(iv) U'∩ A' = ϕ∩ A = ϕ
Q.1. If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y).
Solution:
Given
n (X) = 17
n (Y) = 23
n (X U Y) = 38
We can write it as
n (X U Y) = n (X) + n (Y) – n (X ∩ Y)
Substituting the values
38 = 17 + 23 – n (X ∩ Y)
By further calculation
n (X ∩ Y) = 40 – 38 = 2
So we get
n (X ∩ Y) = 2
Q.2. If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?
Solution:
Given
n (X U Y) = 18
n (X) = 8
n (Y) = 15
We can write it as
n (X U Y) = n (X) + n (Y) – n (X ∩ Y)
Substituting the values
18 = 8 + 15 – n (X ∩ Y)
By further calculation
n (X ∩ Y) = 23 – 18 = 5
So we get
n (X ∩ Y) = 5
Q.3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Solution:
Consider H as the set of people who speak Hindi
E as the set of people who speak English
We know that
n(H ∪ E) = 400
n(H) = 250
n(E) = 200
It can be written as
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
By substituting the values
400 = 250 + 200 – n(H ∩ E)
By further calculation
400 = 450 – n(H ∩ E)
So we get
n(H ∩ E) = 450 – 400
n(H ∩ E) = 50
Therefore, 50 people can speak both Hindi and English.
Q.4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?
Solution:
We know that
n(S) = 21
n(T) = 32
n(S ∩ T) = 11
It can be written as
n (S ∪ T) = n (S) + n (T) – n (S ∩ T)
Substituting the values
n (S ∪ T) = 21 + 32 – 11
So we get
n (S ∪ T)= 42
Therefore, the set (S ∪ T) has 42 elements.
Q.5. If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?
Solution:
We know that
n(X) = 40
n(X ∪ Y) = 60
n(X ∩ Y) = 10
It can be written as
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
By substituting the values
60 = 40 + n(Y) – 10
On further calculation
n(Y) = 60 – (40 – 10) = 30
Therefore, the set Y has 30 elements.
Q.6. In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution:
Consider C as the set of people who like coffee
T as the set of people who like tea
n(C ∪ T) = 70
n(C) = 37
n(T) = 52
It is given that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
Substituting the values
70 = 37 + 52 – n(C ∩ T)
By further calculation
70 = 89 – n(C ∩ T)
So we get
n(C ∩ T) = 89 – 70 = 19
Therefore, 19 people like both coffee and tea.
Q.7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution:
Consider C as the set of people who like cricket
T as the set of people who like tennis
n(C ∪ T) = 65
n(C) = 40
n(C ∩ T) = 10
It can be written as
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
Substituting the values
65 = 40 + n(T) – 10
By further calculation
65 = 30 + n(T)
So we get
n(T) = 65 – 30 = 35
Hence, 35 people like tennis.
We know that,
(T – C) ∪ (T ∩ C) = T
So we get,
(T – C) ∩ (T ∩ C) = Φ
Here
n (T) = n (T – C) + n (T ∩ C)
Substituting the values
35 = n (T – C) + 10
By further calculation
n (T – C) = 35 – 10 = 25
Therefore, 25 people like only tennis.
Q.8. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Solution:
Consider F as the set of people in the committee who speak French
S as the set of people in the committee who speak Spanish
n(F) = 50
n(S) = 20
n(S ∩ F) = 10
It can be written as
n(S ∪ F) = n(S) + n(F) – n(S ∩ F)
By substituting the values
n(S ∪ F) = 20 + 50 – 10
By further calculation
n(S ∪ F) = 70 – 10
n(S ∪ F) = 60
Therefore, 60 people in the committee speak at least one of the two languages
Q.1. If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y).
Solution:
Given
n (X) = 17
n (Y) = 23
n (X U Y) = 38
We can write it as
n (X U Y) = n (X) + n (Y) – n (X ∩ Y)
Substituting the values
38 = 17 + 23 – n (X ∩ Y)
By further calculation
n (X ∩ Y) = 40 – 38 = 2
So we get
n (X ∩ Y) = 2
Q.2. If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?
Solution:
Given
n (X U Y) = 18
n (X) = 8
n (Y) = 15
We can write it as
n (X U Y) = n (X) + n (Y) – n (X ∩ Y)
Substituting the values
18 = 8 + 15 – n (X ∩ Y)
By further calculation
n (X ∩ Y) = 23 – 18 = 5
So we get
n (X ∩ Y) = 5
Q.3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?
Solution:
Consider H as the set of people who speak Hindi
E as the set of people who speak English
We know that
n(H ∪ E) = 400
n(H) = 250
n(E) = 200
It can be written as
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
By substituting the values
400 = 250 + 200 – n(H ∩ E)
By further calculation
400 = 450 – n(H ∩ E)
So we get
n(H ∩ E) = 450 – 400
n(H ∩ E) = 50
Therefore, 50 people can speak both Hindi and English.
Q.4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?
Solution:
We know that
n(S) = 21
n(T) = 32
n(S ∩ T) = 11
It can be written as
n (S ∪ T) = n (S) + n (T) – n (S ∩ T)
Substituting the values
n (S ∪ T) = 21 + 32 – 11
So we get
n (S ∪ T)= 42
Therefore, the set (S ∪ T) has 42 elements.
Q.5. If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?
Solution:
We know that
n(X) = 40
n(X ∪ Y) = 60
n(X ∩ Y) = 10
It can be written as
n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)
By substituting the values
60 = 40 + n(Y) – 10
On further calculation
n(Y) = 60 – (40 – 10) = 30
Therefore, the set Y has 30 elements.
Q.6. In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?
Solution:
Consider C as the set of people who like coffee
T as the set of people who like tea
n(C ∪ T) = 70
n(C) = 37
n(T) = 52
It is given that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
Substituting the values
70 = 37 + 52 – n(C ∩ T)
By further calculation
70 = 89 – n(C ∩ T)
So we get
n(C ∩ T) = 89 – 70 = 19
Therefore, 19 people like both coffee and tea.
Q.7. In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?
Solution:
Consider C as the set of people who like cricket
T as the set of people who like tennis
n(C ∪ T) = 65
n(C) = 40
n(C ∩ T) = 10
It can be written as
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
Substituting the values
65 = 40 + n(T) – 10
By further calculation
65 = 30 + n(T)
So we get
n(T) = 65 – 30 = 35
Hence, 35 people like tennis.
We know that,
(T – C) ∪ (T ∩ C) = T
So we get,
(T – C) ∩ (T ∩ C) = Φ
Here
n (T) = n (T – C) + n (T ∩ C)
Substituting the values
35 = n (T – C) + 10
By further calculation
n (T – C) = 35 – 10 = 25
Therefore, 25 people like only tennis.
Q.8. In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?
Solution:
Consider F as the set of people in the committee who speak French
S as the set of people in the committee who speak Spanish
n(F) = 50
n(S) = 20
n(S ∩ F) = 10
It can be written as
n(S ∪ F) = n(S) + n(F) – n(S ∩ F)
By substituting the values
n(S ∪ F) = 20 + 50 – 10
By further calculation
n(S ∪ F) = 70 – 10
n(S ∪ F) = 60
Therefore, 60 people in the committee speak at least one of the two languages
83 videos237 docs99 tests

1. What is the concept of sets in mathematics? 
2. How are sets represented in set theory? 
3. What are the different types of sets? 
4. How are sets related to Venn diagrams? 
5. What is the importance of sets in reallife applications? 
83 videos237 docs99 tests


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