If A = {0, 1, 2, 3, ..., 8}, B = {3, 5, 7, 9, 11} and C = {0, 5, 10, 20}, find
(i) A ∪ B
(ii) A ∪ C
(iii) B ∪ C
(iv) A ∩ B
(v) A ∩ C
(vi) B ∩ C
Also find the cardinal number of the sets B ∪ C, A ∩ B, A ∩ C and B ∩ C.
Answer
Given,
A = {0, 1, 2, 3, 4, 5, 6, 7, 8},
B = {3, 5, 7, 9, 11},
C = {0, 5, 10, 20}
(i) A ∪ B
A ∪ B - contains all the elements in set A and B.
A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∪ {3, 5, 7, 9, 11}
Hence, A ∪ B = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11}.
(ii) A ∪ C
A ∪ C - contains all the elements in set A and C.
A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∪ {0, 5, 10, 20}
Hence, A ∪ C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10, 20}.
(iii) B ∪ C
B ∪ C - contains all the elements in set B and C.
B ∪ C = {3, 5, 7, 9, 11} ∪ {0, 5, 10, 20}
Hence, B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20}.
(iv) A ∩ B
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∩ {3, 5, 7, 9, 11}
Hence, A ∩ B = {3, 5, 7}.
(v) A ∩ C
A ∩ C - contains all the common elements in set A and C.
A ∩ C = {0, 1, 2, 3, 4, 5, 6, 7, 8} ∩ {0, 5, 10, 20}
Hence, A ∩ C = {0, 5}.
(vi) B ∩ C
B ∩ C - contains all the common elements in set B and C.
B ∩ C = {3, 5, 7, 9, 11} ∩ {0, 5, 10, 20}
Hence, B ∩ C = {5}.
Cardinal numbers :
From part (iii), B ∪ C = {0, 3, 5, 7, 9, 10, 11, 20}.
As B ∪ C has 8 elements, n(B ∪ C) = 8.
From part (iv), A ∩ B = {3, 5, 7}.
As A ∩ B has 3 elements, n(A ∩ B) = 3.
From part (v), A ∩ C = {0, 5}.
As A ∩ C has 2 elements, n(A ∩ C) = 2.
As B ∩ C has only 1 element, n(B ∩ C) = 1.
Hence, n(B ∪ C) = 8, n(A ∩ B) = 3, n(A ∩ C) = 2 and n(B ∩ C) = 1.
Find A′ when A = {0, 1, 4, 7} and ξ = {x | x ∈ W, x ≤ 10}.
Answer
A = {0, 1, 4, 7} and ξ = {x | x ∈ W, x ≤ 10}
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {0, 1, 4, 7}
Hence, A′ = {2, 3, 5, 6, 8, 9, 10}.
Find A′ when A = {consonants} and ξ = {alphabets of English}.
Answer
A = {consonants} and ξ = {alphabets of English}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {alphabets of English} - {consonants}
Hence, A′ = {vowels}.
Find A′ when A = {boys in class VIII of all schools in Bengaluru} and ξ = {students in class VIII of all schools in Bengaluru}.
Answer
A = {boys in class VIII of all schools in Bengaluru} and ξ = {students in class VIII of all schools in Bengaluru}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {students in class VIII of all schools in Bengaluru} - {boys in class VIII of all schools in Bengaluru}
Hence, A′ = {girls in class VIII of all schools in Bengaluru}.
Find A′ when A = {letters of KALKA} and ξ = {letters of KOLKATA}.
Answer
A = {letters of KALKA} and ξ = {letters of KOLKATA}
Writing each element of the set once and only once, we get
A = {K, A, L} and ξ = {K, O, L, A, T}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {K, O, L, A, T} - {K, A, L}
Hence, A′ = {O, T}.
Find A′ when A = {odd natural numbers} and ξ = {whole numbers}.
Answer
A = {odd natural numbers} and ξ = {whole numbers}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {whole numbers} - {odd natural numbers}
Hence, A′ = {even whole numbers}.
If A = {x : x ∈ N and 3 < x < 7} and B = {x : x ∈ W and x ≤ 4}, find
(i) A ∪ B
(ii) A ∩ B
(iii) A - B
(iv) B - A
Answer
A = {x : x ∈ N and 3 < x < 7}
A = {4, 5, 6}
B = {x : x ∈ W and x ≤ 4}
B = {0, 1, 2, 3, 4}
(i) A ∪ B
A ∪ B - contains all the elements in set A and B.
A ∪ B = {4, 5, 6} ∪ {0, 1, 2, 3, 4}
Hence, A ∪ B = {0, 1, 2, 3, 4, 5, 6}.
(ii) A ∩ B
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {4, 5, 6} ∩ {0, 1, 2, 3, 4}
Hence, A ∩ B = {4}.
(iii) A - B
A - B - contains all the elements in set A but not in B.
A - B = {4, 5, 6} - {0, 1, 2, 3, 4}
Hence, A - B = {5, 6}.
(iv) B - A
B - A - contains all the elements in set B but not in A.
B - A = {0, 1, 2, 3, 4} - {4, 5, 6}
Hence, B - A = {0, 1, 2, 3}.
If P = {x : x ∈ W and x < 6} and Q = {x : x ∈ N and 4 ≤ x < 9}, find
(i) P ∪ Q
(ii) P ∩ Q
(iii) P - Q
(iv) Q - P
Also find whether P ∪ Q is a proper superset of P ∩ Q.
Answer
P = {x : x ∈ W and x < 6}
P = {0, 1, 2, 3, 4, 5}
Q = {x : x ∈ N and 4 ≤ x < 9}
Q = {4, 5, 6, 7, 8}
(i) P ∪ Q
P ∪ Q - contains all the elements in set P and Q.
P ∪ Q = {0, 1, 2, 3, 4, 5} ∪ {4, 5, 6, 7, 8}
Hence, P ∪ Q = {0, 1, 2, ..., 8}.
(ii) P ∩ Q
P ∩ Q - contains all the common elements in set P and Q.
P ∩ Q = {0, 1, 2, 3, 4, 5} ∩ {4, 5, 6, 7, 8}
Hence, P ∩ Q = {4, 5}.
(iii) P - Q
P - Q - contains all the elements in set P but not in Q.
P - Q = {0, 1, 2, 3, 4, 5} - {4, 5, 6, 7, 8}
Hence, P - Q = {0, 1, 2, 3}.
(iv) Q - P
Q - P - contains all the elements in set Q but not in P.
Q - P = {4, 5, 6, 7, 8} - {0, 1, 2, 3, 4, 5}
Hence, Q - P = {6, 7, 8}.
Proper superset :
From part (i), P ∪ Q = {0, 1, 2, ..., 8} and from part (ii), P ∩ Q = {4, 5}.
As every element of P ∩ Q = {4, 5} is contained in P ∪ Q = {0, 1, 2, ..., 8} and P ∪ Q ≠ P ∩ Q,
Hence, yes, P ∪ Q is a proper superset of P ∩ Q.
If A = {letters of word INTEGRITY} and B = {letters of word RECKONING}, find
(i) A ∪ B
(ii) A ∩ B
(iii) A - B
(iv) B - A
Also verify that :
(a) n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
(b) n(A - B) = n(A ∪ B) - n(B) = n(A) - n(A ∩ B)
(c) n(B - A) = n(A ∪ B) - n(A) = n(B) - n(A ∩ B)
(d) n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B)
Answer
Writing each element of the set once and only once, we get
A = {letters of word INTEGRITY}
A = {I, N, T, E, G, R, Y}
B = {letters of word RECKONING}
B = {R, E, C, K, O, N, I, G}
(i) A ∪ B
A ∪ B - contains all the elements in set A and B.
A ∪ B = {I, N, T, E, G, R, Y} ∪ {R, E, C, K, O, N, I, G}
Hence, A ∪ B = {I, N, T, E, G, R, Y, C, K, O}.
(ii) A ∩ B
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {I, N, T, E, G, R, Y} ∩ {R, E, C, K, O, N, I, G}
Hence, A ∩ B = {I, N, E, R, G}.
(iii) A - B
A - B - contains all the elements in set A but not in B.
A - B = {I, N, T, E, G, R, Y} - {R, E, C, K, O, N, I, G}
Hence, A - B = {T, Y}.
(iv) B - A
B - A - contains all the elements in set B but not in A.
B - A = {R, E, C, K, O, N, I, G} - {I, N, T, E, G, R, Y}
Hence, B - A = {C, K, O}.
Verification :
Here n(A) = 7, n(B) = 8, n(A ∪ B) = 10, n(A ∩ B) = 5, n(A - B) = 2 and n(B - A) = 3.
(a) n(A) + n(B) - n(A ∩ B) = 7 + 8 - 5 = 10 = n(A ∪ B)
Hence, n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
(b) n(A ∪ B) - n(B) = 10 - 8 = 2 = n(A - B)
and n(A) - n(A ∩ B) = 7 - 5 = 2 = n(A - B)
Hence, n(A - B) = n(A ∪ B) - n(B) = n(A) - n(A ∩ B).
(c) n(A ∪ B) - n(A) = 10 - 7 = 3 = n(B - A)
and n(B) - n(A ∩ B) = 8 - 5 = 3 = n(B - A)
Hence, n(B - A) = n(A ∪ B) - n(A) = n(B) - n(A ∩ B).
(d) n(A - B) + n(B - A) + n(A ∩ B) = 2 + 3 + 5 = 10 = n(A ∪ B)
Hence, n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B).
If ξ = {natural numbers between 10 and 40},
A = {multiples of 5} and B = {multiples of 6}, then
(i) find A ∪ B and A ∩ B.
(ii) verify that n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Answer
ξ = {natural numbers between 10 and 40}
ξ = {11, 12, 13, ..., 39}
A = {multiples of 5}
A = {15, 20, 25, 30, 35}
B = {multiples of 6}
B = {12, 18, 24, 30, 36}
(i) A ∪ B - contains all the elements in set A and B.
A ∪ B = {15, 20, 25, 30, 35} ∪ {12, 18, 24, 30, 36}
A ∪ B = {12, 15, 18, 20, 24, 25, 30, 35, 36}
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {15, 20, 25, 30, 35} ∩ {12, 18, 24, 30, 36}
A ∩ B = {30}
Hence, A ∪ B = {12, 15, 18, 20, 24, 25, 30, 35, 36} and A ∩ B = {30}.
(ii) Here n(A) = 5, n(B) = 5, n(A ∪ B) = 9 and n(A ∩ B) = 1.
⇒ n(A) + n(B) - n(A ∩ B) = 5 + 5 - 1 = 9 = n(A ∪ B)
Hence, n(A ∪ B) = n(A) + n(B) - n(A ∩ B).
If ξ = {1, 2, 3, ..., 9}, A = {1, 2, 3, 4, 6, 7, 8} and B = {4, 6, 8}, then find
(i) A′
(ii) B′
(iii) A ∪ B
(iv) A ∩ B
(v) A - B
(vi) B - A
(vii) (A ∩ B)′
(viii) A′ ∪ B′
Also verify that :
(a) (A ∩ B)′ = A′ ∪ B′
(b) n(A) + n(A′) = n(ξ)
(c) n(A ∩ B) + n((A ∩ B)′) = n(ξ)
(d) n(A - B) + n(B - A) + n(A ∩ B) = n(A ∪ B).
Answer
ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4, 6, 7, 8}
B = {4, 6, 8}
(i) A′
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {1, 2, 3, 4, 6, 7, 8}
Hence, A′ = {5, 9}.
(ii) B′
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {4, 6, 8}
Hence, B′ = {1, 2, 3, 5, 7, 9}.
(iii) A ∪ B
A ∪ B - contains all the elements in set A and B.
A ∪ B = {1, 2, 3, 4, 6, 7, 8} ∪ {4, 6, 8}
Hence, A ∪ B = {1, 2, 3, 4, 6, 7, 8}.
(iv) A ∩ B
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {1, 2, 3, 4, 6, 7, 8} ∩ {4, 6, 8}
Hence, A ∩ B = {4, 6, 8}.
(v) A - B
A - B - contains all the elements in set A but not in B.
A - B = {1, 2, 3, 4, 6, 7, 8} - {4, 6, 8}
Hence, A - B = {1, 2, 3, 7}.
(vi) B - A
B - A - contains all the elements in set B but not in A.
B - A = {4, 6, 8} - {1, 2, 3, 4, 6, 7, 8}
Hence, B - A = φ.
(vii) (A ∩ B)′
(A ∩ B)′ - contains all the elements in universal set which are not in A ∩ B.
(A ∩ B)′ = ξ - (A ∩ B)
(A ∩ B)′ = {1, 2, 3, 4, 5, 6, 7, 8, 9} - {4, 6, 8}
Hence, (A ∩ B)′ = {1, 2, 3, 5, 7, 9}.
(viii) A′ ∪ B′
A′ ∪ B′ - contains all the elements in set A′ and B′.
A′ ∪ B′ = {5, 9} ∪ {1, 2, 3, 5, 7, 9}
Hence, A′ ∪ B′ = {1, 2, 3, 5, 7, 9}.
Verification :
(a) From part (vii), (A ∩ B)′ = {1, 2, 3, 5, 7, 9} ...............(1)
From part (viii), A′ ∪ B′ = {1, 2, 3, 5, 7, 9} ...............(2)
From (1) and (2), we can see
LHS = RHS
Hence, (A ∩ B)′ = A′ ∪ B′.
(b) Here n(A) = 7, n(A′) = 2 and n(ξ) = 9.
n(A) + n(A′) = 7 + 2 = 9 = n(ξ)
Hence, n(A) + n(A′) = n(ξ).
(c) Here n(A ∩ B) = 3, n((A ∩ B)′) = 6 and n(ξ) = 9.
n(A ∩ B) + n((A ∩ B)′) = 3 + 6 = 9 = n(ξ)
Hence, n(A ∩ B) + n((A ∩ B)′) = n(ξ).
(d) Here n(A - B) = 4, n(B - A) = 0, n(A ∩ B) = 3 and n(A ∪ B) = 7.
n(A - B) + n(B - A) + n(A ∩ B) = 4 + 0 + 3 = 7 = n(A ∪ B)
Hence, n(A - B) + n(B - A) + n(A ∩ B) = n(A ∪ B).
If ξ = {x : x ∈ W, x ≤ 10}, A = {x : x ≥ 5} and B = {x : 3 ≤ x < 8}, then verify that (A ∪ B)′ = A′ ∩ B′.
Answer
ξ = {x : x ∈ W, x ≤ 10}
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {x : x ≥ 5}
A = {5, 6, 7, 8, 9, 10}
B = {x : 3 ≤ x < 8}
B = {3, 4, 5, 6, 7}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7, 8, 9, 10}
A′ = {0, 1, 2, 3, 4}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}
B′ = {0, 1, 2, 8, 9, 10}
(A ∪ B)′ = A′ ∩ B′
Taking LHS : (A ∪ B)′
A ∪ B - contains all the elements in set A and B.
A ∪ B = {5, 6, 7, 8, 9, 10} ∪ {3, 4, 5, 6, 7}
A ∪ B = {3, 4, 5, 6, 7, 8, 9, 10}
(A ∪ B)′ = ξ - (A ∪ B)
(A ∪ B)′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7, 8, 9, 10}
(A ∪ B)′ = {0, 1, 2} ...............(1)
Taking RHS : A′ ∩ B′
A′ ∩ B′ - contains all the common elements in set A′ and B′.
A′ ∩ B′ = {0, 1, 2, 3, 4} ∩ {0, 1, 2, 8, 9, 10}
A′ ∩ B′ = {0, 1, 2} ...............(2)
From (1) and (2), we can see
LHS = RHS
Hence, (A ∪ B)′ = A′ ∩ B′.
If ξ = {x : x ∈ W, x ≤ 10}, A = {x : x ≥ 5} and B = {x : 3 ≤ x < 8}, then verify that (A ∩ B)′ = A′ ∪ B′.
Answer
ξ = {x : x ∈ W, x ≤ 10}
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {x : x ≥ 5}
A = {5, 6, 7, 8, 9, 10}
B = {x : 3 ≤ x < 8}
B = {3, 4, 5, 6, 7}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7, 8, 9, 10}
A′ = {0, 1, 2, 3, 4}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}
B′ = {0, 1, 2, 8, 9, 10}
(A ∩ B)′ = A′ ∪ B′
Taking LHS : (A ∩ B)′
A ∩ B - contains all the common elements in set A and B.
A ∩ B = {5, 6, 7, 8, 9, 10} ∩ {3, 4, 5, 6, 7}
A ∩ B = {5, 6, 7}
(A ∩ B)′ = ξ - (A ∩ B)
(A ∩ B)′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7}
(A ∩ B)′ = {0, 1, 2, 3, 4, 8, 9, 10} ...............(1)
Taking RHS : A′ ∪ B′
A′ ∪ B′ - contains all the elements in set A′ and B′.
A′ ∪ B′ = {0, 1, 2, 3, 4} ∪ {0, 1, 2, 8, 9, 10}
A′ ∪ B′ = {0, 1, 2, 3, 4, 8, 9, 10} ...............(2)
From (1) and (2), we can see
LHS = RHS
Hence, (A ∩ B)′ = A′ ∪ B′.
If ξ = {x : x ∈ W, x ≤ 10}, A = {x : x ≥ 5} and B = {x : 3 ≤ x < 8}, then verify that A - B = A ∩ B′.
Answer
ξ = {x : x ∈ W, x ≤ 10}
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {x : x ≥ 5}
A = {5, 6, 7, 8, 9, 10}
B = {x : 3 ≤ x < 8}
B = {3, 4, 5, 6, 7}
B′ - contains all the elements in universal set which are not in B.
B′ = ξ - B
B′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}
B′ = {0, 1, 2, 8, 9, 10}
A - B = A ∩ B′
Taking LHS : A - B
A - B - contains all the elements in set A but not in B.
A - B = {5, 6, 7, 8, 9, 10} - {3, 4, 5, 6, 7}
A - B = {8, 9, 10} ...............(1)
Taking RHS : A ∩ B′
A ∩ B′ - contains all the common elements in set A and B′.
A ∩ B′ = {5, 6, 7, 8, 9, 10} ∩ {0, 1, 2, 8, 9, 10}
A ∩ B′ = {8, 9, 10} ...............(2)
From (1) and (2), we can see
LHS = RHS
Hence, A - B = A ∩ B′.
If ξ = {x : x ∈ W, x ≤ 10}, A = {x : x ≥ 5} and B = {x : 3 ≤ x < 8}, then verify that B - A = B ∩ A′.
Answer
ξ = {x : x ∈ W, x ≤ 10}
ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {x : x ≥ 5}
A = {5, 6, 7, 8, 9, 10}
B = {x : 3 ≤ x < 8}
B = {3, 4, 5, 6, 7}
A′ - contains all the elements in universal set which are not in A.
A′ = ξ - A
A′ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7, 8, 9, 10}
A′ = {0, 1, 2, 3, 4}
B - A = B ∩ A′
Taking LHS : B - A
B - A - contains all the elements in set B but not in A.
B - A = {3, 4, 5, 6, 7} - {5, 6, 7, 8, 9, 10}
B - A = {3, 4} ...............(1)
Taking RHS : B ∩ A′
B ∩ A′ - contains all the common elements in set B and A′.
B ∩ A′ = {3, 4, 5, 6, 7} ∩ {0, 1, 2, 3, 4}
B ∩ A′ = {3, 4} ...............(2)
From (1) and (2), we can see
LHS = RHS
Hence, B - A = B ∩ A′.
If n(A) = 20, n(B) = 16 and n(A ∪ B) = 30, find n(A ∩ B)
Answer
Given:
n(A) = 20
n(B) = 16
n(A ∪ B) = 30
∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Putting the values, we get
⇒ 30 = 20 + 16 - n(A ∩ B)
⇒ n(A ∩ B) = 36 - 30
⇒ n(A ∩ B) = 6
Hence, n(A ∩ B) = 6.
If n(ξ) = 20 and n(A′) = 7, then find n(A)
Answer
Given:
n(ξ) = 20
n(A′) = 7
∵ n(A) + n(A′) = n(ξ)
Putting the values, we get
⇒ n(A) + 7 = 20
⇒ n(A) = 20 - 7
⇒ n(A) = 13
Hence, n(A) = 13.
If n(ξ) = 40, n(A) = 20, n(B′) = 16 and n(A ∪ B) = 32, then find n(B) and n(A ∩ B)
Answer
Given:
n(ξ) = 40
n(A) = 20
n(B′) = 16
n(A ∪ B) = 32
∵ n(B) + n(B′) = n(ξ)
Putting the values, we get
⇒ n(B) + 16 = 40
⇒ n(B) = 40 - 16
⇒ n(B) = 24
∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Putting the values, we get
⇒ 32 = 20 + 24 - n(A ∩ B)
⇒ n(A ∩ B) = 44 - 32
⇒ n(A ∩ B) = 12
Hence, n(B) = 24 and n(A ∩ B) = 12.
If n(ξ) = 32, n(A) = 20, n(B) = 16 and n((A ∪ B)′) = 4, find :
(i) n(A ∪ B)
(ii) n(A ∩ B)
(iii) n(A - B)
Answer
Given:
n(ξ) = 32
n(A) = 20
n(B) = 16
n((A ∪ B)′) = 4
(i) ∵ n(A ∪ B) + n((A ∪ B)′) = n(ξ)
Putting the values, we get
⇒ n(A ∪ B) + 4 = 32
⇒ n(A ∪ B) = 32 - 4
⇒ n(A ∪ B) = 28
Hence, n(A ∪ B) = 28.
(ii) ∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Putting the values, we get
⇒ 28 = 20 + 16 - n(A ∩ B)
⇒ n(A ∩ B) = 36 - 28
⇒ n(A ∩ B) = 8
Hence, n(A ∩ B) = 8.
(iii) ∵ n(A - B) = n(A) - n(A ∩ B)
Putting the values, we get
⇒ n(A - B) = 20 - 8
⇒ n(A - B) = 12
Hence, n(A - B) = 12.
If n(ξ) = 40, n(A′) = 15, n(B) = 12 and n((A ∩ B)′) = 32, find :
(i) n(A)
(ii) n(B′)
(iii) n(A ∩ B)
(iv) n(A ∪ B)
(v) n(A - B)
(vi) n(B - A)
Answer
Given:
n(ξ) = 40
n(A′) = 15
n(B) = 12
n((A ∩ B)′) = 32
(i) ∵ n(A) + n(A′) = n(ξ)
Putting the values, we get
⇒ n(A) + 15 = 40
⇒ n(A) = 40 - 15
⇒ n(A) = 25
Hence, n(A) = 25.
(ii) ∵ n(B) + n(B′) = n(ξ)
Putting the values, we get
⇒ 12 + n(B′) = 40
⇒ n(B′) = 40 - 12
⇒ n(B′) = 28
Hence, n(B′) = 28.
(iii) ∵ n(A ∩ B) + n((A ∩ B)′) = n(ξ)
Putting the values, we get
⇒ n(A ∩ B) + 32 = 40
⇒ n(A ∩ B) = 40 - 32
⇒ n(A ∩ B) = 8
Hence, n(A ∩ B) = 8.
(iv) ∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
Putting the values, we get
⇒ n(A ∪ B) = 25 + 12 - 8
⇒ n(A ∪ B) = 37 - 8
⇒ n(A ∪ B) = 29
Hence, n(A ∪ B) = 29.
(v) ∵ n(A - B) = n(A) - n(A ∩ B)
Putting the values, we get
⇒ n(A - B) = 25 - 8
⇒ n(A - B) = 17
Hence, n(A - B) = 17.
(vi) ∵ n(B - A) = n(B) - n(A ∩ B)
Putting the values, we get
⇒ n(B - A) = 12 - 8
⇒ n(B - A) = 4
Hence, n(B - A) = 4.
If n(A - B) = 12, n(B - A) = 16 and n(A ∩ B) = 5, find :
(i) n(A)
(ii) n(B)
(iii) n(A ∪ B)
Answer
Given:
n(A - B) = 12
n(B - A) = 16
n(A ∩ B) = 5
(i) ∵ n(A - B) = n(A) - n(A ∩ B)
Putting the values, we get
⇒ 12 = n(A) - 5
⇒ n(A) = 12 + 5
⇒ n(A) = 17
Hence, n(A) = 17.
(ii) ∵ n(B - A) = n(B) - n(A ∩ B)
Putting the values, we get
⇒ 16 = n(B) - 5
⇒ n(B) = 16 + 5
⇒ n(B) = 21
Hence, n(B) = 21.
(iii) ∵ n(A ∪ B) = n(A - B) + n(B - A) + n(A ∩ B)
Putting the values, we get
⇒ n(A ∪ B) = 12 + 16 + 5
⇒ n(A ∪ B) = 33
Hence, n(A ∪ B) = 33.