n (A ∪ B) = 50, n (A) = 28, n(B) = 32, where n(x) does notes the cardinal number of the set x.
We know that n (A ∪ E) = n(A) + n(B) - n (A ∩ B)
⟹ 50 = 28 + 32 - n(A ∩ B)
⟹ 50 = 60 - n(A ∩ B)
⟹ n(A ∩ B) = 60 - 50
=10
∴ n(A ∩ B) = 10
We have,
n(P) = 40, n(P ∪ Q) = 60, n(P ∩ Q) = 10, to find n(Q).
We know n (P ∪ Q) = n(P) + n(Q) - n(P ∩ Q)
⟹ 60 - 40 + n (Q) - 10
⟹ 60 - 30 + n(Q)
⟹ n (Q) = 60- 30 = 30
Hence, Q has 30 elements.
Let n(P) denote the number of teachers who teach Physics and n(Q) denote the number of teachers who teach Mathematics.
We have,
n(P or M) = 20
i.e. n(P ∪ M) = 20
n(M) = 12
and n(P ∩ M) = 4
To find: n(P)
We know n (P ∪ M) = n(P) + n(M) – n(P ∩ M)
⟹ 20 = n (P) + 12 – 4
⟹ 20 = n (P) + 8
⟹ n(P) = 20 - 8 = 12
∴ There are 12 Physics teachers.
Let,
n (P) denote the total number of people
n (C) denote the number of people who like coffee and
n (T) denote the number of people who like tea.
Then,
n (P) = 70
n (C) = 37
n(T) = 52
We are given that each person likes at least one of the two drinks, i.e., P = C ∪ T.
To find: n (C ∩ T)
We known n(P) = n(C) + n (T)- n(C ∩ T)
⟹ 70 = 37 + 52 - n(C ∩ T)
⟹ 70 = 89 - n (C ∩ T)
⟹ n (C ∩ T) = 89 - 70
= 19
Hence, 19 people like both coffee and tea.
n (A) = 20, n (A ∪ B) = 42 and n (A ∩ B) = 4, to find: n (B)
We know n (A ∪ B) = n (A) + n (B) - n (A ∩ B)
⟹ 42 = 20 + n(B) - 4
⟹ 42 = 16 + n(B)
⟹ n(B) = 42- 16 = 26
∴ n(B) = 26
To find: n (A - B)
We know that if A and B are disjoint sets, then
A ∩ B = ∮
∴ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
= n(A) + n(B) - n(∮)
⟹ n(A ∪ B) = n (A) + n (B) [∵ n(∮) = 0]
Now,
A = (A - B) ∪ (A ∩ B)
i.e A is the disjoint union of A - 8 and A ∩ B
∴ n(A) = n (A - B) ∪ (A ∩ B)
= n(A - B) + n (A ∩ 8) [∵ A - B and A ∩ B are disjoint]
⟹ 20 = n (A - B) + 4
⟹ n(A - B) = 20 - 4
= 16
∴ n (A - B) = 16
To find: B - A
On a similar lines we have 8 is the disjoint union of B - A and A ∩ B
i.e. B = (B - A) ∪ (A ∩ B)
∴ n(B) = n(B - A) + n (A ∩ B)
⟹ 26 = n (B - A) + 4 [using (i)]
⟹ n(B - A) = 26 - 4
= 22
∴ n (B - A) = 22
Let n(P) denote the total percentage of Indians n(O) denotes
the percentage of Indians who like oranges, and n(B) denotes
the percentage of Indians who like bananas.
Then, n(P) = 100, n(o) = 76 and n(B) = 62
To find: n (0 ∩ B)
Now,
n (P) = n(o) + n (B) - n (O ∩ B)
⟹ 100 = 76 + 62 - n (O ∩ B)
⟹ 100 = 138 - n(O ∩ B)
⟹ n(O ∩ B) = 138 - 100
= 38
∴ 38 % of Indians like both oranges and bananas.
(i) Let, n(P) denote the total number of persons,
n (H) denote the number of persons who speak Hindi and
n (E) denote the number of persons who speak English.
Then,
n (P) = 950, n (H) = 750, n(E) = 460 To find: n (H ∩ E)
n(P) = n(H) + n(E)- n(H ∩ E)
⟹ 950 = 750 + 460 - n(H ∩ E)
⟹ 950 = 2110 - n(H ∩ E)
⟹ n(H ∩ E) = 2110 - 950
= 260
Hence, 260 persons can speak both Hindi and English.
(ii) Clearly H is the disjoint union of H - E & H ∩ E i.e. H = (H- E) ∪ (H ∩ E)
∴ n(H) = n(H - E) + n(H ∩ E) [∵ If A 8c B are disjoint then n (A ∪ B) = n (A) + n(8)]
⟹ 750 = n(H - E) + 260
⟹ n(H - E) = 750 - 260
= 490
Hence, 490 persons can speak Hindi only.
(iii) On a similar lines we have
E = (E - H) ∪ (H ∩ E)
i.e. E is the disjoint union of E - H & H ∩ E
∴ n (E) = n (E - H)+ n(H ∩ E)
⟹ 460 = n (E - H) + 260
⟹ n(E- H) = 460 - 260
= 200
Hence, 200 persons can speak English only.
(i) Let,
n (P) denote the total number of persons,
n (T) denote number of persons who drink tea and
n (C) denote number of persons who drink coffee.
Then, n (P) = 50, n(T-C) = 14, n(T) = 30
To find: n (T ∩ C)
Clearly T is the disjoint union of T- C and T ∩ C
∴ T = (T- C) ∪ (T ∩ C)
∴ n(T) = n(T - C) + n (T ∩ C)
⟹ 30 = 14 + n(T ∩ C)
⟹ n(T ∩ C) = 30 – 14
= 16
Hence, 16 persons drink tea and coffee both.
(ii) To find: C - T
We know n (P) = n (C) + n (T) - n (T ∩ C)
⟹ 50 = n(C) + 30 – 16
⟹ 50 - n (C) + 14
⟹ n(C) = 50 – 14 = 36
New C is the disjoint union of C - T and T ∩ C
∴ C = (C - T) ∪ (C ∩ T)
⟹ n(C) = n (C - T)+ n (C ∩ T)
⟹ 36 = n(C- T)+ 16 [∵ n (T ∩ C) = n (C ∩ T) = 16]
⟹ n(C - T) = 36 - 16
= 20
Hence, 20 persons drink coffee but not tea.