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Chapter 1: Sets – Exercise 1.8

Sets – Exercise 1.8 – Q.1

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

 

Sets – Exercise 1.8 – Q.2

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.

 

Sets – Exercise 1.8 – Q.3

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.

 

Sets – Exercise 1.8 – Q.4

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.

 

Sets – Exercise 1.8 – Q.5(i)

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

 

Sets – Exercise 1.8 – Q.5(ii)

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

 

Sets – Exercise 1.8 – Q.5(iii)

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

 

Sets – Exercise 1.8 – Q.6

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.

 

Sets – Exercise 1.8 – Q.7

(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.

 

Sets – Exercise 1.8 – Q.8

(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.


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