RD Chapter 1- Sets Ex-1.1 |
RD Chapter 1- Sets Ex-1.2 |
RD Chapter 1- Sets Ex-1.3 |
RD Chapter 1- Sets Ex-1.4 |
RD Chapter 1- Sets Ex-1.5 |
RD Chapter 1- Sets Ex-1.6 |
RD Chapter 1- Sets Ex-1.7 |

**Answer
1** :

We have,

n (A ∪ B) = 50

n (A) = 28

n (B) = 32

We know, n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

Substituting the values we get

50 = 28 + 32 – n (A ∩ B)

50 = 60 – n (A ∩ B)

–10 = – n (A ∩ B)

∴ n (A ∩ B) = 10

**Answer
2** :

We have,

n (P) = 40

n (P ∪ Q) = 60

n (P ∩ Q) =10

We know, n (P ∪ Q) = n (P) + n (Q) – n (P ∩ Q)

Substituting the values we get

60 = 40 + n (Q)–10

60 = 30 + n (Q)

N (Q) = 30

∴ Q has 30 elements.

**Answer
3** :

We have,

Teachers teaching physics or math = 20

Teachers teaching physics and math = 4

Teachers teaching maths = 12

Let teachers who teach physics be ‘n (P)’ and for Maths be ‘n (M)’

Now,

20 teachers who teach physics or math = n (P ∪ M) = 20

4 teachers who teach physics and math = n (P ∩ M) = 4

12 teachers who teach maths = n (M) = 12

We know,

n (P ∪ M) = n (M) + n (P) – n (P ∩ M)

Substituting the values we get,

20 = 12 + n (P) – 4

20 = 8 + n (P)

n (P) =12

∴ There are 12 physics teachers.

**Answer
4** :

We have,

A total number of people = 70

Number of people who like Coffee = n (C) = 37

Number of people who like Tea = n (T) = 52

Total number = n (C ∪ T) = 70

Person who likes both would be n (C ∩ T)

We know,

n (C ∪ T) = n (C) + n (T) – n (C ∩ T)

Substituting the values we get

70 = 37 + 52 – n (C ∩ T)

70 = 89 – n (C ∩ T)

n (C ∩ T) =19

∴ There are 19 persons who like both coffee and tea.

Let A and B be two sets such that: n (A) = 20, n (A ∪ B) = 42 and n (A ∩ B) = 4. Find

(i) n (B)

(ii) n (A – B)

(iii) n (B – A)

**Answer
5** :

(i) n (B)

We know,

n (A ∪ B) = n (A) + n (B) – n (A ∩ B)

Substituting the values we get

42 = 20 + n (B) – 4

42 = 16 + n (B)

n (B) = 26

∴ n (B) = 26

(ii) n (A – B)

We know,

n (A – B) = n (A ∪ B) – n (B)

Substituting the values we get

n (A – B) = 42 – 26

= 16

∴ n (A – B) = 16

(iii) n (B – A)

We know,

n (B – A) = n (B) – n (A ∩ B)

Substituting the values we get

n (B – A) = 26 – 4

= 22

∴ n (B – A) = 22

**Answer
6** :

We have,

People who like oranges = 76%

People who like bananas = 62%

Let people who like oranges be n (O)

Let people who like bananas be n (B)

Total number of people who like oranges or bananas = n (O ∪ B) = 100

People who like both oranges and bananas = n (O ∩ B)

We know,

n (O ∪ B) = n (O) + n (B) – n (O ∩ B)

Substituting the values we get

100 = 76 + 62 – n (O ∩ B)

100 = 138 – n (O ∩ B)

n (O ∩ B) = 38

∴ People who like both oranges and banana is 38%.

In a group of 950 persons, 750 can speak Hindi and 460 can speak English. Find:

(i) How many can speak both Hindi and English.

(ii) How many can speak Hindi only.

(iii) how many can speak English only.

**Answer
7** :

We have,

Let, total number of people be n (P) = 950

People who can speak English n (E) = 460

People who can speak Hindi n (H) = 750

(i) How many can speak both Hindi and English.

People who can speak both Hindi and English = n (H ∩ E)

We know,

n (P) = n (E) + n (H) – n (H ∩ E)

Substituting the values we get

950 = 460 + 750 – n (H ∩ E)

950 = 1210 – n (H ∩ E)

n (H ∩ E) = 260

∴ Number of people who can speak both English and Hindi are 260.

(ii) How many can speak Hindi only.

We can see that H is disjoint union of n (H–E) and n (H ∩ E).

(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))

∴ H = n (H–E) ∪ n (H ∩ E)

n (H) = n (H–E) + n (H ∩ E)

750 = n (H – E) + 260

n (H–E) = 490

∴ 490 people can speak only Hindi.

(iii) How many can speak English only.

We can see that E is disjoint union of n (E–H) and n (H ∩ E)

(If A and B are disjoint then n (A ∪ B) = n (A) + n (B))

∴ E = n (E–H) ∪ n (H ∩ E).

n (E) = n (E–H) + n (H ∩ E).

460 = n (H – E) + 260

n (H–E) = 460 – 260 = 200

∴ 200 people can speak only English.

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