# Class 11 Maths NCERT Solutions for Chapter 1 Sets Exercise 1.6

### Sets Exercise 1.6 Solutions

**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**

It is given that:

n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38

n(X ∩ Y) = ?

We know that:

n(X ∪ Y)= n(X) + n(Y) - n(X ∩ Y)

∴ 38 = 17 + 23 - n(X ∩ Y )

⇒ n (X ∩ Y) = 40 - 38 = 2

∴ n(X ∩ Y) = 2

**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**

It is given that :

n(X ∪ Y) = 18, n(X) = 8, n(Y) = 15

n(X ∩ Y) = ?

We know that :

n(X ∪ Y)= n(X) + n(Y) - n(X ∩ Y)

∴ 18 = 8 + 15 - n(X ∩ Y)

⇒ n(X ∩ Y) = 23 - 18 = 5

∴ n(X ∩ Y) = 5

**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**

Let H be the set of people who speak Hindi, and

E be the set of people who speak English

∴ *n*(H ∪ E) = 400, *n*(H) = 250, *n*(E) = 200*n*(H ∩ E) = ?

We know that:*n*(H ∪ E) = *n*(H) + *n*(E) – *n*(H ∩ E)

∴ 400 = 250 + 200 – *n*(H ∩ E)

⇒ 400 = 450 – *n*(H ∩ E)

⇒ *n*(H ∩ E) = 450 – 400

∴ *n*(H ∩ E) = 50

Thus, 50 people can speak both Hindi and English.

**4. If S and T are two sets such that S has 21 elements, T has 32 elements, andS ∩ T has 11 elements, how many elements does S ∪ T have?**

**Solution**

It is given that:

n(S) = 21, n(T) = 32, n(S ∩ T) = 11

We know that:*n* (S ∪ T) = *n* (S) + *n* (T) – *n* (S ∩ T)

∴ *n* (S ∪ T) = 21 + 32 – 11 = 42

Thus, the set (S ∪ T) has 42 elements.

**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**

It is given that:

n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10

We know that:

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

∴ 60 = 40 + n(Y) – 10

∴ n(Y) = 60 – (40 – 10) = 30

Thus, the set Y has 30 elements.

**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**

Let C denote the set of people who like coffee, and

T denote the set of people who like tea*n*(C ∪ T) = 70, *n*(C) = 37, *n*(T) = 52

We know that:*n*(C ∪ T) = *n*(C) + *n*(T) – *n*(C ∩ T)

∴ 70 = 37 + 52 – *n*(C ∩ T)

⇒ 70 = 89 – *n*(C ∩ T)

⇒ *n*(C ∩ T) = 89 – 70 = 19

Thus, 19 people like both coffee and tea.

**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**

Let C denote the set of people who like cricket, and

T denote the set of people who like tennis

∴ *n*(C ∪ T) = 65, *n*(C) = 40, *n*(C ∩ T) = 10

We know that:*n*(C ∪ T) = *n*(C) +* n*(T) – *n*(C ∩ T)

∴ 65 = 40 + *n*(T) – 10

⇒ 65 = 30 + *n*(T)

⇒ *n*(T) = 65 – 30 = 35

Therefore, 35 people like tennis.

Now,

(T – C) ∪ (T ∩ C) = T

Also,

(T – C) ∩ (T ∩ C) = Î¦

∴ *n* (T) = *n* (T – C) + *n* (T ∩ C)

⇒ 35 = *n* (T – C) + 10

⇒ *n* (T – C) = 35 – 10 = 25

Thus, 25 people like only tennis.

**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**

Let F be the set of people in the committee who speak French, and

S be the set of people in the committee who speak Spanish

∴ *n*(F) = 50, *n*(S) = 20, *n*(S ∩ F) = 10

We know that:*n*(S ∪ F) = *n*(S) + *n*(F) – *n*(S ∩ F)

= 20 + 50 – 10

= 70 – 10 = 60

Thus, 60 people in the committee speak at least one of the two languages.