Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 1 Sets Ex 1.6 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 1 Sets Ex 1.6

Question 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:

Since, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y), so

38 = 17 + 23 – n(X ∩ Y)

⇒ n(X ∩ Y) = 17 + 23 – 38

= 40 – 38

= 2.

Question 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:

Here n(X) = 8, n(Y) = 15 and n(X ∪ Y) = 18.

Now, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y).

⇒ 18 = 8 + 15 – n(X ∩ Y)

⇒ n(X ∩ Y) = 8 + 15 – 18

= 5.

Question 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 and E denote the sets of people who can speak Hindi and English respectively.

Then, n(H) = 250, n(E) = 200 and n(H ∪ E) = 400.

Now, n(H ∩ E) = n(H) + n(E) – n(H ∪ E)

= 250 + 200 – 400

= 50.

Thus, 50 people can speak both Hindi and English.

Question 4.

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

Solution:

Here n(S) = 21, ra(T) = 32 and n(S ∩ T) = 11.

Now, n(S ∪ T) = n(S) + n(T) – n(S ∩ T)

= 21 + 32 – 11

= 53 – 11

= 42.

Question 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:

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

Now, n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y).

⇒ 60 = 40 + n(Y) – 10

⇒ n(Y) = 60 – 40 + 10

= 20 + 10

= 30.

∴ Y has 30 elements.

Question 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 like both coffee and tea?

Solution:

Let C be the set of people who like coffee and T be Lie set of people who like tea.

Then, 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)

⇒ n(C ∩ T) = 37 + 52 – 70

= 89 – 70

= 19.

∴ 19 people like both coffee and tea.

Question 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 be the set of people who like cricket and T be the set of people who like tennis.

Then, n(C ∪ T) = 65, n(C) 40, n(C ∩ T) = 10.

We know that

n(C ∪ T) = n(C) + n(T) – (C ∩ T)

So, 65 = 40 + n(T) – 10

⇒ n(T) = 65 – 40 + 10 = 35.

∴ Number of people who like only tennis

= n(T) – n( C ∩ T)

= 35 – 10 = 25.

∴ Number of people who like tennis only and not cricket = 25 and number of people who like tennis is 35.

Question 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 who speak French and S be the set of people who speak Spanish.

Then, n(F) = 50, n(S) = 20 and n(F ∩ S) = 10.

We know that:

n(F ∪ S) = n(F) + n(S) – n(F ∩ S)

= 50 + 20 – 10

= 70 – 10

= 60.

∴ 60 people speak at least one of these two languages.