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NCERT solution class 11 chapter 1 Sets exercise 1.6 mathematics

EXERCISE 1.6


Page No 24:

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).

Answer:

It is given that:

n(X) = 17, n(Y) = 23, n(X ∪ Y) = 38

n(X ∩ Y) = ?

We know that:

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?

Answer:

It is given that:

n(X ∩ Y) = ?

We know that:

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?

Answer:

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.

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?

Answer:

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.

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?

Answer:

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.

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 people like both coffee and tea?

Answer:

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.

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?

Answer:

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) = Φ

∴ 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.

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?

Answer:

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.


 

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