# NCERT Solutions for Class 11th Maths Exercise 1.6 Set

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CBSE NCERT Solutions For Class 11th Maths Chapter 1 : Set. NCERT Solutins For Class 11 Mathematics. Exercise 1.1, Exercise 1.2, Exercise 1.3, Exercise 1.4, Exercise 1.5, Exercise 1.6,  (Miscellaneous Excercise) many more solutions

Exercise 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). 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?

It is given that:

n(X ïf? 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?

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?

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?

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?

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?

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?

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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