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**Sets - NCERT Exercise 1.6 Answers **

**Class 11 Maths**

**Q1: 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: Given,

n( X ) = 17,

n( Y ) = 23 and

n( X ∪ Y ) = 38

We know that, n(X ∪ Y) = n(X) + n(Y) - n (X ∩ Y)

∴ 38 = 17 + 23 - n (X ∩ Y)

⇒ n (X ∩ Y) = 17 + 23 - 38 = 40 - 38 = 2

∴ n (X ∩ Y) = 2.

**Q2: 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: Given,

n( X ) = 8,

n( Y ) = 15 and

n( X ∪ Y ) = 18

∵ n(X ∪ Y) = n(X) + n(Y) - n (X ∩ Y)

∴ 18 = 8 + 15 - n (X ∩ Y)

⇒ n (X ∩ Y) = 8 + 15 - 18 = 23 - 18 = 5

∴ n (X ∩ Y) = 5.

**Q3: 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) = 250

n(E) = 200

n( H ∪ E ) =400

n (H ∩ E) = ?

∵ n(H ∪ E) = n(H) + n(E) - n (H ∩ E)

∴ 400 = 250 + 400 - n(H ∩ E)

∴ n(H ∩ E) = 450 - 400 = 50.

Thus 50 people can speak Hindi and English both.

**Q4: 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: Given

n(S) = 21,

n(T) = 32

n(S ∩ T) = 11

∵ n(S ∪ T) = n(S) + n(T) - n (S ∩ T)

∴ n(S ∪ T) = 21 + 32 - 11 = 42 (answer)

**Q5: 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: Given,

n(X) = 40,

n(Y) = ?

n(X ∪ Y) = 60

X ∩ Y = 10

∵ n(X ∪ Y) = n(X) + n(Y) - n (X ∩ Y)

∴ 60 = 40 + n(Y) - 10

∴ n(Y) = 60 -30 = 30

Thus, the set Y has 30 elements.

**Q6: 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) = 37, n(T) = 52 and n(C ∪ T ) = 70.

∵ n(C ∪ T) = n(C) + n(T) - n (C ∩ T)

∴ 70 = 37 + 52 - n (C ∩ T)

⇒ n (C ∩ T) = 89 - 70 = 19

Thus, 19 people like both coffee and tea.

**Q7: 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(T) = ? and n(C ∩ T ) = 10.

∵ n(C ∪ T) = n(C) + n(T) - n (C ∩ T)

∴ 65 = 40 + n(T) - 10

⇒ n(T) = 65 - 30 = 35

Therefore, 35 people like tennis.

∵ n(T) = n(T - C) + n(T ∩ C)

∴ n(T - C) = n(T) - n(T ∩ C)

= 35 - 10 = 25

Thus, 25 people like only tennis.

**Q8: 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 and n(S ∩ F ) = 10.

∵ n(S ∪ F) = n(S) + n(F) - n (S ∩ F)

∴ n(S ∪ F) = 20 + 50 - 10 = 70 - 10 = 60

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

**☞See also:**

Special Mathematical Constants

SETS (Unit Test Paper)

SETS (VENN DIAGRAMS)

SETS (Operations of Sets)

SETS (NCERT Ex 1.4 Q1-Q5)

SETS (NCERT Ex 1.4 Q6 - Q8)

SETS (NCERT Ex 1.4 Q9 - Q12)

SETS (NCERT Ex 1.5)

Laws of Set Operations

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