Sets – Class XI – Exercise 1.6

  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

Therefore n(X ∩Y) = 2


  1. 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 U Y) = 18, n(X) = 8, n(Y) = 15

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


  1. 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(Hz 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.


  1. 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:

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.


  1. 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.


  1. 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 and  (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.


 

 

 

One thought on “Sets – Class XI – Exercise 1.6

  1. Reblogged this on HarsH ReaLiTy and commented:
    What if my answer is a letter. That’s bad huh… 😉 I have huge respect for people that hack at numbers all day! I look at computer code instead. 😄
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