RBSE Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6 Textbook Exercise Questions and Answers.

RBSE Class 11 Maths Solutions 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).
Answer:
Given
n(X) = 17, n(Y) =23, n(X ∪ F) = 38
n(X ∩ F) = ?
We know that
n(X ∪ Y) = n(X) + n(Y) - n(X ∩ Y)
Thus 38 = 17 + 23 - n(X ∩ Y)
or 38 = 17 + 23 - n(X ∩ F)
n(X ∩ Y) = 40-38
Thus, n(X ∩ F) = 2

Question 2.
If X and Y are two sets such that luF has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have? Solution: Given
n(X ∪ Y) = 18, n(X) = 8, n(Y) = 15 n(X ∩ Y) = ?
∵ n(X ∪ Y) = n(X) + n(Y) - n(X ∩ Y)
18 = 8 + 15 - n(X ∩ Y)
Thus, n(X ∩ Y) = 23 - 18 = 5

Question 3.
In a group of 400 persons, 250 can speak Hindi and 200 can speak English. How many persons can speak both Hindi and English?
Answer:
Let H is set of those persons who speaks Hindi and A is set of those persons who speaks English.
Questionwise, n(H ∪ A ) = 400
n(H) = 250,
n(A) = 200
n(H ∩ A ) = Number of persons who speaks both Hindi and English

Using formula, n(H ∪ A) = n(H) + n(A) - n(H ∩ A)
400 = 250 + 200 - n(H ∩ A)
∴ n(H ∩ A) = 450 - 400
= 50
Thus, 50 person 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 nT has 11 elements how many elements does SuT have?
Answer:
Given,
n(S) = 21, n(T)= 32, n(S ∩ T) = 11 n(S ∪ T) = ?
Using formula n(S ∪ T) = n(S) + n(T) - n(S ∩ T)
= 21 + 32 - 11 = 42

Question 5.
If X and Y are two sets such that X has 40 elements, X ∪ Y has 60 elements and Inf has 10 elements, how many elements does Y have?
Answer:
Given,
n(X) = 40, n(X ∪ Y) - 60, n(X ∩ Y) = 10,
Then n(T) = ?
Using formula n( X ∪ f) = n( X) + n(Y) - n(X ∩ Y)
60 = 40 + n(7) - 10
or 60 + 10 - 40 = n(Y)
n(Y) = 30
Thus, number of elements in Y, n(Y ) = 30

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 be set of persons who like coffee and T be set of persons who like tea. .
Given,
n(C ∪ T)- 70, n(C) = 37, n(T) = 52
n(C ∩ T) = ?
Using formula, n(C ∪ T) = n(C) + n(T) - n(C ∩ T)
70 = 37 + 52 - n(C ∩ T)
H(C ∩ T) = 89 - 70 = 19
Thus, 19 persons 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 be set of persons who like cricket and T be set of persons who like tennis.
Given, '
∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10
Using formula n(C ∪ T) = n(C) + n(T) - n(C ∩ T)
65 = 40 + n(T) -10
⇒ 65 = 30 + n(T)
n(T)= 35
35 persons like tennis.

Number of persons who like tennis but not cricket = n(T) - n(T ∩ C)
= 35 - 10
= 25 persons

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 such persons who can speak French and S be thest of such persons who can speek Spanish. Questionwise,
n(F) = 50, n(S) = 20, n(F ∩ S) = 10 n(F ∪ S) = ?
Using formula, n(F ∪ S) = n(F) + n(S) - n(F ∩ S)
n(F ∪ S) = 50 + 20 - 10
= 70 - 10 =60
Thus, 60 persons can speak at least one of these two languages.