NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6

# NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6

## NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6

NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6 are the part of NCERT Solutions for Class 11 Maths. Here you can find the NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6.

#### Ex 1.6 Class 11 Maths 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).

Solution.
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) = 40 – 38 = 2

#### Ex 1.6 Class 11 Maths 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?

Solution.
Given n(X
Y) = 18, n(X) = 8 and 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

Hence, X ∩ Y has 5 elements.

#### Ex 1.6 Class 11 Maths 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?

Solution.
Let H be the set of people speaking Hindi and E be the set of people speaking English.
n(H) = 250, n(E) = 200 and n(H E) = 400
We know that
n(H
E) = n(H) + n(E) – n(H ∩ E)
400 = 250 + 200 – n(H ∩ E)
n(H ∩ E) = 450 – 400 = 50

Hence, 50 people can speak both Hindi and English.

#### Ex 1.6 Class 11 Maths 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?

Solution.
Given n(S) = 21, n(T) = 32 and n(S ∩ T) = 11
We know that
n(S
T) = n(S) + n(T) – n(S ∩ T)
= 21 + 32 – 11 = 42

Hence, S T has 42 elements.

#### Ex 1.6 Class 11 Maths 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?

Solution.
Given n{X) = 40, n(X
Y) = 60 and 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 – 30 = 30

Hence, Y has 30 elements.

#### Ex 1.6 Class 11 Maths 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?

Solution.
Let C be the set of persons who like coffee and T be the set of persons who like tea.
n(C) = 37, n(T) = 52 and n(C T) = 70
We know that
n(C
T) = n(C) + n(T) – n(C ∩ T)
70 = 37 + 52 – n(C ∩ T)
n(C ∩ T) = 89 – 70 = 19

Hence, 19 people like both coffee and tea.

#### Ex 1.6 Class 11 Maths 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?

Solution.
Let C be the set of people who like cricket and T be the set of people who like tennis. Given, n(C) = 40, n(C ∩ T) = 10 and n(C
T) = 65
We know that
n(C
T) = n(C) + n(T) – n(C ∩ T)
65 = 40 + n(T) – 10
n(T) = 65 – 30 = 35
Number of people who like tennis = 35

Now, the number of people who like tennis only and not cricket
= n(T – C) = n(T) – n(C ∩ T) = 35 – 10 = 25

#### Ex 1.6 Class 11 Maths 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?

Solution.
Let F be the set of the people who speak French and S be the set of the people who speak Spanish.
Given, n(F) = 50, n(S) = 20 and n(F ∩ S) = 10
We know that
n(F
S) = n(F) + n(S) – n(F ∩ S)
n(F
S) = 50 + 20 – 10 = 60
Number of people who speak at least one of these two languages is 60.

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