Sets Exercise 1.6 Solutions – NCERT Class 11 Mathematics Chapter 1

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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)
We know that
n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)
Replacing given values in this formula
38 + n(X ∩ Y) = 17 + 23
38 + n(X ∩ Y) = 40
n(X ∩ Y) = 40 – 38
n(X ∩ Y) = 2


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?
We know that
n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)
Replacing given values in this formula
18 + n(X ∩ Y) = 8 + 15
18 + n(X ∩ Y) = 23
n(X ∩ Y) = 23 – 18
n(X ∩ Y) = 5


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?
Suppose that
X = Set of Hindi speaking people
Y = Set of English speaking people
X ∪ Y = Set of total people

As per question
n(X) = 250
n(Y) = 200
n(X ∪ Y) = 400

We know that
n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)
400 + n(X ∩ Y) = 250 + 200
400 + n(X ∩ Y) = 450
n(X ∩ Y) = 450 – 400 = 50
n(X ∩ Y) = 50

Hence there are 50 people in the group who can speak both English and Hindi.


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?
As per question
n(S) = 21
n(T) = 32
n(S ∩ T) = 11

We know that
n(S ∪ T) + n(S ∩ T) = n(S) + n(T)
n(S ∪ T) + 11 = 21 + 32
n(S ∪ T) + 11 = 53
n(S ∪ T) = 53 – 11 = 42

Hence number of elements in S ∪ T is 42


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?
As per question
n(X) = 40
n(X ∪ Y) = 60
n(X ∩ Y) = 10

We know that
n(X ∪ Y) + n(X ∩ Y) = n(X) + n(Y)
60 + 10 = 40 + n(Y)
70 = 40 + n(Y)
70 – 40 = n(Y)
30 = n(Y)

Hence number of elements in Y are 30


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’s suppose that
C = Set of people who like coffee
T = Set of people who like Tea
C ∪ T = Set of people who likes at least one of two drinks = Set of all people in group

n(C) = 37
n(T) = 52
n(C ∪ T) = 70

We know that
n(C ∪ T) + n(C ∩ T) = n(C) + n(T)
70 + n(C ∩ T) = 37 + 52
70 + n(C ∩ T) = 89
n(C ∩ T) = 89 – 70 = 19
n(C ∩ T) = 19

So out of 70 people in the group, there are 19 people who both likes coffee and tea.


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’s suppose that
C = Set of people who like cricket
T = Set of people who like tennis
C ∪ T = Set of people in the group
C ∩ T = Set of people who like cricket and tennis

As per question
n(C) = 40
n(C ∪ T) = 65
n(C ∩ T) = 10

We know that
n(C ∪ T) + n(C ∩ T) = n(C) + n(T)
65 + 10 = 40 + n(T)
75 = 40 + n(T)
75 – 40 = n(T)
35 = n(T)

So Number of people who like Tennis are 35
Let’s now find out Number of people who like Tennis but not Cricket
In terms of Sets we need to find out n(T ∩ C’)
We know that
n(A ∩ B’) = n(A) – n(A ∩ B) where A, B are two given sets
Similarly
n(T ∩ C’) = n(T) – n(T ∩ C)
Putting in the values
n(T ∩ C’) = 45 – 10 = 35
So number of people who only like Tennis but not Cricket are 35


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?
Suppose that
F = Set of people who speak French
S = Set of people who speak Spanish
F ∩ S = Set of people who speak both French and Spanish

As per question
n(F) = 50
n(S) = 20
n(F ∩ S) = 10

We know that
n(F ∪ S) + n(F ∩ S) = n(F) + n(S)
n(F ∪ S) + 10 = 50 + 20 = 70
n(F ∪ S) + 10 = 70
n(F ∪ S) = 70 – 10 = 60
n(F ∪ S) = 60
So number of people who speak at least one of French or Spanish are 60


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