NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6 is part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 1 Sets Ex 1.6.
- Sets Class 11 Ex 1.1
- Sets Class 11 Ex 1.2
- Sets Class 11 Ex 1.3
- Sets Class 11 Ex 1.4
- Sets Class 11 Ex 1.5
- Sets Class 11 Miscellaneous Exercise
| Board | CBSE |
| Textbook | NCERT |
| Class | Class 11 |
| Subject | Maths |
| Chapter | Chapter 1 |
| Chapter Name | Sets |
| Exercise | Ex 1.6 |
| Number of Questions Solved | 8 |
| Category | NCERT Solutions |
NCERT Solutions for Class 11 Maths 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).
Solution.
Here 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.
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.
Here 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.
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.
Question 4.
If 5 and Tare two sets such that 5 has 21 elements, T has 32 elements, and S ∩T has 11 elements, how many elements does S ∪ T have?
Solution.
Here n(S) = 21, n(T) = 32 and n(S ∩T) = 11
We know that
n(S ∪ T) = n(S) + n(T) – n(S ∩ 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 X ∩ Y has 10 elements, how many elements does X have?
Solution.
Here 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.
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.
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. Here n(Q) = 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 j Now number of people who like tennis only and not cricket
=n(T – C) = n(T) – n(C ∩ T) = 35 – 10 = 25.
Question 8.
In a committee, 50 people speak French, 20 f 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 people who speak French and S be the set of people who speak Spanish.
Here 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 = 60
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