Class 11 1. Sets Exercise 1.6: NCERT Book Solutions
Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams.
NCERT Solutions
All chapters of ncert books Mathematics 1. Sets Exercise 1.6 is solved by exercise and chapterwise for class 11 with questions answers also with chapter review sections which helps the students who preparing for UPSC and other competitive exams and entrance exams.
Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams. - 1. Sets - Exercise 1.6: NCERT Book Solutions for class 11th. All solutions and extra or additional solved questions for 1. Sets : Exercise 1.6 Mathematics class 11th:English Medium NCERT Book Solutions. Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams.
1. Sets : Exercise 1.6 Mathematics class 11th:English Medium NCERT Book Solutions
Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams. - 1. Sets - Exercise 1.6: NCERT Book Solutions for class 11th. All solutions and extra or additional solved questions for 1. Sets : Exercise 1.6 Mathematics class 11th:English Medium NCERT Book Solutions.
Class 11 1. Sets Exercise 1.6: NCERT Book Solutions
NCERT Books Subjects for class 11th Hindi Medium
1. Sets
Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams.
Exercise 1.6
Exercise 1.6
Q1. 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 that
n ( X ) = 17, n ( Y ) = 23 and n ( X ∪ Y ) = 38
n ( X ∩ Y ) = ?
n(X ∪ Y ) = n(X) + n(Y) - n(X ∩ Y)
=> 38 = 17 + 23 - n(X ∩ Y)
=> 38 = 40 - n(X ∩ Y)
=> n(X ∩ Y) = 40 - 38
=> n(X ∩ Y) = 2
Q2. 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 that;
n ( X ) = 8, n ( Y ) = 15 and n ( X ∪ Y ) = 18
n ( X ∩ Y ) = ?
n(X ∪ Y ) = n(X) + n(Y) - n(X ∩ Y)
=> 18 = 8 + 15 - n(X ∩ Y)
=> 18 = 23 - n(X ∩ Y)
=> n(X ∩ Y) = 23 - 18
=> n(X ∩ Y) = 5
Q3. 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 the people can speak hindi be n(H) = 250,
The people can speak English be n(E) = 200 and
and Total people be n(H ∪ E) = 400
and people can speak both Hindi and English be n(H ∩ E) = ?
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
∴ 400 = 250 + 200 – 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.
Q4. 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 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.
Q5. 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 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)
= 60 - 30
= 30
Thus, the set Y has 30 elements.
Q6. 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 then
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.
Q7. 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?
Solutions:
Let C denote the set of people who like cricket, and T denote the set of people who like tennis then-
∴ 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
Also,
(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.
Q8. 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 people in the committee who speak French, and
S be the set of people in the committee who speak Spanish then-
∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10
We know that: n(S ∪ F) = n(S) + n(F) – n(S ∩ F)
= 20 + 50 – 10
= 70 – 10 = 60
Thus, 60 people in the committee speak at least one of the two languages.
Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams.
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Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams. - 1. Sets - Exercise 1.6: NCERT Book Solutions for class 11th. All solutions and extra or additional solved questions for 1. Sets : Exercise 1.6 Mathematics class 11th:English Medium NCERT Book Solutions. Class 11 chapter 1. Sets ncert solution for board exams and term 1 and term 2 exams.
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Mathematics Chapter List
1. Sets
2. Relations and Functions
3. Trigonometric Functions
4. Principle Of Mathematical Induction
5. Complex Numbers and Quadratic Equations
6. Linear Inequalities
7. Permutations and Combinations
8. Binomial Theorem
9. Sequences and Series
10. Straight Lines
11. Conic Sections
12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives
14. Mathematical Reasoning
15. Statistics
16. Probability