Class 11 Maths Chapter 1 Sets Exercise 1.6 Solutions in English Medium

Free NCERT Solutions for Class 11 Maths Chapter 1 Exercise 1.6prepared by expert Mathematics teacher as per CBSE (NCERT) books guidelines.

Free NCERT Solutions for Class 11 Math’s Chapter 1 Sets Exercise 1.1, Exercise 1.2, Exercise 1.3, Exercise 1.4, Exercise 1.5, Exercise 1.6 and Miscellaneous Exercise in English Medium for CBSE.

NCERT Maths Class 11 Sets. Just click on the Exercise wise links given below to practice the Maths solutions for the respective exercise.

SETS	Solutions Link
Exercise 1.1	Click here
Exercise 1.2	Click here
Exercise 1.3	Click here
Exercise 1.4	Click here
Exercise 1.5	Click here
Exercise 1.6	Click here

Class 11 Maths Chapter 1 Sets Exercise 1.6 Solutions in English Medium

EXERCISE : 1.6

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 that,

n(X) = 17

n(Y) = 23

n (X∪ Y ) = 38

n ( X ∩ Y ) = ?

By using the formula,

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

We find that,

n(X ∩ Y) = n(x) + n(Y) – n(X ∪ Y)

= 17 + 23 – 38

= 40 – 38

= 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; haw many elements does X ∩ Y have ?

Solution:-

Given that,

n(X∪ Y) = 18

n(X) = 8

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

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?

Solution:-

Let, H be the set of people who speak Hindi, and E be the set of people who speak English .

n(H∪E) = 400, n(H) = 25, n(E) = 200

n(H∩E) = ?

We know that,

n((H∪E) = n(H) + n(E) – n(H∩E)

400 = 250 + 200 – n(H∩E)

n(H∩E) = 450 – 400

n(H∩E) = 50

Thus, 50 people can speak both Hindi and English.

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 that,

n(S) = 21

n(T) = 32

n(S∪T) = 11

we know that,

n(S∪T) = 21 + 32 – 11 = 4

Thus, the set (S∪T) has 42 elements.

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 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)

n(Y) = 60 – (40-10) = 30

Thus, the set Y has 30 elements.

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 denote the set of people who like coffee,

and T denote the set of people who like tea

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)

n(C∩T) = 89 – 70

n(C∩T) = 19

Thus 19 people like both 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?

Solution:-

Let C denote the set of people who like cricket,

and T denote the set of people who like tennis,

We know that,

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

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.

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 people in the committee who speak French, and S be the set of people in the committee who speak Spanish.

n(F) = 50

n(S) = 20

n(S∩F) = 10

We know that,

n(S∪F) = n(S) + n(F) – n(S∩F)

n(S∪F) = 20 + 50 – 10

n(S∪F) = 70 – 10 = 60

Thus, 60 people in the committee speak at least one the two languages.

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Class 11 Maths Chapter 1 Sets Exercise 1.6 Solutions in English Medium