Chapter 1: Sets Miscellaneous Exercise

According to the question,

Let A and B be two sets such that A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X.

To show, A = B

Proof:

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]

= (A ∩ B) ∪ (A ∩ X) [Distributive law]

= (A ∩ B) ∪ Φ [A ∩ X = Φ]

= A ∩ B (i)

Now, B = B ∩ (B ∪ X)

= B ∩ (A ∪ X) [A ∪ X = B ∪ X]

= (B ∩ A) ∪ (B ∩ X) … [Distributive law]

= (B ∩ A) ∪ Φ [B ∩ X = Φ]

= A ∩ B (i)

Hence, from equations (i) and (ii), we obtain A = B

Solution coming soon.

Let us assume that,

U = The set of all students who took part in the survey

T = The set of students taking tea

C = The set of students taking coffee

Total number of students in a school, n (U) = 600

Number of students taking tea, n (T) = 150

Number of students taking coffee, n (C) = 225

Also, n (T ∩ C) = 100

Now, we have to find the number of students taking neither coffee nor tea, i.e., n (T ∩ C’)

∴ According to the question,

n ( T ∩ C’ )= n( T ∩ C )’

= n (U) – n (T ∩ C)

= n (U) – [n (T) + n(C) – n (T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

∴ Number of students taking neither coffee nor tea = 325 students

Let us assume that,

U = The set of all students in the group

E = The set of students who know English

H = The set of students who know Hindi

∴ H ∪ E = U

Given that,

Number of students who know Hindi n (H) = 100

Number of students who know English, n (E) = 50

Number of students who know both, n (H ∩ E) = 25

We have to find the total number of students in the group, i.e., n (U)

∴ According to the question,

n (U) = n(H) + n(E) – n(H ∩ E)

= 100 + 50 – 25

= 125

∴ Total number of students in the group = 125 students

(i) Let us assume that,

A = The set of people who read newspaper H

B = The set of people who read newspaper T

C = The set of people who read newspaper I

According to the question,

Number of people who read newspaper H, n (A) = 25

Number of people who read newspaper T, n (B) = 26

Number of people who read the newspaper I, n (C) = 26

Number of people who read both newspaper H and I, n (A ∩ C) = 9

Number of people who read both newspaper H and T, n (A ∩ B) = 11

Number of people who read both newspaper T and I, n (B ∩ C) = 8

And, the number of people who read all three newspapers H, T and I, n (A ∩ B ∩ C) = 3

Now, we have to find the number of people who read at least one of the newspapers

∴, we get

= 25 + 26 + 26 – 11 – 8 – 9 + 3

= 80 – 28

= 52

∴ There are a total of 52 students who read at least one newspaper.

(ii) Let us assume that,

a = The number of people who read newspapers H and T only

b = The number of people who read newspapers I and H only

c = The number of people who read newspapers T and I only

d = The number of people who read all three newspapers

According to the question,

D = n(A ∩ B ∩ C) = 3

Now, we have

n(A ∩ B) = a + d

n(B ∩ C) = c + d

And,

n(C ∩ A) = b + d

∴ a + d + c +d + b + d = 11 + 8 + 9

a + b + c + d = 28 – 2d

= 28 – 6

= 22

∴ The number of people who read exactly one newspaper = 52 – 22

= 30 people

Let A, B and C = the set of people who like product A, product B and product C, respectively.

Now, according to the question,

Number of students who like product A, n (A) = 21

Number of students who like product B, n (B) = 26

Number of students who like product C, n (C) = 29

Number of students who like both products A and B, n (A ∩ B) = 14

Number of students who like both products A and C, n (C ∩ A) = 12

Number of students who like both products C and B, n (B ∩ C) = 14

Number of students who like all three products, n (A ∩ B ∩ C) = 8

From the Venn diagram, we get

Number of students who only like product C = {29 – (4 + 8 + 6)}

= {29 – 18}

= 11 students

The other exercise, NCERT Class 11 Maths Solutions Chapter 1, can be accessed using the links given below.

Exercise 1.1 Solutions 6 Questions

Exercise 1.2 Solutions 6 Questions

Exercise 1.3 Solutions 9 Questions

Exercise 1.4 Solutions 12 Questions

Exercise 1.5 Solutions 7 Questions

Exercise 1.6 Solutions 8 Questions