Class 11 NCERT Solutions- Chapter 1 Sets – Miscellaneous Exercise on Chapter 1 | Set 1

Question 1: Decide, among the following sets, which sets are subsets of one and another:

A = {x : x ∈ R and x satisfy x²– 8x + 12 = 0}, B = {2, 4, 6},  C = {2, 4, 6, 8, . . . }, D = {6} 

Solution:

At first, simplifying for set A 
x²– 8x + 12 = 0 
(x-6)(x-2) = 0 
x= 6 or 2 
Now, A = {2,6} 
A set X is said to be a subset of a set Y if every element of X is also an element of Y. 
Therefore, we can write : A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

Question 2: In each of the following, determine whether the statement is true or false.

If it is true, prove it. If it is false, give an example.

(i) If x ∈ A and A ∈ B, then x ∈ B

(ii) If A ⊂ B and B ∈ C, then A ∈ C

(iii) If A ⊂ B and B ⊂ C, then A ⊂ C

(iv) If A ⊄ B and B ⊄ C, then A ⊄ C

(v) If x ∈ A and A ⊄ B, then x ∈ B

(vi) If A ⊂ B and x ∉ B, then x ∉ A

Solution:

(i) False
Example : Let x=1, A={1,2,3} and B = {{1,2,3},{4,5,6}} 
Here x ∈ A and A ∈ B but x ∉ B 
(ii) False
Example : Let A = {1}, B = {1,3} and C = {{1,3},{5,7}} 
Here A ⊂ B and B ∈ C but A ∉ C 
(iii) True
Proof : A ⊂ B : Set B contains all the elements in set A 
B ⊂ C : Set C contains all the elements in set B 
Hence, by transitivity property, set C contains all the elements in set A i.e. A ⊂ C 
(iv) False
Example : Let A = {1,2}, B = {3,4,5} and C = {1,2,6,7} 
Here A ⊄ B and B ⊄ C but A ⊂ C 
(v) False
Example : Let x=1, A = {1,2} and B = {3,4,5} 
Here x ∈ A and A ⊄ B but x ∉ B 
(vi) True
Proof : A ⊂ B : Set B contains all the elements in set A 
So if x ∉ B then also x ∉ A

Question 3: Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.

Solution:

Given : A ∪ B = A ∪ C …(1) 
A ∩ B = A ∩ C …(2) 
A ∪ B = A + B – (A ∩ B) // by principle of inclusion-exclusion 
A ∪ C = A + C – (A ∩ C) // by principle of inclusion-exclusion 
A + B – (A ∩ B) = A + C – (A ∩ C) // from (1) 
A + B – (A ∩ B) = A + C – (A ∩ B) // from (2) 
B = C 

Question 4: Show that the following four conditions are equivalent :

(i) A ⊂ B (ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A

Solution:

//Showing (i)=(ii) 
A ⊂ B means Set B contains all the elements in set A i.e. set A does not have any different element from B 
It means A – B = φ 
//Showing (i)=(iii) 
All elements of set A are there in Set B so A ∪ B = B 
//Showing (i)=(iv) 
All elements of set A are there in Set B so A ∩ B = A 
From above explanation we can say that all above conditions are equivalent.

Question 5: Show that if A ⊂ B, then C – B ⊂ C – A

Solution:

//By taking an example 
Let A = {1,2} and B = {1,2,3,4,5} 
and C = {2,5,6,7,8} 
Set B contains all the elements in set A so A ⊂ B 
Now C-B = {6,7,8} //elements present in C but not in B 
C-A = {5,6,7,8} //elements present in C but not in A 
It is clearly seen that, Set C-A contains all the elements in Set C-B hence C – B ⊂ C – A is proved.

Question 6: Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)

Solution:

(A ∩ B) ∪ (A – B) 
(A ∩ B) ∪ (A ∩ B’) //(A – B) = (A ∩ B’) 
A ∩ (B ∪ B’) 
A ∩ U //B ∪ B’ = U where U represents universal set 
A //by identity property 
Hence, it is proved that A = (A ∩ B) ∪ (A – B)

A ∪ (B – A) 
A ∪ (B ∩ A’) 
//B – A = B ∩ A’ 
(A ∪ B) ∩ (A ∪ A’) 
//distributive law 
(A ∪ B) ∩ U 
//A ∪ A’ = U where U represents universal set 
(A ∪ B) 
Hence, it is proved that A ∪ (B – A) = (A ∪ B)

Question 7: Using properties of sets, show that

(i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A

Solution:

(i) A ∪ (A ∩ B) 
(A ∪ A) ∩ (A ∪ B) //distributive law 
A ∩ (A ∪ B) //A ∪ A = A 
A //by absorption law 

(ii) A ∩ (A ∪ B) 
(A ∩ A) ∪ (A ∩ B) //distributive law 
A ∪ (A ∩ B) //A ∩ A = A 
A //by absorption law

Question 8: Show that A ∩ B = A ∩ C need not imply B = C

Solution:

Let us assume that B ≠ C 
Take A = {1,2}, B ={2,3} and C = {3,4} // here B ≠ C 
A ∩ B = {2} 
A ∩ C = φ 
Here we can see that A ∩ B ≠ A ∩ C 
Therefore, our assumption was wrong 
Hence, B = C is must for A ∩ B = A ∩ C

Question 9: Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B

Solution:

A = A ∩ (A ∪ X) //absorption law 
A = A ∩ (B ∪ X) //given that A ∩ X = B ∩ X 
= (A ∩ B) ∪ (A ∩ X) //distributive law 
= (A ∩ B) ∪ φ //given that A ∩ X = φ 
A = (A ∩ B) …(1) 
Repeating above process by taking B = B ∩ (B ∪ X) 
We get B = (A ∩ B) …(2) 
from (1) and (2) 
A = B

Question 10: Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ

Solution:

(A ∩ B) should be non-empty means atleast one element is common in between them 
… same for (B ∩ C) & (A ∩ C) 
A ∩ B ∩ C = φ means there should not be any element common in all the three sets A, B and C 
Let A = {1,2} B = {2,3} and C = {1,3} 
A ∩ B = {2}
B ∩ C = {3}
A ∩ C = {1}
and A ∩ B ∩ C = φ

Deleted Questions from NCERT

Assume that P (A) = P (B). Show that A = B

Solution:

P(X) represents power set of set X 
To prove A = B we have to prove that A ⊂ B and B ⊂ A 
(eg : if A = {1,2} then P(A) = {φ, {1}, {2}, {1,2}}) 
Power set of any set contains all the possible subsets of it. 
A ∈ P(A) 
as P(A)=P(B) so A ∈ P(B) 
If A is P(B) then clearly A is subset of B. 
A ⊂ B …(1) 
Repeating above process for B ∈ P(B) we get 
B ⊂ A …(2) 
From above equations, 
A = B

Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer

Solution:

It is False
Let A = {1,2} and B = {2,3} 
A ∪ B = {1,2,3} 
P(A) = {φ, {1}, {2}, {1,2}} 
P(B) = {φ, {2}, {3}, {2,3}} 
P(A) ∪ P(B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}} …(1) 
P(A ∪ B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}} …(2) 
P(A) ∪ P(B) ≠ P(A ∪ B) //from (1) and (2)

In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Solution:

There are total 600 students 
Let A and B represents sets of students taking tea and coffee respectively 
n(A) = 150 
n(B) = 225 
Students taking both tea and coffee = n(A ∩ B) = 100 
Students taking either tea or coffee = n(A ∪ B) = ? 
By principle of inclusion-exclusion, 
n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 
= 150 + 225 – 100 
= 275 
Now, 
Number of students who neither take tea nor coffee = total students – Number of Students who either take tea or coffee 
= 600 – 275 
= 325 
There are 325 students who neither take tea nor coffee 

In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Solution:

Let A and B represents sets of students who knows Hindi and English respectively 
n(A) = 100 
n(B) = 50 
Number of students who know both languages = n(A ∩ B) = 25 
It is given that each student knows either Hindi or English 
Hence, Number of Students in a group = n(A ∪ B) 
By principle of inclusion-exclusion, 
n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 
= 100 + 50 – 25 
= 125 
Total students in the group are 125

In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find: 

(i) the number of people who read at least one of the newspapers.

(ii) the number of people who read exactly one newspaper

Solution:

(i) Total number of people in survey = 60 
Let H, I, T represents sets of people reading newspaper H, I and T respectively 
n(H) = 25 
n(I) = 26 
n(T) = 26 
People reading both H and I = n(H ∩ I) = 9 
People reading both H and T = n(H ∩ T) = 11 
People reading both T and I = n(T ∩ I) = 8 
People reading all the three newspapers = n(H ∩ I ∩ T) = 3 
(i) Number of people who read at least one of the newspapers is given by n(H ∪ I ∪ T) 
By principle of inclusion-exclusion, 
n(H ∪ I ∪ T) = n(H) + n(I) + n(T) – n(H ∩ I) – n(H ∩ T) – n(T ∩ I) + n(H ∩ I ∩ T) 
= 25 + 26 + 26 – 9 – 11 – 8 + 3 
= 52 
Number of people who read at least one of the newspapers is 52 

(ii) Take a look at following Venn diagram

Number of people who read exactly one newspaper are represented by green color in above diagram 
= n(H ∪ I ∪ T) – n(H ∩ I) – n(H ∩ T) – n(T ∩ I) + (2 x n(H ∩ I ∩ T)) 
= 52 – 9 – 11 – 8 + (2 x 3) 
= 24 + 6 
= 30 
Number of people who read exactly one newspaper are 30

In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.

Solution:

Let A, B, C represents sets of people who liked product A, B and C respectively 
n(A) = 21 
n(B) = 26 
n(C) = 29 
People who liked product A and B both = n(A ∩ B) = 14 
People who liked product A and C both = n(A ∩ C) = 12 
People who liked product B and C both = n(B ∩ C) = 14 
People who liked all the three products = n(A ∩ B ∩ C) = 8 
Take a look at following Venn diagram

Number of people who only like product C are represented by green colour 
= n(C) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C) 
= 29 – 12 – 14 + 8 
= 11 
Number of people who only like product C are 11