Sets questions with solutions are given here for students to make them understand the concept easily. Practising these problems will help to go through the concept of sets theory. It is an important chapter for Class 11 students, hence we have given the questions based on the NCERT curriculum, with respect to the CBSE syllabus.
A brief introduction for each sub-topic of sets is also provided so that students can solve the questions quickly. Some of the basic definitions related to this concept are given below. Also, read Sets For Class 11, here.
Sets: A collection of well-defined objects. It is denoted by capital letters.
Example: A = {1, 2, 3, 4, 5..}.
Here, set A is a collection of all natural numbers.
Roster form of sets: All elements are written in curly braces { }, separated by commas.
Example: R = {1, 3, 7, 21, 2, 6, 14, 42}.
Set Builder Form: The elements of the set represent a common property.
Example, R = {x : x is a vowel in English alphabet}
First, let us see some questions based on the representation of sets.
Questions on Sets with Solutions
1. Write the solution set of the equation x^{2} – 4=0 in roster form.
Solution: x^{2} – 4 = x^{2} – 2^{2} = (x – 2) (x + 2)
x = 2, -2
Thus, A = {-2, 2}
2. Write the set A = {1, 4, 9, 16, 25, . . . } in set-builder form.
Solution: If we see the pattern here, the numbers are squares of natural numbers, such as:
1^{2} = 1
2^{2} = 4
3^{2} = 9
4^{2} = 16
And so on.
A = {x : x is the square of a natural number}
Or we can write;
A = {x : x = n^{2} , where n ∈ N}
Empty Set: A set with no elements. Also called a void set or null set
Finite & Infinite Set: A set with a definite number of elements (even with zero elements) is a finite set otherwise the set is infinite Equal sets: Two sets that have the same elements are called equal sets. |
3. Write an example of a finite and infinite set in set builder form.
Solution:
Finite set, A = {x : x ∈ N and (x – 1) (x – 2) = 0}
Infinite Set, B = {x : x ∈ N and x is prime}
4. Write an example of equal sets.
Solution: Let there be two sets A and B
A is the set of letters in “ALLOY”
B is the set of letters in “LOYAL”
Hence,
A = {A,L, O,Y}
B = {L,O,Y,A}
Therefore, in both sets, the elements are the same.
So, A = B.
Subsets: A set A is said to be a subset of set B if every element of A is also an element of B
Symbolically, A ⊂ B if a ∈ A ⇒ a ∈ B. Power Set: Collection of all subsets of a set. |
5. Write the subsets of {1,2,3}.
Solution: Let A = {1, 2, 3}
The subsets of A are: φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}
6. Write {x: x ∈ R, 3 ≤ x ≤ 4} as an interval.
Solution: {x: x ∈ R, 3 ≤ x ≤ 4} = [3, 4]
7. Write the interval (6, 12) in set builder form.
Solution:
Let A be the interval (6, 12).
The interval (6, 12) in set builder form is
A = {x: x ∈ R, 6 < x < 12}
8. If set A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}. Then write the universal set for all three sets.
Solution: If U is the universal set for sets A, B and C, then: U = Elements of set A + Elements of set B + Elements of set C
U = {0, 1, 2, 3, 4, 5, 6, 8}
Also, read: Sets Subset And Superset
Union of sets: If A and B are two sets, then A union B will have all the elements of set A and set B. It is represented as A ∪ B.
Intersection of sets: If A and B are two sets, then A intersection B will have common elements of set A and set B. It is represented as A ∩ B. |
9. If A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∪ B.
Solution: A ∪ B = { 2, 4, 6, 8, 10, 12}
10. If A = { 2, 4, 6, 8} and B = { 6, 8, 10, 12}. Find A ∩ B.
Solution: A ∩ B = { 6, 8 }
Also, see: Operation On Sets Intersection Of Sets And Difference Of Two Sets
Properties of Union of Sets | Properties of Intersection of sets |
A ∪ B = B ∪ A (Commutative law) | A ∩ B = B ∩ A (Commutative law) |
(A ∪ B ) ∪ C = A ∪ ( B ∪ C) (Associative law ) | ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) (Associative law) |
A ∪ φ = A (Law of identity element, φ is the identity of ∪) | φ ∩ A = φ, U ∩ A = A (Law of φ and U) |
A ∪ A = A (Idempotent law) | A ∩ A = A (Idempotent law) |
U ∪ A = U (Law of U) | A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C ) (Distributive law ) |
11. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8}. Find A ∪ B ∪ C.
Solution: A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
12. If A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13}, C = {11, 13, 15}. Find A ∩ (B ∪ C).
Solution:
As we know, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
= {7, 9, 11} ∪ {11}
= {7, 9, 11}
Difference of sets: If A and B are two sets, then the difference of set A and set B is a set that has elements of only set A, not B. It is represented as A – B. |
13. If A = { 1, 2, 3, 4, 5, 6}, B = { 2, 4, 6, 8 }. Find A – B and B – A.
Solution: A – B = { 1, 3, 5 }
B – A = { 8 }
Clearly, A – B ≠ B – A
Complement of set: If A is a subset of universal set U, then the complement of a set A is the set that does not have any elements of A. It is denoted as A′.
A′ = {x : x ∈ U and x ∉ A } A′ = U – A |
14. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9}. Find A′.
Solution: A′ = { 2, 4, 6, 8, 10 }
Video Lesson on What are Sets
Practice Questions
Solve the following questions on sets:
- Check whether the given sets are equal sets: A = {1, 2, 3, 4} and B = {2, 4, 1, 3}.
- Write the subsets for the set A = {1, 3, 5, 7}
- Write the set A = {1, 2, 3, 4, 5, …} in set-builder form.
- If A = {1, 3, 5, 7, 9, 11} and B = {1, 2, 3, 13}, the find A – B and B – A.
- Find A ∪ (B ∪ C), if A = {1, 3, 5}, B = {2, 4, 6} and C = {1, 5, 7}.
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A servey was conducted among 300 students . It was found that 125 students like to play cricket . 145 students like to play football . 90 students like to play tennis . 32 students like to play exactly two games out of the three games. How many students like to play exactly one games .?
n(C) = 125, n(F) = 145, n(T) = 90
n(C ⋃ F ⋃ T) = 125 + 145 + 90 – [n(C ⋂ F) + n(F ⋂ T) + n(T ⋂ C)] + n(c ⋂ F ⋂ T)
300 = 360 – [n(C ⋂ F) + n(F ⋂ T) + n(T ⋂ C)] + n(c ⋂ F ⋂ T)
n(C ⋂ F) + n(F ⋂ T) + n(T ⋂ C) = 60 + n(c ⋂ F ⋂ T)….(i)
Also, given that,
n(C ⋂ F) + n(F ⋂ T) + n(T ⋂ C) – 3[n(c ⋂ F ⋂ T)] = 32
n(C ⋂ F) + n(F ⋂ T) + n(T ⋂ C) = 32 + 3[n(c ⋂ F ⋂ T)]….(ii)
From (i) and (ii),
60 + n(c ⋂ F ⋂ T) = 32 + 3[n(c ⋂ F ⋂ T)]
60 – 32 = 3[n(c ⋂ F ⋂ T)] – n(c ⋂ F ⋂ T)
2[n(c ⋂ F ⋂ T)] = 28
n(c ⋂ F ⋂ T) = 14
Number of students like to play exactly one game = n(C) + n(F) + n(T) – 2[n(C ⋂ F) + n(F ⋂ T) + n(T ⋂ C)] + 3[n(c ⋂ F ⋂ T)]
= 125 + 145 + 90 – 2[32 + 3 × 14] + 3 × 14
= 360 – 148 + 42
= 254