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Test: Sets- 2 - Commerce MCQ


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25 Questions MCQ Test Mathematics (Maths) Class 11 - Test: Sets- 2

Test: Sets- 2 for Commerce 2024 is part of Mathematics (Maths) Class 11 preparation. The Test: Sets- 2 questions and answers have been prepared according to the Commerce exam syllabus.The Test: Sets- 2 MCQs are made for Commerce 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Sets- 2 below.
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Test: Sets- 2 - Question 1

If n (A) = 3 and n (B) = 6 and A ⊆ B, then the number of elements in A ∪ B is equal to

Detailed Solution for Test: Sets- 2 - Question 1

n(A) = 3, n(B) = 6
It is given that A is a subset of B which means that all elements of A are present in B.
n(A∪B) = n(A) + n(B) - n(A∩B) ----------- [ As A⊂B ]
= 3 + 6 - 3 = 6

Test: Sets- 2 - Question 2

If n (A) = 3 and n (B) = 6 and A ⊆ B, then the number of elements in A ∩ B is equal to

Detailed Solution for Test: Sets- 2 - Question 2

Since  n(A) = 3   n(B) = 6

A⊆B,A⊂B=B

So, n (A⋂B) =n(A) =3

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Test: Sets- 2 - Question 3

Let A and B be subsets of a set X, Then which of the following is correct

Detailed Solution for Test: Sets- 2 - Question 3

A - B means part of A that does not contain B,
So, A - B = A−(A∩B) 
= A ∩ Bc

Test: Sets- 2 - Question 4

If A , B and C are any three sets, then A – (B ∪ C) is equal to

Detailed Solution for Test: Sets- 2 - Question 4

According to De'Morgan's Law;
A-(BUC) = (A-B)∩(A-C)

Test: Sets- 2 - Question 5

What is the cardinality of the set of odd positive integers less than 10?

Detailed Solution for Test: Sets- 2 - Question 5

Set S of odd positive an odd integer less than 10 is {1, 3, 5, 7, 9}. Then, Cardinality of set S = |S| which is 5.

Test: Sets- 2 - Question 6

If A, B and C are any three sets, then A ∩ (B ∪ C) is equal to

Detailed Solution for Test: Sets- 2 - Question 6

In the problem statement we are taking union of B and C and then taking its intersection with A.
This means A∩(B∪C)will contain elements that are in A and are in either B or C.
∴ A∩(B∪C) is equivalent to taking intersection of A,B and A,C and then taking there union i.e. (A∩B)∪(A∩C)

Test: Sets- 2 - Question 7

If aN = {ax : x ∈ N}, then the set 3N ∩ 7N is

Detailed Solution for Test: Sets- 2 - Question 7

According to the given notation, 
3N = {3x:x∈R} = {3,6,9,12.......}
and 7N = {7x:x∈R} = {7,14,21,28,35,42,.......}
Hence 3N∩7N = {21,42,63}
= {21x:x∈N}
= 21N

Test: Sets- 2 - Question 8

The smallest set A such that A ∪ {1, 2} = {1, 2, 3, 5, 9} is

Detailed Solution for Test: Sets- 2 - Question 8

Let A be the smallest set.
A ∪ {1,2} = {1,2,3,5,9}
A = {3,5,9} or {1,2,3,5,9}
As A is smallest set,we take A = {3,5,9} only
Check:-
{3,5,9} ∪ {1,2} = {1,2,3,5,9}

Test: Sets- 2 - Question 9

If A and B are sets, then A ∩ (B – A) is

Detailed Solution for Test: Sets- 2 - Question 9

As we know that A-B = A
that implies x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
⇒ x ∈ to A-B and x ∈ B
⇒ (x ∈ A and x doesnt belongs to B) and x ∈ to B
So, our assumption is wrong
⇒ A∩B = ϕ

Test: Sets- 2 - Question 10

If A = {0,1,5, 4, 7}. Then the total number subsets of A are

Test: Sets- 2 - Question 11

If A and B be two sets such that n (A) = 70, n (B) = 60, and n (A ∪ B) = 110. Then n (A ∩ B) is equal to

Test: Sets- 2 - Question 12

Which set is the subset of all given sets?

Detailed Solution for Test: Sets- 2 - Question 12

The question should be: “Which set is the subset of all the sets?”
Φ = { } (Empty set) is a subset of all the sets.

Test: Sets- 2 - Question 13

Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in A U B

Detailed Solution for Test: Sets- 2 - Question 13

We know that

n(A∪B)=n(A)+n(B)−n(A∩B)......(i)

n(A∪B)=n(A)+n(B)-n(A∩B)......(i)

Case 1 From (i) , it is clear that n(A∪B)

n(A∪B) will be maximum when n(A∩B)=0

In that case, 

n(A∪B)=n(A)+n(B)=(3+6)=9

∴ Maximum number of elements in 

(A∪B)=9

Case 2 From (i) , it is clear that n(A∪B)

n(A∪B) will be minimum when n(A∩B)=0 maximum ,i.e, when 

n(A∩B)=3

In this case, 

n(A∪B)=n(A)+n(B)−n(A∩B)=(3+6−3)=6

∴ minimum number of elements in 

A∪B=6

Test: Sets- 2 - Question 14

If A = {x : x ≠ x} represents

Test: Sets- 2 - Question 15

If A = {x : x2 − 5x + 6 = 0}, B = {2, 4}, C = {4, 5}, then A × (B ∩ C) is

Detailed Solution for Test: Sets- 2 - Question 15

Clearly, A = {2, 3}, B = {2, 4}, C = {4,5} B ∩ C  = {4}

∴ A × (B ∩ C) = {(2, 4), (3, 4)}.

Test: Sets- 2 - Question 16

Which of the following is a set

Test: Sets- 2 - Question 17

Let A = {x : x ∈ R, |x| < 1}; B = {x : x ∈ R, |x − 1| ≥ 1} and A ∪ B = R − D, then the set D is (R ∈ Set of real numbers)

Detailed Solution for Test: Sets- 2 - Question 17

A = {x : x ∈ R,−1 < x < 1}

B = {x : x ∈ R : x − 1 ≤ −1 or x − 1 ≥ 1]

  = {x : x ∈ R : x ≤ 0 or x ≥ 2}

∴A ∪ B = R − D , where D = [x : x ∈ R,1 ≤ x < 2]

Test: Sets- 2 - Question 18

If A, B and C are non-empty sets, then (A - B) U (B - A) equals 

Detailed Solution for Test: Sets- 2 - Question 18

Use venn diagram approach

Test: Sets- 2 - Question 19

Which of the following statements is true?

Detailed Solution for Test: Sets- 2 - Question 19

Since the curly brackets denote a set, so the other three options are a subset of the set {1, 2, 3} and and only 3 belongs to the set.

Test: Sets- 2 - Question 20

If a set A has n elements, then the total number of subsets of A is

Test: Sets- 2 - Question 21

If A, B, C be any three sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C, then

Detailed Solution for Test: Sets- 2 - Question 21

Given that AUB = AUC
⇒ (AUB) ∩ C = (AUC) ∩C
⇒ (A∩C) U (B∩C) = C           [ ∴(AUC)∩C = C ]
⇒ (A∩B) U (B∩C) = C     ..........(1)  [ ∴ (A∩C) = A∩B ]
Again AUB = AUC
(AUB) ∩ B = (AUC) ∩ B
B = (A∩B) U (C∩B)
= (A∩B) U (B∩C)   ...........(2)
From 1 & 2 we get
B = C

Test: Sets- 2 - Question 22

Given the sets A = {1, 2, 3}, B = {3 , 4}, C = {4 , 5, 6}, then A ∪ (B ∩ C) is

Test: Sets- 2 - Question 23

If A = {a, b}, B = {c, d}, C = {d, e}, then {(a, c), (a, d),(a, e), (b, c), (b, d), (b, e)} is equal to

Test: Sets- 2 - Question 24

Two finite sets have m and n elements. The total number of subsets of the first set is 48 more than the total number of subsets of the second set. The values of m and n are

Detailed Solution for Test: Sets- 2 - Question 24

Total number of subsets of the first set = 2m
Total number of subsets of the second set = 2n
2m = 2n + 48
By hit and trial,
m = 6, n = 4

Test: Sets- 2 - Question 25

Let A = {x : x is a prime factor 240} and B = {x : x is the sum of any two prime factors of 240}. Then

Detailed Solution for Test: Sets- 2 - Question 25

A = {2, 5, 3} and B = {7, 8, 5} and A ∪ B = {2, 3, 5, 7, 8}

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