Test: Sets 1


25 Questions MCQ Test Mathematics (Maths) Class 11 | Test: Sets 1


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This mock test of Test: Sets 1 for JEE helps you for every JEE entrance exam. This contains 25 Multiple Choice Questions for JEE Test: Sets 1 (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Sets 1 quiz give you a good mix of easy questions and tough questions. JEE students definitely take this Test: Sets 1 exercise for a better result in the exam. You can find other Test: Sets 1 extra questions, long questions & short questions for JEE on EduRev as well by searching above.
QUESTION: 1

Which of the following is the empty set?

Solution:

Since x2 + 1 = 0,gives x2 = −1 ⇒ x = ±i. ∴x is not real but x is  real (given).
∴No value of x is possible.

QUESTION: 2

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

Solution:

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.

QUESTION: 3

Consider the following relations
(i) A - B = A - (A ∩ B)
(ii) A = (A ∩ B) ∪ (A - B)
(iii) A - (B ∪ C) = (A - B) ∪ (A - C)
Which of these is/are correct?

Solution:


Using the Venn Diagram,
(i) A - B = A - (A ∩ B) is true
(ii) A = (A ∩ B) ∪ (A - B) is true
(iii) A - (B ∪ C) = (A - B) ∪ (A - C) is false

QUESTION: 4

If A and B be any two sets, then (A ∪ B)‘ is equal to

Solution:

Let A = {1} and B = {1 , 2 } U = {1 , 2 , 3}
so, A U B = {1} U {1 , 2} = {1 , 2}
therefore, (A U B )' = {3}
*option (a):- A' U B' = {2 , 3 } U { 3} = { 2 , 3 }
thus, option (a) is not true
*option (b):- A U B = { 1 , 2}
hence option is not correct
*option (c):- A intersection B = {1} intersection {1,2} = {1}
hence option is not correct
*option (d):- A' intersection B' = {2,3} intersection {3} = {3}
therefore (A U B )' = A' intersection B'

QUESTION: 5

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

Solution:

A ∪ {1, 2} = {1, 2, 3, 5, 9}Now the minimum set of A = {3, 5, 9}A ∪ 1, 2= 1, 2, 3, 5, 9 Now the minimum set of A = 3, 5, 9

QUESTION: 6

In a city 20 percent of the population travels by car, 50 percent travels by bus and 10 percent travels by both car and bus. Then persons travelling by car or bus is

Solution:

n(Persons travelling by car) = 20 = n(C)
n(Persons travelling by bus) = 50 = n(B)
n(Persons travelling by both car and bus) = 10 = n(C ∩ B)
∴ n(C ∪ B) = n(C) + n(B) - n(C ∩ B)
= 20 + 50 - 10
= 60 

QUESTION: 7

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, A = {1, 2, 5}, B = {6, 7}. Then A ∩ B’ is :

Solution:

B' gives us all the elements in U other than 6 and 7. i.e B' = {1, 2, 3, 4, 5, 8, 9, 10} and the intersection of this set with A will be the common elements in both of these (A and B') i.e = {1, 2, 5} with us set A itself.

QUESTION: 8

In rule method the null set is represented by

Solution:

This method involves specifying a rule or condition which can be used to decide whether an object can belong to the set. This rule is written inside a pair of curly braces and can be written either as a statement or expressed symbolically or written using a combination of statements and symbols.
{x :  x ≠ x}
This implies a null set

QUESTION: 9

If A and B are two sets, then A ∪ B = A ∩ B if

Solution:
QUESTION: 10

If  A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12}, C = {4, 5, 6, 12, 14} then (A ∩ B) ∪ (A ∩ C) is equal to

Solution:

∩ = {2, 3, 4, 8, 10} ∩ {3, 4, 5, 10, 12} = {3, 4, 10}
∩ = {4}
∴ (∩ B) ∪ (∩ C) = {3, 4, 10}

QUESTION: 11

The shaded region in the given figure is

Solution:

A - B is A but not B which means that A excluding the intersection of A and B
Similarly in the figure the shaded region is A but not C and B
Which means A excluding B and C. But if B and C both are excluded from A then the intersection of A, B and C gets excluded two times.
So we take B ∪ C to be excluded from A
So the answer is A - (B ∪ C)

QUESTION: 12

Let A and B be two sets containing four and two elements respectively.Then the number of subsets of the set A × B,each having atleast three elements is:

Solution:

Number of elements in Set A = 4
Number of elements in Set B = 2
∴ Number of elements in set (A × B) = 8
∴ Total number of subsets of (A×B) = 28 = 256
Number of subsets having 0 elements = 8C0 = 1
Number of subsets having 1 element = 8C1 = 8
Number of subsets having 2 elements = 8C2 = 28
∴ Number of subsets having atleast 3 elements = 256 - 1 - 8 - 28 = 219

QUESTION: 13

Let A  and B be two sets then (∪ B)′ ∪ (A′ ∩ B) is equal to

Solution:

From Venn-Euler's Diagram,

∴ (∪ B)′ ∪ (A′ ∩ B) = A

QUESTION: 14

The number of elements in the set {(ab) : 2a2 + 3b2 = 35, a∈ Z}, where Z is the set of all integers, is

Solution:

Substitute a=2 and b=3

2(2)^2+3(3)^2=8+27=35

Substitute a=4 and b=1

2(4)^2+3(1)^2=32+3=35

Substitute a=-2 and b=3

2(-2)^2+3(3)^2=8+27=35

Substitute a=-2 and b=-3

2(-2)^2+3(-3)^2=8+27=35

Substitute a=2 and b=-3

2(2)^2+3(-3)^2=8+27=35

Substitute a=-4 and b=1

2(-4)^2+3(1)^2=35

Substitute a=-4 and b=-1

2(-4)^2+3(-1)^2=35

Substitute a=4 and b=-1

2(4)^2+3(-1)^2=32+3=35

{(2,3),(-2,-3),(-2,3),(2,-3),(4,1),(4,-1),(-4,-1),(-4,1)}

 

Therefore, total elements in the set=8

QUESTION: 15

Given n(U) = 20 ,n(A) = 12,n(B) = 9,n(∩ B) = 4, where U is the universal set, A and B are subsets of U, then n((∪ B)c) =

Solution:

n(∪ B) = n(A) + n(B) − n(∩ B) = 12 + 9 − 4 = 17
Now, n((∪ B)c) = n(U) − n(∪ B) = 20 − 17 = 3

QUESTION: 16

In a certain town 25% families own a cell phone,15% families own a scooter and 65% families own neither a cell phone nor a scooter.If 1500 families own both a cell phone and a scooter,then the total number of families in the town is

Solution:

Step-by-step explanation:

Since we have given that

Percentage of families own a cell phone= 25%

Percentage of families own a scooter = 15%

Percentage of families neither own = 65%
So, it becomes,

Hence, the total number of families in the town is 30000.

QUESTION: 17

If A = {2, 3, 4, 8, 10}, B = {3, 4, 5, 10, 12} and C = {4, 5, 6, 12,14}, then (A ∪ B) ∩ (A ∪ C) is equal to

Solution:

A∩B = elements common in sets A and B. 

A∩C = elements common in sets A and C

A∩C={4} 

(A∩B)∪(A∩C)=elements in (A∩B) and (A∩C)

(A∩B)∪(A∩C)={3,4,10}∪{4}

(A∩B)∪(A∩C)=3,4,10

QUESTION: 18

If= {2, 4, 5}, = {7, 8, 9}, then n(× B) is equal to

Solution:

A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4,8), (4, 9), (5, 7), (5, 8), (5, 9)}
n(A × B) = n(A).n(B) = 3 × 3 = 9

QUESTION: 19

If A and B are disjoint, then n (A ∪ B) is equal to

Solution:

As n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

But in case of disjoint sets n(A ∩ B) = 0

∴ n(A ∪ B) = n(A) + n(B)

QUESTION: 20

The number of non-empty subsets of the set {1, 2, 3, 4} is

Solution:

The number of non- empty subsets = 2n − 1 = 24 − 1 = 16 − 1 = 15

QUESTION: 21

The set of intelligent students in a class is

Solution:

As the opinions of different person's is differentabout the intelligent student. So, we can't exactly have the same student's name in a set, hence it is not a well defined collection.

QUESTION: 22

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

Solution:

∩ (− A) = φ,[∵ ∈ − ⇒ x∉ A]

QUESTION: 23

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

Solution:

Since⊆ B,∴∪ B. So,n(∪ B) = n(B) = 6

QUESTION: 24

In a class of 100 students, 55 students have passed in Mathematics and 67 students have passed in Physics. Then the number of students who have passed in Physics only is

Solution:

Let A denote no. of student who pass in physics
let B denote no. of student who pass in mathematics
According to question
n(A)= 55
n(B)=67
n(A union B)=100
n(A intersection B)= n(A) +n(B)-n(A union B)
55+67-100
122-100⇒22
no. of student who passed only in physics is given by n(A)-n(A intersection B)⇒55-22=33

QUESTION: 25

If A = {1, 2, 3, 4}, B = {4, 5, 6, 7}, then A - B =

Solution:

A - B is nothing but the elements which are present in set A but not present in set B. In other words it is elements of set A excluding the elements which are in set B. Here A = {1, 2, 3, 4} and B = {4, 5, 6, 7} so, A - B = {1, 2, 3}.

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