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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:

⇒ x^{2 }+ 1 = 0

⇒ x^{2 }= −1

⇒ x = ± √-1

i.e., **x** is an **imaginary number**.

Hence {x : x is a real number and x^{2 }+ 1 = 0} is an **empty set**.

QUESTION: 2

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

Solution:

- Set S of odd positive integer less than 10 is {1, 3, 5, 7, 9}.
- Then, the
**cardinality**of set S, i.e., the number of elements in the set, 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:

By applying De Morgan's Law,

(A U B)' = A' ∩ 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}

If A = {3, 5, 9}

Then, A ∪ {1, 2} = {1, 2, 3, 5, 9}

So, the minimum set of A = **{3, 5, 9}**

QUESTION: 6

In a city, 20 per cent of the population travels by car, 50 per cent travels by bus and 10 per cent 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}
- The
**intersection**of this set with A will be the common elements in both of these (A and B') i.e., = {1, 2, 5} which is**set A**itself.

QUESTION: 8

In the 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:

If A ∪ B = A ∩ B then A = B.

**Example:** Let A = {1, 2, 3, 4} & B = {1, 2, 3, 4}

(AUB) = (A⋂B)

⇒{1,2,3,4} = {1,2,3,4}

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:

A = {2, 3, 4, 8, 10}

B = {3, 4, 5, 10, 12}

C = {4, 5, 6, 12, 14}

► (A ∩ B) = {3,4,10}

► (A ∩ C) = {4}

► (A ∩ B) U (A ∩ C) = {3,4,10}

QUESTION: 11

The shaded region in the given figure is:

Solution:

- (A - B) is A but not B which means 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 get excluded two times.
- So, we take B ∪ C to be excluded from A. Therefore 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) = 2^{8} = 256

⇒ Number of subsets having 0 elements = ^{8}C_{0 }= 1

⇒ Number of subsets having 1 element = ^{8}C_{1} = 8

⇒ Number of subsets having 2 elements = ^{8}C_{2} = 28

∴ Number of subsets having atleast 3 elements = 256 - 1 - 8 - 28 = 219

QUESTION: 13

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

Solution:

From Venn-Euler's Diagram

∴ (A ∪ B)′ ∪ (A′ ∩ B) = A′

QUESTION: 14

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

Solution:

- Since 3b
^{2 }= 35 - 2a^{2}is odd and at most 35, so b must be odd and b^{2 }< 12. Thus, b= ± 1 or ± 3. - If b= ± 1, then a= ± 4.

If b= ± 3, then a= ± 2. - This will give us 8 solutions:

{(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(A ∩ B) = 4, where U is the universal set, A and B are subsets of U, then n((A ∪ B)^{c}) =

Solution:

- n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 12 + 9 − 4 = 17
- Now, n((A ∪ B)
^{c}) = n(U) − n(A ∪ 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,

**P(C ∪ S) ^{' }= 1 - P(C ∪ S)**

0.65 = 1 - P(C ∪ S)

1 - 0.65 = P(C ∪ S)

P(C ∪ S) = 0.35

So, it becomes,

0.35 = 0.25 + 0.15 - x

0.35 = 0.40 - x

x = 0.05

Thus, 0.05 x Total number of families in town = 1500

Total number of families in the town =

1500 / 0.05 = 30,000

Hence, the total number of families in the town is

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 = {2,3,4,8,10}, B = {3,4,5,10,12} and C = {4,5,6,12,14}

⇒ (AUB) = {2,3,4,5,8,10,12}

⇒ (AUC) = {2,3,4,5,6,8,10,12,14}

⇒ (AUB) ⋂ (AUC) = {2,3,4,5,8,10,12}

QUESTION: 18

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

Solution:

n(A × B) = n(A).n(B)

⇒ 3 × 3 = 9

A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4,8), (4, 9), (5, 7), (5, 8), (5, 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 = **2 ^{n} − 1 **= 2

[{1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,4}, {1,2,3}, {1,3,4} ,{2,3,4}, {1,2,3,4}]

QUESTION: 21

The set of intelligent students in a class is:

Solution:

As the opinions of different person is different about 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 two sets, then A ∩ (B − A) is:

Solution:

A ∩ (B − A) = ψ,[∵ x ∈ B − A ⇒ x∉ A]

QUESTION: 23

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

Solution:

Since A ⊆ B

∴ A ∪ B = B

So, n(A ∪ 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
**M**denote the sets of students who have passed in**Mathematics**and**P**denote the sets of students who have passed in**Physics**. - Thus from the question we get:

► n(M∪P) = 100

► n(M) = 55

► n(P) = 67 **n(M∩P) = n(M) + n(P) - n(M∪P)**

► n(M∩P) = 55 + 67 - 100

► n(M∩P) = 22- ∴ The no. of who passed in physics only is given by:

= n(P) - n(M∩P) = 67 - 22 =**45**

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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