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SSC > Set Theory

Explore popular questions from Set Theory for SSC. This collection covers Set Theory previous year SSC questions hand picked by experienced teachers.

Q 1.

Correct2

Incorrect-0.5

{tex} A- \left(B \cap C\right) {/tex} is equal to

A

{tex} \left(A-B\right)\cap \left(A-C\right) {/tex}

B

{tex} \left(A-B\right)\cup \left(A-C\right) {/tex}

C

{tex} \left(A\cap B\right)-C {/tex}

None of these

Explanation

From Venn diagram

{tex} \left(A \cap B\right)^{c} {/tex} = Portion exterior to

{tex} \left(A \cap B\right)^{c} \cap A {/tex} = Portion showing both shadings = A-B

Q 2.

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If A = {tex} \{ x : x^{2} = 1 \} {/tex} and B {tex} \{ x : x^{4} = 1 \} {/tex}, then {tex} A \cup B {/tex} is equal to:

{i, -i}

B

{-1, 1}

C

{-1, 1, i. -i}

D

None of these

Explanation

Given that {tex} A = \{ x:x^{2}=1 \},\ B=\{ x:x^{4}=1 \} {/tex}

=> {tex} A = \{ -1,\ 1 \},\ B = \{ -1,\ 1,\ -i,\ i \} {/tex}

Now, {tex} A - B = \phi,\ B-A = \{ -i,\ i \} {/tex}

Therefore, {tex} \left(A-B\right) \cup \left(B-A\right) = \{ -i,\ i \} {/tex}

Q 3.

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Incorrect-0.5

If A = {tex} \{ x : x = 4n+1, 2 \le n \le 5 \} {/tex} then number of subsets of A is:

A

16

15

C

4

D

none of these

Explanation

Given that {tex} A = \{ x:x=4n+1;\ 2 \le n \le 5 \} {/tex}

Number of elements in set A is 4 ,

So, number of proper subsets = {tex} 2^{4} - 1 = 15 {/tex}.

Q 4.

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Incorrect-0.5

Let R and S be two relations on a set A. Then which is not correct?

R and S are transitive, then R u S is also transitive.

B

R and S are transitive, then R n S is also transitive.

C

R and S are reflexive, then R n S is also reflexive.

D

R and S are symmetric, then R U S is also symmetric

Explanation

---

Q 5.

Correct2

Incorrect-0.5

The group of beautiful girls is:

A

a null set

B

a finite set

C

a singleton set

not a set

Explanation

Beautiful is relative term so, it is not well defined term. Therefore, it is not a set.

Q 6.

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R is a relation over the set of real numbers and it is given by {tex} nm \ge 0 {/tex}. Then R is:

A

symmetric and transitive

B

reflexive and symmetric

C

a partial order relation

an equivalence relation

Explanation

---

Q 7.

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In a city of 55 students, the number of students studying different subjects are 23 in mathematics, 24 in physics, 19 in chemistry, 12 in mathematics and physics, 9 in mathematics and chemistry, 7 in physics and chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is:

A

6

B

9

7

D

all of these

Explanation

---

Q 8.

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If {tex} N_{a}=\{ an : n \epsilon N \} {/tex}, then {tex} N_{3} \cap N_{4} {/tex} is equal to:

A

{tex} N_{7} {/tex}

{tex} N_{12} {/tex}

C

{tex} N_{3} {/tex}

D

{tex} N_{4} {/tex}

Explanation

Given that {tex} N_{a}=\{ an:n \epsilon N \} {/tex}

{tex} N_{3} \cap N_{4} = \{ 3,\ 6,\ 9.\ 12,\ 15, ..... \} \cap \{ 4,\ 8,\ 12,\ 16,\ 20, ... \} {/tex}

= {tex} \{ 12,\ 24,\ 36, ... \} = N_{12} {/tex}

Q 9.

Correct2

Incorrect-0.5

The relation "Congruence modulo m" is:

A

reflexive only

B

transitive only

C

symmetric only

an equivalence relation

Explanation

{tex} x=3 \left(mod\ 7\right) => x-3 = 7p,\ \left(p\ \epsilon\ I\right) {/tex}

=> {tex} x=7p+3,\ p\ \epsilon\ I\ i.e.,\ \{ 7p+3 : p\ \epsilon\ z \} {/tex}

Therefore, Solution set of x is {tex} \{ 7p + 3 : p\ \epsilon\ I \} {/tex}.

Q 10.

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Set A has 3 elements and set B has 4 elements. The number of injections that can be defined from A to B is:

A

144

B

12

24

D

30

Explanation

Given that {tex} n \left(A\right)=3 {/tex} and {tex} n \left(B\right)=4 {/tex}, the number of injections or one-one mapping is given by.

{tex} ^{4}p_{3} \frac{4 !}{ \left(4-3\right)!}= 4 \times 3 \times 2 \times 1 = 24 {/tex}

Q 11.

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Incorrect-0.5

Which of the four statements given below is different from the other?

A

{tex} f:A \rightarrow B {/tex}

{tex} f:x \rightarrow f \left(x\right) {/tex}

C

f is a mapping from A to B

D

f is a function from A to B

Explanation

---

Q 12.

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Incorrect-0.5

Which of the following is correct?

A

{tex} A \cap B \subset A \cup B {/tex}

{tex} A \cap B \subseteq A \cup B {/tex}

C

{tex} A \cup B \subset A \cap B {/tex}

D

None of these

Explanation

If {tex} A = B,\ then\ A\cap B = A \cup B {/tex}

{tex} A \cap B \subseteq A \cup B {/tex}

Q 13.

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Incorrect-0.5

Let {tex} f:N \rightarrow R:f \left(x\right)=\frac{ \left(2x-1\right) }{2} {/tex} and {tex} g:Q \rightarrow R:g \left(x\right)=x+2 {/tex} be two functions then {tex} \left(gof\right) \left(\frac{3}{2}\right) {/tex}

3

B

1

C

{tex} \frac{7}{2} {/tex}

D

None of these

Explanation

{tex} f \left(x\right)=\frac{2x-1}{2}\ and\ g \left(x\right)=x+2 {/tex}

{tex} \left(gof \left(x\right) \right)=g \left(\frac{1}{2} \left(2x-1\right) \right) {/tex}

= {tex} x-\frac{1}{2}+2 = x+\frac{3}{2} {/tex}

{tex} \therefore \left(gof\right) \left(\frac{3}2{}\right)=\frac{3}{2}+\frac{3}{2}=3 {/tex}

But, {tex} \frac{3}{2} \notin N{/tex}

Q 14.

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Incorrect-0.5

If N be the set of all natural numbers, consider {tex} f:N \rightarrow N:f \left(x\right)=2x \forall x \epsilon N {/tex}, then f is:

A

one-one onto

one-one into

C

many-one

D

one of these

Explanation

Let {tex} x_{1},x_{2} \epsilon N {/tex}, then {tex} f \left(x_{1}\right)=f \left(x_{2}\right) {/tex}

=> {tex} 2x_{1}=2x_{2} => x_{1}=x_{2} {/tex}

Let y = 2x, then {tex} x = \frac{y}{2} \notin N {/tex}

Now, if we put y = 5, then {tex} x = \frac{5}{2} \notin N {/tex}

This shows that {tex} 5 \epsilon N {/tex}, has no pre image in n

So, f is into

Hence, f is one-one into

Q 15.

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N is the set of natural numbers. The relation R is defined on {tex} N \times N {/tex} as follows: {tex} \left(a,\ b\right)R \left(c,\ d\right) \Leftrightarrow a+d=b+c {/tex} is:

A

reflexive

B

symmetric

C

transitive

all of these

Explanation

(a) {tex} \left(a,\ b\right) R \left(a,\ b\right) \Leftrightarrow \ a+b=b+a {/tex}

Therefore, R is reflexive

(b) {tex} \left(a,\ b\right) R \left(c,\ d\right) => a+d = b+c {/tex}

=> {tex} c+b = d+a {/tex}

=> {tex} \left(c,\ d\right) R \left(a,\ b\right) {/tex}

Therefore, R is symmetric

(c) {tex} \left(a,\ b \right) R \left(c,\ d\right)\ and\ \left(c,\ d\right) R \left(e,\ f \right) {/tex}

=> {tex} a+d=b+c\ and \ c + f=d+e {/tex}

=> {tex} a+b+c+f = b+c+d+e {/tex}

=> {tex} a+f = b+e {/tex}

=> {tex} \left(a,\ b\right) R \left(e,\ f\right) {/tex}

Therefore, R is transitive

Thus R is an equivalence relation {tex} N \times N {/tex}

Q 16.

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If {tex} E=\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \}; {/tex} {tex} A=\{ 1, 3, 5, 7 \},\ B=\{ 2, 4, 6, 8 \} {/tex} then find the value of {tex} \left(A\cup B\right)' {/tex} and {tex} A'\cap B' {/tex}

A

{tex} \{ 6 \} {/tex}

{tex} \{ 9 \} {/tex}

C

{tex} \{ \} {/tex}

D

None of these

Explanation

{tex} E = \{ 1,2,3,4,5,6,7,8,9,10 \} {/tex}

{tex} A = \{ 1,3,5,7 \} {/tex}

{tex} B = \{ 2,4,6,8,10 \} {/tex}

{tex} A\cup B = \{ 1,2,3,4,5,6,7,8,9,10 \} {/tex}

{tex} \left(A\cup B\right)' = \{ 9 \} {/tex}

{tex} A' = \{ 2,4,6,8,9,10 \} {/tex}

{tex} B' = \{ 1,3,5,7,9 \} {/tex}

{tex} A'\cap B' = \{ 9 \} {/tex}

Q 17.

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{tex} A=\{ 4, 9, 16, 25, 36 \},\ B=\{ 9, 25, 49, 81 \},\ C=\{ 16, 81, 256 \} {/tex} then find {tex} A- \left(B\cap C\right) {/tex}

A

{tex} \{ 81 \} {/tex}

{tex} \{ 4, 9, 16, 25, 36 \} {/tex}

C

{tex} \{ 4, 36 \} {/tex}

D

{tex} \{ 4, 16, 36 \} {/tex}

Explanation

{tex} A=\{ 4,\ 9,\ 16,\ 25,\ 36 \},\ B=\{ 9,\ 25,\ 49,\ 81 \},\ C=\{ 16,\ 81,\ 256 \} {/tex}

{tex} B\cap C = \{ 81 \} {/tex}

{tex} A- \left(B\cap C\right) = \{ 4,\ 9,\ 16,\ 25,\ 36 \}-\{ 81 \} {/tex}

={tex} \{ 4,\ 9,\ 16,\ 25,\ 36 \} {/tex}

Q 18.

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If {tex} n \left(A\right)=6, n \left(B\right)=5, n \left(A\cap B\right)=3 {/tex} then find {tex} n \left(A\cup B\right) {/tex}

A

11

B

9

8

D

None of these

Explanation

{tex} n \left(A\cup B\right) = n \left(A\right)+n \left(B\right)-n \left(A\cap B\right) {/tex}

= 6+5-3 = 8

Q 19.

Correct2

Incorrect-0.5

lf {tex} A=\{ 1, 2, 3, 4, 5, 6 \},\ B=\{ 4, 5, 6 \} {/tex}, then {tex} n \left(A\cap B\right) {/tex}

3

B

6

C

4

D

2

Explanation

{tex} A = \{ 1,2,3,4,5,6 \} {/tex}

{tex} B = \{ 4,5,6 \} {/tex}

{tex} A\cap B = \{ 4,5,6 \} {/tex}

{tex} n \left(A\cap B\right) = 3 {/tex}

Q 20.

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Incorrect-0.5

If {tex} n \left(p \left(A\right)=16 \right) {/tex} then what is the value of {tex} n \left(A\right) {/tex}

A

12

4

C

8

D

2

Explanation

{tex} nP \left(A\right) = 16 = 2^{4} {/tex}

The value of {tex} n \left(A\right) = 4 {/tex}

Q 21.

Correct2

Incorrect-0.5

If {tex} A=\{ 4, 5, 6, 7, 8, 9, 10 \},\ A\cap B=\{ 6, 7, 8 \},\ A\cup B=\{ 4, 5, 6, 7, 8, 9, 10, 11, 12 \} {/tex}, then find B.

A

{tex} \{ 4, 5, 6, 7, 8 \} {/tex}

{tex} \{ 6, 7, 8, 11, 12 \} {/tex}

C

{tex} \{ 4, 5, 7, 10, 11, 12 \} {/tex}

D

None of these

Explanation

{tex} A \cup B = \{ 4,5,6,7,8,9,10,11,12 \} {/tex}

{tex} A = \{ 4,5,67,8,9,10 \} {/tex}

{tex} A\cap B = \{ 6,7,8 \} {/tex}

{tex} B = \{ 6,7,8,11,12 \} {/tex}

Q 22.

Correct2

Incorrect-0.5

If {tex} E=\{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \},\ A=\{1, 2, 3, 4, 5 \},\ B=\{2, 4, 6, \} {/tex} then find {tex} \left(A\cup B\right)' {/tex}

{tex} \{ 8, 9 10 \} {/tex}

B

{tex} \{ 7, 8, 9 10 \} {/tex}

C

{tex} \{ 6, 7, 8, 9, 10 \} {/tex}

D

cannot be found

Explanation

{tex} E = \{ 1,2,3,4,5,6,7,8,9,10 \} {/tex}

{tex} A = \{ 1,2,3,5 \},\ B = \{ 2,4,6,7 \} {/tex}

{tex} A\cup B = \{ 1,2,3,4,5,6,7 \} {/tex}

{tex} \left(A\cup B\right)' = \{ 8,9,10 \} {/tex}

Q 23.

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Incorrect-0.5

{tex} n \left(A\cup B\right)=8,\ n \left(A\right)=6,\ n \left(B\right)=4 {/tex} find {tex} n \left(A\cap B\right) {/tex}

A

10

B

7

2

D

4

Explanation

{tex} n \left(A\cup B\right)=8,\ n \left(A\right)=6,\ n \left(B\right)=4 {/tex}

{tex} n \left(A\cup B\right)=n \left(A\right)+n \left(B\right)-n \left(A\cap B\right) {/tex}

{tex} 8 = 6+4-n \left(A\cap B\right) {/tex}

{tex} n \left(A\cap B\right) = 10-8=2 {/tex}

Q 24.

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In an examination {tex} 77 ^{\circ} {/tex} pupils passed in English {tex} 65 ^{\circ} {/tex} students passed in Mathematics and {tex} 50 ^{\circ} {/tex} students passed both in English and Mathematics. Then how many percentage of students fail in both the subjects.

A

16

8

C

12

D

20

Explanation



Let the total number of students be 100

Passed in Maths = 77

Passed in English = 65

Passed in both subjects 50

Total Number of passed students = 27 + 50 + 15 = 92

[77 + 68 - 50]

Failed in both the subjects = 100 - 92 = 8

Q 25.

Correct2

Incorrect-0.5

In an examination, {tex} 45 ^{\circ} {/tex} students failed in Science and {tex} 56 ^{\circ} {/tex} failed in Mathematics if {tex} 16 ^{\circ} {/tex} failed in both Science and Mathematics the percentage of those who passed in both the subject is.

15

B

31

C

49

D

42

Explanation



Let the number of students appeared for the examination be 100.

Let the circle A represent who failed 1n Science and

B represent the students who failed in Mathematics respectively.

Number of students failed in Science only = 45 - 16 = 29.

Number of students failed in Mathematics only = 56 - 16 = 40

Total No of students failed = Number of students failed in science + Number of students failed in Mathematics + failed in both.

= 29 + 40 + 16

Number of students passed = 100 - 85 = 15