SSC CGL SET THEORY MCQ ENGLISH

Question – 1 : If A = {x ∈ R: x is the sum of divisors of 2}, B = {x ∈ R: x2 = 2} and C = {x ∈ R: x2 – 5x + 6 = 0}. Find (A ∪ B) ∩ C.

Answer – (A) : {3}

Answer – (B) : {2,−2}

Answer – (C) : {3}

Answer – (D) : None of these

View Answer

Correct Answer : (C)

Explanation : Given: A = {x ∈ R: x is the sum of divisors of 2}, B = {x ∈ R: x2 = 2} and C = {x ∈ R: x2 – 5x + 6 = 0}The divisors of 2 are: 1 and 2∴ The sum of divisors of 2 is 3⇒ A = {3}∵ x2 – 2 = 0⇒x=±2⇒B={2,−2}Similarly, x2 – 5x + 6 = 0⇒ x2 – 2x – 3x + 6 = 0⇒ (x – 2) (x – 3) = 0⇒ x = 2 or 3⇒ C = {2, 3}⇒(A∪B)={2,−2,3}⇒(A∪B)∩C={2,−2,3}∩{2,3}={3}

Question – 2 : A function f:X→Y is said to be Inverse function if:

Answer – (A) : f−1(y)=x

Answer – (B) : f′(x)=y

Answer – (C) : f−1(y)=y

Answer – (D) : f−1(x)=x

View Answer

Correct Answer : (A)

Explanation : If f:X→Y is one-one onto function, then the pre-image of y∈Y is called the Inverse image of y under the function fThe function f:Y→X which associates to each y∈Y its inverse image x∈X given by f(x)=y , is called the inverse function of f:X→Y . Clearly, f−1(y)=x⇔f(x)=y,y∈Y,x∈X

Question – 3 : There are 35 teachers in a school, who teach either Mathematics or Physics. Of these, 23 teach Mathematics and 19 teach Physics. How many teach both Mathematics and Physics?

Answer – (A) : 7

Answer – (B) : 7

Answer – (C) : 9

Answer – (D) : Cannot be determined

View Answer

Correct Answer : (B)

Explanation : Let A be the set of teachers who teach Mathematics and B be the set of those who teach Physics.Given,n(A ⋃ B) = 35, n(A) = 23 and n(B) = 19By using the relation n(A ⋃ B) = n(A) + n(B) – n(A ⋂ B), we get35 = 23 + 19 – n(A ⋂ B)⇒ n(A ⋂ B) = 42 – 35 = 7∴ The number of teachers who teach both the subjects is 7.

Question – 4 : What is the domain and range of the function (x)= function f(x)=(16−x2)?

Answer – (A) : [−4,4],[0,4]

Answer – (B) : [0,4],[−4,4]

Answer – (C) : [−4,4],[0,4]

Answer – (D) : [−4,4],[−4,4]

View Answer

Correct Answer : (C)

Explanation : Given function is f(x)=(16−x2)The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes. For domain, f(x)≥0⇒16−x2≥0⇒16≥x2⇒x2≤16⇒−4≤x≤4So, domain of f(x)=[−4,4]For Range,f(x) is maximum at x=0 i.e. f(0)=4f(x) is minimum at x=4 i.e. f(4)=0So, Range of f(x)=[0,4]Therefore, the domain and range of the function (x)= of the function f(x)=(16−x2) are [−4,4],[0,4] .

Question – 5 : Let A and B be subsets of X and C = (A ∩ B’) ∪ (A’ ∩ B), where A’ and B’ are complements of A and B respectively in X. What is C equal to ?

Answer – (A) : (A ∪ B) – (A ∩ B)

Answer – (B) : (A’ ∪ B) – (A’ ∩ B)

Answer – (C) : (A ∪ B) – (A ∩ B)

Answer – (D) : (A’ ∪ B’) – (A’ ∩ B’)

View Answer

Correct Answer : (C)

Explanation : C = (A ∩ B’) ∪ (A’ ∩ B)Let us draw the Venn diagram and compare it with options.This also represents (A ∪ B) – (A ∩ B)

Question – 6 : The union of the sets { 1,2,5 } and { 1,2,6 } is the set __________

Answer – (A) : { 1,2,5,6 }

Answer – (B) : { 1,2,5,6 }

Answer – (C) : { 1,2,1,2 }

Answer – (D) : { 1,5,6,3 }

View Answer

Correct Answer : (B)

Explanation : The union of the sets A and B is the set that contains those elements that are either in A or in B . Let, A={1,2,5} and B={1,2,6}So, A∪B={1,2,5,6}

Question – 7 : Study the following information carefully and answer the questions based on it.In the figure, how many of them are only young and foreigners?

Answer – (A) : 144

Answer – (B) : 151

Answer – (C) : 280

Answer – (D) : 234

View Answer

Correct Answer : (A)

Explanation : Here the highlighted portion indicates the total number of young foreigners.

Question – 8 : If n (X) = 300, n (Y) = 400 and n (X ∪ Y) = 500, then n (X – Y) is equal to:

Answer – (A) : 100

Answer – (B) : 140

Answer – (C) : 150

Answer – (D) : 100

View Answer

Correct Answer : (D)

Explanation : Given: n (X) = 300, n (Y) = 400 and n (X ∪ Y) = 500As we know that, for any two finite sets A and B, n (A ∪ B) = n (A) + n (B) – n (A ∩ B)⇒ n (X ∪ Y) = n (X) + n (Y) – n (X ∩ Y)⇒ 500 =300 + 400 – n (X ∩ Y)⇒ n (X ∩ Y) = 200As we know that, for any two finite sets A and B, n (A – B) = n (A) – n (A ∩ B)⇒ n (X – Y) = n (X) – n (X ∩ Y) = 300 – 200 = 100

Question – 9 : Which of the following is an example of a non-empty set?

Answer – (A) : Set of all even prime numbers

Answer – (B) : (x:x2−2=0 and x is rational)

Answer – (C) : {x:x is a positive integer, x<8 and also x>12}

Answer – (D) : { x:x is a common point of any two parallel lines }

View Answer

Correct Answer : (A)

Explanation : Option (A): set of all even prime number S={2} this is non empty set.Option (B):x2−2=0x2=2x=±2x=2 or −2 is irrational number so this set is empaty setOption (C): if x<8 then x=1,2,3,4,5,6,7 and simultaneously x>12so x=13,14,15…… .This not possible so, this is empty set.Option (D): two parallel line never intersect so no common point exist, this is also empty set.

Question – 10 : If S = {x : x2 + 1 = 0, x is real}, then S is:

Answer – (A) : An empty set

Answer – (B) : {0}

Answer – (C) : {1}

Answer – (D) : An empty set

View Answer

Correct Answer : (D)

Explanation : Given that,S = {x : x2 + 1 = 0, x is real}⇒ x2 + 1 = 0⇒ x2 = – 1⇒x=±(−1)⇒ x = ± i imaginaryBut given that set only contain real value of x but after solving x2 + 1 we will get only complex value so both are contradict to each other, So we can say that set does not have any value in it. it is an empty set.

Question – 11 : If A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B ?

Answer – (A) : 26

Answer – (B) : 26

Answer – (C) : 25

Answer – (D) : None of these

View Answer

Correct Answer : (B)

Explanation : Given,A = {x, y, z} and B = {1, 2}⇒ n(A) = 3 and n(B) = 2As we know that,If A and B are two non-empty sets such that n(A) = p and n(B) = q then the number of relations that can be defined from A to B = 2pqHere, p = 3 and q = 2So, the number of relations from A to B = 26

Question – 12 : If A and B are two given sets, then A ∩ (A ∩ B)’ is equal to:

Answer – (A) : A ∩ B’

Answer – (B) : A

Answer – (C) : φ

Answer – (D) : B

View Answer

Correct Answer : (A)

Explanation : If A and B are two given sets then,⇒ A ∩ ( A ∩ B )’ = A ∩ (A’ ∪ B’ ) (De Morgan’s Law)⇒ (A ∩ A’) ∪ ( A ∩ B’ ) = ϕ ∪ ( A ∩ ϕB’ )∴ A ∩ ( A ∩ B)’ = A ∩ B’

Question – 13 : Which of the following is an empty set?

Answer – (A) : {x ∈ R: x2 = – 1}

Answer – (B) : {x ∈ R: x3 = 1}

Answer – (C) : {x ∈ R: x2 = – 1}

Answer – (D) : None of these

View Answer

Correct Answer : (C)

Explanation : Option (A):Given: {x ∈ R : x2 – 4x + 4 = 0}x2 – 4x + 4 = 0⇒ x2 – 2x – 2x + 4 = 0⇒ (x – 2)(x – 2) = 0⇒ x = 2So, 2 is an element of {x ∈ R : x2 – 4x + 4 = 0} ⇒ {x ∈ R : x2 – 4x + 4 = 0} is not an empty set.Option (B):Given: {x ∈ R: x3 = 1}x3 = 1⇒ x3 – 1 = 0⇒ (x – 1) (x2 + x + 1) = 0⇒ x = 1 or ω or ω2, where ω=−1+i32 and ω2=−1−i32∵ x ∈ R ⇒ only x = 1 is an element of {x ∈ R: x3 = 1}So, the given set {x ∈ R: x3 = 1} is not an empty set.Option (C):Given: {x ∈ R: x2 = – 1}⇒ x2 = -1⇒ x = ± i ∉ RSo, the given set {x ∈ R: x2 = -1} is an empty set.

Question – 14 : From the following Venn diagram, find A ∩ (B ∪ C).

Answer – (A) : {6, 8}

Answer – (B) : {6, 8}

Answer – (C) : {3, 5, 6, 7, 8, 9}

Answer – (D) : {2, 4, 6, 8}

View Answer

Correct Answer : (B)

Explanation : From the given diagram we can see that:A = {2,4,6,8}B = {3,5,6,7}C = {7,8,9}Therefore,A ∩ (B ∪ C) = {6,8}

Question – 15 : If A={x∣x2−5x−6=0} and B={y∣y2−7y−8=0} . Then find A∪(A∩B)?

Answer – (A) : {−1,6}

Answer – (B) : {−1,6}

Answer – (C) : {−1,8}

Answer – (D) : None of these

View Answer

Correct Answer : (B)

Explanation : Here, A={x∣x2−5x−6=0}x2−5x−6=0⇒x2−6x+x−6=0⇒x(x−6)+1(x−6)=0⇒(x+1)(x−6)=0∴x=−1,6Which can be represented in roster form as: A={−1,6} . Similarly, B={y∣y2−7y−8=0}y2−7y−8=0⇒y2−8y+y−8=0⇒y(y−8)+1(y−8)=0⇒(y+1)(y−8)=0∴y=−1,8Which can be represented in roster from as: B={−1,8} . Let, C=(A∩B)={−1}Now,A∪(A∩B)=A∪C={−1,6}

Question – 16 : In a class, 54 students are good in Hindi only, 63 students are good in Mathematics only and 41 students are good in English only. There are 18 students who are good in both Hindi and Mathematics. 10 students are good in all three subjects. What is the number of students who are good in Hindi and Mathematics but not in English?

Answer – (A) : 8

Answer – (B) : 12

Answer – (C) : 10

Answer – (D) : 8

View Answer

Correct Answer : (D)

Explanation : Let H,M,E denote the set of students studying Hindi, Mathematics and English.n(H∩M)−n(H∩M∩E)=18−10 =8

Question – 17 : In a class of 25 students, 12 have taken mathematics, 8 have taken mathematics and computer science. The number of students who have taken computer science but NOT mathematics is:

Answer – (A) : 13

Answer – (B) : 8

Answer – (C) : 4

Answer – (D) : 17

View Answer

Correct Answer : (A)

Explanation : Let A and B be the sets of students who have taken Mathematics and Computer Science respectively.Given,n(A ⋃ B) = 25, n(A) = 12 and n(A ⋂ B) = 8n(A ⋃ B) = n(A) + n(B) – n(A ⋂ B)⇒ 25 = 12 + n(B) – 8⇒ 25 = 4 + n(B)⇒ n(B) = 25 – 4 = 21And the number of students who have taken computer science but NOT mathematics is,n(B – A) = n(B) – n(A ⋂ B) = 21 – 8 = 13

Question – 18 : Consider the proper subsets of {1, 2, 3, 4}. How many of these proper subsets are a superset of the set {3}?

Answer – (A) : 7

Answer – (B) : 6

Answer – (C) : 7

Answer – (D) : 8

View Answer

Correct Answer : (C)

Explanation : A proper subset is one that contains few elements of the original set.A superset is one which contain all the element including elements of the original set.The proper subsets of {1, 2, 3, 4} and supperset of set {3} are{1, 3}, {2, 3}, {3, 4}, {1, 2, 3}, {2, 3, 4}, {1, 3, 4}, {1, 2, 3, 4}

Question – 19 : If A = {x ∈ R: x2 – 10x + 9 = 0}, B = {y ∈ R: y2 – 3y + 2 = 0} and U = {z ∈ N: 1 ≤ z ≤ 10}. Find the set representing the shaded region.

Answer – (A) : {3, 4, 5, 6, 7, 8, 10}

Answer – (B) : {3, 4, 5, 6, 7, 8}

Answer – (C) : {3, 4, 5, 6, 7, 10}

Answer – (D) : None of these

View Answer

Correct Answer : (A)

Explanation : Given: A = {x ∈ R: x2 – 10x + 9 = 0}, B = {y ∈ R: y2 – 3y + 2 = 0} and U = {z ∈ N: 1 ≤ z ≤ 10}⇒ x2 – 10x + 9 = 0 (Given)⇒ x2 – x – 9x + 9 = 0⇒ (x – 1) (x – 9) = 0⇒ x = 1 or 9⇒ A = {1, 9}Similarly, y2 – 3y + 2 = 0⇒ y2 – y – 2y + 2 = 0⇒ (y – 1) (y – 2) = 0⇒ y = 1 or 2⇒ B = {1, 2}U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ——-(Given)The shaded region represents (A ∪ B)’.= A ∪ B = {1, 2, 9}= (A ∪ B)’ = U – (A ∪ B) = {3, 4, 5, 6, 7, 8, 10}

Question – 20 : If X = {a, e, i}, Y = {b, d} and U = {a, b, d, e, i, o, u} where U is the universal set for X and Y. Find X’ ∩ Y’?

Answer – (A) : {o, u}

Answer – (B) : {d, o, u}

Answer – (C) : {o, u}

Answer – (D) : None of these

View Answer

Correct Answer : (C)

Explanation : Given: X = {a, e, i}, Y = {b, d} and U = {a, b, d, e, i, o, u}Here, we have to find the value of X’ ∩ Y’As we know that, (A ∪ B)’ = A’ ∩ B’⇒ X ∪ Y = {a, e, i} ∪ {b, d} = {a, b, d, e, i}As we know that, A’ = U – A⇒ X’ ∩ Y’ = (X ∪ Y)’ = U – (X ∪ Y) = {a, b, d, e, i, o, u} – {a, b, d, e, i} = {o, u}

Question – 21 : The difference of {1,2,3} and {1,2,5} is the set___________.

Answer – (A) : {3}

Answer – (B) : {5}

Answer – (C) : {3}

Answer – (D) : {2}

View Answer

Correct Answer : (C)

Explanation : The difference of the sets A and B denoted by (A−B) is the set containing those elements that are in A not in B . Let, A={1,2,3} and B={1,2,5}So, A−B={3}Hence, the correct option is (C)

Question – 22 : 22 students play cricket and 11 play football from the class of 25, then how many of them play both cricket and football. Given each student play at least one of the sports.

Answer – (A) : 8

Answer – (B) : 6

Answer – (C) : 7

Answer – (D) : 5

View Answer

Correct Answer : (A)

Explanation : Set A is the students who play cricket, so n(A) = 22Set B is the students who play football, so n(B) = 11Set of all students A ∪ B, so n(A ∪ B) = 25∵ n(A ∪ B) = n(A) + n(B) – n(A ∩ B)⇒ 25 = 22 + 11 – n(A ∩ B)⇒ n(A ∩ B) = 33 – 25 = 8

Question – 23 : Let A and B be two finite sets such that n(A) = 20, n(B) = 28 and n(A ∪ B) = 36, find n(A ∩ B).

Answer – (A) : 12

Answer – (B) : 13

Answer – (C) : 14

Answer – (D) : 15

View Answer

Correct Answer : (A)

Explanation : Using the formula n(A ∪ B) = n(A) + n(B) – n(A ∩ B)Then n(A ∩B) = n(A) + n(B) – n(A ∪B)= 20 + 28 – 36= 48 – 36n(A ∩ B) = 12

Question – 24 : If A={3,5,7,9,11},B={7,9,11,13},C={11,13,15} . Find A∩(B∪C) .

Answer – (A) : {7,9,11}

Answer – (B) : {7,9,13}

Answer – (C) : {5,9,13}

Answer – (D) : {5,7,13}

View Answer

Correct Answer : (A)

Explanation : Given:A={3,5,7,9,11},B={7,9,11,13},C={11,13,15}we have to find the value of A∩(B∪C)A∩(B∪C)=(A∩B)∪(A∩C)(A∩B)={7,9,11} , (A∩C)={11}A∩(B∪C) ={7,9,11}∪{11}={7,9,11}

Question – 25 : Let P = Set of all integral multiples of 3Q = Set of all integral multiples of 4R = Set of all integral multiples of 6Consider the following relations:I. P ∪ Q = RII. P ⊂ RIII. R ⊂ (P ∪ Q)Which of the relation(s) given above is/are correct?

Answer – (A) : Only III

Answer – (B) : Only II

Answer – (C) : Only III

Answer – (D) : II and III

View Answer

Correct Answer : (C)

Explanation : Given:P = Set of all integral multiples of 3Q = Set of all integral multiples of 4R = Set of all integral multiples of 6Concept:Set symbols:Union: A ∪ B i.e. in A or B.Proper Subset: A ⊂ B i.e. every element of A is in B.Not a Subset: A ⊈ B i..e. A is not a subset of B.Explanation:P = Set of all integral multiples of 3⇒ P = {3, 6, 9, 12, 15, 18, ……},Q = Set of all integral multiples of 4⇒ Q = {4, 8, 12, 16, 20, ……}R = Set of all integral multiples of 6⇒ R = {6, 12, 18, 24, 30, ……}Considering the relation,I. P ∪ Q = RP ∪ Q = {3, 4, 6, 8, 9, 12, 15, 16,….} ≠ RII. P ⊂ RAll the elements of P are not in R so P ⊈ RIII. R ⊂ (P ∪ Q)P ∪ Q = {3, 4, 6, 8, 9, 12, 15, 16,….}Therefore,All the elements of R are in P ∪ Q⇒ R ⊂ (P ∪ Q)∴ Only (III) statement is correct.

Question – 26 : If A={1,4},B={2,3},C={3,5} then (A×B)∩(A×C) is equal to:

Answer – (A) : {(1,3),(4,3)}

Answer – (B) : {(1,3),(2,5)}

Answer – (C) : {(1,3),(1,5),(2,5)}

Answer – (D) : None of these

View Answer

Correct Answer : (A)

Explanation : Given: A={1,4},B={2,3},C={3,5}Here, we have to find (A×B)∩(A×C)As we know that, A×B={(a,b)∣a∈A and b∈B}⇒A×B={(1,2),(1,3),(4,2),(4,3)} ……(1)⇒A×C={(1,3),(1,5),(4,3),(4,5)} ……(2)From (1) and (2) we can say that,⇒(A×B)∩(A×C)={(1,3),(4,3)}

Question – 27 : Let n(U)=700,n(A)=200,n(B)=300 and n(A∩B)=100 , then n(Ac∩Bc)= ?

Answer – (A) : 300

Answer – (B) : 600

Answer – (C) : 300

Answer – (D) : 200

View Answer

Correct Answer : (C)

Explanation : Given:n(U)=700n(A)=200n(B)=300n(A∩B)=100As we know,n(Ac∩Bc)=n(U)−n(A∪B)⇒n(Ac∩Bc)=700−n(A∪B)∵n(A∪B)=n(A)+n(B)−n(A∩B)⇒n(A∪B)=200+300−100⇒n(A∪B)=400⇒n(Ac∩Bc)=700−n(A∪B)⇒n(Ac∩Bc)=700−400=300

Question – 28 : The number of cosets of H in G is {where, G = (Z, +) and H = (4Z, +)}

Answer – (A) : 4

Answer – (B) : 1

Answer – (C) : 0

Answer – (D) : 2

View Answer

Correct Answer : (A)

Explanation : Given: G = (Z, +) and H = (4Z, +) is a subgroup of G.G = (Z, +) is an abelian groupAs we know that, if G is an abelian group then every subgroup of G is a normal subgroup.∴ H is a normal subgroupSo, all the left and right cosets of H are same.Now,let’s find out the cosets of H in G.The distinct cosets of H in G are:0 + H = {4n : n ∈ Z} = H1 + H = {4n + 1: n ∈ Z}2 + H = {4n + 2: n ∈ Z}3 + H = {4n + 3: n ∈ Z}So, H, 1 + H, 2 + H, 3 + H are the cosets of H in G.

Question – 29 : In a school, out of 50 students, 25 have scooters and 35 have cycles for coming to school. The number of students who have scooter and cycle both are:

Answer – (A) : 10

Answer – (B) : 18

Answer – (C) : 10

Answer – (D) : 12

View Answer

Correct Answer : (C)

Explanation : Given,Out of 50 students in a school, 25 have scooters and 35 have cycles for coming to school.Let A be the number of students who have scooter for coming to school and B be the number of students who are having cycles for coming to school⇒ n(A) = 25, n(B) = 35 and n(A ∪ B) = 50As we know that,n(A ∪ B) = n(A) + n(B) – n(A ∩ B)⇒ 50 = 25 + 35 – n(A ∩ B)⇒ n(A ∩ B) = 60 – 50 = 10So, there are 10 students in school who have both scooter and cycle for coming to school.

Question – 30 : Let x ϵ {2, 3, 4} and y ϵ {4, 6, 9, 10}. If A be the set of all order pairs (x, y) such that x is a factor of y. Then, how many elements does the set A contain.

Answer – (A) : 6

Answer – (B) : 10

Answer – (C) : 7

Answer – (D) : 6

View Answer

Correct Answer : (D)

Explanation : Given:x ϵ {2, 3, 4} and,y ϵ {4, 6, 9, 10}To find the set of all ordered pairs (x,y) such that x is a factor of y.Let, A = { (x,y) : x is a factor of y}Here,2 is a factor of 4 implies (2, 4) ∈ A2 is a factor of 6 implies (2, 6) ∈ A2 is a factor of 10 implies (2, 10) ∈ A3 is a factor of 6 implies (3, 6) ∈ A3 is a factor of 9 implies (3, 9) ∈ A4 is a factor of 4 implies (4, 4) ∈ ASo,A = {(2, 4), (2, 6), (2, 10), (3, 6), (3, 9), (4, 4) }∴ Set A contains 6 elements.

SSC CGL SET THEORY MCQ ENGLISH

SSC CGL SET THEORY MCQ ENGLISH