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The CORRECT ANSWER is (c) 125

The explanation: Given, n(H)=100, n(E)=50, n(H∩E) = 25

We know, n (H∪E) = n (H) + n (E) – n(H∩E)

n (H∪E) = 100+50-25=125

Since each of the student knows either HINDI or English, so n(U) = n (H∪E) = 125.

Right CHOICE is (a) U

The best I can explain: In the Venn DIAGRAM, region a denotes A and region b denotes A’. Region a and b together form UNIVERSAL set so, A∪A’=U.

Right OPTION is (c) A∪B

Easiest explanation: (A∪B) ∩ (B∪A) = (A∪B) ∩ (A∪B) (COMMUTATIVE law)

= A∪B (A∩A=A).

The correct ANSWER is (a) A

Easiest explanation: SET which contains all ODD NATURAL NUMBERS as well as all even natural numbers is set of natural numbers. So, B∪C=A.

The CORRECT option is (b) False

To EXPLAIN: If A={1,2,3} and B={3,4,5}, then A-B={1,2,3} – {3,4,5} = {1,2} and B-A={3,4,5} – {1,2,3} = {4,5}. They are not equal so,

A-B and B-A are not always equal.

The correct OPTION is (c) Ф⊆A

Easiest explanation: A null SET is a subset of EVERY set HENCE Ф is a subset of A

The correct answer is (c) May be equal

To elaborate: Equal sets are the sets having same number of ELEMENTS with VALUE of each element SET A and B to be equal. So, if the two sets have same number of elements, then they may or may not be equal.

Right option is (b) False

The best I can explain: SET A has ONE element 0 while set B has no element in it and is EMPTY so they both can’t be equal.

Right choice is (b) {-1, -2}

EXPLANATION: Solving the EQUATION:

X^2+2X+X+2=0

(x+2) (X+1) = 0

X= -2 and X = -1

Correct choice is (b) False

The best explanation: Given, n(U)=1000, n(F)=700, n(C)=600, n(F∩C) =100

n (F∪C) = n (F) + n (C) – n(F∩C)

n (F∪C) = 700+600-100=1200

We KNOW, n (A∪O) ≤ n(U). Here this is not valid so, data is incorrect.

The correct option is (b) B

Best explanation: Intersection of SET A and B is a set that contains elements which is common to both set A and set B.

A = {a, E, i, o, U}

B = {a, e, u}

A∩B = {a, e, u} = B.

The correct choice is (C) SET of Rectangle

The explanation: Set of rectangles is CONSIDERED as universal set for set of SQUARES because all squares are rectangles.

The CORRECT choice is (d) {1,2,3,4,5,6}

To EXPLAIN: UNIVERSAL set is the set which is superset of all basic sets of that TYPE.

{1,2,3,4,5,6} is the set which contains all the elements of set A, B, C.

The correct choice is (a) SET of isosceles TRIANGLES

To explain: Set of isosceles triangles can be CONSIDERED as UNIVERSAL set for set of equilateral triangles because all equilateral triangles are isosceles.

Correct option is (a) Φ∈A

The best I can explain: Null set is subset of EVERY set so, Φ⊂P(A) and Φ⊂A. Since Φ⊂A and POWER set of set A is set of all SUBSETS of set A so, Φ∈P(A). Hence Φ∈A is INCORRECT.

Right answer is (B) False

Explanation: A= {a, b, C}. Possible SUBSETS of SET A are φ, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}. So, P(A) = {φ, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}}.

Correct option is (a) TRUE

Explanation: The roots of x^3-6x^2+11x-6=0 are 1,2,3 hence A and B are the same sets so the STATEMENT is true.

Correct option is (a) X⊂Y

Easiest explanation: If every element of set X is in set Y then X is CALLED subset of Y.X⊂Y.

But every element of Y MAY or may not be the element of X so we can’t say Y⊂X and HENCE we can’t DECIDE EQUALITY.

Right option is (c) Set of mothers in a family

To EXPLAIN I would SAY: Set of points in a LINE, set of prime NUMBERS and natural numbers are uncountable hence BELONG to an infinite set. Here the number of mothers in a family can be counted.

Right answer is (c) A SET of points in a LINE

The EXPLANATION is: Since girls in a college, PLAYERS in a team and number of edges in a square are all fixed QUANTITIES they come under finite set and points on a line are infinite hence they come under infinite set.

The CORRECT option is (a) True

The explanation is: A number less than 5 cannot be greater than 7. If we plot points less than 5 on the number LINE and then plot numbers greater than 7 on it we find, no point is COMMON to both so, it is a null set.

The CORRECT ANSWER is (C) {{2,3}}

To EXPLAIN: POWER set of set A is set of all subsets of set A. Each element of power set is subset of the given set. Subsets of {2,3} is Φ, {2}, {3}, {2,3}.

Right ANSWER is (a) ZERO

Easiest EXPLANATION: Null set or empty set is a set which does not contain any ELEMENT. So, the number of ELEMENTS in a null set is zero.

The correct answer is (b) 2

The best I can explain: We know, N (T ∪ C) = n (T) + n (C) – n (T ∩ C)

GIVEN, n (T ∪ C) = 7, n(T)=4, n(C)=5

7 = 4 + 5 – n (T ∩ C)

n (T ∩ C) = 2.

Hence, 2 members like both tea and coffee.

The correct answer is (a) True

Easiest explanation: Complement of a SET A is a set which contains elements of U which are not the elements of A. A’ = U-A = {1, 2, 3, 4, 5, 6} – {1,4} = {2,3,5,6} and B’ = U-B = {1, 2, 3, 4, 5, 6} – {2,3,5} = {1,4,6}. A’∩B’= {2,3,5,6} ∩ {1,4,6} = {6}.

A∪B = {1,2,3,4,5} => (A∪B)’={6}.

Hence, A’∩B’=(A∪B)’.

The CORRECT option is (a) True

Explanation: If A⊂B then EVERY element of A is in set B. SINCE x is an element of A so, x also belong to B. x∈B is true.

The correct option is (c) Set of irrational numbers

The BEST I can EXPLAIN: There are two categories in real numbers which are rational and irrational. The real NUMBER which is not rational is irrational so, R-S DENOTES the set of irrational numbers.

The CORRECT answer is (c) {1,2}

EXPLANATION: A-B is the set of elements that belongs to A but not to B. Here, 1 and 2 belongs to A but not to B. So,

A = {1,2,3,4}

B = {3,4,5,6}

A – B = {1,2,3,4} – {3,4,5,6} = {1,2}.

The correct answer is (a) True

The best EXPLANATION: LET A = {1,2} and B = {2,3}. Here A∩B = {2} = B∩A.

A∩B or B∩A is same set that contains elements which are COMMON to both set A and B.

The CORRECT option is (b) A⊆B

The EXPLANATION is: 1 and 2 are both AVAILABLE in the set B Hence A is a subset of B.

Correct CHOICE is (b) False

Easiest EXPLANATION: Let X = {1,2}. A= {{1,2},3}, B= {{1,2},3,4}. Since ELEMENTS of X does not belongs to set B so, X is not a subset of B. X⊂B is false.

Right option is (d) Equal

Explanation: Prime numbers less than 6 are 2,3,5. A={2,3,5}

30=2*3*5 => 2,3,5 are prime factors of 30. B={2,3,5}

Since EVERY element of A is in B and every element of B is in A so they are equal sets.

Right choice is (b) An EMPTY set is a FINITE set

For explanation: An empty has 0 elements hence it is COUNTABLE and finite, Therefore, we SAY that empty set is a finite set. An infinite set has an uncountable number of elements.

The correct answer is (B) {X: x = 2^n where n ∈ N}

For EXPLANATION: The SEQUENCE is a geometric PROGRESSION with base 2 hence 2^n is the correct answer.

Right option is (d) {x: x = 2n-1 where n ∈ N}

EASY EXPLANATION: The given SEQUENCE is an odd number sequence of the format 2n+1 or 2n-1, but since n is given as a natural number so 2n+1 is not valid as 0 is not a natural number hence 2n-1 is the CORRECT ANSWER.

The correct ANSWER is (a) {3}

The best explanation: SINCE x is given as natural NUMBER so x can be positive only.x^2-9=0 => (x-3)(x+3)=0=>x=3,-3.

Here, -3 is not a natural number so, the SET {x : x is a natural number and x^2-9=0} can be written as {3}.

The correct OPTION is (a) Honest persons

Explanation: A SET is a collection of well defined objects but honesty has no precise definition.

Prime number is a number which can be divisible only by 1 and itself.

Even number is a number which can be divisible by 2.

Set of letters forming the word SCHOOL {S,C,H,O,L}.

Right choice is (d) 17

The EXPLANATION: n(U)=100, n(A)=72, n(O)=45

n (A∪O) = n (A) + n (O) – n(A∩O)

n (A∪O) ≤ n(U) => n (A∪O) ≤ 100

Greatest value of n (A∪O) is 100 which gives least value of n(A∩O)

100=72+45 – n(A∩O)

n(A∩O) = 17 least value of n(A∩O).

The CORRECT choice is (c) {6}

For EXPLANATION: Complement of a SET A is a set which contains elements of U which are not the elements of A. A’ = U-A = {1, 2, 3, 4, 5, 6} – {1,4} = {2,3,5,6} and B’ = U-B = {1, 2, 3, 4, 5, 6} – {2,3,5} = {1,4,6}. A’∩B’={2,3,5,6} ∩ {1,4,6} = {6}.

The correct ANSWER is (a) {2,3,5,6}

Explanation: Complement of a set A is a set which contains elements of U which are not the elements of A. So, A’ = U – A = {1, 2, 3, 4, 5, 6} – {1,4} = {2,3,5,6}.