Which of the following statement is true for all A? (a) {ϕ} ∈ P(P(A)) (b) ϕ ∈ A (c) {ϕ} ⊂ A (d) {{ϕ}} ⊂ A

Correct! Wrong!

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Option A is not correct as ϕ is not always a element of A. Option B is not correct as {ϕ} is not always a subset of A. Option C is not correct as {ϕ} is not always a element of A. Option D is correct, as ϕ is always an element of P(A) and so {ϕ} is always an element of P(P(A)).

Of the options given, ____ is an empty set.

Correct! Wrong!

e ∈ A p ∈ B e ∈ D But as no letter can be both vowel and a consonant, C is a empty set.

The set A = {2 x2 − 3x + 1 = 0 x∈ N} is equal to ____.

Correct! Wrong!

The roots of the equation 2 x2- 3x + 1 = 0 are 1/2 and 1. But as x ∈ N, 1 is the only element of A. The set A =x: 2 x2- 3x + 1 = 0 x ∈ N is equal to {1}.

If A = {1, 2, 3, 5, 8}, B = {3, 5, 8}, then the proper subsets of A which are super sets of B are ____.

Correct! Wrong!

Option B is not correct as {3, 5, 8} in not included. Option C is not correct as {1, 2, 5, 8} is not a super set of B Option D is not correct as {1, 2, 5, 3} is not a super set of B The correct answer is A - {3, 5, 8}, {1, 3, 5, 8}, {2, 3, 5, 8}

If A and B are two sets, then A ∩ ( A U B) equals to ____.

Correct! Wrong!

As A ⊂ A ∪ B always A ∩ (A ∪ B) = A ∵ A⊂ B ⇒ A ∩ B = A]

B - (A ∩ B) is equal to ____.

Correct! Wrong!

B - (A ∩ B) = B ∩ (A ∩ B)C [because A - B = A ∩ B'] = B ∩ (AC U BC) [ By De Morgan's law] = (B ∩ AC) U (B ∩ BC) [By distributive law] = (B ∩ AC)U ϕ = B ∩ AC = B – A

If A is empty set, then number of elements in P (P(A)) is ____ [P (A) = denotes the power set of A.]

Correct! Wrong!

A is an empty set i.e A = ϕ P (A) = {ϕ} P (P (A)) = {ϕ, {ϕ}} So P (P (A)) has 2 elements.

Of the options given, ____ is a single tone set.

Correct! Wrong!

5, 6, 7 ∈ A, hence A is not the correct option. 5, 1/2, 5and 1/3 ∈ C, hence C is not the correct option. 6, 7 ∈ D, hence D is not the correct option. B = {6}, hence B is a sing tone set.

A market research group conducted a survey of 200 consumers and reported that 1240 like product A and 960 liked product B. The least number that must have liked both the products is ____.

Correct! Wrong!

The trick in the question lies in the fact that there are consumers who like neither product A nor B. Let S denote the set of consumers questioned, i.e the universal set , M be the set who liked product A and N be the set who liked product B Given that n(S) = 2000 n(M) = 1240 n(N) = 960 n(M U N) = n(M) + n(N) - n(M ∩ N) = 1240 + 960 - n(M ∩ N) = 2200 - n(M ∩ N) -------(1) But, n(M U N) is maximum when n(M ∩ N) is least. Now, M ∪ N ⊆ S ⇒ nM ∪ N ≤ nS ⇒nM∪N ≤ 2000 So the maximum value of n(M U N) is 2000 So by (1) 2200 - n(M ∩ N) < 2000 Therefore, 200 < n(M ∩ N) Hence, the least number of consumers of who liked both product A and B is 200.

If A and B are finite sets such that A⊂B, then n(A ∩ B) is equal to ____.

Correct! Wrong!

As A ⊂ B ⇒A ∩ B = A ⇒n(A ∩ B) = n(A)

Two finite sets have m and n elements respectively. The total number of subset of first set is 48 more than the total number of the second set. The values of m and n respectively are ___.

Correct! Wrong!

Let A and B be such sets, i.e n(A) = m and n(B) = n So, n(P(A)) = 2m and n(P(B)) = 2n 2m = 2n + 48 (This implies m > n) ⇒ 2m - 2n= 48 i.e. 2n (2m - n - 1) = 48 = 24× 3 ⇒ n = 4 and 2m - n -1 = 3 ⇒ 2m - n = 4 = 22 ⇒ m - n = 2 ⇒ m = n + 2 = 4 + 2 = 6 So, m = 6 and n = 4

The set A = {all elements in a week except Sunday} in roaster form is ____.

Correct! Wrong!

The elements in the set should contain all days, except Sunday. Hence, A = {Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}

Given A = {0, 1, 2, 3} and B =x ∈ R, │0 ≤ x ≤ 3, then ____.

Correct! Wrong!

A = {0, 1, 2, 3} and B = [0, 3] 0, 1, 2, 3 ∈ B ⇒ A ⊂ B Option B is not correct as B ⊄ A as 1/2 ∈ B but 1/2∉ A Option C is not correct as ϕ ∈ A ∩ B ⇒ A ∩ B ≠ ϕ Option D is not correct as A'⊄ B as 4 ∈ A' but 4 ∈ B

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} Given the sets A = {2, 5, 6}, B = {5, 6, 7}, C = {1, 2, 5, 7, 8}, then (i) (B U C')' is ____ and (ii) (A - B')' is ____.

Correct! Wrong!

C' = {3, 4, 6, 9, 10} B U C' = {3, 4, 5, 6, 7, 9, 10} (B U C')' = {1, 2, 8} B' = {1, 2, 3, 4, 8, 9, 10} A - B' = {5, 6} (A - B')' = {1, 2, 3, 4, 7, 8, 9, 10}

If A = {1, 3} B = {1, 3, 4}, then ____. [P(X) denotes the power set of X]

Correct! Wrong!

P(A) = {ϕ, {1}, {3}, {1, 3}} P(B) = {ϕ, {1}, {3}, {4}, {1, 3}, {1, 4}, {3, 4}, {1, 3, 4}} So, ϕ ∈ P(B) {1} ∈ P(B) {3} ∈ P(B) {1, 3} ∈ PB i.e every element of P(A) is an element of P(B) Therefore, P(A) ⊆ P(B)

If A = {2, 4, 5, 7, 8}, B = {3, 5, 7, 9, 10} and C = {1, 2, 3, 4, 5}, then (A ∪ B) ∩ C = ____. .

Correct! Wrong!

Given, A = {2, 4, 5, 7, 8}, B = {3, 5, 7, 9, 10} and C = {1, 2, 3, 4, 5} (A ∪ B) ∩ C A ∪ B = {2, 4, 5, 7, 8} ∪ {3, 5, 7, 9, 10} = {2, 3, 4, 5, 7, 8, 9, 10} (A ∪ B) ∩ C = {2, 3, 4, 5, 7, 8, 9, 10} ∩ {1, 2, 3, 4, 5} = {2, 3, 4, 5}

Given A = {2, 5, 7, 8} B = {1, 5, 6, 8} C = {5, 8, 10, 12} Then, the statement that is true is ____.

Correct! Wrong!

B U C = {1, 5, 6, 8, 10, 12} A - (B U C) = {2, 5, 7, 8} - {1, 5, 6, 8, 10, 12} = {2, 7} A - B = {2, 5, 7, 8, } - {1, 5, 6, 8} = {2, 7} A - C = {2, 5, 7, 8} - {5, 8, 10, 12} = {2, 7} (A - B) ∩ (A - C) = {2, 7} ∩ {2, 7} = {2, 7} Therefore, A - (B U C) = (A - B) ∩ (A - C) Option C is true. Option A is not correct as 2 ∈ A and 2 ∉ B and C Option B is not correct as 2∈ A - B, but 2 ∉ B - C Option D is not correct as A - (B ∩ C) = {2, 5, 7, 8} - {5, 8} = (2, 7} A ∩ B ∩C = {5, 8}

The set A = {.....-9, -6 -3, 0, 3, 6, 9, 12 ...} in set-builder form is ____.

Correct! Wrong!

The elements in the set are of the form 3 × 3. Hence, the correct answer is {x: x is a multiple of 3}.

The power set of A = {1, {2, 3}, {4}} is ____.

Correct! Wrong!

Option A in not correct as {2,3} and {4} are not subsets of A. (in fact, they are elements of A) Option B is not correct as {1,4} and {2,3,4} are not subsets of A Option C is not correct, as ϕ is not included in the power set. The right answer is Option D - {ϕ, {1}, {{2, 3}}, {{4}}, {1, {2, 3}} {{2, 3}}, {4}}, {1, {4}}, {1,{2,3},{4}}}.

From the sets given below, select equal sets. P = {1, 2, 3, 5}, Q = {2, 7, 8, 5}, R = {a, b}, S = {1 − 3}, T = {7, 2, 8, 5}, U = {−3, 1}

Correct! Wrong!

We know that, two sets are said to be equal if they have exactly the same elements. P = {1, 2, 3, 5}, Q = {2, 7, 8, 5}, R = {a, b}, S = {1 − 3}, T = {7, 2, 8, 5}, U = {−3, 1} Q = T and S = U are the equal sets, because all the elements of set Q are elements of set T and vice-versa and all the elements of set S are elements of set U and vice-versa.

Which of the statement is false?

Correct! Wrong!

Let's take A = {1, 2, 3}, B = {2, 3, 4} B - A = {4} {2, 3, 4} ⊈ {4} Hence, B ⊂ B - A is false.

U = {a, b, c, d, e, 1, 2, 5} = Universal set If A = {a, 2, 5}, B = {a, b, 5} and C = {b, 2, 1, d}, then (AC U B) ∩ C is equal to ____.

Correct! Wrong!

AC = {b, c, d, e, 1} AC U B = {a, b, c, d, e, 1, 5} (AC U B) ∩ C = {b, d, 1}

In a school, there are 30 students who have taken Mathematics or Physics. Of these, 18 students have taken Mathematics and 7 taken both Mathematics and Physics. The number who has taken Physics is ____.

Correct! Wrong!

Let M denote the set of students who have taken Mathematics and P denote for Physics. Therefore, n(M U P) = 30 and n(M ∩ P) = 7, n(M) = 18 Now, n(M U P) = n(M) + n(P) - n(M ∩ P) 30 = 18 + n(P) – 7 n(P) = 30 - 18 + 7 n(p) = 30 - 18 + 7 = 19 Therefore, 19 students have taken physics.

(A ∩ C')' U (B ∩ C) is equal to ____.

Correct! Wrong!

(A ∩ C')' U (B ∩ C) = (A' U (C')') U (B ∩ C) [By De Morgan's law] = (A' U C) U (B ∩ C) = [(A' U C) U B] ∩ [(A' U C) U C] = [A' U B U C] ∩ [A' U C] = A' ∩ C [∵A'∪ C ⊂ A'∪ B ∪ C and A ⊂ B ⇒ A ∩ B = A]

A U (B - A) is equal to ____.

Correct! Wrong!

A U (B - A) = A U (B ∩ AC) = (A U B) ∩ (A U AC) [Distributive law] = (A U B) ∩ (U) [U universal set] = A U B [because A ∩ U = A, for all sets A]

CBSE Class 11th Math 1 - Sets MCQs
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