12th Grade > Mathematics
SETS MCQs
Sets(11th And 12th Grade)
Total Questions : 60
| Page 6 of 6 pages
Answer: Option C. -> 2n
:
C
Number of subsets of A =nC0 +nC1 + .............+nCn =2n
:
C
Number of subsets of A =nC0 +nC1 + .............+nCn =2n
Answer: Option B. -> [x : 1 ≤ x < 2]
:
B
A = {x:x∈ R, -1 < x < 1}
B = {x:x∈ R:x-1≤ -1 or x-1≥ 1}
= {x : x∈ R:x≤ 0 or x≥ 2}
∴ A∪ B = R - D, where D = {x : x∈ R, 1≤ x < 2}.
:
B
A = {x:x∈ R, -1 < x < 1}
B = {x:x∈ R:x-1≤ -1 or x-1≥ 1}
= {x : x∈ R:x≤ 0 or x≥ 2}
∴ A∪ B = R - D, where D = {x : x∈ R, 1≤ x < 2}.
Answer: Option C. -> 300
:
C
From de-Morgan's law of complementation, we have A′∩B′=(A∪B)′.
⇒n(A′∩B′)=n((A∪B)′)
But,n((A∪B)′)=n(U)−n(A∪B) by definition of complement of a set.
∴n(A′∩B′)=n(U)−n(A∪B)
=n(U)−[n(A)+n(B)−n(A∩B)]
=700−(200+300−100)
=300
:
C
From de-Morgan's law of complementation, we have A′∩B′=(A∪B)′.
⇒n(A′∩B′)=n((A∪B)′)
But,n((A∪B)′)=n(U)−n(A∪B) by definition of complement of a set.
∴n(A′∩B′)=n(U)−n(A∪B)
=n(U)−[n(A)+n(B)−n(A∩B)]
=700−(200+300−100)
=300
Answer: Option A. -> 80
:
A
Let the numberof people who do not play NFS be n(N') = 50. (Given)
Similarly, the number of people who do not play Dota be n(D') = 40. (Given)
And, thenumberof people who do not play any game be n((N ∪ D)')=n(N'∩ D') = 10. (Given) (de-Morgan's law)
We have to find the numberof people who do not play both the games = n(N∩ D)'.
We know from de-Morgan's law that for sets A and B,
(A∩B)′=A′∪B′.
So, n((N∩D)′)=n(N′∪D′)=n(N′)+n(D′)−n(N′∩D′)
=50+40−10
=80
:
A
Let the numberof people who do not play NFS be n(N') = 50. (Given)
Similarly, the number of people who do not play Dota be n(D') = 40. (Given)
And, thenumberof people who do not play any game be n((N ∪ D)')=n(N'∩ D') = 10. (Given) (de-Morgan's law)
We have to find the numberof people who do not play both the games = n(N∩ D)'.
We know from de-Morgan's law that for sets A and B,
(A∩B)′=A′∪B′.
So, n((N∩D)′)=n(N′∪D′)=n(N′)+n(D′)−n(N′∩D′)
=50+40−10
=80
Answer: Option A. -> A
:
A
If A and B are any two sets, then A∩B⊆A.
Also, A⊆A∪(A∩B)
⇒A⊆A∪(A∩B)⊆A
⇒A∪(A∩B)=A
:
A
If A and B are any two sets, then A∩B⊆A.
Also, A⊆A∪(A∩B)
⇒A⊆A∪(A∩B)⊆A
⇒A∪(A∩B)=A
Question 56. In a random survey 250 people participated. Out of 250 people who took part in the survey, 40 people have listened to Pink Floyd. 30 people have listened to metallica and 20 people have listened to John Denver. If 10 people have listened to all three then find the no. of people who have listened only Pink Floyd.
Answer: Option C. -> 30
:
C
Let the no. of people who have listened to Pink Floyd be n(P) = 40
Similarly, the no. of people who have listened to Metallica be n(M) = 30
and the no. of people who have listened to John Denver be= n(J) = 20
Also given that the no. of people who listen to all three i.e. n(P∩M∩J)=10
We have to find the no. of people who have listened to only Pink floyd.
Let that no. be x.
Now, from Demorgan's second law, we know that
P−(M∪J)=(P–M)∩(P–J)
n(P−(M∪J))=n(P–M)+n(P–J)–n((P–M)∪(P–J))
⇒n(P−(M∪J))=30+30−30
⇒n(P−(M∪J))=30
Thus, 30 people listen to only Pink Floyd.
:
C
Let the no. of people who have listened to Pink Floyd be n(P) = 40
Similarly, the no. of people who have listened to Metallica be n(M) = 30
and the no. of people who have listened to John Denver be= n(J) = 20
Also given that the no. of people who listen to all three i.e. n(P∩M∩J)=10
We have to find the no. of people who have listened to only Pink floyd.
Let that no. be x.
Now, from Demorgan's second law, we know that
P−(M∪J)=(P–M)∩(P–J)
n(P−(M∪J))=n(P–M)+n(P–J)–n((P–M)∪(P–J))
⇒n(P−(M∪J))=30+30−30
⇒n(P−(M∪J))=30
Thus, 30 people listen to only Pink Floyd.
Answer: Option D. -> A ∩ (Bc).
:
D
A∩(A∩B)c) = A∩ (Ac∪Bc)
= (A∩(Ac )∪ (A∩(Bc)
=∅∪ (A∩(Bc) = A∩(Bc).
:
D
A∩(A∩B)c) = A∩ (Ac∪Bc)
= (A∩(Ac )∪ (A∩(Bc)
=∅∪ (A∩(Bc) = A∩(Bc).
Answer: Option C. -> A∩B
:
C
AB = A - B = {3,4,5,6}
=A(AB) = {1,2}
:
C
AB = A - B = {3,4,5,6}
=A(AB) = {1,2}
Answer: Option B. -> 7
:
B
A set of n elements has 2n subsets.
Every set is a subset of itself.
Thus, the number of proper subsets of a set is 2n−1 subsets.
The set {1,2,3} has 3 elements and hence, 23−1=7 proper subsets.
:
B
A set of n elements has 2n subsets.
Every set is a subset of itself.
Thus, the number of proper subsets of a set is 2n−1 subsets.
The set {1,2,3} has 3 elements and hence, 23−1=7 proper subsets.
Answer: Option B. -> {x:x is a real number and x2 +1 = 0}
:
B
Sincex2 + 1 = 0, givesx2 = -1
⇒ x =± i
∴ x is not real but x is real (given)
∴ No value of x is possible.
:
B
Sincex2 + 1 = 0, givesx2 = -1
⇒ x =± i
∴ x is not real but x is real (given)
∴ No value of x is possible.