12th Grade > Mathematics
SETS MCQs
Sets(11th And 12th Grade)
Total Questions : 60
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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 has to be is real (given)
∴ No real value of x is possible in this case.
:
B
Sincex2 + 1 = 0, givesx2=−1
⇒ x=±i
∴ x is not real but x has to be is real (given)
∴ No real value of x is possible in this case.
Answer: Option B. -> Y
:
B
Since
4n−3n−1=(3+1)n−3n−1=3n+nC13n−1+nC23n−2+.....+nCn−13+nCn−3n−1=nC232+nC3.33+...+nCn3n,(nC0=nCn,nCn−1=nC1.....soon.)=9[nC2+nC3(3)+......+nC43n−1]
∴4n−3n−1is a multiple of 9 for n≥2.
Forn=1,4n−3n−1=4−3−1=0Forn=2,4n−3n−1=16−6−1=9∴4n−3n−1 is a multiple of 9 for all nϵN
∴X contains elements, which are multiples of 9, and clearly Y contains all multiples of 9.
∴X⊂Yi.e.,X∪Y=Y
:
B
Since
4n−3n−1=(3+1)n−3n−1=3n+nC13n−1+nC23n−2+.....+nCn−13+nCn−3n−1=nC232+nC3.33+...+nCn3n,(nC0=nCn,nCn−1=nC1.....soon.)=9[nC2+nC3(3)+......+nC43n−1]
∴4n−3n−1is a multiple of 9 for n≥2.
Forn=1,4n−3n−1=4−3−1=0Forn=2,4n−3n−1=16−6−1=9∴4n−3n−1 is a multiple of 9 for all nϵN
∴X contains elements, which are multiples of 9, and clearly Y contains all multiples of 9.
∴X⊂Yi.e.,X∪Y=Y
Answer: Option B. -> 5
:
B
Given that n(A−B)=25+x,n(B−A)=2x and n(A∩B)=2x.
We know that n(A)=n(A−B)+n(A∩B).=25+x+2x=3x+25
Similarly, n(B)=n(B−A)+n(A∩B)
=2x+2x = 4x.
n(A)=2(n(B))⇒3x+25 = 2×4x
⇒5x=25⇒x=5
:
B
Given that n(A−B)=25+x,n(B−A)=2x and n(A∩B)=2x.
We know that n(A)=n(A−B)+n(A∩B).=25+x+2x=3x+25
Similarly, n(B)=n(B−A)+n(A∩B)
=2x+2x = 4x.
n(A)=2(n(B))⇒3x+25 = 2×4x
⇒5x=25⇒x=5
Answer: Option B. -> N12
:
B
Na={an:n∈N}
N3∩N4={3,6,9,12,15……}∩{4,8,12,16,20,……}
={12,24,36……}=N12
[∵ 3,4 are relatively prime numbers ]
∴N3∩N4 = N12
:
B
Na={an:n∈N}
N3∩N4={3,6,9,12,15……}∩{4,8,12,16,20,……}
={12,24,36……}=N12
[∵ 3,4 are relatively prime numbers ]
∴N3∩N4 = N12
Answer: Option B. -> F2∪F3∪F4
:
B
Since every rectangle, rhombus and square is parallelogram so
F1=F2∪F3∪F4
:
B
Since every rectangle, rhombus and square is parallelogram so
F1=F2∪F3∪F4
Answer: Option B. -> 3300
:
B
n(A) = 40% of 10,000 = 4,000
n(B) = 20% of 10,000 = 2,000
n(C) = 10% of 10,000 = 1,000
n(A∩ B) = 5% of 10,000 = 500
n(B∩ C) = 3% of 10,000 = 300
n(C∩ A) = 4% of 10,000 = 400
n(A∩ B∩ C) = 2% of 10,000 = 200
We want to find the number of families which buy only A= n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B∩ C)]
=4000 - [500 + 400 - 200] = 4000 - 700 = 3300
:
B
n(A) = 40% of 10,000 = 4,000
n(B) = 20% of 10,000 = 2,000
n(C) = 10% of 10,000 = 1,000
n(A∩ B) = 5% of 10,000 = 500
n(B∩ C) = 3% of 10,000 = 300
n(C∩ A) = 4% of 10,000 = 400
n(A∩ B∩ C) = 2% of 10,000 = 200
We want to find the number of families which buy only A= n(A) - [n(A ∩ B) + n(A ∩ C) - n(A ∩ B∩ C)]
=4000 - [500 + 400 - 200] = 4000 - 700 = 3300
Answer: Option C. -> 19
:
C
n(A∪B) = n(A)+n(B)−n(A∩B)
We have, 70 = 37 + 52 - n(A∩B)
n(A∩B) = 19.
:
C
n(A∪B) = n(A)+n(B)−n(A∩B)
We have, 70 = 37 + 52 - n(A∩B)
n(A∩B) = 19.
Answer: Option B. -> A ∩ ¯B .
:
B
Try taking some values of A and B
We'll see that A - B = A ∩¯B .
:
B
Try taking some values of A and B
We'll see that A - B = A ∩¯B .
Answer: Option B. -> 5
:
B
Given that n(A−B)=25+x,n(B−A)=2x and n(A∩B)=2x.
We know that n(A)=n(A−B)+n(A∩B).=25+x+2x=3x+25
Similarly, n(B)=n(B−A)+n(A∩B)
=2x+2x = 4x.
n(A)=2(n(B))⇒3x+25 = 2×4x
⇒5x=25⇒x=5
:
B
Given that n(A−B)=25+x,n(B−A)=2x and n(A∩B)=2x.
We know that n(A)=n(A−B)+n(A∩B).=25+x+2x=3x+25
Similarly, n(B)=n(B−A)+n(A∩B)
=2x+2x = 4x.
n(A)=2(n(B))⇒3x+25 = 2×4x
⇒5x=25⇒x=5
Answer: Option D. -> not a well defined collection
:
D
Intelligence cannot be defined for students in a class. Hence, the group of intelligent students is nota well defined collection.
:
D
Intelligence cannot be defined for students in a class. Hence, the group of intelligent students is nota well defined collection.