Sets(12th Grade > Mathematics ) Questions and answers for exam Preparation

12th Grade > Mathematics

SETS MCQs

Sets(11th And 12th Grade)

Total Questions : 60 | Page 1 of 6 pages

Question 1. Which of the following is an empty set

{x:x is a real number and x2−1=0}
{x:x is a real number and x2+1=0}
{x:x is a real number and x2−9=0}
{x:x is a real number and x2=x+2}

Discuss Question

Answer: Option B. -> {x:x is a real number and x2+1=0}
:
B
Since

x^{2}

+ 1 = 0, gives

x^{2} = - 1

⇒

x = \pm i

∴

x

is not real but

x

has to be is real (given)
∴ No real value of

x

is possible in this case.

Question 2. If X = {

4^{n}

- 3n - 1 : n ∈ N} and Y = { 9(n-1) : n ∈ N}, then X ∪ Y is equal to

X
Y
N
None of these

Discuss Question

Answer: Option B. -> Y
:
B
Since

4^{n} - 3 n - 1 = (3 + 1)^{n} - 3 n - 1 = 3^{n} +^{n} C_{1} 3^{n - 1} +^{n} C_{2} 3^{n - 2} + . . . . . +^{n} C_{n - 1} 3 +^{n} C_{n} - 3 n - 1 =^{n} C_{2} 3^{2} +^{n} C_{3} {.3}^{3} + . . . +^{n} C_{n} 3^{n}, (^{n} C_{0} =^{n} C_{n},^{n} C_{n - 1} =^{n} C_{1} . . . . . s o o n .) = 9 [^{n} C_{2} +^{n} C_{3} (3) + . . . . . . +^{n} C_{4} 3^{n - 1}]

∴ 4^{n} - 3 n - 1

is a multiple of 9 for

n \geq 2

F o r n = 1, 4^{n} - 3 n - 1 = 4 - 3 - 1 = 0 F o r n = 2, 4^{n} - 3 n - 1 = 16 - 6 - 1 = 9 ∴ 4^{n} - 3 n - 1

is a multiple of 9 for all

n ϵ N

∴ X

contains elements, which are multiples of 9, and clearly

Y

contains all multiples of 9.

∴ X \subset Y i . e ., X \cup Y = Y

Question 3. Let

n (A - B) = 25 + x, n (B - A) = 2 x

and

n (A \cap B) = 2 x .

n (A) = 2 (n (B)),

then

x

= ___.

Discuss Question

Answer: Option B. -> 5
:
B
Given that

n (A - B) = 25 + x, n (B - A) = 2 x

and

n (A \cap B) = 2 x .

We know that

n (A) = n (A - B) + n (A \cap B) .

= 25 + x + 2 x = 3 x + 25

Similarly,

n (B) = n (B - A) + n (A \cap B)

= 2 x + 2 x

4 x .

n (A) = 2 (n (B)) \Rightarrow 3 x + 25

2 \times 4 x

\Rightarrow 5 x = 25 \Rightarrow x = 5

Question 4. If

N_{a} = {a n : n \in N}

, then

N_{3} \cap N_{4}

Discuss Question

Answer: Option B. -> N12
:
B

N_{a} = {a n : n \in N}

N_{3} \cap N_{4} = {3, 6, 9, 12, 15 \dots \dots} \cap {4, 8, 12, 16, 20, \dots \dots}

= {12, 24, 36 \dots \dots} = N_{12}

[

∵

3,4 are relatively prime numbers ]

∴

N_{3} \cap N_{4}

N_{12}

Question 5. Let

F_{1}

be the set of all parallelograms,

F_{2}

the set of rectangles,

F_{3}

the set of rhombuses,

F_{4}

the set of squares and

F_{5}

the set of trapeziums in a plane, then

F_{1}

is equal to

F2∩F3
F2∪F3∪F4
F3∩F4
F2∩F1

Discuss Question

Answer: Option B. -> F2∪F3∪F4
:
B
Since every rectangle, rhombus and square is parallelogram so

F_{1}

F_{2} \cup F_{3} \cup F_{4}

Question 6. In a town of 10,000 families it was found that 40% family buy newspaper A, 20% buy newspaper B and 10% families buy newspaper C, 5% families buy A and B, 3% buy B and C and 4% buy A and C. If 2% families buy all the three newspapers, then number of families which buy A only is

3100
3300
2900
1400

Discuss Question

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

Question 7. In a group of 70 people, 37 like coffee, 52 like tea and each person like at least one of the two drinks. The number of persons liking both coffee and tea is:

Discuss Question

Answer: Option C. -> 19
:
C

n (A \cup B)

n (A) + n (B) - n (A \cap B)

We have, 70 = 37 + 52 -

n (A \cap B)

n (A \cap B)

= 19.

Question 8. If A = [ x:x is a multiple of 3] and B = [x:x is a multiple of 5] , then A - B is (

¯ A

means complement of A)

¯A ∩ B
A ∩ ¯B .
¯A ∩ ¯B .
A∩B

Discuss Question

Answer: Option B. -> A ∩ ¯B .
:
B
Try taking some values of A and B
We'll see that A - B = A

\cap

¯ B

Question 9. Let

n (A - B) = 25 + x, n (B - A) = 2 x

and

n (A \cap B) = 2 x .

n (A) = 2 (n (B)),

then

x

= ___.

Discuss Question

Answer: Option B. -> 5
:
B
Given that

n (A - B) = 25 + x, n (B - A) = 2 x

and

n (A \cap B) = 2 x .

We know that

n (A) = n (A - B) + n (A \cap B) .

= 25 + x + 2 x = 3 x + 25

Similarly,

n (B) = n (B - A) + n (A \cap B)

= 2 x + 2 x

4 x .

n (A) = 2 (n (B)) \Rightarrow 3 x + 25

2 \times 4 x

\Rightarrow 5 x = 25 \Rightarrow x = 5

Question 10. The group of intelligent students in a class is __________.

a null set
a finite set
a well defined collection
not a well defined collection

Discuss Question

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.