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

12th Grade > Mathematics

SETS MCQs

Sets(11th And 12th Grade)

Total Questions : 60 | Page 4 of 6 pages

Question 31. If A={1,2,3,4,5,6}, B={1,2}, then

\frac{A}{(\frac{A}{B})}

is equal to

A
N
A∩B
A∪B

Discuss Question

Answer: Option C. -> A∩B
:
C

\frac{A}{B}

= A - B = {3,4,5,6}
=

\frac{A}{(\frac{A}{B})}

= {1,2}

Question 32. 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.

Question 33. If a set A has n elements, then the total number of subsets of A is

Discuss Question

Answer: Option C. -> 2n
:
C
Number of subsets of A =

^{n} C_{0}

^{n} C_{1}

+ .............+

^{n} C_{n}

2^{n}

Question 34. 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 35.

If n (\cup) = 700, n (A) = 200, n (B) = 300

and

n (A \cap B) = 100,

then

n (A^{^{'}} \cap B^{^{'}}) =

___.

Discuss Question

Answer: Option C. -> 300
:
C
From de-Morgan's law of complementation, we have

A^{'} \cap B^{'} = (A \cup B)^{'} .

\Rightarrow n (A^{'} \cap B^{'}) = n ((A \cup B)^{'})

But,

n ((A \cup B)^{'}) = n (U) - n (A \cup B)

by definition of complement of a set.

∴ n (A^{'} \cap B^{'}) = n (U) - n (A \cup B)

= n (U) - [n (A) + n (B) - n (A \cap B)]

= 700 - (200 + 300 - 100)

= 300

Question 36. In a city 20 percent of the population travels by car, 50 percent travels by bus and 10 percent travels by both car and bus. Then the percentage of population travelling by car or bus is

80 percent
40 percent
60 percent
70 percent

Discuss Question

Answer: Option C. -> 60 percent
:
C
n(C) = 20, n(B) = 50, n(C∩ B) = 10
Now n(C∪ B) = n(C) + n(B) - n(C∩ B)
= 20 + 50 - 10 = 60.

Question 37. If X =

{8^{n} - 7 n - 1 : n \in N} a n d Y = {49 (n - 1) : n \in N}

, then

X ⊆ Y
Y ⊆ X
X = Y
None of these

Discuss Question

Answer: Option A. -> X ⊆ Y
:
A
Since

8^{n} - 7 n - 1 = (7 + 1)^{n} - 7 n - 1

7^{n} +^{n} C_{1} 7^{n - 1} +^{n} C_{2} 7^{n - 2} + . . . . . +^{n} C_{n - 1} 7 +^{n} C_{n} - 7 n - 1

^{n} C_{2} 7^{2} +^{n} C_{3} 7^{3} + . . . +^{n} C_{n} 7^{n}, (^{n} C_{0} =^{n} C_{n}^{n} C_{1} =^{n} C_{n - 1} e t c,)

= 49[

^{n} C_{2} +^{n} C_{3} (7) + . . . . . . . . +^{n} C_{n} 7^{n - 2}]

∴

8^{n} - 7 n - 1

is a multiple of 49 for n

\geq

2
For n = 1 ,

8^{n} - 7 n - 1 = 8 - 7 - 1 = 0;

For n = 2,

8^{n} - 7 n - 1 = 64 - 14 - 1 = 49

∴

8^{n} - 7 n - 1

is a multiple of 49 for n

\geq

∴

X contains elements which are multiples of 49 and clearly

γ

contains all multiplies of 49.

∴

\subseteq

Question 38. 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 39. In a committee 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. The number of persons speaking at least one of these two languages is

Discuss Question

Answer: Option A. -> 60
:
A