Free Objective Test 02 Practice Test - 11th and 12th
Question 1
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
SOLUTION
Solution : B
Try taking some values of A and B
We'll see that A - B = A ∩ ¯B .
Question 2
If A = [(x,y): x2+y2=25] And B = [(x,y): x2+9y2=144], then A∩B contains
One point
Three points
Two points
Four points
SOLUTION
Solution : D
A = Set of all values (x,y) : x2+y2=25=52
B = x2144+y216=1 i.e., x2(12)2+y2(4)2=1
Clearly , A ∩ B consists of four points.
Question 3
Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in (A U B )
3
6
9
18
SOLUTION
Solution : B
n(A ∪ B) = n(A) + n(B) - n(A ∩B) = 3 + 6 - 1 (A∩B)
Since maximum number of elements in A ∩ B = 3
∴ Minimum number of elements in A ∪ B = 9 - 3 = 6 .
Question 4
If Na={an:n∈N} , then N3∩N4 =
N7
N12
N3
N4
SOLUTION
Solution : 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
Question 5
If X = {8n−7n−1:n∈N} and Y={49(n−1):n∈N} , then
X ⊆ Y
Y ⊆ X
X = Y
None of these
SOLUTION
Solution : A
Since 8n−7n−1=(7+1)n−7n−1
= 7n+nC17n−1+nC27n−2+.....+nCn−17+nCn−7n−1
= nC272+nC373+...+nCn7n,(nC0=nCnnC1=nCn−1etc,)
= 49[nC2+nC3(7)+........+nCn7n−2]
∴ 8n−7n−1 is a multiple of 49 for n ≥ 2
For n = 1 , 8n−7n−1=8−7−1=0;
For n = 2, 8n−7n−1=64−14−1=49
∴ 8n−7n−1 is a multiple of 49 for n ≥ N
∴ X contains elements which are multiples of 49 and clearly γ
contains all multiplies of 49. ∴ X ⊆ Y.
Question 6
If n(∪)=700, n(A)=200, n(B)=300 and n(A∩B)=100, then n(A′∩B′)= ___.
400
240
300
500
SOLUTION
Solution : 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
Question 7
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
60
40
38
58
SOLUTION
Solution : A
n(S∪F) = n(S)+n(F)−n(S∩F)
⇒n(S∪F) = 20 + 50 -10 = 60
Question 8
If A={1,2,3,4,5,6}, B={1,2}, then A(AB) is equal to
A
N
A∩B
A∪B
SOLUTION
Solution : C
AB = A - B = {3,4,5,6}
=A(AB) = {1,2}
Question 9
Let n(A−B)=25+x, n(B−A)=2x and n(A∩B)=2x. If n(A)=2(n(B)), then x = ___.
4
5
6
7
SOLUTION
Solution : 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
Question 10
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:
16
13
19
20
SOLUTION
Solution : C
n(A∪B) = n(A)+n(B)−n(A∩B)
We have, 70 = 37 + 52 - n(A∩B)
n(A∩B) = 19.
Question 11
Let F1 be the set of all parallelograms, F2 the set of rectangles, F3 the set of rhombuses, F4 the set of squares and F5 the set of trapeziums in a plane, then F1 is equal to
F2∩F3
F2∪F3∪F4
F3∩F4
F2∩F1
SOLUTION
Solution : B
Since every rectangle, rhombus and square is parallelogram so
F1=F2∪F3∪F4
Question 12
Of the 200 candidates who were interviewed for a position at a call center, 100 had a two-wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both two-wheeler and credit card, 30 had both credit card and mobile phone and 60 had both two wheeler and mobile phone and 10 had all three. How many candidates had none of the three?
0
10
20
18
SOLUTION
Solution : B
Number of candidates who had none of the three = Total number of candidates - number of candidates who had at least one of three devices.
Total number of candidates = 200 =n(U), where U is the universal set
Number of candidates who had at least one of the three =n(A∪B∪C), where A is the set of those who have a two wheeler, B the set of those who have a credit card and C the set of those who have a mobile phone.
We know that n(A∪B∪C)=n(A)+n(B)+n(C)
−[n(A∩B)+n(B∩C)+n(C∩A)]+n(A∩B∩C)
Therefore, n(A∪B∪C)=100+70+140−{40+30+60}+10
Or n(A∪B∪C)=190.
As 190 candidates who attended the interview had at least one of the three gadgets, n(U)−n(A∪B∪C)= 200 - 190 = 10 candidates had none of three.
Question 13
A={x:x∈R, x2=16 and 2x=6} can be represented in the roster form as _________ .
A = {}
A = { 14, 3, 4 }
A = { 3 }
A = { 4 }
SOLUTION
Solution : A
x2=16⇒x=±4
2x=6⇒x=3There is no value of x which satisfies both the given equations. The set A is an empty set or a null set.
Thus, A = {}.
Question 14
If a set A has n elements, then the total number of subsets of A is
n
n2
2n
2n
SOLUTION
Solution : C
Number of subsets of A = nC0 + nC1 + .............+ nCn = 2n
Question 15
The group of intelligent students in a class is __________.
a null set
a finite set
a well defined collection
not a well defined collection
SOLUTION
Solution : D
Intelligence cannot be defined for students in a class. Hence, the group of intelligent students is not a well defined collection.