Sets, Relations and Functions Test

Result

Sets, Relations and Functions Test - 5

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
2 / -0.83

If n (P) = 5, n(Q) = 12 and n(P U Q) = 14 then n(P ∩Q) =

A
3

B
4

C
5

D
7

Solution

n (P) = 5, n(Q) = 12 and n(PUQ) = 14
n(PUQ) = n(P) + n(Q) - n(P ∩Q)
14 = 5 + 12 - n(P ∩Q)
n(P ∩Q) = 3
Question 2
2 / -0.83

In a class of 50 students, 20 play Hockey, 15 play Cricket and 11 play Football. 7 play both Hockey and Cricket, 4 play Cricket and Football and 5 play Hockey and football. If 18 students do not play any of these given sports, how many students play exactly two of these sports?

A
12

B
10

C
11

D
15

Solution
Question 3
2 / -0.83

Let A and B be two finite disjoint sets such that n(A ∪B) = 475 and n(A) = 435, then n(B) is

A
75

B
35

C
900

D
40

Solution

As a and b are disjoint sets, they have no elements in common. Thus,
n(a ∪b) = n(a) + n(b)
→n(b) = n(a ∪b) −n(a)
= 475 −435
= 40
Question 4
2 / -0.83

If n(A) + n(B) + n(C) = n(A U B U C) then what are the sets A, B, C are called?

A
Joint

B
Disjoint

C
Equivalent

D
Equal

Solution

A and B are disjoint set so their is no common element between A and B.
n(AuBuC) = n(A)+n(B)+ n(C)-n(AnB)-n(Anc)-n(BnC)+n(AnBnC)
As we know n(AuB)=n(A)+ n(B)-n(AnB)
Here when we add A and B we are adding (AnB) two times , as it is present in both A and B. So we subtract (AnB) one time from A and B.
Which means we need to take the area of A and B set but no area should repeat.
When we take three sets A,B and C
We take area of A+ Area of B + area of C ————eq
But they are intersecting so the area we are taking is greater than actual area. We have again added (AnB), (AnC) and (BnC) twice. Therefore we subtract the intersecting area from eq one time as done above
A+B+C-(AnB)-(AnC)-(BnC)

The intersection of the sets a and b is the set of all the elements which belong to both A and B. It is denoted by A ∩B (“A intersection B ”).
•If A and B do not have any element in common then A ∩B= a null set = Ø.
Question 5
2 / -0.83

If in class XI of a certain school, 24 students got distinction in mathematics, 19 students got distinction in physics and 15 students got distinction in both the subjects, then the number of students in class XI is:

A
43

B
58

C
28

D
39

Solution

r :- c
Explanation:- Students got distinction in mathematics = 24
Student got distinction in physics = 19
Students got distinction in both= 15
Total number of students = (24+19)-15
= 28
Question 6
2 / -0.83

There is a group of 80 persons who can drive scooter or car or both. Out of these, 35 can drive scooter and 60 can drive car. Find how many can drive both scooter and car? How many can drive scooter only? How many can drive car only?(Respectively)

A
20, 45, 15

B
15, 20, 45

C
35, 35, 10

D
25, 10, 45

Solution

Let S = {Persons who drive scooter}
C = {Persons who drive car}
Given, n(S ∪C) = 80, n(S) = 35, n(C) = 60
Therefore, n(S ∪C) = n(S) + n(C) - n(S ∩C)
80 = 35 + 60 - n(S ∩C)
80 = 95 - n(S ∩C)
Therefore, n(S ∩C) = 95 –80 = 15
Therefore, 15 persons drive both
scooter and car.
Therefore, the number of persons who
drive a scooter only = n(S) - n(S ∩C)
= 35 –15
= 20
Also, the number of persons who drive car only = n(C) - n(S ∩C)
= 60 - 15
= 45
Question 7
2 / -0.83

If P and Q are two sets such that n(P) = 120, n(Q) = 50 and n(P ∪Q) = 140 then, n(P ∩Q) is:

A
20

B
30

C
70

D
90

Solution

n(P ∪Q) = n(P) + n(Q) - n(P ∩Q)
140 = 120 + 50 - x
140 = 170 -x
x = 30
Question 8
2 / -0.83

If two sets A and B are having 99 elements in common, then the number of elements common to each of the setsA × B andB × A are

A
2⁹⁹

B
100

C
18

D
99²

Solution

Correct option is D.
To find A ×B we take one element from set A and one from set B.
Given that 99 elements are common to both set A and set B.
Suppose these common elements are N₁ ,N₂ ,N₃ ,...N₉₉ .
Select an ordered pair for A ×B such that both are selected out of these common elements. Examples: (N₁ ,N₂ ),(N₃ ,N₅ )
All these will also be elements of B ×A. Hence number of elements common to A ×B and B ×A is 99 ×99=99² ( first element in ordered pair can be selected in 99 ways; second element can also be selected in 99 ways)
n[(A ×B)∩(B ×A)]=n[(A ∩B)∩(B ∩A)]=(99)(99)=99²
Question 9
2 / -0.83

If U = 100, n(A) = 30, n(B) = 40, n(A ∩B) = 10, then n( A '∩B ') =

A
50

B
30

C
40

D
90

Solution

By demorgan 's law
n( A '∩B ')=n(A ∪B)'= n(U) - n(A ∪B)
=100 - [ n(A) + n(B) - n(A ∩B) ]
=100 - 60 = 40
Question 10
2 / -0.83

The number of non-empty subsets of the set {1, 2, 3, 4} is

A
16

B
14

C
15

D
12

Solution

The number of non- empty subsets =
2 ⁿ −1 = 2⁴ −1 = 16 −1 = 15
Question 11
2 / -0.83

A group of 120 students, 90 take mathematics and 72 take economics. If 10 students take neither of the two, how many students take both:

A
20

B
62

C
52

D
30

Solution

P(U) = 120
Mathematics = P(A) = 90
Economics = P(B) = 72
P(A ∪B)'= 10
P(A ∪ B) = P(U) - P(A ∪ B)'
= 120 - 10 = 110
P(A ∪ B) = P(A) + P(B) - P(A ∪ B)
110 = 90 + 72 -P(A ∩ B)
P( A ∩ B) = 162 - 110 = 52
Question 12
2 / -0.83

In a class of 50 students, 35 students play cricket and 30 students play football; identify the students playing both the games.

A
20

B
115

C
15

D
65
Question 13
2 / -0.83

In a class of 40 students 14 take physics and 29 take chemistry. If 5 students take both, how many students take neither of the subjects?

A
5

B
14

C
38

D
2

Solution

Total students who took any of the subjects = 40 - x
Let x be the no. of students who took neither of the subjects
Then, 40 - x = 14 + 29 - 5
⇒x = 2
Question 14
2 / -0.83

If A = {1, 4, 9, 16} then n (A) = _____

A
1

B
2

C
3

D
4

Solution

N(A) indicates the number of elements in set A, there are 4 elements.
Question 15
2 / -0.83

Find n (A –B) from the figure

A
1

B
2

C
4

D
3

Solution

A={4,6,10} B={6,8}
A-B={4,10}
= n(A-B) = 2 Elements
Question 16
2 / -0.83

A,B,C are three sets such that n(A) = 25, n(B) = 20, n(c) = 27, n(A ∩B) = 5, n(B ∩C ) = 7 and A ∩C = ∅then n(A ∪B ∪C) is equal to

A
67

B
65

C
60

D
72

Solution

Here A ∩C = Ø ⇒ A ∩B ∩C = Ø
n (A ∪B ∪C) = n(A) + n(B) + n(C) - n(A ∩B) - n(B ∩C) - n(C ∩A) + n(A ∩B ∩C)
= 25 + 20 + 27 - 5 - 7 - 0 - 0
= 60
Question 17
2 / -0.83

In a factory 70% of the workers like oranges and 64% like apples. If x% like both oranges and apples, then

A
x ≤34

B
x ≥64

C
34 ≤x ≤64

D
None of these

Solution

let the total number of workers be 100. A = who likes oranges , B = who likes apples
So n(A ) = 70 and n(B) = 64, n(A ∩B) = x
Also n(A U B) ≤100
n(A) + n(B) -n(A ∩B) ≤100
70 + 64 - x ≤100
x ≥34
and n(A ∩B) ≤n(B) ⇒x ≤64
then 34 ≤x ≤64

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Sets, Relations and Functions Test - 5

Result

Sets, Relations and Functions Test - 5

Score

-

Rank

-

SHARING IS CARING

Question 1

3

4

5

7

Solution

Question 2

12

10

11

15

Solution

Question 3

75

35

900

40

Solution

Question 4

Joint

Disjoint

Equivalent

Equal

Solution

Question 5

43

58

28

39

Solution

Question 6

20, 45, 15

15, 20, 45

35, 35, 10

25, 10, 45

Solution

Question 7

20

30

70

90

Solution

Question 8

299

100

18

992

Solution

Question 9

50

30

40

90

Solution

Question 10

16

14

15

12

Solution

Question 11

20

62

52

30

Solution

Question 12

20

115

15

65

Question 13

2⁹⁹

99²