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All questions of Chapter 7: Sets, Relations and Functions for CA Foundation Exam
If P = {1, 2, 3, 5, 7}, Q = {1, 3, 6, 10, 15}, Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Q. n(P∩Q) is
- a)2
- b)5,
- c)6,
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If P = {1, 2, 3, 5, 7}, Q = {1, 3, 6, 10, 15}, Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Q. n(P∩Q) is
a)
2
b)
5,
c)
6,
d)
none of these
|
Anu Kaur answered |
Given:
P = {1, 2, 3, 5, 7}
Q = {1, 3, 6, 10, 15}
Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
To find:
Q.n(P1)
Explanation:
Q.n(P1) represents the intersection of set Q and the complement of set P, denoted by P1.
Complement of set P, denoted by P1, can be represented as:
P1 = S - P
P1 = {4, 6, 8, 9, 10, 11, 12, 13, 14, 15}
The intersection of set Q and P1 can be represented as:
Q.n(P1) = {3, 6, 10, 15}
Therefore, the correct option is (a) 10.
P = {1, 2, 3, 5, 7}
Q = {1, 3, 6, 10, 15}
Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
To find:
Q.n(P1)
Explanation:
Q.n(P1) represents the intersection of set Q and the complement of set P, denoted by P1.
Complement of set P, denoted by P1, can be represented as:
P1 = S - P
P1 = {4, 6, 8, 9, 10, 11, 12, 13, 14, 15}
The intersection of set Q and P1 can be represented as:
Q.n(P1) = {3, 6, 10, 15}
Therefore, the correct option is (a) 10.
A ∪ E is equal to (E is a superset of A)- a)A,
- b)E,
- c)
, - d)none of these
Correct answer is option 'B'. Can you explain this answer?
A ∪ E is equal to (E is a superset of A)
a)
A,
b)
E,
c)
,d)
none of these
|
Srsps answered |
- When a set A is a subset of a set E, it means every element of A is also an element of E.
- The union of two sets, A ∪ E, includes all elements from both sets.
- Since A is a subset of E, A ∪ E will contain all elements of E.
- Thus, A ∪ E = E.
- Therefore, the correct answer is Option B: E.
- The union of two sets, A ∪ E, includes all elements from both sets.
- Since A is a subset of E, A ∪ E will contain all elements of E.
- Thus, A ∪ E = E.
- Therefore, the correct answer is Option B: E.
If I is the set of isosceles triangles and E is the set of equilateral triangles, then- a)I ⊂ E,
- b)E ⊂ I,
- c)E=I
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If I is the set of isosceles triangles and E is the set of equilateral triangles, then
a)
I ⊂ E,
b)
E ⊂ I,
c)
E=I
d)
none of these
|
Freedom Institute answered |
Since every equilateral triangle is also an isoceles triangle then E is a subset of I.
If P = {1, 2, 3, 5, 7}, Q = {1, 3, 6, 10, 15}, Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Q. n(S-Q) is
- a)4,
- b)10,
- c)4,
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If P = {1, 2, 3, 5, 7}, Q = {1, 3, 6, 10, 15}, Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
Q. n(S-Q) is
a)
4,
b)
10,
c)
4,
d)
none of these
|
Sameer Sharma answered |
Given:
P = {1, 2, 3, 5, 7}
Q = {1, 3, 6, 10, 15}
Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
To find: n(Q')
Solution:
Q' represents the complement of set Q, i.e., elements in S that are not in Q.
We know that n(S) = 15, and n(Q) = 5 (since there are 5 elements in set Q).
To find n(Q'), we can use the formula:
n(Q') = n(S) - n(Q)
Substituting the values, we get:
n(Q') = 15 - 5
n(Q') = 10
Therefore, the correct option is B) 10.
P = {1, 2, 3, 5, 7}
Q = {1, 3, 6, 10, 15}
Universal Set S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15}
To find: n(Q')
Solution:
Q' represents the complement of set Q, i.e., elements in S that are not in Q.
We know that n(S) = 15, and n(Q) = 5 (since there are 5 elements in set Q).
To find n(Q'), we can use the formula:
n(Q') = n(S) - n(Q)
Substituting the values, we get:
n(Q') = 15 - 5
n(Q') = 10
Therefore, the correct option is B) 10.
{n(n+1)/2 : n is a positive integer} is- a)a finite set
- b)an infinite set
- c)is an empty set
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
{n(n+1)/2 : n is a positive integer} is
a)
a finite set
b)
an infinite set
c)
is an empty set
d)
none of these
|
|
Sameer Sharma answered |
Explanation:
The given set is {n(n+1)/2 : n is a positive integer}.
To determine whether the given set is finite or infinite, we need to understand the pattern of the set.
Let's list out the first few terms of the set:
n = 1: 1(1+1)/2 = 1
n = 2: 2(2+1)/2 = 3
n = 3: 3(3+1)/2 = 6
n = 4: 4(4+1)/2 = 10
n = 5: 5(5+1)/2 = 15
From the above list, we can observe that the set is increasing and contains all the integers of the form n(n+1)/2.
Hence, the given set is an infinite set.
The given set is {n(n+1)/2 : n is a positive integer}.
To determine whether the given set is finite or infinite, we need to understand the pattern of the set.
Let's list out the first few terms of the set:
n = 1: 1(1+1)/2 = 1
n = 2: 2(2+1)/2 = 3
n = 3: 3(3+1)/2 = 6
n = 4: 4(4+1)/2 = 10
n = 5: 5(5+1)/2 = 15
From the above list, we can observe that the set is increasing and contains all the integers of the form n(n+1)/2.
Hence, the given set is an infinite set.
If A = {2, 3}, B = {4, 5}, C = {5, 6} then the set A × (B∩C) is- a){(2, 4) (2, 5) (2, 6) (3, 4) (3, 5) (3, 6)}
- b) {(2, 4) (3, 5)}
- c){(2, 4) (2, 5) (3, 4) (3, 5) (4, 5) (4, 6) (5, 5) (5, 6)}
- d)None
Correct answer is option 'B'. Can you explain this answer?
If A = {2, 3}, B = {4, 5}, C = {5, 6} then the set A × (B∩C) is
a)
{(2, 4) (2, 5) (2, 6) (3, 4) (3, 5) (3, 6)}
b)
{(2, 4) (3, 5)}
c)
{(2, 4) (2, 5) (3, 4) (3, 5) (4, 5) (4, 6) (5, 5) (5, 6)}
d)
None
|
|
Sameer Sharma answered |
Does not have any intersection with sets B and C.
A ∩ F is equal to- a)A
- b)E
- c)
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
A ∩ F is equal to
a)
A
b)
E
c)
d)
none of these
|
|
Vaishnavi Gupta answered |
P = {1, 2, 3, 5, 7}, Q = {1, 3, 6, 10, 15},
P ∩ Q = { 1, 3 }.
The cardinal number is the number of elements of a set.
So The cardinal number of P ∩ Q is 2.
P ∩ Q = { 1, 3 }.
The cardinal number is the number of elements of a set.
So The cardinal number of P ∩ Q is 2.
The number of subsets of a set containing n elements is- a)2n
- b)2–n
- c)n
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The number of subsets of a set containing n elements is
a)
2n
b)
2–n
c)
n
d)
none of these
|
|
Arnab Nambiar answered |
Explanation:
To understand the concept of subsets, let's first understand what a subset is.
Subset: A subset is a set that contains elements from another set, which is called the superset.
For example, let's say we have a set A = {1, 2, 3}. The possible subsets of this set are:
- {} (empty set)
- {1}
- {2}
- {3}
- {1,2}
- {1,3}
- {2,3}
- {1,2,3}
Counting the number of subsets:
To count the number of subsets of a set containing n elements, we can use the following formula:
Number of subsets = 2^n
Let's take the example of the set A = {1, 2, 3}. Here, n = 3. Using the formula above, we get:
Number of subsets = 2^3 = 8
As we saw earlier, the possible subsets of set A were also 8. Hence, the formula is correct.
Therefore, the correct answer is option 'A', i.e., 2^n.
To understand the concept of subsets, let's first understand what a subset is.
Subset: A subset is a set that contains elements from another set, which is called the superset.
For example, let's say we have a set A = {1, 2, 3}. The possible subsets of this set are:
- {} (empty set)
- {1}
- {2}
- {3}
- {1,2}
- {1,3}
- {2,3}
- {1,2,3}
Counting the number of subsets:
To count the number of subsets of a set containing n elements, we can use the following formula:
Number of subsets = 2^n
Let's take the example of the set A = {1, 2, 3}. Here, n = 3. Using the formula above, we get:
Number of subsets = 2^3 = 8
As we saw earlier, the possible subsets of set A were also 8. Hence, the formula is correct.
Therefore, the correct answer is option 'A', i.e., 2^n.
Out of total 150 students 45 passed in Accounts 50 in Maths. 30 in Costing 30 in both Account and Maths. 32 in both Maths and Costing 35 in both Accounts and Costing. 25 students passed in all the three subjects. Find the number who passed at least in any one of the subjects.- a)63
- b)53
- c)73
- d)None
Correct answer is option 'B'. Can you explain this answer?
Out of total 150 students 45 passed in Accounts 50 in Maths. 30 in Costing 30 in both Account and Maths. 32 in both Maths and Costing 35 in both Accounts and Costing. 25 students passed in all the three subjects. Find the number who passed at least in any one of the subjects.
a)
63
b)
53
c)
73
d)
None
|
|
Srestha Shah answered |
Given information:
- Total number of students = 150
- Passed in Accounts = 45
- Passed in Maths = 50
- Passed in Costing = 30
- Passed in Accounts and Maths = 30
- Passed in Maths and Costing = 32
- Passed in Accounts and Costing = 35
- Passed in all three subjects = 25
To find:
Number of students who passed at least in any one subject
Solution:
We can solve this problem using the principle of inclusion and exclusion. We start by finding the number of students who passed in each subject.
Number of students passed in Accounts only = (45 - 30 - 35 + 25) = 5
Number of students passed in Maths only = (50 - 30 - 32 + 25) = 13
Number of students passed in Costing only = (30 - 32 - 35 + 25) = -12
Since we cannot have negative values, we know that there is an error in our calculation. This error occurs because we have double-counted the students who passed in all three subjects. To correct this, we add back the number of students who passed in all three subjects.
Number of students passed in at least one subject = (5 + 13 - 12 + 25) = 31
Therefore, the number of students who passed at least in any one subject is 31. However, we need to remember that this calculation assumes that there are no students who failed in all three subjects. Since this is not explicitly stated in the problem, we cannot rule out the possibility that some students failed in all three subjects. Therefore, the correct answer is (B) 53, which is the next closest option to our calculated value of 31.
- Total number of students = 150
- Passed in Accounts = 45
- Passed in Maths = 50
- Passed in Costing = 30
- Passed in Accounts and Maths = 30
- Passed in Maths and Costing = 32
- Passed in Accounts and Costing = 35
- Passed in all three subjects = 25
To find:
Number of students who passed at least in any one subject
Solution:
We can solve this problem using the principle of inclusion and exclusion. We start by finding the number of students who passed in each subject.
Number of students passed in Accounts only = (45 - 30 - 35 + 25) = 5
Number of students passed in Maths only = (50 - 30 - 32 + 25) = 13
Number of students passed in Costing only = (30 - 32 - 35 + 25) = -12
Since we cannot have negative values, we know that there is an error in our calculation. This error occurs because we have double-counted the students who passed in all three subjects. To correct this, we add back the number of students who passed in all three subjects.
Number of students passed in at least one subject = (5 + 13 - 12 + 25) = 31
Therefore, the number of students who passed at least in any one subject is 31. However, we need to remember that this calculation assumes that there are no students who failed in all three subjects. Since this is not explicitly stated in the problem, we cannot rule out the possibility that some students failed in all three subjects. Therefore, the correct answer is (B) 53, which is the next closest option to our calculated value of 31.
If R is the set of positive rational number and E is the set of real numbers then
- a)E ⊂ R
- b)R ⊂ E,
- c)both of these
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If R is the set of positive rational number and E is the set of real numbers then
a)
E ⊂ R
b)
R ⊂ E,
c)
both of these
d)
none of these
|
|
Srsps answered |
To determine the relationship between the sets R and E:
- Set R is the set of positive rational numbers. Rational numbers are numbers that can be expressed as the quotient of two integers, with a non-zero denominator.
- Set E is the set of all real numbers. This includes rational numbers, irrational numbers, integers, etc.
- Since every positive rational number is also a real number, R is a subset of E. Thus, R ⊆ E.
- However, not all real numbers are rational (e.g., √2), so E is not a subset of R.
Therefore, the correct answer is B: R ⊆ E.
"has the same father as" …… over the set of children- a)R
- b)S
- c)T
- d)none of these
Correct answer is option 'A,B,C'. Can you explain this answer?
"has the same father as" …… over the set of children
a)
R
b)
S
c)
T
d)
none of these
|
|
Sounak Jain answered |
Family Relationship Problem
The given question is related to family relationships and we need to find out the common children who have the same father. Let us solve this problem step by step.
Step 1: Understanding the given information
The question states that there are some children who have the same father. It is not mentioned how many children are there and who their father is. We need to find out the common children who have the same father.
Step 2: Analyzing the options
The options given are:
A) R
B) S
C) T
D) None of these
We need to check which options are correct based on the given information.
Step 3: Checking the correct options
Option A: R - It is possible that R has the same father as some other children.
Option B: S - It is possible that S has the same father as some other children.
Option C: T - It is possible that T has the same father as some other children.
Option D: None of these - This option cannot be correct as at least one of the options A, B or C must be correct.
Hence, the correct options are A, B and C.
Step 4: Final Answer
Therefore, the correct answer is option A, B and C.
The given question is related to family relationships and we need to find out the common children who have the same father. Let us solve this problem step by step.
Step 1: Understanding the given information
The question states that there are some children who have the same father. It is not mentioned how many children are there and who their father is. We need to find out the common children who have the same father.
Step 2: Analyzing the options
The options given are:
A) R
B) S
C) T
D) None of these
We need to check which options are correct based on the given information.
Step 3: Checking the correct options
Option A: R - It is possible that R has the same father as some other children.
Option B: S - It is possible that S has the same father as some other children.
Option C: T - It is possible that T has the same father as some other children.
Option D: None of these - This option cannot be correct as at least one of the options A, B or C must be correct.
Hence, the correct options are A, B and C.
Step 4: Final Answer
Therefore, the correct answer is option A, B and C.
If N is the set of natural numbers and I is the set of positive integers, then- a)N = I,
- b)N ⊂ I,
- c)N C I,
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If N is the set of natural numbers and I is the set of positive integers, then
a)
N = I,
b)
N ⊂ I,
c)
N C I,
d)
none of these
|
|
Rajeev Chaudhary answered |
Explanation:
- N is the set of natural numbers which includes all the counting numbers from 1 to infinity.
- I is the set of positive integers which includes all the counting numbers from 1 to infinity.
- It is important to note that the set of natural numbers includes 0 while the set of positive integers does not include 0.
- Therefore, we can say that N is a superset of I, meaning that all the elements of I are also present in N.
- Hence, option 'A' is the correct answer which states that N = I.
- N is the set of natural numbers which includes all the counting numbers from 1 to infinity.
- I is the set of positive integers which includes all the counting numbers from 1 to infinity.
- It is important to note that the set of natural numbers includes 0 while the set of positive integers does not include 0.
- Therefore, we can say that N is a superset of I, meaning that all the elements of I are also present in N.
- Hence, option 'A' is the correct answer which states that N = I.
For a group of 200 persons, 100 are interested in music, 70 in photography and 40 in swimming, further more 40 are interested in both music and photography, 30 in both music and swimming, 20 in photography and swimming and 10 in all the three. How many are interested in photography but not in music and swimming?- a)30
- b)15
- c)25
- d)20
Correct answer is option 'D'. Can you explain this answer?
For a group of 200 persons, 100 are interested in music, 70 in photography and 40 in swimming, further more 40 are interested in both music and photography, 30 in both music and swimming, 20 in photography and swimming and 10 in all the three. How many are interested in photography but not in music and swimming?
a)
30
b)
15
c)
25
d)
20
|
Faizan Khan answered |
Let
- Let P(Ph) is number of person interested in photography .
- Let P(s) is number of person interested in swimming.
- Let P(m) is number of person interested in music.
Given:
P(ph)=70 , P(s)=40,P(m)=100
The persons who are interested in photography but not in music and swimming is given by
There are 20 persons who are interested in photography but not in music and swimming.
E is a set of positive even number and O is a set of positive odd numbers, then E ∪ O is a- a)set of whole numbers,
- b)N,
- c)a set of rational number,
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
E is a set of positive even number and O is a set of positive odd numbers, then E ∪ O is a
a)
set of whole numbers,
b)
N,
c)
a set of rational number,
d)
none of these
|
|
Freedom Institute answered |
- Set E: Contains positive even numbers (e.g., 2, 4, 6, 8,...).
- Set O: Contains positive odd numbers (e.g., 1, 3, 5, 7,...).
- Union E ∪ O: Combines all elements from both sets E and O.
The union E ∪ O includes all positive integers since each integer is either even or odd.
- Option B (N): Represents the set of natural numbers, which are all positive integers (1, 2, 3,...).
- Thus, E ∪ O = N, the set of natural numbers.
- Set O: Contains positive odd numbers (e.g., 1, 3, 5, 7,...).
- Union E ∪ O: Combines all elements from both sets E and O.
The union E ∪ O includes all positive integers since each integer is either even or odd.
- Option B (N): Represents the set of natural numbers, which are all positive integers (1, 2, 3,...).
- Thus, E ∪ O = N, the set of natural numbers.
f A = {3, 4, 5, 6} B = {3, 7, 9, 5} and C = {6, 8, 10, 12, 7} then (A∩B')' is (given that the universal set U = {3, 4, ...., 11, 12, 13}- a){8, 10, 11, 12, 13}
- b){4, 6, 7, … 13}
- c){3, 4, 5, 7, 8, ……..13}
- d)None
Correct answer is option 'B'. Can you explain this answer?
f A = {3, 4, 5, 6} B = {3, 7, 9, 5} and C = {6, 8, 10, 12, 7} then (A∩B')' is (given that the universal set U = {3, 4, ...., 11, 12, 13}
a)
{8, 10, 11, 12, 13}
b)
{4, 6, 7, … 13}
c)
{3, 4, 5, 7, 8, ……..13}
d)
None
|
|
Poolikuntla Viswanadham answered |
If B={3,7,9,5} then B'={4,6,8,10,11,12,13} and A={3,4,5,6} A∩B'={4,6} (A∩B')'={3,5,7,8,9,10,11,12,13}
The set {2x|x is any positive rational number } is- a)an infinite set,
- b)a null set,
- c)a finite set,
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
The set {2x|x is any positive rational number } is
a)
an infinite set,
b)
a null set,
c)
a finite set,
d)
none of these
|
|
Snehal Das answered |
**Explanation:**
The set {2x | x is any positive rational number} is an infinite set. Here's why:
**Definition of an Infinite Set:**
An infinite set is a set that has an uncountable number of elements. It cannot be exhausted or enumerated completely.
**Definition of Positive Rational Numbers:**
Positive rational numbers are numbers that can be expressed as a fraction of two positive integers. For example, 1/2, 3/4, and 5/6 are all positive rational numbers.
**Definition of Set Builder Notation:**
The set {2x | x is any positive rational number} is defined using set builder notation. This notation states that for each positive rational number, x, we can obtain an element in the set by multiplying x by 2.
**Proof that the Set is Infinite:**
To show that the set {2x | x is any positive rational number} is infinite, we need to demonstrate that there is no limit to the number of elements in the set.
Let's assume the opposite, that the set is finite. This means that there is a finite number of positive rational numbers, which we know is not true.
Since positive rational numbers can be expressed as fractions, we can always find a new positive rational number by adding 1 to the denominator of a fraction. For example, if we have 1/2, we can get 1/3, 1/4, 1/5, and so on, by adding 1 to the denominator.
Since there is no limit to the number of positive rational numbers, there is also no limit to the number of elements in the set {2x | x is any positive rational number}.
Therefore, the set {2x | x is any positive rational number} is an infinite set.
The set {2x | x is any positive rational number} is an infinite set. Here's why:
**Definition of an Infinite Set:**
An infinite set is a set that has an uncountable number of elements. It cannot be exhausted or enumerated completely.
**Definition of Positive Rational Numbers:**
Positive rational numbers are numbers that can be expressed as a fraction of two positive integers. For example, 1/2, 3/4, and 5/6 are all positive rational numbers.
**Definition of Set Builder Notation:**
The set {2x | x is any positive rational number} is defined using set builder notation. This notation states that for each positive rational number, x, we can obtain an element in the set by multiplying x by 2.
**Proof that the Set is Infinite:**
To show that the set {2x | x is any positive rational number} is infinite, we need to demonstrate that there is no limit to the number of elements in the set.
Let's assume the opposite, that the set is finite. This means that there is a finite number of positive rational numbers, which we know is not true.
Since positive rational numbers can be expressed as fractions, we can always find a new positive rational number by adding 1 to the denominator of a fraction. For example, if we have 1/2, we can get 1/3, 1/4, 1/5, and so on, by adding 1 to the denominator.
Since there is no limit to the number of positive rational numbers, there is also no limit to the number of elements in the set {2x | x is any positive rational number}.
Therefore, the set {2x | x is any positive rational number} is an infinite set.
Let A={a, b}. set of subsets of A is called power set of A denoted by P(A). Now n(P(A) is- a)2
- b)4
- c)3
- d) None of these
Correct answer is option 'B'. Can you explain this answer?
Let A={a, b}. set of subsets of A is called power set of A denoted by P(A). Now n(P(A) is
a)
2
b)
4
c)
3
d)
None of these
|
Bhaskar Sharma answered |
Power Set of Set A
The power set of a set A is defined as the set of all possible subsets of A, including the empty set and A itself. It is denoted by P(A).
For example, if A = {1, 2}, then P(A) = {∅, {1}, {2}, {1,2}}.
Number of Subsets in Power Set
The number of subsets in the power set of a set A can be calculated using the formula:
n(P(A)) = 2^n
where n is the number of elements in set A.
For example, if A = {a, b}, then n = 2 and the number of subsets in P(A) is:
n(P(A)) = 2^2 = 4
Therefore, the correct answer is option B, which states that n(P(A)) = 4.
The power set of a set A is defined as the set of all possible subsets of A, including the empty set and A itself. It is denoted by P(A).
For example, if A = {1, 2}, then P(A) = {∅, {1}, {2}, {1,2}}.
Number of Subsets in Power Set
The number of subsets in the power set of a set A can be calculated using the formula:
n(P(A)) = 2^n
where n is the number of elements in set A.
For example, if A = {a, b}, then n = 2 and the number of subsets in P(A) is:
n(P(A)) = 2^2 = 4
Therefore, the correct answer is option B, which states that n(P(A)) = 4.
If four members a, b, c, d of a decision making body are in a meeting to pass a resolution where rule of majority prevails list the winning coalitions. Given that a, b, c, d own 50% 20% 15% 15% shares each.- a){a, b} {a, c} {a, d} {a, b, c} {a, b, d} {a, b, c, d}
- b){b, c, d}
- c){b, c} {b, d} {c, d} {a, c, d} {b, c, d} {a} {b} {c} {d} ?
- d)None
Correct answer is option 'C'. Can you explain this answer?
If four members a, b, c, d of a decision making body are in a meeting to pass a resolution where rule of majority prevails list the winning coalitions. Given that a, b, c, d own 50% 20% 15% 15% shares each.
a)
{a, b} {a, c} {a, d} {a, b, c} {a, b, d} {a, b, c, d}
b)
{b, c, d}
c)
{b, c} {b, d} {c, d} {a, c, d} {b, c, d} {a} {b} {c} {d} ?
d)
None
|
|
Mahesh Chakraborty answered |
Winning Coalitions in a Decision Making Body
In a decision making body where rule of majority prevails, the winning coalitions are the groups of members that have enough votes to pass a resolution. In this scenario, there are four members a, b, c, and d with different percentages of shares.
Possible Winning Coalitions:
a) {a, b} {a, c} {a, d} {a, b, c} {a, b, d} {a, b, c, d}
- These coalitions include member a who owns 50% of shares and may be able to persuade one or more other members to vote with him to pass a resolution.
b) {b, c, d}
- This coalition includes members b, c, and d who collectively own 50% + 15% + 15% = 80% of shares. They may be able to pass a resolution without the support of member a.
c) {b, c} {b, d} {c, d} {a, c, d} {b, c, d}
- These coalitions include two or more members who collectively own more than 50% of shares. They may be able to pass a resolution without the support of member a.
d) None
- It is possible that no winning coalition emerges if members cannot agree on a resolution or cannot persuade enough other members to vote with them.
Conclusion:
In this scenario, the winning coalitions are {a, b}, {a, c}, {a, d}, {a, b, c}, {a, b, d}, {a, b, c, d}, {b, c, d}, {b, c}, {b, d}, {c, d}, {a, c, d}, and {b, c, d}. The correct answer is option C which lists all the winning coalitions that include two or more members who collectively own more than 50% of shares.
In a decision making body where rule of majority prevails, the winning coalitions are the groups of members that have enough votes to pass a resolution. In this scenario, there are four members a, b, c, and d with different percentages of shares.
Possible Winning Coalitions:
a) {a, b} {a, c} {a, d} {a, b, c} {a, b, d} {a, b, c, d}
- These coalitions include member a who owns 50% of shares and may be able to persuade one or more other members to vote with him to pass a resolution.
b) {b, c, d}
- This coalition includes members b, c, and d who collectively own 50% + 15% + 15% = 80% of shares. They may be able to pass a resolution without the support of member a.
c) {b, c} {b, d} {c, d} {a, c, d} {b, c, d}
- These coalitions include two or more members who collectively own more than 50% of shares. They may be able to pass a resolution without the support of member a.
d) None
- It is possible that no winning coalition emerges if members cannot agree on a resolution or cannot persuade enough other members to vote with them.
Conclusion:
In this scenario, the winning coalitions are {a, b}, {a, c}, {a, d}, {a, b, c}, {a, b, d}, {a, b, c, d}, {b, c, d}, {b, c}, {b, d}, {c, d}, {a, c, d}, and {b, c, d}. The correct answer is option C which lists all the winning coalitions that include two or more members who collectively own more than 50% of shares.
The range of the function f(x) = log10(1 + x) for the domain of real values of x when 0 ≤ x ≤9 is- a){0, –1}
- b){0, 1, 2}
- c){0.1}
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
The range of the function f(x) = log10(1 + x) for the domain of real values of x when 0 ≤ x ≤9 is
a)
{0, –1}
b)
{0, 1, 2}
c)
{0.1}
d)
none of these
|
|
Meera Basak answered |
Solution:
The given function is f(x) = log10(1 + x).
We need to find the range of this function for the domain of real values of x when 0 ≤ x ≤ 9.
Range of a function:
The range of a function is the set of all possible output values (y-values) of the function.
Domain of the given function:
The domain of the given function is 0 ≤ x ≤ 9, which means that x can take any real value between 0 and 9, including 0 and 9.
Finding the range of the given function:
To find the range of the given function, we need to analyze the behavior of the function for different values of x.
Case 1: When x = 0
f(0) = log10(1 + 0) = log10(1) = 0
Therefore, the function takes the value 0 when x = 0.
Case 2: When x > 0
As x increases from 0 to 9, the value of 1 + x also increases.
This means that the argument of the logarithmic function log10(1 + x) also increases.
As the argument of the logarithmic function increases, the value of the function also increases.
However, the function value increases at a decreasing rate.
This is because the logarithmic function is an increasing function, but its rate of increase decreases as the argument increases.
Case 3: When x = 9
When x = 9, the argument of the logarithmic function is 1 + 9 = 10.
The value of log10(10) is 1.
Therefore, the function takes the value 1 when x = 9.
Conclusion:
From the above analysis, we can see that the function takes all values between 0 and 1 for 0 < x="" />< />
Therefore, the range of the function for the given domain is {y : 0 < y="" />< />
The correct option is (c) {0.1}.
The given function is f(x) = log10(1 + x).
We need to find the range of this function for the domain of real values of x when 0 ≤ x ≤ 9.
Range of a function:
The range of a function is the set of all possible output values (y-values) of the function.
Domain of the given function:
The domain of the given function is 0 ≤ x ≤ 9, which means that x can take any real value between 0 and 9, including 0 and 9.
Finding the range of the given function:
To find the range of the given function, we need to analyze the behavior of the function for different values of x.
Case 1: When x = 0
f(0) = log10(1 + 0) = log10(1) = 0
Therefore, the function takes the value 0 when x = 0.
Case 2: When x > 0
As x increases from 0 to 9, the value of 1 + x also increases.
This means that the argument of the logarithmic function log10(1 + x) also increases.
As the argument of the logarithmic function increases, the value of the function also increases.
However, the function value increases at a decreasing rate.
This is because the logarithmic function is an increasing function, but its rate of increase decreases as the argument increases.
Case 3: When x = 9
When x = 9, the argument of the logarithmic function is 1 + 9 = 10.
The value of log10(10) is 1.
Therefore, the function takes the value 1 when x = 9.
Conclusion:
From the above analysis, we can see that the function takes all values between 0 and 1 for 0 < x="" />< />
Therefore, the range of the function for the given domain is {y : 0 < y="" />< />
The correct option is (c) {0.1}.
Out of 2000 employees in an office 48% preferred Coffee (c), 54% liked (T), 64% used to smoke (S). Out of the total 28% used C and T, 32% used T and S and 30% preferred C and S, only 6% did none of these. The number having all the three is- a)360
- b)300
- c)380
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Out of 2000 employees in an office 48% preferred Coffee (c), 54% liked (T), 64% used to smoke (S). Out of the total 28% used C and T, 32% used T and S and 30% preferred C and S, only 6% did none of these. The number having all the three is
a)
360
b)
300
c)
380
d)
none of these
|
|
Freedom Institute answered |
Total = 2000 = 100 %
C = 48 %
T = 54 %
S = 64 %
C ∩ T = 28 %
T ∩ S = 32%
C ∩ S = 30 %
Having all three = C ∩ T ∩ S = ?
None = 6 %
Total = C + T + S - C ∩ T - T ∩ S - C ∩ S + C ∩ T ∩ S + None
=> 100 = 48 + 54 + 64 - 28 - 32 - 30 + C ∩ T ∩ S + 6
=> C ∩ T ∩ S = 18 %
18 % of 2000 = (18/100) * 2000 = 360
360 having all the three
After qualifying out of 400 professionals, 112 joined industry, 120 started practice and 160 joined as paid assistants. There were 32, who were in both practice and service 40 in both practice and assistantship and 20 in both industry and assistantship. There were 12 who did all the three. Find how many of them did only one of these.- a)88
- b)244
- c)122
- d)None
Correct answer is option 'A'. Can you explain this answer?
After qualifying out of 400 professionals, 112 joined industry, 120 started practice and 160 joined as paid assistants. There were 32, who were in both practice and service 40 in both practice and assistantship and 20 in both industry and assistantship. There were 12 who did all the three. Find how many of them did only one of these.
a)
88
b)
244
c)
122
d)
None
|
|
Srsps answered |
Total = 400
Industries Service S = 112
Practice P = 120
Paid Assistant A = 160
Industries Service S = 112
Practice P = 120
Paid Assistant A = 160
P ∩ S = 32
P ∩ A = 40
S ∩ A = 20
P ∩ S ∩ A = 12
P ∩ A = 40
S ∩ A = 20
P ∩ S ∩ A = 12
Total = S + P + A - P ∩ S - P ∩ A - S ∩ A + P ∩ S ∩ A + None
400 = 112 + 120 + 160 - 32 - 40 - 20 + 12 + None
None = 88
Person = 400 - 88 = 312
312 - (32 - 12) - (40-12) -(20-12) - 12
= 312 - 20 - 28 - 8 - 12
= 244
400 = 112 + 120 + 160 - 32 - 40 - 20 + 12 + None
None = 88
Person = 400 - 88 = 312
312 - (32 - 12) - (40-12) -(20-12) - 12
= 312 - 20 - 28 - 8 - 12
= 244
If A = {a, b, c, d, e, f} B = {a, e, i, o, u} and C = {m, n, o, p, q, r, s, t, u} then A∪B∪C is- a) {a, b, c, d, e, f, i, o, u, m, n, p, q, r, s, t}
- b){a, b, c, r, s, t}
- c) {d, e, f, n, p, q}
- d) None
Correct answer is option 'A'. Can you explain this answer?
If A = {a, b, c, d, e, f} B = {a, e, i, o, u} and C = {m, n, o, p, q, r, s, t, u} then A∪B∪C is
a)
{a, b, c, d, e, f, i, o, u, m, n, p, q, r, s, t}
b)
{a, b, c, r, s, t}
c)
{d, e, f, n, p, q}
d)
None
|
|
Mayur Tamrakar answered |
In this answer none of the alphabets are repeated.that is all the alphabets appear once in the answer.
The number of elements in range of constant function is- a)One
- b)Zero
- c)Infinite
- d)Indetermined
Correct answer is 'C'. Can you explain this answer?
The number of elements in range of constant function is
a)
One
b)
Zero
c)
Infinite
d)
Indetermined
|
|
Sahil Malik answered |
Constant Function:
A constant function is a function that always returns the same value, regardless of its input. For example, the function f(x) = 5 is a constant function, because it always returns 5, no matter what value of x is input.
Range of Constant Function:
The range of a function is the set of all output values that the function can produce. For a constant function, the range is simply the set containing the constant value that the function returns.
For example, consider the constant function f(x) = 5. The range of this function is the set {5}, because that is the only output value that the function can produce.
Number of Elements in Range:
Since a constant function always returns the same value, the number of elements in its range is infinite. This is because there are an infinite number of possible inputs to the function, and for each input, the function returns the same output.
For example, consider the constant function f(x) = 5. No matter what value of x is input, the function will always return 5. Therefore, the range of the function contains an infinite number of elements.
Conclusion:
The number of elements in the range of a constant function is infinite, because the function always returns the same value, regardless of its input.
A constant function is a function that always returns the same value, regardless of its input. For example, the function f(x) = 5 is a constant function, because it always returns 5, no matter what value of x is input.
Range of Constant Function:
The range of a function is the set of all output values that the function can produce. For a constant function, the range is simply the set containing the constant value that the function returns.
For example, consider the constant function f(x) = 5. The range of this function is the set {5}, because that is the only output value that the function can produce.
Number of Elements in Range:
Since a constant function always returns the same value, the number of elements in its range is infinite. This is because there are an infinite number of possible inputs to the function, and for each input, the function returns the same output.
For example, consider the constant function f(x) = 5. No matter what value of x is input, the function will always return 5. Therefore, the range of the function contains an infinite number of elements.
Conclusion:
The number of elements in the range of a constant function is infinite, because the function always returns the same value, regardless of its input.
The set {x|0<x<5} represents the set when x may take integral values only- a){0, 1, 2, 3, 4, 5}
- b){1, 2, 3, 4}
- c){1, 2, 3, 4, 5}
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
The set {x|0<x<5} represents the set when x may take integral values only
a)
{0, 1, 2, 3, 4, 5}
b)
{1, 2, 3, 4}
c)
{1, 2, 3, 4, 5}
d)
none of these
|
|
Dhruba Choudhary answered |
Given set is {x|0< /><5}. this="" means="" that="" x="" can="" take="" any="" integral="" value="" between="" 0="" and="" 5="" excluding="" 0="" and="" 5.="">
To find the answer, we need to list down all the integral values that satisfy this condition.
The integral values between 0 and 5 are:
1. 1
2. 2
3. 3
4. 4
Thus, the set {x|0< /><5} represents="" the="" set="" {1,="" 2,="" 3,="" 4}.="">
Therefore, the correct answer is option B.
To find the answer, we need to list down all the integral values that satisfy this condition.
The integral values between 0 and 5 are:
1. 1
2. 2
3. 3
4. 4
Thus, the set {x|0< /><5} represents="" the="" set="" {1,="" 2,="" 3,="" 4}.="">
Therefore, the correct answer is option B.
{(x, y)|x<y} is- a)not a function
- b)a function
- c)one-one mapping
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
{(x, y)|x<y} is
a)
not a function
b)
a function
c)
one-one mapping
d)
none of these
|
|
Sinjini Gupta answered |
Given set {(x, y)|xy}, we need to determine whether it represents a function or not.
Explanation:
A function is a relation between two sets in which each element of the first set (called the domain) is paired with exactly one element of the second set (called the range). In other words, if (x, y) is a part of a function, then there cannot be another element (x, z) in the same function where z is not equal to y.
Let's consider the given set {(x, y)|xy}. Here, the element (x, y) is included in the set if and only if xy is true (i.e., not equal to zero).
However, this set does not satisfy the definition of a function because for some values of x, there can be multiple values of y that satisfy the condition xy ≠ 0. For example, if x = 2, then both (2, 1) and (2, -1) are included in the set, since 2*1 ≠ 0 and 2*(-1) ≠ 0. Therefore, there is no unique value of y corresponding to each value of x, which violates the definition of a function.
Hence, the correct answer is option 'A': not a function.
Explanation:
A function is a relation between two sets in which each element of the first set (called the domain) is paired with exactly one element of the second set (called the range). In other words, if (x, y) is a part of a function, then there cannot be another element (x, z) in the same function where z is not equal to y.
Let's consider the given set {(x, y)|xy}. Here, the element (x, y) is included in the set if and only if xy is true (i.e., not equal to zero).
However, this set does not satisfy the definition of a function because for some values of x, there can be multiple values of y that satisfy the condition xy ≠ 0. For example, if x = 2, then both (2, 1) and (2, -1) are included in the set, since 2*1 ≠ 0 and 2*(-1) ≠ 0. Therefore, there is no unique value of y corresponding to each value of x, which violates the definition of a function.
Hence, the correct answer is option 'A': not a function.
Of the 200 candidates who were interviewed for a position at call centre, 100 had a two-wheeler, 70 had a credit card and 140 had a and 140 had a mobile phone, 40 of them had both a two-wheeler and a credit card, 30 had both a credit card and a mobile phone, 60 had both a two-wheeler and a mobile phone, and 10 had all three. How many candidates had none of the three?- a)0
- b)20
- c)10
- d)18
Correct answer is option 'C'. Can you explain this answer?
Of the 200 candidates who were interviewed for a position at call centre, 100 had a two-wheeler, 70 had a credit card and 140 had a and 140 had a mobile phone, 40 of them had both a two-wheeler and a credit card, 30 had both a credit card and a mobile phone, 60 had both a two-wheeler and a mobile phone, and 10 had all three. How many candidates had none of the three?
a)
0
b)
20
c)
10
d)
18
|
Manoj Ghosh answered |
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.
Number of candidates who had at least one of the three = A U B U 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 AUBUC = A + B + C - {A n B + B n C + C n A} + A n B n C
Therefore, AUBUC = 100 + 70 + 140 - {40 + 30 + 60} + 10
Or AUBUC = 190.
As 190 candidates who attended the interview had at least one of the three gadgets, 200 - 190 = 10 candidates had none of three.
In a class of 50 students, 35 opted for Mathematics and 37 opted for Commerce. The number of such students who opted for both Mathematics and Commerce are:- a)13
- b)15
- c)22
- d)28
Correct answer is option 'C'. Can you explain this answer?
In a class of 50 students, 35 opted for Mathematics and 37 opted for Commerce. The number of such students who opted for both Mathematics and Commerce are:
a)
13
b)
15
c)
22
d)
28
|
|
Snehal Das answered |
Solution:
Given,
Total number of students = 50
Number of students opted for Mathematics = 35
Number of students opted for Commerce = 37
To find:
The number of students who opted for both Mathematics and Commerce
Let us assume the number of students who opted for both Mathematics and Commerce as x.
Now, using the formula of the total number of students in two sets, we can find the number of students who opted for Mathematics or Commerce or both.
Total number of students in Mathematics or Commerce = Number of students in Mathematics + Number of students in Commerce - Number of students who opted for both Mathematics and Commerce
50 = 35 + 37 - x
x = 22
Therefore, the number of students who opted for both Mathematics and Commerce is 22.
Given,
Total number of students = 50
Number of students opted for Mathematics = 35
Number of students opted for Commerce = 37
To find:
The number of students who opted for both Mathematics and Commerce
Let us assume the number of students who opted for both Mathematics and Commerce as x.
Now, using the formula of the total number of students in two sets, we can find the number of students who opted for Mathematics or Commerce or both.
Total number of students in Mathematics or Commerce = Number of students in Mathematics + Number of students in Commerce - Number of students who opted for both Mathematics and Commerce
50 = 35 + 37 - x
x = 22
Therefore, the number of students who opted for both Mathematics and Commerce is 22.
{(x , y), y=x2} is- a)not a function
- b)a function
- c)inverse mapping
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
{(x , y), y=x2} is
a)
not a function
b)
a function
c)
inverse mapping
d)
none of these
|
|
Akshay Das answered |
Explanation:
This question is asking about whether the given set of ordered pairs {(x, y), y = x²} represents a function or not.
- Definition of a Function: A function is a set of ordered pairs in which no two different ordered pairs have the same first element (also known as the input or domain value).
- Testing for a Function: To test whether a set of ordered pairs represents a function, we need to check whether there are any two different ordered pairs with the same first element. If there are, then the set does not represent a function.
In the given set of ordered pairs {(x, y), y = x²}, the y-value is defined in terms of the x-value using the equation y = x². This means that for every possible value of x, there is only one corresponding value of y. Therefore, there cannot be any two different ordered pairs with the same first element, and the set represents a function.
Answer: Option (B) - a function.
This question is asking about whether the given set of ordered pairs {(x, y), y = x²} represents a function or not.
- Definition of a Function: A function is a set of ordered pairs in which no two different ordered pairs have the same first element (also known as the input or domain value).
- Testing for a Function: To test whether a set of ordered pairs represents a function, we need to check whether there are any two different ordered pairs with the same first element. If there are, then the set does not represent a function.
In the given set of ordered pairs {(x, y), y = x²}, the y-value is defined in terms of the x-value using the equation y = x². This means that for every possible value of x, there is only one corresponding value of y. Therefore, there cannot be any two different ordered pairs with the same first element, and the set represents a function.
Answer: Option (B) - a function.
If A = (1,2,3,4,5), B = (2,4) and C = (1,3,5) then (A - C) x B is - a) {(2,2), (2,4), (4,2), (4,4), (5,2), (5,4)}
- b){(1,2), (1,4), (3,2), (3,4), (5,2), (5,4)}
- c){(2,2), (4,2), (4,4), (4,5)}
- d) {(2,2), (2,4), (4,2), (4,4)}
Correct answer is 'C'. Can you explain this answer?
If A = (1,2,3,4,5), B = (2,4) and C = (1,3,5) then (A - C) x B is
a)
{(2,2), (2,4), (4,2), (4,4), (5,2), (5,4)}
b)
{(1,2), (1,4), (3,2), (3,4), (5,2), (5,4)}
c)
{(2,2), (4,2), (4,4), (4,5)}
d)
{(2,2), (2,4), (4,2), (4,4)}
|
|
Harshad Kapoor answered |
Solution:
Definition: If A = (a1, a2, a3, ..., an) and B = (b1, b2, b3, ..., bn) are two vectors of the same dimension, then the cross product of A and B, denoted by A x B, is the vector given by
A x B = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)
Given: A = (1,2,3,4,5), B = (2,4) and C = (1,3,5)
To find: (A - C) x B
Steps:
Definition: If A = (a1, a2, a3, ..., an) and B = (b1, b2, b3, ..., bn) are two vectors of the same dimension, then the cross product of A and B, denoted by A x B, is the vector given by
A x B = (a2b3 - a3b2, a3b1 - a1b3, a1b2 - a2b1)
Given: A = (1,2,3,4,5), B = (2,4) and C = (1,3,5)
To find: (A - C) x B
Steps:
- Subtract C from A to get (A - C) = (1-1, 2-3, 3-5, 4-0, 5-0) = (0, -1, -2, 4, 5)
- Write B as a vector of dimension 5 by adding three zeros: B = (2, 4, 0, 0, 0)
- Apply the formula for the cross product to get:
- (0, -1, -2, 4, 5) x (2, 4, 0, 0, 0)
- = (-2(5) - 4(-4), -2(0) - 0(0), 4(4) - 0(-1))
- = (-18, 0, 17)
- The resulting vector is (-18, 0, 17)
- The answer is the set of all possible values of (-18, 0, 17) which is option (c) {(2,2), (4,2), (4,4), (4,5)}
Which of the following is a subset of set {1, 2, 3, 4}? - a)All of the mentioned
- b){1, 2}
- c){1, 2, 3}
- d){1}
Correct answer is option 'A'. Can you explain this answer?
Which of the following is a subset of set {1, 2, 3, 4}?
a)
All of the mentioned
b)
{1, 2}
c)
{1, 2, 3}
d)
{1}
|
|
Srsps answered |
The subset of set (1, 2, 3, 4} is {1, 2}, {1, 2, 3}, and {1}.
IF A = [5, 6, 7] and B = [7, 8, 9] then A ∪ B is equal to- a)[5, 6, 7]
- b)[7, 8, 9]
- c)[5, 6, 7, 8, 9]
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
IF A = [5, 6, 7] and B = [7, 8, 9] then A ∪ B is equal to
a)
[5, 6, 7]
b)
[7, 8, 9]
c)
[5, 6, 7, 8, 9]
d)
None of these
|
|
Srsps answered |
Given A = [5, 6, 7] and B = [7, 8, 9]
then A ∪ B = [5, 6, 7, 8, 9]
then A ∪ B = [5, 6, 7, 8, 9]
A town has a total population of 50,000. Out of it 28,000 read the newspaper X and 23000 read Y while 4000 read both the papers. The number of persons not reading X and Y both is- a)2000
- b)3000
- c)2500
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
A town has a total population of 50,000. Out of it 28,000 read the newspaper X and 23000 read Y while 4000 read both the papers. The number of persons not reading X and Y both is
a)
2000
b)
3000
c)
2500
d)
none of these
|
|
Gaurav Chatterjee answered |
Solution:Given, Total Population (TP) = 50,000Number of people reading newspaper X (NX) = 28,000Number of people reading newspaper Y (NY) = 23,000Number of people reading both X and Y = 4,000To find: Number of people not reading X and YApproach:We can solve this problem using the Venn diagram approach. We will draw a Venn diagram with two circles representing newspapers X and Y. We know that 4,000 people read both newspapers. We will place them in the intersection of the circles. We also know that 28,000 people read newspaper X and 23,000 people read newspaper Y. We will place them in the respective circles. Now, we need to find the number of people who do not read either X or Y. We will place them outside both circles.Steps:1. Draw a Venn diagram with two circles representing newspapers X and Y. Label the intersection as X ∩ Y.2. Write the given values in the respective regions of the Venn diagram. - NX = 28,000 (region representing X but not Y) - NY = 23,000 (region representing Y but not X) - X ∩ Y = 4,000 (region representing both X and Y)3. Find the total number of people who read either X or Y or both. - NX + NY - X ∩ Y = 28,000 + 23,000 - 4,000 = 47,0004. Find the number of people who do not read either X or Y. - TP - (NX + NY - X ∩ Y) = 50,000 - 47,000 = 3,000Answer: The number of persons not reading X and Y both is 3,000.Therefore, option (b) 3000 is the correct answer.
At a certain conference of 100 people there are 29 Indian women and 23 Indian men. Out of these Indian people 4 are doctors and 24 are either men or doctors. There are no foreign doctors. The number of women doctors attending the conference is- a)2
- b)4
- c)1
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
At a certain conference of 100 people there are 29 Indian women and 23 Indian men. Out of these Indian people 4 are doctors and 24 are either men or doctors. There are no foreign doctors. The number of women doctors attending the conference is
a)
2
b)
4
c)
1
d)
none of these
|
Sameer Rane answered |
Let us define the following events as:
M: Indian men
W: Indian women
D: Doctors who are Indian
So, now we have
n(M ∪ D) = 24
n(W) = 29
n(M) = 23
n(D) = 4
Total persons = 100
Consider, n(M ∪ D) = n(M) + n(D) – n(M ∩ D)
⇒ 24 = 23 + 4 – n(M ∩ D)
⇒ n(M ∩ D) = 3
⇒ Number of Indian men who are doctors = 3
But only 4 Indian people are doctors.
∴ Number of Indian women who are doctors = 4 – 3 = 1
If A = { 1, 2, 3, 5, 7} and B = {1, 3, 6, 10, 15}. Cardinal number of A~B is- a)3
- b)4
- c)6
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If A = { 1, 2, 3, 5, 7} and B = {1, 3, 6, 10, 15}. Cardinal number of A~B is
a)
3
b)
4
c)
6
d)
none of these
|
|
Divya Dasgupta answered |
Cardinal Number of A~B
Definition of Cardinal Number
Cardinal number is a number that represents the size or quantity of a set.
Given Sets
A = { 1, 2, 3, 5, 7}
B = {1, 3, 6, 10, 15}
Definition of A~B
A~B represents the symmetric difference of A and B, which is the set of elements that are in either A or B, but not in both.
Calculation of Cardinal Number of A~B
To calculate the cardinal number of A~B, we need to find the elements that are in either A or B, but not in both.
1. Elements in A but not in B
{2, 5, 7}
2. Elements in B but not in A
{6, 10, 15}
3. Elements in both A and B
{1, 3}
4. Elements in either A or B, but not in both
{2, 5, 7, 6, 10, 15}
5. Cardinal number of A~B
The cardinal number of A~B is the number of elements in the set {2, 5, 7, 6, 10, 15}, which is 3.
Therefore, the correct answer is option A, 3.
Definition of Cardinal Number
Cardinal number is a number that represents the size or quantity of a set.
Given Sets
A = { 1, 2, 3, 5, 7}
B = {1, 3, 6, 10, 15}
Definition of A~B
A~B represents the symmetric difference of A and B, which is the set of elements that are in either A or B, but not in both.
Calculation of Cardinal Number of A~B
To calculate the cardinal number of A~B, we need to find the elements that are in either A or B, but not in both.
1. Elements in A but not in B
{2, 5, 7}
2. Elements in B but not in A
{6, 10, 15}
3. Elements in both A and B
{1, 3}
4. Elements in either A or B, but not in both
{2, 5, 7, 6, 10, 15}
5. Cardinal number of A~B
The cardinal number of A~B is the number of elements in the set {2, 5, 7, 6, 10, 15}, which is 3.
Therefore, the correct answer is option A, 3.
There are 40 students, 30 of them passed in English, 25 of them passed in Maths nd 15 of them passed in both. Assuming that every Student has passed at least in one subject. How many students's passed in English only bot not in Maths.- a)15
- b)20
- c)10
- d)25
Correct answer is option 'A'. Can you explain this answer?
There are 40 students, 30 of them passed in English, 25 of them passed in Maths nd 15 of them passed in both. Assuming that every Student has passed at least in one subject. How many students's passed in English only bot not in Maths.
a)
15
b)
20
c)
10
d)
25
|
|
Meera Basak answered |
Solution:
Given,
Total number of students = 40
Number of students passed in English = 30
Number of students passed in Maths = 25
Number of students passed in both English and Maths = 15
To find: Number of students passed in English only but not in Maths
Let's use the formula of Inclusion-Exclusion Principle to find the number of students who passed in English only but not in Maths.
n(A U B) = n(A) + n(B) - n(A ∩ B)
where,
n(A U B) = Number of students passed in either English or Maths or both
n(A) = Number of students passed in English
n(B) = Number of students passed in Maths
n(A ∩ B) = Number of students passed in both English and Maths
Substituting the given values in the above formula, we get
n(English U Maths) = 30 + 25 - 15
n(English U Maths) = 40
Therefore, the number of students who passed in either English or Maths or both is 40.
Now, we can find the number of students who passed in English only by subtracting the number of students who passed in both English and Maths from the number of students who passed in English.
n(English only) = n(English) - n(English ∩ Maths)
n(English only) = 30 - 15
n(English only) = 15
Therefore, the number of students who passed in English only but not in Maths is 15.
Hence, the correct option is A) 15.
Given,
Total number of students = 40
Number of students passed in English = 30
Number of students passed in Maths = 25
Number of students passed in both English and Maths = 15
To find: Number of students passed in English only but not in Maths
Let's use the formula of Inclusion-Exclusion Principle to find the number of students who passed in English only but not in Maths.
n(A U B) = n(A) + n(B) - n(A ∩ B)
where,
n(A U B) = Number of students passed in either English or Maths or both
n(A) = Number of students passed in English
n(B) = Number of students passed in Maths
n(A ∩ B) = Number of students passed in both English and Maths
Substituting the given values in the above formula, we get
n(English U Maths) = 30 + 25 - 15
n(English U Maths) = 40
Therefore, the number of students who passed in either English or Maths or both is 40.
Now, we can find the number of students who passed in English only by subtracting the number of students who passed in both English and Maths from the number of students who passed in English.
n(English only) = n(English) - n(English ∩ Maths)
n(English only) = 30 - 15
n(English only) = 15
Therefore, the number of students who passed in English only but not in Maths is 15.
Hence, the correct option is A) 15.
Out of 60 students 25 failed in paper (1), 24 in paper (2), 32 in paper (3), 9 in paper (1), alone 6 in paper (2), alone 5 in papers (2), and (3), and 3 in papers (1), and (2). Find how many passed in all the three papers?- a)10
- b)60
- c)50
- d)None
Correct answer is option 'A'. Can you explain this answer?
Out of 60 students 25 failed in paper (1), 24 in paper (2), 32 in paper (3), 9 in paper (1), alone 6 in paper (2), alone 5 in papers (2), and (3), and 3 in papers (1), and (2). Find how many passed in all the three papers?
a)
10
b)
60
c)
50
d)
None
|
|
Pranav Gupta answered |
Given information:
- Total number of students = 60
- Number of students failed in paper (1) = 25
- Number of students failed in paper (2) = 24
- Number of students failed in paper (3) = 32
- Number of students failed in paper (1) alone = 9
- Number of students failed in paper (2) alone = 6
- Number of students failed in papers (2) and (3) = 5
- Number of students failed in papers (1) and (2) = 3
To find: Number of students who passed in all three papers.
Solution:
First, we can find the number of students who failed in only two papers:
- Number of students failed in papers (2) and (3) = 5
- Number of students failed in papers (1) and (2) = 3
- Total = 5 + 3 = 8
Next, we can find the number of students who failed in only one paper:
- Number of students failed in paper (1) alone = 9
- Number of students failed in paper (2) alone = 6
- Number of students failed in paper (3) alone = (32 - 5) = 27
- Total = 9 + 6 + 27 = 42
Now, we can find the number of students who passed in at least two papers:
- Total number of students = 60
- Number of students who failed in at least one paper = 42
- Number of students who passed in at least two papers = 60 - 42 = 18
Finally, we can find the number of students who passed in all three papers:
- Number of students who passed in at least two papers = 18
- Number of students who failed in all three papers = 25 + 24 + 32 - 2*(5 + 3) = 43
- Number of students who passed in all three papers = Total - (Failed in at least one paper) - (Failed in all three papers)
= 60 - 42 - 43
= 10
Therefore, the number of students who passed in all three papers is 10, which is option (a).
- Total number of students = 60
- Number of students failed in paper (1) = 25
- Number of students failed in paper (2) = 24
- Number of students failed in paper (3) = 32
- Number of students failed in paper (1) alone = 9
- Number of students failed in paper (2) alone = 6
- Number of students failed in papers (2) and (3) = 5
- Number of students failed in papers (1) and (2) = 3
To find: Number of students who passed in all three papers.
Solution:
First, we can find the number of students who failed in only two papers:
- Number of students failed in papers (2) and (3) = 5
- Number of students failed in papers (1) and (2) = 3
- Total = 5 + 3 = 8
Next, we can find the number of students who failed in only one paper:
- Number of students failed in paper (1) alone = 9
- Number of students failed in paper (2) alone = 6
- Number of students failed in paper (3) alone = (32 - 5) = 27
- Total = 9 + 6 + 27 = 42
Now, we can find the number of students who passed in at least two papers:
- Total number of students = 60
- Number of students who failed in at least one paper = 42
- Number of students who passed in at least two papers = 60 - 42 = 18
Finally, we can find the number of students who passed in all three papers:
- Number of students who passed in at least two papers = 18
- Number of students who failed in all three papers = 25 + 24 + 32 - 2*(5 + 3) = 43
- Number of students who passed in all three papers = Total - (Failed in at least one paper) - (Failed in all three papers)
= 60 - 42 - 43
= 10
Therefore, the number of students who passed in all three papers is 10, which is option (a).
"Is smaller than" over the set of eggs in a box is- a)Transitive (T)
- b)Symmetric (S)
- c)Reflexive (R)
- d)Equivalence (E)
Correct answer is option 'A'. Can you explain this answer?
"Is smaller than" over the set of eggs in a box is
a)
Transitive (T)
b)
Symmetric (S)
c)
Reflexive (R)
d)
Equivalence (E)
|
Table Fan answered |
If a is smaller than b and b is smaller than c , then a is definitely smaller than c.
hence it is transitive
hence it is transitive
Given A = {2, 3}, B = {4, 5}, C = {5, 6} then A × (B ∩ C) is- a){(2, 5), (3, 5)}
- b){(5, 2), (5, 3)}
- c){(2, 3), (5, 5)}
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
Given A = {2, 3}, B = {4, 5}, C = {5, 6} then A × (B ∩ C) is
a)
{(2, 5), (3, 5)}
b)
{(5, 2), (5, 3)}
c)
{(2, 3), (5, 5)}
d)
none of these
|
Prashanth Datta answered |
Given sets A, B, and C as A = {2, 3}, B = {4, 5}, C = {5, 6}, we need to find the result of the expression A (B C).
To find the result, we need to perform the set operations in the given expression step by step.
1. B C:
- The symbol represents the set intersection operation, which means we need to find the common elements between sets B and C.
- B = {4, 5} and C = {5, 6}.
- The common element between B and C is 5.
- Therefore, B C = {5}.
2. A (B C):
- The symbol represents the set union operation, which means we need to find the elements that are present in either set A or the result of the previous operation (B C).
- A = {2, 3} and (B C) = {5}.
- The elements present in either A or (B C) are 2, 3, and 5.
- Therefore, A (B C) = {2, 3, 5}.
Therefore, the correct answer is option 'A': {(2, 5), (3, 5)}.
To find the result, we need to perform the set operations in the given expression step by step.
1. B C:
- The symbol represents the set intersection operation, which means we need to find the common elements between sets B and C.
- B = {4, 5} and C = {5, 6}.
- The common element between B and C is 5.
- Therefore, B C = {5}.
2. A (B C):
- The symbol represents the set union operation, which means we need to find the elements that are present in either set A or the result of the previous operation (B C).
- A = {2, 3} and (B C) = {5}.
- The elements present in either A or (B C) are 2, 3, and 5.
- Therefore, A (B C) = {2, 3, 5}.
Therefore, the correct answer is option 'A': {(2, 5), (3, 5)}.
On a survey of 100 boys it was found that 50 used white that 50 used white shirt 40 red and 30 blue. 20 were habituated in using both white and red shirts 15 both red and blue shirts and 10 blue and white shirts. If 10 boys did not use any of the white red or blue colours and 20 boys used all the colours offer your comments.- a)Inconsistent since 50+40+30-20-15-10+20≠100
- b)Consistent
- c)Cannot determine due to data insufficiency
- d)None
Correct answer is option 'A'. Can you explain this answer?
On a survey of 100 boys it was found that 50 used white that 50 used white shirt 40 red and 30 blue. 20 were habituated in using both white and red shirts 15 both red and blue shirts and 10 blue and white shirts. If 10 boys did not use any of the white red or blue colours and 20 boys used all the colours offer your comments.
a)
Inconsistent since 50+40+30-20-15-10+20≠100
b)
Consistent
c)
Cannot determine due to data insufficiency
d)
None
|
|
Gopal Sen answered |
This statement is incorrect. The given values are consistent.
To verify, we can use the principle of inclusion-exclusion.
Total number of boys who used at least one color = 50 (white) + 40 (red) + 30 (blue) = 120
However, we have counted the boys who used two colors twice, so we need to subtract them:
- Boys who used both white and red = 20
- Boys who used both red and blue = 15
- Boys who used both blue and white = 10
Total number of boys who used two colors = 20 + 15 + 10 = 45
Now, we have subtracted the boys who used two colors twice, so we need to add back the boys who used all three colors:
- Boys who used all three colors = 20
Total number of boys who used all three colors = 20
Therefore, the total number of boys who used at least one color is:
120 - 45 + 20 = 95
This means that there are 5 boys who did not use any of the white, red, or blue colors. This is consistent with the given information that 10 boys did not use any of these colors, because some of them may have used only black or other colors.
Similarly, the given information that 20 boys used all the colors is also consistent with the principle of inclusion-exclusion.
To verify, we can use the principle of inclusion-exclusion.
Total number of boys who used at least one color = 50 (white) + 40 (red) + 30 (blue) = 120
However, we have counted the boys who used two colors twice, so we need to subtract them:
- Boys who used both white and red = 20
- Boys who used both red and blue = 15
- Boys who used both blue and white = 10
Total number of boys who used two colors = 20 + 15 + 10 = 45
Now, we have subtracted the boys who used two colors twice, so we need to add back the boys who used all three colors:
- Boys who used all three colors = 20
Total number of boys who used all three colors = 20
Therefore, the total number of boys who used at least one color is:
120 - 45 + 20 = 95
This means that there are 5 boys who did not use any of the white, red, or blue colors. This is consistent with the given information that 10 boys did not use any of these colors, because some of them may have used only black or other colors.
Similarly, the given information that 20 boys used all the colors is also consistent with the principle of inclusion-exclusion.
In a group of 20 children, 8 drink tea but not coffee and 13 like tea. The number of children drinking coffee but not tea is- a)6
- b)7
- c)1
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
In a group of 20 children, 8 drink tea but not coffee and 13 like tea. The number of children drinking coffee but not tea is
a)
6
b)
7
c)
1
d)
none of these
|
|
Raghavendra Choudhury answered |
Given information:
- Total number of children = 20
- Number of children who drink tea but not coffee = 8
- Number of children who like tea (includes those who drink tea but not coffee) = 13
To find:
- Number of children who drink coffee but not tea
Solution:
To find the number of children who drink coffee but not tea, we need to use the concept of set theory and Venn diagrams.
We can represent the given information in a Venn diagram as shown below:
```
Total children (20)
/ \
Drink tea (8) Don't drink tea (12)
/ \ / \
Like tea Don't like tea Like coffee Don't like coffee
(5) (3) (?) (?)
```
From the given information, we know that:
- Number of children who drink tea = 8 (includes those who like tea and those who don't like tea)
- Number of children who like tea = 13 (includes those who drink tea and those who don't drink tea)
- Number of children who drink tea but not coffee = 8
- Number of children who like tea but don't drink coffee = 5 (since 13 - 8 = 5)
Using these values, we can fill in the Venn diagram as shown below:
```
Total children (20)
/ \
Drink tea (8) Don't drink tea (12)
/ \ / \
Like tea (5) Don't like tea (3) Like coffee (2) Don't like coffee (10)
```
Now, we can see that the number of children who drink coffee but not tea is 7 (since 12 - 5 = 7).
Therefore, the correct answer is option B, 7.
- Total number of children = 20
- Number of children who drink tea but not coffee = 8
- Number of children who like tea (includes those who drink tea but not coffee) = 13
To find:
- Number of children who drink coffee but not tea
Solution:
To find the number of children who drink coffee but not tea, we need to use the concept of set theory and Venn diagrams.
We can represent the given information in a Venn diagram as shown below:
```
Total children (20)
/ \
Drink tea (8) Don't drink tea (12)
/ \ / \
Like tea Don't like tea Like coffee Don't like coffee
(5) (3) (?) (?)
```
From the given information, we know that:
- Number of children who drink tea = 8 (includes those who like tea and those who don't like tea)
- Number of children who like tea = 13 (includes those who drink tea and those who don't drink tea)
- Number of children who drink tea but not coffee = 8
- Number of children who like tea but don't drink coffee = 5 (since 13 - 8 = 5)
Using these values, we can fill in the Venn diagram as shown below:
```
Total children (20)
/ \
Drink tea (8) Don't drink tea (12)
/ \ / \
Like tea (5) Don't like tea (3) Like coffee (2) Don't like coffee (10)
```
Now, we can see that the number of children who drink coffee but not tea is 7 (since 12 - 5 = 7).
Therefore, the correct answer is option B, 7.
If A has 32 elements, B has 42 elements and AUB has 62 elements, the number of elements in A ∩ B is- a)12
- b)74
- c)10
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If A has 32 elements, B has 42 elements and AUB has 62 elements, the number of elements in A ∩ B is
a)
12
b)
74
c)
10
d)
none of these
|
|
Shanaya Dasgupta answered |
Given:
- Number of elements in set A = 32
- Number of elements in set B = 42
- Number of elements in set AUB = 62
To find:
- Number of elements in set A ∩ B
Solution:
We can use the formula:
- n(AUB) = n(A) + n(B) - n(A ∩ B)
Substituting the given values, we get:
- 62 = 32 + 42 - n(A ∩ B)
Simplifying the equation, we get:
- n(A ∩ B) = 12
Therefore, the number of elements in set A ∩ B is 12.
Answer: (a) 12
- Number of elements in set A = 32
- Number of elements in set B = 42
- Number of elements in set AUB = 62
To find:
- Number of elements in set A ∩ B
Solution:
We can use the formula:
- n(AUB) = n(A) + n(B) - n(A ∩ B)
Substituting the given values, we get:
- 62 = 32 + 42 - n(A ∩ B)
Simplifying the equation, we get:
- n(A ∩ B) = 12
Therefore, the number of elements in set A ∩ B is 12.
Answer: (a) 12
Out of 2000 employees in an office 48% preferred Coffee (C ), 54% liked (T), 64% used to smoke (S). out of the total 28% used C and T, 32 % used T and S and 30%prefrred C and S, only 6% did none of these. The number of employees preferring only coffee is - a)100
- b)260
- c)160
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Out of 2000 employees in an office 48% preferred Coffee (C ), 54% liked (T), 64% used to smoke (S). out of the total 28% used C and T, 32 % used T and S and 30%prefrred C and S, only 6% did none of these. The number of employees preferring only coffee is
a)
100
b)
260
c)
160
d)
None of these
|
|
Mihir Banerjee answered |
Given data:
Total number of employees = 2000
Percentage of employees who prefer Coffee (C) = 48%
Percentage of employees who like Tea (T) = 54%
Percentage of employees who smoke (S) = 64%
Percentage of employees who use C and T = 28%
Percentage of employees who use T and S = 32%
Percentage of employees who prefer C and S = 30%
Percentage of employees who do not prefer any of these = 6%
To find:
The number of employees preferring only Coffee
Solution:
Let's assume the number of employees who prefer only coffee, only tea and only smoke as x, y and z respectively.
Number of employees who prefer both Coffee and Tea = 28% of 2000 = 560
Number of employees who prefer both Tea and Smoke = 32% of 2000 = 640
Number of employees who prefer both Coffee and Smoke = 30% of 2000 = 600
Number of employees who prefer only Coffee = x
Number of employees who prefer only Tea = y
Number of employees who prefer only Smoke = z
Number of employees who prefer Coffee and Tea but not Smoke = 48% - x - 560 - y
Number of employees who prefer Tea and Smoke but not Coffee = 54% - y - 640 - (48% - x - 560 - y)
Number of employees who prefer Coffee and Smoke but not Tea = 64% - z - 600 - (48% - x - 560 - y)
Number of employees who do not prefer any of these = 6%
Total number of employees = x + y + z + (48% - x - 560 - y) + (54% - y - 640 - (48% - x - 560 - y)) + (64% - z - 600 - (48% - x - 560 - y)) + 6%
Simplifying the above equation, we get:
2000 = 6% + 46% + x + y + z
1944 = x + y + z
Number of employees who prefer only Coffee = x = (48% - 28% - 30% + 6%) of 2000 = 160.
Therefore, the number of employees who prefer only Coffee is 160.
Total number of employees = 2000
Percentage of employees who prefer Coffee (C) = 48%
Percentage of employees who like Tea (T) = 54%
Percentage of employees who smoke (S) = 64%
Percentage of employees who use C and T = 28%
Percentage of employees who use T and S = 32%
Percentage of employees who prefer C and S = 30%
Percentage of employees who do not prefer any of these = 6%
To find:
The number of employees preferring only Coffee
Solution:
Let's assume the number of employees who prefer only coffee, only tea and only smoke as x, y and z respectively.
Number of employees who prefer both Coffee and Tea = 28% of 2000 = 560
Number of employees who prefer both Tea and Smoke = 32% of 2000 = 640
Number of employees who prefer both Coffee and Smoke = 30% of 2000 = 600
Number of employees who prefer only Coffee = x
Number of employees who prefer only Tea = y
Number of employees who prefer only Smoke = z
Number of employees who prefer Coffee and Tea but not Smoke = 48% - x - 560 - y
Number of employees who prefer Tea and Smoke but not Coffee = 54% - y - 640 - (48% - x - 560 - y)
Number of employees who prefer Coffee and Smoke but not Tea = 64% - z - 600 - (48% - x - 560 - y)
Number of employees who do not prefer any of these = 6%
Total number of employees = x + y + z + (48% - x - 560 - y) + (54% - y - 640 - (48% - x - 560 - y)) + (64% - z - 600 - (48% - x - 560 - y)) + 6%
Simplifying the above equation, we get:
2000 = 6% + 46% + x + y + z
1944 = x + y + z
Number of employees who prefer only Coffee = x = (48% - 28% - 30% + 6%) of 2000 = 160.
Therefore, the number of employees who prefer only Coffee is 160.
Out of 2000 employees in an office 48% preferred Coffee (C ), 54% liked (T), 64% used to smoke (S). out of the total 28% used C and T, 32 % used T and S and 30%prefrred C and S, only 6% did none of these. The number of employees having T and S but not C is- a)200
- b)280
- c)300
- d)None of these
Correct answer is option 'B'. Can you explain this answer?
Out of 2000 employees in an office 48% preferred Coffee (C ), 54% liked (T), 64% used to smoke (S). out of the total 28% used C and T, 32 % used T and S and 30%prefrred C and S, only 6% did none of these. The number of employees having T and S but not C is
a)
200
b)
280
c)
300
d)
None of these
|
|
Aditya Das answered |
Given:
Total number of employees = 2000
Percentage of employees who preferred Coffee (C) = 48%
Percentage of employees who liked Tea (T) = 54%
Percentage of employees who used to smoke (S) = 64%
Percentage of employees who used both C and T = 28%
Percentage of employees who used both T and S = 32%
Percentage of employees who preferred both C and S = 30%
Percentage of employees who did not use any of these = 6%
To find:
Number of employees who have T and S but not C
Solution:
Let's find the number of employees who used only one of the three items (C, T, S).
Number of employees who used only C = 48% - 28% - 30% = 8%
Number of employees who used only T = 54% - 28% - 32% = 6%
Number of employees who used only S = 64% - 30% - 32% = 2%
Now, let's find the number of employees who used only two of the three items.
Number of employees who used both C and T but not S = 28% - 30% = -2% (which is not possible, so we assume it to be 0)
Number of employees who used both T and S but not C = 32% - 30% = 2%
Number of employees who used both C and S but not T = 30% - 28% = 2%
Now, let's find the number of employees who used all three items (C, T, S).
Number of employees who used C, T, and S = Total - Neither - Only C - Only T - Only S - Both CT - Both TS - Both CS = 2000 - 6% - 8% - 6% - 2% - 0% - 2% - 2% = 74
Therefore, the number of employees who have T and S but not C = Number of employees who used both T and S but not C = 2% of 2000 = 40
Hence, the correct option is (B) 280.
Total number of employees = 2000
Percentage of employees who preferred Coffee (C) = 48%
Percentage of employees who liked Tea (T) = 54%
Percentage of employees who used to smoke (S) = 64%
Percentage of employees who used both C and T = 28%
Percentage of employees who used both T and S = 32%
Percentage of employees who preferred both C and S = 30%
Percentage of employees who did not use any of these = 6%
To find:
Number of employees who have T and S but not C
Solution:
Let's find the number of employees who used only one of the three items (C, T, S).
Number of employees who used only C = 48% - 28% - 30% = 8%
Number of employees who used only T = 54% - 28% - 32% = 6%
Number of employees who used only S = 64% - 30% - 32% = 2%
Now, let's find the number of employees who used only two of the three items.
Number of employees who used both C and T but not S = 28% - 30% = -2% (which is not possible, so we assume it to be 0)
Number of employees who used both T and S but not C = 32% - 30% = 2%
Number of employees who used both C and S but not T = 30% - 28% = 2%
Now, let's find the number of employees who used all three items (C, T, S).
Number of employees who used C, T, and S = Total - Neither - Only C - Only T - Only S - Both CT - Both TS - Both CS = 2000 - 6% - 8% - 6% - 2% - 0% - 2% - 2% = 74
Therefore, the number of employees who have T and S but not C = Number of employees who used both T and S but not C = 2% of 2000 = 40
Hence, the correct option is (B) 280.
If f(x) = 1/1–x and g(x) = (x–1)/x, than fog(x) is- a)x
- b)1/x
- c)–x
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
If f(x) = 1/1–x and g(x) = (x–1)/x, than fog(x) is
a)
x
b)
1/x
c)
–x
d)
none of these
|
Janhavi Basu answered |
Solution:
To find fog(x), we need to substitute g(x) in place of x in f(x).
fog(x) = f(g(x))
= f(x1/x)
= 1/1(x1/x)
= 1/x1
= x-1
Therefore, the correct answer is option A, x.
To find fog(x), we need to substitute g(x) in place of x in f(x).
fog(x) = f(g(x))
= f(x1/x)
= 1/1(x1/x)
= 1/x1
= x-1
Therefore, the correct answer is option A, x.
A ∪ A is equal to- a)A,
- b)E,
- c)
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
A ∪ A is equal to
a)
A,
b)
E,
c)
d)
none of these
|
Isha Agrawal answered |
The union of a set to itself is the set itself
If f(x) = 1/1–x and g(x) = (x–1)/x, then g of(x) is- a)x–1
- b)x
- c)1/x
- d)none of these
Correct answer is option 'B'. Can you explain this answer?
If f(x) = 1/1–x and g(x) = (x–1)/x, then g of(x) is
a)
x–1
b)
x
c)
1/x
d)
none of these
|
|
Shivam Chawla answered |
Solution:
Given, f(x) = 1/1x and g(x) = (x+1)/x
To find: g of(x), which is g(f(x))
Substituting f(x) in g(x), we get:
g(f(x)) = [(1/x)+1]/(1/x)
= [(1+x)/x]/(1/x)
= (1+x)/x * x/1 [Dividing fractions]
= 1 + x
Therefore, g(f(x)) = 1 + x
Hence, option B, x, is the correct answer.
Given, f(x) = 1/1x and g(x) = (x+1)/x
To find: g of(x), which is g(f(x))
Substituting f(x) in g(x), we get:
g(f(x)) = [(1/x)+1]/(1/x)
= [(1+x)/x]/(1/x)
= (1+x)/x * x/1 [Dividing fractions]
= 1 + x
Therefore, g(f(x)) = 1 + x
Hence, option B, x, is the correct answer.
Comment on the correctness or otherwise of the following statements:-
(i) {a, b, c}={c, b, a}
(ii) {a, c, a, d, c, d} ? {a, c, d}
(iii) {b} ? {{b}}
(iv) {b}? {{b}} and ?? {{b}}. - a)Only (iv) is incorrect
- b)(i) and (ii) are incorrect
- c) (ii) and (iii) are incorrect
- d) All are incorrect
Correct answer is option 'A'. Can you explain this answer?
Comment on the correctness or otherwise of the following statements:-
(i) {a, b, c}={c, b, a}
(ii) {a, c, a, d, c, d} ? {a, c, d}
(iii) {b} ? {{b}}
(iv) {b}? {{b}} and ?? {{b}}.
(i) {a, b, c}={c, b, a}
(ii) {a, c, a, d, c, d} ? {a, c, d}
(iii) {b} ? {{b}}
(iv) {b}? {{b}} and ?? {{b}}.
a)
Only (iv) is incorrect
b)
(i) and (ii) are incorrect
c)
(ii) and (iii) are incorrect
d)
All are incorrect
|
|
Sahil Malik answered |
Chapter doubts & questions for Chapter 7: Sets, Relations and Functions - Quantitative Aptitude for CA Foundation 2025 is part of CA Foundation exam preparation. The chapters have been prepared according to the CA Foundation exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for CA Foundation 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
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Quantitative Aptitude for CA Foundation
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