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If n(A) = 35, n(B) = 30, and n(A $\rm \cap$ B) = 20, find n (A $\cup$ B) and no(A) by using a Venn diagram.
Atith Adhikari
·
1 year ago
Solution
Given
$\rm n(A) = 35, n(B) = 30, n(A \cap B) = 20$
To find: $\rm n(A \cup B) = ?$ and $\rm n_{o}(A) = ?$
We plot this information in a Venn diagram.
By using the Venn diagram, we find that:
$\rm n(A \cup B) = n_{o}(A) + n(A \cap B) + n_{o}(B) = 45$
$\rm n_{o} (A) = 15$
If n(X) = 40, n(Y) = 60 and n(X $\rm \cup$ Y) = 85, then find n(X $\rm \cap$ Y) and no(Y) by using formula.
Atith Adhikari
·
1 year ago
Solution
Given
$\rm n(X) = 40, n(Y) = 60, n(X \cup Y) = 85$
By using the formula for the union of two sets, we get
$\rm n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$
$\rm or, n(X \cap Y) = n(X) + n(Y) - n(X \cup Y)$
$\rm or, n(X \cap Y) = 40 + 60 - 85$
$\rm \therefore n(X \cap Y) = 15$
Now, we show the above information in a Venn diagram.
If P = {Multiples of 2 up to 20}, Q = {Multiples of 3 up to 24} and U = {integers from 1 to 25}, find no(P) and n(P $\rm \cup$ Q) by using a Venn diagram.
Atith Adhikari
·
1 year ago
Solution
Given
P = {Multiples of 2 up to 20}; P = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
Q = {Multiples of 3 up to 24}; Q = {3, 6, 9, 12, 15, 18, 21, 24}
$\rm (P \cap Q)$ = {6, 12, 18}
$\rm (P \cup Q)$ = {2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21, 24}
U = {Integers from 1 to 25}; U = {1, 2, 3, 4, 5, …, 21, 22, 23, 24, 25}
Showing the above information in a Venn diagram.
Hence,
$\rm n_{o} (P) = 7$
$\rm n(A \cup B) = n_{o}(P) + n(P \cap Q) + n_{o}(Q)$
$\rm \therefore n(A \cup B) = 7 + 3 + 5 = 15$
If A = {factors of 18}, B = {multiples of 3 up to 27}, and U = {integers from 1 to 30}, find the cardinality of no(B) and n(A$\rm \cup$B) by using a Venn diagram.
Atith Adhikari
·
1 year ago
Solution
Given
A = {factors of 18}; A = {1, 2, 3, 6, 9, 18}
B = {multiples of 3 up to 27}; B = {3, 6, 9, 12, 15, 18, 21, 24, 27}
$\rm (A \cap B)$ = {3, 6, 9, 18}
$\rm (A \cup B)$ = {1, 2, 3, 6, 9, 12, 15, 18, 21, 24, 27}
U = {integers from 1 to 30}; U = {1, 2, 3, 4, 5, …, 26, 27, 28, 29, 30}
Showing the above information in a Venn diagram.
Hence, the cardinality of the required sets are given below:
$\rm n_{o} (A) = 2$
$\rm n(A \cup B) = n_{o}(A) + n(A \cap B) + n_{o}(B)$
$\rm \therefore n(A \cup B) = 2 + 4+ 5 = 11$
In sets A and B, A has 40 members, B has 50 members, and $\rm (A \cup B)$ has 60 members. By how many elements $\rm (A \cap B)$ is formed?
Atith Adhikari
·
1 year ago
Solution
Let the cardinality of sets A and B be represented by n(A) and n(B), respectively.
According to the question,
$\rm n(A) = 40, n(B) = 50, n(A \cup B) = 60$
To find: $\rm n (A \cap B) =?$
We use the formula for the union of two sets to find the cardinality of the intersection of those sets, we get,
$\rm n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$\rm or, n (A \cap B) = n(A) + n(B) - n(A \cup B)$
$\rm or, n(A \cap B) = 40 + 50 - 60$
$\rm \therefore n(A \cap B) = 30$
Hence, 30 elements. form the set $\rm (A \cap B)$.
In sets A and B, A has 50 members, B has 60 members, and 30 members are the same in both sets. By how many elements $\rm (A \cup B)$ is formed?
Atith Adhikari
·
1 year ago
Solution
Let the cardinality of sets A and B be represented by n(A) and n(B), respectively.
According to the question,
$\rm n(A) = 50, n(B) = 60, n(A \cap B) = 30$
To find: $\rm n(A \cup B) = ?$
We use the formula for the union of two sets to find the cardinality of the intersection of those sets, we get,
$\rm n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$\rm or, n(A \cup B) = 50 + 60 - 30$
$\rm n(A \cup B) = 110 - 30$
$\rm \therefore n(A \cup B) = 80$
Hence, 80 members form the set $\rm n(A \cup B)$.
If n(A) = 40, n(B) = 60, and n(A $\rm \cup$ B) = 80,
- Find the value of n (A $\rm \cap$ B).
- Draw a Venn diagram of the above information.
Atith Adhikari
·
1 year ago
Solution
Given
$\rm n(A) = 40, n(B) = 60, n(A \cup B) = 80$
By using the formula for the union of two sets, we get
$\rm n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$\rm or, n(A \cap B) = n(A) + n(B) - n(A \cup B)$
$\rm or, n(A \cap B) = 40 + 60 - 80$
$\rm \therefore n(A \cap B) = 20$
Now, we show the above information in a Venn diagram.
Out of 90 civil servants, 65 were working in the office, 50 were working in the field and 35 were working in both the premises (sites).
- Let O and F represent the set of civil servants working in office and field respectively, then find $\rm n(F)$ and $\rm n(O \cap F)$.
- Represent the given information in a Venn diagram.
- How many civil servants were absent?
- How many civil servants were working in the field only?
Atith Adhikari
·
1 year ago
Solution
Let O represent the set of civil servants working in the office. Similarly, let F represent the set of civil servants working in the field.
According to the question, 65 civil servants work in the office, i.e., $\rm n(O) = 65$, and 50 work in the field, i.e., $\rm n(F) = 50$.
Also, there were a total of 90 civil servants, $\rm n(U) = 90$. 35 of them worked in both office and field, $\rm n(O \cap F) = 35$. A few civil servants may be absent, $\rm n (\overline{O \cup F})$.
1) The required values of $\rm n(F)$ and $\rm n(O \cap F)$ are $\rm n(F) = 50$ and $\rm n(O \cap F) = 35$.
2) Representing the above information in a Venn diagram.
3) Let $\rm n (\overline{O \cup F})$ represent the set of civil servants who were absent. Using the formula
$\rm n(U) = n(O) + n(F) - n(O \cap F) + n (\overline{O \cup F})$
$\rm or, n (\overline{O \cup F}) = n(U) - n(O) - n(F) + n(O \cap F)$
$\rm or, n (\overline{O \cup F}) = 90 - 65 - 50 + 35$
$\rm \therefore n (\overline{O \cup F}) = 10$
Hence, 10 civil servants were absent.
4) We can obtain the number of civil servants working in fields only by using the formula $\rm n_o{F} = n(F) - n(O \cap F)$
$\rm \therefore n_o{F} = 50 - 35 = 15$
Out of 100 students, 80 passed in Science, 71 in Mathematics, 10 failed in both subjects, and 7 did not appear in an examination.
- Let S and M represent the sets of students who were passed in science and Maths respectively, then find n(S) and n(M).
- Find the number of students who passed in either Science or Mathematics.
- Find the number of students who passed in both subjects.
- Represent the above information in a Venn diagram.
Atith Adhikari
·
1 year ago
Solution
Given,
- Out of 100 students, 80 passed in Science, 71 in Mathematics.
- 10 failed in both subjects.
- 7 did not appear in an examination.
Let
- \( S \) represent the set of students who passed in Science.
- \( M \) represent the set of students who passed in Mathematics.
Required Values:
- \( n(S) = 80 \)
- \( n(M) = 71 \)
- \( n\overline{(S \cup M)} = 10 \)
- \( n(U) = 100 \)
The total number of students is the sum of students who passed, students who failed, and students who did not appear in the examination.
\( \rm n(U) = n( S \cup M) + n \overline{n(S \cup M)}+ 7 \)
\( \rm or, n(U) = n(S) + n(M) - n(S \cap M) + n \overline{n(S \cup M)} + 7 \)
\( \rm or, 100 = 80 + 71 - n (S \cap M) + 10 + 7\)
\( \rm or, n(S \cap M) = 168 - 100 \)
\( \rm \therefore n(S \cap M) = 68 \)
The number of students who passed in either Science or Math are those students who did not fail in both subjects. The number of such students (say x) is equal to the difference in total students who attended the examination and students who failed both subjects.
\( \rm x = n(U) - 7 - n \overline{n(S \cup M)} \)
\( \rm x = 100 - 7 - 10 \)
\( \rm x = 83 \)
Hence,
\( n(S) = 80 \) and \( n(M) = 71 \).
Number of students who passed in either Science or Maths is 83.
Number of students who passed in both subjects: \( n(S \cap M) = 68 \).
Representing the above information in a Venn diagram
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