Cardinality of the Two Sets

Question 2

In the given Venn-diagram, 80 people are in set M, 90 people are in set E and 15 people are not in both the sets. Determine the cardinality of following sets.

Solutions:

1. $n_o(M)$ (Number of people in M only)
   $n_o(M) = 20$
2. $n_o(E)$ (Number of people in E only)
   $n_o(E) = 30$
3. $n(M)$ (Total number of people in M)
   $n(M) = n_o(M) + n(M \cap E) = 20 + 60 = 80$
4. $n(E)$ (Total number of people in E)
   $n(E) = n_o(E) + n(M \cap E) = 30 + 60 = 90$
5. $n(M \cup E)$ (Number of people in M or E or both)
   $n(M \cup E) = n_o(M) + n_o(E) + n(M \cap E) = 20 + 30 + 60 = 110$

   Alternatively:

   $n(M \cup E) = n(M) + n(E) - n(M \cap E) = 80 + 90 - 60 = 110$
6. $n(M \cap E)$ (Number of people in both M and E)
   $n(M \cap E) = 60$
7. $n(\overline{M \cup E})$ (Number of people in neither M nor E)
   $n(\overline{M \cup E}) = 15$
8. $n(U)$ (Total number of people in the universal set)
   $n(U) = n(M \cup E) + n(\overline{M \cup E}) = 110 + 15 = 125$

   Alternatively:

   $n(U) = n_o(M) + n_o(E) + n(M \cap E) + n(\overline{M \cup E}) = 20 + 30 + 60 + 15 = 125$

Question 3

a) If $n(U) = 200$, $n_o(M) = 2x$, $n_o(E) = 3x$, $n(M \cap E) = 60$ and $n(\overline{M \cup E}) = 40$, find the value of x.

Solution:

We use the formula for the universal set:

$n(U) = n_o(M) + n_o(E) + n(M \cap E) + n(\overline{M \cup E})$

Substitute the given values into the formula:

$200 = 2x + 3x + 60 + 40$

Simplify the equation:

$200 = 5x + 100$

Subtract 100 from both sides:

$200 - 100 = 5x$ $100 = 5x$

Divide by 5 to solve for x:

or, $\frac{100}{5} = x$

$x = \frac{100}{5} = 20$

Therefore, the value of x is 20.

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b) If $n(U) = 350$, $n(A) = 200$, $n(B) = 220$ and $n(A \cap B) = 120$, then find $n(A \cup B)$ and $n(\overline{A \cup B})$.

Solution:

Finding $n(A \cup B)$:

We use the formula:

$n(A \cup B) = n(A) + n(B) - n(A \cap B)$

Substitute the given values:

$n(A \cup B) = 200 + 220 - 120 = 300$

Finding $n(\overline{A \cup B})$:

We use the formula:

$n(\overline{A \cup B}) = n(U) - n(A \cup B)$

Substitute the values:

$n(\overline{A \cup B}) = 350 - 300 = 50$

Therefore, $n(A \cup B) = 300$ and $n(\overline{A \cup B}) = 50$.

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c) If $n(A) = 35$ and $n(\overline{A}) = 25$, then find the value of $n(U)$.

Solution:

We know that the universal set consists of all elements in A and all elements not in A:

$n(U) = n(A) + n(\overline{A})$

Substitute the given values:

$n(U) = 35 + 25 = 60$

Therefore, the value of $n(U) = 60$.

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d) Out of two sets P and Q, there are 40 elements in P, 60 elements in $(P \cup Q)$ and 10 elements in $(P \cap Q)$. How many elements are there in Q? Find.

Solution:

Given:

• $n(P) = 40$
• $n(P \cup Q) = 60$
• $n(P \cap Q) = 10$

We use the formula:

$n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)$

Substitute the known values:

$60 = 40 + n(Q) - 10$

Simplify:

$60 = 30 + n(Q)$

Solve for $n(Q)$:

$n(Q) = 60 - 30 = 30$

Therefore, there are 30 elements in set Q.

Question 4

a) In a survey of 180 students of a school, 45 like Nepali only and 60 like English only but 15 like none of the subjects. Based on this information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find the number of students who like both the subjects.
3. Find the number of students who like at least one subject.

Solution:

i) Venn Diagram:

ii) Finding the number of students who like both subjects:

Given information:

• Total students: $n(U) = 180$
• Students who like Nepali only: $n_o(N) = 45$
• Students who like English only: $n_o(E) = 60$
• Students who like none: $n(\overline{N \cup E}) = 15$

Using the universal set formula:

$n(U) = n_o(N) + n_o(E) + n(N \cap E) + n(\overline{N \cup E})$

Substitute the given values:

$180 = 45 + 60 + n(N \cap E) + 15$

Simplify:

$180 = 120 + n(N \cap E)$

Solve for $n(N \cap E)$:

$n(N \cap E) = 180 - 120 = 60$

Therefore, 60 students like both subjects.

iii) Finding the number of students who like at least one subject:

Students who like at least one subject means $n(N \cup E)$.

Method 1: Using the complement

$n(N \cup E) = n(U) - n(\overline{N \cup E}) = 180 - 15 = 165$

Method 2: Using the addition principle

$n(N \cup E) = n_o(N) + n_o(E) + n(N \cap E) = 45 + 60 + 60 = 165$

Therefore, 165 students like at least one subject.

b) In a survey among the 1200 students of a school, 100 like Mathematics only and 200 like Science only but 700 like neither of the subjects. Based on the information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find the number of students who like both the subjects.
3. Find the number of students who like at least one subject.

Solution:

i) Venn Diagram:

ii) Finding the number of students who like both subjects:

Given information:

• Total students: $n(U) = 1200$
• Students who like Mathematics only: $n_o(M) = 100$
• Students who like Science only: $n_o(S) = 200$
• Students who like neither: $n(\overline{M \cup S}) = 700$

Using the universal set formula:

$n(U) = n_o(M) + n_o(S) + n(M \cap S) + n(\overline{M \cup S})$

Substitute the given values:

$1200 = 100 + 200 + n(M \cap S) + 700$

Simplify:

$1200 = 1000 + n(M \cap S)$

Solve for $n(M \cap S)$:

$n(M \cap S) = 1200 - 1000 = 200$

Therefore, 200 students like both subjects.

iii) Finding the number of students who like at least one subject:

Students who like at least one subject means $n(M \cup S)$.

Method 1: Using the complement

$n(M \cup S) = n(U) - n(\overline{M \cup S}) = 1200 - 700 = 500$

Method 2: Using the addition principle

$n(M \cup S) = n_o(M) + n_o(S) + n(M \cap S) = 100 + 200 + 200 = 500$

Therefore, 500 students like at least one subject.

c) In a survey among 60 students, 10 play football only and 20 play volleyball only but 12 play neither of the games. Based on the information, answer the following questions:

1. Show the above information in a Venn-diagram.
2. Find the number of students who play both the games.
3. Find the number of students who play at least one game.

Solution:

i) Venn Diagram:

ii) Finding the number of students who play both games:

Given information:

• Total students: $n(U) = 60$
• Students who play football only: $n_o(F) = 10$
• Students who play volleyball only: $n_o(V) = 20$
• Students who play neither game: $n(\overline{F \cup V}) = 12$

Using the universal set formula:

$n(U) = n_o(F) + n_o(V) + n(F \cap V) + n(\overline{F \cup V})$

Substitute the given values:

$60 = 10 + 20 + n(F \cap V) + 12$

Simplify:

$60 = 42 + n(F \cap V)$

Solve for $n(F \cap V)$:

$n(F \cap V) = 60 - 42 = 18$

Therefore, 18 students play both games.

iii) Finding the number of students who play at least one game:

Students who play at least one game means $n(F \cup V)$.

Method 1: Using the complement

$n(F \cup V) = n(U) - n(\overline{F \cup V}) = 60 - 12 = 48$

Method 2: Using the addition principle

$n(F \cup V) = n_o(F) + n_o(V) + n(F \cap V) = 10 + 20 + 18 = 48$

Therefore, 48 students play at least one game.