Unit - 1, Sets

Exercise - 1.1

 1. a) Present the cardinality of sets with examples and show it to your teacher.

Answer👉 Present the cardinality of sets with examples

Cardinality refers to the number of elements in a set. For a set $A$, the cardinality is denoted as $n(A)$.

Examples:

1. If $A = \{1, 2, 3, 4\}$, then $n(A) = 4$ (4 elements).
2. If $B = \{a, b, c\}$, then $n(B) = 3$ (3 elements).

To present to your teacher, you can list these sets and their cardinalities clearly.

b) Fortwo sets A and B, A c B, find the values of n(AUB) and n(AMB). IFA and B are overlapping sets, state the formula for n(AUB).

Answer👉 

For two sets $A$ and $B$, if $A \subseteq B$:

1. Values of $n(A \cup B)$:
   The union of two sets $A$ and $B$ combines all elements in both sets without duplication.
   If $A \subseteq B$, then $A \cup B = B$.
   Thus, $n(A \cup B) = n(B)$.

2. Values of $n(A \cap B)$:
   The intersection of $A$ and $B$ is all elements common to both sets.
   If $A \subseteq B$, then $A \cap B = A$.
   Thus, $n(A \cap B) = n(A)$.

Formula for $n(A \cup B)$ if $A$ and $B$ are overlapping sets:

For overlapping sets, the cardinality of the union is calculated as:

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

This formula adjusts for elements that are counted twice in both sets.

d) There are 12 and 8 elements in the sets A and B respectively, Find the ‘minimum number of elements that would be in the set n(Au B).

Answer👉 

Minimum number of elements in $n(A \cup B)$:

Given $n(A) = 12$ and $n(B) = 8$, the minimum number of elements in $A \cup B$ occurs when $A$ and $B$ overlap entirely, i.e., $A \cap B = B$ (all elements of $B$ are in $A$).

Using the formula:

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

If $n(A \cap B) = n(B) = 8$:

$$n(A \cup B) = 12 + 8 - 8 = 12$$

Thus, the minimum number of elements in $A \cup B$ is 12.

2. Inthe 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. ”

Answer👉 From the Venn diagram provided:

• $n(M) = 80$: Total people in $M$.
• $n(E) = 90$: Total people in $E$.
• $n(M \cap E) = 60$: People in both $M$ and $E$.
• $n(U) = n(M \cup E) + n(\text{not in both sets}) = (n(M) + n(E) - n(M \cap E)) + 15$.

Subquestions:

a) $n(M) = 80$
b) $n(E) = 90$
c) $n(M \cup E) = n(M) + n(E) - n(M \cap E) = 80 + 90 - 60 = 110$
d) $n(M \cap E) = 60$
e) $n(M \cup E) + 15 = 110 + 15 = 125$ (Cardinality of Universal set $n(U)$).
f) $n(M \cap \text{not } E) = n(M) - n(M \cap E) = 80 - 60 = 20$.
g) $n(E \cap \text{not } M) = n(E) - n(M \cap E) = 90 - 60 = 30$.
h) $n(U) = n(M \cup E) + 15 = 125$.

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Question 3

Answer👉 

a)

• Given $n(U) = 200$, $n(M) = 2x$, $n(E) = 3x$, $n(M \cap E) = 60$, $n(M \cup E) = 40$.
• Formula: $n(M \cup E) = n(M) + n(E) - n(M \cap E)$.

Substitute the given values:

$$40 = 2x + 3x - 60$$ $$40 = 5x - 60$$ $$5x = 100 \quad \implies \quad x = 20$$

So:

$$n(M) = 2x = 40, \quad n(E) = 3x = 60.$$

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b)

• $n(U) = 350$, $n(A) = 200$, $n(B) = 220$, $n(A \cap B) = 120$.
• Formula: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.

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

So:

$$n(A \cup B) = 300.$$

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c)

• Given $n(A) = 35$, $n(\bar{A}) = 25$.
• Universal set: $n(U) = n(A) + n(\bar{A})$.

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

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d)

• $n(P) = 40$, $n(P \cup Q) = 60$, $n(P \cap Q) = 10$.
• Formula: $n(P \cup Q) = n(P) + n(Q) - n(P \cap Q)$.

Substitute the given values:

$$60 = 40 + n(Q) - 10$$ $$n(Q) = 30.$$

Thus, $n(Q) = 30$.

b) In a survey among 1200 students of a school:

• 100 students like Mathematics only.
• 200 students like Science only.
• 700 students like neither of the subjects.

Subquestions:

i) Show the above information in a Venn diagram.
We can denote the sets:

• $M$: Students who like Mathematics.
• $S$: Students who like Science.

Using the formula for total students:

$$n(U) = n(M \cup S) + n(\text{none}) = 1200 \quad \implies \quad n(M \cup S) = 1200 - 700 = 500.$$

Let $x$ be the number of students who like both subjects.

$$n(M \cup S) = n(M) + n(S) - n(M \cap S).$$

Substitute the known values:

$$500 = 100 + 200 - x \quad \implies \quad x = 300 - 500 = 100.$$

So, 100 students like both subjects.

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ii) Find the number of students who like both subjects.

$$n(M \cap S) = 100$$