Set Operations, Venn Diagrams & Applications

(i) Solution:
Combine all distinct elements of $A$ and $B$:
$A \cup B = \{1, 2, 3, 7, 9\}$

(ii) Solution:
Combine alphabetic characters:
$A \cup B = \{a, b, d, e, i, o, u\}$

(iii) Solution:
Write the sets in roster form first:
• $A = \{6, 7, 8, 9, \dots\}$
• $B = \{1, 2, 3, 4\}$
Combining them:
$A \cup B = \{1, 2, 3, 4, 6, 7, 8, 9, \dots\}$ or $\{x \in \mathbb{N} \mid x \neq 5\}$

(iv) Solution:
The union of any set with the empty set is the set itself:
$A \cup B = \{0, 1, 2, \alpha\}$

(i) Solution: The common elements are 1 and 2.
$\{1, 2\}$

(ii) Solution: The first set contains odd integers, and the second contains even integers. They share no common elements.
$\emptyset$ or $\{\}$

(iii) Solution: The common letters are 'a' and 't'.
$\{a, t\}$

(iv) Solution: Since every negative integer is also an integer (negative integers are a subset of integers), their intersection is the set of negative integers itself.
$\{x \mid x \text{ is a negative integer}\}$

Recall that two sets are disjoint if $A \cap B = \emptyset$.

(i) Solution: Not Disjoint
The multiples of 2 and 3 share common multiples (specifically, multiples of 6, like $6, 12, 18, \dots$). Since their intersection is not empty, they are not disjoint.

(ii) Solution: Not Disjoint
Both sets contain the elements $e$ and $\pi$. Their intersection is $\{e, \pi\} \neq \emptyset$. So they are not disjoint.

(iii) Solution: Not Disjoint
Every irrational number is also a real number. Their intersection is the set of irrational numbers, which is non-empty. So they are not disjoint.

Recall that $A - B$ contains elements that are in $A$ but not in $B$.

(i) Solution:
Remove common elements 3 and 8 from $A$:
$\{1, 5\}$

(ii) Solution:
Since $A \subseteq B$, all elements of $A$ are removed:
$\emptyset$

(iii) Solution:
$A$ and $B$ share no common elements. Removing elements of $B$ from $A$ changes nothing:
$\{2, 6\}$

Based on the elements in the Venn diagram:

• Universal set $U = \{1, 2, 4, 5, 6, 7, 8, 9, 10\}$
• Set $A = \{1, 2, 4, 5\}$
• Set $B = \{4, 5, 6, 7, 8\}$
• Set $C = \{5, 8, 9\}$

Solutions:

(i) $A$: $\{1, 2, 4, 5\}$

(ii) $B$: $\{4, 5, 6, 7, 8\}$

(iii) $(A \cap B)'$: $A \cap B = \{4, 5\}$. So, $(A \cap B)' = U - \{4, 5\} = \{1, 2, 6, 7, 8, 9, 10\}$.

(iv) $A - B$: Elements in $A$ but not $B$: $\{1, 2\}$.

(v) $A \cup B \cup C$: Combined elements: $\{1, 2, 4, 5, 6, 7, 8, 9\}$.

(vi) $A \cap (B \cup C)$: $B \cup C = \{4, 5, 6, 7, 8, 9\}$. Intersecting with $A$: $\{4, 5\}$.

Let's verify this step-by-step using a Venn diagram of two overlapping sets $A$ and $B$:

1. Left Hand Side ($A - B$): This region contains elements in $A$ that are strictly outside circle $B$. It corresponds to the left-crescent of circle $A$.
2. Right Hand Side ($A \cap B'$):
   • $B'$ is the entire region outside circle $B$.
   • When we intersect $B'$ with $A$ ($A \cap B'$), we look for elements that lie inside $A$ AND outside $B$.
   • This is exactly the left-crescent region of circle $A$.

Since both operations describe the exact same region (elements in $A$ only, excluding the overlap $A \cap B$), we have verified that:
$A - B = A \cap B'$

Let $M$ be Mathematics and $S$ be Science.

We know that: $n(M \cup S) = n(M) + n(S) - n(M \cap S)$
$\Rightarrow n(M \cap S) = n(M) + n(S) - n(M \cup S)$

(a) Solution:
To find the minimum possible value of $n(M \cap S)$, we must maximize the union $n(M \cup S)$. Since percentages are capped at 100%, the maximum possible value of $n(M \cup S)$ is 100%.
$\Rightarrow \text{Min } n(M \cap S) = 85\% + 75\% - 100\% = 160\% - 100\% = \mathbf{60\%}$.

(b) Solution:
If 10% opted for neither, then the union of students taking these tests is fixed at:
$n(M \cup S) = 100\% - 10\% = 90\%$.
Substituting this fixed value into our equation:
$\Rightarrow n(M \cap S) = 85\% + 75\% - 90\% = 160\% - 90\% = \mathbf{70\%}$.
The minimum percentage of students who opted for both tests increases to 70%.

Let $E$ represent English speakers and $H$ represent Hindi speakers.

We are given:
• $n(E \cup H) = 50$ (since everyone speaks at least one language)
• $n(E) = 35$
• $n(H) = 25$

Step 1: Find the number of people who speak both languages:
$n(E \cup H) = n(E) + n(H) - n(E \cap H)$
$\Rightarrow 50 = 35 + 25 - n(E \cap H)$
$\Rightarrow 50 = 60 - n(E \cap H) \Rightarrow n(E \cap H) = 10$.

Step 2: Find the number of people who speak only English ($k$):
$k = n(E - H) = n(E) - n(E \cap H)$
$\Rightarrow k = 35 - 10 = \mathbf{25}$.

Thus, the value of $k$ is 25.

Let $O$ = Origami, $I$ = Instrumental music, and $F$ = Fine arts.

We are given:
• $n(O \cup I \cup F) = 56$
• $n(O) = 17$,  $n(I) = 28$,  $n(F) = 25$
• $n(O \cap I \cap F) = 4$

Step 1: Write down the 3-set cardinality formula:
$n(O \cup I \cup F) = n(O) + n(I) + n(F) - [n(O \cap I) + n(I \cap F) + n(F \cap O)] + n(O \cap I \cap F)$

Let $S_2 = n(O \cap I) + n(I \cap F) + n(F \cap O)$ represent the sum of double intersections.

$\Rightarrow 56 = 17 + 28 + 25 - S_2 + 4$
$\Rightarrow 56 = 74 - S_2 \Rightarrow S_2 = 18$.

Step 2: Relate $S_2$ to the number of students receiving certificates for exactly two activities:
$S_2 = (\text{exactly two}) + 3 \times n(O \cap I \cap F)$
$\Rightarrow 18 = (\text{exactly two}) + 3(4)$
$\Rightarrow 18 = (\text{exactly two}) + 12 \Rightarrow \text{exactly two} = 18 - 12 = \mathbf{6}$.

Thus, 6 students received certificates for exactly two activities.

Let's calculate the values for each region of the 3-circle Venn diagram starting from the center:

1. All three ($E \cap G \cap S$): $25$
2. Exactly two languages:
   • English and German only: $n(E \cap G) - 25 = 40 - 25 = 15$
   • German and Spanish only: $n(G \cap S) - 25 = 30 - 25 = 5$
   • English and Spanish only: $n(E \cap S) - 25 = 25 - 25 = 0$
3. Exactly one language:
   • English only: $n(E) - (15 + 25 + 0) = 60 - 40 = 20$
   • German only: $n(G) - (15 + 25 + 5) = 50 - 45 = 5$
   • Spanish only: $n(S) - (0 + 25 + 5) = 35 - 30 = 5$
4. Union of all speakers ($n(E \cup G \cup S)$):
   $20 (\text{E only}) + 5 (\text{G only}) + 5 (\text{S only}) + 15 (\text{E\&G}) + 5 (\text{G\&S}) + 0 (\text{E\&S}) + 25 (\text{all 3}) = \mathbf{75}$.
5. None of the languages:
   $n(U) - n(E \cup G \cup S) = 100 - 75 = \mathbf{25}$.

Final Answers:

(a) At least two languages: (exactly two + all three) = $15 + 5 + 0 + 25 = \mathbf{45}$ students.

(b) At most one language: (exactly one + none) = $20 (\text{E only}) + 5 (\text{G only}) + 5 (\text{S only}) + 25 (\text{none}) = \mathbf{55}$ students.

(c) Not speak any: $\mathbf{25}$ students.