Set Theory Questions in English

(B) Two sets are said to be equivalent if they have the same number of elements,i.e.,their cardinal numbers are equal.
In option $A$: $n(A) = 4$ and $n(B) = 4$. Since $n(A) = n(B)$,they are equivalent.
In option $B$: $n(A) = 3$ and $n(B) = 4$. Since $n(A) \neq n(B)$,they are not equivalent.
In option $C$: $n(A) = 0$ and $n(B) = 0$. Since $n(A) = n(B)$,they are equivalent.
In option $D$: $n(A) = 3$ and $n(B) = 3$. Since $n(A) = n(B)$,they are equivalent.
Therefore,the pair in option $B$ is not equivalent.

(B) To find the cardinal number of the set,we first list the unique letters present in the word '$ASSASSINATION$'.
The letters are: $A, S, S, A, S, S, I, N, A, T, I, O, N$.
The distinct letters are: $A, S, I, N, T, O$.
The set is $\{A, S, I, N, T, O\}$.
The number of elements in this set is $6$.
Therefore,the cardinal number is $6$.

(C) Let $A$ be the set of natural numbers $\leq 30$ that are divisible by $7$ or $11$.
Numbers divisible by $7$ up to $30$ are: $7, 14, 21, 28$.
Numbers divisible by $11$ up to $30$ are: $11, 22$.
Combining these,the set $A = \{7, 11, 14, 21, 22, 28\}$.
The cardinal number of a set is the number of elements in the set.
Counting the elements in set $A$,we get $n(A) = 6$.

(B) The set is defined as $S = \{x: x = 2n, n \in N, 4 \leq x \leq 11\}$.
Since $n \in N$ (natural numbers),we test values of $n$ such that $4 \leq 2n \leq 11$.
Dividing the inequality by $2$,we get $2 \leq n \leq 5.5$.
Since $n$ must be a natural number,the possible values for $n$ are $2, 3, 4, 5$.
For $n = 2, x = 2(2) = 4$.
For $n = 3, x = 2(3) = 6$.
For $n = 4, x = 2(4) = 8$.
For $n = 5, x = 2(5) = 10$.
The set is $S = \{4, 6, 8, 10\}$.
The cardinal number of a set is the number of elements in the set.
Therefore,the cardinal number is $4$.

(C) To determine which set is finite,we analyze each option:
$A$: The set of all prime numbers is infinite.
$B$: There are infinitely many quadrilaterals that can be drawn on a plane.
$C$: We are given the condition $x \in N$ (natural numbers) and $x^2 - 25 \leq 0$. Solving $x^2 \leq 25$ gives $-5 \leq x \leq 5$. Since $x \in N$,the possible values for $x$ are ${1, 2, 3, 4, 5}$. This set contains $5$ elements,so it is finite.
$D$: The set of multiples of $3$ in natural numbers is ${3, 6, 9, 12, \dots}$,which is infinite.
Therefore,the correct set is ${x: x \in N \text{ and } x^2 - 25 \leq 0}$.

(C) Two sets $A$ and $B$ are said to be equivalent if they have the same number of elements,i.e.,$n(A) = n(B)$.
For option $A$: $n(A) = 26$ (letters in English alphabet) and $n(B) = 24$. Since $26 \neq 24$,they are not equivalent.
For option $B$: $n(A) = 3$. Set $B$ is an infinite set as $n \in N$,so $n(B) = \infty$. They are not equivalent.
For option $C$: $n(A) = 3$. Set $B$ contains ordered pairs,$n(B) = 3$. Since $n(A) = n(B) = 3$,sets $A$ and $B$ are equivalent.
For option $D$: $n(A) = 4$ (for $n = 0, 1, 2, 3$) and $n(B) = 3$. Since $4 \neq 3$,they are not equivalent.
Thus,the correct case is $C$.

(B) Two sets $A$ and $B$ are said to be equal if they contain exactly the same elements.
Option $A$: $A=\{12, 14, 16\}$ and $B=\{16, 18, 20\}$. Since $12 \in A$ but $12 \notin B$,$A \neq B$.
Option $B$: $A=\phi$ (empty set) and $B=\{\}$ (empty set). Both sets contain no elements,so $A=B$.
Option $C$: $A=\{x: x \in W \text{ and } x < 1\}$. Since $W$ (whole numbers) starts from $0$,the only element is $0$. Thus $A=\{0\}$,while $B=\phi$. Since $\{0\} \neq \phi$,$A \neq B$.
Option $D$: $A=\{x: x \text{ is a day of the week beginning with } S\} = \{\text{Sunday, Saturday}\}$. $B=\{\text{Sunday}\}$. Since $\{\text{Sunday, Saturday}\} \neq \{\text{Sunday}\}$,$A \neq B$.
Therefore,both $B$ and $C$ (if interpreted as empty) are technically equal,but $A=\phi$ and $B=\{\}$ is the standard definition of equality for empty sets.

(A) Let $C$ be the set of students who play cricket and $F$ be the set of students who play football.
Given:
$n(C) = 50$
$n(F) = 20$
$n(C \cap F) = 10$
We need to find the number of students who play at least one of these two games,which is $n(C \cup F)$.
Using the formula: $n(C \cup F) = n(C) + n(F) - n(C \cap F)$
$n(C \cup F) = 50 + 20 - 10 = 60$
Alternatively,from the Venn diagram:
Students playing only cricket = $50 - 10 = 40$
Students playing only football = $20 - 10 = 10$
Students playing both = $10$
Total students playing at least one game = $40 + 10 + 10 = 60$.

(D) Let $A = {0}.$ The power set of a set $A,$ denoted by $P(A),$ is the set of all possible subsets of $A.$
For the set $A = {0},$ the subsets are the empty set $\phi$ and the set itself ${0}.$
Therefore,the power set $P(A) = {\phi, {0}}.$

(D) Let $A = \{\{a, b\}, c\}$.
To determine the power set $P(A)$,we note that the set $A$ contains two elements: $x = \{a, b\}$ and $y = c$.
The number of elements in the power set $P(A)$ is given by $2^n$,where $n$ is the number of elements in $A$.
Here,$n = 2$,so $P(A)$ will contain $2^2 = 4$ elements.
The elements of the power set $P(A)$ are the subsets of $A$,which are:
$1$. The empty set: $\phi$
$2$. The set containing the first element: $\{\{a, b\}\}$
$3$. The set containing the second element: $\{c\}$
$4$. The set $A$ itself: $\{\{a, b\}, c\}$
Therefore,$P(A) = \{\phi, \{\{a, b\}\}, \{c\}, A\}$.

(B) Two sets $A$ and $B$ are said to be comparable if $A \subseteq B$ or $B \subseteq A$.
$(a)$ $1 \in A$ but $1 \notin B$ and $6 \in B$ but $6 \notin A$. Thus,$A \not\subseteq B$ and $B \not\subseteq A$. Therefore,they are not comparable.
$(b)$ $A = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ and $B = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$. Since every element of $A$ is in $B$,$A \subset B$. Therefore,$A$ and $B$ are comparable.
$(c)$ $\{4, 5\} \in A$ but $\{4, 5\} \notin B$ and $4 \in B$ but $4 \notin A$. Thus,$A \not\subseteq B$ and $B \not\subseteq A$. Therefore,they are not comparable.

(C) Given the set $A = \{\phi, \{\phi\}, 1, \{1, \phi\}, 7\}$.
$1$. $\phi \in A$: The empty set $\phi$ is an element of $A$,so this is true.
$2$. $\{\phi\} \in A$: The set containing the empty set $\{\phi\}$ is an element of $A$,so this is true.
$3$. $\{1\} \in A$: The set $\{1\}$ is not an element of $A$. The elements of $A$ are $\phi, \{\phi\}, 1, \{1, \phi\},$ and $7$. Since $\{1\}$ is not listed as an element,this statement is false.
$4$. $\{7, \phi\} \subset A$: Since both $7 \in A$ and $\phi \in A$,the set $\{7, \phi\}$ is a subset of $A$,so this is true.
Therefore,the false statement is $\{1\} \in A$.

(B) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
$1$. The elements of set $A$ are $1$,$2$,${3, 4}$,and $5$.
$2$. Since ${3, 4}$ is listed as an element within the set $A$,the statement ${3, 4} \in A$ is true.
$3$. The statement ${3, 4} \subset A$ is false because the elements of the set ${3, 4}$ are $3$ and $4$,which are not elements of $A$.
$4$. The statement $1 \subset A$ is false because $1$ is an element of $A$,not a subset ($1 \in A$ is true).
$5$. The statement ${1, 2, 5} \in A$ is false because the set ${1, 2, 5}$ is not an element of $A$ (though it is a subset of $A$).

(D) Given sets are $A = \{1, 3, 5\}$ and $B = \{x : x \text{ is an odd natural number } < 6\}$.
Since the odd natural numbers less than $6$ are $1, 3, 5$,we have $B = \{1, 3, 5\}$.
Comparing the two sets,we observe that $A = B$.
By the definition of subsets,if $A = B$,then $A \subseteq B$ and $B \subseteq A$. Since $A$ and $B$ are equal,the statements $A \subset B$ (improperly used as subset),$B \subset A$,and $A = B$ are all mathematically consistent with the set equality.
Therefore,none of the given options $A, B, C$ are false.
Thus,the correct choice is $D$.

(B, C) Given the set $A = \{1, 2, \{3, 4\}, 5\}$.
$1$. The elements of $A$ are $1$,$2$,$\{3, 4\}$,and $5$.
$2$. Since $\{3, 4\}$ is an element of $A$,the statement $\{3, 4\} \in A$ is true.
$3$. Since $\{3, 4\}$ is an element of $A$,the set containing this element,which is $\{\{3, 4\}\}$,is a subset of $A$. Therefore,$\{\{3, 4\}\} \subset A$ is also true.
$4$. The elements $3$ and $4$ are not individual elements of $A$; they are part of the set $\{3, 4\}$. Thus,$\{3, 4\} \subset A$ is false,and $\{1, 3, 5\} \subset A$ is false.

(A) The power set of a set $A$,denoted by $P(A)$,is the set of all possible subsets of $A$.
Given the set $A = \{8, 9\}$.
The number of elements in $A$ is $n = 2$.
The number of elements in the power set $P(A)$ is $2^n = 2^2 = 4$.
The subsets of $A$ are: $\phi$ (the empty set),$\{8\}$,$\{9\}$,and $\{8, 9\}$.
Therefore,the power set is $P(A) = \{\phi, \{8\}, \{9\}, \{8, 9\}\}$.

(B) The power set of a set $C$,denoted by $P(C)$,is the set of all subsets of $C$.
Given $C = \{1, \{2\}\}$.
The elements of $C$ are $1$ and $\{2\}$.
The subsets of $C$ are:
$1$. The empty set: $\phi$
$2$. The set containing the first element: $\{1\}$
$3$. The set containing the second element: $\{\{2\}\}$
$4$. The set containing both elements: $\{1, \{2\}\}$
Therefore,the power set $P(C) = \{\phi, \{1\}, \{\{2\}\}, \{1, \{2\}\}\}$.

(A) Given the set $A = \{x : x = \frac{n-1}{n+1},\ n \in W,\ n \le 10\}$.
Since $n \in W$ (whole numbers), the possible values for $n$ are:
$\{0,1,2,3,4,5,6,7,8,9,10\}$
For $n = 0$:
$x = \frac{0-1}{0+1} = -1$
For $n = 1$:
$x = \frac{1-1}{1+1} = 0$
Since $1 \in W$ and $1 \le 10$, we get $x = 0$ is obtained.
Therefore, $0 \in A$ is a correct statement.

(D) First,we list the elements of each set by identifying the unique letters in the given words:
$A = \{B, O, W, L\}$
$B = \{E, L, B, O, W\}$
$C = \{B, E, L, O, W\}$
Comparing the sets,we see that $A = \{B, O, W, L\}$ and $B = \{B, O, W, L, E\}$. Since all elements of $A$ are in $B$,$A \subset B$ is true.
Comparing $B$ and $C$,we see that $B = \{B, E, L, O, W\}$ and $C = \{B, E, L, O, W\}$. Thus,$B = C$.
Since $B = C$,the statement $B \supset C$ is true (as every set is a subset of itself).
The statement $B$ is a proper subset of $C$ is false because $B$ is equal to $C$,and a set cannot be a proper subset of itself.

(A) set is called finite if it has a finite number of elements. If a set $A$ is finite,then any subset $B$ of $A$ (where $B \subseteq A$) must also contain a finite number of elements,because the number of elements in $B$ cannot exceed the number of elements in $A$. Therefore,every subset of a finite set is finite. Option $B$ is false because a subset of an infinite set can be finite (e.g.,the set ${1}$ is a subset of the infinite set of natural numbers $\mathbb{N}$). Option $C$ is false because a subset of an infinite set can also be infinite (e.g.,the set of even numbers is a subset of $\mathbb{N}$). Option $D$ is false because a proper subset of a finite set always has fewer elements than the set itself,so it cannot be equivalent to the set.

(D) Given sets are:
$A = \{2, 4, 6, 8, 10, 12, \dots\}$
$B = \{5, 10, 15, 20, 25, \dots\}$
$C = \{10, 20, 30, 40, \dots\}$
First,find the intersection of $A$ and $B$:
$A \cap B$ represents the set of numbers that are multiples of both $2$ and $5$. Since the least common multiple of $2$ and $5$ is $10$,$A \cap B = \{10, 20, 30, \dots\}$.
Comparing this with set $C$,we see that $A \cap B = C$.
Now,find $(A \cap B) \cap C$:
$(A \cap B) \cap C = C \cap C = C$.
Therefore,the set $(A \cap B) \cap C$ is equal to $C$.

(C) Given sets are $A = \{2, 4, 6, 8, 10, \dots\}$,$B = \{5, 10, 15, 20, \dots\}$,and $C = \{10, 20, 30, \dots\}$.
Since $C \subset B$,the union $B \cup C$ is simply $B$.
Therefore,$B \cup C = \{5, 10, 15, 20, 25, 30, \dots\} = B$.
Now,we find the intersection $A \cap (B \cup C) = A \cap B$.
$A \cap B$ represents the set of numbers that are multiples of both $2$ and $5$.
The least common multiple of $2$ and $5$ is $10$.
Thus,$A \cap B = \{10, 20, 30, \dots\} = C$.

(A) Step $1$: Find $A \cap U$.
Since $A = \{2, 4, 7\}$ and $U = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$,the intersection $A \cap U$ is the set of elements common to both,which is $\{2, 4, 7\}$.
Step $2$: Find $B \cup C$.
Since $B = \{3, 5, 7, 9, 11\}$ and $C = \{7, 8, 9, 10, 11\}$,the union $B \cup C$ is the set of all elements present in either $B$ or $C$,which is $\{3, 5, 7, 8, 9, 10, 11\}$.
Step $3$: Compute $(A \cap U) \cap (B \cup C)$.
We need to find the intersection of $\{2, 4, 7\}$ and $\{3, 5, 7, 8, 9, 10, 11\}$.
The only common element is $7$.
Therefore,$(A \cap U) \cap (B \cup C) = \{7\}$.

(A) Given $U = \{a, b, c, d, e, f\}$ and $A = \{a, b, c\}$.
First,find the complement of $A$,denoted as $A^{\prime}$.
$A^{\prime} = U - A = \{d, e, f\}$.
Now,find the union of $U$ and $A^{\prime}$.
$U \cup A^{\prime} = \{a, b, c, d, e, f\} \cup \{d, e, f\}$.
Since $\{d, e, f\}$ is a subset of $U$,the union is simply $U$.
$U \cup A^{\prime} = \{a, b, c, d, e, f\} = U$.

(D) First,find the union of sets $A$ and $B$:
$A \cup B = \{a, b, c\} \cup \{c, d, e, f\} = \{a, b, c, d, e, f\}$
Next,find the union of the result with set $C$:
$(A \cup B) \cup C = \{a, b, c, d, e, f\} \cup \{c, d, e\}$
Combining all unique elements from both sets,we get:
$(A \cup B) \cup C = \{a, b, c, d, e, f\}$
Since this set is equal to the universal set $U$,the final answer is $U$.

(A) First,find the intersection of sets $A$ and $B$:
$A \cap B = \{a, b, c\} \cap \{c, d, e, f\} = \{c\}$
Next,find the intersection of sets $A$ and $C$:
$A \cap C = \{a, b, c\} \cap \{c, d, e\} = \{c\}$
Finally,find the union of the two resulting sets:
$(A \cap B) \cup (A \cap C) = \{c\} \cup \{c\} = \{c\}$

(C) Two sets are disjoint if their intersection is an empty set,i.e.,they have no common elements.
$(i)$ The set ${x : x in mathbb{N}, 4 leq x leq 6} = {4, 5, 6}$. The intersection of ${1, 2, 3, 4}$ and ${4, 5, 6}$ is ${4}$. Since they have a common element,they are not disjoint.
$(ii)$ The sets ${a, e, i, o, u}$ and ${c, d, e, f}$ have the element $e$ in common. Thus,they are not disjoint.
$(iii)$ The set of even integers and the set of odd integers have no common elements because an integer cannot be both even and odd simultaneously. Therefore,their intersection is $emptyset$. Thus,they are disjoint sets.

(D) The set $A - B$ consists of elements that belong to $A$ but do not belong to $B$.
Given $A = \{3, 5, 7, 11\}$ and $B = \{7, 8, 9, 10, 11\}$.
Removing elements of $B$ from $A$,we get $A - B = \{3, 5\}$.
The complement $(A - B)'$ is defined as $U - (A - B)$.
$U = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$.
$(A - B)' = \{2, 3, 4, 5, 6, 7, 8, 9, 10, 11\} - \{3, 5\} = \{2, 4, 6, 7, 8, 9, 10, 11\}$.
Since this result is not present in the given options,the correct choice is $D$.

(A) Let $E$, $M$, and $C$ represent the sets of students who passed in English, Maths, and Commerce, respectively.
Given:
Students passed in English only $= 46$
Students passed in Maths only $= 46$
Students passed in Commerce only $= 58$
Students passed in English and Maths (including all three) $= 16$. So, English and Maths only $= 16 - 7 = 9$.
Students passed in Maths and Commerce (including all three) $= 24$. So, Maths and Commerce only $= 24 - 7 = 17$.
Students passed in English and Commerce (including all three) $= 26$. So, English and Commerce only $= 26 - 7 = 19$.
Students passed in all three subjects $= 7$.
Total students who passed in at least one subject $= (\text{Only } E) + (\text{Only } M) + (\text{Only } C) + (\text{Only } E \cap M) + (\text{Only } M \cap C) + (\text{Only } E \cap C) + (E \cap M \cap C)$
$= 46 + 46 + 58 + 9 + 17 + 19 + 7 = 202$.
Since the total number of students is $100$, and the calculated sum of students who passed is $202$, which exceeds the total class size, there is an inconsistency in the provided problem data. However, based on the provided Venn diagram image, the values are: Only $E=11$, Only $M=13$, Only $C=15$, $E \cap M \text{ only}=9$, $M \cap C \text{ only}=17$, $E \cap C \text{ only}=19$, and All three $=7$.
Total passed $= 11 + 13 + 15 + 9 + 17 + 19 + 7 = 91$.
Students who failed in all subjects $= 100 - 91 = 9$.

(A) We are given that $n(X \cup Y) = 18$,$n(X) = 8$,and $n(Y) = 15$.
Using the set theory formula for the union of two sets:
$n(X \cup Y) = n(X) + n(Y) - n(X \cap Y)$
Rearranging the formula to solve for $n(X \cap Y)$:
$n(X \cap Y) = n(X) + n(Y) - n(X \cup Y)$
Substituting the given values:
$n(X \cap Y) = 8 + 15 - 18$
$n(X \cap Y) = 23 - 18$
$n(X \cap Y) = 5$
Therefore,the number of elements in $X \cap Y$ is $5$.

(B) Given that $n(A) = 40$,$n(A \cup B) = 60$,and $n(A \cap B) = 10$.
We use the formula for the number of elements in the union of two sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Substituting the given values into the formula:
$60 = 40 + n(B) - 10$
$60 = 30 + n(B)$
$n(B) = 60 - 30$
$n(B) = 30$
Therefore,set $B$ has $30$ elements.

(C) Given:
$n(S) = 21$
$n(T) = 32$
$n(S \cap T) = 11$
We use the formula for the number of elements in the union of two sets:
$n(S \cup T) = n(S) + n(T) - n(S \cap T)$
Substituting the given values:
$n(S \cup T) = 21 + 32 - 11$
$n(S \cup T) = 53 - 11$
$n(S \cup T) = 42$
Therefore,the set $S \cup T$ has $42$ elements.

(A) Let $H$ be the set of people who speak Hindi and $E$ be the set of people who speak English.
Given: $n(H \cup E) = 1000$,$n(H) = 750$,$n(E) = 400$.
Using the formula: $n(H \cup E) = n(H) + n(E) - n(H \cap E)$.
Substituting the values: $1000 = 750 + 400 - n(H \cap E)$.
$1000 = 1150 - n(H \cap E)$.
$n(H \cap E) = 1150 - 1000 = 150$.
Number of people who can speak Hindi only is given by $n(H) - n(H \cap E)$.
$= 750 - 150 = 600$.

(A) Let $M$ be the set of students who opted for Mathematics and $B$ be the set of students who opted for Biology.
Given: $n(M \cup B) = 50$,$n(M) = 35$,$n(B) = 37$.
Using the formula $n(M \cup B) = n(M) + n(B) - n(M \cap B)$:
$50 = 35 + 37 - n(M \cap B)$
$50 = 72 - n(M \cap B)$
$n(M \cap B) = 72 - 50 = 22$.
So,$22$ students have opted for both Mathematics and Biology.
Number of students who have opted for only Mathematics $= n(M) - n(M \cap B) = 35 - 22 = 13$.

(A) Let $A$ be the set of people who like coffee and $B$ be the set of people who like tea.
Given that each person likes at least one of the two drinks,the total number of people is represented by the union of the two sets: $n(A \cup B) = 70$.
We are given $n(A) = 37$ and $n(B) = 52$.
We use the formula for the union of two sets: $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Substituting the known values into the formula:
$70 = 37 + 52 - n(A \cap B)$
$70 = 89 - n(A \cap B)$
$n(A \cap B) = 89 - 70 = 19$.
Therefore,$19$ people like both coffee and tea.

(B) Let $E$ be the set of people who are egg-eaters and $M$ be the set of people who are meat-eaters.
We are given:
$n(U) = 5000$ (Total population)
$n(E) = 3200$
$n(M) = 2500$
$n(E \cap M) = 1500$
Using the formula for the union of two sets:
$n(E \cup M) = n(E) + n(M) - n(E \cap M)$
$n(E \cup M) = 3200 + 2500 - 1500$
$n(E \cup M) = 5700 - 1500 = 4200$
This represents the number of people who eat either egg or meat or both.
The number of pure vegetarians is the total population minus those who eat egg or meat:
$\text{Pure vegetarians} = n(U) - n(E \cup M)$
$= 5000 - 4200 = 800$

(C) The Cartesian product $A \times B$ is defined as the set of all ordered pairs $(a, b)$ such that $a \in A$ and $b \in B$.
Given $A = \{1, 2\}$ and $B = \{2, 3\}$.
$A \times B = \{1, 2\} \times \{2, 3\}$
$= \{(1, 2), (1, 3), (2, 2), (2, 3)\}$

(D) First,find the intersection of sets $B$ and $C$:
$(B \cap C) = \{2, 3, 5, 6, 7\} \cap \{5, 6, 7, 8, 9\} = \{5, 6, 7\}$.
Next,find the Cartesian product of set $A$ and the resulting set $(B \cap C)$:
$A \times (B \cap C) = \{a, b\} \times \{5, 6, 7\}$.
By pairing each element of $A$ with each element of the intersection set,we get:
$A \times (B \cap C) = \{(a, 5), (a, 6), (a, 7), (b, 5), (b, 6), (b, 7)\}$.

(C) First,find the union of sets $B$ and $C$:
$(B \cup C) = \{b, c, e\} \cup \{b, c, f\} = \{b, c, e, f\}$.
Now,calculate the Cartesian product $A \times (B \cup C)$:
$A \times (B \cup C) = \{a, d\} \times \{b, c, e, f\} = \{(a, b), (a, c), (a, e), (a, f), (d, b), (d, c), (d, e), (d, f)\}$.
According to the distributive property of Cartesian product over union,$A \times (B \cup C) = (A \times B) \cup (A \times C)$.
Therefore,the correct option is $(A \times B) \cup (A \times C)$.

(B) First,find the intersection of sets $B$ and $C$:
$(B \cap C) = \{b, c, e\} \cap \{b, c, f\} = \{b, c\}$
Next,find the Cartesian product of $A$ and $(B \cap C)$:
$A \times (B \cap C) = \{a, d\} \times \{b, c\} = \{(a, b), (a, c), (d, b), (d, c)\}$
According to the distributive property of Cartesian product over intersection,$A \times (B \cap C) = (A \times B) \cap (A \times C)$.
Thus,the correct option is $(A \times B) \cap (A \times C)$.

(A) Given sets are $A = \{a, d\}, B = \{b, c, e\}$ and $C = \{b, c, f\}$.
First,calculate the set difference $(B - C)$:
$(B - C) = \{b, c, e\} - \{b, c, f\} = \{e\}$.
Now,find the Cartesian product $A \times (B - C)$:
$A \times (B - C) = \{a, d\} \times \{e\} = \{(a, e), (d, e)\}$.
Next,calculate $(A \times B) - (A \times C)$:
$A \times B = \{(a, b), (a, c), (a, e), (d, b), (d, c), (d, e)\}$.
$A \times C = \{(a, b), (a, c), (a, f), (d, b), (d, c), (d, f)\}$.
$(A \times B) - (A \times C) = \{(a, e), (d, e)\}$.
Since both results are equal,$A \times (B - C) = (A \times B) - (A \times C)$.

(B) Given sets are $A=\{1,2,3\}, B=\{2,3,4\}, C=\{1,3,4\},$ and $D=\{2,4,5\}.$
First,we find the Cartesian products:
$(A \times B) = \{(1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,2), (3,3), (3,4)\}$
$(C \times D) = \{(1,2), (1,4), (1,5), (3,2), (3,4), (3,5), (4,2), (4,4), (4,5)\}$
Now,find the intersection of these two sets:
$(A \times B) \cap (C \times D) = \{(1,2), (1,4), (3,2), (3,4)\}$
Next,evaluate the expression in option $B$:
$(A \cap C) = \{1,3\} \cap \{1,3,4\} = \{1,3\}$
$(B \cap D) = \{2,3,4\} \cap \{2,4,5\} = \{2,4\}$
$(A \cap C) \times (B \cap D) = \{1,3\} \times \{2,4\} = \{(1,2), (1,4), (3,2), (3,4)\}$
Since both results are identical,the correct option is $(A \cap C) \times (B \cap D).$

(C) Let $A \cap B$ be the set of common elements between $A$ and $B$. Given that $A$ and $B$ have $n$ elements in common,the number of elements in $A \cap B$ is $n$,i.e.,$|A \cap B| = n$.
An ordered pair $(x, y)$ belongs to $(A \times B) \cap (B \times A)$ if and only if $(x, y) \in A \times B$ and $(x, y) \in B \times A$.
This implies $x \in A$ and $y \in B$,and also $x \in B$ and $y \in A$.
Therefore,$x \in A \cap B$ and $y \in A \cap B$.
Since there are $n$ choices for $x$ from $A \cap B$ and $n$ choices for $y$ from $A \cap B$,the total number of such ordered pairs $(x, y)$ is $n \times n = n^{2}$.
Thus,$A \times B$ and $B \times A$ have $n^{2}$ elements in common.

(C) The formula for the number of elements in the union of two sets is given by:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Given values:
$n(A) = 40$
$n(B) = 26$
$n(A \cap B) = 16$
Substituting these values into the formula:
$n(A \cup B) = 40 + 26 - 16$
$n(A \cup B) = 66 - 16$
$n(A \cup B) = 50$

(C) Let the total number of students be $100$.
Let $H$ be the set of students who failed in Hindi and $E$ be the set of students who failed in English.
Given: $n(H) = 30$,$n(E) = 45$,and $n(H \cap E) = 20$.
We need to find the number of students who failed in at least one subject,which is given by the union $n(H \cup E)$.
Using the formula: $n(H \cup E) = n(H) + n(E) - n(H \cap E)$.
$n(H \cup E) = 30 + 45 - 20 = 55$.
So,$55 \%$ of students failed in at least one subject.
The percentage of students who passed in both subjects is the complement of those who failed in at least one subject.
Percentage passed $= 100 \% - 55 \% = 45 \%$.

(D) Let $H$ be the set of students who failed in Hindi and $E$ be the set of students who failed in English.
Given: $n(H) = 40 \%$,$n(E) = 50 \%$,and $n(H \cap E) = 21 \%$.
The percentage of students who failed in Hindi is $40 \%$.
Therefore,the percentage of students who passed in Hindi is $100 \% - n(H) = 100 \% - 40 \% = 60 \%$.

(C) Let $H$ be the set of people who speak Hindi and $E$ be the set of people who speak English.
Given:
Total number of people $n(H \cup E) = 50$
Number of people who speak Hindi $n(H) = 35$
Number of people who speak both Hindi and English $n(H \cap E) = 25$
First, find the number of people who speak Hindi only:
$n(H \text{ only}) = n(H) - n(H \cap E) = 35 - 25 = 10$
Since all people speak either Hindi or English or both, the total number of people is given by:
$n(H \cup E) = n(H \text{ only}) + n(E \text{ only}) + n(H \cap E)$
$50 = 10 + n(E \text{ only}) + 25$
$50 = 35 + n(E \text{ only})$
$n(E \text{ only}) = 50 - 35 = 15$
Therefore, the number of people who speak English only is $15$.

(B) Let the total number of workers be $x$.
Let $C$ be the set of workers who prefer cold drink and $T$ be the set of workers who prefer tea.
Given: $n(C) = 72 \% \text{ of } x$,$n(T) = 44 \% \text{ of } x$.
Since every worker prefers at least one drink,$n(C \cup T) = 100 \% \text{ of } x$.
We know that $n(C \cup T) = n(C) + n(T) - n(C \cap T)$.
Substituting the values: $100 \% = 72 \% + 44 \% - n(C \cap T) \%$.
$100 \% = 116 \% - n(C \cap T) \%$.
Therefore,$n(C \cap T) = 116 \% - 100 \% = 16 \%$.
It is given that $40$ workers like both,so $16 \% \text{ of } x = 40$.
$x = \frac{40 \times 100}{16} = 250$.
Thus,the total number of workers is $250$.

(C) Let $A$ be the set of people who watch news on $T.V.$ and $B$ be the set of people who read a newspaper.
Given: $n(A) = 65 \%$,$n(B) = 40 \%$,and $n(A \cap B) = 25 \%$.
The percentage of people who watch news on $T.V.$ or read a newspaper is given by the union formula:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$n(A \cup B) = 65 \% + 40 \% - 25 \% = 80 \%$.
The percentage of people who neither watch news on $T.V.$ nor read a newspaper is:
$100 \% - n(A \cup B) = 100 \% - 80 \% = 20 \%$.

(B) Total number of families $= 80$.
Number of families owning a car $= 20 \% \text{ of } 80 = \frac{20}{100} \times 80 = 16$.
Remaining families $= 80 - 16 = 64$.
Number of families owning a motorcycle $= 50 \% \text{ of the remaining families} = \frac{50}{100} \times 64 = 32$.
Total number of families owning a vehicle $= 16 + 32 = 48$.
Number of families not owning any vehicle $= 80 - 48 = 32$.