Venn Diagram and Operation on Sets Questions in English

(B) Step $1$: Find the intersection of sets $B$ and $C$.
$B \cap C = \{3, 4\} \cap \{4, 5, 6\} = \{4\}$.
Step $2$: Find the union of set $A$ with the result from Step $1$.
$A \cup (B \cap C) = \{1, 2, 3\} \cup \{4\} = \{1, 2, 3, 4\}$.

(C) To find the intersection $A \cap B$,we need to find the points $(x, y)$ that satisfy both equations:
$y = \frac{1}{x}$ and $y = -x$.
Substituting the second equation into the first,we get:
$-x = \frac{1}{x}$
$-x^2 = 1$
$x^2 = -1$
Since $x$ must be a real number $(x \in R)$,there is no real value of $x$ that satisfies $x^2 = -1$.
Therefore,there are no common points between sets $A$ and $B$.
Hence,$A \cap B = \phi$.

(C) The expression $(A - B) \cup (B - A)$ represents the symmetric difference of sets $A$ and $B$,denoted as $A \Delta B$.
By definition,$A - B = A \cap B^c$ and $B - A = B \cap A^c$.
Thus,$(A - B) \cup (B - A) = (A \cap B^c) \cup (B \cap A^c)$.
Using the distributive law,this simplifies to $(A \cup B) - (A \cap B)$,which contains all elements that are in either $A$ or $B$ but not in both.
Therefore,the correct option is $(C)$.

(A) Given the expression $R \times (P^c \cup Q^c)^c$.
Using De Morgan's Law,we know that $(P^c \cup Q^c)^c = (P^c)^c \cap (Q^c)^c = P \cap Q$.
Substituting this back into the expression,we get $R \times (P \cap Q)$.
Using the distributive property of the Cartesian product over intersection,$R \times (P \cap Q) = (R \times P) \cap (R \times Q)$.
Thus,the correct option is $A$.

(A) First,find the intersection $A \cap B$:
$A \cap B = \{2, 3, 4, 8, 10\} \cap \{3, 4, 5, 10, 12\} = \{3, 4, 10\}$.
Next,find the intersection $A \cap C$:
$A \cap C = \{2, 3, 4, 8, 10\} \cap \{4, 5, 6, 12, 14\} = \{4\}$.
Finally,find the union of these two sets:
$(A \cap B) \cup (A \cap C) = \{3, 4, 10\} \cup \{4\} = \{3, 4, 10\}$.

(A) By the distributive law of sets,$A \cap (A \cup B) = (A \cap A) \cup (A \cap B)$.
Since $A \cap A = A$,the expression becomes $A \cup (A \cap B)$.
Since $(A \cap B) \subseteq A$,the union of $A$ and a subset of $A$ is simply $A$.
Therefore,$A \cap (A \cup B) = A$.

(A) First,find the union of sets $B$ and $C$:
$B \cup C = \{b, c, d\} \cup \{a, b, d, e\} = \{a, b, c, d, e\}$.
Now,find the intersection of set $A$ with the resulting set:
$A \cap (B \cup C) = \{a, b, c\} \cap \{a, b, c, d, e\} = \{a, b, c\}$.
Therefore,the correct option is $A$.

(A) By the definition of set difference,$B - A$ is the set of elements that are in $B$ but not in $A$.
Therefore,any element $x \in (B - A)$ satisfies $x \notin A$.
Since $A \cap (B - A)$ consists of elements that are in both $A$ and $(B - A)$,and no element can be in both $A$ and $(B - A)$ simultaneously,the intersection is the empty set.
Thus,$A \cap (B - A) = \phi$.

(D) The shaded region represents the elements that are in set $A$ but are not in set $B$ and not in set $C$.
This means the region belongs to $A$ and excludes the union of sets $B$ and $C$.
Therefore,the shaded region is represented by $A - (B \cup C)$.

(A) From the Venn diagram,the set $A$ is composed of elements in $(A - B)$ and $(A \cap B)$.
Similarly,the set $B$ is composed of elements in $(B - A)$ and $(A \cap B)$.
The union of these disjoint regions is $(A - B) \cup (B - A) \cup (A \cap B) = A \cup B$.

(A) Using De Morgan's Law,$(A \cup B)' = A' \cap B'$.
So,the expression becomes $(A' \cap B') \cup (A' \cap B)$.
By the distributive law,this is equivalent to $A' \cap (B' \cup B)$.
Since $B' \cup B = U$ (the universal set),we have $A' \cap U = A'$.
Thus,$(A \cup B)' \cup (A' \cap B) = A'$.

(C) We use the formula for the union of two sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Substituting the given values:
$0.25 = 0.16 + 0.14 - n(A \cap B)$
Simplifying the equation:
$0.25 = 0.30 - n(A \cap B)$
Solving for $n(A \cap B)$:
$n(A \cap B) = 0.30 - 0.25 = 0.05$

(C) Since $A$ and $B$ are disjoint,$A \cap B = \phi$.
Therefore,$n(A \cap B) = 0$.
Using the formula for the union of two sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Substituting the value of $n(A \cap B)$:
$n(A \cup B) = n(A) + n(B) - 0 = n(A) + n(B)$.

(B) For any two sets $A$ and $B$,the number of elements in their union is given by the inclusion-exclusion principle:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Since $A$ and $B$ are not disjoint,$n(A \cap B) \neq 0$,which confirms the formula.

(A) The difference of two sets $A$ and $B$,denoted by $A - B$,is defined as the set of elements which belong to $A$ but do not belong to $B$.
Mathematically,$A - B = \{x : x \in A \text{ and } x \notin B\}$.
Since $x \notin B$ is equivalent to $x \in B^c$,we have $A - B = \{x : x \in A \text{ and } x \in B^c\}$.
Therefore,$A - B = A \cap B^c$.

(A) Using the set difference property,$A - (B \cap C) = A \cap (B \cap C)^c$.
By De Morgan's Law,$(B \cap C)^c = B^c \cup C^c$.
So,$A \cap (B^c \cup C^c) = (A \cap B^c) \cup (A \cap C^c)$.
Since $A \cap B^c = A - B$ and $A \cap C^c = A - C$,we get $(A - B) \cup (A - C)$.

(B) According to the Distributive Law of sets,the intersection of a set with the union of two other sets is given by:
$A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
Therefore,the correct option is $B$.

(A) First,find the union of sets $A$ and $B$:
$A \cup B = \{1, 2, 3, 4, 5\} \cup \{2, 4, 6\} = \{1, 2, 3, 4, 5, 6\}$.
Next,find the intersection of $(A \cup B)$ with set $C$:
$(A \cup B) \cap C = \{1, 2, 3, 4, 5, 6\} \cap \{3, 4, 6\} = \{3, 4, 6\}$.
Thus,the correct option is $A$.

(B) Given sets are $A = \{4, 8, 12, 16, 20, 24, \dots\}$ and $B = \{6, 12, 18, 24, 30, \dots\}$.
The intersection $A \cap B$ contains elements that are common to both sets $A$ and $B$.
Since $A$ contains multiples of $4$ and $B$ contains multiples of $6$,the common elements are multiples of the least common multiple $(LCM)$ of $4$ and $6$.
$LCM(4, 6) = 12$.
Thus,$A \cap B = \{12, 24, 36, \dots\}$,which represents all multiples of $12$.

(C) Let $M$,$P$,and $C$ represent the sets of students opting for Mathematics,Physics,and Chemistry,respectively.
We are given:
$n(M) = 100$,$n(P) = 70$,$n(C) = 40$
$n(M \cap P) = 30$,$n(M \cap C) = 28$,$n(P \cap C) = 23$
$n(M \cap P \cap C) = 18$
The number of students who have opted for Mathematics alone is given by the formula:
$n(M \text{ only}) = n(M) - n(M \cap P) - n(M \cap C) + n(M \cap P \cap C)$
Substituting the values:
$n(M \text{ only}) = 100 - 30 - 28 + 18$
$n(M \text{ only}) = 100 - 58 + 18$
$n(M \text{ only}) = 42 + 18 = 60$
Thus,$60$ students have opted for Mathematics alone.

(D) For $(1)$: $A - B$ is the set of elements in $A$ but not in $B$. $A \cap B$ is the set of elements common to $A$ and $B$. Thus,$A - (A \cap B)$ also represents elements in $A$ that are not in $B$. So,$(1)$ is correct.
For $(2)$: The set $A$ can be partitioned into two disjoint sets: elements in $A$ that are also in $B$ $(A \cap B)$ and elements in $A$ that are not in $B$ $(A - B)$. Their union is $A$. So,$(2)$ is correct.
For $(3)$: By De Morgan's Law,$A - (B \cup C) = A \cap (B \cup C)^c = A \cap (B^c \cap C^c) = (A \cap B^c) \cap (A \cap C^c) = (A - B) \cap (A - C)$. The given relation uses $\cup$ instead of $\cap$. So,$(3)$ is incorrect.
Therefore,$(1)$ and $(2)$ are correct.

(A) By De Morgan's Law,we know that $\bar{E}_1 \cap \bar{E}_2 = \overline{E_1 \cup E_2}$.
Let $A = E_1 \cup E_2$. Then the expression becomes $P(A \cap \bar{A})$.
Since $A \cap \bar{A} = \phi$ (the empty set),we have $P(A \cap \bar{A}) = P(\phi) = 0$.
Since $0 < \frac{1}{4}$,the correct option is $A$.

(A) The statement "All students are hard working" implies that every element of the set $S$ is also an element of the set $H$.
This is mathematically represented as $S \subseteq H$.
In terms of Venn diagrams,this means the circle representing set $S$ must be entirely contained within the circle representing set $H$ inside the universal set $U$.
This corresponds to the diagram in option $A$.

(A) The statement "No child is naughty" implies that there is no intersection between the set of children $(C)$ and the set of naughty persons $(N)$.
Mathematically,this is represented as $C \cap N = \phi$.
This means that the sets $C$ and $N$ are disjoint,which is correctly represented by the Venn diagram where the two circles do not overlap,as shown in option $A$.

(A) The statement "No policeman is a thief" implies that the set of policemen $(P)$ and the set of thieves $(T)$ have no elements in common.
This is represented by disjoint sets in a Venn diagram,where $P \cap T = \phi$.
Therefore,the correct representation is the Venn diagram where the circles for $P$ and $T$ do not overlap.

(C) The statement "Some teenagers are not dreamers" implies that there exists a set of teenagers $(T)$ who are not part of the set of dreamers $(D)$.
In set theory,this corresponds to the region $T - D$ (or $T \cap D^c$),which represents elements that are in $T$ but not in $D$.
Looking at the options:
- Option $A$ shows disjoint sets.
- Option $B$ shows the intersection $T \cap D$.
- Option $C$ shows the region $T$ excluding the intersection $T \cap D$,which is exactly $T - D$.
Therefore,the correct Venn diagram is represented by option $C$.

(C) The statement "All mothers are women" means that every element of the set $M$ (mothers) is also an element of the set $W$ (women).
This is represented by the subset relation $M \subseteq W$.
In a Venn diagram,this means the circle representing set $M$ must be entirely contained within the circle representing set $W$.
This corresponds to the diagram shown in option $C$.

(B) Given $X = \{ 4^n - 3n - 1 : n \in N \}$.
For $n=1, 4^1 - 3(1) - 1 = 0$.
For $n=2, 4^2 - 3(2) - 1 = 16 - 6 - 1 = 9$.
For $n=3, 4^3 - 3(3) - 1 = 64 - 9 - 1 = 54$.
So,$X = \{ 0, 9, 54, \dots \}$.
Given $Y = \{ 9(n - 1) : n \in N \}$.
For $n=1, 9(1-1) = 0$.
For $n=2, 9(2-1) = 9$.
For $n=3, 9(3-1) = 18$.
So,$Y = \{ 0, 9, 18, 27, \dots \}$.
Since every element of $X$ is a multiple of $9$ and $X \subset Y$,it follows that $X \cup Y = Y$.

(A) The set $A$ contains all $x$ such that $f(x) = 0$.
The set $B$ contains all $x$ such that $g(x) = 0$.
The intersection $A \cap B$ contains all $x$ such that $f(x) = 0$ $AND$ $g(x) = 0$.
If $f(x) = 0$ and $g(x) = 0$,then the sum of their squares must also be zero,i.e.,${[f(x)]^2} + {[g(x)]^2} = 0$.
Thus,$A \cap B = \{x : {[f(x)]^2} + {[g(x)]^2} = 0\}$.

(B) First,find the intersection of sets $B$ and $C$:
$B \cap C = \{3, 4\} \cap \{4, 5, 6\} = \{4\}$.
Next,find the union of set $A$ with the result:
$A \cup (B \cap C) = \{1, 2, 3\} \cup \{4\} = \{1, 2, 3, 4\}$.
Thus,the correct option is $B$.

(C) Since $A \subseteq B$,all elements of $A$ are contained within $B$.
Therefore,the union of $A$ and $B$ is simply the set $B$,i.e.,$A \cup B = B$.

(C) To find the intersection $A \cap B$,we solve the system of equations:
$y = \frac{1}{x}$ and $y = -x$
Substituting $y = -x$ into the first equation,we get:
$-x = \frac{1}{x}$
$-x^2 = 1$
$x^2 = -1$
Since $x$ is a real number $(x \in R)$,$x^2$ cannot be negative.
Therefore,there is no real value of $x$ that satisfies both equations.
Hence,$A \cap B = \phi$.

(C) The set $A$ represents the graph of the function $f(x) = e^x$.
The set $B$ represents the graph of the function $g(x) = x$.
For any real number $x$,it is a known property that $e^x > x$ for all $x \in R$.
Since the curves $y = e^x$ and $y = x$ never intersect,there are no common points $(x, y)$ between set $A$ and set $B$.
Therefore,$A \cap B = \phi$.

(C) The expression $(A - B) \cup (B - A)$ represents the symmetric difference of sets $A$ and $B$,denoted as $A \Delta B$.
By definition,$A - B$ contains elements in $A$ but not in $B$,and $B - A$ contains elements in $B$ but not in $A$.
The union of these two disjoint sets is the set of all elements that are in either $A$ or $B$,excluding the elements that are common to both $A$ and $B$.
Therefore,$(A - B) \cup (B - A) = (A \cup B) - (A \cap B)$.

(A) The intersection of two sets $A$ and $B$,denoted by $A \cap B$,is the set of all elements that are common to both $A$ and $B$.
Given $A = \{1, 2, 3, 4\}$ and $B = \{2, 3, 4, 5, 6\}$.
The common elements in both sets are $2, 3,$ and $4$.
Therefore,$A \cap B = \{2, 3, 4\}$.

(A) Given sets are $A = \{a, b, c\}$,$B = \{b, c, d\}$,and $C = \{a, b, d, e\}$.
First,find the union of sets $B$ and $C$:
$B \cup C = \{b, c, d\} \cup \{a, b, d, e\} = \{a, b, c, d, e\}$.
Now,find the intersection of set $A$ with the resulting set $(B \cup C)$:
$A \cap (B \cup C) = \{a, b, c\} \cap \{a, b, c, d, e\}$.
The common elements in both sets are $a, b,$ and $c$.
Therefore,$A \cap (B \cup C) = \{a, b, c\}$.

(A) Given sets are $A = \{2, 3, 4, 8, 10\}$,$B = \{3, 4, 5, 10, 12\}$,and $C = \{4, 5, 6, 12, 14\}$.
First,find $(A \cup B) = \{2, 3, 4, 5, 8, 10, 12\}$.
Next,find $(A \cup C) = \{2, 3, 4, 5, 6, 8, 10, 12, 14\}$.
Now,find the intersection $(A \cup B) \cap (A \cup C) = \{2, 3, 4, 5, 8, 10, 12\} \cap \{2, 3, 4, 5, 6, 8, 10, 12, 14\}$.
The common elements are $\{2, 3, 4, 5, 8, 10, 12\}$.
Alternatively,using the distributive law,$(A \cup B) \cap (A \cup C) = A \cup (B \cap C)$.
$B \cap C = \{4, 5, 12\}$.
$A \cup (B \cap C) = \{2, 3, 4, 8, 10\} \cup \{4, 5, 12\} = \{2, 3, 4, 5, 8, 10, 12\}$.

(C) Given that ${N_a} = \{an : n \in N\}$.
This represents the set of all multiples of $a$.
Therefore,${N_6} = \{6, 12, 18, 24, 30, 36, 42, 48, \dots\}$ and ${N_8} = \{8, 16, 24, 32, 40, 48, \dots\}$.
The intersection ${N_6} \cap {N_8}$ consists of all numbers that are multiples of both $6$ and $8$.
The common multiples of $6$ and $8$ are the multiples of their least common multiple,$LCM(6, 8)$.
$LCM(6, 8) = 24$.
Thus,${N_6} \cap {N_8} = \{24, 48, 72, \dots\} = {N_{24}}$.

(D) The shaded region represents the part of set $A$ that does not overlap with set $B$ or set $C$.
This region consists of elements that are in $A$ but not in $B$ and not in $C$.
Mathematically,this is expressed as $A - (B \cup C)$.

(A) The set $A$ contains all values of $x$ such that $f(x) = 0$.
The set $B$ contains all values of $x$ such that $g(x) = 0$.
The intersection $A \cap B$ contains all values of $x$ such that both $f(x) = 0$ and $g(x) = 0$ are satisfied simultaneously.
For any real numbers $a$ and $b$,$a^2 + b^2 = 0$ if and only if $a = 0$ and $b = 0$.
Therefore,$[f(x)]^2 + [g(x)]^2 = 0$ is equivalent to $f(x) = 0$ and $g(x) = 0$.
Thus,$A \cap B = \{x : [f(x)]^2 + [g(x)]^2 = 0\}$.

(C) We use the formula for the union of two sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Given:
$n(A) = 0.16$
$n(B) = 0.14$
$n(A \cup B) = 0.25$
Substituting the values into the formula:
$0.25 = 0.16 + 0.14 - n(A \cap B)$
$0.25 = 0.30 - n(A \cap B)$
$n(A \cap B) = 0.30 - 0.25$
$n(A \cap B) = 0.05$

(C) Two sets $A$ and $B$ are said to be disjoint if they have no common elements,i.e.,$A \cap B = \emptyset$.
For any two finite sets $A$ and $B$,the formula for the number of elements in their union is given by:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Since $A$ and $B$ are disjoint,$n(A \cap B) = 0$.
Therefore,$n(A \cup B) = n(A) + n(B) - 0 = n(A) + n(B)$.

(D) The set difference $A - B$ is defined as the set of all elements that are in $A$ but not in $B$.
Mathematically,this is represented as $A - B = \{x : x \in A \text{ and } x \notin B\}$.
Since $x \notin B$ is equivalent to $x \in B^c$,where $B^c$ is the complement of $B$ with respect to $X$,we can write:
$A - B = \{x : x \in A \text{ and } x \in B^c\}$.
This is equivalent to the intersection of $A$ and $B^c$,which is $A \cap B^c$.

(A) The difference of two sets $A$ and $B$,denoted by $A - B$,is defined as the set of all elements that are in $A$ but not in $B$.
Mathematically,$A - B = \{x : x \in A \text{ and } x \notin B\}$.
Since $x \notin B$ is equivalent to $x \in B^c$ (where $B^c$ is the complement of $B$ with respect to the universal set $U$),
we can write $A - B = \{x : x \in A \text{ and } x \in B^c\}$.
Therefore,$A - B = A \cap B^c$.

(A) By using De Morgan's Law for sets,we know that the complement of an intersection is the union of the complements: $(B \cap C)^c = B^c \cup C^c$.
Given the expression $A - (B \cap C)$,we can rewrite this as $A \cap (B \cap C)^c$.
Substituting the De Morgan's Law result,we get $A \cap (B^c \cup C^c)$.
Using the distributive property of sets,this becomes $(A \cap B^c) \cup (A \cap C^c)$.
Since $A \cap B^c = A - B$ and $A \cap C^c = A - C$,the expression simplifies to $(A - B) \cup (A - C)$.

(A) Given sets are $A = \{1, 2, 4\}, B = \{2, 4, 5\}, C = \{2, 5\}$.
First,find the set difference $(A - B)$:
$A - B = \{x : x \in A \text{ and } x \notin B\} = \{1\}$.
Next,find the set difference $(B - C)$:
$B - C = \{x : x \in B \text{ and } x \notin C\} = \{4\}$.
Finally,find the Cartesian product $(A - B) \times (B - C)$:
$(A - B) \times (B - C) = \{1\} \times \{4\} = \{(1, 4)\}$.

(A) Given sets are $A = \{1, 2, 3\}$ and $B = \{3, 8\}$.
First,find the union of sets $A$ and $B$: $A \cup B = \{1, 2, 3, 8\}$.
Next,find the intersection of sets $A$ and $B$: $A \cap B = \{3\}$.
Now,calculate the Cartesian product $(A \cup B) \times (A \cap B) = \{1, 2, 3, 8\} \times \{3\}$.
This results in the set of ordered pairs: $\{(1, 3), (2, 3), (3, 3), (8, 3)\}$.

(D) The shaded region represents the set of all elements that are in set $C$ but are not in set $A$ and not in set $B$.
This is equivalent to removing the intersection of $C$ with $A$ and the intersection of $C$ with $B$ from set $C$.
Mathematically,this is expressed as $C - (A \cup B)$.

(C) The shaded region consists of elements that are in $A$ but not in $B$,elements in $C$ but not in $B$,and the intersection $A \cap B \cap C$.
This can be expressed as $(A \setminus B) \cup (C \setminus B) \cup (A \cap B \cap C)$.
Using set identities,$(A \setminus B) = A \cap B^C$ and $(C \setminus B) = C \cap B^C$.
Thus,the region is $(A \cap B^C) \cup (C \cap B^C) \cup (A \cap B \cap C)$.
Simplifying this,we get $(A \cup C) \cap B^C \cup (A \cap B \cap C)$.
Evaluating the given options,option $C$ represents the correct set operation for the shaded region.