JEE Practice Questions (Single Choice Type)

Hint

(d) Since every rectangle ( \(F_2\) ), rhombus ( \(F_3\) ), and square ( \(F_4\) ) is a type of parallelogram, we can say:
\(
F_2 \subseteq F_1, \quad F_3 \subseteq F_1, \quad F_4 \subseteq F_1
\)
However, not every trapezium is a parallelogram, hence:
\(
F_5 \nsubseteq F_1
\)
The union of the sets \(F_2, F_3\), and \(F_4\) can be expressed as:
\(
F_1=F_1 \cup F_2 \cup F_3 \cup F_4
\)
This indicates that all rectangles, rhombuses, and squares are included within the set of parallelograms.
From the analysis, we conclude that the set of all parallelograms ( \(F_1\) ) can indeed be represented as the union of itself and the other sets of quadrilaterals that are also parallelograms.
Thus, the correct answer is option (d): \(F_1=F_1 \cup F_2 \cup F_3 \cup F_4\)

Hint

(c)

(c) Since \(A \subseteq B, A \cup B=B\)
So, \(n(A \cup B)=n(B)=6\)

Explanation:
Since \(\boldsymbol{A} \subseteq \boldsymbol{B}\), every element in \(\boldsymbol{A}\) is also in \(\boldsymbol{B}\).
Because \(\boldsymbol{A}\) is a subset of \(\boldsymbol{B}\), the union of \(\boldsymbol{A}\) and \(\boldsymbol{B}\) is simply \(\boldsymbol{B}\). \(A \cup B=B\).
The number of elements in \(A \cup B\) is equal to the number of elements in \(B\).
\(
n(A \cup B)=n(B)
\)
Substitute the given value: \(n(A \cup B)=6\).

Hint

(c) Set \(\boldsymbol{A}\) is defined by the equation \(\boldsymbol{y}=\frac{1}{\boldsymbol{x}}\), where \(\boldsymbol{x} \neq \mathbf{0}\).
Set \(\boldsymbol{B}\) is defined by the equation \(\boldsymbol{y}=-\boldsymbol{x}\).
The intersection \(\boldsymbol{A} \cap \boldsymbol{B}\) consists of the points that are in both sets.
\(
\begin{aligned}
&\text { For } A \cap B \text {, we have }\\
&\begin{array}{ll}
& -x=\frac{1}{x} \\
\Rightarrow & -x^2=1 \\
\Rightarrow & x^2=-1, \text { which is not possible } \\
\therefore & A \cap B=\phi
\end{array}
\end{aligned}
\)

Note: To find the intersection \(A \cap B\), we need to determine if there are any points that satisfy both conditions:
From Set A: \(y=\frac{1}{x}\)
From Set B: \(y=-x\)
Setting these equal to each other: \(\frac{1}{x}=-x\)
\(x^2=-1\)
Since there are no real solutions to this equation, we conclude that:
\(A \cap B=\emptyset\)

Hint

(b) According to the question,
\(
\begin{aligned}
& 2^m-2^n=112 \\
\Rightarrow \quad & 2^n\left(2^{m-n}-1\right)=2^4 \times 7
\end{aligned}
\)
Comparing exponents on both sides, we get
\(
2^n=2^4 \text { and } 2^{m-n}-1=7
\)
Now, \(2^n=2^4\)
\(
\Rightarrow \quad n=4
\)
And \(2^{m-n}=8\)
\(
\begin{array}{ll}
\Rightarrow & 2^{m-n}=2^3 \\
\Rightarrow & m-n=3 \\
\Rightarrow & m-4=3 \\
\therefore & m=7
\end{array}
\)
\(
\text { The value of } m n \text { is }=28
\)

Hint

(d) Understanding the Given Information:
We have two non-empty subsets A and B of a set X.
It is stated that A is not a subset of B. This means there exists at least one element in A that is not in B.
Finding Elements in A and B :
Let’s assume some elements for A and B to illustrate the concept.
For example, let \(A=\{2,5,7\}\) and \(B=\{1,5,6\}\).
Here, we can see that \(A\) contains the elements 2 and 7 which are not in \(B\).
Checking Subset Relationships:
Since \(A\) is not a subset of \(B\), we confirm that there are elements in \(A\) that do not belong to B.
We also check if \(B\) is a subset of \(A\). In our example, \(B\) contains the element 1 which is not in \(A\), so \(B\) is also not a subset of \(A\).
Exploring the Complement of B :
The complement of B , denoted as \(B^{\prime}\), consists of all elements in X that are not in \(B\).If \(X=\{1,2,3,4,5,6,7\}\), then \(B^{\prime}=\{2,3,4,7\}\).
Analyzing the Relationship Between \(\mathbf{A}\) and \(\mathbf{B}^{\prime}\) :
We need to check if \(A\) is a subset of \(B^{\prime}\) or if \(A\) and \(B^{\prime}\) are disjoint.
In our example, \(A\) has elements 2 and 7 which are also in \(B^{\prime}\). Therefore, \(A\) is not a subset of \(B^{\prime}\) since it contains the element 5 which is not in \(B^{\prime}\).
Conclusion:
Since \(A\) is not a subset of \(B\) and we have found elements that are common between \(A\) and \(B^{\prime}\), we conclude that \(A\) and \(B^{\prime}\) are non-disjoint.
Thus, we can summarize that \(A\) and \(B\) have some common elements, and \(A\) is not a subset of \(B\).
The correct conclusion is that A and \(B^{\prime}\) are non-disjoint.

Hint

(c)

\(
\begin{aligned}
&4 N=\{4,8,12,16,20,24, \ldots\} \text { and } 6 N=\{6,12,18,24, \ldots\}\\
&\text { Hence, } 4 N \cap 6 N=\{12,24,36, \ldots\}=12 N
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
\left(A \cap B^{\prime}\right)^{\prime} \cup(B \cap C) & =\left(\left(A^{\prime} \cup\left(B^{\prime}\right)^{\prime}\right) \cup(B \cap C)\right. \\
& =\left(A^{\prime} \cup B\right) \cup(C \cap B) \\
& =A^{\prime} \cup(B \cup(C \cap B)) \\
& =A^{\prime} \cup B
\end{aligned}
\)

Hint

(a)
\(
(A \cup B)^{\prime} \cup\left(A^{\prime} \cap B\right)
\)
\(
\begin{aligned}
& =\left(A^{\prime} \cap B^{\prime}\right) \cup\left(A^{\prime} \cap B\right) \\
& =\left(A^{\prime} \cup A^{\prime}\right) \cap\left(A^{\prime} \cup B\right) \cap\left(B^{\prime} \cup A^{\prime}\right) \cap\left(B^{\prime} \cup B\right) \\
& =A^{\prime} \cap\left\{A^{\prime} \cup\left(B \cap B^{\prime}\right)\right\} \cap U \\
& =A^{\prime} \cap\left\{A^{\prime} \cup \phi\right\} \cap U \\
& =A^{\prime} \cap A^{\prime} \cap U \\
& =A^{\prime} \cap U=A^{\prime}
\end{aligned}
\)

Hint

(d) \(\text { Shaded region in the figure is clearly } A-(B \cup C) \text {. }\)

Hint

(c)

\(
\begin{aligned}
{\left[\left(A^{\prime}\right.\right.} & \left.\left.\cap B^{\prime} \cap C\right)\right] \cup[(B \cap C) \cup(A \cap C)] \\
& \left.=\left[(A \cup B)^{\prime} \cap C\right)\right] \cup[(B \cup A) \cap C] \\
& =\left[(A \cup B)^{\prime} \cup(B \cup A)\right] \cap C \\
& =X \cap C \\
& =C
\end{aligned}
\)

Hint

(b) Families with at least one: 1003 (total) – 63 (neither) = 940 families have at least a radio or a TV.
Families with both: 794 (radio) + 187 (TV) – 940 (at least one) = 41 families have both.

Hint

(b)

\(
\begin{aligned}
&\begin{aligned}
& n(A)=40 \% \text { of } 10,000=4,000 \\
& n(B)=20 \% \text { of } 10,000=2,000 \\
& n(C)=10 \% \text { of } 10,000=1,000 \\
& n(A \cap B)=5 \% \text { of } 10,000=500 \\
& n(B \cap C)=3 \% \text { of } 10,000=300 \\
& n(C \cap A)=4 \% \text { of } 10,000=400 \\
& n(A \cap B \cap C)=2 \% \text { of } 10,000=200
\end{aligned}\\
&\text { Number of families which buy newspaper } A \text { only }\\
&\begin{aligned}
& =n\left(A \cap B^c \cap C^c\right) \\
& =n\left[A \cap(B \cup C)^c\right] \\
& =n(A)-n[A \cap(B \cup C)] \\
& =n(A)-n[(A \cap B) \cup(A \cap C)] \\
& =n(A)-[n(A \cap B)+n(A \cap C)-n(A \cap B \cap C)] \\
& =4000-[500+400-200] \\
& =4000-700 \\
& =3300
\end{aligned}
\end{aligned}
\)

Hint

(d) Let \(H\) be the set of people who speak Hind, and \(E\) be the set of people who speak English.
\(
\therefore \quad n(H \cup E)=500, n(H)=350, n(E)=300
\)
We have to find \(n(H \cap E)\).
Now, \(n(H \cup E)=n(H)+n(E)-n(H \cap E)\)
\(
\begin{array}{ll}
\therefore & 500=350+300-n(H \cap E) \\
\Rightarrow & 500=650-n(H \cap E) \\
\Rightarrow & n(H \cap E)=650-500 \\
\therefore & n(H \cap E)=150
\end{array}
\)
Thus, 150 people can speak both Hindi and English.
\(
\begin{aligned}
\text { Number of people who speak Hindi only } & =n(H)-n(H \cap E) \\
& =350-150=200
\end{aligned}
\)
\(
\begin{aligned}
\text { Number of people who speak English only } & =n(E)-n(H \cap E) \\
& =300-150=150
\end{aligned}
\)

Hint

(a)

Let \(F, E\) and \(S\) be the sets of students who learn French, English and Sanskrit, respectively.
Then.
\(
\begin{aligned}
& n(U)=50, n(F)=17, n(E)=13, n(S)=15, \\
& n(F \cap E)=9, n(E \cap S)=4, n(F \cap S)=5, \\
& n(F \cap E \cap S)=3 .
\end{aligned}
\)
From the figure, we have
\(
\begin{aligned}
& a=n(E \cap F \cap S)=3 \text { and } a+d=n(F \cap S)=5 \Rightarrow d=2 \\
& a+b=n(F \cap E)=9 \Rightarrow b=6 \\
& a+c=n(E \cap S)=4 \Rightarrow c=1 \\
& a+b+d+e=n(F)=17 \Rightarrow 3+6+2+e=17 \Rightarrow e=6 \\
& a+b+c+g=n(E)=13 \Rightarrow 3+6+1+g=13 \Rightarrow g=3 \\
& a+c+d+f=n(S)=15 \Rightarrow 3+1+2+f=15 \Rightarrow f=9
\end{aligned}
\)
(i) Number of students learning French only \(=e=6\)
(ii) Number of students learning English only \(=g=3\)
(iii) Number of students learning Sanskrit only \(=f=9\)
(iv) Number of students learning English and Sanskrit but not French \(=c=1\)
(v) Number of students learning French and Sanskrit but not English \(=d=2\)
(vi) Number of students learning French and English but not Sanskrit \(=b=6\)
(vii) Number of students learning at least one of the three languages \(=a+b+c+d+e+f+g=30\)
(viii) Number of students learning none of the three languages \(=50-30=20\)
(ix) Number of students learning exactly one language \(=e+f+g=6+3+9=18\)
(x) Number of students learning exactly two languages \(=b+c+d=6+2+1=9\)

Hint

(c)
\(
\begin{array}{rlr}
B & =B \cap(A \cup B) & \\
& =B \cap(A \cup C) & (\because A \cup B=A \cup C) \\
& =(B \cap A) \cup(B \cap C) & \\
& =(A \cap B) \cup(B \cap C) & \\
& =(A \cap C) \cup(B \cap C) & (\because A \cap B=A \cap C) \\
& =(A \cup B) \cap C & \\
& =(A \cup C) \cap C & \\
& =C &
\end{array}
\)

Hint

(d) Let \(U\) be the set of all students who took part in the survey, \(T\) be the set of students who like tea, and \(C\) be the set of students who like coffee.
According to the question,
\(
n(U)=800, n(T)=250, n(C)=300, n(T \cap C)=150
\)
To find the number of students taking neither tea nor coffee i.e., we have to find \(n\left(T^{\prime} \cap C^{\prime}\right)\)
\(
\begin{aligned}
n\left(T^{\prime} \cap C^{\prime}\right) & =n\left((T \cup C)^{\prime}\right) \\
& =n(U)-n(T \cup C) \\
& =n(U)-[n(T)+n(C)-n(T \cap C)] \\
& =800-[250+300-150]=400
\end{aligned}
\)
Hence, 400 students like neither tea nor coffee.

Alternate Method:

We can use Venn diagram as shown in the figure.
In the above figure regions are
\(r \rightarrow\) students liking tea and coffee both \(=n(T \cap C)\)
\(p \rightarrow\) students liking tea only \(=n(T)-n(T \cap C)\)
\(q \rightarrow\) students liking coffee only \(=n(C)-n(T \cap C)\)
\(s \rightarrow\) students liking neither tea nor coffee \(=n\left(T^{\prime} \cap C^{\prime}\right)\)
Here, \(r=150\) (given)
Now, \(p+r=\) students liking tea \(=250\) (given)
\(
\Rightarrow \quad p=250-150=100
\)
\(q+r=\) students liking coffee \(=300\) (given)
\(
\begin{aligned}
\Rightarrow \quad q & =300-150=150 \\
p+q+r & =\text { students liking at least one of tea and coffee } \\
& =100+150+150=400
\end{aligned}
\)
\(s=\) students liking neither tea nor coffee \(=800-400=400\)

Hint

(c) Let \(M\) be the set of students who passed in Mathematics, \(E\) be the set of students who passed in English and \(S\) be the set of students who passed in Science.

\(
\begin{aligned}
&\text { Given }\\
&\begin{aligned}
& n(U)=100 \\
& n(E)=15, n(M)=12, n(S)=8 \\
& n(E \cap M)=6, n(M \cap S)=7 \\
& n(E \cap S)=4, \text { and } n(E \cap M \cap S)=4
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { From the Venn diagram, we have }\\
&\begin{array}{ll}
a=n(E \cap M \cap S)=4 \text { and } & a+d=n(M \cap S)=7 \Rightarrow d=3 \\
a+b=n(M \cap E)=6 & \Rightarrow b=2 \\
a+c=n(S \cap E)=4 & \Rightarrow c=0 \\
a+b+d+e=n(M)=12 &
\end{array}
\end{aligned}
\)
\(
\begin{array}{lll}
\Rightarrow & 4+2+3+\mathrm{e}=12 & \Rightarrow e=3 \\
& a+b+c+g=15 & \\
\Rightarrow & 4+2+0+g=n(E)=15 & \Rightarrow g=9 \\
& a+c+d+f=n(S)=8 & \\
\Rightarrow & 4+0+3+f=8 & \Rightarrow f=1
\end{array}
\)
Number of students passed in
(a) English and Mathematics but not in Science \(=b=2\)
(b) Mathematics and Science but not in English \(=d=3\)
(c) Mathematics only \(=e=3\)
(d) more than one subject \(=a+b+c+d=4+2+0+3=9\)

Hint

(b)
(i) The factors of 21 are \(1,3,7\) and 21. So, 21 is an element of the set.
Hence, the given statement is true.
(ii) The factors of 64 are \(1,2,4,8,16,32\) and 64.
\(\therefore\) Sum of factors \(=1+2+4+8+16+32+64\) \(=127 \neq 2 \times 64\)
So, 64 is not an element of the set.
Hence, the given statement is false.
(iii) We have, \(x^4-3 x^3+4 x^2-5 x+6=0\) For \(x=2,(2)^4-3(2)^3+4(2)^2-5(2)+6=0\)
\(\Rightarrow 16-24+16-10+6=0\) \(\Rightarrow 4=0\); which is not true.
So, 2 is not an element of the set.
Hence, the given statement is true.
(iv) We have number, 23562
Sum of the digits of the number is \(2+3+5+6+2=18\); which is divisible by 9.
So, 23562 is an element of the set.
Hence, the given statement is false.

Hint

(d)

(i)
\(
\begin{aligned}
A= & \{x \mid x \text { is prime factor of } 6\}=\{2,3\} \\
B= & \left\{x \mid x \text { is solution of } x^2-5 x+6=0\right\} \\
& x^2-5 x+6=0 \\
\text { or } & (x-2)(x-3)=0 \\
\text { or } & x=2,3 \\
\therefore & B=\{2,3\}
\end{aligned}
\)
Therefore, \(A=B\).
(ii)
\(
\begin{aligned}
A & =\{x \mid x \text { is a letter in the word REPLACED }\} \\
& =\{\mathrm{R}, \mathrm{E}, \mathrm{P}, \mathrm{~L}, \mathrm{~A}, \mathrm{C}, \mathrm{D}\} \\
B & =\{y \mid y \text { is a letter in the word PARCELED }\} \\
& =\{\mathrm{P}, \mathrm{~A}, \mathrm{R}, \mathrm{C}, \mathrm{E}, \mathrm{~L}, \mathrm{D}\}
\end{aligned}
\)
Therefore, \(A=B\).
(iii)
\(
\begin{aligned}
& A=\{x \mid x \text { is natural number, } x>1\}=\{2,3,4, \ldots\} \\
& B=\{x \mid x \text { is natural number, } x \geq 1\}=\{1,2,3, \ldots\}
\end{aligned}
\)
We observe that \(1 \in B\) but \(1 \notin A\). Therefore, \(A \neq B\). 

Hint

(b) Let \(A=\{1,2\}, B=\{0,6,8\}\), and \(C=\{0,1,2,6,9\}\)
Here, \(A \not \subset B\) and \(B \not \subset C\), but \(A \subset C\). So, the given statement is false.

Hint

(b) Let \(A=\{3,5,7\}\) and \(B=\{3,4,6\}\)
Now, \(5 \in A\) and \(A \not \subset B\)
However, \(5 \notin B\)
So, the given statement is false.

Hint

(a) We have, \(\quad n(A \cup B)=n(A)+n(B)-n(A \cap B)\), \(n(A \cup B)\) is minimum or maximum according as \(n(A \cap B)\) is maximum or minimum, respectively.
Case I: If \(n(A \cap B)\) is minimum i.e., \(n(A \cap B)=0\) such that
\(
\begin{aligned}
A & =\{a, b, c, d, e, f\} \text { and } B=\{g, h, i\} \\
\therefore \quad n(A \cup B) & =n(A)+n(B)=6+3=9
\end{aligned}
\)\(
Case II: If [latex]n(A \cap B)\) is maximum i.e., \(n(A \cap B)=3\), such that
\(
\begin{aligned}
A & =\{a, b, c, d, e, f\} \text { and } B=\{d, a, c\} \\
\therefore n(A \cup B) & =n(A)+n(B)-n(A \cap B)=6+3-3=6
\end{aligned}
\)

Hint

(c) Let \(S, P\) and \(C\) denote the sets of rivers polluted by sulphur compounds, by phosphates and by crude oil respectively, and let \(a, b, c, d, e, f, g\) denote the elements (impurities) contained in the bounded region as shown in the diagram.

\(
\begin{aligned}
&\text { Then, }\\
&\begin{aligned}
& a+d+e+g=520 \\
& c+d+f+g=425 \\
& b+e+f+g=335 \Rightarrow d+g=100
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
e+g & =180 \Rightarrow f+g=150 \\
g & =28
\end{aligned}
\)
After solving, we get
\(
g=28, f=122, e=152, b=33, d=72, c=203 \text { and } a=268
\)
The number of rivers were polluted by atleast one of the three impurities
\(
=(a+b+c+d+e+f+g)=878
\)
and the number of rivers were polluted by exactly one of the three impurities,
\(
a+b+c=268+33+203=504
\)

Hint

(b) Let \(H\) and \(B\) be the set of those people who can speak Hindi and Bengali respectively, then according to the problem, we have
\(
\begin{aligned}
n(H \cup B) & =1000, \\
n(H) & =750, n(B)=400
\end{aligned}
\)
We know that,
\(
\begin{aligned}
n(H \cup B) & =n(H)+n(B)-n(H \cap B) \\
1000 & =750+400-n(H \cap B) \\
\therefore \quad n(H \cap B) & =150
\end{aligned}
\)
\(\therefore\) Number of people speaking Hindi and Bengali both is 150.
Also, \(\quad n\left(H \cap B^{\prime}\right)=n(H)-n(H \cap B)\)
\(
\begin{aligned}
& =750-150 \\
& =600
\end{aligned}
\)
Thus, number of people speaking Hindi only is 600.
Again, \(n\left(B \cap H^{\prime}\right)=n(B)-n(B \cap H)=400-150=250\)
Thus, number of people speaking Bengali only is 250.

Hint

(c) Let \(F, H\) and \(B\) be the sets of television watchers who watch Football, Hockey and Basketball, respectively.
Then, according to the problem, we have
\(
\begin{aligned}
& \qquad \begin{aligned}
n(U) & =500, n(F)=285, n(H)=195, \\
n(B) & =115, n(F \cap B)=45, \\
n(F \cap H) & =70, n(H \cap B)=50 \\
\text { and } \quad n\left(F^{\prime}\right. & \left.\cup H^{\prime} \cup B^{\prime}\right)=50,
\end{aligned}
\end{aligned}
\)
where \(U\) is the set of all the television watchers.
Since, \(n\left(F^{\prime} \cup H^{\prime} \cup B^{\prime}\right)=n(U)-n(F \cup H \cup B)\)
\(
\begin{array}{lrl}
\Rightarrow & & 50=500-n(F \cup H \cup B) \\
\Rightarrow & & n(F \cup H \cup B)=450
\end{array}
\)
\(
\begin{aligned}
&\text { We know that, }\\
&\begin{aligned}
& n(F \cup H \cup B)=n(F)+n(H)+n(B)-n(F \cap H) \\
&-n(H \cap B)-n(B \cap F)+n(F \cap H \cap B)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
450= & 285+195+115-70-50-45+n(F \cap H \cap B) \\
& n(F \cap H \cap B)=20
\end{aligned}
\)
which is the number of those who watch all the three games. Also, number of persons who watch football only
\(
\begin{aligned}
& =n\left(F \cap H^{\prime} \cap B^{\prime}\right) \\
& =n(F)-n(F \cap H)-n(F \cap B)+n(F \cap H \cap B) \\
& =285-70-45+20=190
\end{aligned}
\)
The number of persons who watch hockey only
\(
\begin{aligned}
& =n\left(H \cap F^{\prime} \cap B^{\prime}\right) \\
& =n(H)-n(H \cap F)-n(H \cap B)+n(H \cap F \cap B) \\
& =195-70-50+20=95
\end{aligned}
\)
and the number of persons who watch basketball only
\(
\begin{aligned}
& =n\left(B \cap H^{\prime} \cap F^{\prime}\right) \\
& =n(B)-n(B \cap H)-n(B \cap F)+n(H \cap F \cap B) \\
& =115-50-45+20=40
\end{aligned}
\)
Hence, required number of those who watch exactly one of the three games
\(
=190+95+40=325
\)

Hint

(a) Let \(P, C\) and \(M\) denotes the sets of students studying Physics, Chemistry and Mathematics, respectively.
Let \(a, b, c, d, e, f, g\) denote the elements (students) contained in the bounded region as shown in the diagram.

Then,
\(
\begin{aligned}
a+d+e+g & =170 \\
c+d+f+g & =100 \\
b+e+f+g & =46 \\
d+g & =30 \\
e+g & =23 \\
f+g & =28 \\
g & =18
\end{aligned}
\)
After solving, we get \(g=18, f=10, e=5, d=12, a=35\), \(b=13\) and \(c=60\)
\(
\therefore \quad a+b+c+d+e+f+g=153
\)
So, the number of students who have not offered any of these three subjects \(=175-153=22\)
Number of students studying Mathematics only, \(c=60\)
Number of students studying Physics only, \(a=35\)
Number of students studying Chemistry only, \(b=13\)

Hint

(b) Given, \(X=\left\{4^n-3 n-1: n \in N\right\}\), where \(N\) is the set of natural numbers so \(N=1,2,3, \ldots\).
After putting all these values of N, we will get the elements of X.
\(
\Rightarrow X=\{0,9,54, \ldots .\}
\)
Similarly, \(Y=\{9(n-1): n \in N\}\), where \(N\) is the set of natural numbers,
\(
\Rightarrow \mathrm{Y}=\{0,9,18,27, \ldots .\}
\)
Here, all elements of X is contained in set Y.
Since \(X\) is a subset of \(Y\), the union of the two sets is simply:
\(
X \cup Y=Y
\)

Hint

(d) The intersection of \(N_5\) and \(N_7\), denoted as \(N_5 \cap N_7\), is equal to \(N_{35}\).
Explanation:
Definition of \(N_a\) :
The set \(N_a\) is defined as the set of all multiples of the number ‘ a ‘, where ‘ a ‘ is a positive integer. So, \(N_5=\{5 n \mid n \in N\}=\{5,10,15,20, \ldots\}\) and
\(
N_7=\{7 n \mid n \in N\}=\{7,14,21,28, \ldots\}
\)
Intersection: The intersection of two sets contains only the elements that are common to both sets. Therefore, \(N_5 \cap N_7\) will contain all numbers that are multiples of both 5 and 7.
Common Multiples: The numbers that are multiples of both 5 and 7 are multiples of the least common multiple (LCM) of 5 and 7 , which is 35.
Result: Therefore, \(N_5 \cap N_7\) is the set of all multiples of 35 , which is
\(
N_{35}=\{35 n \mid n \in N\}=\{35,70,105, \ldots\}
\)

Hint

(c) \(\phi\).
We need to find \(A \cap(A \cup B)^{\prime}\).
Using De Morgan’s laws, \((A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\).
Therefore, \(A \cap\) \((A \cup B)^{\prime}=A \cap\left(A^{\prime} \cap B^{\prime}\right)=\left(A \cap A^{\prime}\right) \cap B^{\prime}=\phi \cap B^{\prime}=\phi\).

Hint

(b)

\(A-B \rightarrow\) Regions 3 and 5
\(B-C \rightarrow\) Regions 2 and 6
\(C-A \rightarrow\) Regions 4 and 7
\(\therefore(A-B) \cup(B-C) \cup(C-A) \rightarrow\) Regions 2,3,4,5,6,7
\(\therefore [(A-B) \cup(B-C) \cup(C-A)]^{\prime}\) Region 1 which is \(A \cap B \cap C\)

Hint

(a)

Hint

(c) The intersection of two sets, \(\boldsymbol{A} \cap \boldsymbol{B}\), contains elements present in both sets.
The least common multiple (LCM) of two numbers is the smallest positive integer that is a multiple of both numbers.
Find the least common multiple (LCM) of 4 and 6.
Find the prime factorization of each number.
Prime factorization of 4 is \(2^2\).
Prime factorization of 6 is \(2 \times 3\).
Calculate the LCM:
The LCM is found by taking the highest power of each prime factor present in either factorization.
\(
\operatorname{LCM}(4,6)=2^2 \times 3^1=4 \times 3=12 .
\)
Determine the elements of \(A \cap B\).
\(A \cap B\) consists of all multiples of the \(\operatorname{LCM}(4,6)\).
Therefore, \(A \cap B\) consists of all multiples of 12. 

Hint

(d) Let the original set contain \((2 n+1)\) elements, then the number of subsets containing more than \(n\) elements, that is, subsets containing ( \(n+1\) ) elements, ( \(n+2\) ) elements, \(\ldots,(2 n+1)\) elements will be required to be computed.
\(\therefore\) Required number of subsets
\(
\begin{aligned}
& ={ }^{2 n+1} C_{n+1}+{ }^{2 n+1} C_{n+2}+\cdots+{ }^{2 n+1} C_{2 n}+{ }^{2 n+1} C_{2 n+1} \\
= & { }^{2 n+1} C_n+{ }^{2 n+1} C_{n+1}+\cdots+{ }^{2 n+1} C_1+{ }^{2 n+1} C_0 \\
= & { }^{2 n+1} C_0+{ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\cdots+{ }^{2 n+1} C_{n-1}+{ }^{2 n+1} C_n \\
= & \frac{1}{2}\left[(1+1)^{2 n+1}\right]=\frac{1}{2}\left[2^{2 n+1}\right]=2^{2 n}
\end{aligned}
\)

Hint

(c) Key Concept:
If \(A=\{a, b\}\)
\(\Rightarrow P(A)=\{\phi\},\{a\},\{b\},\{a, b\}\)

In the given case, we get \(P(A)=\{\phi,\{\phi,\{\{\phi\}\}, A\}\)

Hint

(a) The number of elements in the universal set \(U\) is \(n(U)=20\).
The number of elements in set \(A\) is \(n(A)=12\).
The number of elements in set \(\boldsymbol{B}\) is \(\boldsymbol{n}(\boldsymbol{B})=9\).
The number of elements in the intersection of \(A\) and \(B\) is \(n(A \cap B)=4\).
The Principle of Inclusion-Exclusion for two sets states
\(
n(A \cup B)=n(A)+n(B)-n(A \cap B)
\)
The complement of a set \(X\) in a universal set \(U\) is given by \(n\left(X^{\prime}\right)=n(U)-n(X)\).
Use the formula: \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\).
Substitute the given values: \(n(A \cup B)=12+9-4\).
Calculate the sum: \(n(A \cup B)=21-4\).
\(
n(A \cup B)=17
\)
Use the formula for the complement: \(n\left((A \cup B)^{\prime}\right)=n(U)-n(A \cup B)\).
Substitute the values: \(n\left((A \cup B)^{\prime}\right)=20-17\).
\(
n\left((A \cup B)^{\prime}\right)=3
\)

Hint

(c) The percentage of Indians who like cheese is 63%.
The percentage of Indians who like apples is \(76 \%\).
The percentage of Indians who like both cheese and apples is \(x \%\).
The total percentage of people cannot exceed \(100 \%\).
The percentage of people liking both cannot exceed the percentage of people liking either item individually.
Let C denote the percentage of people who like cheese and and A denote the percentage of people who like apples.
Given, \(\mathrm{n}(\mathrm{C})=63, \mathrm{n}(\mathrm{A})=76\) and \(\mathrm{n}(\mathrm{C} \cap \mathrm{A})={x}\).
We know that,
\(
\begin{aligned}
& n(C \cup A)=n(C)+n(A)-n(C \cap A) \\
& \Rightarrow n(C \cup A)=63+76-x
\end{aligned}
\)
\(
\Rightarrow x=139-n(C \cup A) \dots(i)
\)
\(
\begin{aligned}
&\text { Now, } n(C \cup A) \leq 100 \text { [Since, total people surveyed is 100] }\\
&\begin{aligned}
& \Rightarrow 139-n(C \cup A) \geq 139-100=39[\text { From }(\mathrm{i})] \\
& \Rightarrow x \geq 39
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\begin{aligned}
& \text { Also, } n(C \cap A) \leq n(C) \text { and } n(C \cap A) \leq n(A) \\
& \Rightarrow x \leq 63 \\
& \therefore 39 \leq x \leq 63 \text {. }
\end{aligned}\\
&\text { Thus, the value of } x \text { lies between } 39 \text { and } 63 \text { also; the value of } x \text { can be } 39 \text { or } 63 \text {. }
\end{aligned}
\)

Hint

(d)

Here, region 1 represents \(M \cap P\) which means students studying Mathematics and Physics.
Region 2 represents \(M \cap C\) which means students studying Mathematics and Chemistry.
Region 3 represents \(P \cap C\) which means students studying Physics and Chemistry.
Region 4 represents \(M \cap P \cap C\) which means students studying Mathematics, Physics as well as Chemistry.
First of all we will find the number of students studying only mathematics, which is given as:
\(
n(M)-n(M \cap P)-n(M \cap C)+n(M \cap P \cap C) \dots(1)
\)
Here, \(n(M)=\) total number of students studying mathematics = 23
\(n(M \cap P)=\) number of students studying Mathematics and Physics = 12
\(n(M \cap C)=\) number of students studying Mathematics and Chemistry = 9
\(n(M \cap P \cap C)=\) number of students studying Mathematics, Physics as well as Chemistry = 4
We have to add \(n(M \cap P \cap C)\) once because ( \(M \cap P \cap C\) ) is contained in ( \(M \cap P\) ) as well as in \((M \cap C)\) and thus it gets subtracted twice.
On putting the respective values in equation (1), we get:
Number of students studying only Mathematics \(=23-12-9+4=6\)
Similarly, number of students studying only Chemistry is given as:
\(
\begin{aligned}
& =n(C)-n(C \cap M)-n(C \cap P)+n(M \cap P \cap C) \\
& =19-9-7+4=7
\end{aligned}
\)
And, the number of students studying only Physics is given as:
\(
\begin{aligned}
& =n(P)-n(P \cap C)-n(P \cap M)+n(M \cap P \cap C) \\
& =24-7-12+4=9
\end{aligned}
\)
So, the number of students who study exactly one subject is \(=6+7+9=22\).

Hint

(b) The first set has \(m\) elements.
The second set has \(n\) elements.
The total number of subsets of the first set is 56 more than the total number of subsets of the second set.
The number of subsets of a set with \(x\) elements is \(2^x\).
The number of subsets of the first set is \(2^m\).
The number of subsets of the second set is \(2^n\).
The given condition is \(2^m-2^n=56\).
Factor out \(2^n\) from the left side: \(2^n\left(2^{m-n}-1\right)=56=2^3 \times 7\).
\(
\text { This implies } 2^n=2^3 \text { and } 2^{m-n}-1=7 \text {. }
\)
\(
\text { From } 2^n=2^3 \text {, we get } n=3 \text {. }
\)
From \(2^{m-n}-1=7\), we get \(2^{m-n}=8\).
Since \(8=2^3\), we have \(2^{m-n}=2^3\).
This implies \(m-n=3\).
Substitute \(n=3\) : \(m-3=3\).
Therefore, \(m=6\).
The values of \(m\) and \(n\) are 6 and 3 respectively.

Hint

(a) \(a N=\{a x: x \in N\}\) represents the set of multiples of \(a\).
\(b N \cap c N=d N\) means the intersection of multiples of \(b\) and \(c\) is the set of multiples of \(d\).
\(b, c \in N\) are relatively prime.
The intersection of two sets of multiples, \(b N \cap c N\), represents the set of common multiples of \(b\) and \(c\).
If \(b\) and \(c\) are relatively prime, their least common multiple (LCM) is their product, \(b c\).
\(b N\) is the set of all multiples of \(b\). \(c N\) is the set of all multiples of \(c\). \(b N \cap c N\) is the set of common multiples of \(b\) and \(c\).
Since \(b N \cap c N=d N, d N\) represents the set of common multiples of \(b\) and \(c\).
The smallest positive common multiple is the least common multiple (LCM).
Therefore, \(d\) must be the LCM of \(b\) and \(c\).
Given that \(b\) and \(c\) are relatively prime, their LCM is their product.
\(
\operatorname{LCM}(b, c)=b c
\)
Since \(d=\operatorname{LCM}(b, c)\), and \(\operatorname{LCM}(b, c)=b c\), then \(d=b c\).

Hint

(b) 

\(
\begin{gathered}
n(A)=40 \% \text { of } 10000=4000 \\
n(B)=20 \% \text { of } 10000=2000 \\
n(C)=10 \% \text { of } 10000=1000
\end{gathered}
\)
\(
\begin{aligned}
& n(A \cap B)=5 \% \text { of } 10000=500 \\
& n(B \cap C)=3 \% \text { of } 10000=300 \\
& n(C \cap A)=4 \% \text { of } 10000=400 \\
& n(A \cap B \cap C)=2 \% \text { of } 10000=200
\end{aligned}
\)
\(
\begin{aligned}
&\text { We want to find } n\left(A \cap B^c \cap C^c\right)=n\left[A \cap(B \cup C)^c\right]\\
&\begin{aligned}
& =n(A)-n[A \cap(B \cup C)]=n(A)-n[(A \cap B) \cup(A \cap C)] \\
& =n(A)-[n(A \cap B)+n(A \cap C)-n(A \cap B \cap C)] \\
& =4000-[500+400-200]=4000-700=3300
\end{aligned}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
\begin{aligned}
A \times B =\{2,3\} \times\{4,5\} \\
 =\{(2,4),(2,5),(3,4),(3,5)\}
\end{aligned} \\
\text { and } \begin{aligned}
B \times C  =\{4,5\} \times\{5,6\} \\
=\{(4,5),(4,6),(5,5),(5,6)\}
\end{aligned} \\
\begin{aligned}
\therefore(A \times B) & \cup(B \times C)=\{(2,4),(2,5),(3,4),(3,5), \\
& (4,5),(4,6),(5,5),(56)\}
\end{aligned} \\
\text { Now, } n\{(A \times B) \cup(B \times C)\}=8
\end{aligned}
\)

Hint

(a) The union of set \(Y\) and set \(\{1,2\}\) is \(\{1,2,3,5,9\}\) This can be written as \(Y \cup\{1,2\}=\{1,2,3,5,9\}\).
The elements \(\{3,5,9\}\) are in \(\{1,2,3,5,9\}\) but not in \(\{1,2\}\). Therefore, \(Y\) must contain \(\{3,5,9\}\).
The smallest set \(Y\) contains only the necessary elements. So, the smallest set \(Y\) is \(\{3,5,9\}\).
The largest set \(Y\) can include elements from \(\{1,2\}\) without changing the union.
If \(Y\) contains \(\{1,2,3,5,9\}\), then \(Y \cup\{1,2\}=\{1,2,3,5,9\}\).
So, the largest set \(Y\) is \(\{1,2,3,5,9\}\).
The smallest set \(Y\) is \(\{3,5,9\}\) and the largest set \(Y\) is \(\{1,2,3,5,9\}\).

Hint

(b)

\(
\begin{aligned}
2 \cos ^2 \theta+\sin \theta & \leq 2 \\
2\left(1-\sin ^2 \theta\right)+\sin \theta & \leq 2 \\
2 \sin ^2 \theta-\sin \theta & \geq 0 \\
\sin \theta(2 \sin \theta-1) & \geq 0
\end{aligned}
\)
\(
\begin{aligned}
&\begin{array}{ll}
\Rightarrow & \sin \theta\left(\sin \theta-\frac{1}{2}\right) \geq 0 \\
\therefore & \sin \theta \leq 0 \text { and } \sin \theta \geq \frac{1}{2}
\end{array}\\
&\text { Now, the values of } \theta \text { which lie in the interval } \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2} \text {. }\\
&\left[\because B=\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\right\}\right]
\end{aligned}
\)
So, \(\theta\) satisfy \(\sin \theta \leq 0\) in the interval \(\pi \leq \theta \leq \frac{3 \pi}{2}\) and \(\theta\) satisfy \(\sin \theta \geq \frac{1}{2}\) in the interval \(\frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6}\).
\(
\begin{aligned}
& \therefore \quad A \cap B=\left\{\theta: \pi \leq \theta \leq \frac{3 \pi}{2}\right\} \\
& \text { and } \quad \begin{aligned}
A \cap B & =\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6}\right\} \\
\text { Hence, } A \cap B & =\left\{\theta: \frac{\pi}{2} \leq \theta \leq \frac{5 \pi}{6} \text { or } \pi \leq \theta \leq \frac{3 \pi}{2}\right\} \\
& =\left\{\theta: \theta \in\left[\frac{\pi,}{2} \frac{5 \pi}{6}\right] \cup\left[\pi, \frac{3 \pi}{2}\right]\right\}
\end{aligned}
\end{aligned}
\)

Hint

(d) Given, \(M, C\) and \(T\) are the sets of drinks; milk, coffee and tea, respectively. Let us denote the number of drinks (students) contained in the bounded region as shown in the diagram by \(a, b, c, d, e, f\) and \(g\), respectively.

Then,
\(
\begin{aligned}
g & =10 \\
g+f & =20 \Rightarrow f=10 \quad[\because g=10] \\
g+e & =30 \Rightarrow e=20 \\
d+g & =25 \Rightarrow d=15 \\
a=12, b & =5, c=8
\end{aligned}
\)
Thus, total number of students taking drinks \(M\) or \(C\) or \(T\)
\(
\begin{aligned}
& =a+b+c+d+e+f+g \\
& =12+5+8+15+20+10+10=80
\end{aligned}
\)
Hence, the number of students taking none of them drinks
\(
=100-80=20
\)

Hint

(a) Let \(C=\) Set of people who read paper \(A\)
and \(D=\) Set of people who read paper \(B\)
Given, \(\quad n(C)=25, n(D)=20, n(C \cap D)=8\)
\(
\begin{aligned}
\therefore \quad n\left(C \cap D^{\prime}\right) & =n(C)-n(C \cap D) \\
& =25-8=17
\end{aligned}
\)
But total number of people who read \(A\) but not \(B=30 \%\)
\(\therefore\) Percentage of people reading \(A\) but not \(B=30 \%\) of 17
\(
=\frac{30 \times 17}{100}=\frac{51}{10}
\)
and \(n\left(C^{\prime} \cap D\right)=n(D)-n(C \cap D)=20-8=12\)
Also, total number of people who read \(B\) but not \(A=40 \%\)
\(\therefore\) Percentage of people reading \(B\) but not \(A=40 \%\) of 12
\(
=\frac{40 \times 12}{100}=\frac{24}{5}
\)
and given total people who read \(A\) and \(B=50 \%\)
\(\therefore\) Total number of people who read \(A\) and \(B=50 \%\) of 8
\(
=\frac{50 \times 8}{100}=4
\)
\(\therefore\) Percentage of people reading an advertisement
\(
=\frac{51}{10}+\frac{24}{5}+4=13.9 \%
\)

Hint

(c)

\(
\begin{aligned}
\text { Let } \quad E & =\text { Set of people having injuries in eyes } \\
\therefore \quad n(E) & =30 \\
A & =\text { Set of people having injuries in arms } \\
\therefore n(A) & =50 \\
\text { and } \quad L & =\text { Set of people having injuries in legs } \\
\therefore \quad n(L) & =70
\end{aligned}
\)
Let us denote the number of injuries contained in the bounded region as shown in the diagram by \(a, b, c, d, e, f\) and \(g\), respectively.

Then,
\(
\begin{array}{r}
b+e+f+g=30 \dots(i) \\
a+d+e+g=50 \dots(ii) \\
c+d+f+g=70 \dots(iii) \\
d+e+f=44 \dots(iv)
\end{array}
\)
and \(a+b+c+d+e+f+g=100 \dots(v)\)
On adding Eqs. (i), (ii) and (iii), we get
\(
\begin{aligned}
a+b+c+2(d+e+f)+3 g & =150 \\
\Rightarrow 100-d-e-f-g+2(d+e+f)+3 g & =150 \text { [from Eq. (v)] }
\end{aligned}
\)
\(
\begin{aligned}
\Rightarrow & d+e+f+2 g & =50 & \\
\Rightarrow & 44+2 g & =50 & \text { [from Eq. (iv)] } \\
\therefore & g & =3 &
\end{aligned}
\)
Hence, there are three claims involved loss or injury to all the three.

Hint

(a) By Venn diagram

\(
\text { It is clear that } A \cap(A \cup B)=A
\)

Example: Let’s say \(A=\{1,2,3\}\) and \(B=\{3,4,5\}\).
\(A \cup B=\{1,2,3,4,5\}\)
\(A \cap(A \cup B)=\{1,2,3\} \cap\{1,2,3,4,5\}=\{1,2,3\}=A\)

Hint

(d) De Morgan’s Law: The complement of an intersection of two sets equals the union of their complements. Therefore, \((\mathbf{Y} \cap \mathbf{X})^{\prime}\) is equivalent to \(\mathbf{Y}^{\prime} \cup \mathbf{X}^{\prime}\). Substituting this into the original expression gives: \(\mathbf{X} \cap\left(\mathbf{Y}^{\prime} \cup \mathbf{X}^{\prime}\right)\)
Distributive Law: The distributive law for sets is similar to the distributive property in algebra. The intersection of a set with the union of two other sets is equal to the union of the intersections. Applying this to the expression: \(\left(\mathbf{X} \cap \mathbf{Y}^{\prime}\right) \cup\left(\mathbf{X} \cap \mathbf{X}^{\prime}\right)\)
Complement Law: The intersection of a set with its complement results in the empty set. \(\mathbf{X} \cap \mathbf{X}^{\prime}\) equals the empty set, denoted by \(\phi\). Substituting this into the expression: \(\left(\mathbf{X} \cap \mathbf{Y}^{\prime}\right) \cup \phi\)
Identity Law: The union of a set with the empty set is the set itself. Therefore: \(X \cap Y^{\prime}\)

Hint

(b) The number of elements in the power set is equal to the total number of subsets.
Therefore, the number of elements in the power set is \(2^n\).

Let set \(A\) contains \(n\) elements.
Power set of \(A\) is the set of all subsets.
\(\therefore\) Number of subsets of \(A={ }^n C_o+{ }^n C_1+{ }^n C_2+\ldots+{ }^n C_n=2^n\)
\(\therefore\) Power set of \(A\) contains \(2^n\) elements.

Hint

(c)

(a) \(A-B \subseteq A\)
If an element \(\boldsymbol{x}\) is in \(\boldsymbol{A}-\boldsymbol{B}\), it means \(\boldsymbol{x}\) is in \(\boldsymbol{A}\) and not in \(\boldsymbol{B}\). Therefore, \(\boldsymbol{x}\) is in \(\boldsymbol{A}\). So, \(\boldsymbol{A}-\boldsymbol{B}\) is a subset of \(\boldsymbol{A}\). This statement is true.

(b) \(B^{\prime}-A^{\prime} \subseteq A\)
Using the property \(\boldsymbol{X}-\boldsymbol{Y}=\boldsymbol{X} \cap \boldsymbol{Y}^{\prime}\), this statement can be rewritten as: \(B^{\prime}-A^{\prime}=B^{\prime} \cap\left(A^{\prime}\right)^{\prime}=B^{\prime} \cap A\).
So, the statement becomes \(\boldsymbol{B}^{\prime} \cap \boldsymbol{A} \subseteq \boldsymbol{A}\). If an element \(\boldsymbol{x}\) is in \(\boldsymbol{B}^{\prime} \cap \boldsymbol{A}\), it means \(\boldsymbol{x}\) is in \(\boldsymbol{B}^{\prime}\) and \(\boldsymbol{x}\) is in \(\boldsymbol{A}\). Therefore, \(\boldsymbol{x}\) is in \(\boldsymbol{A}\). So, \(\boldsymbol{B}^{\prime} \cap \boldsymbol{A}\) is a subset of \(\boldsymbol{A}\). This statement is true.

(c) \(A \subseteq A-B\)
For \(\boldsymbol{A}\) to be a subset of \(\boldsymbol{A}-\boldsymbol{B}\), every element in \(\boldsymbol{A}\) must also be in \(\boldsymbol{A}-\boldsymbol{B}\). This means that for every \(\boldsymbol{x} \in \boldsymbol{A}, \boldsymbol{x}\) must not be in \(\boldsymbol{B}\). This implies that the intersection of \(\boldsymbol{A}\) and \(\boldsymbol{B}\) must be empty, i.e., \(\boldsymbol{A} \cap \boldsymbol{B}=\phi\). However, this is not always true for any sets \(A\) and \(B\). For example, if \(\boldsymbol{A}=\{1,2\}\) and \(\boldsymbol{B}=\{2,3\}\), then \(\boldsymbol{A}-\boldsymbol{B}=\{1\}\). In this case, \(\boldsymbol{A}=\{1,2\}\) is not a subset of \(\boldsymbol{A}-\boldsymbol{B}=\{1\}\) because \(2 \in A\) but \(2 \notin A-B\). Therefore, this statement is not true.

(d) \(A \cap B^{\prime} \subseteq A\)
If an element \(\boldsymbol{x}\) is in \(\boldsymbol{A} \cap \boldsymbol{B}^{\prime}\), it means \(\boldsymbol{x}\) is in \(\boldsymbol{A}\) and \(\boldsymbol{x}\) is in \(\boldsymbol{B}^{\prime}\). Therefore, \(\boldsymbol{x}\) is in \(\boldsymbol{A}\). So, \(\boldsymbol{A} \cap \boldsymbol{B}^{\prime}\) is a subset of \(\boldsymbol{A}\). This statement is true.
Therefore, the statement that is not true is (c) \(\boldsymbol{A} \subseteq \boldsymbol{A}-\boldsymbol{B}\).

Hint

(b) 

\(
\begin{aligned}
A=\{1,2,3\} & \\
B & =\{3,8\} \\
A \cup B & =\{1,2,3,8\} \\
A \cap B & =\{3\} \\
(A \cup B) \times(A \cap B) & =\{1,2,3,8\} \times\{3\} \\
& =\{(1,3),(2,3),(3,3)(8,3)\}
\end{aligned}
\)

Hint

(c)
\(
\begin{aligned}
A & =\{x: x \text { is a multiple of } 3\} \\
A & =\{x: x=3 m, m \in N\} \\
B & =\{x: x \text { is a multiple of } 5\} \\
B & =\{x: x=5 n, n \in N\}
\end{aligned}
\)
The intersection of two sets, \(\boldsymbol{A} \cap \boldsymbol{B}\), contains elements common to both sets. A number that is a multiple of both 3 and 5 is a multiple of their least common multiple.
The LCM of 3 and 5 is \(3 \times 5=15\).
Elements in \(A \cap B\) are multiples of the LCM, which is 15 . So, \(A \cap B=\{15,30,45, \ldots\}\).
\(
\begin{gathered}
A \cap B=\{x: x \text { is a multiple of both } 3 \text { and } 5\} \\
=\{15,30,45, \ldots\}
\end{gathered}
\)

Hint

(b) We observe that there is no real number \(x\) such that \(x^2+1=0\)
\(
x= \pm i
\)
\(\therefore \quad\left\{x: x\right.\) is a real number and \(\left.x^2+1=0\right\}=\phi\)

Hint

(d) Understand the definitions. A set \(A\) is said to be a subset of another set \(B\) (denoted \(A \subseteq B\) ) if every element of \(A\) is also an element of \(B\). The complement of \(B\), denoted \(B^{\prime}\), consists of all elements in the universal set \(X\) that are not in \(B\)

Given that \(A\) is not a subset of \(B\) (i.e., \(A \nsubseteq B\) ), this means there exists at least one element in \(A\) that is not in \(B\). Therefore, this element must belong to the complement of \(B\)

Since there is at least one element in \(A\) that is also in \(B^{\prime}\) (the complement of \(B\) ), it follows that \(A\) and \(B^{\prime}\) share at least one common element. This implies that \(A\) and \(B^{\prime}\) are not disjoint, as disjoint sets have no elements in common.

Consequently, we can conclude that the statement ” A and the complement of B are non-disjoint” is always true when \(A\) is not a subset of \(B\)

Thus, the correct choice is that \(A\) and the complement of \(B\) are non-disjoint.

Hint

(b)

\(
\begin{aligned}
&\text { Given, }\\
&\begin{array}{r}
a=20 \dots(i) \\
e+f+g=15 \dots(ii) \\
b+d+f+g=22 \dots(iii) \\
c=11 \dots(iv)
\end{array}
\end{aligned}
\)

\(
\begin{aligned}
&\Rightarrow \quad \begin{aligned}
c+d+e+g & =25 \dots(v) \\
d+e+g & =14 \dots(vi) \\
e & =7 \dots(vii) \\
f & =3 \dots(viii)
\end{aligned}\\
&\text { From Eqs. (vii), (viii) and (ix), }\\
&\begin{aligned}
e+g & =12 \dots(ix) \\
e & =7, f=3, g=5
\end{aligned}
\end{aligned}
\)
From Eq. (vi), \(d=2\)
From Eq. (iii) \(b+2+3+5=22\)
\(
\therefore \quad b=12
\)
Hence, \(\quad a=20, b=12, c=11, d=2, e=7, f=3, g=5\)
Number of children play all the three games \(=g=5\)
Number of children play cricket and hockey but not football
\(
=d=2
\)
Number of children play hockey only \(=b=12\)
Total number of children in the group
\(
=a+b+c+d+e+f+g=60
\)

Hint

(d)

\(
\begin{aligned}
a+f+e+g & =21 \dots(i) \\
b+d+f+g & =26 \dots(ii) \\
c+d+e+g & =29 \dots(iii) \\
f+g & =14 \dots(iv)\\
g+d & =15 \dots(v) \\
e+g & =12 \dots(vi) \\
g & =8 \dots(vii)
\end{aligned}
\)
From Eqs. (vii) and (vi), \(e=4\)
Form Eqs. (vii) and (v), \(d=7\)
From Eqs. (vii) and (iv), \(f=6\)
From Eq. (iii), \(c+7+4+8=29 \Rightarrow c=29-19=10=c\)
From Eq. (ii), \(b+7+6+8=26 \Rightarrow b=26-21 \Rightarrow b=5\)
From Eq. (i), \(a+6+4+8=21 \Rightarrow a=21-18 \Rightarrow a=3\)
\(
\begin{aligned}
n(B)+n(H)+n(F) & =a+b+c+d+e+f+g \\
& =3+5+10+7+4+6+8=43
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
a+e+f+g & =120 \dots(i) \\
b+d+f+g & =90 \dots(ii) \\
e+f+c+d & =70 \dots(iii) \\
g+f & =40 \dots(iv) \\
f+d & =30 \dots(v) \\
e+f & =50 \dots(vi)
\end{aligned}
\)

\(
\begin{aligned}
& & U-(a+b+c+d+e+f+g) & =20 \\
\Rightarrow & & a+b+c+d+e+f+g & =180 \dots(vii)
\end{aligned}
\)
From Eqs. (i) and (iv), \(\quad a+e=80 \dots(viii)\)
From Eqs. (ii) and (iv), \(\quad b+d=50 \dots(ix)\)
\(
\begin{aligned}
& \text { From Eqs. (iii) and (v), } \quad e+c=40 \dots(x) \\
& \text { from Eqs (viii), (ix) \& (x), } a+b+c+d+e+e=197 \dots(xi) \\
& \text { from (xi), (vii) and (iv), } \begin{aligned}
197-e+40 & =180 \\
170-e+40 & =180 \\
e & =210-180=30
\end{aligned} \\
& \text { From Eq. (vi), } e+f=50 \\
& \Rightarrow \quad 30+f=50 \\
& \Rightarrow \quad f=20
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
b+e+f+g & =205 \dots(i) \\
a+d+f+g & =210 \dots(ii) \\
c+d+e+g & =120 \dots(iii) \\
f+g & =100 \dots(iv) \\
e+g & =800 \dots(v) \\
d+g & =35 \dots(vi) \\
g & =20 \dots(vii)
\end{aligned}
\)

From Eqs. (vi) and (vii), \(d=15\)
From Eqs. (vii) and (v), \(e=60\)
From Eqs. (vii) and (iv), \(f=80\)
From Eq. (i), \(b+60+80+20=205 \Rightarrow b=205-160\)
\(\Rightarrow b=45=\) Can speak English but not Hindi or Tamil.
From Eq. (ii) \(a+15+80+20=210\)
\(
\Rightarrow \quad a+115=210 \Rightarrow a=95
\)
From Eq. (iii),
\(
\begin{aligned}
c+15+60+20 & =120 \\
c & =120-95 \Rightarrow c=25
\end{aligned}
\)
People who can speak neither \(E\) nor \(H\) nor \(T\)
\(
\begin{aligned}
& =450-(95+45+25+15+60+80+20) \\
& =450-340=110
\end{aligned}
\)

Hint

(d)

A\(
c+f+g+e=42 \dots(i)
\)
\(
\begin{aligned}
b+d+g+e & =36 \dots(ii) \\
a+f+d+g & =30 \dots(iii) \\
g+e & =15 \dots(iv) \\
f+g & =10 \dots(v) \\
d & =4 \dots(vi) \\
e & =11 \dots(vii)
\end{aligned}
\)

From (iv) and (vii), \(g+11=15 \Rightarrow g=4 \dots(viii)\)
From (v) and (viii), \(f+4=10 \Rightarrow f=6 \dots(ix) \)
From (i), \(c+6+4+11=42 \Rightarrow c=21 \dots(x) \)
From (ii), \(b+4+4+11=36 \Rightarrow b=17 \dots(xi) \)
From (iii), \(a+6+4+4=30 \Rightarrow a=16 \dots(xii)\)
Number of required persons
\(
\begin{aligned}
& =123-(16+17+21+4+11+6+4) \\
& =123-79 \\
& =44
\end{aligned}
\)

Hint

 (a) The collection of all girls in your class.
Explanation:
A set is a well-defined collection of distinct objects. This means that we can determine with certainty whether or not an object belongs to the set. “Girl” is a clearly defined term. We can identify all the girls in a class with certainty. Therefore, (a) represents a set.

Why other options are incorrect:
(b), (c), and (d) are not sets because the terms “intelligent,” “beautiful,” and “tall” are subjective. What one person considers intelligent, beautiful, or tall may not be the same for another. There is no objective, universally accepted measure for these qualities. Therefore, these collections are not well-defined and do not form sets.

Hint

(a) The set is \(\left\{(x, y): x^2+y^2 \leq 1 \leq x+y, x, y \in \mathbb{R}\right\}\). \(x, y \in \mathbb{R}\) means \(x\) and \(y\) can be any real numbers.
The conditions define a region within the unit circle, which contains infinitely many real points.
Therefore, set (a) is infinite.
The set is \(\left\{(x, y): x^2+y^2 \leq 1 \leq x+y, x, y \in \mathbb{Z}\right\}\).
\(x, y \in \mathbb{Z}\) means \(x\) and \(y\) must be integers.
Integer pairs satisfying \(x^2+y^2 \leq 1\) are limited to \(( \pm 1,0),(0, \pm 1),(0,0)\).
Checking \(x+y \geq 1:(1,0)\) satisfies \(1 \geq 1,(0,1)\) satisfies \(1 \geq 1\).
Other pairs like \((-1,0),(0,-1),(0,0)\) do not satisfy \(x+y \geq 1\).
Thus, set (b) contains only two elements: \((1,0)\) and \((0,1)\), making it finite.
The set is \(\left\{(x, y): x^2 \leq y \leq|x|, x, y \in \mathbb{Z}\right\}\). \(x, y \in \mathbb{Z}\) means \(x\) and \(y\) must be integers.
If \(x=0\), then \(0 \leq y \leq 0\), so \(y=0\). Pair: \((0,0)\).
If \(x=1\), then \(1 \leq y \leq 1\), so \(y=1\). Pair: \((1,1)\).
If \(x=-1\), then \(1 \leq y \leq 1\), so \(y=1\). Pair: \((-1,1)\).
For \(|x|>1, x^2>|x|\), so no integer \(y\) can satisfy \(x^2 \leq y \leq|x|\).
Thus, set (c) contains only three elements: \((0,0),(1,1),(-1,1)\), making it finite.
The set is \(\left\{(x, y): x^2+y^2=1, x, y \in \mathbb{Z}\right\}\). \(x, y \in \mathbb{Z}\) means \(x\) and \(y\) must be integers.
Integer pairs satisfying \(x^2+y^2=1\) are \(( \pm 1,0)\) and \((0, \pm 1)\).
Thus, set (d) contains only four elements: \((1,0),(-1,0),(0,1),(0,-1)\), making it finite.

Hint

(d) The term “intelligent” is not a well-defined term. So, the given collection does not form a set.

Hint

(a)

\(
\begin{aligned}
& n=1.2,3 \ldots .(n \in N) \\
& X(\text { at } n=1)=8-7-1=0 \\
& X(\text { at } n=2)=64-14-1=49 \\
& X(\text { at } n=3)=512-21-1=490 \\
& X=\{0,49,490 \ldots \ldots .\}
\end{aligned}
\)
\(
\begin{aligned}
& Y(\text { at } n=1)=49-49=0 \\
& Y(\text { at } n=2)=98-49=49 \\
& Y(\text { at } n=3)=147-49=98 \\
& Y=\{0,49,98,147 \ldots . .490 \ldots\}
\end{aligned}
\)
Here we can conclude that every element of X can be found in Y but vice versa is not true.
So, \(X \subset Y\).

Explanation: We have,
\(
\begin{aligned}
8^n-7 n-1 & =(1+7)^n-7 n-1 \\
& =\left({ }^n C_0+{ }^n C_1 \times 7+{ }^n C_2 \times 7^2+\ldots+{ }^n C_n \times 7^n\right)-(7 n+1) \\
& =7^2\left({ }^n C_2+{ }^n C_3 \times 7+\ldots+{ }^n C_n \times 7^{n-2}\right)
\end{aligned}
\)
\(\therefore \quad X\) contains some multiples of 49.
Clearly, \(Y\) contains all multiples of 49 including 0.
\(
\therefore \quad X \subset Y
\)

Hint

(b) Set \(A=\{x: x=2 n+1, n \in Z\}\).
Set \(B=\{x: x=2 n, n \in Z\}\).
\(Z\) represents the set of all integers.
Set \(A\) consists of all numbers of the form \(2 n+1\) where \(n\) is an integer. These are all odd integers: \(\ldots,-3,-1,1,3, \ldots\).
Set \(\boldsymbol{B}\) consists of all numbers of the form \(2 \boldsymbol{n}\) where \(\boldsymbol{n}\) is an integer. These are all even integers: \(\ldots,-4,-2,0,2,4, \ldots\).
The union \(\boldsymbol{A} \cup \boldsymbol{B}\) includes all elements present in either \(\boldsymbol{A}\) or \(\boldsymbol{B}\).
Combining all odd and even integers results in the set of all integers.
\(
A \cup B=\{\ldots,-4,-3,-2,-1,0,1,2,3,4, \ldots\}
\)
The set of all integers is denoted by \(\boldsymbol{Z}\).
The union of set \(\boldsymbol{A}\) (odd integers) and set \(\boldsymbol{B}\) (even integers) is the set of all integers, \(\boldsymbol{Z}\).

Hint

(c) Set \(A\) is defined as \(\{x: x=4 n, n \in Z\}\).
Set \(B\) is defined as \(\{x: x=6 n, n \in Z\}\).
The intersection of two sets contains elements common to both sets.
The least common multiple (LCM) is the smallest positive integer divisible by both numbers.
Set \(A\) contains all multiples of 4.
Set \(B\) contains all multiples of 6.
The intersection \(A \cap B\) consists of elements that are multiples of both 4 and 6. This means \(x\) must be a common multiple of 4 and 6.
Find the LCM of 4 and 6.
Multiples of \(4: 4,8,12,16, \ldots\)
Multiples of 6: 6, 12, 18, 24, …
The LCM of 4 and 6 is 12.
Since \(x\) must be a multiple of 12 , the intersection is \(\{x: x=12 n, n \in Z\}\).
The intersection \(A \cap B\) is the set of all multiples of 12.

Hint

(c) It is evident from the Venn-diagram that

\(
(A-B) \cup(B-A)=(A \cup B)-(A \cap B)
\)

Hint

(a) It is evident from the Venn-diagram that
\(
\begin{array}{ll}
& (A-B) \cup(B-A)=(A \cup B)-(A \cap B) \\
\therefore \quad & ((A-B) \cup(B-A)) \cup(A \cap B) \\
& =((A \cup B)-(A \cap B)) \cup(A \cap B)=A \cup B .
\end{array}
\)

Hint

(a) We have
\(
\begin{aligned}
& A \cup X=B \cup X \text { for some set } X \\
& A \cap(A \cup X)=A \cap(B \cup X) \\
& A=(A \cap B) \cup(A \cap X) [\because A \cap(A \cup X)=A] \\
& A=(A \cap B) \cup \emptyset [\because A \cap X=\phi] \\
& A=A \cap B \dots(i)
\end{aligned}
\)
Again
\(
\begin{aligned}
& A \cup X=B \cup X \\
& B \cup(A \cup X)=B \cap(B \cup X) \\
& (B \cap A) \cup(B \cap X)=B \\
& (B \cap A) \cup \emptyset=B \\
& A \cap B=B \dots(ii)
\end{aligned}
\)
\(
\begin{aligned}
&\text { From (i) and (ii), we have }\\
&A=B
\end{aligned}
\)

Hint

(a) Set \(A\) is defined by \(|x-3|<1\) and \(|y-3|<1\).
Set \(B\) is defined by \(4 x^2+9 y^2-32 x-54 y+109 \leq 0\).
The condition \(|x-3|<1\) implies \(-1<x-3<1\).
This simplifies to \(2<x<4\).
The condition \(|y-3|<1\) implies \(-1<y-3<1\).
This simplifies to \(2<y<4\).
Set \(A\) represents the interior of a square with vertices at \((2,2),(4,2),(4,4)\), and \((2,4)\). Thus, \(A\) is the set of all points \((x, y)\) lying inside the square formed by the lines \(x=2, x=4, y=2\) and \(y=4\).
We have,
\(
\begin{aligned}
& 4 x^2+9 y^2-32 x-54 y+109 \leq 0 \\
& 4\left(x^2-8 x\right)+9\left(y^2-6 y\right)+109 \leq 0 \\
& 4(x-4)^2+9(y-3)^2 \leq 36 \\
& \frac{(x-4)^2}{3^2}+\frac{(y-3)^2}{2^2} \leq 1
\end{aligned}
\)
Thus, \(B\) is the set of all points lying inside the ellipse having its centre at \((4,3)\) and major and minor axes of lengths 3 and 2 units.
The square \(A\) has corners \((2,2),(4,2),(4,4),(2,4)\).
The ellipse \(\boldsymbol{B}\) is centered at \((4,3)\).
The ellipse extends from \(4-3=1\) to \(4+3=7\) in \(x\) and \(3-2=1\) to \(3+2=5\) in \(y\).
The square \(A\) is entirely contained within the ellipse \(B\).
For example, the point \((2,2)\) is in \(\boldsymbol{A}\).
For \(\boldsymbol{B}\) : \(\frac{(2-4)^2}{9}+\frac{(2-3)^2}{4}=\frac{4}{9}+\frac{1}{4}=\frac{16+9}{36}=\frac{25}{36} \leq 1\), so \((2,2)\) is in \(\boldsymbol{B}\).
The point \((4,4)\) is in \(A\). For \(B: \frac{(4-4)^2}{9}+\frac{(4-3)^2}{4}=0+\frac{1}{4}=\frac{1}{4} \leq 1\), so \((4,4)\) is in B.
Since all points in \(\boldsymbol{A}\) satisfy the condition for \(\boldsymbol{B}, \boldsymbol{A} \subset \boldsymbol{B}\).
Set \(A\) is a square region, and set \(B\) is an elliptical region, with \(A\) being a subset of \(B\).

Hint

(c)
\(
\text { The expression } x^2+y^2 \leq r^2 \text { represents points inside or on a circle with radius } r \text {. }
\)
Clearly, \(A\) is the set of all points lying inside or on the circle with centre at the origin and radius 1 and \(B\) is the set of all points lying inside or on the circle with centre at the origin and radius 2 units. Clearly, \(A \subset B\). Therefore, \(A-B=\phi\) and \(B-A \neq \phi\)

Hint

(b)
Let M = Set of teachers teaching Mathematics.
\(
P=\text { Set of teachers teaching Physics. }
\)
We have,
\(
\begin{array}{ll}
& n(M \cup P)=20, n(M)=12 \text { and } n(M \cap P)=4 \\
\therefore & n(M \cup P)=n(M)+n(P)-n(M \cap P) \\
\Rightarrow & 20=12+n(P)-4 \\
\Rightarrow & n(P)=12
\end{array}
\)
So, the required number \(=n(P)-n(M \cap P)\)
\(
=12-4=8
\)

Hint

(a) Let \(U\) be the set of all consumers who were questioned, \(A\) be the set of consumers who lied product \(P_1\) and \(B\) be the set of consumers who liked product \(P_2\). It is given that
\(
\begin{array}{ll}
& n(U)=2000, n(A)=1720 \text { and } n(B)=1450 \\
\therefore & n(A \cup B)=n(A)+n(B)-n(A \cap B) \\
\Rightarrow & n(A \cup B)=1720+1450-n(A \cap B) \\
\Rightarrow & n(A \cup B)=3170-n(A \cap B)
\end{array}
\)
Now,
\(
\begin{array}{ll}
& A \cup B \subset U \\
\Rightarrow & n(A \cup B) \leq n(U) \\
\Rightarrow & 3170-n(A \cap B) \leq 2000 \\
\Rightarrow & n(A \cap B) \geq 1170
\end{array}
\)
Hence, the least number of consumers who like both the products is 1170.

Hint

Let \(F, B\) and \(C\) denote the sets of students who received medals in Football, Basketball and Cricket respectively. Then, we have
\(
\begin{aligned}
& n(F)=37, n(B)=15, n(C)=20 \\
& n(F \cup B \cup C)=58 \text { and } n(F \cap B \cap C)=3
\end{aligned}
\)
Now,
\(
\begin{aligned}
& n(F \cup B \cup C)=n(F)+n(B)+n(C)-n(F \cap B)-n(B \cap C) \\
& -n(C \cap F)+n(F \cap D \cap C) \\
\Rightarrow \quad & 58=38+15+20-{n(F \cap B)+n(B \cap C)+n(C \cap F)}+3 \\
\Rightarrow \quad & n(F \cap B)+n(B \cap C)+n(C \cap F)=18
\end{aligned}
\)
Hence, the number of students who received medals in exactly two of the three sports
\(
\begin{aligned}
& =n(F \cap B)+n(B \cap C)+n(C \cap F)-3 n(F \cap B \cap C) \\
& =18-3 \times 3=9
\end{aligned}
\)

Hint

(d) Clearly, \(A\) is the set of all points on the circle \(x^2+y^2=25\) and \(B\) is the set of all points on the ellipse \(x^2+9 y^2=144\). These two intersect at four points \(P, Q, R\) and \(S\). Hence, \(A \cap B\) contains four points.

Explanation:

From \(x^2+y^2=25\), we get \(x^2=25-y^2\).
Substitute this into \(x^2+9 y^2=144\).
\(
\left(25-y^2\right)+9 y^2=144
\)
\(
\begin{aligned}
& 25+8 y^2=144 \\
& 8 y^2=144-25 \\
& 8 y^2=119 \\
& y^2=\frac{119}{8} \\
& y= \pm \sqrt{\frac{119}{8}}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Substitute } y^2=\frac{119}{8} \text { into } x^2=25-y^2 \text {. }\\
&\begin{aligned}
& x^2=25-\frac{119}{8} \\
& x^2=\frac{200-119}{8} \\
& x^2=\frac{81}{8} \\
& x= \pm \sqrt{\frac{81}{8}}
\end{aligned}
\end{aligned}
\)
Since there are two possible values for \(x\) (positive and negative) and two possible values for \(y\) (positive and negative), there are \(2 \times 2=4\) distinct intersection points.
The intersection \(\boldsymbol{A} \cap \boldsymbol{B}\) contains four points.

Hint

(a) since \(A, B, C\) are disjoint sets.
\(
\begin{array}{ll}
\therefore & n(A \cup B \cup C)=n(A)+n(B)+n(C) \\
\Rightarrow & n(A \cup B \cup C)=10+6+5=21
\end{array}
\)

Hint

(c) Let there be \(x\) families in the town. Let \(P\) and \(C\) denote the set of families using phone and car respectively. Then,
\(
n(P)=\frac{25 x}{100}=\frac{x}{4}, n(C)=\frac{15 x}{100}=\frac{3 x}{20}
\)
and, \(n(\bar{P} \cap \bar{C})=\frac{65 x}{100}=\frac{13 x}{20}\)
Also, \(n(P \cap C)=2000\).
Now,
\(
\begin{aligned}
& n(\bar{P} \cap \bar{C})=\frac{13 x}{20} \\
& n(\overline{P \cup C})=\frac{13 x}{20} \\
& n(U)-n(P \cup C)=\frac{13 x}{20} \\
& x-{n(P)+n(C)-n(P \cap C)}=\frac{13 x}{20} \\
& x-\left(\frac{x}{4}+\frac{3 x}{20}-2000\right)=\frac{13 x}{20} \\
& \frac{x}{20}=2000 \Rightarrow x=40000
\end{aligned}
\)
Thus, statement 3 is true.
\(10 \%\) of the total families in the town is 4000 and it is given that 2000 families own both a car and a phone. So, statement 1 is not true.
Now,
\(
\begin{array}{ll}
& n(P \cup C)=n(P)+n(C)-n(P \cap C) \\
\Rightarrow & n(P \cup C)=\frac{x}{4}+\frac{3 x}{20}-2000 \\
\Rightarrow & n(P \cup C)=10000+6000-2000=14000 \\
\Rightarrow & n(P \cup C)=35 \% \text { of } 40000
\end{array}
\)
Thus, statement 2 is correct.

Hint

(c)
We have,
\(
\begin{aligned}
A & =B \cap C \text { and } B=C \cap A \\
\Rightarrow \quad A & =(C \cap A) \cap C \Rightarrow A=C \cap A=B
\end{aligned}
\)

Hint

(a) We have,
\(
\begin{aligned}
& \left(A \cap B^{\prime} \cap C^{\prime}\right)^{\prime}=A^{\prime} \cup B \cup C \\
& (A \cup B \cup C) \cap\left(A \cap B^{\prime} \cap C^{\prime}\right)^{\prime} \\
& =(A \cup B \cup C) \cap\left(A^{\prime} \cup B \cup C\right)
\end{aligned}
\)
\(
=\left(A \cap A^{\prime}\right) \cup(B \cup C) \quad[\text { By distributivity of } \cup \text { over } \cap]
\)
\(
=\phi \cup(B \cup C)=B \cup C
\)
\(
\begin{aligned}
&\text { Hence, }\\
&\begin{aligned}
& (A \cup B \cup C) \cap\left(A \cap B^{\prime} \cap C^{\prime}\right)^{\prime} \cup C^{\prime} \\
& =(B \cup C) \cap C^{\prime} \\
& =\left(B \cap C^{\prime}\right) \cup\left(C \cap C^{\prime}\right) \\
& =B \cap C^{\prime}
\end{aligned}
\end{aligned}
\)

Hint

a) Given: We have given the percentage of combatants who lost at least one limb which are \(70 \%\) lost their one eye, \(80 \%\) lost their an ear, \(75 \%\) lost their an arm, \(85 \%\) lost their a leg and we have to find the percentage of combatants who lost all the four limbs.
First, we will consider that there are a total of 100 combatants.
Then, if 70% combatants lost one eye which implies that 30 do not lose any eye.
If \(80 \%\) combatants lost any ear which implies that 20 do not lose any ear.
If \(75 \%\) combatants lost any arm which implies that 25 do not lose any arm.
If \(85 \%\) combatants lost any leg which implies that 15 do not lose any leg.
Now, we will add up the number of combatants which lost at least one limb.
Now, Number of combatants who do not lost any limb \(=30+20+25+15=90\)
Now, to find the number of combatants who lost their all four limbs, we will subtract the number of combatants from the total which is 100.
Number of combatants who lost all the four limbs is equal to \(100-90=10\)
Which means the value of \(x\) is 10.
Therefore, the minimum percentage of combatants who lost all their 4 limbs is \(10 \%\).

Hint

(b)

\(
\begin{aligned}
&\text { We have, }\\
&\begin{aligned}
& (A \cup B) \cap B^{\prime}=A \\
& \therefore\left((A \cup B) \cap B^{\prime}\right) \cup A^{\prime}=A \cup A^{\prime}=N
\end{aligned}
\end{aligned}
\)

Hint

(c) Let \(A=\left\{(x, y): y=e^x, x \in R\right\}\)
i.e., A denotes the set of exponential function.
Graph of exponential function is given by and \(B=\{(x, y): y=x, x \in R\}\)
i.e. \(B\) denotes the set of identity function (equation is a straight line).
Graph of identity function is given by
Then, there is nothing common between the sets A and B.
\(
\therefore A \cap B=\phi
\)

Hint

(b) Given: Set A of all matrices of order \(3 \times 3\) with entries 0 or 1 only.
Let \(B\) be the subset of \(A\) consisting of all matrices whose determinant is 1. \(\Rightarrow B \in(0,1)\) and \(|B|=1\)
Assume \(B=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|\)
Now, \(|B|=1(1-0)-0+0\)
\(
\therefore|B|=1
\)
Let \(C\) be the subset of \(A\) consisting of all matrices whose determinant is -1.
\(
\Rightarrow C \in(0,1) \text { and }|C|=-1
\)
Assume \(C=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right|\)
Now, \(|\mathrm{C}|=1(0-1)-0+0\)
\(
\therefore|C|=-1
\)
Here, number of elements in \(B\) and \(C\) are equal.

Hint

(b) The cardinality of set \(A\) is \(n(A)=200\).
The cardinality of set \(\boldsymbol{B}\) is \(\boldsymbol{n}(\boldsymbol{B})=300\).
The cardinality of the intersection of \(A\) and \(B\) is \(n(A \cap B)=100\).
The cardinality of the intersection of the complements of \(A\) and \(B\) is \(n\left(A^{\prime} \cap B^{\prime}\right)=300\)
De Morgan’s Law states that \((A \cup B)^{\prime}=A^{\prime} \cap B^{\prime}\).
The cardinality of the complement of a set \(X\) is \(n\left(X^{\prime}\right)=n(U)-n(X)\).
Use the formula: \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\).
Substitute the given values: \(n(A \cup B)=200+300-100\).
Calculate the result: \(n(A \cup B)=400\).
Apply De Morgan’s Law: \(\boldsymbol{A}^{\prime} \cap \boldsymbol{B}^{\prime}=(A \cup \boldsymbol{B})^{\prime}\).
Use the complement rule: \(n\left((A \cup B)^{\prime}\right)=n(U)-n(A \cup B)\). So, \(n\left(A^{\prime} \cap B^{\prime}\right)=n(U)-n(A \cup B)\).
Substitute the known values into the equation from Step 2: \(300=n(U)-400\).
Rearrange the equation to find \(n(U): n(U)=300+400\).
Calculate the final value: \(n(U)=700\).
\(
\text { The cardinality of the universal set } n(U) \text { is } 700 \text {. }
\)

Hint

(a) Total number of boys: 800
Boys who played cricket: 224
Boys who played hockey: 240
Boys who played basketball: 336
Boys who played both basketball and hockey: 64
Boys who played both cricket and basketball: 80
Boys who played both cricket and hockey: 40
Boys who played all three games: 24
Calculate the number of boys who played at least one game:
Use the formula:
\(
n(C \cup H \cup B)=n(C)+n(H)+n(B)-n(C \cap H)-n(C \cap B)-n(H \cap B)+n(C \cap H \cap B)
\)
Substitute the given values:
\(n(C \cup H \cup B)=224+240+336-40-80-64+24\)
\(n(C \cup H \cup B)=800-184+24\)
\(n(C \cup H \cup B)=640\)
Calculate the number of boys who did not play any game.
Subtract the number of boys who played at least one game from the total number of boys:
Number of boys not playing any game \(=800-n(C \cup H \cup B)\)
Number of boys not playing any game \(=800-640\)
Number of boys not playing any game \(=160\)

Hint

(c) Set \(A\) is defined by the equation \(a^2+3 b^2=28\), where \(a\) and \(b\) are integers.
Set \(B\) is defined by the condition \(a>b\), where \(a\) and \(b\) are integers.
First, find all integer pairs \((a, b)\) that satisfy the equation for set \(A\), then filter these pairs to satisfy the condition for set \(\boldsymbol{B}\).
Since \(3 b^2 \leq 28\), we have \(b^2 \leq \frac{28}{3} \approx 9.33\).
Possible integer values for \(b\) are \(0, \pm 1, \pm 2, \pm 3\).
If \(b=0, a^2=28\), so \(a= \pm \sqrt{28}\), which is not an integer.
If \(b= \pm 1, a^2+3( \pm 1)^2=28 \Longrightarrow a^2=25\), so \(a= \pm 5\).
If \(b= \pm 2, a^2+3( \pm 2)^2=28 \Longrightarrow a^2=16\), so \(a= \pm 4\).
If \(b= \pm 3, a^2+3( \pm 3)^2=28 \Longrightarrow a^2=1\), so \(a= \pm 1\).
\(
A=\{(5,1),(-5,1),(5,-1),(-5,-1),(4,2),(-4,2),(4,-2),(-4,-2),(1,3),(-1,3),(1,-3),(-1,-3)\}
\)
\(
\begin{aligned}
&\text { Filter elements of } A \text { that satisfy } a>b \text { (elements of } A \cap B \text { ). }\\
&\begin{aligned}
& (5,1) \text { since } 5>1 \\
& (5,-1) \text { since } 5>-1 \\
& (4,2) \text { since } 4>2 \\
& (4,-2) \text { since } 4>-2 \\
& (1,-3) \text { since } 1>-3 \\
& (-1,-3) \text { since }-1>-3
\end{aligned}
\end{aligned}
\)
There are 6 elements in \(A \cap B\).

Hint

(c) You are determining the number of intersection points between a hyperbola and a circle.
Set \(A\) is defined by the equation \(y=\frac{4}{x}\), where \(x \neq 0\).
Set \(B\) is defined by the equation \(x^2+y^2=8\).
Substitute the equation for \(y\) from set \(A\) into the equation for set \(B\) and solve for \(x\).
Substitute \(y\) into the circle equation.
Substitute \(y=\frac{4}{x}\) into \(x^2+y^2=8\).
\(
\begin{aligned}
& x^2+\left(\frac{4}{x}\right)^2=8 \\
& x^2+\frac{16}{x^2}=8
\end{aligned}
\)
Multiply by \(x^2\) to clear the denominator.
\(
x^4+16=8 x^2
\)
Rearrange into a quadratic form in terms of \(x^2\).
\(
x^4-8 x^2+16=0
\)
Factor the quadratic.
\(
\left(x^2-4\right)^2=0
\)
Solve for \(x^2\).
\(
x^2=4
\)
Solve for \(x\).
\(
x= \pm 2
\)
For \(x=2\) :
\(
y=\frac{4}{2}=2
\)
For \(x=-2\) :
\(
y=\frac{4}{-2}=-2
\)
The intersection points are \((2,2)\) and \((-2,-2)\).
\(A \cap B\) contains 2 points only.

Hint

(b) You are determining the correct set relationship between two circles defined by their equations.
Set \(A\) represents a circle centered at the origin with radius \(\sqrt{4}=2\).
Set \(B\) represents a circle centered at the origin with radius \(\sqrt{9}=3\).
For a point ( \(x, y\) ) to be in \(A \cap B\), it must satisfy both \(x^2+y^2=4\) and \(x^2+y^2=9\). Since \(4 \neq 9\), no point can satisfy both equations simultaneously.
Therefore, \(A \cap B=\phi\).
(a) \(A-B=\phi\) : This is incorrect because \(A-B=A\) since \(A \cap B=\phi\).
(b) \(\boldsymbol{B}-\boldsymbol{A}=\boldsymbol{B}\) : This is correct because \(\boldsymbol{B}-\boldsymbol{A}=\boldsymbol{B}\) since \(\boldsymbol{A} \cap \boldsymbol{B}=\phi\).
(c) \(A \cap B \neq \phi\) : This is incorrect because \(A \cap B=\phi\).
(d) \(A \cap B=A\) : This is incorrect because \(A \cap B=\phi \neq A\).
The correct statement is \(B-A=B\).

Hint

(c) Set \(A\) is defined by the equation \(y^2=x\).
Set \(\boldsymbol{B}\) is defined by the equation \(y=|x|\).
Substitute \(y=|x|\) into \(y^2=x\).
This yields \((|x|)^2=x\).
\((|x|)^2=x\) simplifies to \(x^2=x\).
Rearrange to \(x^2-x=0\).
Factor out \(x: x(x-1)=0\).
This gives two possible values for \(x\) : \(x=0\) or \(x=1\).
For \(x=0, y=|0|=0\).
For \(x=1, y=|1|=1\).
The intersection points of sets \(A\) and \(B\) are \((0,0)\) and \((1,1)\).

Hint

(a) Set \(A=\left\{\theta: \cos \theta>-\frac{1}{2}, 0 \leq \theta \leq \pi\right\}\).
Set \(B=\left\{\theta: \sin \theta>\frac{1}{2}, \frac{\pi}{3} \leq \theta \leq \pi\right\}\).
Determine the intervals for \(\theta\) for each set and then find their intersection and union.
\(\cos \theta>-\frac{1}{2}\) in the interval \(0 \leq \theta \leq \pi\).
The value \(\cos \theta=-\frac{1}{2}\) occurs at \(\theta=\frac{2 \pi}{3}\).
Since \(\cos \theta\) is decreasing in \(0 \leq \theta \leq \pi, \cos \theta>-\frac{1}{2}\) implies \(\theta<\frac{2 \pi}{3}\).
So, \(A=\left\{\theta: 0 \leq \theta<\frac{2 \pi}{3}\right\}\).
\(\sin \theta>\frac{1}{2}\) in the interval \(\frac{\pi}{3} \leq \theta \leq \pi\).
The value \(\sin \theta=\frac{1}{2}\) occurs at \(\theta=\frac{\pi}{6}\) and \(\theta=\frac{5 \pi}{6}\).
In the given range, \(\sin \theta>\frac{1}{2}\) implies \(\frac{\pi}{6}<\theta<\frac{5 \pi}{6}\).
Considering the given domain for \(\boldsymbol{B}, \frac{\pi}{3} \leq \theta \leq \pi\), the interval for \(\boldsymbol{B}\) is \(\left\{\theta: \frac{\pi}{3} \leq \theta<\frac{5 \pi}{6}\right\}\).
\(
\begin{aligned}
&A \cap B=\left\{\theta: 0 \leq \theta<\frac{2 \pi}{3}\right\} \cap\left\{\theta: \frac{\pi}{3} \leq \theta<\frac{5 \pi}{6}\right\} .\\
&\text { The intersection is the common interval: }\left\{\theta: \frac{\pi}{3} \leq \theta<\frac{2 \pi}{3}\right\} \text {. }
\end{aligned}
\)
\(
\begin{aligned}
&A \cup B=\left\{\theta: 0 \leq \theta<\frac{2 \pi}{3}\right\} \cup\left\{\theta: \frac{\pi}{3} \leq \theta<\frac{5 \pi}{6}\right\}\\
&\text { The union is the combined interval: }\left\{\theta: 0 \leq \theta<\frac{5 \pi}{6}\right\} \text {. }
\end{aligned}
\)
\(
\text { The correct option is (a) } A \cap B=\{\theta: \pi / 3 \leq \theta \leq 2 \pi / 3\} \text {. }
\)

Hint

(c) The rule method, or set-builder notation, is used to describe a property that no element can satisfy when representing the null set.
\(\{x: x \neq x\}\) This reads as “the set of all x such that x is not equal to x “. Since no element can be not equal to itself, this set will have no elements, thus representing the null set. This is the correct representation using the rule method.

Hint

(c) 

\(
\begin{aligned}
& n\left(A^{\prime} \cap B^{\prime}\right)=n(A \cup B)^{\prime}=n(U)-n(A \cup B) \\
& =n(U)-[n(A)+n(B)-n(A \cap B)] \\
& =700-[200+300-100]=300
\end{aligned}
\)

Hint

(c) Let \(M\) and \(E\) denote the sets of students who have taken Mathematics and Economics respectively. Then, we have
\(
n(M \cup E)=35, n(M)=17 \text { and } n\left(M \cap E^{\prime}\right)=10
\)
Now,
\(
\begin{aligned}
& n\left(M \cap E^{\prime}\right)=n(M)-n(M \cap E) \\
& \Rightarrow 10=17-n(M \cap E) \Rightarrow n(M \cap E)=7
\end{aligned}
\)
Now,
\(
\begin{aligned}
& n(M \cup E)=n(M)+n(E)-n(M \cap E) \\
& \Rightarrow 35=17+n(E)-7 \Rightarrow n(E)=25 \\
& \therefore n\left(E \cap M^{\prime}\right)=n(E)-n(E \cap M)=25-7=18
\end{aligned}
\)

Hint

(c) Let \(c\) be the event of traveling by car.
Let \(B\) be the event of traveling by bus.
\(
\begin{aligned}
& P(C)=20 \% \\
& P(B)=50 \% \\
& P(C \cap B)=10 \%
\end{aligned}
\)
The percentage of people traveling by car or bus is \(P(C \cup B)\).
\(
P(C \cup B)=P(C)+P(B)-P(C \cap B)
\)
\(
\begin{aligned}
& P(C \cup B)=20 \%+50 \%-10 \% \\
& P(C \cup B)=70 \%-10 \% \\
& P(C \cup B)=60 \%
\end{aligned}
\)

Hint

(b) It is given that; the total number of students in the class is 45.
The number of students who can speak Hindi is 22.
The number of students who can speak English is 12.
We have to find the number of students who can speak both Hindi and English.
So, as per the given information
\(\)
\begin{aligned}
& n(H \cup E)=45 \\
& n(H)=22 \\
& n(E)=12
\end{aligned}
latex]
Let us consider the number of students who can speak both Hindi and English is \(\)xlatex] that is
\(\)
n(H \cap E)=x
latex]
We have to find the value of \(\)n(H \cap E)latex].
We know that,
\(\)
\Rightarrow n(H \cup E)=n(H)+n(E)+n(H \cap E)
latex]
Substitute the values in the above formula we get,
\(\)
\Rightarrow 45=22+12+x
latex]
Simplifying we get,
\(\)
\Rightarrow x=11
latex]
Hence, the number of students who can speak both Hindi and English is 11.

Hint

(c) Let’s say the number of families that own a cell phone is \(A\), number of families that own a scooter \(B\), the number of families that own neither of them is \(C\) and the total number of families in that town is \(x\).
According to set theory we can equate an equation as
\(
A+B-(A \cap B)+C=x
\)
Here, \(A=\frac{25 x}{100}, B=\frac{15 x}{100}, C=\frac{65 x}{100}\) and \((A \cap B)=1500\)
Substituting the above values in the formed equation we get
\(
\Rightarrow \frac{25 x}{100}+\frac{15 x}{100}-1500+\frac{65 x}{100}=x
\)
Subtracting \(x\) from both the sides of the above equation we get
\(
\Rightarrow \frac{25 x}{100}+\frac{15 x}{100}-1500+\frac{65 x}{100}-x=0
\)
Adding 1500 to both the sides of the above equation we get
\(
\Rightarrow \frac{25 x}{100}+\frac{15 x}{100}+\frac{65 x}{100}-x=1500
\)
Further simplifying the above equation, we get
\(
\begin{aligned}
& \Rightarrow \frac{105 x}{100}-x=1500 \\
& \Rightarrow \frac{105 x-100 x}{100}=1500 \\
& \Rightarrow \frac{5 x}{100}=1500
\end{aligned}
\)
Multiplying both the sides of the above equation to 100 we get
\(
\Rightarrow 5 x=150000
\)
Dividing both the sides of the above equation by 5 we get
\(
\Rightarrow x=30000
\)
Therefore, we conclude that the total number of families in that town is 30000.

Hint

(c) The number of subsets of a set with \(n\) elements is \(2^n\).
We need to find the number of subsets of the remaining elements after fixing 3 as an element in the subset.
Since element 3 must be in every subset, consider the remaining elements in set \(A\).
The number of remaining elements is \(n^{\prime}=3\).
The number of subsets formed by these \(n^{\prime}\) elements is \(2^{n^{\prime}}\).
\(
2^3=8
\)
Each of these 8 subsets, when combined with element 3 , forms a unique subset of \(A\) containing 3.
Thus, the number of subsets of set \(A\) containing element 3 is 8.

Hint

(c) The total number of elements in the universal set is \(n(U)=100\).
The number of elements in set \(A\) is \(n(A)=50\).
The number of elements in set \(\boldsymbol{B}\) is \(\boldsymbol{n}(\boldsymbol{B})=20\).
The number of elements in the intersection of sets \(A\) and \(B\) is \(n(A \cap B)=10\).
The number of elements in the union of two sets is
\(
n(A \cup B)=n(A)+n(B)-n(A \cap B)
\)
The number of elements in the complement of a set \(X\) is \(n\left(X^C\right)=n(U)-n(X)\).
First, calculate the number of elements in the union of sets \(A\) and \(B\), then subtract this value from the total number of elements in the universal set.
Use the formula: \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\).
Substitute the given values: \(n(A \cup B)=50+20-10\).
Calculate the result: \(n(A \cup B)=60\).
Use the complement formula: \(n\left\{(A \cup B)^C\right\}=n(U)-n(A \cup B)\).
Substitute the values: \(n\left\{(A \cup B)^C\right\}=100-60\).
Calculate the final answer: \(n\left\{(A \cup B)^C\right\}=40\).

Hint

(c) The inequality \(|x-3|<4\) can be written as \(-4<x-3<4\).
Add 3 to all parts: \(-4+3<x<4+3\).
This simplifies to \(-1<x<7\).
Since \(x \in Z\), the first set is \(\{0,1,2,3,4,5,6\}\).
The inequality \(|x-4|<5\) can be written as \(-5<x-4<5\).
Add 4 to all parts: \(-5+4<x<5+4\).
This simplifies to \(-1<x<9\).
Since \(x \in Z\), the second set is \(\{0,1,2,3,4,5,6,7,8\}\).
The intersection of \(\{0,1,2,3,4,5,6\}\) and \(\{0,1,2,3,4,5,6,7,8\}\) is the set of common elements.
The common elements are \(\{0,1,2,3,4,5,6\}\).

Hint

(b) List the sets \(A_n\) for \(n\) from 2 to 10 and then find their union.
The first 10 prime numbers are \(2,3,5,7,11,13,17,19,23,29\).
Define the sets \(A_n\).
\(
\begin{aligned}
& A_2=\{2,3\} \\
& A_3=\{2,3,5\} \\
& A_4=\{2,3,5,7\} \\
& A_5=\{2,3,5,7,11\} \\
& A_6=\{2,3,5,7,11,13\} \\
& A_7=\{2,3,5,7,11,13,17\} \\
& A_8=\{2,3,5,7,11,13,17,19\} \\
& A_9=\{2,3,5,7,11,13,17,19,23\} \\
& A_{10}=\{2,3,5,7,11,13,17,19,23,29\}
\end{aligned}
\)
Since \(A_2 \subset A_3 \subset \ldots \subset A_{10}\), the union is the largest set.
\(
\begin{aligned}
& \bigcup_{n=2}^{10} A_n=A_{10} \\
& A_{10}=\{2,3,5,7,11,13,17,19,23,29\}
\end{aligned}
\)
The union of the sets \(A_n\) from \(n=2\) to \(n=10\) is the set of the first 10 prime numbers.

Hint

(b) List the sets \(A_n\) for \(n\) from 2 to 10 and then find their intersection.
The first 10 prime numbers are \(2,3,5,7,11,13,17,19,23,29\).
Define the sets \(A_n\).
\(
\begin{aligned}
& A_3=\{2,3,5\} \\
& A_4=\{2,3,5,7\} \\
& A_5=\{2,3,5,7,11\} \\
& A_6=\{2,3,5,7,11,13\} \\
& A_7=\{2,3,5,7,11,13,17\} \\
& A_8=\{2,3,5,7,11,13,17,19\} \\
& A_9=\{2,3,5,7,11,13,17,19,23\} \\
& A_{10}=\{2,3,5,7,11,13,17,19,23,29\}
\end{aligned}
\)
Since \(A_3 \subset \ldots \subset A_{10}\), the intersection is the smallest set.
\(
\begin{aligned}
& \bigcap_{n=3}^{10} A_n=A_{3} \\
& A_{3}=\{2,3,5\}
\end{aligned}
\)

Hint

(c) Understand the Subset Relationship: The condition \(A_1 \subset A_2 \subset A_3 \subset \ldots \subset A_{100}\) means each set is contained within the next set in the sequence. \(\boldsymbol{A}_1\) is a subset of \(\boldsymbol{A}_2, \boldsymbol{A}_2\) is a subset of \(\boldsymbol{A}_3\), and so on, up to \(\boldsymbol{A}_{100}\).
The Union of Nested Sets: When sets are nested in this way, the union of all the sets is the largest set in the sequence. The largest set \(\left(\boldsymbol{A}_{100}\right)\) contains all the elements present in all preceding sets ( \(\boldsymbol{A}_1\) through \(\boldsymbol{A}_{99}\) ).
Determine the Number of Elements in the Union: Since \(\bigcup_{i=1}^{100} A_i=A_{100}\), the number of elements in the union equals the number of elements in \(\boldsymbol{A}_{100}\), which is \(\boldsymbol{n}\left(\boldsymbol{A}_{100}\right)\).
Calculate \(\boldsymbol{n}\left(\boldsymbol{A}_{100}\right)\) : The formula \(\boldsymbol{n}\left(\boldsymbol{A}_{\boldsymbol{i}}\right)=\boldsymbol{i}+\mathbf{1}\) is given. Substituting \(\boldsymbol{i}=\mathbf{1 0 0}\) into this formula gives:
\(
n\left(A_{100}\right)=100+1=101
\)

Hint

(a) You are solving for the greatest possible value of the symmetric difference of two sets, \(n(A \Delta B)\).
What’s given in the problem
The cardinality of set \(A\) is \(n(A)=7\).
The cardinality of set \(\boldsymbol{B}\) is \(\boldsymbol{n}(\boldsymbol{B})=6\).
The intersection of sets \(\boldsymbol{A}\) and \(\boldsymbol{B}\) is not empty, \((\boldsymbol{A} \cap \boldsymbol{B}) \neq \phi\).
The formula for the cardinality of the symmetric difference is \(n(A \Delta B)=n(A)+n(B)-2 n(A \cap B)\).
To maximize \(n(A \Delta B)\), we need to minimize \(n(A \cap B)\).
Minimize the intersection of the sets to maximize the symmetric difference.
Determine the minimum possible value of \(n(A \cap B)\).
Since \((A \cap B) \neq \phi\), the minimum value for \(n(A \cap B)\) is 1.
So, \(n(A \cap B)_{\text {min }}=1\).
Calculate the greatest possible value of \(n(A \Delta B)\).
Substitute the values into the formula: \(n(A \Delta B)=n(A)+n(B)-2 n(A \cap B)\).
\(
\begin{aligned}
& n(A \Delta B)_{\max }=7+6-2(1) \\
& n(A \Delta B)_{\max }=13-2 \\
& n(A \Delta B)_{\max }=11
\end{aligned}
\)
The greatest possible value of \(n(A \Delta B)\) is 11.

Hint

(a) You are finding the least possible value of the symmetric difference of two sets, \(n(A \Delta B)\).
What’s given in the problem
The number of elements in set \(A\) is \(n(A)=7\).
The number of elements in set \(\boldsymbol{B}\) is \(\boldsymbol{n}(\boldsymbol{B})=6\).
The intersection of sets \(A[latex] and [latex]B\) is not empty, \(n(A \cap B) \neq \phi\).
The formula for the symmetric difference of two sets is
\(
n(A \Delta B)=n(A)+n(B)-2 n(A \cap B)
\)
To minimize \(n(A \Delta B)\), we need to maximize \(n(A \cap B)\).
How to solve
Determine the maximum possible value of the intersection and substitute it into the symmetric difference formula.
Determine the maximum possible value of \(n(A \cap B)\).
Since \(n(A \cap B) \neq \phi, n(A \cap B) \geq 1\).
The maximum number of common elements is limited by the smaller set.
So, \(n(A \cap B) \leq \min (n(A), n(B))\).
\(
n(A \cap B) \leq \min (7,6)
\)
Thus, the maximum value of \(n(A \cap B)\) is 6.
Use the formula: \(n(A \Delta B)=n(A)+n(B)-2 n(A \cap B)\).
Substitute the given values and the maximum \(n(A \cap B)\) :
\(
\begin{aligned}
& n(A \Delta B)=7+6-2 \times 6 \\
& n(A \Delta B)=13-12 \\
& n(A \Delta B)=1
\end{aligned}
\)
Calculate the least possible value of \(n(A \Delta B)\).
The least possible value of \(n(A \Delta B)\) is 1.

Hint

(c) We have \(n\left(A_i\right)=i+2\)
\(
\therefore \mathrm{n}\left(\mathrm{~A}_1\right)=3, \mathrm{n}\left(\mathrm{~A}_2\right)=4, \mathrm{n}\left(\mathrm{~A}_3\right)=5
\)
Similarly \(n\left(A_{100}\right)=102\)
given \(\mathrm{A}_1 \subset \mathrm{~A}_2 \subset \mathrm{~A}_3 \ldots \ldots \ldots \subset \mathrm{~A}_{100}\)
As \(A_3\) is a subset of all the successive sets
\(
\begin{aligned}
& \therefore \mathrm{A}_3 \cap \mathrm{~A}_4 \cap \mathrm{~A}_5 \cap \ldots \ldots \ldots \cap \mathrm{~A}_{100}=\mathrm{A}_3 \\
& \therefore \mathrm{~A}_3=\mathrm{A} \\
& \therefore \mathrm{n}(\mathrm{~A})=\mathrm{n}\left(\mathrm{~A}_3\right)=5
\end{aligned}
\)

Hint

(a) The formula for the number of elements in the union of three sets is:
\(
n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C)
\)
Apply the given condition: \(A\) and \(B\) are disjoint
Disjoint sets have no common elements, which means their intersection is the empty set. Therefore:
\(
n(A \cap B)=0
\)
Substituting this into the formula from step 1:
\(
n(A \cup B \cup C)=n(A)+n(B)+n(C)-0-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C)
\)
Simplifying:
\(
n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C)
\)
Apply the given condition: \(n(A)=n(A \cap C)\)
This condition implies that all elements of \(\boldsymbol{A}\) are also elements of \(\boldsymbol{C}\). If \(\boldsymbol{n}(\boldsymbol{A})=\boldsymbol{n}(\boldsymbol{A} \cap \boldsymbol{C})\), then \(\boldsymbol{A}\) must be a subset of \(\boldsymbol{C}\), i.e., \(\boldsymbol{A} \subseteq \boldsymbol{C}\).
If \(A \subseteq C\), then:
\(A \cap C=A\)
\(\boldsymbol{A} \cap \boldsymbol{B} \cap \boldsymbol{C}=(\boldsymbol{A} \cap \boldsymbol{B}) \cap \boldsymbol{C}=\varnothing \cap \boldsymbol{C}=\varnothing\) (since \(\boldsymbol{A}\) and \(\boldsymbol{B}\) are disjoint, \(\boldsymbol{A} \cap \boldsymbol{B}=\varnothing\) )
Substitute these into the formula:
Replace \(\boldsymbol{n}(\boldsymbol{A} \cap \boldsymbol{C})\) with \(\boldsymbol{n}(\boldsymbol{A})\) and \(\boldsymbol{n}(\boldsymbol{A} \cap \boldsymbol{B} \cap \boldsymbol{C})\) with 0 in the simplified formula
\(
n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A)-n(B \cap C)+0
\)
Simplify the expression:
The \(\boldsymbol{n}(\boldsymbol{A})\) terms cancel out:
\(
n(A \cup B \cup C)=n(B)+n(C)-n(B \cap C)
\)
Conclusion
Given the conditions, \(\boldsymbol{n}(\boldsymbol{A} \cup \boldsymbol{B} \cup \boldsymbol{C})\) is necessarily equal to \(\boldsymbol{n}(\boldsymbol{B})+\boldsymbol{n}(\boldsymbol{C})-\boldsymbol{n}(\boldsymbol{B} \cap \boldsymbol{C})\) . This can be further simplified to \(\boldsymbol{n}(\boldsymbol{B} \cup \boldsymbol{C})\), as the formula for the union of two sets is \(n(B \cup C)=n(B)+n(C)-n(B \cap C)\).
The final answer is \(\boldsymbol{n}(\boldsymbol{B} \cup \boldsymbol{C})\).

Hint

(d) We have,
\(
\frac{n^3+5 n^2+2}{n}=n^2+5 n+\frac{2}{n}
\)
\(\therefore \frac{n^3+5 n^2+2}{n}\) is an integer, if \(\frac{2}{n}\) is an integer
\(
\Rightarrow n= \pm 1, \pm 2
\)
\(\Rightarrow\) A consists of four elements viz. \(-1,1,-2,2\)
The number of elements in set \(A\) is 4.

Hint

(a)

\(
\begin{aligned}
& p=\frac{7 n^2+3 n+3}{n} \\
& p=\frac{7 n^2}{n}+\frac{3 n}{n}+\frac{3}{n} \\
& p=7 n+3+\frac{3}{n}
\end{aligned}
\)
Since \(p\) must be a natural number, \(\frac{3}{n}\) must be an integer.
This implies that \(n\) must be a divisor of 3.
As \(n \in N\), possible values for \(n[latex] are 1 and 3.
[latex]
\begin{aligned}
&\text { For } n=1 \text { : }\\
&p=7(1)+3+\frac{3}{1} =13
\end{aligned}
\)
\(
\begin{aligned}
&\text { For } n=3 \text { : }\\
&p=7(3)+3+\frac{3}{3}=25
\end{aligned}
\)
For \(p=13: 13\) is a prime number.
For \(p=25: 25\) is not a prime number since \(25=5 \times 5\).
The only prime number found is 13. So, \(A=\{13\}\).
The number of elements in the set \(A\) is 1.

Hint

(c) If \(A \subset B\) and \(B \subset C\), then these sets is represented in Venn diagram as

Clearly,
\(
A \cup B=B
\)
and \(B \cap C=B\)
Hence, \(A \cup B=B \cap C\)

Hint

(b) 

\(
\begin{aligned}
&\text { The number of subsets containing } 2,3 \text {, and } 5 \text { is } 2^n \text {. }\\
&2^3=8
\end{aligned}
\)