Past JEE Main Entrance Paper

Hint

(b) Product of two even numbers is always even andthe  product of two odd numbers is always odd.
\(\therefore\) Number of required subsets
\(=\) Total number of subsets – Total number of subsets having only odd numbers
\(
=2^{100}-2^{50}=2^{50}\left(2^{50}-1\right)
\)

Explanation:

You are solving for the number of non-empty subsets of \(S=\{1,2,3, \ldots, 100\}\) where the product of elements in each subset is even.
What’s given in the problem
The set is \(S=\{1,2,3, \ldots, 100\}\).
We need to find the number of non-empty subsets.
The product of elements in the subset must be even.
A product of integers is even if and only if at least one of the integers is even.
A product of integers is odd if and only if all of the integers are odd.
How to solve?
Calculate the total number of non-empty subsets and subtract the number of non-empty subsets containing only odd numbers.

Find the total number of non-empty subsets of \(S\):
The total number of elements in \(S\) is \(n=100\).
The total number of subsets is \(2^n=2^{100}\).
The total number of non-empty subsets is \(2^{100}-1\).

Find the number of non-empty subsets containing only odd numbers:
The number of odd elements in \(S\) is 50 (i.e., \(\{1,3, \ldots, 99\}\) ).
The number of subsets formed only by odd numbers is \(2^{50}\).
The number of non-empty subsets formed only by odd numbers is \(2^{50}-1\).

Calculate the number of non-empty subsets with an even product:
Subtract the number of non-empty subsets with an odd product from the total number of non-empty subsets.
Number of subsets with even product \(=\left(2^{100}-1\right)-\left(2^{50}-1\right)\).
Number of subsets with even product \(=2^{100}-2^{50}\).
This can be factored as \(2^{50}\left(2^{50}-1\right)\).

Hint

(b) You are solving for the set S containing non-negative real numbers x that satisfy the given equation.
What’s given in the problem
The set \(S\) is defined as \(S=\{x \in R: x \geq 0\) and \(2|\sqrt{x}-3|+\sqrt{x}(\sqrt{x}-6)+6=0\}\).
How to solve?
Solve the equation by substituting \(y=\sqrt{x}\) and considering two cases based on the absolute value.

Step 1: Substitute \(\mathrm{y}=\sqrt{\mathrm{x}}\)
Let \(\mathrm{y}=\sqrt{\mathrm{x}}\).
The equation becomes \(2|y-3|+y(y-6)+6=0\).
Expand the term: \(2|y-3|+y^2-6 y+6=0\).

Step 2: Consider Case 1: \(y \geq 3\)
If \(y \geq 3\), then \(|y-3|=y-3\).
Substitute into the equation: \(2(y-3)+y^2-6 y+6=0\).
Simplify: \(2 y-6+y^2-6 y+6=0\).
Combine terms: \(y^2-4 y=0\).
Factor: \(\mathrm{y}(\mathrm{y}-4)=0\).
Solutions for y are \(\mathrm{y}=0\) or \(\mathrm{y}=4\).
Since \(\mathrm{y} \geq 3\), we accept \(\mathrm{y}=4\).
Step 3: Consider Case 2: \(y<3\)
If \(y<3\), then \(|y-3|=-(y-3)=3-y\).
Substitute into the equation: \(2(3-y)+y^2-6 y+6=0\).
Simplify: \(6-2 y+y^2-6 y+6=0\).
Combine terms: \(y^2-8 y+12=0\).
Factor: \((y-2)(y-6)=0\).
Solutions for \(y\) are \(y=2\) or \(y=6\).
Since \(\mathrm{y}<3\), we accept \(\mathrm{y}=2\).

Step 4: Convert back to x
For \(y=4: \sqrt{x}=4 \Longrightarrow x=4^2=16\).
For \(y=2: \sqrt{x}=2 \Longrightarrow x=2^2=4\).
Both \(x=16\) and \(x=4\) satisfy \(x \geq 0\).
The set \(S\) contains exactly two elements, \(S=\{4,16\}\).

Hint

(a) \(f(x)+2 f\left(\frac{1}{x}\right)=3 x \text { ..(i) }\)
\(f\left(\frac{1}{x}\right)+2 f(x)=\frac{3}{x} \text { ..(ii) }\)
Adding (i) and (ii)
\(
\Rightarrow f(x)+f\left(\frac{1}{x}\right)=x+\frac{1}{x} \text { ..(iii) }
\)
Substracting (i) from (ii)
\(
\Rightarrow f(x)-f\left(\frac{1}{x}\right)=\frac{3}{x}-3 x \text { ..(iv) }
\)
On adding (iii) and (iv)
\(
\begin{aligned}
&\Rightarrow f(x)=\frac{2}{x}-x \\
&f(x)=f(-x) \Rightarrow \frac{2}{x}-x=\frac{-2}{x}+x \Rightarrow x=\frac{2}{x} \\
&x^{2}=2 \quad \text { or } x=\sqrt{2},-\sqrt{2}
\end{aligned}
\)

Hint

(d) \(\mathrm{P}=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}\)
\(
\Rightarrow \sin \theta=(\sqrt{2}+1) \cos \theta \Rightarrow \tan \theta=\sqrt{2}+1
\)
Now, \(Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}\)
\(
\begin{aligned}
\Rightarrow \cos \theta=(\sqrt{2}-1) \sin \theta \Rightarrow \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\
\Rightarrow \tan \theta=\sqrt{2}+1 \\
\therefore \quad \mathrm{P}=\mathrm{Q}
\end{aligned}
\)

Hint

(d) \(\mathrm{S}=\{1,2,3,4\}\)
Let \(\mathrm{A}\) and \(\mathrm{B}\) be disjoint subsets of \(\mathrm{S}\)
Now for any element \(a \in S\), has got three possibilities either, it is in A or B
or none
\(\Rightarrow\) For every element out of 4 elements there are three choices
\(\therefore\) Total options \(=3^{4}=81\)
Here \(\mathrm{A} \neq \mathrm{B}\) except when \(\mathrm{A}=\mathrm{B}=\phi\)
\(\therefore 81-1=80\) ordered pairs \((\mathrm{A}, \mathrm{B})\) are there for which \(\mathrm{A} \neq \mathrm{B}\)
Hence total number of unordered pairs of disjoint subsets \(=\frac{80}{2}+1=41\)

Hint

(b) \(2^{m}=112+2^{n} \Rightarrow 2^{m}-2^{n}=112\)
\(
\Rightarrow 2^{n}\left(2^{m-n}-1\right)=2^{4}\left(2^{3}-1\right)
\)
\(
\therefore m=7, n=4 \Rightarrow m n=28
\)

Hint

(d) Here set S contains 5 odd and 4 even numbers. Since each of \(\mathrm{N}_{\mathrm{K}}\) containing five elements out of which exactly are odd.
\(\begin{aligned}
&\therefore \mathrm{N}_{1}={ }^{5} \mathrm{C}_{1} \times{ }^{4} \mathrm{C}_{4}=5 \\
&\mathrm{~N}_{2}={ }^{5} \mathrm{C}_{2} \times{ }^{4} \mathrm{C}_{3}=40 \\
&\mathrm{~N}_{3}={ }^{5} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{2}=60 \\
&\mathrm{~N}_{4}={ }^{5} \mathrm{C}_{4} \times{ }^{4} \mathrm{C}_{1}=20 \\
&\mathrm{~N}_{5}={ }^{5} \mathrm{C}_{5}=1 \\
&\therefore \mathrm{N}_{1}+\mathrm{N}_{2}+\mathrm{N}_{3}+\mathrm{N}_{4}+\mathrm{N}_{5}=126
\end{aligned}\)

Hint

(b) \(C \rightarrow\) person like coffee
\(T \rightarrow\) person like Tea

\(
\begin{aligned}
& n(C)=73 \\
& n(T)=65 \\
& n(C \cup T) \leq 100 \\
& n(C)+n(T)-n(C \cap T) \leq 100 \\
& 73+65-x \leq 100 \\
& x \geq 38 \\
& 73-x \geq 0 \Rightarrow x \leq 73 \\
& 65-x \geq 0 \Rightarrow x \leq 65 \\
& 38 \leq x \leq 65
\end{aligned}
\)

\(
\Rightarrow x \neq 36 .
\)

Hint

(d) Let \(n(U)=100\), then \(n(A)=63, n(B)=76\)
\(
n(A \cap B)=x
\)
Now,
\(
\begin{aligned}
&n(A \cup B)=n(A)+n(B)-n(A \cap B) \leq 100 \\
&=63+76-x \leq 100 \\
&\Rightarrow x \geq 139-100 \Rightarrow x \geq 39 \\
&\because n(A \cap B) \leq n(A) \Rightarrow x \leq 63 \\
&\therefore 39 \leq x \leq 63
\end{aligned}
\)

\(
\text { From options possible value of } x=55 \text {. }
\)

Hint

Explanation: \(\cup_{i=1}^{50} X_i=\mathrm{X}_1, \mathrm{X}_2, \ldots . ., \mathrm{X}_{50}=50\) sets. Given each sets having 10 elements.
So total elements \(=50 \times 10\)
\(\cup_{i=1}^n Y_i=\mathrm{Y}_1, \mathrm{Y}_2, \ldots . ., \mathrm{Y}_{\mathrm{n}}=\mathrm{n}\) sets. Given each sets having 5 elements.
So total elements \(=5 \times n\)
Now each element of set T contains exactly 20 of sets \(\mathrm{X}_{\mathrm{i}}\).
So number of effective elements in set \(T=\frac{50 \times 10}{20}\)
Also each element of set T contains exactly 6 of sets \(\mathrm{Y}_{\mathrm{i}}\).
So number of effective elements in set \(T=\frac{50 \times n}{6}\)
\(
\begin{aligned}
& \therefore \frac{50 \times 10}{20}=\frac{5 \times n}{6} \\
& \Rightarrow \mathrm{n}=30
\end{aligned}
\)

Hint

\(
\begin{aligned}
A=\{x \in Z, \quad& 2^{(x+2)\left(x^{2}-5 x+6\right)}=2^{0} \\
&(x+2)\left(x^{2}-5 x+6\right)=0 \\
& x=-2,2,3
\end{aligned}
\)

\(
\begin{aligned}
B=\{x \in Z, \quad&-3<2 x-1<9 \\
\Rightarrow &-2<2 x<10 \\
\Rightarrow &-1<x<5
\end{aligned}
\)

\(
\begin{aligned}
&x=0,1,2,3,4 \\
&A \times B=\{(-2,0),(-2,1)(-2,2)(-2,3)(-2,4),(2,0),(2,1)(2,2),(2,3),(2,4)(3,0),(3,1) \\
&(3,2)(3,3),(3,4)] \\
&A \times B \text { elements }=15 \\
&\text { total subsets }=2^{15}
\end{aligned}
\)

Hint

(d) Let \(\mathrm{n}(\mathrm{A})=\) number of students opted Mathematics \(=70\),
\(\mathrm{n}(\mathrm{B})=\) number of students opted Physics \(=46\),
\(\mathrm{n}(\mathrm{C})=\) number of students opted Chemistry \(=28\),
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B})=23\)
\(\mathrm{n}(\mathrm{B} \cap \mathrm{C})=9\)
\(\mathrm{n}(\mathrm{A} \cap \mathrm{C})=14\),
\(\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})=4\)
Nown( \(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\)
\(=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})+\mathrm{n}(\mathrm{C})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})-\mathrm{n}(\mathrm{B} \cap \mathrm{C})-\mathrm{n}(\mathrm{A} \cap \mathrm{C})+\mathrm{n}(\mathrm{A} \cap \mathrm{B} \cap \mathrm{C})\) \(=70+46+28-23-9-14+4=102\)

So number of students not opted for any course \(=\) Total \(-\mathrm{n}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})\) \(=140-102=38\).

Hint

(d) Let \(A=\{1,2,3,4\}\)
\(B=\{3,4,5,6\}\)
\(C=\{1,2,3,4,7,8\}\)
Here \(A \cap B=\{3,4\} \subseteq C\)
\(A-C=\phi \subseteq B\)
but \(\mathrm{A} \nsubseteq \mathrm{B}\)
So not true (wrong) statement is d, If \(A-C \subseteq B\) then \(A \subseteq B\)

Hint

(c) Step 1: Define the Variables
Let \(x\) be the total number of families in the town.

Step 2: Identify the Percentages
– Families owning a phone: \(25 \%\) of \(x\)
– Families owning a car: \(15 \%\) of \(x\)
– Families owning neither a phone nor a car: \(65 \%\) of \(x\)
– Families owning both a car and a phone: 2000 families

Step 3: Calculate Families Owning Either a Phone or a Car
Since \(65 \%\) of families own neither a phone nor a car, the percentage of families that own either a phone or a car is:
\(
100 \%-65 \%=35 \%
\)
Thus, the number of families owning either a phone or a car is:
\(
0.35 x
\)
Step 4: Use the Principle of Inclusion-Exclusion
According to the principle of inclusion-exclusion for two sets:
\(
n(P \cup C)=n(P)+n(C)-n(P \cap C)
\)
Where:
\(-n(P)=0.25 x\) (families with a phone)
\(-n(C)=0.15 x\) (families with a car)
\(-n(P \cap C)=2000\) (families with both)
Substituting the values we have:
\(
0.35 x=0.25 x+0.15 x-2000
\)
Step 5: Simplify the Equation
Combine like terms:
\(
0.35 x=0.40 x-2000
\)
Rearranging gives:
\(
\begin{aligned}
& 0.35 x-0.40 x=-2000 \\
& -0.05 x=-2000
\end{aligned}
\)
Step 6: Solve for \(x\)
Dividing both sides by -0.05 :
\(
x=\frac{2000}{0.05}=40000
\)
Step 7: Verify Each Statement
1. Statement A: \(5 \%\) of families own both a car and a phone.
– Calculation:
\(
\frac{2000}{40000} \times 100=5 \%
\)
– True.
2. Statement B: \(35 \%\) of families own either a car or a phone.
– Calculation:
\(0.35 \times 40000=14000\) families
– True.
3. Statement C: 40,000 families live in the town.
– True.

Conclusion
All three statements (A, B, and C) are correct.

Hint

(c) We have \(\mathrm{X} \cap(\mathrm{Y} \cup \mathrm{X})^{\prime}=\mathrm{X} \cap\left(\mathrm{Y}^{\prime} \cap \mathrm{X}^{\prime}\right)\)
\(
\begin{aligned}
&=\mathrm{X} \cap \mathrm{Y}^{\prime} \cap \mathrm{X}^{\prime}=\mathrm{X} \cap \mathrm{X}^{\prime} \cap \mathrm{Y}^{\prime} \\
&=\phi \cap \mathrm{Y}^{\prime}=\phi
\end{aligned}
\)

Hint

(c) From the given conditions,
\(
\begin{aligned}
&n(A)=25, n(B)=7 \text { and } n(A \cap B)=3 \\
&n(A \cup B)=n(A)+n(B)-n(A \cap B)=25+7-3=29
\end{aligned}
\)

Hint

(c) Let the number of newspapers which are read be \(\mathrm{n}\). Then \(60 n=\) \((300)(5) \Rightarrow n=25\)

Hint

(b) Given that \(\bigcup_{i=1}^{30} A_{i}=\cup_{j=1}^{n} B_{j}=S \text { …(i) }\)
and each \(A_{i}^{\text {s }}\) contain 5 elements
So, total number of elements in \(A_{i}=5 \times 30=150\).
Since each element of \(S\) belongs to exactly ten of the \(A_{i}\) ‘s.
\(
\therefore n(S)=n\left[\bigcup_{i=1}^{30} A_{i}\right]=\frac{150}{10}=15 \text { …(ii) }
\)
Now, each \(\mathrm{B}_{\mathrm{j}}^{\text {s }}\) contain 3 elements
So, total number of elements in \(B_{j}=3 \times n=3 n\).
Since each element of \(\mathrm{S}\) belongs to exactly nine of the \(\mathrm{B}_{\mathrm{j}}{ }^{\prime {\text {s }}}\).
\(
\therefore n(S)=n\left[\bigcup_{j=1}^{n} B_{j}\right]=\frac{3 n}{9} \text { …(iii) }
\)
from (ii) and (iii)
\(
\frac{3 n}{9}=15 \Rightarrow n=45 \text {. }
\)

Hint

Concept:
If \(A\) and \(B\) are two sets then, \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)
Calculation:
Given: \(n(A)=3\) and \(n(B)=6\).
As we know that, if \(A\) and \(B\) are two sets then, \(n(A \cup B)=n(A)+n(B)-n(A \cap B)\)
\(
\Rightarrow n(A \cup B)=3+6-n(A \cap B)
\)
In order to minimize \(n(A \cup B)\) we have to maximize \(n(A \cap B)\).
If \(A\) is a subset of \(B\), then \(A \cap B=A \Rightarrow n(A \cap B)=n(A)=3\)
\(
\Rightarrow n(A \cup B)=3+6-3=6 \text {. }
\)
Hence, the minimum number of elements that \((A \cup B)\) can have is 6 .

Hint

(d) We have
\(
\begin{aligned}
&n(U)=100, \text { where } U \text { stands for universal set } \\
&n(M \cap C \cap T)=d=10 ; n(M \cap C)=b+d=20 ; \\
&n(C \cap \mathrm{T})=d+f=30 ; n(M \cap T)=d+e=25 \\
&\quad \Rightarrow b=10, f=20 \text { and } e=15 \\
&\quad n(\text { only } M)=a=12 ; n(\text { only } C)=c=5 ; n(\text { only } T)=\mathrm{g}=8
\end{aligned}
\)
Filling all the entries we obtain the Venn diagram as shown:

\(\begin{gathered}
\quad \therefore \quad n(M \cap C \cup T)=\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}+\mathrm{g} \\
=12+10+5+15+10+20+8=80 \\
\therefore \quad n(M \cup C \cup T)^{\prime}=100-80=20
\end{gathered}\)

Hint

\(
\text { (b) } A =\{x:|x|<3, x \in Z\}
\)
\(
\begin{aligned}
& A =\{-2,-1,0,1,2\} \\
& R =\{(x, y): y=|x|, x \neq-1\} \\
& R =\{(-2,2),(0,0),(1,1),(2,2)\}
\end{aligned}
\)
\(R\) has four elements
Number of elements in the power set of \(R\)
\(
=2^4=16
\)

Hint

(b) Let \(X=\{1,2,3,4,5\}\)
Total no. of elements \(=5\)
Each element has 3 options. Either set \(Y\) or set \(Z\) or none. \((\because Y \cap Z=\phi)\)
So, number of ordered pairs \(=3^5\)

Explanation: The set \(X=\{1,2,3,4,5\}\).
\(Y \subseteq X\).
\(Z \subseteq X\).
\(Y \cap Z=\phi\).
How to solve?
Analyze the possible placements for each element of \(X\) to satisfy the given conditions.
For each element \(x \in X\), there are three possibilities:
\(x \in Y\) and \(x \notin Z\).
\(\quad x \notin Y\) and \(x \in Z\).
\(x \notin Y\) and \(x \notin Z\).
An element cannot be in both \(Y\) and \(Z\) because \(Y \cap Z=\phi\).
Since there are 5 elements in \(X\), and each element has 3 independent choices:
The total number of ordered pairs ( \(Y, Z\) ) is \(3^5\).
\(
3^5=243
\)

Hint

(b) Let \(x \in A\) and \(x \in B \Leftrightarrow x \in A \cup B\)
\(
\begin{aligned}
& \Leftrightarrow x \in A \cup C \quad(\because A \cup B=A \cup C) \\
& \Leftrightarrow x \in C
\end{aligned}
\)
\(
\begin{aligned}
& \therefore \quad B=C . \\
& \text { Let } x \in A \text { and } x \in B \Leftrightarrow x \in A \cap B \\
& \Leftrightarrow x \in A \cap C \quad(\because A \cap B=A \cap C) \\
& \Leftrightarrow x \in C \\
& \therefore B=C
\end{aligned}
\)

Hint

For disoint sets, \(A \cap B=\phi\)
Each element in either \(A\) or \(B\) or neither.
\(\therefore\) Total ways \(=3^4=81 ; A=B\) iff \(A=B=\phi\)
Otherwise, \(A\) and \(B\) are interchangable
\(\therefore\) Number of unordered pair for disoint subsets of
\(
S=\frac{3^4+1}{2}=41
\)

Hint

\(\because A \times B\) has 8 elements.
\(\therefore\) Number of subsets \(=2^8=256\)
Number of subsets with zero element \(={ }^8 C_0=1\)
Number of subsets with one element \(={ }^8 C_1=8\)
Number of subsets with one elements \(={ }^8 C_2=28\)
Hence, Number of subsets of \(A \times B\) having 3 or more elements
\(
=256-(1+8+28)=256-37=219
\)

Hint

Since, \(4^n-3 n-1=(1+3)^n-3 n-1\)
\(
\begin{aligned}
& =\left(1+{ }^n C_1 \cdot 3+{ }^n C_2 \cdot 3^2+{ }^n C_3 \cdot 3^3+\ldots+{ }^n C_n \cdot 3^n\right)-3 n-1 \\
& =3^2\left({ }^n C_2+{ }^n C_3 \cdot 3+\ldots+{ }^n C_n \cdot 3^{n-2}\right)
\end{aligned}
\)
\(\Rightarrow 4^n-3 n-1\) is a multiple of 9 for \(n \geq 2\)
For \(n=1,4^n-3 n-1=4-3-1=0\)
For \(n=2,4^n-3 n-1=16-6-1=9\)
\(\therefore 4^n-3 n-1\) is multiple of 9 for all \(n \in N\).
It is clear that \(X\) contains elements, which are multiples of 9 and \(Y\) contains all multiples of 9 .
\(
\therefore X \subseteq Y \text { i.e., } X \cup Y=Y
\)

Hint

\(
n(A)=4, n(B)=2 \Rightarrow n(A \times B)=8
\)
The number of subsets of \(A \times B\) having at least three elements
\(
\begin{aligned}
& ={ }^8 C_3+{ }^8 C_4+{ }^8 C_5+\ldots+{ }^8 C_8 \\
& =2^8-\left({ }^8 C_0+{ }^8 C_1+{ }^8 C_2\right) \\
& =256-(1+8+28)=219
\end{aligned}
\)

Explanation: Cardinality of \(A \times B\) : Since set \(A\) has 4 elements and set \(B\) has 2 elements, their Cartesian product \(A \times B\) has \(4 \times 2=8\) elements.
Total number of subsets: A set with 8 elements has \(2^8=256\) subsets.
Subsets with fewer than 3 elements: We need to subtract the number of subsets with 0,1 , or 2 elements.
Subsets with 0 elements: 1 (the empty set)
Subsets with 1 element: \(\binom{8}{1}=8\)
Subsets with 2 elements: \(\binom{8}{2}=\frac{8 \times 7}{2 \times 1}=28\)
Subsets with 3 or more elements: Subtract the unwanted subsets from the total: \(256-1-8-28=219\).

Note:

\(
\begin{aligned}
&\text { Combination Formula }\\
&{ }^n C_r=\frac{n!}{(n-r)!r!}
\end{aligned}
\)

Hint

\(
\begin{array}{ll}
x+2 y+3 z=42, & x, y, z \geq 0 \\
z=0 & x+2 y=42 \Rightarrow 22 \\
z=1 & x+2 y=39 \Rightarrow 20 \\
z=2 & x+2 y=36 \Rightarrow 19 \\
z=3 & x+2 y=33 \Rightarrow 17 \\
z=4 & x+2 y=30 \Rightarrow 16 \\
z=5 & x+2 y=27 \Rightarrow 14 \\
z=6 & x+2 y=24 \Rightarrow 13 \\
z=7 & x+2 y=21 \Rightarrow 11 \\
z=8 & x+2 y=18 \Rightarrow 10 \\
z=9 & x+2 y=15 \Rightarrow 8
\end{array}
\)
\(
\begin{array}{ll}
z=10 & x+2 y=12 \Rightarrow 7 \\
z=11 & x+2 y=9 \Rightarrow 5 \\
z=12 & x+2 y=6 \Rightarrow 4 \\
z=13 & x+2 y=3 \Rightarrow 2 \\
z=14 & x+2 y=0 \Rightarrow 1
\end{array}
\)
Total : 169

Hint

Sum of elements in \(A \cap B\)
\(
\begin{aligned}
= & \underbrace{(2+4+6+\ldots+200)}_{\text {Multiple of } 2}-\underbrace{(6+12+\ldots+198)}_{\text {Multiple of } 2 \text { & } 3 \text { i.e. } 6} \\
& -\underbrace{(10+20+\ldots+200)}_{\text {Multiple of } 5 \text { & } 2 \text { i.e. } 10}+\underbrace{(30+60+\ldots+180)}_{\text {Multiple of } 2,5 \text { & 3 i.e. } 30} \\
= & 5264
\end{aligned}
\)

Hint

\(
\begin{aligned}
& A =\{2,3,4,5, \ldots,, 30\} \\
& ( a , b ) \simeq( c , d ) \quad \Rightarrow ad = bc \\
& (4,3) \simeq(c, d) \quad \Rightarrow \quad 4 d=3 c \\
& \Rightarrow \frac{4}{3}=\frac{c}{d} \\
& \frac{c}{d}=\frac{4}{3} \quad \& c, d \in\{2,3, \ldots \ldots, 30\} \\
& ( c , d )=\{(4,3),(8,6),(12,9),(16,12),(20, \\
& 15),(24,18),(28,21)\}
\end{aligned}
\)
No. of ordered pair \(=7\)

Hint

Let the types of games be denoted as \(A, B, C\).
In all the given Venn diagrams we can observe that there is at least one region where all \(A, B, C\) intersect i.e. \(A \cap B \cap C\)
Thus none of the given Venn diagrams represents the condition ‘Some of the students play two types of games, but none play all three games’

Hint

(a)

\(
\begin{aligned}
&\text { We have, } B_1=A_1 \Rightarrow B_1 \subset A_1\\
&\begin{aligned}
& B_2=A_2-A_1 \Rightarrow B_2 \subset A_2 \\
& B_3=A_3-\left(A_1 \cup A_2\right) \Rightarrow B_3 \subset A_3 \\
& B_1 \cup B_2 \cup B_3 \subset A_1 \cup A_2 \cup A_3 \\
& \because \quad A_1 \cup A_2 \cup A_3 \supset B_1 \cup B_2 \cup B_3
\end{aligned}
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
& A \cup B=\{1,2,3,4,5\} \cup\{2,4,6\} \\
& A \cup B=\{1,2,3,4,5,6\}
\end{aligned}
\)
\(
\begin{aligned}
&\text { The intersection }(A \cup B) \cap C \text { contains elements common to both }(A \cup B) \text { and } C \text {. }\\
&\begin{aligned}
& (A \cup B) \cap C=\{1,2,3,4,5,6\} \cap\{3,4,6\} \\
& (A \cup B) \cap C=\{3,4,6\}
\end{aligned}
\end{aligned}
\)

Hint

(b) The statement \(x y \in A\) is true. If \(x\) and \(y\) are elements of \(A\), which is the set of squares of natural numbers, then \(x=m^2\) and \(y=n^2\) for some natural numbers \(m\) and \(n\). Therefore, \(x y=m^2 n^2=(m n)^2\), and since \(m\) and \(n\) are natural numbers, \(m n\) is also a natural number, meaning \(x y\) is the square of a natural number and thus an element of \(A\).
The other options are not necessarily true. For example, \(5^2-3^2=25-9=16\), which is in \(A\), but \(3^2-5^2=9-25=-16\), which is not in \(A\). Similarly, \(x+y\) and \(\frac{x}{y}\) are not always in \(A\). For example, \(1^2+2^2=1+4=5\), which is not a perfect square. And \(\frac{4^2}{2^2}=\frac{16}{4}=4\), which is in \(A\), but \(\frac{2^2}{4^2}=\frac{4}{16}=\frac{1}{4}\), which is not a natural number.

Hint

(b)

Given the sets \(A=\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}:|\alpha-1| \leq 4\) and \(|\beta-5| \leq 6\}\) and \(B=\left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}: 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}\).
Let’s analyze set A first:
The condition \(|\alpha-1| \leq 4\) implies \(-4 \leq \alpha-1 \leq 4\), which means \(-3 \leq \alpha \leq 5\).
The condition \(|\beta-5| \leq 6\) implies \(-6 \leq \beta-5 \leq 6\), which means \(-1 \leq \beta \leq 11\).
So, set A is a rectangle defined by the inequalities \(-3 \leq \alpha \leq 5\) and \(-1 \leq \beta \leq 11\)
Now let’s analyze set \(B\) :
The inequality \(16(\alpha-2)^2+9(\beta-6)^2 \leq 144\) can be rewritten by dividing by 144:
\(
\begin{aligned}
& \frac{16(\alpha-2)^2}{144}+\frac{9(\beta-6)^2}{144} \leq 1 \\
& \frac{(\alpha-2)^2}{9}+\frac{(\beta-6)^2}{16} \leq 1
\end{aligned}
\)
This is the equation of an ellipse centered at \((2,6)\) with semi-major axis \(\boldsymbol{b}=\mathbf{4}\) (along the \(\boldsymbol{\beta}\)-axis, since \(\mathbf{1 6 > 9}\) ) and semi-minor axis \(\boldsymbol{a}=\mathbf{3}\) (along the \(\boldsymbol{\alpha}\) axis).
To determine the relationship between \(A\) and \(B\), it is necessary to check if one set is a subset of the other. Consider the boundary points of ellipse \(B\) and check if they fall within the bounds of rectangle \(A\).
The center of the ellipse is \((\mathbf{2 , 6})\).
The \(\alpha\)-values for the ellipse range from \(2-3=-1\) to \(2+3=5\). The \(\beta\)-values for the ellipse range from \(6-4=2\) to \(6+4=10\).
Comparing these ranges with the dimensions of rectangle A:
For \(\boldsymbol{\alpha}\) : The rectangle spans from -3 to 5. The ellipse spans from -1 to 5. The ellipse’s \(\boldsymbol{\alpha}\)-range is completely contained within the rectangle’s \(\boldsymbol{\alpha}\)-range.
For \(\boldsymbol{\beta}\) : The rectangle spans from -1 to 11. The ellipse spans from 2 to 10. The ellipse’s \(\boldsymbol{\beta}\)-range is completely contained within the rectangle’s \(\boldsymbol{\beta}\)-range.
Since both the \(\boldsymbol{\alpha}\) and \(\boldsymbol{\beta}\) ranges of the ellipse (set B) are contained within the corresponding ranges of the rectangle (set A), all points belonging to the ellipse are also contained within the rectangle.
Therefore, \(\boldsymbol{B} \subset \boldsymbol{A}\).

Hint

(b) Set A is defined as \(\left\{x \in(0, \pi)-\left\{\frac{\pi}{2}\right\}: \log _{(2 / \pi)}|\sin x|+\log _{(2 / \pi)}|\cos x|=2\right\}\).
Set B is defined as \(\{x \geqslant 0: \sqrt{x}(\sqrt{x}-4)-3|\sqrt{x}-2|+6=0\}\).
The logarithm property \(\log _b M+\log _b N=\log _b(M N)\).
The definition of absolute value: \(|y|=y\) if \(y \geq 0\) and \(|y|=-y\) if \(y<0\).
How to solve?
Solve for the elements in set A and set B separately, then find the number of elements in their union.
Use the logarithm property: \(\log _{(2 / \pi)}(|\sin x||\cos x|)=2\).
Convert to exponential form: \(|\sin x \cos x|=\left(\frac{2}{\pi}\right)^2=\frac{4}{\pi^2}\). Use the identity \(\sin (2 x)=2 \sin x \cos x:\left|\frac{\sin (2 x)}{2}\right|=\frac{4}{\pi^2}\). This simplifies to \(|\sin (2 x)|=\frac{8}{\pi^2}\).
\(
\text { Since } \pi \approx 3.14, \pi^2 \approx 9.86 \text {, so } \frac{8}{\pi^2} \approx \frac{8}{9.86}<1 \text {. }
\)
In the domain \((0, \pi)-\left\{\frac{\pi}{2}\right\}, 2 x \in(0,2 \pi)-\{\pi\}\).
For \(|\sin (2 x)|=\frac{8}{\pi^2}\), there are four solutions for \(2 x\) in ( \(0,2 \pi\) ).
These solutions for \(2 x\) are \(\alpha, \pi-\alpha, \pi+\alpha, 2 \pi-\alpha\) where \(\sin (\alpha)=\frac{8}{\pi^2}\).
Thus, \(x=\frac{\alpha}{2}, \frac{\pi-\alpha}{2}, \frac{\pi+\alpha}{2}, \frac{2 \pi-\alpha}{2}\).
All these values are distinct and within \((0, \pi)-\left\{\frac{\pi}{2}\right\}\).
So, \(n(A)=4\).
Let \(y=\sqrt{x}\). The equation becomes \(y(y-4)-3|y-2|+6=0\).
This simplifies to \(y^2-4 y-3|y-2|+6=0\).
Case 1: \(y-2 \geq 0 \Longrightarrow y \geq 2\).
\(y^2-4 y-3(y-2)+6=0 \Longrightarrow y^2-7 y+12=0\).
\((y-3)(y-4)=0 \Longrightarrow y=3\) or \(y=4\).
Both \(y=3\) and \(y=4[latex] satisfy [latex]y \geq 2\).
So \(\sqrt{x}=3 \Longrightarrow x=9\) and \(\sqrt{x}=4 \Longrightarrow x=16\).

Case 2: \(y-2<0 \Longrightarrow 0 \leq y<2\).
\(y^2-4 y-3(-(y-2))+6=0 \Longrightarrow y^2-4 y+3 y-6+6=0\).
\(y^2-y=0 \Longrightarrow y(y-1)=0\).
\(y=0\) or \(y=1\).
Both \(y=0\) and \(y=1\) satisfy \(0 \leq y<2\).
So \(\sqrt{x}=0 \Longrightarrow x=0\) and \(\sqrt{x}=1 \Longrightarrow x=1\).

Thus, \(B=\{0,1,9,16\}\), so \(n(B)=4\).
The elements of A are of the form \(\frac{k \pi \pm \alpha}{2}\), which are irrational multiples of \(\pi\) or sums/differences involving \(\pi\).
The elements of \(B\) are integers \(\{0,1,9,16\}\).
Therefore, \(\mathrm{A} \cap \mathrm{B}=\phi\).
\(
n(A \cup B)=n(A)+n(B)-n(A \cap B)=4+4-0=8
\)

Hint

(b) Set \(A=\{1,2,3, \ldots, 10\}\).
Set \(B=\left\{\frac{m}{n}: m, n \in A, m<n\right.\) and \(\left.\operatorname{gcd}(m, n)=1\right\}\).
How to solve?
To solve this, list all possible pairs \((m, n)\) from set \(A\) such that \(m<n\) and \(\operatorname{gcd}(m, n)=1\), then count the number of such pairs.
List all possible values for \(n\) from set \(A\).
\(
n \in\{2,3,4,5,6,7,8,9,10\} \text { since } m<n \text { and } m \geq 1 \text {. }
\)
For each value of \(n\), list all possible values for \(m\) such that \(m \in A, m<n\), and \(\operatorname{gcd}(m, n)=1\).
For \(n=2: m=1 . \operatorname{gcd}(1,2)=1\).
For \(n=3: m=1,2 . \operatorname{gcd}(1,3)=1, \operatorname{gcd}(2,3)=1\).
For \(n=4: m=1,3 . \operatorname{gcd}(1,4)=1, \operatorname{gcd}(3,4)=1\).
For \(n=5: m=1,2,3,4 . \operatorname{gcd}(1,5)=1, \operatorname{gcd}(2,5)=1, \operatorname{gcd}(3,5)=1, \operatorname{gcd}(4,5)=1\).
For \(n=6: m=1,5 . \operatorname{gcd}(1,6)=1, \operatorname{gcd}(5,6)=1\).
For \(n=7: m=1,2,3,4,5,6 . \operatorname{gcd}(m, 7)=1\) for all \(m<7\).
For \(n=8: m=1,3,5,7 . \operatorname{gcd}(1,8)=1, \operatorname{gcd}(3,8)=1, \operatorname{gcd}(5,8)=1, \operatorname{gcd}(7,8)=1\).
For \(n=9: m=1,2,4,5,7,8 . \operatorname{gcd}(1,9)=1, \operatorname{gcd}(2,9)=1, \operatorname{gcd}(4,9)=1, \operatorname{gcd}(5,9)=1\), \(\operatorname{gcd}(7,9)=1, \operatorname{gcd}(8,9)=1\).
For \(n=10: m=1,3,7,9 . \operatorname{gcd}(1,10)=1, \operatorname{gcd}(3,10)=1, \operatorname{gcd}(7,10)=1, \operatorname{gcd}(9,10)=1\).
Count the total number of valid pairs \((m, n)\).
\(
\text { Total count }=1+2+2+4+2+6+4+6+4=31 .
\)
The number of elements in set \(\boldsymbol{B}\) is 31.

Alternate explanation:

To find the number of elements in set \(B\), we consider pairs \(\left(\frac{m}{n}\right)\) where \(m, n \in A\) with \(m<n\) and \(\operatorname{gcd}(m, n)=1\).
Here’s the breakdown for each possible \(m\) :
For \(m=1\) :
Possible values for \(n\) are \(2,3,4,5,6,7,8,9,10\).
Total pairs: 9.
For \(m=2\) :
Possible values for \(n\) are \(3,5,7,9\) (since these have \(\operatorname{gcd}(2, n)=1\) ).
Total pairs: 4.
For \(m=3\) :
Possible values for \(n\) are \(4,5,7,8,10\).
Total pairs: 5.
For \(m=4\) :
Possible values for \(n\) are \(5,7,9\).
Total pairs: 3.
For \(m=5\) :
Possible values for \(n\) are \(6,7,8,9\).
Total pairs: 4.
For \(m=6\) :
Possible value for \(n\) is 7.

Total pairs: 1.
For \(m=7\) :

Possible values for \(n\) are \(8,9,10\).
Total pairs: 3.
For \(m=8\) :
Possible value for \(\boldsymbol{n}\) is \(\mathbf{9}\).

Total pairs: 1.
For \(m=9\) :
Possible value for \(n\) is 10.

Total pairs: 1.
Adding all these up, the total number of elements in set \(B\) is:
\(
9+4+5+3+4+1+3+1+1=31
\)

Hint

(a) You are finding the number of integers in the range \([100,700]\) that are neither multiples of 3 nor multiples of 4.
What’s given in the problem
The set \(A\) contains integers \(n\) such that \(100 \leq n \leq 700\).
\(n\) is not a multiple of 3.
\(n\) is not a multiple of 4.
The number of integers in a range \([a, b]\) is \(b-a+1\).
The Principle of Inclusion-Exclusion states \(|X \cup Y|=|X|+|Y|-|X \cap Y|\).
Calculate the total numbers, then subtract those divisible by 3 or 4 using the Inclusion-Exclusion Principle.
Calculate the total number of integers in the range:\(\text { Total numbers }=700-100+1=601\)
Find the count of numbers divisible by 3:
Smallest multiple of \(3 \geq 100\) is 102.
Largest multiple of \(3 \leq 700\) is 699.
Count of multiples of \(3=\frac{699-102}{3}+1=\frac{597}{3}+1=199+1=200\).
Find the count of numbers divisible by 4:
Smallest multiple of \(4 \geq 100\) is 100.
Largest multiple of \(4 \leq 700\) is 700.
Count of multiples of \(4=\frac{700-100}{4}+1=\frac{600}{4}+1=150+1=151\).
Find the count of numbers divisible by both 3 and 4 (i.e., by 12).
Smallest multiple of \(12 \geq 100\) is 108.
Largest multiple of \(12 \leq 700\) is 696.
Count of multiples of \(12=\frac{696-108}{12}+1=\frac{588}{12}+1=49+1=50\).
Apply the Inclusion-Exclusion Principle:
Numbers divisible by 3 or \(4=(\) Count of multiples of 3\()+(\) Count of multiples of 4\()-(\) Count of multiples of 12\()\). Numbers divisible by 3 or \(4=200+151-50=301\).
Number of elements in \(A=\) (Total numbers) – (Numbers divisible by 3 or 4 ).
Number of elements in \(A=601-301=300\).

Hint

(c) This is a problem involving sets and the Principle of Inclusion-Exclusion.
Here’s how to solve this problem:
Define the sets and given information:
Let \(\boldsymbol{A}\) represent the set of men who received medals in event \(\boldsymbol{A}\).
Let \(\boldsymbol{B}\) represent the set of men who received medals in event \(\boldsymbol{B}\).
Let \(\boldsymbol{C}\) represent the set of men who received medals in event \(\boldsymbol{C}\).
Given information:
\(|A|=48\)
\(|B|=25\)
\(|C|=18\)
\(|A \cup B \cup C|=60\) (total men)
\(|\boldsymbol{A} \cap \boldsymbol{B} \cap \boldsymbol{C}|=5\) (men who received medals in all three events)
Apply the Principle of Inclusion-Exclusion:
The formula for the union of three sets is:
\(
|A \cup B \cup C|=|A|+|B|+|C|-(|A \cap B|+|A \cap C|+|B \cap C|)+|A \cap B \cap C|
\)
Substitute the given values into the formula:
\(
60=48+25+18-(|A \cap B|+|A \cap C|+|B \cap C|)+5
\)
Solve for the sum of the pairwise intersections:
\(
\begin{aligned}
& 60=91-(|A \cap B|+|A \cap C|+|B \cap C|)+5 \\
& 60=96-(|A \cap B|+|A \cap C|+|B \cap C|) \\
& |A \cap B|+|A \cap C|+|B \cap C|=96-60 \\
& |A \cap B|+|A \cap C|+|B \cap C|=36
\end{aligned}
\)
Calculate the number of men who received medals in exactly two events:
The sum of the pairwise intersections (36) includes the men who received medals in all three events multiple times. Each man who received medals in all three events is counted three times in this sum (once for each pair). Therefore, to find the number of men who received medals in exactly two events, subtract three times the number of men who received medals in all three events:
Exactly two events \(=(|\boldsymbol{A} \cap \boldsymbol{B}|+|\boldsymbol{A} \cap \boldsymbol{C}|+|\boldsymbol{B} \cap \boldsymbol{C}|)-3 \times|\boldsymbol{A} \cap \boldsymbol{B} \cap \boldsymbol{C}|\)
Exactly two events \(=36-3 \times 5\)
Exactly two events \(=\mathbf{3 6 – 1 5}\)
Exactly two events \(=21\)
Therefore, 21 men received medals in exactly two of the three events.

Alternative,ly usinga  Venn diagram:

\(
\begin{aligned}
& n(A)=48, n(B)=25, n(C)=18 \\
& \text { Total men }=60=n(A \cup B \cup C) \\
& n(A \cap B \cap C)=5
\end{aligned}
\)
\(
n(A \cup B \cup C)=n(A)+n(B)+n(C)-n(A \cap B)-n(A \cap C)-n(B \cap C)+n(A \cap B \cap C)
\)
\(
60=48+25+18-[n(A \cap B)+n(B \cap C)+n(A \cap C)]+5 \Rightarrow n(A \cap B)+n(A \cap C)+n(B \cap C)=36
\)
\(
\begin{aligned}
&\text { Exactly two medals received in } 3 \text { events }\\
&=n(A \cap B)+n(B \cap C)+n(A \cap C)-3 n(A \cap B \cap C)=36-3(5)=21
\end{aligned}
\)

Hint

(c) Let H and L be the set of people suffering from heart ailment and lungs infections respectively.
\(
\begin{aligned}
& n (H)=89 \%, n (L)=98 \% \\
& n (H \cup L) \leq 100 \Rightarrow n (H)+n (L)-n (H \cap L) \leq 100
\end{aligned}
\)
Now \(100 \geq 89+98-n (H \cap L) \Rightarrow 87 \leq n(H \cap L)\)
Now \(n (H \cap L)=\min (n (H), n (L)) \leq 89\)
\(
\Rightarrow 87 \leq n (H \cap L) \leq 89
\)
\(\therefore n (H \cap L) \notin 79,81,83,85\)

Hint

(b) You are finding the number of real solutions for the given equation \((|x|-3)|x+4|=6\).
What’s given in the problem
The equation is \((|x|-3)|x+4|=6\).
We are looking for real solutions, \(x \in \mathbb{R}\).
How to solve?
Solve the equation by considering different cases based on the absolute value definitions.
Consider the case \(x \leq-4\).
In this case, \(|x|=-x\) and \(|x+4|=-(x+4)\).
The equation becomes \((-x-3)(-(x+4))=6\).
This simplifies to \((x+3)(x+4)=6\).
Expanding gives \(x^2+7 x+12=6\).
So, \(x^2+7 x+6=0\).
Factoring yields \((x+1)(x+6)=0\).
Solutions are \(x=-1\) or \(x=-6\).
Since we assumed \(x \leq-4\), only \(x=-6\) is a valid solution.
Consider the case \(-4<x<0\).
In this case, \(|x|=-x\) and \(|x+4|=x+4\).
The equation becomes \((-x-3)(x+4)=6\).
Expanding gives \(-x^2-7 x-12=6\).
So, \(x^2+7 x+18=0\).
Calculate the discriminant \(D=b^2-4 a c=7^2-4(1)(18)=49-72=-23\).
Since \(D<0\), there are no real solutions in this case.
Consider the case \(x \geq 0\).
In this case, \(|x|=x\) and \(|x+4|=x+4\).
The equation becomes \((x-3)(x+4)=6\).
Expanding gives \(x^2+x-12=6\).
So, \(x^2+x-18=0\).
Use the quadratic formula \(x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}\).
\(
\begin{aligned}
& x=\frac{-1 \pm \sqrt{1^2-4(1)(-18)}}{2(1)} \\
& x=\frac{-1 \pm \sqrt{1+72}}{2} \\
& x=\frac{-1 \pm \sqrt{73}}{2}
\end{aligned}
\)
Since we assumed \(x \geq 0\), only \(x=\frac{-1+\sqrt{73}}{2}\) is a valid solution. Note that \(\sqrt{73} \approx 8.5\), so \(\frac{-1+8.5}{2}>0\).
The number of elements in the set is 2.

Hint

(d) As roots are real so, \(D \geq 0\)
\(
\begin{aligned}
& (m+1)^2-4(m+4) \geq 0 \\
& \Rightarrow m^2-2 m-15 \geq 0 \\
& \Rightarrow(m-5)(m+3) \geq 0
\end{aligned}
\)
\(
m \in (-\infty,-3] \cup[5, \infty)
\)

\(
\mathrm{A}=\{(-\infty,-3] \cup[5, \infty)\}
\)
\(
\text { Given } B=[-3,5)
\)
\(
A-B=(-\infty,-3) \cup[5, \infty)
\)

Hint

(d) Set \(A=\{x \in R:|x|<2\}\).
Set \(B=\{x \in R:|x-2| \geq 3\}\).
How to solve
First, express sets \(A\) and \(B\) as intervals, then perform the set operations and compare with the given options.
Express set \(A\) as an interval.
The inequality \(|x|<2\) means that \(x\) is between -2 and 2 . \(A=(-2,2)\).

\(
\begin{aligned}
&\text { The inequality }|x-2| \geq 3 \text { means } x-2 \leq-3 \text { or } x-2 \geq 3 \text {. }\\
&\begin{aligned}
& x-2 \leq-3 \Longrightarrow x \leq-1 \\
& x-2 \geq 3 \Longrightarrow x \geq 5 \\
& B=(-\infty,-1] \cup[5, \infty)
\end{aligned}
\end{aligned}
\)

\(
\begin{aligned}
&\begin{aligned}
& B-A=B \cap A^c . \\
& A^c=(-\infty,-2] \cup[2, \infty) . \\
& B-A=((-\infty,-1] \cup[5, \infty)) \cap((-\infty,-2] \cup[2, \infty)) . \\
& B-A=(-\infty,-2] \cup[5, \infty) . \\
& R-(-2,5)=(-\infty,-2] \cup[5, \infty) .
\end{aligned}\\
&\text { Option (d) states } B-A=R-(-2,5) \text {, which is correct. }
\end{aligned}
\)

Hint

(d) Let the population of the city be 100.

Then, \(n(A)=25, n(B)=20\) and \(n(A \cap B)=8\)
So, \(n(A \cap \bar{B})=17\) and \(n(\bar{A} \cap B)=12\)
According to the question, Percentage of the population who look into advertisement is
\(
\begin{aligned}
& =\left[\frac{30}{100} \times n(A \cap \bar{B})\right]+\left[\frac{40}{100} \times n(\bar{A} \cap B)\right]+\left[\frac{50}{100} \times n(A \cap B)\right] \\
& =\left(\frac{30}{100} \times 17\right)+\left(\frac{40}{100} \times 12\right)+\left(\frac{50}{100} \times 8\right) \\
& =5.1+4.8+4=13.9
\end{aligned}
\)

Hint

(c) Set \(A\) is defined by the inequalities: \(|a-5|<1\) and \(|b-5|<1\).
Set \(B\) is defined by the inequality: \(4(a-6)^2+9(b-5)^2 \leq 36\).
The inequality \(|x-c|<r\) describes an open interval \((c-r, c+r)\).
The inequality \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} \leq 1\) describes an ellipse centered at \((h, k)\) with semi-axes \(a\) and \(b\).
How to solve?
Analyze the geometric shapes represented by sets \(A\) and \(B\) and compare their regions.

Analyze set \(A\):
The inequality \(|a-5|<1\) means \(-1<a-5<1\), which simplifies to \(4<a<6\).
The inequality \(|b-5|<1\) means \(-1<b-5<1\), which simplifies to \(4<b<6\).
Set \(A\) represents an open square region with vertices at \((4,4),(6,4),(6,6)\), and \((4,6)\).

Analyze set \(B\):
Divide the inequality \(4(a-6)^2+9(b-5)^2 \leq 36\) by 36.
This gives \(\frac{4(a-6)^2}{36}+\frac{9(b-5)^2}{36} \leq \frac{36}{36}\).
Simplify to \(\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4} \leq 1\).
Set \(B\) represents an ellipse centered at \((6,5)\) with a horizontal semi-axis of \(\sqrt{9}=3\) and a vertical semi-axis of \(\sqrt{4}=2\).

Compare the regions:
The square \(A\) has \(a\) values from 4 to 6 and \(b\) values from 4 to 6.
The ellipse \(B\) has \(a\) values from \(6-3=3\) to \(6+3=9\) and \(b\) values from \(5-2=3\) to \(5+2=7\)
Hence, we can say entire set \(A\) is inside of the set \(B\).
\(
\text { Set } A \text { is a subset of set } B \text {. }
\)