Word Problem - Set Theory Questions in English

(B) Given $A = \{x : x \in R, -1 < x < 1\} = (-1, 1).$
Given $B = \{x : x \in R, x - 1 \le -1 \text{ or } x - 1 \ge 1\} = \{x : x \le 0 \text{ or } x \ge 2\} = (-\infty, 0] \cup [2, \infty).$
Now,$A \cup B = (-1, 1) \cup (-\infty, 0] \cup [2, \infty).$
Combining these intervals,we get $A \cup B = (-\infty, 1) \cup [2, \infty).$
We are given $A \cup B = R - D.$
Therefore,$D = R - (A \cup B) = R - ((-\infty, 1) \cup [2, \infty)) = [1, 2).$
Thus,$D = \{x : x \in R, 1 \le x < 2\}.$

(C) By De Morgan's Law,$n(A^c \cap B^c) = n((A \cup B)^c)$.
Using the property of complements,$n((A \cup B)^c) = n(U) - n(A \cup B)$.
We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Substituting the given values: $n(A \cup B) = 200 + 300 - 100 = 400$.
Therefore,$n(A^c \cap B^c) = 700 - 400 = 300$.

(B) Given total families $N = 10,000$.
$n(A) = 40\% \text{ of } 10,000 = 4,000$
$n(B) = 20\% \text{ of } 10,000 = 2,000$
$n(C) = 10\% \text{ of } 10,000 = 1,000$
$n(A \cap B) = 5\% \text{ of } 10,000 = 500$
$n(B \cap C) = 3\% \text{ of } 10,000 = 300$
$n(A \cap C) = 4\% \text{ of } 10,000 = 400$
$n(A \cap B \cap C) = 2\% \text{ of } 10,000 = 200$
To find the number of families that buy newspaper $A$ only,we use the formula:
$n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$
$n(A \text{ only}) = 4,000 - 500 - 400 + 200$
$n(A \text{ only}) = 4,000 - 700 = 3,300$

(C) Let $C$ be the set of people travelling by car and $B$ be the set of people travelling by bus.
Given: $n(C) = 20$,$n(B) = 50$,and $n(C \cap B) = 10$.
We need to find the percentage of people travelling by car or bus,which is $n(C \cup B)$.
Using the formula: $n(C \cup B) = n(C) + n(B) - n(C \cap B)$.
Substituting the values: $n(C \cup B) = 20 + 50 - 10 = 60$.
Hence,the required percentage is $60\%$.

(D) Let $M$,$P$,and $C$ represent the sets of students studying Mathematics,Physics,and Chemistry respectively.
Given: $n(M) = 23, n(P) = 24, n(C) = 19$.
$n(M \cap P) = 12, n(M \cap C) = 9, n(P \cap C) = 7$.
$n(M \cap P \cap C) = 4$.
The number of students studying exactly one subject is given by:
$n(\text{Only } M) = n(M) - n(M \cap P) - n(M \cap C) + n(M \cap P \cap C) = 23 - 12 - 9 + 4 = 6$.
$n(\text{Only } P) = n(P) - n(P \cap M) - n(P \cap C) + n(P \cap M \cap C) = 24 - 12 - 7 + 4 = 9$.
$n(\text{Only } C) = n(C) - n(C \cap M) - n(C \cap P) + n(C \cap M \cap P) = 19 - 9 - 7 + 4 = 7$.
Total number of students who have taken exactly one subject $= 6 + 9 + 7 = 22$.

(B) Given the equation $A \cup \{1, 2\} = \{1, 2, 3, 5, 9\}$.
To find the smallest set $A$,we must include all elements present in the result set that are not in the set $\{1, 2\}$.
The elements in the result set are $\{1, 2, 3, 5, 9\}$.
The elements already present in the union are $\{1, 2\}$.
Therefore,the set $A$ must contain at least the elements $\{3, 5, 9\}$.
Thus,the smallest set $A$ is $\{3, 5, 9\}$.

(C) Given that $A \subseteq B$,which means every element of set $A$ is also an element of set $B$.
Therefore,the union of $A$ and $B$ is simply set $B$,i.e.,$A \cup B = B$.
Thus,$n(A \cup B) = n(B)$.
Since $n(B) = 6$,we have $n(A \cup B) = 6$.

(A) Let the total number of combatants be $100$.
Let $A, B, C, D$ be the sets of combatants who lost an eye,an ear,an arm,and a leg respectively.
Given: $n(A) = 70, n(B) = 80, n(C) = 75, n(D) = 85$.
The number of people who did $NOT$ lose these parts are:
$n(A^c) = 100 - 70 = 30$
$n(B^c) = 100 - 80 = 20$
$n(C^c) = 100 - 75 = 25$
$n(D^c) = 100 - 85 = 15$
The number of people who lost all four limbs is at least $100 - (n(A^c) + n(B^c) + n(C^c) + n(D^c))$.
Minimum $x = 100 - (30 + 20 + 25 + 15) = 100 - 90 = 10$.

(D) Let $C, H, B$ represent the sets of boys who play cricket,hockey,and basketball,respectively.
Given: $n(C) = 224, n(H) = 240, n(B) = 336$.
Intersections: $n(H \cap B) = 64, n(C \cap B) = 80, n(C \cap H) = 40, n(C \cap H \cap B) = 24$.
Using the Principle of Inclusion-Exclusion,the number of boys who play at least one game is:
$n(C \cup H \cup B) = n(C) + n(H) + n(B) - [n(C \cap H) + n(H \cap B) + n(C \cap B)] + n(C \cap H \cap B)$
$n(C \cup H \cup B) = 224 + 240 + 336 - (40 + 64 + 80) + 24$
$n(C \cup H \cup B) = 800 - 184 + 24 = 640$.
The number of boys who did not play any game is:
$n(U) - n(C \cup H \cup B) = 800 - 640 = 160$.

(C) Let $A$ denote the set of Americans who like cheese and $B$ denote the set of Americans who like apples.
Let the total population be $100$.
Then $n(A) = 63$ and $n(B) = 76$.
We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Since $n(A \cup B) \le 100$,we have $n(A) + n(B) - n(A \cap B) \le 100$.
$63 + 76 - n(A \cap B) \le 100 \implies 139 - n(A \cap B) \le 100 \implies n(A \cap B) \ge 39$.
Also,$A \cap B \subseteq A$ and $A \cap B \subseteq B$,so $n(A \cap B) \le n(A)$ and $n(A \cap B) \le n(B)$.
Thus,$n(A \cap B) \le 63$ and $n(A \cap B) \le 76$,which implies $n(A \cap B) \le 63$.
Combining these,we get $39 \le n(A \cap B) \le 63$.
Therefore,$39 \le x \le 63$.

(A) Let $M$ be the set of teachers who teach mathematics and $P$ be the set of teachers who teach physics.
Given that $n(M \cup P) = 20$,$n(M) = 12$,and $n(M \cap P) = 4$.
Using the formula $n(M \cup P) = n(M) + n(P) - n(M \cap P)$,
$20 = 12 + n(P) - 4$
$20 = 8 + n(P)$
$n(P) = 20 - 8 = 12$.
Thus,the number of teachers teaching physics is $12$.

(A) Let $C, H, F$ denote the sets of members who are on the cricket team,hockey team,and football team respectively.
We are given $n(C) = 21, n(H) = 26, n(F) = 29$.
Also,$n(H \cap C) = 14, n(H \cap F) = 15, n(F \cap C) = 12$ and $n(C \cap H \cap F) = 8$.
We need to find the total number of members,which is $n(C \cup H \cup F)$.
Using the principle of inclusion-exclusion:
$n(C \cup H \cup F) = n(C) + n(H) + n(F) - n(C \cap H) - n(H \cap F) - n(F \cap C) + n(C \cap H \cap F)$
$n(C \cup H \cup F) = (21 + 26 + 29) - (14 + 15 + 12) + 8$
$n(C \cup H \cup F) = 76 - 41 + 8 = 43$.
Thus,there are $43$ members in total.

(D) Let $M$ be the set of students who passed in Mathematics and $P$ be the set of students who passed in Physics.
Given: $n(M) = 55$,$n(P) = 67$,and $n(M \cup P) = 100$.
Using the formula: $n(M \cup P) = n(M) + n(P) - n(M \cap P)$.
$100 = 55 + 67 - n(M \cap P)$.
$100 = 122 - n(M \cap P)$.
$n(M \cap P) = 122 - 100 = 22$.
The number of students who passed in Physics only is given by $n(P \setminus M) = n(P) - n(M \cap P)$.
$n(P \setminus M) = 67 - 22 = 45$.

(A) Let $N, P, H$ be the sets of students taking needle work,physics,and history respectively.
Given $n(N) = 12, n(P) = 16, n(H) = 18$ and $n(N \cup P \cup H) = 30$.
Since no one takes all three,$n(N \cap P \cap H) = 0$.
The formula for the union of three sets is:
$n(N \cup P \cup H) = n(N) + n(P) + n(H) - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + n(N \cap P \cap H)$
Substituting the values:
$30 = 12 + 16 + 18 - [n(N \cap P) + n(P \cap H) + n(N \cap H)] + 0$
$30 = 46 - [n(N \cap P) + n(P \cap H) + n(N \cap H)]$
$n(N \cap P) + n(P \cap H) + n(N \cap H) = 46 - 30 = 16$.
The number of students taking exactly two subjects is given by:
$n(N \cap P) + n(P \cap H) + n(N \cap H) - 3n(N \cap P \cap H)$
$= 16 - 3(0) = 16$.

(B) We are given that $n(A) = 3$ and $n(B) = 6$.
The formula for the number of elements in the union of two sets is $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
To minimize $n(A \cup B)$,we must maximize $n(A \cap B)$.
The maximum possible value for $n(A \cap B)$ is the number of elements in the smaller set,which is $3$ (since $A \subset B$ is possible).
Therefore,the minimum number of elements in $A \cup B$ is $3 + 6 - 3 = 6$.

(D) Let $S$ be the sum of integers from $1$ to $100$: $S = \frac{100}{2}(1 + 100) = 5050$.
Let $S_1$ be the sum of integers divisible by $3$: $S_1 = 3 + 6 + ... + 99 = 3(1 + 2 + ... + 33) = 3 \times \frac{33 \times 34}{2} = 1683$.
Let $S_2$ be the sum of integers divisible by $5$: $S_2 = 5 + 10 + ... + 100 = 5(1 + 2 + ... + 20) = 5 \times \frac{20 \times 21}{2} = 1050$.
Let $S_3$ be the sum of integers divisible by both $3$ and $5$ (i.e.,divisible by $15$): $S_3 = 15 + 30 + ... + 90 = 15(1 + 2 + ... + 6) = 15 \times \frac{6 \times 7}{2} = 315$.
The sum of integers divisible by $3$ or $5$ is $S_1 + S_2 - S_3 = 1683 + 1050 - 315 = 2418$.
The required sum is $S - (S_1 + S_2 - S_3) = 5050 - 2418 = 2632$.

(D) Let $S$ be the set of the first $100$ natural numbers,so $n(S) = 100$.
Let $A$ be the event that the number is even. The even numbers are $2, 4, 6, \dots, 100$. Thus,$n(A) = 50$.
Let $B$ be the event that the number is divisible by $5$. The numbers divisible by $5$ are $5, 10, 15, \dots, 100$. Thus,$n(B) = 20$.
The intersection $A \cap B$ represents numbers that are both even and divisible by $5$,i.e.,divisible by $10$. These are $10, 20, 30, \dots, 100$. Thus,$n(A \cap B) = 10$.
Using the formula for the union of two sets: $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 50 + 20 - 10 = 60$.
The required probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{60}{100} = \frac{3}{5}$.

(B) Total number of balls $= 30$.
Let $A$ be the set of numbers that are multiples of $5$ between $1$ and $30$:
$A = \{5, 10, 15, 20, 25, 30\}$,so $n(A) = 6$.
Let $B$ be the set of numbers that are multiples of $7$ between $1$ and $30$:
$B = \{7, 14, 21, 28\}$,so $n(B) = 4$.
The intersection $A \cap B$ contains numbers that are multiples of both $5$ and $7$ (i.e.,multiples of $35$):
$A \cap B = \emptyset$,so $n(A \cap B) = 0$.
The number of favorable outcomes is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 6 + 4 - 0 = 10$.
The probability $P(A \cup B) = \frac{n(A \cup B)}{\text{Total outcomes}} = \frac{10}{30} = \frac{1}{3}$.

(D) Let $S$ be the set of numbers from $1$ to $90$. The total number of outcomes is $n(S) = 90$.
Let $A$ be the set of numbers divisible by $6$. The number of elements is $n(A) = \lfloor \frac{90}{6} \rfloor = 15$.
Let $B$ be the set of numbers divisible by $8$. The number of elements is $n(B) = \lfloor \frac{90}{8} \rfloor = 11$.
The numbers divisible by both $6$ and $8$ are divisible by $\text{lcm}(6, 8) = 24$. The number of elements is $n(A \cap B) = \lfloor \frac{90}{24} \rfloor = 3$.
Using the inclusion-exclusion principle,the number of elements divisible by $6$ or $8$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 15 + 11 - 3 = 23$.
The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{23}{90}$.

(D) Let $E$ be the set of people who read the English newspaper and $H$ be the set of people who read the Hindi newspaper.
Given: $P(E) = 20\%$,$P(H) = 40\%$,and $P(E \cap H) = 5\%$.
We need to find the percentage of people who do not read either paper,which is $P(E^c \cap H^c) = 1 - P(E \cup H)$.
Using the addition theorem: $P(E \cup H) = P(E) + P(H) - P(E \cap H)$.
$P(E \cup H) = 20\% + 40\% - 5\% = 55\%$.
Therefore,the percentage of people who do not read either paper is $100\% - 55\% = 45\%$.

(A) Let $M$ be the set of students who passed in Mathematics and $S$ be the set of students who passed in Statistics.
Given: $n(U) = 125$,$n(M) = 70$,$n(S) = 55$,and $n(M \cap S) = 30$.
Students who passed in only Mathematics = $n(M) - n(M \cap S) = 70 - 30 = 40$.
Students who passed in only Statistics = $n(S) - n(M \cap S) = 55 - 30 = 25$.
Total students who passed in exactly one subject = $40 + 25 = 65$.
The probability that a student selected at random passed in only one subject is $\frac{65}{125} = \frac{13}{25}$.

(D) The inclusion-exclusion principle states that $P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)$.
Substituting the given values: $P(A \cup B \cup C) = 0.3 + 0.4 + 0.8 - 0.08 - x - 0.28 + 0.09 = 1.23 - x$.
Given $P(A \cup B \cup C) \ge 0.75$,we have $1.23 - x \ge 0.75$,which implies $x \le 0.48$.
Also,for any events $B$ and $C$,$P(BC) \ge P(ABC) = 0.09$.
Furthermore,$P(B \cup C) = P(B) + P(C) - P(BC) = 0.4 + 0.8 - x = 1.2 - x \le 1$,so $x \ge 0.2$.
Additionally,$P(BC) \ge P(ABC) + P(AB) - P(A) = 0.09 + 0.08 - 0.3 = -0.13$ (always true).
However,the condition $P(BC) \ge P(ABC) + P(AC) - P(A) = 0.09 + 0.28 - 0.3 = 0.07$ must hold.
Also,$P(B \cap C \cap A^c) = P(BC) - P(ABC) = x - 0.09 \ge 0$,so $x \ge 0.09$.
Since $P(B \cup C) \le 1$,$x \ge 0.2$. Combining these,$0.2 \le x \le 0.48$. None of the options match this range.

(B) Let $M, P$,and $C$ be the events of passing in Mathematics,Physics,and Chemistry respectively.
Given:
$P(M \cup P \cup C) = \frac{75}{100} = \frac{3}{4}$
$P(\text{at least two}) = P(M \cap P) + P(P \cap C) + P(M \cap C) - 2P(M \cap P \cap C) = \frac{50}{100} = \frac{1}{2}$
$P(\text{exactly two}) = P(M \cap P) + P(P \cap C) + P(M \cap C) - 3P(M \cap P \cap C) = \frac{40}{100} = \frac{2}{5}$
Subtracting the two equations:
$P(M \cap P \cap C) = \frac{1}{2} - \frac{2}{5} = \frac{1}{10}$
Using the inclusion-exclusion principle:
$P(M \cup P \cup C) = (m + p + c) - (mp + pc + mc) + mpc = \frac{3}{4}$
From the 'at least two' condition:
$(mp + pc + mc) - 2mpc = \frac{1}{2} \Rightarrow (mp + pc + mc) = \frac{1}{2} + 2(\frac{1}{10}) = \frac{7}{10}$
Substituting these into the inclusion-exclusion equation:
$(m + p + c) - \frac{7}{10} + \frac{1}{10} = \frac{3}{4}$
$m + p + c = \frac{3}{4} + \frac{6}{10} = \frac{15 + 12}{20} = \frac{27}{20}$.

(B) Let $A$ be the set of integers divisible by $2$ and $B$ be the set of integers divisible by $5$ in the range $[1, 100]$.
We need to find the sum of elements in $A \cup B$,which is given by $S(A \cup B) = S(A) + S(B) - S(A \cap B)$.
$S(A)$ is the sum of multiples of $2$: $2, 4, 6, \dots, 100$. This is an $A$.$P$. with $n=50$,$a=2$,$l=100$. $S(A) = \frac{50}{2}(2 + 100) = 25 \times 102 = 2550$.
$S(B)$ is the sum of multiples of $5$: $5, 10, 15, \dots, 100$. This is an $A$.$P$. with $n=20$,$a=5$,$l=100$. $S(B) = \frac{20}{2}(5 + 100) = 10 \times 105 = 1050$.
$S(A \cap B)$ is the sum of multiples of $10$ ($LCM$ of $2$ and $5$): $10, 20, 30, \dots, 100$. This is an $A$.$P$. with $n=10$,$a=10$,$l=100$. $S(A \cap B) = \frac{10}{2}(10 + 100) = 5 \times 110 = 550$.
Therefore,$S(A \cup B) = 2550 + 1050 - 550 = 3050$.

(C) Let $S$ be the set $A$ with $|A| = 2n + 1$. We want to find the number of subsets with at least $n$ elements. This is given by the sum: $\sum_{k=n}^{2n+1} \binom{2n+1}{k}$.
We know that the total number of subsets is $\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$.
Let $X = \sum_{k=n}^{2n+1} \binom{2n+1}{k}$.
Using the property $\binom{m}{r} = \binom{m}{m-r}$,we have $\sum_{k=0}^{n-1} \binom{2n+1}{k} = \sum_{k=n+2}^{2n+1} \binom{2n+1}{k}$.
Also,$\sum_{k=0}^{2n+1} \binom{2n+1}{k} = \binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n} + \binom{2n+1}{n+1} + \dots + \binom{2n+1}{2n+1} = 2^{2n+1}$.
Since $\binom{2n+1}{n} = \binom{2n+1}{n+1}$,the sum of the first $n+1$ terms is equal to the sum of the last $n+1$ terms.
Let $S_1 = \sum_{k=0}^{n} \binom{2n+1}{k}$ and $S_2 = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k}$.
Since $S_1 + S_2 = 2^{2n+1}$ and $S_1 = S_2$ (by symmetry),we have $S_2 = \frac{2^{2n+1}}{2} = 2^{2n}$.
Thus,the number of subsets with at least $n+1$ elements is $2^{2n}$.
The question asks for at least $n$ elements,which is $\binom{2n+1}{n} + S_2 = \binom{2n+1}{n} + 2^{2n}$.

(D) Given: $P(A^c) = 0.3 \implies P(A) = 1 - 0.3 = 0.7$.
$P(B) = 0.4 \implies P(B^c) = 1 - 0.4 = 0.6$.
$P(A \cap B^c) = 0.5$.
We know that $P(A) = P(A \cap B) + P(A \cap B^c)$,so $0.7 = P(A \cap B) + 0.5 \implies P(A \cap B) = 0.2$.
We need to find $P[B / (A \cup B)^c]$. By De Morgan's Law,$(A \cup B)^c = A^c \cap B^c$.
$P[B / (A^c \cap B^c)] = \frac{P(B \cap (A^c \cap B^c))}{P(A^c \cap B^c)}$.
Since $B \cap B^c = \emptyset$,the numerator $P(B \cap A^c \cap B^c) = P(\emptyset) = 0$.
Thus,$P[B / (A \cup B)^c] = 0$.

(A) Let $P(A)$ be the probability of failing in Physics and $P(B)$ be the probability of failing in Mathematics.
Given $P(A) = 20\% = 0.2$ and $P(B) = 10\% = 0.1$.
Assuming the events are independent,the probability of failing in at least one subject is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Since the events are independent,$P(A \cap B) = P(A) \times P(B) = 0.2 \times 0.1 = 0.02$.
Therefore,$P(A \cup B) = 0.2 + 0.1 - 0.02 = 0.3 - 0.02 = 0.28$.
Converting to percentage,$0.28 \times 100 = 28\%$.

(D) The formula for the union of two events is $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = 0.5 + 0.7 - 0.6 = 0.6$.
However,it is important to note that the given values $P(A) = 0.5$ and $P(A \cap B) = 0.6$ are mathematically inconsistent because $P(A \cap B)$ cannot be greater than $P(A)$ since $(A \cap B) \subseteq A$ implies $P(A \cap B) \leq P(A)$.

(D) Let $S$ be the set of the first $200$ positive integers,so $n(S) = 200$.
Let $A$ be the set of integers divisible by $6$. Then $n(A) = \lfloor 200/6 \rfloor = 33$.
Let $B$ be the set of integers divisible by $8$. Then $n(B) = \lfloor 200/8 \rfloor = 25$.
The intersection $A \cap B$ contains integers divisible by $\text{lcm}(6, 8) = 24$. Then $n(A \cap B) = \lfloor 200/24 \rfloor = 8$.
Using the inclusion-exclusion principle,the number of integers divisible by $6$ or $8$ is $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 33 + 25 - 8 = 50$.
The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{50}{200} = \frac{1}{4}$.

(C) Let $|S|$ be the number of elements in the set $S$.
Given that each of the $30$ sets $A_i$ has $5$ elements,the sum of the number of elements in all $A_i$ is $30 \times 5 = 150$.
Since each element of $S$ belongs to exactly $10$ of the $A_i$'s,we have $|S| \times 10 = 150$,which implies $|S| = 15$.
Similarly,for the sets $B_j$,each of the $n$ sets has $3$ elements,so the sum of the number of elements in all $B_j$ is $n \times 3 = 3n$.
Since each element of $S$ belongs to exactly $9$ of the $B_j$'s,we have $|S| \times 9 = 3n$.
Substituting $|S| = 15$,we get $15 \times 9 = 3n$.
$135 = 3n \Rightarrow n = 45$.

(C) Let $P$ be the set of families owning a phone and $C$ be the set of families owning a car.
Given: $n(P) = 25\%$,$n(C) = 15\%$,and $n(P^c \cap C^c) = 65\%$.
We know that $n(P^c \cap C^c) = n((P \cup C)^c) = 100\% - n(P \cup C)$.
Therefore,$n(P \cup C) = 100\% - 65\% = 35\%$. This confirms statement $2$ is correct.
Using the formula $n(P \cup C) = n(P) + n(C) - n(P \cap C)$:
$35\% = 25\% + 15\% - n(P \cap C)$.
$n(P \cap C) = 40\% - 35\% = 5\%$.
Since $5\%$ of the total families $= 2000$,the total number of families $= \frac{2000 \times 100}{5} = 40,000$. This confirms statement $3$ is correct.
Statement $1$ claims $10\%$ own both,which is incorrect as we calculated $5\%$.
Thus,statements $2$ and $3$ are correct.

(B) Let $N_i$ be the number of students who gave exactly $i$ wrong answers.
Since no student gave more than $k$ wrong answers,the total number of wrong answers is $\sum_{i=1}^{k} i \cdot N_i$.
We are given that $a_i$ is the number of students who gave at least $i$ wrong answers.
Thus,$a_i = N_i + N_{i+1} + \dots + N_k$.
This implies $N_i = a_i - a_{i+1}$ for $i < k$ and $N_k = a_k$.
The total number of wrong answers is $1 \cdot N_1 + 2 \cdot N_2 + \dots + k \cdot N_k$.
Substituting the values of $N_i$:
$= 1(a_1 - a_2) + 2(a_2 - a_3) + \dots + (k-1)(a_{k-1} - a_k) + k(a_k)$
$= a_1 + (2-1)a_2 + (3-2)a_3 + \dots + (k-(k-1))a_k$
$= a_1 + a_2 + a_3 + \dots + a_k$.

(C) Let the number of newspapers be $n$.
Since every student reads $5$ newspapers,the total number of readings is $300 \times 5 = 1500$.
Since every newspaper is read by $60$ students,the total number of readings is also $60 \times n$.
Equating the two,we get $60n = 1500$.
Therefore,$n = \frac{1500}{60} = 25$.

(B) The formula for the number of elements in the union of two sets is $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
Given $n(A) = 3$ and $n(B) = 6$,we have $n(A \cup B) = 3 + 6 - n(A \cap B) = 9 - n(A \cap B)$.
To minimize $n(A \cup B)$,we must maximize $n(A \cap B)$.
The maximum number of elements in $A \cap B$ is $n(A) = 3$ (since $A \subseteq B$ is possible).
Therefore,the minimum number of elements in $A \cup B$ is $9 - 3 = 6$.

(D) We use the formula for the union of two sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
Substituting the given values:
$110 = 70 + 60 - n(A \cap B)$
$110 = 130 - n(A \cap B)$
$n(A \cap B) = 130 - 110$
$n(A \cap B) = 20$

(B) Given total families $N = 10,000$.
$n(A) = 40\% \text{ of } 10,000 = 4,000$
$n(B) = 20\% \text{ of } 10,000 = 2,000$
$n(C) = 10\% \text{ of } 10,000 = 1,000$
$n(A \cap B) = 5\% \text{ of } 10,000 = 500$
$n(B \cap C) = 3\% \text{ of } 10,000 = 300$
$n(A \cap C) = 4\% \text{ of } 10,000 = 400$
$n(A \cap B \cap C) = 2\% \text{ of } 10,000 = 200$
We need to find the number of families that buy only newspaper $A$,which is given by $n(A \cap B^c \cap C^c)$.
Using the formula: $n(A \text{ only}) = n(A) - n(A \cap B) - n(A \cap C) + n(A \cap B \cap C)$
$n(A \text{ only}) = 4,000 - 500 - 400 + 200$
$n(A \text{ only}) = 4,000 - 900 + 200 = 3,300$.

(C) Let $C$ be the set of people who travel by car and $B$ be the set of people who travel by bus.
Given: $n(C) = 20\%$,$n(B) = 50\%$,and $n(C \cap B) = 10\%$.
We need to find the percentage of people who travel by car or bus,which is $n(C \cup B)$.
Using the formula: $n(C \cup B) = n(C) + n(B) - n(C \cap B)$.
$n(C \cup B) = 20 + 50 - 10 = 60$.
Therefore,the percentage of people who travel by car or bus is $60\%$.

(A) Let $M, P, C$ represent the sets of students studying Mathematics,Physics,and Chemistry respectively.
Given: $n(M) = 23, n(P) = 24, n(C) = 19$
$n(M \cap P) = 12, n(M \cap C) = 9, n(P \cap C) = 7$
$n(M \cap P \cap C) = 4$
Number of students studying only Mathematics:
$n(M \text{ only}) = n(M) - n(M \cap P) - n(M \cap C) + n(M \cap P \cap C) = 23 - 12 - 9 + 4 = 6$
Number of students studying only Physics:
$n(P \text{ only}) = n(P) - n(P \cap M) - n(P \cap C) + n(M \cap P \cap C) = 24 - 12 - 7 + 4 = 9$
Number of students studying only Chemistry:
$n(C \text{ only}) = n(C) - n(C \cap M) - n(C \cap P) + n(M \cap P \cap C) = 19 - 9 - 7 + 4 = 7$
Total students studying exactly one subject:
$6 + 9 + 7 = 22$

(B) Given that $A \cup \{1, 2\} = \{1, 2, 3, 5, 9\}$.
This implies that the elements $3, 5,$ and $9$ must be present in set $A$ because they are in the union but not in the set $\{1, 2\}$.
The elements $1$ and $2$ may or may not be in set $A$.
Thus,$A$ must contain $\{3, 5, 9\}$ as a subset.
Looking at the options,option $B$ is $\{3, 5, 9\}$,which satisfies the condition $A \cup \{1, 2\} = \{1, 2, 3, 5, 9\}$.

(C) Let $T$ be the set of families with a telephone and $C$ be the set of families with a car.
Given: $n(T) = 25\%$,$n(C) = 15\%$,and $n(T^c \cap C^c) = 65\%$.
By De Morgan's Law,$n(T \cup C)^c = 65\%$,so $n(T \cup C) = 100\% - 65\% = 35\%$.
Using the formula $n(T \cup C) = n(T) + n(C) - n(T \cap C)$:
$35\% = 25\% + 15\% - n(T \cap C)$
$35\% = 40\% - n(T \cap C)$
$n(T \cap C) = 5\%$.
Statement $1$ says $10\%$ have both,which is false $(5\% \neq 10\%)$.
Statement $2$ says $35\%$ have either a car or a telephone,which is $n(T \cup C) = 35\%$. This is true.
Statement $3$: Since $5\%$ of total families $= 2000$,let $N$ be the total number of families.
$0.05 \times N = 2000 \implies N = \frac{2000}{0.05} = 40,000$. This is true.
Thus,statements $2$ and $3$ are true.

(D) Let $C$,$H$,and $B$ represent the sets of boys who play cricket,hockey,and basketball,respectively.
Given:
$n(U) = 800$
$n(C) = 224, n(H) = 240, n(B) = 336$
$n(B \cap H) = 64, n(C \cap B) = 80, n(C \cap H) = 40$
$n(C \cap H \cap B) = 24$
Using the Principle of Inclusion-Exclusion:
$n(C \cup H \cup B) = n(C) + n(H) + n(B) - [n(C \cap H) + n(H \cap B) + n(C \cap B)] + n(C \cap H \cap B)$
$n(C \cup H \cup B) = 224 + 240 + 336 - [40 + 64 + 80] + 24$
$n(C \cup H \cup B) = 800 - 184 + 24 = 640$
The number of boys who do not play any game is $n(U) - n(C \cup H \cup B) = 800 - 640 = 160$.

(C) Let $C$ be the set of people who like cheese and $A$ be the set of people who like apples.
Given: $n(C) = 63\%$ and $n(A) = 76\%$.
We know that $n(C \cup A) = n(C) + n(A) - n(C \cap A)$.
Since $n(C \cup A) \le 100\%$,we have $63 + 76 - x \le 100$,which implies $139 - x \le 100$,so $x \ge 39$.
Also,the number of people who like both cannot exceed the number of people who like either individual item,so $x \le n(C)$ and $x \le n(A)$.
Thus,$x \le 63$ and $x \le 76$,which means $x \le 63$.
Combining these,we get $39 \le x \le 63$.

(B) Let $M$ be the set of teachers who teach Mathematics and $P$ be the set of teachers who teach Physics.
Given that the total number of teachers is $n(M \cup P) = 20$.
The number of teachers who teach Mathematics is $n(M) = 12$.
The number of teachers who teach both subjects is $n(M \cap P) = 4$.
We know that $n(M \cup P) = n(M) + n(P) - n(M \cap P)$.
Substituting the values: $20 = 12 + n(P) - 4$.
$20 = 8 + n(P)$,which gives $n(P) = 12$.
The number of teachers who teach only Physics is $n(P) - n(M \cap P) = 12 - 4 = 8$.

(A) Let $C$,$H$,and $F$ represent the sets of students in the cricket,hockey,and football teams,respectively.
Given:
$n(C) = 21$,$n(H) = 26$,$n(F) = 29$
$n(H \cap C) = 14$,$n(H \cap F) = 15$,$n(F \cap C) = 12$
$n(H \cap C \cap F) = 8$
We need to find the total number of students,which is $n(C \cup H \cup F)$.
Using the Principle of Inclusion-Exclusion:
$n(C \cup H \cup F) = n(C) + n(H) + n(F) - [n(H \cap C) + n(H \cap F) + n(F \cap C)] + n(H \cap C \cap F)$
$n(C \cup H \cup F) = 21 + 26 + 29 - [14 + 15 + 12] + 8$
$n(C \cup H \cup F) = 76 - 41 + 8$
$n(C \cup H \cup F) = 35 + 8 = 43$
Thus,the total number of students is $43$.

(D) Let $M$ be the set of students who passed in Mathematics and $P$ be the set of students who passed in Physics.
Given: $n(M \cup P) = 100$,$n(M) = 55$,$n(P) = 67$.
We know that $n(M \cup P) = n(M) + n(P) - n(M \cap P)$.
$100 = 55 + 67 - n(M \cap P)$.
$100 = 122 - n(M \cap P)$.
$n(M \cap P) = 122 - 100 = 22$.
Students who passed only in Physics = $n(P) - n(M \cap P)$.
$= 67 - 22 = 45$.

(C) Let $n$ be the number of newspapers.
Total number of readings by students = $300 \times 5 = 1500$.
Total number of readings provided by newspapers = $n \times 60$.
Since both represent the total number of readings,we have:
$60n = 1500$
$n = \frac{1500}{60}$
$n = 25$.
Thus,the number of newspapers is $25$.

(B) For each element $x \in S$,the condition $x \in A \cup B$ implies that $x$ must belong to at least one of the sets $A$ or $B$.
There are $3$ possibilities for each element $x \in S$:
$1. x \in A$ and $x \notin B$
$2. x \notin A$ and $x \in B$
$3. x \in A$ and $x \in B$
Since there are $4$ elements in set $S$,and each element has $3$ independent choices,the total number of ordered pairs $(A, B)$ is $3 \times 3 \times 3 \times 3 = 3^4 = 81$.

(A) Given $2n(A \setminus B) = n(B \setminus A)$,we have $2(n(A) - n(A \cap B)) = n(B) - n(A \cap B)$.
This simplifies to $2n(A) - 2n(A \cap B) = n(B) - n(A \cap B)$,which gives $n(A \cap B) = 2n(A) - n(B)$.
Also given $5n(A \cap B) = n(A) + 3n(B)$.
Substituting $n(A \cap B)$,we get $5(2n(A) - n(B)) = n(A) + 3n(B)$.
$10n(A) - 5n(B) = n(A) + 3n(B) \implies 9n(A) = 8n(B) \implies \frac{n(A)}{n(B)} = \frac{8}{9}$.
Let $n(A) = 8k$ and $n(B) = 9k$. Then $n(A \cap B) = 2(8k) - 9k = 7k$.
We know $n(A \cup B) = n(A) + n(B) - n(A \cap B) = 8k + 9k - 7k = 10k$.
Given $n(A \cup B) \leq 10$,so $10k \leq 10 \implies k = 1$.
Thus,$n(A) = 8, n(B) = 9, n(A \cap B) = 7$.
The required value is $\frac{8 \cdot 9 \cdot 7}{8} = 9 \cdot 7 = 63$.