Word Problem - Set Theory Questions in English

(D) Given: $P(A) = \frac{2}{5}$,$P(B) = \frac{1}{4}$,$P(A \cup B) = \frac{1}{2}$.
By De Morgan's Law,$P(A' \cup B') = P((A \cap B)') = 1 - P(A \cap B)$.
First,find $P(A \cap B)$ using the formula $P(A \cup B) = P(A) + P(B) - P(A \cap B)$:
$\frac{1}{2} = \frac{2}{5} + \frac{1}{4} - P(A \cap B)$
$\frac{1}{2} = \frac{8+5}{20} - P(A \cap B)$
$\frac{1}{2} = \frac{13}{20} - P(A \cap B)$
$P(A \cap B) = \frac{13}{20} - \frac{10}{20} = \frac{3}{20}$.
Now,$P(A' \cup B') = 1 - \frac{3}{20} = \frac{17}{20}$.

(C) For each of the $5$ subjects,a candidate has $2$ possibilities: either pass or fail.
Since there are $5$ subjects,the total number of possible outcomes (combinations of pass/fail) is $2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32$.
The candidate passes the examination only if they pass in all $5$ subjects. There is only $1$ such way (Pass,Pass,Pass,Pass,Pass).
Therefore,the number of ways in which the candidate can fail (i.e.,fail in at least one subject) is the total number of outcomes minus the case where they pass all subjects.
Number of ways to fail $= 2^5 - 1 = 32 - 1 = 31$.

(B) Given that $n(X)=200, n(A)=90, n(B)=80$ and $n(A' \cap B')=40$.
By De Morgan's Law,$n(A' \cap B') = n((A \cup B)') = n(X) - n(A \cup B)$.
So,$40 = 200 - n(A \cup B)$,which implies $n(A \cup B) = 160$.
We know that $n(A \cup B) = n(A) + n(B) - n(A \cap B)$.
$160 = 90 + 80 - n(A \cap B) \implies 160 = 170 - n(A \cap B) \implies n(A \cap B) = 10$.
Now,$n(A \cap B') = n(A) - n(A \cap B) = 90 - 10 = 80$.

(C) Given set $A = \{1, 2, 3, 4, 5, 6\}$.
The total number of subsets of $A$ is $2^n$,where $n = 6$,so $2^6 = 64$.
The subsets containing at least two elements are the total subsets minus the subsets with zero elements (empty set) and the subsets with exactly one element.
Number of subsets with zero elements = ${}^6C_0 = 1$.
Number of subsets with one element = ${}^6C_1 = 6$.
Therefore,the number of subsets with at least two elements = $64 - (1 + 6) = 64 - 7 = 57$.

(D) Let $n(U) = 60$ be the total number of students.
Let $C$ be the set of students who play cricket,$n(C) = 25$.
Let $T$ be the set of students who play tennis,$n(T) = 20$.
Given $n(C \cap T) = 10$.
The number of students who play at least one game is $n(C \cup T) = n(C) + n(T) - n(C \cap T)$.
$n(C \cup T) = 25 + 20 - 10 = 35$.
The number of students who play neither game is $n(U) - n(C \cup T)$.
$60 - 35 = 25$.

(C) Given $|A| = 3$ and $|B| = 6$.
For $A \cup B$,the formula is $|A \cup B| = |A| + |B| - |A \cap B|$.
To minimize $|A \cup B|$,we must maximize $|A \cap B|$. The maximum value of $|A \cap B|$ is $\min(|A|, |B|) = 3$.
Thus,$\min |A \cup B| = 3 + 6 - 3 = 6$. So,Statement $(I)$ is true.
For $A \cap B$,the maximum value is $\min(|A|, |B|) = 3$.
Thus,$\max |A \cap B| = 3$. So,Statement $(II)$ is true.

(D) Given,$n(A \times B) = 35$ and $B \subset A$.
Since $n(A \times B) = n(A) \times n(B) = 35$,the possible factors of $35$ are $(35, 1)$ or $(7, 5)$.
Since $B \subset A$ and $A, B$ are non-singleton sets,$n(A) > n(B) > 1$.
Thus,$n(A) = 7$ and $n(B) = 5$.
Now,calculating ${}^{n(A)}C_{n(B)} = {}^{7}C_{5}$.
Using the property ${}^{n}C_{r} = {}^{n}C_{n-r}$,we get ${}^{7}C_{5} = {}^{7}C_{2}$.
${}^{7}C_{2} = \frac{7 \times 6}{2 \times 1} = 21$.

(B) Given: $n(U)=100$,$n(A)=50$,$n(B)=60$,$n(A \cap B)=20$.
Using the formula for the union of two sets:
$n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$n(A \cup B) = 50 + 60 - 20 = 90$.
By De Morgan's Law,$A^{\prime} \cap B^{\prime} = (A \cup B)^{\prime}$.
Therefore,$n(A^{\prime} \cap B^{\prime}) = n(U) - n(A \cup B)$.
$n(A^{\prime} \cap B^{\prime}) = 100 - 90 = 10$.

(B) Let the total number of families be $x$.
Let $A$ be the set of families that own cell phones,so $n(A) = \frac{65}{100}x$.
Let $B$ be the set of families that own scooters,so $n(B) = 15000$.
The number of families that own both is $n(A \cap B) = \frac{15}{100}x$.
Since every family owns at least one,$n(A \cup B) = x$.
Using the formula $n(A \cup B) = n(A) + n(B) - n(A \cap B)$,we have:
$x = \frac{65x}{100} + 15000 - \frac{15x}{100}$
$x = \frac{50x}{100} + 15000$
$x = 0.5x + 15000$
$0.5x = 15000$
$x = \frac{15000}{0.5} = 30000$.
Thus,the total number of families in the town is $30000$.

(B) Let $A$ be the set of integers from $1$ to $1000$ divisible by $2$,and $B$ be the set of integers from $1$ to $1000$ divisible by $3$.
We need to find $|A \cup B| = |A| + |B| - |A \cap B|$.
$|A| = \lfloor \frac{1000}{2} \rfloor = 500$.
$|B| = \lfloor \frac{1000}{3} \rfloor = 333$.
$|A \cap B|$ is the number of integers divisible by both $2$ and $3$,i.e.,divisible by $\text{lcm}(2, 3) = 6$.
$|A \cap B| = \lfloor \frac{1000}{6} \rfloor = 166$.
Therefore,$|A \cup B| = 500 + 333 - 166 = 667$.

(C) Let $S = \{-7, -5, -3, -2, 2, 4, 6, 13\}$. The sum of all elements is $S_{total} = -7-5-3-2+2+4+6+13 = 8$.
Let $x = a+b+c+d$ and $y = e+f+g+h$.
Since $a, b, c, d, e, f, g, h$ are distinct elements of $S$,we have $x+y = 8$.
We want to minimize $x^2+y^2$.
We know that $x^2+y^2 = \frac{1}{2}((x+y)^2 + (x-y)^2)$.
Since $(x+y)^2 = 8^2 = 64$ is constant,we minimize $x^2+y^2$ by minimizing $|x-y|$.
We have $x+y = 8$,so $y = 8-x$.
Then $x-y = x-(8-x) = 2x-8 = 2(x-4)$.
We want $x$ to be as close to $4$ as possible.
The sum of four elements from $S$ that is closest to $4$ is $x = -2-3+6+4 = 5$ or $x = -7+2+4+6 = 5$.
If $x=5$,then $y = 8-5 = 3$.
Then $x^2+y^2 = 5^2+3^2 = 25+9 = 34$.
Alternatively,if $x=4$,then $y=4$. This is impossible as elements are distinct.
If $x=6$,then $y=2$. $x^2+y^2 = 36+4 = 40$.
If $x=3$,then $y=5$. $x^2+y^2 = 9+25 = 34$.
Thus,the minimum value is $34$.

(C) Let $n(M)$ be the number of members proficient in Mathematics and $n(S)$ be the number of members proficient in Statistics.
Given $n(M \cup S) = 25$,$n(M) = 19$,and $n(S) = 16$.
Using the formula $n(M \cup S) = n(M) + n(S) - n(M \cap S)$:
$25 = 19 + 16 - n(M \cap S)$
$25 = 35 - n(M \cap S)$
$n(M \cap S) = 35 - 25 = 10$.
The probability that a person selected at random is proficient in both is $P(M \cap S) = \frac{n(M \cap S)}{n(M \cup S)} = \frac{10}{25} = \frac{2}{5}$.

(A) Let $E$ be the set of students who passed in English and $M$ be the set of students who passed in Mathematics.
Given:
Total students $= 205$
$n(E) = 105$
$n(M) = 70$
$n(E \cap M) = 30$
We need to find the number of students who passed in at least one subject,which is $n(E \cup M)$.
Using the formula: $n(E \cup M) = n(E) + n(M) - n(E \cap M)$
$n(E \cup M) = 105 + 70 - 30 = 145$
Now,the number of students who failed in both subjects is the total number of students minus those who passed in at least one subject.
Students failed in both $= 205 - 145 = 60$

(C) Let $S = \{1, 2, 3, \ldots, 1000\}$. The total number of outcomes is $n(S) = 1000$.
Let $A$ be the set of perfect cubes in $S$. Since $10^3 = 1000$,$A = \{1^3, 2^3, \ldots, 10^3\}$,so $n(A) = 10$.
Let $B$ be the set of natural numbers having an odd number of divisors. $A$ number has an odd number of divisors if and only if it is a perfect square. The largest perfect square $\le 1000$ is $31^2 = 961$. Thus,$B = \{1^2, 2^2, \ldots, 31^2\}$,so $n(B) = 31$.
The intersection $A \cap B$ consists of numbers that are both perfect cubes and perfect squares,i.e.,perfect sixth powers. These are $1^6 = 1$,$2^6 = 64$,and $3^6 = 729$. Thus,$n(A \cap B) = 3$.
Using the inclusion-exclusion principle,$n(A \cup B) = n(A) + n(B) - n(A \cap B) = 10 + 31 - 3 = 38$.
The probability is $P(A \cup B) = \frac{n(A \cup B)}{n(S)} = \frac{38}{1000} = \frac{19}{500}$.

(B) Given that $P(\bar{A}) = \frac{2}{3}$,we can find $P(A)$ using the complement rule: $P(A) = 1 - P(\bar{A}) = 1 - \frac{2}{3} = \frac{1}{3}$.
We know the formula for the union of two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $\frac{3}{4} = \frac{1}{3} + P(B) - \frac{1}{4}$.
Rearranging to solve for $P(B)$: $P(B) = \frac{3}{4} + \frac{1}{4} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3}$.
We need to find $P(\bar{A} \cap B)$. Using the set theory property,$P(\bar{A} \cap B) = P(B) - P(A \cap B)$.
Substituting the values: $P(\bar{A} \cap B) = \frac{2}{3} - \frac{1}{4} = \frac{8 - 3}{12} = \frac{5}{12}$.
Thus,the correct option is $B$.

(B) We know that the probability of the union of two events is given by $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
Substituting the given values: $P(A \cup B) = 0.5 + 0.4 - 0.3 = 0.6$.
The probability that neither $A$ nor $B$ occurs is given by $P(A^c \cap B^c) = 1 - P(A \cup B)$.
Therefore,$P(A^c \cap B^c) = 1 - 0.6 = 0.4$.

(D) Given data:
$P(A \text{ fails}) = 0.2$
$P(A \cap B \text{ fail}) = 0.15$
To find the probability that $A$ fails alone, we subtract the probability that both $A$ and $B$ fail from the total probability that $A$ fails.
$P(A \text{ fails alone}) = P(A \text{ fails}) - P(A \cap B \text{ fail})$
$P(A \text{ fails alone}) = 0.2 - 0.15 = 0.05$
Thus, the correct option is $D$.

(B) We are given the expression $[x - |x|] = 5$.
Case $1$: If $x \geq 0$,then $|x| = x$.
Substituting this into the expression,we get $[x - x] = [0] = 0$.
Since $0 \neq 5$,there is no solution for $x \geq 0$.
Case $2$: If $x < 0$,then $|x| = -x$.
Substituting this into the expression,we get $[x - (-x)] = [2x] = 5$.
For $[2x] = 5$,we must have $5 \leq 2x < 6$,which implies $2.5 \leq x < 3$.
However,this contradicts our assumption that $x < 0$.
Therefore,there is no real value of $x$ that satisfies the given equation.
Thus,the set is the empty set,$\phi$.

(B) Given $S = \{1, 2, 3, \ldots, 50\}$,so $n(S) = 50$.
For set $A$,we solve $n + \frac{50}{n} > 27$.
$n^2 - 27n + 50 > 0$.
$(n - 25)(n - 2) > 0$.
This holds for $n < 2$ or $n > 25$.
Since $n \in S$,$n=1$ or $n \in \{26, 27, \ldots, 50\}$.
Thus,$A = \{1, 26, 27, \ldots, 50\}$,so $n(A) = 1 + 25 = 26$.
For set $B$,primes in $S$ are $\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\}$,so $n(B) = 15$.
For set $C$,squares in $S$ are $\{1, 4, 9, 16, 25, 36, 49\}$,so $n(C) = 7$.
Probabilities are $P(A) = \frac{26}{50}$,$P(B) = \frac{15}{50}$,$P(C) = \frac{7}{50}$.
Therefore,$P(A) > P(B) > P(C)$.

(D) For each element $x \in A$,there are three mutually exclusive possibilities regarding its membership in subsets $P$ and $Q$ such that $P \cap Q = \phi$:
$1$. $x \in P$ and $x \notin Q$
$2$. $x \notin P$ and $x \in Q$
$3$. $x \notin P$ and $x \notin Q$
Since there are $n$ elements in $A$ and each element has $3$ independent choices,the total number of ways to choose subsets $P$ and $Q$ is $3 \times 3 \times \dots \times 3$ ($n$ times) = $3^n$.

(A) Let $S$,$D$,and $P$ denote the sets of persons who like singing,dancing,and painting,respectively.
Given: $n(S \cup D \cup P) = 265$,$n(S) = 200$,$n(D) = 110$,$n(P) = 55$,$n(S \cap D) = 60$,$n(S \cap P) = 30$,and $n(S \cap D \cap P) = 10$.
Using the principle of inclusion-exclusion:
$n(S \cup D \cup P) = n(S) + n(D) + n(P) - n(S \cap D) - n(S \cap P) - n(D \cap P) + n(S \cap D \cap P)$
$265 = 200 + 110 + 55 - 60 - 30 - n(D \cap P) + 10$
$265 = 285 - n(D \cap P)$
$n(D \cap P) = 285 - 265 = 20$
The number of persons who like only dancing and painting is given by $n(D \cap P) - n(S \cap D \cap P)$.
Therefore,$20 - 10 = 10$.

(C) Let $M, P, B$ be the sets of students who failed in Mathematics,Physics,and Biology respectively. Let the regions in the Venn diagram be represented by $a, b, c, d, e, f, g$ as shown,and $h$ be the number of students who passed in all subjects.
Total students = $100$,so $a+b+c+d+e+f+g+h = 100$. Given $h = 1$,so $a+b+c+d+e+f+g = 99$.
Given:
$1) \text{ Failed in Mathematics: } a+b+d+e = 50$
$2) \text{ Failed in Physics: } b+c+d+f = 45$
$3) \text{ Failed in Biology: } d+e+f+g = 40$
$4) \text{ Failed in exactly two subjects: } b+e+f = 32$
Summing equations $(1), (2), (3)$:
$(a+b+d+e) + (b+c+d+f) + (d+e+f+g) = 50 + 45 + 40 = 135$
$(a+b+c+d+e+f+g) + (b+d+e+f) + d = 135$
$99 + (b+e+f) + 2d = 135$
$99 + 32 + 2d = 135$
$131 + 2d = 135$
$2d = 4 \implies d = 2$.
Thus,the number of students failing in all three subjects is $2$.

(A) Given $(A \cap C) \cup (B \cap C^{\prime}) = \phi$.
Since the union of two sets is the empty set $\phi$,each set must be empty.
Therefore,$A \cap C = \phi$ and $B \cap C^{\prime} = \phi$.
From $B \cap C^{\prime} = \phi$,we have $B \subseteq C$.
Since $A \cap C = \phi$ and $B \subseteq C$,it follows that $A \cap B = \phi$.

(C) We have $A = 5^{n} - 4n - 1 = (1 + 4)^{n} - 4n - 1$.
Using the binomial expansion,$(1 + 4)^{n} = {}^{n}C_{0} + {}^{n}C_{1}(4) + {}^{n}C_{2}(4^{2}) + \dots + {}^{n}C_{n}(4^{n})$.
So,$A = (1 + 4n + 16({}^{n}C_{2} + {}^{n}C_{3}(4) + \dots + {}^{n}C_{n}(4^{n-2}))) - 4n - 1$.
$A = 16({}^{n}C_{2} + {}^{n}C_{3}(4) + \dots + {}^{n}C_{n}(4^{n-2}))$.
This shows that every element of $A$ is a multiple of $16$.
For $n=1$,$5^{1}-4(1)-1 = 0$.
For $n=2$,$5^{2}-4(2)-1 = 16$.
For $n=3$,$5^{3}-4(3)-1 = 112 = 16 \times 7$.
Thus,$A = \{0, 16, 112, \dots\}$.
$B = \{16(n-1) : n \in N\} = \{0, 16, 32, 48, \dots\}$.
Since every element of $A$ is a multiple of $16$,$A \subseteq B$.

(A) Let $A$ be the set of families who own a car and $B$ be the set of families who own a house.
Given: $P(A) = 0.60$,$P(B) = 0.30$,and $P(A \cap B) = 0.20$.
We need to find the probability that a family owns a car or a house but not both,which is represented by the symmetric difference $P(A \Delta B)$.
The formula for symmetric difference is $P(A \Delta B) = P(A \cup B) - P(A \cap B)$.
First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.60 + 0.30 - 0.20 = 0.70$.
Now,$P(A \Delta B) = P(A \cup B) - P(A \cap B) = 0.70 - 0.20 = 0.50$.
Thus,the probability is $0.5$.

(C) Given that $P(A \cup B) = 0.6$ and $P(A \cap B) = 0.3$.
We know that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
So,$P(A) + P(B) = P(A \cup B) + P(A \cap B) = 0.6 + 0.3 = 0.9$.
We need to find $P(A') + P(B')$.
Since $P(A') = 1 - P(A)$ and $P(B') = 1 - P(B)$,
$P(A') + P(B') = (1 - P(A)) + (1 - P(B)) = 2 - (P(A) + P(B))$.
Substituting the value,$P(A') + P(B') = 2 - 0.9 = 1.1$.

(C) Given $|x^{2} - 10| \le 6$,we have $-6 \le x^{2} - 10 \le 6$.
Adding $10$ to all sides,we get $4 \le x^{2} \le 16$.
Taking the square root,$x \in [-4, -2] \cup [2, 4]$,so $A = [-4, -2] \cup [2, 4]$.
Given $|x - 2| > 1$,we have $x - 2 > 1$ or $x - 2 < -1$.
This implies $x > 3$ or $x < 1$,so $B = (-\infty, 1) \cup (3, \infty)$.
Now,$A \cap B = ([-4, -2] \cup [2, 4]) \cap ((-\infty, 1) \cup (3, \infty)) = [-4, -2] \cup (3, 4]$.
Thus,option $C$ is correct.

(A) The set $S = \{1, 2, 3, . . . , 11\}$ contains $6$ odd numbers $\{1, 3, 5, 7, 9, 11\}$ and $5$ even numbers $\{2, 4, 6, 8, 10\}$.
The total number of subsets of $S$ is $2^{11} = 2048$.
$A$ subset $B$ has an odd product if and only if all elements of $B$ are odd. The number of such subsets is $2^6 = 64$.
$A$ subset $B$ has an even product if it contains at least one even number. The number of such subsets is $2^{11} - 2^6 = 2048 - 64 = 1984$.
The condition $n(B) \ge 2$ excludes subsets with $0$ or $1$ element.
Subsets with $0$ elements: $\emptyset$ (product is not defined or considered odd,so it is excluded).
Subsets with $1$ element: $\{1\}, \{3\}, \{5\}, \{7\}, \{9\}, \{11\}$ (all odd products) and $\{2\}, \{4\}, \{6\}, \{8\}, \{10\}$ (all even products).
We must exclude the $5$ subsets that contain exactly one even number (since their product is even but $n(B) < 2$).
Therefore,the required number of subsets is $1984 - 5 = 1979$.