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Given, $(a, b)$ belongs to relation $R$ if $gcd(a, b) = 1$, $2a \neq b$. Here $gcd$ means greatest common divisor. $gcd$ of two numbers is the largest number that divides both of them. (1) For Reflexive, In $aRa$, $gcd(a, a) = a$. Therefore, this relation is not reflexive. (2) For Symmetric: Take $a = 2, b = 1 \Rightarrow gcd(2, 1) = 1$. Also $2a = 4 \neq b$. Now $gcd(b, a) = 1 \Rightarrow gcd(1, 2) = 1$ and $2b$ should not be equal to $a$. But here, $2b = 2 = a \Rightarrow R$ is not Symmetric. (3) For Transitive: Let $a = 14, b = 19, c = 21$ $gcd(a, b) = 1, 2a \neq b$ $gcd(b, c) = 1, 2b \neq c$ $gcd(a, c) = 7, 2a \neq c$ Hence not transitive. $\Rightarrow R$ is neither symmetric nor transitive.

Given that $aRb$ if $2a + 3b = 5m$, $m \in I$. (1) $(a, a) \in R$ as $2a + 3a = 5a$, $a \in N$. Hence, R is reflexive. (2) If $(a, b) \in R$ then $2a + 3b = 5m$. Now, $5(a + b) = 5n$. $3a + 2b + 2a + 3b = 5n$ $\therefore 3a + 2b = 5(n - m)$. $\therefore (b, a) \in R$. $\therefore R$ is symmetric. (3) If $(a, b) \in R$ and $(b, c) \in R$ then $2a + 3b = 5m$, $2b + 3c = 5n$. $\Rightarrow 2a + 5b + 3c = 5(m + n)$. $\Rightarrow 2a + 3c = 5(m + n - b)$. $\therefore (a, c) \in R$. $\therefore R$ is transitive. Hence, R is an equivalence relation.

To make $R$ symmetric, we need to add $(b, a)$ and $(c, b)$ to $R$. So, $R = \{(a, b), (b, c), (b, a), (c, b)\}$. To make $R$ transitive: Since $(a, b) \in R$ and $(b, c) \in R$, we need to add $(a, c)$ to $R$. Now, $R = \{(a, b), (b, c), (b, a), (c, b), (a, c)\}$. Since $(b, a) \in R$ and $(a, c) \in R$, we need to add $(b, c)$ to $R$. But $(b, c)$ is already in $R$. Since $(c, b) \in R$ and $(b, a) \in R$, we need to add $(c, a)$ to $R$. Now, $R = \{(a, b), (b, c), (b, a), (c, b), (a, c), (c, a)\}$. Since $(a, b) \in R$ and $(b, a) \in R$, we need to add $(a, a)$ to $R$. Since $(b, c) \in R$ and $(c, b) \in R$, we need to add $(b, b)$ to $R$. Since $(c, a) \in R$ and $(a, c) \in R$, we need to add $(c, c)$ to $R$. So, $R = \{(a, b), (b, c), (b, a), (c, b), (a, c), (c, a), (a, a), (b, b), (c, c)\}$. The elements that must be added are $(b, a), (c, b), (a, c), (c, a), (a, a), (b, b), (c, c)$. Therefore, the minimum number of elements that must be added is $7$.

Given $(a, b) R (c, d) \implies ad(b - c) = bc(a - d)$. Symmetric: $(c, d) R (a, b) \implies cb(d - a) = da(c - b)$. This is symmetric. Reflexive: $(a, b) R (a, b) \implies ab(b - a) \neq ba(a - b)$. Thus, not reflexive. Transitive: $(2, 3) R (3, 2)$ and $(3, 2) R (5, 30)$ but $((2, 3), (5, 30)) \notin R$. Thus, not transitive. Therefore, R is symmetric but neither reflexive nor transitive.

For relation $T$: if $(a, b) \in T$, then $a^2 - b^2 = I$ where $I \in Z$. Then, $(b, a)$ on relation $T$ means $b^2 - a^2 = -I$. Since $-I \in Z$, $T$ is symmetric. For relation $S = {(a, b) : a, b \in R - {0}, 2 + \frac{a}{b} > 0}$, $2 + \frac{a}{b} > 0 \implies \frac{a}{b} > -2$. If $(b, a) \in S$ then $2 + \frac{b}{a}$ is not necessarily positive. Therefore, $S$ is not symmetric.

$S = \{1, 2, 3, ..., 10\}$ $P(S) =$ power set of $S$ $AR_1B \implies (A \cap B^c) \cup (A^c \cap B) = \emptyset$ $R_1$ is reflexive, symmetric For transitive: $(A \cap B^c) \cup (A^c \cap B) = \emptyset$; $\emptyset = \emptyset \implies A = B$ $(B \cap C^c) \cup (B^c \cap C) = \emptyset \implies B = C$ $\therefore A = C \implies R_1$ is an equivalence relation. $R_2 \equiv A \cup B^c = A^c \cup B$ $R_2 \implies$ Reflexive, symmetric For transitive: $A \cup B^c = A^c \cup B \implies A = B$ $B \cup C^c = B^c \cup C \implies B = C$ $\therefore A = C \implies A \cup C^c = A^c \cup C \implies R_2$ is an equivalence relation.

Given sets are $A = {2, 3, 4}$ and $B = {8, 9, 12}$. We want to find the number of elements of the form $((a_1, b_1), (a_2, b_2))$ such that $a_1$ divides $b_2$ and $a_2$ divides $b_1$. For the first condition, $a_1$ divides $b_2$, with $a_1 \in A$ and $b_2 \in B$, we can list the pairs: $(a_1, b_2) \in {(2, 8), (2, 12), (3, 9), (3, 12), (4, 8), (4, 12)}$. This gives $6$ pairs. For the second condition, $a_2$ divides $b_1$, we can similarly list the pairs. This again has $6$ valid pairs. Now, for every pair from the first condition, we can have any pair from the second condition. This leads to $6 \times 6 = 36$ relations.

Let $|A|$ be the number of medals in event A, $|B|$ be the number of medals in event B, and $|C|$ be the number of medals in event C. We are given $|A| = 48$, $|B| = 25$, and $|C| = 18$. We are also given that the total number of people who received medals is $60$, so $|A \cup B \cup C| = 60$, and the number of people who received medals in all three events is $5$, so $|A \cap B \cap C| = 5$. We want to find the number of people who received medals in exactly two of the three events. Let $x$ be the number of people who received medals in exactly two events. Using the Principle of Inclusion-Exclusion, we have: $|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |A \cap C| + |B \cap C|) + |A \cap B \cap C|$ $60 = 48 + 25 + 18 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$ $60 = 91 - (|A \cap B| + |A \cap C| + |B \cap C|) + 5$ $|A \cap B| + |A \cap C| + |B \cap C| = 91 + 5 - 60 = 36$ Now, let $N_{2}$ be the number of people who received medals in exactly two events. Then: $|A \cap B| + |A \cap C| + |B \cap C| = N_{2} + 3|A \cap B \cap C|$ $36 = N_{2} + 3(5)$ $N_{2} = 36 - 15 = 21$ Therefore, the number of people who received medals in exactly two of the three events is $21$.

Let's analyze the relation $R$. $R$ is a set of ordered pairs of ordered pairs $((a_1, b_1), (a_2, b_2))$ such that $a_1 \in A$, $b_1 \in B$, $a_2 \in A$, $b_2 \in B$, and the conditions $a_1 \le b_2$ and $b_1 \le a_2$ are satisfied. We need to find the number of such ordered pairs. However, the question is formulated wrongly, since we are looking at a relation of pairs. A proper relation is given by $R = {(a_1, b_1) : a_1 \in A, b_1 \in B}$. The number of elements in the relation is not a relation between ordered pairs, it is simply asking to find pairs $(a, b)$ such that $a \le b$. So let $S = {(a,b): a \in A, b \in B, a \le b}$ and find the number of elements in $S$. Now we enumerate: - If $a=1$, then $b$ can be 2, 4, 5, 8, 10. (5 options) - If $a=3$, then $b$ can be 4, 5, 8, 10. (4 options) - If $a=4$, then $b$ can be 4, 5, 8, 10. (4 options) - If $a=6$, then $b$ can be 8, 10. (2 options) - If $a=9$, then $b$ can be 10. (1 option) So the total number of elements in $S$ is $5+4+4+2+1 = 16$. However, we are looking for elements that obey BOTH rules. That is: Number of pairs satisfying BOTH $a_1 \le b_2$ and $b_1 \le a_2$. Let's re-read the provided solution: - For each $a_1$ in $A = {1, 3, 4, 6, 9}$, we find the possible $b_2$ values in $B = {2, 4, 5, 8, 10}$ such that $a_1 \le b_2$. - $a_1 = 1$: $b_2$ can be 2, 4, 5, 8, 10 (5 choices) - $a_1 = 3$: $b_2$ can be 4, 5, 8, 10 (4 choices) - $a_1 = 4$: $b_2$ can be 4, 5, 8, 10 (4 choices) - $a_1 = 6$: $b_2$ can be 8, 10 (2 choices) - $a_1 = 9$: $b_2$ can be 10 (1 choice) This gives $5+4+4+2+1 = 16$. - For each $b_1$ in $B = {2, 4, 5, 8, 10}$, we find the possible $a_2$ values in $A = {1, 3, 4, 6, 9}$ such that $b_1 \le a_2$. - $b_1 = 2$: $a_2$ can be 3, 4, 6, 9 (4 choices) - $b_1 = 4$: $a_2$ can be 4, 6, 9 (3 choices) - $b_1 = 5$: $a_2$ can be 6, 9 (2 choices) - $b_1 = 8$: $a_2$ can be 9 (1 choice) - $b_1 = 10$: no $a_2$ works (0 choices) This gives $4+3+2+1+0 = 10$. So, $16 \times 10 = 160$. Thus the final number of elements is 160.

Let $S = {1, 2, 3, …, 10}$. $R = {(A, B): A∩B ≠ 𝜙; A, B∈ M}$. For Reflexive, $M$ is subset of 'S'. So $𝜙 ∈ M$ for $𝜙 ∩ 𝜙 = 𝜙 ⇒$ but relation is $A ∩ B ≠ 𝜙$. So it is not reflexive. For symmetric, $A R B ⇒ A ∩ B ≠ 𝜙, ⇒ B R A ⇒ B ∩ A ≠ 𝜙$, So it is symmetric. For transitive, If $A = {(1, 2), (2, 3)}, B = {(2, 3), (3, 4)}, C = {(3, 4), (5, 6)}$. $A R B$ & $B R C$ but $A$ does not relate to $C$. So it not transitive.

Given that the number of subsets of set $A$ is 56 more than that of set $B$, we have: $2^m - 2^n = 56$ $2^n(2^{m-n} - 1) = 56$ $2^n(2^{m-n} - 1) = 2^3 \times 7$ Comparing both sides, we have: $2^n = 2^3$ and $2^{m-n} - 1 = 7$ $n = 3$ and $2^{m-n} = 8$ $n = 3$ and $2^{m-n} = 2^3$ $n = 3$ and $m-n = 3$ $n = 3$ and $m = 6$ So the coordinates of point P are $(6,3)$. The coordinates of point Q are $(-2,-3)$. The distance between $P$ and $Q$ is: $PQ = \sqrt{(6 - (-2))^2 + (3 - (-3))^2}$ $PQ = \sqrt{(6+2)^2 + (3+3)^2}$ $PQ = \sqrt{8^2 + 6^2}$ $PQ = \sqrt{64 + 36}$ $PQ = \sqrt{100}$ $PQ = 10$ Therefore, the distance between $P(6,3)$ and $Q(-2,-3)$ is 10.

Reflexive: Since $a*b - a*b = 0$ is divisible by $5$, $(a, b)R(a, b)$ holds. Symmetric: If $(a, b)R(c, d)$, then $ad - bc$ is divisible by $5$. Thus, $bc - ad$ is divisible by $5$, so $(c, d)R(a, b)$ holds. Not Transitive: Consider $(3, 1)R(10, 5)$ because $3*5 - 1*10 = 5$ is divisible by $5$. Consider $(10, 5)R(1, 1)$ because $10*1 - 5*1 = 5$ is divisible by $5$. But $(3, 1)$ is not related to $(1, 1)$ because $3*1 - 1*1 = 2$ is not divisible by $5$. Hence, $R$ is reflexive and symmetric but not transitive.

Given the set ${1, 2, 3, 4}$, and that $R$ is an equivalence relation. Since ${((1, 2), (1, 3))} \subset R$, $R$ must contain the pairs $(1, 2)$ and $(1, 3)$. An equivalence relation must be reflexive, symmetric, and transitive. Reflexive property: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4) \in R$. Symmetric property: Since $(1, 2) \in R$, $(2, 1) \in R$. Since $(1, 3) \in R$, $(3, 1) \in R$. Transitive property: Since $(1, 2) \in R$ and $(1, 3) \in R$, it follows that $(2, 3) \in R$. By symmetric property, $(3, 2) \in R$. Therefore, the elements in $R$ are: $(1, 1)$, $(2, 2)$, $(3, 3)$, $(4, 4)$, $(1, 2)$, $(2, 1)$, $(1, 3)$, $(3, 1)$, $(2, 3)$, $(3, 2)$. Thus, the number of elements in $R$ is $10$.

To determine if the given relations $R_1$ and $R_2$ are equivalence relations, we need to check whether each of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and transitivity. For $R_1$: Reflexivity: For $R_1$ to be reflexive, $aR_1a$ must hold for all $a \in R$. This means $a^2 + a^2 = 1$, or $2a^2 = 1$, which implies $a^2 = \frac{1}{2}$. Since this is not true for every real number $a$, $R_1$ is not reflexive. Symmetry: If $aR_1b$, then $a^2 + b^2 = 1$. This also means $b^2 + a^2 = 1$, so $bR_1a$. Thus, $R_1$ is symmetric. Transitivity: If $aR_1b$ and $bR_1c$, then $a^2 + b^2 = 1$ and $b^2 + c^2 = 1$. Adding these, we get $a^2 + 2b^2 + c^2 = 2$. There is no guarantee that $a^2 + c^2 = 1$. Therefore, $R_1$ is not transitive. $R_1$ is not an equivalence relation. For $R_2$: Reflexivity: For any $(a, b) \in N \times N$, $a + b = b + a$. Therefore, $(a, b)R_2(a, b)$, and $R_2$ is reflexive. Symmetry: If $(a, b)R_2(c, d)$, then $a + d = b + c$. This can be reordered to $c + b = d + a$, so $(c, d)R_2(a, b)$. Thus, $R_2$ is symmetric. Transitivity: If $(a, b)R_2(c, d)$ and $(c, d)R_2(e, f)$, then $a + d = b + c$ and $c + f = d + e$. Adding the two equations, we get $a + d + c + f = b + c + d + e$. Simplifying, we get $a + f = b + e$, so $(a, b)R_2(e, f)$. $R_2$ is reflexive, symmetric, and transitive; therefore, it is an equivalence relation. Therefore, only $R_2$ is an equivalence relation.

Given the relation $(x_1, y_1) R (x_2, y_2)$ if $x_1 \le x_2$ or $y_1 \le y_2$. For reflexivity, we need to check if $(x_1, y_1) R (x_1, y_1)$. Since $x_1 \le x_1$ and $y_1 \le y_1$, the condition $x_1 \le x_1$ or $y_1 \le y_1$ is true. Therefore, $R$ is reflexive. For symmetry, if $(x_1, y_1) R (x_2, y_2)$, then $x_1 \le x_2$ or $y_1 \le y_2$. For $R$ to be symmetric, $(x_2, y_2) R (x_1, y_1)$ must also be true, which means $x_2 \le x_1$ or $y_2 \le y_1$. This is not necessarily true. For example, $(1, 2) R (3, 4)$ since $1 \le 3$ or $2 \le 4$, but $(3, 4) R (1, 2)$ is false because neither $3 \le 1$ nor $4 \le 2$ is true. Thus, $R$ is not symmetric. For transitivity, consider pairs $(3, 9)$, $(4, 6)$, and $(2, 7)$. $(3, 9) R (4, 6)$ because $3 \le 4$. $(4, 6) R (2, 7)$ because $6 \le 7$. However, $(3, 9) R (2, 7)$ is false because neither $3 \le 2$ nor $9 \le 7$ is true. Therefore, $R$ is not transitive. Since $R$ is reflexive but not symmetric, statement (I) is correct. Since $R$ is not transitive, statement (II) is incorrect. Thus, only statement (I) is correct.

Let's find the total number of integers in the range $[100, 700]$. The total number of integers is $700 - 100 + 1 = 601$. Thus, the total number of integers is $601$. Let $n(3)$ be the number of multiples of 3 in the given range. The first multiple of 3 is $102 = 3 \times 34$ and the last multiple of 3 is $699 = 3 \times 233$. So, $n(3) = 233 - 34 + 1 = 200$. Let $n(4)$ be the number of multiples of 4 in the given range. The first multiple of 4 is $100 = 4 \times 25$ and the last multiple of 4 is $700 = 4 \times 175$. So, $n(4) = 175 - 25 + 1 = 151$. Let $n(12)$ be the number of multiples of 12 in the given range. The first multiple of 12 is $108 = 12 \times 9$ and the last multiple of 12 is $696 = 12 \times 58$. So, $n(12) = 58 - 9 + 1 = 50$. The number of elements in A is given by the total number of elements minus (multiples of 3 + multiples of 4) + (multiples of 12) $n(A) = 601 - (200 + 151) + 50 = 601 - 351 + 50 = 250 + 50 = 300$. Therefore, the number of elements in A is $300$.

Given the set $A = {1, 2, 3, 4, 5}$ and the relation $xRy$ if and only if $4x \le 5y$. We need to find the number of elements in R (denoted by m) and the minimum number of elements that need to be added to R to make it symmetric (denoted by n). First, let's find the relation R: $4x \le 5y \implies \frac{x}{y} \le \frac{5}{4} = 1.25$ Now, let's find the pairs $(x, y)$ that satisfy this condition: If $x = 1$, then $y$ can be $1, 2, 3, 4, 5$. If $x = 2$, then $y$ can be $2, 3, 4, 5$. If $x = 3$, then $y$ can be $3, 4, 5$. If $x = 4$, then $y$ can be $4, 5$. If $x = 5$, then $y$ can be $4, 5$. So, $R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 2), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5), (5, 4), (5, 5)}$. The number of elements in $R$ is $m = 16$. Now, we need to find the elements to be added to R to make it symmetric. A relation is symmetric if $(x, y) \in R$ implies $(y, x) \in R$. Currently, the elements $(1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (5,4) \in R$. The symmetric pairs that are missing are: $(2, 1), (3, 1), (4, 1), (5, 1), (3, 2), (4, 2), (5, 2), (4, 3), (5, 3)$. The number of elements to be added is $n = 9$. Therefore, $m + n = 16 + 9 = 25$.