Let $A = {1, 2, 3, 4, 5}$. Let $R$ be a relation on $A$ defined by $xRy$ if and only if $4x \le 5y$. Let $m$ be the number of elements in $R$ and $n$ be the minimum number of elements from $A \times A$ that are required to be added to $R$ to make it a symmetric relation. Then $m + n$ is equal to :
Correct Answer:
C
Explanation
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, $m + n = 16 + 9 = 25$.