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

<p>$$\begin{aligned} & A=\{1,2,3,4,5\} \\ & x R y \Leftrightarrow 4 x \leq 5 y \\ & 4 x \leq 5 y \quad \Rightarrow \quad \frac{x}{y} \leq \frac{5}{4} \quad \Rightarrow \frac{x}{y} \leq 1.25 \end{aligned}$$</p> <p>$$\begin{aligned} & R=\{(1,2),(1,3),(1,4),(1,5),(1,1),(2,2),(2,3),(2,4), \\ & (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\} \\ & \therefore \quad n(R)=m=16 \end{aligned}$$</p> <p>Elements to be added to $R$ to make it symmetric</p> <p>$(1,2) \in R \quad \Rightarrow \quad(2,1)$ should be added</p> <p>Similarly, $(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3), (5,3)$</p> <p>$\therefore 9$ elements should be added</p> <p>$$\begin{array}{ll} \therefore & n=9 \\ \therefore & m+n=25 \end{array}$$</p>