Let $$A = \{1, 2, 3, 4, 5\}$$. Let $$R$$ be a relation on $$A$$ defined by $$xRy$$ if and only if $$4x \leq 5y$$. 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 :

The relation $$R$$ on $$A = \{1,2,3,4,5\}$$ is defined by $$xRy$$ iff $$4x \leq 5y$$, i.e., $$y \geq 4x/5$$.

Count elements m: For $$x=1$$: $$y \geq 0.8$$, so $$y \in \{1,2,3,4,5\}$$ (5 pairs). $$x=2$$: $$y \geq 1.6$$ (4). $$x=3$$: $$y \geq 2.4$$ (3). $$x=4$$: $$y \geq 3.2$$ (2). $$x=5$$: $$y \geq 4$$ (2). Total $$m = 16$$.

All 5 diagonal elements $$(i,i)$$ are in $$R$$. Off-diagonal pairs in R: 11.

Check symmetry: For each off-diagonal $$(x,y) \in R$$, check if $$(y,x) \in R$$. The only symmetric off-diagonal pair is $$\{4,5\}$$ (both $$(4,5)$$ and $$(5,4)$$ are in $$R$$). The remaining 9 off-diagonal pairs lack their reverse.

So $$n = 9$$ elements must be added.

$$m + n = 16 + 9 = 25$$.

The correct answer is Option 1: $$25$$.