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5 5 votes

Let $\mathrm{A}=\{1,2,3,4,5\}$. Let R be a relation on A defined by $x \mathrm{R} y$ if and only if $4 x \leq 5 \mathrm{y}$. Let m be the number of elements in R and 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 :

  1. 23
  2. 26
  3. 25
  4. 24

3 Answers

2 2 votes
$$
\begin{aligned}
& A=\{1,2,3,4,5\} \\
& x R y \Leftrightarrow 4 x \leq 5 y \\
& 4 x \leq 5 y \quad \Rightarrow \quad \frac{x}{y} \leq \frac{5}{4} \quad \Rightarrow \frac{x}{y} \leq 1.25 \\
& R=\{(1,2),(1,3),(1,4),(1,5),(1,1),(2,2),(2,3),(2,4), \\
& (2,5),(3,3),(3,4),(3,5),(4,4),(4,5),(5,4),(5,5)\} \\
& \therefore \quad n(R)=m=16
\end{aligned}
$$

Elements to be added to $R$ to make it symmetric
$(1,2) \in R \quad \Rightarrow \quad(2,1)$ should be added
Similarly, $(3,1),(4,1),(5,1),(3,2),(4,2),(5,2),(4,3),(5,3)$
$\therefore 9$ elements should be added

$$
\begin{array}{ll}
\therefore & n=9 \\
\therefore & m+n=25
\end{array}
$$
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