Explanation

(c)
Since, the relation $R$ is defined as
$R = \{(x,\ y)|x,\ y$ are real numbers and $x = wy$ for some rational number $w$}
(i) Reflexive $xRx \Rightarrow x = wx$
$\therefore\ \ \ \ \ \ \ \ w = 1 \in$ Rational number
$\Rightarrow$ The relation $R$ is reflexive.
(ii) Symmetric $xRy \Rightarrow yRx$
As $0R1$
$\Rightarrow \ \ 0 = 0\ (1)$ but $1R0 \Rightarrow 1 = w.(0)$,
Which is not true for any rational number
$\Rightarrow$ The relation $R$ is not symmetric
Thus, $R$ is not equivalent relation.
Now, for the relation $S$ is defined as
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$m,\ n,\ p$ and $q \in$ integers such that $n,\ q \neq 0$ and $qm = pn\}$
(i) Reflexive $\frac{m}{n}S\frac{m}{n} \Rightarrow mn = mn$ (True)
$\Rightarrow$ The relation $S$ is reflexive
(ii) Symmetric $\frac{m}{n}S\frac{p}{q} \Rightarrow mq = np$
$\Rightarrow np = mq \Rightarrow \frac{p}{q}S\frac{m}{n}$
$\Rightarrow$ The relation $S$ is symmetric.
(iii) Transitive $\frac{m}{n}S\frac{p}{q}$ and $\frac{p}{q}S\frac{r}{s}$
$\Rightarrow mq = np$ and $ps = rq$
$\Rightarrow mq.ps = np.rq$
$\Rightarrow ms = nr\ \ \ \Rightarrow \frac{m}{n} = \frac{r}{s} \Rightarrow \frac{m}{n}S\frac{r}{s}$
$\Rightarrow$ The relation $S$ is transitive
$\Rightarrow$ The relation $S$ is equivalent relation.