x 1 y and when x=40 then y=16. If x=10, find y. - Vidyadip

Question

$x \propto \frac{1}{\sqrt{y}}$ and when $\mathrm{x}=40$ then $\mathrm{y}=16$. If $\mathrm{x}=10$, find $\mathrm{y}$.

✓

Answer

$
\begin{aligned}
& x \propto \frac{1}{\sqrt{y}} \\
& \therefore x=\mathrm{k} \times \frac{1}{\sqrt{y}}
\end{aligned}
$
where, $\mathrm{k}$ is the constant of variation.
$
\begin{aligned}
& \therefore \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k} \ldots \text { (i) } \\
& \text { When } \mathrm{x}=40, \mathrm{y}=16 \\
& \therefore \text { Substituting } \mathrm{x}=40 \text { and }=16 \text { in (i), we get } \\
& \mathrm{x} \times \sqrt{ } \mathrm{y}=\mathrm{k} \\
& \therefore 40 \times \sqrt{ } 16=\mathrm{k} \\
& \therefore \mathrm{k}=40 \times 4 \\
& \therefore \mathrm{k}=160
\end{aligned}
$
Substituting $k=160$ in (i), we get
$
\begin{aligned}
& x \times \sqrt{ } y=k \\
& \therefore x \times \sqrt{ } y=160 \text {...(ii) }
\end{aligned}
$
This is the equation of variation.
When $\mathrm{x}=10, \mathrm{y}=$ ?
$\therefore$ Substituting, $x=10$ in (ii), we get
$
\begin{aligned}
& x \times \sqrt{ } y=160 \\
& \therefore 10 \times \sqrt{y}=160 \\
& \therefore \sqrt{y}=\frac{160}{10} \\
& \therefore \sqrt{y}=16
\end{aligned}
$
$\therefore \mathrm{y}=256 \ldots$ [Squaring both sides]

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