Question

If $x\text{and}y$ vary inversely, then find $(\frac{2p}{q}+\frac{q}{p})$.

$x$	$\frac{1}{5}$	$\frac{3}{5}$	$p$
$y$	$\frac{3}{2}$	$q$	$\frac{8}{3}$

• A. $\frac{7}{6}$
• B. $\frac{223}{180}$
• C. $\frac{881}{180}$
• D. $\frac{5}{9}$

Answer 1

The correct option is C $\frac{881}{180}$

Given, $x\text{and}y$ vary inversely.
$\therefore \frac{x_1}{x_2}=\frac{y_2}{y_1}$

$\left(i\right)$
Let,
$x_1=\frac{1}{5},y_1=\frac{3}{2},x_2=\frac{3}{5}$
$y_2=q$
$\Rightarrow y_2=\frac{x_1\times y_1}{x_2}$
$\Rightarrow q=\frac{\frac{1}{5}\times \frac{3}{2}}{\frac{3}{5}}=\frac{1}{2}$

$\left(ii\right)$
Let,
$x_1=\frac{1}{5},y_1=\frac{3}{2},x_2=p$
$y_2=\frac{8}{3}$
$\Rightarrow x_2=\frac{x_1\times y_1}{y_2}$
$\Rightarrow p=\frac{\frac{1}{5}\times \frac{3}{2}}{\frac{8}{3}}=\frac{9}{80}$
Therefore,
$(\frac{2p}{q}+\frac{q}{p})=\frac{2\times \frac{9}{80}}{\frac{1}{2}}+\frac{\frac{1}{2}}{\frac{9}{80}}$
$=\frac{9}{40}\times \frac{2}{1}+\frac{1}{2}\times \frac{80}{9}$
$=\frac{9}{20}+\frac{40}{9}$
$=\frac{881}{180}$