Question

$P$ varies directly as $Q$ and inversely as $R$; also $P=\frac{2}{3}$ when $Q=\frac{3}{7}$ and $R=\frac{9}{14}$ : find $Q$ when $P=\sqrt{48}$ and $R=\sqrt{75}$.

Answer 1

As $P$ varies directly as $Q$ and inversely as $R$ then let,
AR

$P=K\frac{Q}{R}$

where $K$ is a constant and $P=\frac{2}{3},Q=\frac{3}{7},R=\frac{9}{14}$ satisfies the equation so,

$\frac{2}{3}=K\frac{\frac{3}{7}}{\frac{9}{14}}=\frac{2}{3}K\Rightarrow K=1$

So,$P=\frac{Q}{R}$

For $P=\sqrt{48},R=\sqrt{75}$ we have

$\sqrt{48}=\frac{Q}{\sqrt{75}}$

$Q=\sqrt{48\times 75}=\sqrt{3600}$

$Q=60$