Question

The cube root of $x$ varies inversely as the square of $y$. If $x=8$ when $y=3$, then find $x$ when $y=1\frac{1}{2}$.

Answer 1

Given that, the cube root of $x$ varies inversely as square of $y$

$⟹\sqrt[3]{3x}\propto \frac{1}{y^2}$ or

$\sqrt[3]{3x}=k.\frac{1}{y^2}\dots \left(1\right)$, where $k$ is a constant.

Given that, $x=8$ and $y=3$

$\therefore \sqrt[3]{38}=k\times \frac{1}{(3{)}^2}$

$2=k\times \frac{1}{9}$

$⟹k=18\dots \left(2\right)$

Now, if $y=1\frac{1}{2}=\frac{3}{2}$

From $\left(1\right)$ and $\left(2\right)$:

$\sqrt[3]{3x}=18.\frac{1}{y^2}$

$\sqrt[3]{3x}=18\times \frac{1}{{\left(\frac{3}{2}\right)}^2}$

$\sqrt[3]{3x}=18\times \frac{4}{9}$

$\sqrt[3]{3x}=8$

$x=512$