Question

If $y$ varies directly as $\sqrt{x}$ and $y=24$ when $x=3$, find $y$ when $x=16$

Answer 1

The statement " $x$ varies directly as $y$ ," means that when $y$ increases, $x$ increases by the same factor. In other words, $y$ and $x$ always have the same ratio that is:

$x=ky$

where $k$ is the constant of variation.

Here, it is given that $y$ varies as $\sqrt{x}$ if $y=24$ and $x=3$, therefore,

$y=k\sqrt{x}\Rightarrow 24=k\times \sqrt{3}\Rightarrow \sqrt{3}k=24\Rightarrow k=\frac{24}{\sqrt{3}}$

Thus, the general equation is $y=\frac{24}{\sqrt{3}}\sqrt{x}$.

Since $x=16$, therefore,

$y=\frac{24}{\sqrt{3}}\sqrt{x}\Rightarrow y=\frac{24}{\sqrt{3}}\sqrt{16}\Rightarrow y=\frac{24}{\sqrt{3}}\times 4\Rightarrow y=\frac{96\times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\Rightarrow y=\frac{96\sqrt{3}}{3}\Rightarrow y=32\sqrt{3}$

Hence, $y=32\sqrt{3}$ when $x=16$.