The value of $y$ varies inversely as $\sqrt x$ and when $x=24$, $y=15$.
What is $x$ when $y=3$?
I'm having trouble on this and I don't get why it's not $\frac{2\sqrt6\cdot15}{3}=10\sqrt6$?
Am I misinterpreting the problem? This is how I learned inverse proportion so I'm really unsure.
Your mistake is you got $\sqrt x=10\sqrt6$ and not $x=10\sqrt6$. You must square to get $x=600$.
For your note, here is an elaborative answer.
Two variable quantities $x$ and $y$ are said to be inversely proportional if and only if their product is a constant. Symbolically, $$x\propto\frac1y\iff xy=k$$ for some constant $k$.
Now, for the given problem, we should have $y\sqrt x=k$. Now, putting the given values, $k=15\sqrt{24}$. Finally, for $y=3$, $$3\sqrt x =15\sqrt{24}\implies\boxed{x=600}$$
Hope this helps. Ask anything if not clear :)
Notice your approach is right but the value you get $10\sqrt{6}$ is the value of $\sqrt x$ not $x$. Therefore $$\sqrt x=10\sqrt6$$ $$x=(10\sqrt{6})^2=600$$
Alternatively,
The value of y varies inversely as $\sqrt x$ i.e.
$$y\propto\frac{1}{\sqrt x}$$$$x\propto \frac{1}{y^2}$$$$ xy^2=\text{constant}$$ $$\therefore x_1y_1^2=x_2y_2^2$$ $$\implies 24\cdot 15^2=x_2\cdot 3^2$$$$x_2=600$$