# Help with inverse proportions

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 $$x=600$$.

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 :)

• ohhh thank you!!! I get it now. – random guy 2000 Apr 7 at 17:09
• @randomguy2000: Then click on the tick button below the vote score on my post. It is a mark to ensure that the question is answered. :) – ultralegend5385 Apr 8 at 2:37

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.

$$y\propto\frac{1}{\sqrt x}$$$$x\propto \frac{1}{y^2}$$$$xy^2=\text{constant}$$ $$\therefore x_1y_1^2=x_2y_2^2$$ $$\implies 24\cdot 15^2=x_2\cdot 3^2$$$$x_2=600$$