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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.

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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 :)

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  • $\begingroup$ ohhh thank you!!! I get it now. $\endgroup$ – random guy 2000 Apr 7 at 17:09
  • $\begingroup$ @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. :) $\endgroup$ – ultralegend5385 Apr 8 at 2:37
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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.

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

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