How can I find the percentage decrease? (Algebra)

**PythagorasNeophyte** · September 24th 2010, 02:23 AM

Hello everyone. I have an algebra question:

Given that y varies directly as the cube of x, find the percentage decrease in y if x is halved.

Since $y$ is in direct proportion with $x^3$ ,
let $y=kx^3$ , where k is a constant.

If x is halved, it would be $\dfrac{x^3}{2}$ .

But it doesn't solve the problem.

I wonder how the percentage can be calculated when there are no values for $x$ and $y$ ...

**Prove It** · September 24th 2010, 02:34 AM

Actually, if $x$ is halved then you need to replace $x$ with $\frac{x}{2}$ , which gives

$y = k\left(\frac{x}{2}\right)^3$

$= \frac{kx^3}{8}$ .

Since you end up with $\frac{1}{8}$ or $12.5\%$ of what you started with, that means the percentage decrease is $87.5\%$ .

**PythagorasNeophyte** · September 24th 2010, 06:50 PM

Hello Prove It,

many thanks for your reply, however, I still need clarifications about finding the percentage decrese.

Do you mean that $\dfrac{kx^2}{8}$ can be written as $\dfrac{1}{8}$ ?

How can $\dfrac{kx^2}{8}$ be known as $\dfrac{1}{8}$ ?

**Prove It** · September 24th 2010, 06:54 PM

Dividing by $8$ is the same as finding $\frac{1}{8}$ of that quantity.

So $\frac{kx^2}{8} = \frac{1}{8}kx^2$ .

**PythagorasNeophyte** · September 24th 2010, 07:09 PM

I see...so I was wrong when I wrote $\dfrac{x^3}{2}$ , when it should be $y = k\left(\frac{x}{2}\right)^2$ .

**Prove It** · September 24th 2010, 07:19 PM

Originally Posted by PythagorasNeophyte

I see...so I was wrong when I wrote $\dfrac{x^3}{2}$ , when it should be $y = k\left(\frac{x}{2}\right)^2$ .

Correct.

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