# Thread: How can I find the percentage decrease? (Algebra)

1. ## How can I find the percentage decrease? (Algebra)

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

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

But it doesn't solve the problem.

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

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

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

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

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

3. Hello Prove It,

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

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

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

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

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

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