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

  1. #1
    Junior Member PythagorasNeophyte's Avatar
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    Question 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 $\displaystyle y$...
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  2. #2
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    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 $\displaystyle 87.5\%$.
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  3. #3
    Junior Member PythagorasNeophyte's Avatar
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    Hello Prove It,

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

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

    So $\displaystyle \frac{kx^2}{8} = \frac{1}{8}kx^2$.
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  5. #5
    Junior Member PythagorasNeophyte's Avatar
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    I see...so I was wrong when I wrote $\displaystyle \dfrac{x^3}{2}$, when it should be $\displaystyle y = k\left(\frac{x}{2}\right)^2$.
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    Quote Originally Posted by PythagorasNeophyte View Post
    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.
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