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Math Help - Percent Greater than Problem?

  1. #1
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    Percent Greater than Problem?

    If q is 10% greater than p and r is 10% greater than y, qr is what percent greater than py?
    I set up the following equations, Q=10/100(p)+p
    R=10/100(y)+y
    QR=x/100(py)+py
    How do I solve them? I would appreciate any response.
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  2. #2
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    Re: Percent Greater than Problem?

    Hello, Sewilliams13!

    \text{If }q\text{ is 10\% greater than }p\text{, and }r\text{ is 10\% greater than }y,
    qr\text{ is what percent greater than }py\,?

    I prefer simpler fractions.


    [1]\;q \:=\:p + \tfrac{1}{10}p \quad\Rightarrow\quad q \:=\:\tfrac{11}{10}p

    [2]\;r \:=\:y + \tfrac{1}{10}y \quad\Rightarrow\quad r \:=\:\tfrac{11}{10}y


    Multiply [1] and [2]: . qr \:=\:\left(\tfrac{11}{10}\right)^2py \;=\;\tfrac{121}{100}py \;=\;\left(1 + \tfrac{21}{100}\right)py

    . . . . . . . . . . . . . . . . qr \;=\;py + \tfrac{21}{100}py


    \text{Therefore, }qr\text{ is 21\% greater than }py.

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  3. #3
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    Re: Percent Greater than Problem?

    In [1] & [2] why did you take a way the p and the y in the next equation
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  4. #4
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    Re: Percent Greater than Problem?

    They were not "taken away"; they were simply added together; like, 1 + 1/10 = 10/10 + 1/10 = 11/10
    Thanks from Sewilliams13
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  5. #5
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    Re: Percent Greater than Problem?

    -_-Ahh...Thank you so much!
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