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Math Help - Percentage change question

  1. #1
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    Percentage change question

    I'll just set out the numbers first then say what I would like to know.

    351 195 65 39 650
    54% 30% 10% 6% 100%

    would there be a way to figure out the smallest possible whole number values for A-D that would need to be added to make the numbers represent these percentages.

    A+351 B+195 C+65 D+39 A+B+C+D+650
    35% 32% 22% 11% 100%

    please note I'm not a very mathy person so intricate explanations will be wasted on me :P. I'm really just concerned if its possible and how simple/difficult it is to do.

    Thanks.
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  2. #2
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    Quote Originally Posted by clovis View Post
    I'll just set out the numbers first then say what I would like to know.

    351 195 65 39 650
    54% 30% 10% 6% 100%

    would there be a way to figure out the smallest possible whole number values for A-D that would need to be added to make the numbers represent these percentages.

    A+351 B+195 C+65 D+39 A+B+C+D+650
    35% 32% 22% 11% 100%

    please note I'm not a very mathy person so intricate explanations will be wasted on me :P. I'm really just concerned if its possible and how simple/difficult it is to do.

    Thanks.
    \frac{x}{351} = 0.54

    x= 190.

    \frac{190}{A+352} = 0.35
    Now solve for A.
    Repeat the same thing for other numbers.
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  3. #3
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    Lexington, MA (USA)
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    Hello, clovis!

    If you are "not a very mathy person", you'll hate this approach.
    Perhaps others will be amused by the Herculean task I've suggested.


    I'll just set out the numbers first, then say what I would like to know.

    . . \begin{array}{|c|c|c|c||c|}351	& 195 & 65 & 39 & 650 \\ \hline <br />
54\% & 30\% & 10\% & 6\%	& 100\% \end{array}

    Would there be a way to figure out the smallest possible whole number values for A\text{ - }D
    that would need to be added to make the numbers represent these percentages.

    . . \begin{array}{|c|c|c|c||c|}A+351 & B+195 & C+65 & D+39 & A+B+C+D+650 \\ \hline<br />
35\% & 32\% & 22\% & 11\% & 100\% \end{array}

    From the first column, we have: . \frac{A+351}{A+B+C+D+650} \:=\:0.35

    . . A+351 \:=\:0.35(A+B+C+D+650) \quad\Rightarrow . A+351 \:=\:0.35A + 0.35B + 0.35C + 0.35D + 227.5

    . . -0.65A + 0.35B + 0.35C + 0.35D \:=\:123.5 \quad\Rightarrow\quad -13A + 7B + 7C + 7D \:=\:2470 .[1]


    Second column: . \frac{B+195}{A+B+C+D+650} \:=\:0.32 \quad\Rightarrow\quad 8A - 17B + 8C + 8D \:=\:-325 .[2]

    Third column: . \frac{C+65}{A+B+C+D+650} \:=\:0.22 \quad\Rightarrow\quad 11A + 11B - 39C + 11D \:=\:-3900 .[3]

    Fourth column: . \frac{D+39}{A+B+C+D+650} \:=\:0.11 \quad\Rightarrow\quad 11A + 11B + 11C - 89D \:=\:-3250 .[4]


    Now all you have to do is solve the system of equations.


    I'll wait in the car . . .
    .
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