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Thread: An acceptable answer for a question that relates to simultaneous equations?

  1. #1
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    An acceptable answer for a question that relates to simultaneous equations?

    I recently happened upon a problem. I have produced a solution, however, I am not sure whether it is acceptable or not. Here is the problem:

    $\displaystyle a$ apples weigh the same as $\displaystyle b$ bananas. $\displaystyle B$ bananas weigh the same as $\displaystyle C$ cherries. How many cherries weigh the same as $\displaystyle A$ apples?

    I believe the solution to be $\displaystyle A$ apples$\displaystyle =\frac{AbC}{aB}$ cherries. The problem is that I am not sure whether it is acceptable to have $\displaystyle A$ on both sides of the equation. I have asked a few individuals and opinions vary. Is the solution acceptable?

    Thank you. All help on this problem is appreciated.
    Last edited by JRichardson1729; Oct 18th 2010 at 11:30 AM.
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  2. #2
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    Quote Originally Posted by JRichardson1729 View Post
    I believe the solution to be $\displaystyle A$ apples$\displaystyle =\frac{AbC}{aB}$ cherries.
    If A = AbC / aB
    then aB = AbC / A
    so aB = bC

    I'll leave it at that, since the problem makes no sense to me...
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    Quote Originally Posted by Wilmer View Post
    If A = AbC / aB
    then aB = AbC / A
    so aB = bC

    I'll leave it at that, since the problem makes no sense to me...
    Unfortunately, the question asks how many cherries weigh the same as $\displaystyle A$ apples, so I think that it is necessry to have an A somewhere in the solution.

    This is my method:

    Posit that the weights of an apple, banana and cherry are $\displaystyle x, y, z$ respectively.

    $\displaystyle ax=by, By=Cz$

    $\displaystyle aBx=bBy, bBy=bCz$

    $\displaystyle aBx=bCz$

    $\displaystyle x=\frac{bC}{aB}z$

    $\displaystyle Ax=\frac{AbC}{aB}z$

    I'm relatively sure that you need the last step, or at least something like it, but I am not sure about how to only get $\displaystyle A$ on one side (with $\displaystyle x$ i.e. the weight of an apple).
    Last edited by JRichardson1729; Oct 18th 2010 at 11:39 AM. Reason: Organisation
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    Bump.
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  5. #5
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    I think it's good.

    If you remove the A on both sides, you'll get the number of cheries that will weigh the same as one apple. So, your solution is good. At least, that's what I think.
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