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Thread: puzzle problems

  1. #1
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    puzzle problems

    Qu: If a pounds of sugar and b pounds of coffee together cost c cents, while d pounds of sugar and e ppunds of coffee together cost f cents. what is the price of one pound of each?

    I am trying to use simultaneous equations to do it but i cannot find a way to do it.

    $ a + b = c$
    $ d + e = f$

    the problem is that if i do this:
    $ae + be = ce$
    $db + be = fb$

    I can only eliminate be but i am stuck with this:
    $ae - db = ce - bf$

    is there something i am not seeing.
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  2. #2
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    Re: puzzle problems

    Quote Originally Posted by bigmansouf View Post
    Qu: If a pounds of sugar and b pounds of coffee together cost c cents, while d pounds of sugar and e ppunds of coffee together cost f cents. what is the price of one pound of each?

    I am trying to use simultaneous equations to do it but i cannot find a way to do it.

    $ a + b = c$
    $ d + e = f$

    the problem is that if i do this:
    $ae + be = ce$
    $db + be = fb$

    I can only eliminate be but i am stuck with this:
    $ae - db = ce - bf$

    is there something i am not seeing.
    I would say that you are not seeing anything. The equations you write make no sense at all! "a", "b", "d", and "e" are all weights in pounds while "c" and "f" are amounts of money in cents. You cannot add "pounds" and get "cents". Also you are asked to find the price per pound but there is nothing in your equations representing those amounts.

    The total cost of "a" pounds of something is "ax" where x is the cost per pound. Letting "x" be the cost per pound of sugar, and "y" the cost per pound of coffee, your equations should be
    ax+ by= c and dx+ ey= f. Solve those equations for x and y.
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  3. #3
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    Re: puzzle problems

    Geezzz BigMan, looks like you're making
    up problems after too many beers !

    To start, you need weight variables:
    u = sugar cost per pound
    v = coffee cost per pound

    au + bv = c
    du + ev = f

    Can you follow this:
    au + bv + du + ev = c + f

    u(a + d) = c + f - v(b - e)

    u = [c + f - v(b - e)] / (a + d)
    That gives the price of 1 pound of sugar.
    You can similarly solve for v.

    Try it using an example, like:
    a=3, b=5, c=38
    d=7, e=4, f=58
    Solve for u and v.
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  4. #4
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    Re: puzzle problems

    Quote Originally Posted by DenisB View Post
    Geezzz BigMan, looks like you're making
    up problems after too many beers !

    To start, you need weight variables:
    u = sugar cost per pound
    v = coffee cost per pound

    au + bv = c
    du + ev = f

    Can you follow this:
    au + bv + du + ev = c + f

    u(a + d) = c + f - v(b - e)

    u = [c + f - v(b - e)] / (a + d)
    That gives the price of 1 pound of sugar.
    You can similarly solve for v.

    Try it using an example, like:
    a=3, b=5, c=38
    d=7, e=4, f=58
    Solve for u and v.
    thanks
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