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Math Help - 1 problem only

  1. #1
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    Post 1 problem only

    Hi! Please help. I will go to class in a few minutes. I don't have an equation for it. Thanks!!!


    A merchant has 3 items on sale: namely, a radio for $50, a clock for $30, and a flashlight for $1. At the end of the day, she has sold a total of 100 of the 3 items and has taken exactly $1000 on the total sales. How many radios did he sell?
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  2. #2
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    "A merchant has 3 items on sale: namely, a radio for $50, a clock for $30, and a flashlight for $1. At the end of the day, she has sold a total of 100 of the 3 items and has taken exactly $1000 on the total sales. How many radios did he sell?"

    This is not a particularly simple problem, as it is over-defined. It turns out to be a test of integer programming. However, this does not mean it should be approached initially any different than any other problem.

    Step #1 - Name Stuff.

    R = # of radios sold
    C = # of clocks sold
    F = # of flashlights sold

    Thus,

    R + C + F = 100
    50*R + 30*C + 1*F = 1000

    The most logical substitution would be F = 100 - R - C, giving

    50*R + 30*C + (100 - R - C) = 1000

    or

    49*R + 29*C = 900

    If this were a continuous problem, there would be infinitely many solutions. It's not a continuous problem. Radios and Clocks don't sell in half-units.

    One thing that will help is noticing that (900 - 49*R)/29 must be an integer.

    Good luck. Let's see what you get.
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  3. #3
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    Joined
    Dec 2007
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    Guess and check

    Use a table instead of a equation.
    $1000-100
    $50-radio $30-clocks $1-flashlights
    50 radios 20 clocks 30 flashlights =$3130.
    The number is too big, so you must decrease the number of radios and increase the number of clocks and/or flashlights.
    Hope this helps
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