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Thread: Another Quadratic Equation

  1. #1
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    Another Quadratic Equation

    I buy a number of cheap calculators for a total of $240, mark each of them up by $2 and sell all but four of them for a total of $240. How many calculators did I buy?

    Okay, so this is what I have done so far:
    Let y represent the cost per calculator, and let x represent the number of calculator.
    Therefore, xy = 240
    And with the addition of the $2.00, minus the four pencils:
    (x-4)(y+2) = 240
    xy+2x-4y-8 = 240

    and now im stumped. what do i do?
    please explain + thanks in advance!

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  2. #2
    Member great_math's Avatar
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    Let $\displaystyle x$ be the no. of calculators. Cost per calculator=$$\displaystyle \frac {240}{x}$

    $\displaystyle (x-4)\left(\frac {240}{x}+2\right)=240$

    $\displaystyle \Rightarrow (x-4)\left(\frac{240+2x}{x}\right)=240$

    $\displaystyle \Rightarrow (x-4)(240+2x)=240x$

    $\displaystyle \Rightarrow 240x+2x^2-960-8x=240x$

    $\displaystyle \Rightarrow 2x^2-8x-960=0$

    $\displaystyle \Rightarrow x^2-4x-480=0$

    $\displaystyle \Rightarrow (x+20)(x-24)=0$

    $\displaystyle \Rightarrow x=-20$(Impossible) or $\displaystyle x=24$

    So $\displaystyle 24$ is the required answer
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  3. #3
    Member great_math's Avatar
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    Quote Originally Posted by foreverbrokenpromises View Post
    I buy a number of cheap calculators for a total of $240, mark each of them up by $2 and sell all but four of them for a total of $240. How many calculators did I buy?

    Okay, so this is what I have done so far:
    Let y represent the cost per calculator, and let x represent the number of calculator.
    Therefore, xy = 240
    And with the addition of the $2.00, minus the four pencils:
    (x-4)(y+2) = 240
    xy+2x-4y-8 = 240

    and now im stumped. what do i do?
    please explain + thanks in advance!

    Now you have to substitute $\displaystyle xy=240$

    $\displaystyle 240+2x-4y-8=240$

    $\displaystyle \Rightarrow 2x-4y=8$

    $\displaystyle \Rightarrow x-2y=4$

    $\displaystyle \Rightarrow x=4+2y$

    Now $\displaystyle xy=240$

    $\displaystyle \Rightarrow (4+2y)y=240$

    $\displaystyle \Rightarrow 2y^2+4y-240=0$

    $\displaystyle \Rightarrow y^2+2y-120=0$

    $\displaystyle \Rightarrow (y+12)(y-10)=0$

    $\displaystyle \Rightarrow y=-12$(not possible ) or $\displaystyle y=10$

    $\displaystyle \Rightarrow x=4+2\times10=24$
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