Question 193790
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If <i>n</i> cups cost <i>x</i> each and the total cost was $48, then the cost per cup must be *[tex \LARGE x = \frac{48}{n}] dollars.


The revenue from the sales of all but two of the cups given a $3 mark-up must then be:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{48}{n} + 3\right)(n - 2)]


And if the profit is $22, then the total revenue must be $48 + $22 = $70, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{48}{n} + 3\right)(n - 2) = 70]


Multiply:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 48 - \frac{96}{n} + 3n - 6 = 70]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \  - \frac{96}{n} + 3n - 28 = 0]


Multiply by <i>n</i>


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3n^2 - 28n - 96 = 0]


The positive root will be the number of cups purchased.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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