Question 361495
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Let *[tex \Large x] represent the original cost of the shorts.  Let *[tex \Large y] represent the original cost of the tops.


Then the amount saved on shorts would be *[tex \Large 0.15x] and the amount saved on tops would be *[tex \Large 0.20y].  Likewise, the amount spent on shorts would be *[tex \Large 0.85x] and the amount spent on tops *[tex \Large 0.80y].  So we can write:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.15x\ +\ 0.20y\ =\ 24]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.85x\ +\ 0.80y\ =\ 135]


Multiply the first equation by -4 and then add the equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.25x\ +\ 0.00y\ =\ 39]


Solve for *[tex \Large x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 156]


Substituting in Equation 1 and solving for *[tex \Large y]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y\ =\ 3]


Which was the cost of the tops at regular price, hence she spent:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.8\,\cdot\,3.00\ =\ 2.40]


Sounds like a fire sale at Ross to me.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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