Question 497088
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Can it be right?  Well, let's see:  Your claim is that you need to mix 31.5 gallons of 75% solution with 30 gallons of 25% solution to make 61.5 gallons of 60%.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.75\ \times\ 31.5\ =\ 23.625]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.25\ \times\ 30\ =\ 7.5]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 23.625\ +\ 7.5\ =\ 31.125]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.60\ \times\ 61.5\ =\ 36.9]


And the answer to the question you posed is a big resounding NO.


Now, for the treasure map to the answer to the real question:


Let *[tex \Large x] represent the amout of 75% solution, then the resulting amount of 60% solution must be *[tex \Large x] plus the 30 gallons of 25% solution.  Therefore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.75x\ +\ 0.25(30)\ =\ 0.6(x\ +\ 30)]


Solve for *[tex \Large x].


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