Question 654637
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Let *[tex \LARGE x] represent the amount of 45% solution required and then *[tex \LARGE 160\ -\ x] must be the amount of 25% percent solution required since the grand total has to be 160 gallons.  The amount of pure fertilizer in the 45% solution must then be *[tex \LARGE 0.45x], the amount of pure fertilizer in the 25% solution must be *[tex \LARGE 0.25(160\ -\ x)], and these two amounts must add up to the amount of pure fertilizer in 160 gallons of 30% solution, namely *[tex \LARGE 0.30(160)\ =\ 48]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 0.45x\ +\ 0.25(160\ -\ x)\ =\ 48]


Solve for *[tex \LARGE x], then calculate *[tex \LARGE 160\ -\ x]


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