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Sagot :
we know that A has 60% of salt, so if way we had "x" amount of ounces of A solution the amount in it will be 60% of "x", or namely (60/100)*x = 0.6x.
Likewise for solution B if we had "y" ounces of it, the amount of salt in it will be (75/100) * y or 0.75y, thus
[tex]\begin{array}{lcccl} &\stackrel{solution}{quantity}&\stackrel{\textit{\% of }}{amount}&\stackrel{\textit{oz of }}{amount}\\ \cline{2-4}&\\ A&x&0.60&0.6x\\ B&y&0.75&0.75y\\ \cline{2-4}&\\ mixture&60&0.65&39 \end{array}~\hfill \begin{cases} x+y=60\\ 0.6x+0.75y=39 \end{cases} \\\\\\ \stackrel{\textit{since we know that}}{x+y=60}\implies y = 60 - x~\hfill \stackrel{\textit{substituting on the 2nd equation}}{0.6x+0.75(60-x)=39}[/tex]
[tex]0.6x+45-0.75x=39\implies -0.15x+45=39\implies -0.15x=-6 \\\\\\ x = \cfrac{-6}{-0.15}\implies \boxed{x = 40} \\\\\\ \stackrel{\textit{we know that}}{y = 60 - x}\implies y = 60-40\implies \boxed{y = 20}[/tex]
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