Question

A 10​% solution of fertilizer is to be mixed with a ​60% solution of fertilizer in order to get 125 gallons of a 50​% solution. How many gallons of the 10​% solution and 60​% solution should be​ mixed?

Answer 1

`x=\textit{gallons of solution at 10\%}\\\\ ~~~~~~ 10\%~of~x\implies \cfrac{10}{100}(x)\implies 0.10 (x) \\\\\\ y=\textit{gallons of solution at 60\%}\\\\ ~~~~~~ 60\%~of~y\implies \cfrac{60}{100}(y)\implies 0.6 (y) \\\\\\ \textit{150 gallons of solution at 50\%}\\\\ ~~~~~~ 50\%~of~150\implies \cfrac{50}{100}(150)\implies 75 \\\\[-0.35em] ~\dotfill`

`\begin{array}{lcccl} &\stackrel{gallons}{quantity}&\stackrel{\textit{\% of gallons that is}}{\textit{fertilizer only}}&\stackrel{\textit{gallons of}}{\textit{fertilizer only}}\\ \cline{2-4}&\\ \textit{Sol'n of 10\%}&x&0.10&0.10x\\ \textit{Sol'n of 60\%}&y&0.60&0.60y\\ \cline{2-4}&\\ mixture&150&0.5&75 \end{array} \\\\\\ \begin{cases} x + y = 150\\\\ 0.10x+0.60y=75 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=150}\implies y=150-x \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{using the 2nd equation}}{0.10x+0.60y=75}\implies \stackrel{\textit{substituting on the 2nd equation from above}}{0.10x+0.60(150-x)=75} \\\\\\ 0.10x+90-0.60x=75\implies 90-0.50x=75\implies 90=0.50x+75 \\\\\\ 15=0.50x\implies \cfrac{15}{0.50}=x\implies \boxed{30=x}\hspace{5em}\stackrel{ 150~~ - ~~30 }{\boxed{y=120}}`