Q:

A chemical company makes two brands of antifreeze. The first brand is 65% pure antifreeze, and the second brand is 90% pure antifreeze. In order to obtain 40 gallons of a mixture that contains 80% pure antifreeze, how many gallons of each brand of antifreeze must be used?

Accepted Solution

A:
[tex]\bf \begin{array}{lcccl} &\stackrel{solution}{gallons}&\stackrel{\textit{\% of }}{antifreeze}&\stackrel{\textit{gallons of }}{antifreeze}\\ \cline{2-4}&\\ \textit{1st brand}&x&0.65&0.65x\\ \textit{2nd brand}&y&0.90&0.9y\\ \cline{2-4}&\\ mixture&40&0.8&32 \end{array}~\hfill \to \begin{cases} x+y&=40\\ \boxed{x}=40-y\\ \cline{1-2} 0.65x+0.9y&=32 \end{cases} \\\\[-0.35em] ~\dotfill[/tex][tex]\bf \stackrel{\textit{substituting in the 2nd equation}}{0.65\left( \boxed{40-y} \right)+0.9y=32}\implies 26-0.65y+0.9y=32 \\\\\\ 26+0.25y=32\implies 0.25y=6\implies y=\cfrac{6}{0.25}\implies \blacktriangleright y=24 \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{since we know that}}{x=40-y\implies }x=40-24\implies \blacktriangleright x=16 \blacktriangleleft[/tex]