Question

a metallurgist has one alloy containing 30% aluminum and another containing 54% aluminum. How many pounds of alloy must he use to make 48 pounds of a third alloy containing 52% aluminum? (round to two decimal places if necessary).

Answer 1

`x=\textit{lbs of aluminum at 30\%}\\\\ ~~~~~~ 30\%~of~x\implies \cfrac{30}{100}(x)\implies 0.3 (x) \\\\\\ y=\textit{lbs of aluminum at 54\%}\\\\ ~~~~~~ 54\%~of~y\implies \cfrac{54}{100}(y)\implies 0.54 (y) \\\\\\ \textit{48 lbs of aluminum at 52\%}\\\\ ~~~~~~ 52\%~of~48\implies \cfrac{52}{100}(48)\implies 0.52 (48)\implies 24.96 \\\\[-0.35em] ~\dotfill`

`\begin{array}{lcccl} &\stackrel{lbs}{quantity}&\stackrel{\textit{\% of lbs that is}}{\textit{AL only}}&\stackrel{\textit{lbs of}}{\textit{AL only}}\\ \cline{2-4}&\\ \textit{1st alloy}&x&0.3&0.3x\\ \textit{2nd alloy}&y&0.54&0.54y\\ \cline{2-4}&\\ mixture&48&0.52&24.96 \end{array}~\hfill \begin{cases} x + y = 48\\\\ 0.3x+0.54y=24.96 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=48\implies y=48-x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation from above}}{0.3x~~ + ~~0.54(48-x)~~ = ~~24.96}\implies 0.3x+25.92-0.54x=24.96 \\\\\\ 0.30x-0.54x=-0.96\implies -0.24x=-0.96\implies x=\cfrac{-0.96}{-0.24} \\\\\\ \boxed{x=4}\hspace{9em}y=48-x\implies \boxed{y=44}`