Question

A metallurgist needs to make 12.4 lb. of an 10) alloy containing 50% gold. He is going to melt and combine one metal that is 60% gold with another metal that is 40% gold.How much of each should he use?

Answer 1

now, let's say, we add "x" lbs of the 60% gold alloy, so.. how much gold is in it? well, is just 60%, so (60/100) * x, or 0.6x.

likewise, if we use "y" lbs of the 40% alloy, how much gold is in it? well, 40% of y, or (40/100) * y, or 0.4y.

now, whatever "x" and "y" are, their sum must be 12.4 lbs.

we also know that the gold amount in each added up, must equal that of the 50% resulting alloy.


`\bf \begin{array}{lccclll} &\stackrel{lbs}{amount}&\stackrel{gold~\%}{quantity}&\stackrel{gold}{quantity}\\ &------&------&------\\ \textit{60\% alloy}&x&0.6&0.6x\\ \textit{40\% alloy}&y&0.4&0.4y\\ ------&------&------&------\\ \textit{50\% alloy}&12.4&0.50&6.2 \end{array}`


`\bf \begin{cases} x+y=12.4\implies \boxed{y}=12.4-x\\ 0.6x+0.4y=6.2\\ -------------\\ 0.6x+0.4\left( \boxed{12.4-x} \right)=6.2 \end{cases} \\\\\\ 0.6x-0.4x+4.96=6.2\implies 0.2x=1.24\implies x=\cfrac{1.24}{0.2} \\\\\\ x=\stackrel{lbs}{6.2}`

how much of the 40% alloy? well, y = 12.4 - x.