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Helen [10]
3 years ago
7

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?
Mathematics
1 answer:
sergeinik [125]3 years ago
5 0
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.
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