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lapo4ka [179]
3 years ago
15

How many ounces of a 35% alcohol solution must be mixed with 10ounces of 40% alcohol solution to make a 37% alcohol solution?

Mathematics
1 answer:
spayn [35]3 years ago
5 0
First off, let's change the format of the percentage to decimal.. so 35% is just 35/100 or 0.35 and 40% is 40/100 or 0.4 and so on.

\bf \begin{array}{lccclll}
&amount&concentration&
\begin{array}{llll}
concentrated\\
amount
\end{array}\\
&-----&-------&-------\\
\textit{35\% sol'n}&x&0.35&0.35x\\
\textit{40\% sol'n}&10&0.40&4.00\\
-----&-----&-------&-------\\
mixture&y&0.37&0.37y
\end{array}

now, whatever "x" amount is, we know that x + 10 must add up to "y". x + 10 =  y, because both quantities added will give the mixture amount.

and whatever 0.35x + 4 = 0.37y, because both concentrated amounts must give the 37% of the mixture.

\bf \begin{cases}
x+10=\boxed{y}\\
0.35x+4=0.37y\\
----------\\
0.35x+4=0.37\left( \boxed{x+10} \right)
\end{cases}

solve for "x", to see how much of the 35% solution will be needed.

what about "y"? well, x + 10 = y.
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