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How many gallons each of
25% alcohol and
10% alcohol should be mixed to obtain 15 gal of
12% alcohol?

Gallons of

Solution

Percent

(as a decimal)

Gallons of

Pure

Alcohol

x

25%=0.25

y

10%=0.1

15

12%=

How many gallons of
25% alcohol should be in the mixture?
◻ gal

How many gallons each of $25 \%$ alcohol and $10 \%$ alcohol should be mixed to obtain $15$ gal of $12 \%$ alcohol? $\newline$ \begin{tabular}{|c|c|c|} $\newline$ \hline \begin{tabular}{c} $\newline$ Gallons of \\ $\newline$ Solution $\newline$ \end{tabular} & \begin{tabular}{c} $\newline$ Percent \\ $\newline$ (as a decimal) $\newline$ \end{tabular} & \begin{tabular}{c} $\newline$ Gallons of \\ $\newline$ Pure \\ $\newline$ Alcohol $\newline$ \end{tabular} \\ $\newline$ \hline $x$ & $25 \%=0.25$ & \\ $\newline$ \hline $y$ & $10 \%=0.1$ & \\ $\newline$ \hline $15$ & $12 \%=$ & \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ How many gallons of $25 \%$ alcohol should be in the mixture? $\square$ gal

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Weighted averages: word problems

Full solution

Q. How many gallons each of

25 \%

alcohol and

10 \%

alcohol should be mixed to obtain

15

gal of

12 \%

alcohol?

\newline

\begin{tabular}{|c|c|c|}

\newline

\hline \begin{tabular}{c}

\newline

Gallons of \\

\newline

Solution

\newline

\end{tabular} & \begin{tabular}{c}

\newline

Percent \\

\newline

(as a decimal)

\newline

\end{tabular} & \begin{tabular}{c}

\newline

Gallons of \\

\newline

Pure \\

\newline

Alcohol

\newline

\end{tabular} \\

\newline

\hline

x

&

25 \%=0.25

& \\

\newline

\hline

y

&

10 \%=0.1

& \\

\newline

\hline

15

&

12 \%=

& \\

\newline

\hline

\newline

\end{tabular}

\newline

How many gallons of

25 \%

alcohol should be in the mixture?

\square

gal

Define Variables: Let $x$ be the gallons of $25\%$ alcohol and $y$ be the gallons of $10\%$ alcohol. We know that $x + y = 15$ because we want $15$ gallons of the final mixture.
Calculate Total Alcohol: The total amount of pure alcohol in the final mixture should be $15$ gallons times $12\%$ , which is $15 \times 0.12 = 1.8$ gallons.
Alcohol Content Equation: The amount of pure alcohol in the $25\%$ solution is $0.25x$ and in the $10\%$ solution is $0.1y$ . So, the equation for the alcohol content is $0.25x + 0.1y = 1.8$ .
Solve System of Equations: We have two equations now: $x + y = 15$ and $0.25x + 0.1y = 1.8$ . We can solve this system of equations by substitution or elimination. Let's use substitution.
Express $y$ in terms of $x$ : From the first equation, we can express $y$ in terms of $x$ : $y = 15 - x$ .
Substitute $y$ into Equation: Substitute $y = 15 - x$ into the second equation: $0.25x + 0.1(15 - x) = 1.8$ .
Solve for x: Now, let's solve for x: $0.25x + 1.5 - 0.1x = 1.8$ .
Combine Like Terms: Combine like terms: $0.15x + 1.5 = 1.8$ .
Subtract $1$ . $5$ : Subtract $1.5$ from both sides: $0.15x = 0.3$ .
Divide to Find x: Divide by $0.15$ to find $x$ : $x = \frac{0.3}{0.15}.$

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