Question
A solution containing $12\ \%\ $ alcohol is to be mixed with a solution containing $4\ \%\ $ alcohol to make $20$ gallons of solution containing $9\ \%\ $ alcohol. How much of each solution should be used?
✓
Answer
Let $x$ gallons of $12\ \% \ $ alcohol and $y$ gallons of $4\ \% \ $ alcohol be mixed.
Then, we have
$ x+y=20 \dots....(i) $
And, $12\ \% \ $ of $x+4\ \% \ $ of $y=9\ \% \ $ of $20$
$ \Rightarrow \frac{12}{100} \times+\frac{4}{100} y$
$=\frac{9}{100} \times 20$
$\Rightarrow 12 x+4 y=180$
$\Rightarrow 3 x+y=45 \ldots .(i i) $
Subtracting eqn. $(i)$ from eqn. $(ii),$ we get
$ 2 x=25$
$\Rightarrow x=12.5$
$\Rightarrow 12.5+y=20$
$\Rightarrow y=7.5 $
Hence $,12.5$ gallons of $12\ \% \ $ alcohol and $7.5$ gallons of $4\ \% \ $ alcohol should be used.
Then, we have
$ x+y=20 \dots....(i) $
And, $12\ \% \ $ of $x+4\ \% \ $ of $y=9\ \% \ $ of $20$
$ \Rightarrow \frac{12}{100} \times+\frac{4}{100} y$
$=\frac{9}{100} \times 20$
$\Rightarrow 12 x+4 y=180$
$\Rightarrow 3 x+y=45 \ldots .(i i) $
Subtracting eqn. $(i)$ from eqn. $(ii),$ we get
$ 2 x=25$
$\Rightarrow x=12.5$
$\Rightarrow 12.5+y=20$
$\Rightarrow y=7.5 $
Hence $,12.5$ gallons of $12\ \% \ $ alcohol and $7.5$ gallons of $4\ \% \ $ alcohol should be used.
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