# Sharon has two solutions available in the lab, one solution with 6% alcohol and another with 11% alcohol. How much of each should she mix together to obtain 10 gallons of a solution that contains 7% alcohol?

Dec 5, 2017

8 gallons at 6%
2 gallons at 11%

#### Explanation:

Let the solution measure of 6% concentration be ${S}_{6}$
Let the solution measure of 11% concentration be ${S}_{11}$

For concentrations we have:

$\left[{S}_{6} \times \frac{6}{100}\right] + \left[{S}_{11} \times \frac{11}{100}\right] = 10 \times \times \frac{7}{100}$

$\frac{6 {S}_{6}}{100} + \frac{11 {S}_{11}}{100} = \frac{7}{10} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . E q u a t i o n \left(1\right)$

For volume we have:

${S}_{6} + {S}_{11} = 10$

Thus ${S}_{6} = 10 - {S}_{11} \text{ } \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . E q u a t i o n \left(2\right)$
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Use $E q n \left(2\right)$ to substitute for ${S}_{6}$ in $E q n \left(1\right)$

color(green)( (6color(red) (S_6))/100+(11S_11)/100=7/10 color(white)("d")->color(white)("dd")(6(color(red) (10-S_11)))/100+(11S_11)/100=7/10

$\textcolor{w h i t e}{\text{dddddddddddddddd")->color(white)("ddd")-(6S_11)/100color(white)("d}} + \frac{11 {S}_{11}}{100} = \frac{7}{10} - \frac{6}{10}$

$\textcolor{w h i t e}{\text{dddddddddddddddd")->color(white)("dddddddddddddd}} \frac{5 {S}_{11}}{100} = \frac{1}{10}$

color(white)("dddddddddddddddd")->color(white)("dddddd")S_11=1/10xx100/5=2" gallons"

From this ${S}_{6} = 10 - 2 = 8 \text{ gallons}$