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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?

1 Answer

Dec 5, 2017

8 gallons at 6%
2 gallons at 11%

Explanation:

Let the solution measure of 6% concentration be $S_{6}$ $S_6$
Let the solution measure of 11% concentration be $S_{11}$ $S_11$

For concentrations we have:

$[S_{6} \times \frac{6}{100}] + [S_{11} \times \frac{11}{100}] = 10 \times \times \frac{7}{100}$ $[S_6xx6/100]+[S_11xx11/100]=10xxxx7/100$

$(6S_6)/100+(11S_11)/100=7/10" "......................Equation(1)$

For volume we have:

$S_6+S_11=10$

Thus $S_6=10-S_11" ".......................Equation(2)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Use $Eqn(2)$ to substitute for $S_6$ in $Eqn(1)$

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

$color(white)("dddddddddddddddd")->color(white)("ddd")-(6S_11)/100color(white)("d")+(11S_11)/100 =7/10-6/10$

$color(white)("dddddddddddddddd")->color(white)("dddddddddddddd")(5S_11)/100=1/10$

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

From this $S_6=10-2=8" gallons"$

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