Question 514144: How many ounces of a 10% alcohol solution must be added to 40 ounces of a 25% solution to reduce it to a 15% solution? Found 2 solutions by stanbon, Maths68:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! How many ounces of a 10% alcohol solution must be added to 40 ounces of a 25% solution to reduce it to a 15% solution?
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Equation:
alcohol + alcohol = alcohol
0.10x + 0.25*40 = 0.15(x+40)
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Multiply thru by 100:
10x + 25*40 = 15x + 15*40
5x = 10*40
x = 80 oz (amt. of 10% solution needed)
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Cheers,
Stan H.
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You can put this solution on YOUR website! Solution A
Amount =x ounce
Concentration =10% = 0.1
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Solution B
Amount = 40 ounce
Concentration =25% = 0.25
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Resultant Solution
Amount =(40+x) ounce
Concentration =15%=0.15
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[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
(x*0.1)+(40)(0.25)=(40+x)(0.15)
0.1x+10=6+0.15x
0.1x-0.15x=6-10
-0.05x=-4
-0.05x/0.05=-4/0.05
x=80
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Result
80 ounces of a 10% alcohol solution must be added to 40 ounces of a 25% solution to reduce it to a 15% solution?
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