Question 511334: How many gallons of water must be added to 40 gallons of 10% alcohol solution to produce an 8% alcohol solution? Found 3 solutions by josmiceli, stanbon, Maths68:Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website! Let = gallons of water to be added is gallons of alcohol in solution
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10 gallons of water must be added
You can put this solution on YOUR website! How many gallons of water must be added to 40 gallons of 10% alcohol solution to produce an 8% alcohol solution?
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Equation:
water + water = water
x + 0.90*40 = 0.92(x+40)
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Multiply thru by 100 to get:
100x + 90*40 = 92x + 92*40
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8x = 2*40
x = 10 gallons (amt. of water to add)
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cheers,
stan H.
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You can put this solution on YOUR website! Water
Amount = x
Concentration = 0% =0 (Water is pure therefore alcohal concentration will be 0%)
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Alcohol
Amount =40
Concentration =10% = 0.10
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Mixture
Amount = (40+x) gallons
Concentration = 8%=0.08
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[Water Amount *Concentration of Water] + [Alcohol Amount *Alcohol Concentration ] = Amount of Mixture * Concentration of Mixture
(x*0)+(40)*(0.10)=(0.08)*(40+x)
0+4=3.2+0.08x
4-3.2=0.08x
0.8=0.08x
0.8/0.08=0.08x/0.08
10=x
x=10
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Water
Amount = x =10 gallons
Concentration = 0% =0
Alcohol
Amount =40 gallons
Concentration =10% = 0.10
Mixture
Amount = (40+x) = (40+10) = 50 gallons
Concentration = 8%=0.08
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10 gallons of water must be added to 40 gallons of 10% alcohol solution to produce a 50 gallons of 8% alcohol solution.
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