Question 566596: How many gallons of 20% alcohol must be mixed with 3 gallons of 70% alcohol to get a mixture that is 30% alcohol Answer by Maths68(1474) (Show Source):
You can put this solution on YOUR website! Solution A
Amount =x gallons
Concentration =20% = 0.2
Solution B
Amount = 3 gallons
Concentration =70% = 0.70
Resultant Solution
Amount =(x+3)gallons
Concentration = 30% = 0.3
[Amount Solution A * Concentration A] + [Amount Solution B * Concentration of B] = Amount of Resultant * Concentration of resultant
(x)(0.2)+(3)(0.7)=(x+3)(0.3)
0.2x+2.1=0.3x+0.9
0.2x-0.3x=0.9-2.1
-0.1x=-1.2
-0.1x/-0.1=-1.2/-0.1
x=12
Solution A
Amount = x = 12 gallons
Concentration =20% = 0.2
12 gallons of 20% alcohol must be mixed with 3 gallons of 70% alcohol to get a 15 gallons of mixture that is 30% alcohol
