SOLUTION: How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 40% antifreeze solution?

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Question 1036653: How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 40% antifreeze solution?
Found 3 solutions by Boreal, MathTherapy, ikleyn:
Answer by Boreal(15235)   (Show Source):
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.10x+(12-x)=.40(12). The amount of 10% antifreeze is .10x, pure antifreeze is 12 gallons- what is x, and the end result is 40% antifreeze.
.10x+12-x=4.8
-.90x=-7.2
x=8 gallons of .10
12-x=4 gallons of pure.
2/3 is 10%, and the resultant is 2/3s of the way from 100% down to 10%.

Answer by MathTherapy(10552)   (Show Source):
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How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 40% antifreeze solution?

Answer by ikleyn(52781)   (Show Source):
You can put this solution on YOUR website!
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How much pure antifreeze must be added to 12 gallons of 10% antifreeze to make a 40% antifreeze solution?
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Let "x" be the volume of pure antifreeze to be added, in gallons. Then the total volume of the solution will be 12+x gallons. The amount of the pure antifreeze in the new solution will be 0.1*12 + x gallons. The percentage concentration equation is = 0.4. Simplify it: 1.2 + x = 0.4*12 + 0.4x, 1.2 + x = 4.8 + 0.4x, x - 0.4x = 4.8 - 1.2 0.6x = 3.6, x = = 6. Answer. 6 gallons of pure antifreeze is needed.