Question 1036653
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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

{{{(0.1*12 +x)/(12+x)}}} = 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 = {{{3.6/0.6}}} = 6.

<U>Answer</U>. 6 gallons of pure antifreeze is needed.
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