SOLUTION: how much pure antifreeze must be added to 12 gallons of 10% to make a 80% antifreeze solution?

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Question 538882: how much pure antifreeze must be added to 12 gallons of 10% to make a 80% antifreeze solution?
Answer by unlockmath(1688) About Me  (Show Source):
You can put this solution on YOUR website!
Hello,
Best way to solve this is draw 3 containers. Put a plus between the first two and an equal sign between the second and third container.
(X gallons) + (12 gallons ) = (Y gallons)
100% (X) + 10% (12) = 80% (y)

We know the first container has pure 100% antifreeze and the second container has 12 gallons of 10%. When combined it needs to make Y gallons of 80%.
Follow so far?
Make 2 equations:
x+12=y
1x+.1(12)=.8y (Change the percent to decimals)
Use substitution to solve.
1x+.1(12)=.8(x+12)
Expand out to get:
x+1.2=.8x + 9.6
Subtract .8x and subtract 1.2 to get:
.2x=8.4
Divide by .2:
x= 42 gallons needed of pure antifreeze.
Make sense?
RJ
www.math-unlock.com