SOLUTION: A radiator contains 6 liters of 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution? So far, I know that t

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Question 98496This question is from textbook College Algebra and Trigonometry
: A radiator contains 6 liters of 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution?
So far, I know that the first solution contains 6 liters and I need to multiply that by 25%. Then since we are "replacing" pure antifreeze to the mixture that is 100% that I need to multiply to the unknown mixture of x+6 (which is my second solution) so that I can produce a 33% solution. Does this make any sense? I'm not sure how to do this. Thank You This question is from textbook College Algebra and Trigonometry

Found 2 solutions by ankor@dixie-net.com, stanbon:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
A radiator contains 6 liters of 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution?
:
It looks like they want you to start with 6 and end up with 6.
We could do it like this:
:
Let x = amt to be drained and replaced
:
The % antifreeze equation:
:
.25(6) - .25(x) + 1.0(x) = .33(6)
1.5 - .25x + 1x = 1.98
-.25x + 1x = 1.98 - 1.5
.75x = .48
x = .48/.75
x = .64 liters of pure antifreeze to replace .64 liters of the 25% solution
:
:
Check our solution for x.
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25% amt is now: 6 - .64 = 5.36
:
.25(5.36) + 1.0(.64) = .33(6)
1.34 + .64 = 1.98; confirms our solution
:
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Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website!
A radiator contains 6 liters of 25% antifreeze solution. How much should be drained and replaced with pure antifreeze to produce a 33% antifreeze solution?
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Note: Keep track of the active ingredient.
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Start: 25%*6 liters = (3/2)liters of active
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You subtract x liters but it is still 25% active:
You now have: (3/2)-(x/4) active;
You add x amount of 100% active:
You now have (3/2)-(x/4)+x active
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That amount of active is 33% of the 6 liters you now have:
EQUATION:
(3/2)-(x/4)+x = 33%*6
(3/2) + (3/4)x = 2
(3/4)x = 1/2
x = (1/2)(4/3)
x = 2/3 liters
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That is the amount of 100% alcohol you should add.
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Cheers,
Stan H.