Question 98496
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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It looks like they want you to start with 6 and end up with 6.
We could do it like this:
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Let x = amt to be drained and replaced
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The % antifreeze equation:
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.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
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Check our solution for x.
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25% amt is now: 6 - .64 = 5.36
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.25(5.36) + 1.0(.64) = .33(6)
1.34 + .64 = 1.98; confirms our solution
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