SOLUTION: What formula would I use to solve this problem?
A radiator contains 6 liters of a 25% anti-freeze solution. How much should be drained and replaced with pure anti-freeze to prod
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A radiator contains 6 liters of a 25% anti-freeze solution. How much should be drained and replaced with pure anti-freeze to prod
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Question 288228: What formula would I use to solve this problem?
A radiator contains 6 liters of a 25% anti-freeze solution. How much should be drained and replaced with pure anti-freeze to produce a 33% solution?
I've been working on a similar problem that I know the answer to in order to figure out how to solve this problem, but I can't seem to find the correct formula for either one.
My most recent attempt was:
6(.25 -x) + 1x = 6(.33)
6.25-6x+x=1.98
4.23-5x=0
4.23=5x
x=.846
This didn't work for the other problem, so I know this can't be correct here, either.
Could someone explain the correct formula to me?
My book doesn't cover percent mixture problems very in depth, just touches on one type and then moves on. Thanks for any help given! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A radiator contains 6 liters of a 25% anti-freeze solution.
How much should be drained and replaced with pure anti-freeze to produce a 33% solution?
:
You almost got it:
:
Let x = amt of 25% to be drained, and
x = amt of pure antifreeze to be added
:
.25(6-x) + x = .33(6)
1.5 - .25x + x = 1.98
-.25x + x = 1.98 - 1.5
.75x = .48
x =
x = .64 liters of 25% removed and .64 liters of pure antifreeze to be added
