Question 805678: Please help me solve this: How much water should be added to 30 quarts of juice that is 70% juice to get a diluted mixture that is 60% juice?
My work: Let w = the amount of water needed
30+w= the total amount of the mixture
100w + 2100 = 60(30+w)
100w + 2100 = 1800 +60w
-60w + 100w + 2100 = 1800 + 60w + (-60w)
40w + 2100 =1800
-2100 + 40w + 2100 = 1800 + -2100
40w = -300
1/40(40w)=-300(1/40)
w = -7.5
How can you add negative amounts of water?
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! Wrong equation.
The amount of juice is  quarts (70% of 30 quarts) in the undiluted juice.
That plus  quarts of juice in the added water has to add to the amount of juice in the final dilution, which should be  quarts of juice
The equation is
 quarts.
The mistake:
Using 100, 70 and 60 for 100%, 70% and 60% means that the quantities in your equation are not in quarts, but in fancy units of your invention.
That makes it difficult to relate equations with the "real" situation, but we could work with that, if we prefer large numbers to decimals.
However, if we accept your units,
 represents the amount of juice in  quarts of final diluted mixture,
 represents the amount of pure, undiluted juice in the initial  quarts of original "juice (drink) that is 70% juice".
However, adding to that  a  term makes no sense.
Maybe could represent the amount of water in the  quarts of 100% pure water added, but it does not make sense to add the initial amount of pure juice plus the initial amount of pure water to find the final amount of pure juice.
It would make more sense to add  , representing the amount of juice in the quarts of pure water (containing 0% juice) added.
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