SOLUTION: A tub contains 300 l of a 32% salt solution. How much water must be added to reduce it to a 20% solution?

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Question 123564: A tub contains 300 l of a 32% salt solution. How much water must be added to reduce it to a 20% solution?
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let w = amount of water to be added
In words:
salt / (original solution + added water) = 20%
.32%2A300+=+96l of salt
96+%2F+%28300+%2B+w%29+=+.2
96+=+.2%28300+%2B+w%29
96+=+60+%2B+.2w
.2w+=+36
w+=+180l of water need to be added
check:
96+%2F+%28300+%2B+w%29+=+.2
96+%2F+%28300+%2B+180%29+=+.2
96+%2F+480+=+.2
.2+=+.2
OK