SOLUTION: Should this problem be worded differently - the word water evaporated changed to water added? How much water must be evaporated from 50 liters of a 65% salt solution to produce a

Question 99099: Should this problem be worded differently - the word water evaporated changed to water added?
How much water must be evaporated from 50 liters of a 65% salt solution to produce a solution that is a 40% salt solution.
.65(50) = .40y
50 - x = y
where x= water evaporated and y = new amount of the solution with 40% salt.
Substituting in 50 - x for y:
32.5 = .40(50 - x)
32.5 = 20 - .40x
12.5 = .40x
-31.25 = x
When I solve this problem I get x = water evaported = -31.25.
To decrease the percentage of salt, you need to add water, correct?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
YOU ARE CORRECT. THE WORDING SHOULD BE "ADDED" NOT "EVAPORATED". THE MORE WATER EVAPORATED, THE STRONGER THE SOLUTION. IF WE LOOK AT THE LIMITING CASE: WHEN WE EVAPORATE ALL THE WATER, WE HAVE PURE SALT LEFT.
Let x=amount of water that needs to be added
Now we know that the pure salt in the 50 liter solution (0.65*50) plus the pure salt in the water added (0) has to equal the pure salt in the final mixture(0.40)(50+x). So our equation to solve is:
0.65*50=0.40(50+x) get rid of the parens
32.5=20+0.40x subtract 20 from both sides
32.5-20=20-20+0.40x collect like terms
12.5=0.40x divide both sides by 0.40
x=31.25 liters -------------amount of pure water needed
CK
0.65*50=0.40*81.25
32.5=32.5

Hope this helps----ptaylor

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