SOLUTION: A chemist has 20 liters of a solution which is 35% salt. How much water must be added to make the solution 28% salt. My calculations are not given me the answer that my book ha

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Question 5309: A chemist has 20 liters of a solution which is 35% salt. How much water must be added to make the solution 28% salt.
My calculations are not given me the answer that my book has. Please help me.
.35(20x) = .28(20+x)
7.0x=5.6+.28x
-.28x=5.6
6.72x=5.6
x=1.2 or 5
Am I correct?
Thank you
Found 2 solutions by rapaljer, longjonsilver:
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = amount of water to be added
20 = amount of 35% solution
x+20 = amount of 28% solution.

Equatiaon: Amount of salt before at 35% = Amount of salt after dilution at 28%
.35x = .28 (x + 20)
.35x = .28x + 5.6
.07x = 5.6
x=+5.6%2F.07+=+560%2F7+=+80 liters.

See if that looks right.

R^2 at SCC

Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
i disagree with the other solution - i think the answer is 5 litres. THat also looks more sense too...

Let x = amount of water to be added.

In general, we have the formula (amount of salt)/(total volume)*100 = % salt

so, initially, %28a%2F20%29%2A100+=+35 --> a = 7

Finally (remembering that the amount of salt does not change), %28a%2F%2820%2Bx%29%29%2A100+=+28
--> 700/28 = 20+x
25 = 20+x
so x = 5litres

jon.