SOLUTION: a chemist had a solution of 60% acid. she added some distilled water, reducing the acid concentration to 40%. she then added 1 liter of water to further reduce the acid concentrati

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Question 188891: a chemist had a solution of 60% acid. she added some distilled water, reducing the acid concentration to 40%. she then added 1 liter of water to further reduce the acid concentration to 30%. how much of the 30% solution did she then have?
Found 2 solutions by Alan3354, ptaylor:
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
1 liter of water changed the concentration from 40% to 30%.
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0.4*x = 0.3*(x+1)
4x = 3x + 3
x = 3 liters or 40%
4 liters of 30%
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To check:
40% of 3 l = 1.2 l of acid
30% of 4 l = 1.2 l of acid

Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website!
Let x=amount of the 40% solution
and let (x+1)=amount of 30% solution
Since we are only adding water, the amount of pure acid in the 40% solution (0.40x) has to equal the amount of pure acid in the 30% solution (0.30(x+1)), so:
0.40x=0.30(x+1) get rid of parens
0.40x=0.30x+0.30 subtract 0.30x from each side
0.40x-0.30x=0.30 collect like terms
0.10x=0.30 divide each side by 0.10
x=3 liters---------------amount of 40% solution
x=1=3+1=4 liters---------------------amount of 30% solution that she has
CK
0.40*3=0.30(3+1)
1.2=1.2
Hope this helps---ptaylor