SOLUTION: How much 70% acid solution should be mixed with 7 liters of 20% acid solution to end up with a 30% acid solution?

Question 96352: How much 70% acid solution should be mixed with 7 liters of 20% acid solution to end up with a 30% acid solution?
Found 2 solutions by checkley71, ptaylor:
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
.7X+.2*7=.3(7+X)
.7X+1.4=2.1+.3X
.7X-.3X=2.1-1.4
.4X=.7
X=.7/.4
X=1.75 L OF 80% ACID IS NEEDS TO BE ADDED.
PROOF
.7*1.75+.2*7=.3(7+1.75)
1.225+1.4=2.1+.525
2.625=2.625
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 70% acid solution needed

Now we know that the amount of pure acid in the 70% solution (0.70x) plus the amount of pure acid in the 20% solution (0.20(7)) has to equal the amount of pure acid in the final mixture (0.30*(7+x)). So our equation to solve is:
0.70x+0.20*7=0.30(7+x) get rid of parens
0.70x+1.4=2.10+0.30x subtract 0.30x and also 1.4 from both sides
0.70x+1.4-1.4-0.30x=2.1-1.4+0.30x-0.30x collect like terms
0.40x=0.70 divide both sides by 0.40

x=1.75 liters---------------amount of 70% solution needed
CK
0.70*1.75+0.20*7=0.30*8.75
1.225+1.4=2.625
2.625=2.625
Hope this helps----ptaylor

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