SOLUTION: A chemist wants to mix a 5% acid solution with a 25% acid solution to obtain 50 liters of 20% acid solution. how many liters of each solution should be used?

Question 267958: A chemist wants to mix a 5% acid solution with a 25% acid solution to obtain 50 liters of 20% acid solution. how many liters of each solution should be used?
Answer by ptaylor(2198) (Show Source): You can put this solution on YOUR website!
Let x=amount of 25% solution needed
Then 50-x=amount of 5% solution needed
Now we know that the amount of pure acid in the 25% solution (0.25x) plus the amount of pure acid on the 5% solution (0.05(50-x)) has to equal the amount of pure acid in the final mixture (0.20*50). So, our equation to solve is:
0.25x+0.05(50-x)=0.20*50 simplify
0.25x+2.5-0.05x=10 subtract 2.5 from each side
0.25x+2.5-2.5-0.05x=10-2.5 collect like terms
0.20x=7.5 divide each side by 0.20
x=37.5 liters-----------------------amount of 25% solution needed
50-x=50-37.5=12.5 liter---------------------amount of 5% solution needed
CK
0.25*37.5+0.05*12.5=0.20*50
9.375+0.625=10
10=10
You could work this problem using two equations and two unknowns:
x+y=50-------------------------eq1
0.25x+0.05y=0.20*50 or
0.25x+0.05y=10------------------eq2
Does this help?----ptaylor
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