SOLUTION: How many gallons of a 9% salt solution must be mixed with 35 gallons of a 11% solution to obtain a 10% solution?

Question 987424: How many gallons of a 9% salt solution must be mixed with 35 gallons of a 11% solution to obtain a 10% solution?
Found 2 solutions by josgarithmetic, macston:
Answer by josgarithmetic(39625) (Show Source): You can put this solution on YOUR website!
This very typical mixture problem exercise should be well understood as explained in this lesson:
http://www.algebra.com/my/Two-Part-Mixture-with-one-material-quantity-unknown.lesson?content_action=show_dev
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
.
x=gallons of 9% solution needed.
.
0.09x+0.11(35gal)=0.10(35gal+x)
0.09x+3.85gal=3.5gal+0.10x
0.35gal=0.01x
35gal=x
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ANSWER: 35 gallons of 9% solution should be used.
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CHECK:
0.09x+0.11(35gal)=0.10(35gal+x)
0.09(35gal)+0.11(35gal)=0.10(35gal+35gal)
3.15gal+3.65gal=0.10(70gal)
7.0gal=7.0gal

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