SOLUTION: How many milliliters of water must be mixed with 26 milliliters of a 34% sulfuric acid solution to obtain a 22% sulfuric acid solution?

Question 988181: How many milliliters of water must be mixed with 26 milliliters of a 34% sulfuric acid solution to obtain a 22% sulfuric acid solution?
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
You essentially want to understand this:
One unknown quantity, two-part mixtures
http://www.algebra.com/my/Two-Part-Mixture-with-one-material-quantity-unknown.lesson?content_action=show_dev
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

How many milliliters of water must be mixed with 26 milliliters of a 34% sulfuric acid solution to obtain a 22% sulfuric acid solution?

Let amount of water be W
Solve using any of the two equations:

Equation 1
0W + .34(26) = .22(W + 26) 8.84 = .22W + 5.72 .22W = 8.84 - 5.72 .22W = 3.12 Solve for W, the amount of water

OR

Equation 2
W + .66(26) = .78(W + 26) W + 17.16 = .78W + 20.28 W - .78W = 20.28 - 17.16 .22W = 3.12 Solve for W, the amount of water

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