SOLUTION: How many ounces of a 25% solution must be mixed with 14 ounces of a 30% alcohol solution to make a 26% alcohol solution?

Question 703561: How many ounces of a 25% solution must be mixed with 14 ounces of a 30% alcohol solution to make a 26% alcohol solution?
Found 2 solutions by nshah11, josgarithmetic:
Answer by nshah11(47) (Show Source): You can put this solution on YOUR website!
Assume x oz. of a 25% solution must be mixed.
(0.25x + 0.3(14))/(14 + x) = 0.26
0.25x + 4.2 = 3.64 + 0.26x
0.01x = 0.56
x = 56 oz.
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Volume of low percent mixture =V[b], a variable unknown
Concentration of low percent mixture as decimal fraction = P[b]
Volume of high percent mixture = V[a]
Concentration of high percent mixture decimal fraction = P[a]
Target concentraation percent as decimal fraction = P[t]

is unknown and we need to find.
= 0.25
= 14 "ounces"
= 0.30
= 0.26

Setup rational equation representing that we want to add to get as a result:

Solve for , and then substitute the known values.
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