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?

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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) About Me  (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) About Me  (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]

V%5Bb%5D is unknown and we need to find.
P%5Bb%5D = 0.25
V%5Ba%5D = 14 "ounces"
P%5Ba%5D = 0.30
P%5Bt%5D = 0.26

Setup rational equation representing V%5Bb%5D that we want to add to get P%5Bt%5D as a result:
%28V%5Bb%5D%2AP%5Bb%5D%2BV%5Ba%5D%2AP%5Ba%5D%29%2F%28V%5Bb%5D%2BV%5Ba%5D%29=P%5Bt%5D

Solve for V%5Bb%5D, and then substitute the known values.