SOLUTION: how many gallons of a 10% alcohol solution be mixed with 20 gallons of a 10% acid solution to obtain an 8% acid solution? The answer is 40/3. How do you solve the problem?

Question 309994: how many gallons of a 10% alcohol solution be mixed with 20 gallons of a 10% acid solution to obtain an 8% acid solution? The answer is 40/3. How do you solve the problem?
Found 2 solutions by stanbon, richwmiller:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how many gallons of a 10% alcohol solution be mixed with 20 gallons of a 10% acid solution to obtain an 8% acid solution? The answer is 40/3. How do you solve the problem?
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Equation:
alcohol + alcohol = alcohol
0.10x + 0.10*20 = 0.8(x+20)
Multiply thru by 100 to get:
10x + 10*20 = 8x + 8*20
2x = -2*20
x = -20
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The answer is unrealistic because the
stated problem is unrealistic. If
you add 10% solution to 10% solution
you can never get 8% solution.
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Also, note you seem to be adding alcohol
and acid. Hmmmm!!!
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Cheers,
Stan H.
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
This problem doesn't have enough info.
First we are dealing with alcohol and acid.We don't know how much acid is in the alcohol solution nor how much alcohol is in the acid solution.
I suspect you are copying the problem incorrectly.

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