SOLUTION: how many liters of pure alcohol must be mixed with 20 liters of 25% alcohol solution to get 85% mixture?

Question 473567: how many liters of pure alcohol must be mixed with 20 liters of 25% alcohol solution to get 85% mixture?
Found 2 solutions by stanbon, oberobic:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
how many liters of pure alcohol must be mixed with 20 liters of 25% alcohol solution to get 85% mixture?
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Eqution:
alcohol + alcohol = alcohol
x + 0.25*20 = 0.85(x+20)
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Multiply thru 100 to get:
100x + 25*20 = 85x + 85*20
15x = 60*20
x = 5 liters (amt. of pure alcohol needed)
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Cheers,
Stan H.
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Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
When solving mixture problems, try to figure out how "pure" stuff you need.
In this case, we're told 20 liters of 25% alcohol, which means there are 5 liters of pure alcohol and 15 liters of water.
We're also told we need to obtain an unknown amount of pure alcohol to obtain an 85% solution.
Let x be the unknown.
So when we add that to the 20 liters, that will be (20+x).
.
20*25% + x*100% = 85% * (20+x)
20*.25 + x*1.00 = .85 (20+x)
5 + x = 17 + .85x
.
subtract .85 x from both sides
subtract 5 from both sides
.
.15x = 12
x = 80
,
Does this make sense?
The best way to tell is to check your answer.
.
Adding 80 liters of pure alcohol to the 20 liters of 25% alcohol yields 100 liters.
We know from above that 20 liters of 25% alcohol contain 5 liters of pure alcohol.
80+5 = 85
So we have 85 liters of pure alcohol in the 100 liters of solution, which is 85% alcohol.
.
Answer: Add 80 liters of pure alcohol.
.
Done.
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