Question 1143037: How many gallons each of 25% alcohol and 5% alcohol should be mixed to obtain 20 gal of 16% alcohol?
Found 3 solutions by Boreal, greenestamps, josgarithmetic:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! x gal of 25%
20-x gal of 5%
pure alcohol is x(.25)+(20-x)(.05)=20*.16=3.2
so .25x+1-.05x=3.2 gallons of 25%
.20x=2.2 gallons of 5%
x=11 gallons of 25% or 2.75 gallons of pure alcohol
20-x=9 gallons of 5% or 0.45 gallons of pure
That equals 3.2 gallons pure, checks.
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a non-algebraic way to solve mixture problems like this that will get you to the answer much faster and with far less work than the standard algebraic method show by the other tutor.
Picture the problem as starting with 5% alcohol and adding 25% alcohol and stopping when the mixture is 16% alcohol. The problem is easily solved by modeling the percentages on a number line and seeing what fraction of the way from 5% to 25% you go before you stop at 16%.
On the number line, the distance from 5 to 25 is 20; the distance from 5 to 16 is 11. That means 16% is 11/20 of the way from 5% to 25%.
And that means 11/20 of the mixture is the 25% alcohol that you are adding.
ANSWER: 11/20 of 20 gallons, or 11 gallons, is the 25% alcohol; the other 9 gallons is the 5% alcohol.
Answer by josgarithmetic(39617)  (Show Source):
You can put this solution on YOUR website! ---------------------------------------------------------------------------
How many gallons each of 25% alcohol and 5% alcohol should be mixed to obtain 20 gal of 16% alcohol?
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H, 25%
L, 5%
T, 16%
M, 20 gallons
v, unknown volume of the H concentration alcohol
M-v, unknown vol of the L conc alc
Solve for v, substitute the given values...
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