SOLUTION: Please help me with this problem as I am terrible at math. It comes from a worksheet. A pharmacist has 1 liter of a solution that is 22% alcohol. How much pure alcohol must be a

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: Please help me with this problem as I am terrible at math. It comes from a worksheet. A pharmacist has 1 liter of a solution that is 22% alcohol. How much pure alcohol must be a      Log On

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Question 85501: Please help me with this problem as I am terrible at math. It comes from a worksheet.
A pharmacist has 1 liter of a solution that is 22% alcohol. How much pure alcohol must be added to bring the solution up to 27% alcohol

Found 3 solutions by Nate, stanbon, ankor@dixie-net.com:
Answer by Nate(3500) About Me  (Show Source):
You can put this solution on YOUR website!
It's like an averaged equation:
[x1y1 + x2y2]/(x1 + x2) = Final Percentage
x1 is the amount of solution with y1 percentage
x2 is the amount of a different solution with y2 percentage
[1(22%) + x2(100%)]/(1 + x2) = 27%
[0.22 + x2]/(1 + x2) = 0.27
0.22 + x2 = 0.27 + 0.27x2
0.73x2 = 0.05
x2 = 5/73

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A pharmacist has 1 liter of a solution that is 22% alcohol. How much pure alcohol must be added to bring the solution up to 27% alcohol
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Rule: Follow the active ingredient.
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22% solution DATA:
Amount = 1 liter ; Amt of active ingredient = 0.22*1=0.22 liters
----------------
Pure Alcohol DATA:
Amount = x liters ; Amt of active ingredient = 100%*x = x liters
----------------
27% Mixture DATA:
Amount = 1+x liters ; Amt of active ingredient = 0.27(1+x)=0.27+0.27x liters
---------------
EQUATION:
active + active = active
0.22 + x = 0.27+0.27x
0.05 = 0.73x
x= 0.0685 liters (Amount of 100% alcohol that must be added)
===============
Cheers,
Stan H.

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A pharmacist has 1 liter of a solution that is 22% alcohol. How much pure alcohol must be added to bring the solution up to 27% alcohol
:
Let x = the amt of pure alcohol to be added;
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Then the final amt = (x + 1)
:
Write a per cent equation:
100% (pure) alcohol is 1.0(x)
:
.22(1) + 1.0(x) = .27(x+ 1)
:
.22 + x = .27x + .27
:
x - .27x = .27 - .22
:
.73x = .05
:
x = .05/.73
:
x = .068 liters of pure alcohol required to get a 27% solution
:
:
Check our answer: resulting amt would be 1.068:
.22 + .068 = .27(1.068)
.288 = .288; proves our solution
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