SOLUTION: Hi. I need help with mixtures I went over this topic last year in school but completely forgot how to do it. Here is the problem. a pharmacist needs to strengthen a 20% alcohol

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Question 894881: Hi. I need help with mixtures I went over this topic last year in school but completely forgot how to do it. Here is the problem.
a pharmacist needs to strengthen a 20% alcohol solution so that it contains 36% alcohol. how much pure alcohol should be added to 240 milliliters of the 20% solution?
Found 3 solutions by stanbon, MathTherapy, richwmiller:
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
a pharmacist needs to strengthen a 20% alcohol solution so that it contains 36% alcohol. how much pure alcohol should be added to 240 milliliters of the 20% solution?
----------------
Equation:
Quantity:: x + y = 240 ml
Alcohol:: 0.20x + y = 0.36*240
-------
Subtract the lower equation
from the upper equation to get ::
0.80x = 0.64*240
x = 192 ml (amt. of 20% solution needed)
240-x = 48 ml (amt. of pure alcohol needed)
-----------------------
Cheers,
Stan H.
-----------------------
Answer by MathTherapy(10553)   (Show Source): You can put this solution on YOUR website!
Hi. I need help with mixtures I went over this topic last year in school but completely forgot how to do it. Here is the problem.
a pharmacist needs to strengthen a 20% alcohol solution so that it contains 36% alcohol. how much pure alcohol should be added to 240 milliliters of the 20% solution?


Let amount of pure alcohol be A

Percentage of alcohol in initial solution, plus amount of PUR alcohol to be added, equals

percentage of alcohol in resultant solution, OR in mathematical terminology:

.2(240) + A = .36(240 + A)

     48 + A = 86.4 + .36A

   A - .36A = 86.4 – 48

       .64A = 38.4

A, or amount of alcohol to add = , or  mL

You can do the check!! 



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Answer by richwmiller(17219)   (Show Source): You can put this solution on YOUR website!
 .2*240+x=.36(240+x)

48+x=86.4+.36x

.64x=38.4

x=60 mL 

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