SOLUTION: In an experiment, a 100 ml of alcoholic solution must have a 20% alcohol in it. If 40ml of water is added in the solution, how much alcohol must be added, to maintain the 20% alcoh

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: In an experiment, a 100 ml of alcoholic solution must have a 20% alcohol in it. If 40ml of water is added in the solution, how much alcohol must be added, to maintain the 20% alcoh      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 847856: In an experiment, a 100 ml of alcoholic solution must have a 20% alcohol in it. If 40ml of water is added in the solution, how much alcohol must be added, to maintain the 20% alcohol solution?
Found 2 solutions by josgarithmetic, lwsshak3:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The concentration of alcohol in the 100 ml. alcohol solution is still unknown. You do not then have a way to find how much water to add to make 20% alcohol.

Assuming percents are vol/vol, and you have 100 ml. of some c percent alcohol, and you want to add 40 ml. of water; AND also some v amount of 100% alcohol to make 20% solution, ...

Amount of pure alcohol = 100%2Ac%2B100%2Av
Amount of solution result = 100%2B40%2Bv
Equation for adjustments to be 20% alcohol: %28100c%2B100v%29%2F%28100%2B40%2Bv%29=20;

Simplified: %28100%28c%2Bv%29%29%2F%28140%2Bv%29=20
further simplifiable based on the convenient numbers,
%2810%28c%2Bv%29%29%2F%28140%2Bv%29=2
OR
highlight%285%28c%2Bv%29%2F%28140%2Bv%29=1%29----simplified equation to solve for v or c.

Recheck your problem description for better/more precise problem description or for any missing information omitted from your help request.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
In an experiment, a 100 ml of alcoholic solution must have a 20% alcohol in it. If 40ml of water is added in the solution, how much alcohol must be added, to maintain the 20% alcohol solution?
***
let x=amount of alcohol to add
concentration of original solution=20%=20/100
concentration of new solution=20%=(20+x)/140
20/100=(20+x)/140
2800=2000+100x
100x=800
x=8
amount of alcohol to add=8ml